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Bookshelves/General_Chemistry/The_Video_Textbook_of_General_Chemistry_(Farmer)/20%3A_Solubility_Products/20.6_Removal_of_an_Ion_from_Solution_Using_Precipitation_(Video)
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This project was preformed to supply Libretext Authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students . Video Topics A metal can almost be completely removed from the solution by precipitation though the formation of an insoluble salt. This video contains examples, which show how this process can occur. Link to Video Removal of an Ion from Solution Using Precipitation: https://youtu.be/VW0bhcs84I4 Attribution Prof. Steven Farmer ( Sonoma State University )
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Courses/BridgeValley_Community_and_Technical_College/Consumer_Chemistry/11%3A_The_Chemical_Equation/11.05%3A_Composition%2C_Decomposition%2C_and_Combustion_Reactions
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Learning Objectives Recognize composition, decomposition, and combustion reactions. Predict the products of a combustion reaction. Three classifications of chemical reactions will be reviewed in this section. Predicting the products in some of them may be difficult, but the reactions are still easy to recognize. A composition reaction (sometimes also called a combination reaction or a synthesis reaction ) produces a single substance from multiple reactants. A single substance as a product is the key characteristic of the composition reaction. There may be a coefficient other than one for the substance, but if the reaction has only a single substance as a product, it can be called a composition reaction. In the reaction \[\ce{2H_2(g) + O_2(g) → 2H_2O(ℓ)}\nonumber \] water is produced from hydrogen and oxygen. Although there are two molecules of water being produced, there is only one substance—water—as a product. So this is a composition reaction. A decomposition reaction starts from a single substance and produces more than one substance; that is, it decomposes. The key characteristics of a decomposition reaction are: one substance as a reactant and more than one substance as the products. For example, in the decomposition of sodium hydrogen carbonate (also known as sodium bicarbonate): \[\ce{2NaHCO_3(s) → Na_2CO_3(s) + CO_2(g) + H_2O(ℓ) }\nonumber \] sodium carbonate, carbon dioxide, and water are produced from the single substance sodium hydrogen carbonate. Composition and decomposition reactions are difficult to predict; however, they should be easy to recognize. Example \(\PageIndex{1}\): Identifying Reactions Identify each equation as a composition reaction, a decomposition reaction, or neither. \(\ce{Fe2O3 + 3SO3 → Fe2(SO4)3}\) \(\ce{NaCl + AgNO3 → AgCl + NaNO3}\) \(\ce{(NH4)2Cr2O7 → Cr2O3 + 4H2O + N2}\) Solution In this equation, two substances combine to make a single substance. This is a composition reaction. Two different substances react to make two new substances. This does not fit the definition of either a composition reaction or a decomposition reaction, so it is neither. In fact, you may recognize this as a double-replacement reaction. A single substance reacts to make multiple substances. This is a decomposition reaction. Exercise \(\PageIndex{1}\) Identify the equation as a composition reaction, a decomposition reaction, or neither. \[\ce{C3H8 → C3H4 + 2H2} \nonumber \] Answer decomposition A combustion reaction occurs when a reactant combines with oxygen, many times from the atmosphere, to produce oxides of all other elements as products; any nitrogen in the reactant is converted to elemental nitrogen, N 2 . Many reactants, called fuels , contain mostly carbon and hydrogen atoms, reacting with oxygen to produce CO 2 and H 2 O. For example, the balanced chemical equation for the combustion of methane, CH 4 , is as follows: \[\ce{CH_4 + 2O_2 → CO_2 + 2H_2O}\nonumber \] Kerosene can be approximated with the formula \(\ce{C_{12}H_{26}}\), and its combustion equation is: \[\ce{2C_{12}H_{26} + 37O_2 → 24CO-2 + 26H_2O}\nonumber\] Sometimes fuels contain oxygen atoms, which must be counted when balancing the chemical equation. One common fuel is ethanol, \(\ce{C_2H_5OH}\), whose combustion equation is: \[\ce{C_2H_5OH + 3O_2 → 2CO_2 + 3H_2O}\nonumber \] If nitrogen is present in the original fuel, it is converted to \(\ce{N_2}\), not to a nitrogen-oxygen compound. Thus, for the combustion of the fuel dinitroethylene, whose formula is \(\ce{C_2H_2N_2O_4}\), we have: \[\ce{2C_2H_2N_2O_4 + O_2 → 4CO_2 + 2H_2O + 2N_2}\nonumber \] Example \(\PageIndex{2}\): Combustion Reactions Complete and balance each combustion equation. the combustion of propane (\(\ce{C3H8}\)) the combustion of ammonia (\(\ce{NH3}\)) Solution The products of the reaction are CO 2 and H 2 O, so our unbalanced equation is \[\ce{C3H8 + O2 → CO2 + H2O} \nonumber \] Balancing (and you may have to go back and forth a few times to balance this), we get \[\ce{C3H8 + 5O2 → 3CO2 + 4H2O} \nonumber \] The nitrogen atoms in ammonia will react to make N 2 , while the hydrogen atoms will react with O 2 to make H 2 O: \[\ce{NH3 + O2 → N2 + H2O} \nonumber \] To balance this equation without fractions (which is the convention), we get \[\ce{4NH3 + 3O2 → 2N2 + 6H2O} \nonumber \] Exercise \(\PageIndex{2}\) Complete and balance the combustion equation for cyclopropanol (\ce{C3H6O}\)). Answer \[\ce{C_3H_6O + 4O_2 → 3CO_2 + 3H_2O}\nonumber \] Key Takeaways A composition reaction produces a single substance from multiple reactants. A decomposition reaction produces multiple products from a single reactant. Combustion reactions are the combination of some compound with oxygen to make oxides of the other elements as products (although nitrogen atoms react to make N 2 ).
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Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Contextual_Modules/Developing_an_Analytical_Method_for_the_Analysis_of_a_Medicinal_Plant/10_Instructors_Guide/02_Part_II._Separating_and_Analyzing_Mixtures_Using_HPLC
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Investigation 2 For this study we will use a reverse-phase HPLC equipped with a UV detector to monitor absorbance. What is a reverse-phase separation and how is it different from a normal-phase separation? How does the choice between a reverse-phase separation and a normal-phase separation affect the order in which analytes elute from an HPLC? In a reverse-phase HPLC separation, the stationary phase is non-polar and the mobile phase is polar. For a normal-phase separation, the stationary phase is polar and the mobile phase is non-polar. Separations in HPLC depend on a difference in the solubility of the analytes in the mobile phase and in the stationary phase. In a reverse-phase separation, for example, analytes of lower polarity are more soluble in the non-polar stationary phase, spending more time in the stationary phase and eluting at a later time than more polar analytes. In a normal-phase separation, the order of elution is reversed, with less polar analytes spending more time in the mobile phase and eluting before more polar analytes. Investigation 3 Using the data in Figure 1 determine each analyte’s retention time. Based on your answers to Investigation 1 and Investigation 2, does the relative order of elution order make sense? Why or why not? The retention times for the analytes are: hydrophilic compounds tr (min) lipophilic compounds tr (min).1 danshensu 4.81 dihydrotanshinone 50.50 rosmarinic acid 27.93 cryptotanshinone 55.21 lithospermic acid 29.44 tanshinone I 56.47 salvianolic acid A 35.80 tanshinone IIA 62.60 As seen in Investigation 2, for a reverse-phase HPLC separation, we expect more polar compounds to elute earlier than less polar compounds, a trend we see here as all four hydrophilic compounds elute before the four lipophilic compounds. The trend in retention times within each group is harder to discern, particularly given the changing composition of the mobile phase; however, danshensu is significantly more soluble in water than the other hydrophilic compounds and elutes much earlier. Note: The data used to create Figure 1 are not drawn directly from the original paper. Instead, the retention times and the relationships between peak height and analyte concentrations, in μg/mL, were determined using the HPLC data in Figure 8b and the corresponding extraction yields, in mg/g, from the first row of Table 3, obtained using a 1.00-g sample of Danshen and 35.0 mL of solvent. The resulting values for k in the equation A = kC were used to generate the data for this chromatogram and for all subsequent chromatograms. Details on the standard used to generate Figure 1 are included in Investigation 7. Although the original paper reports peak areas instead of peak heights, the latter is used in this exercise as it is easier for students to measure. Investigation 4 Based on Figure 2, are there features in these UV spectra that distinguish Danshen’s hydrophilic compounds from its lipophilic compounds? What wavelength should we choose if our interest is the hydrophilic compounds only? What wavelength should we choose if our interest is the lipophilic compounds only? What is the best wavelength for detecting all of Danshen’s constituents? The UV spectra for the lipophilic compounds cryptotanshinone and tanshinone I show a single strong absorption band between 240 nm and 270 nm. The hydrophilic compounds danshensu and salvianolic acid A, on the other hand, have strong adsorption bands at wavelengths below 240 nm and at wavelengths above 270 nm. Clearly choosing a single wavelength for this analysis requires a compromise. Any wavelength in the immediate vicinity of 280 nm is an appropriate choice as the absorbance value for salvianolic acid A is strong, and the absorbance values for tanshinone I, cryptotanshinone, and danshensu are similar in magnitude. At wavelengths greater than 285 nm the absorbance of tanshinone I, cryptotanshinone, and danshensu decrease in value, and the absorbance of danshensu decreases toward zero as the wavelength approaches 250 nm. All four compounds absorb strongly at wavelengths below 230 nm, but interference from the many other constituents of Danshen extracts may present problems. The data in the figures that follow were obtained using a wavelength of 280 nm. Note: The data for Figure 2 are not drawn from the original paper. The UV spectra for cryptotanshinone and for tanshinone I are adapted from “Analysis of Protocatechuic Acid, Protocatechuic Aldehyde and Tanshinones in Dan Shen Pills by HPLC,” the full reference for which is Huber, U. Agilent Publication Number 5968-2882E (released 12/98 and available at https://www.chem.agilent.com/Library...s/59682882.pdf ), and the UV spectra for danshensu and for salvianolic acid A are adapted from “Simultaneous detection of seven phenolic acids in Danshen injection using HPLC with ultraviolet detector,” the full reference for which is Xu, J.; Shen, J.; Cheng, Y.; Qu, H. J. Zhejiang Univ. Sci. B. 2008, 9, 728-733 (DOI:10.1631/jzus.B0820095). These sources also provide UV spectra for tanshinone IIA and for rosmarinic acid, but not for dihydrotanshinone nor for lithospermic acid. Investigation 5 For a UV detector, what is the expected relationship between peak height and the analyte’s concentration in μg/mL? For the results in Figure 1, can you assume the analyte with the smallest peak height is present at the lowest concentration? Why or why not? For a UV detector, we expect the absorbance to follow Beer’s law, A = kC , where A is the analyte’s absorbance, C is the analyte’s concentration, and k is a proportionality constant that accounts for the analyte’s wavelength-dependent absorptivity and the detector’s pathlength. Because each analyte has a different value for k , we cannot assume that the analyte with the smallest peak height is also the analyte present at the lowest concentration. Investigation 6 Calculate the concentration, in μg/mL, for each analyte in the standard sample whose chromatogram is shown in Figure 1. Using this standard sample as a single-point external standard, calculate the proportionality constant for each analyte that relates its absorbance to its concentration in μg/mL. Do your results support your answer to Investigation 5? Why or why not? The table below shows the absorbance values (in mAU) for each analyte from Figure 1, the analyte’s concentration in the standard sample, and its value for k . analyte absorbance (mAU) C (μg/mL) k (mAU•mL/μg) danshensu 96.3 60.0 1.605 rosmarinic acid 125.6 143.1 0.878 lithospermic acid 71.4 133.1 0.536 salvianolic acid A 66.1 41.7 1.585 dihydrotanshinone 442.9 15.1 2.841 cryptotanshinone 54.4 28.9 1.882 tanshinone I 59.5 37.2 1.599 tanshinone IIA 105.2 71.7 1.467 Using danshensu as an example, concentrations are derived from the data for the stock standard, accounting for its dilution and converting from mg to μg \[C= \mathrm{\dfrac{6.00\: mg}{10.00\: mL}\times\dfrac{1.00\: mL}{10.00\: mL}\times\dfrac{1000\: μg}{mg}=60.0\: μg/mL}\nonumber\] and k is calculated as \[k=\dfrac{A}{C}=\mathrm{\dfrac{96.3\: mAU}{60.0\: μg/mL}=1.605\: mAU•mL/μg}\nonumber\] Although dihydrotanshinone is present at the lowest concentration and has the smallest peak height, it has the largest value for k and is the strongest absorbing analyte. If, for example, dihydrotanshinone is present at a concentration of 25.0 μg/mL (a concentration smaller than the other seven compounds), its absorbance of 71.0 mAU will be greater than that for lithospermic acid, salvianolic acid A, cryptotanshinone, and tanshinone I. This is consistent with our expectations from Investigation 5. Note: See the comments for Investigation 3 for details on the data used in this investigation.
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Courses/Solano_Community_College/Introductory_Chemistry_at_Solano_College_-_Updated_2023_04_03/11%3A_States_of_Matter/11.14%3A_Melting
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Have you ever gone ice skating? In the winter, many people find the snow and ice beautiful. They enjoy getting out to ski or ice-skate. Others don’t find that time of year to be so much fun. When the snow melts, the roads get very sloppy and messy. Those people look forward to spring when all the ice and snow are gone and the weather is warmer. Melting Point Solids are similar to liquids in that both are condensed states, with particles that are far closer together than those of a gas. However, while liquids are fluid, solids are not. The particles of most solids are packed tightly together in an orderly arrangement. The motion of individual atoms, ions, or molecules in a solid is restricted to vibrational motion about a fixed point. Solids are almost completely incompressible and are the densest of the three states of matter. As a solid is heated, its particles vibrate more rapidly as it absorbs kinetic energy. Eventually, the organization of the particles within the solid structure begins to break down, and the solid starts to melt. The melting point is the temperature at which a solid changes into a liquid. At its melting point, the disruptive vibrations of the particles of the solid overcome the attractive forces operating within the solid. As with boiling points, the melting point of a solid is dependent on the strength of those attractive forces. Sodium chloride \(\left( \ce{NaCl} \right)\) is an ionic compound that consists of a multitude of strong ionic bonds. Sodium chloride melts at \(801^\text{o} \text{C}\). Ice (solid \(\ce{H_2O}\)) is a molecular compound whose molecules are held together by hydrogen bonds. Though hydrogen bonds are the strongest of the intermolecular forces, the strength of hydrogen bonds is much less than that of ionic bonds. The melting point of ice is 0 °C. The melting point of a solid is the same as the freezing point of the liquid. At that temperature, the solid and liquid states of the substance are in equilibrium. For water, this equilibrium occurs at \(0^\text{o} \text{C}\). \[\ce{H_2O} \left( s \right) \rightleftharpoons \ce{H_2O} \left( l \right)\nonumber \] We tend to think of solids as those materials that are solid at room temperature. However, all materials have melting points of some sort. Gases become solids at extremely low temperatures, and liquids will also become solid if the temperature is low enough. The table below gives the melting points of some common materials. Materials Melting Point (°C) Hydrogen -259 Oxygen -219 Diethyl ether -116 Ethanol -114 Water 0 Pure silver 961 Pure gold 1063 Iron 1538 Simulation Explore how heat and temperature relate to phase change in this simulation: Does the temperature of water rise while it is boiling? Summary The melting point is the temperature at which a solid changes into a liquid. Intermolecular forces have a strong influence on melting point. Review Define melting point. What happens when a material melts? Would you expect ethane (C 2 H 6 ) to have a higher or lower melting point than water? Explain your answer.
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Bookshelves/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/III%3A_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_1/02%3A_Ligand_Binding_in_Coordination_Complexes_and_Organometallic_Compounds
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2.1: Introduction 2.2: How Tightly Do Ligands Bind? 2.3: Electron Counting in Transition Metal Complexes 2.4: Chelation Monodentate ligands bind through only one donor atom. Bidentate ligands bind through two donor sites. Bidentate binding allows a ligand to bind more tightly. Tridentate ligands, which bind through three donors, can bind even more tightly, and so on. This phenomenon is generally called the "chelate effect". 2.5: Pi Coordination- Donation from Alkenes 2.6: Hapticity The term used to describe the participation of multiple atoms simultaneously during pi coordination is hapticity. 2.7: Hard and Soft Acid and Base Concepts Not all metals form coordination complexes with all possible ligands. Some metals are more likely to form compounds with certain ligands. This observation has eventually led to a classification system called Hard and Soft Acids and Bases (HSAB). 2.8: Ligand Field Theory 2.9: Ligand Field Stabilization Energy We can use the relative energy levels of the d orbitals in a given complex to calculate whether the overall energy would be higher or lower in a high-spin vs. a low-spin case, for example. The calculation provides us with a value that is called the ligand field stabilization energy. Although we have been thinking of bonding in transition metal complexes in terms of molecular orbital ideas, ligand field stabilization energy actually has its roots in a separate approach called crystal field theory 2.10: Spectrochemical Series The d orbital energy splitting is influenced by how strongly the ligand interacts with the metal. Ligands that interact only weakly produce little change in the d orbital energy levels, whereas ligands that interact strongly produce a larger change in d orbital energy levels. The spectrochemical series is a list of ligands based on the strength of their interaction with metal ions. 2.11: Ligand Lability 2.12: Jahn-Teller Distortion In some cases, the d-electron count can have a subtle influence on the geometry of a complex. For example, an octahedral complex might be distorted, either stretched along one axis or else compressed. In Jahn-Teller Distortion, this effect arises from unequally-distributed electrons in the same level (degeneracy). Although this phenomenon is structural, it can sometimes exert an influence on the stability of complexes that translates into accelerated ligand substitutions. 2.13: Multiple Bonds in Coordination Complexes 2.14: Solutions to Selected Problems 2.15: More Solutions to Selected Problems
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Courses/Madera_Community_College/Concepts_of_Physical_Science/14%3A_Nuclear_Radiation/14.01%3A_Introduction_to_Radioactivity
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There is an ongoing quest to find substructures of matter. At one time, it was thought that atoms would be the ultimate substructure, but just when the first direct evidence of atoms was obtained, it became clear that they have a substructure and a nucleus . The nucleus itself has spectacular characteristics. For example, certain nuclei are unstable, and their decay emits radiations with energies millions of times greater than atomic energies. Some of the mysteries of nature, such as why the core of the earth remains molten and how the sun produces its energy, are explained by nuclear phenomena. The exploration of radioactivity and the nucleus revealed fundamental and previously unknown particles, forces, and conservation laws. That exploration has evolved into a search for further underlying structures, such as quarks. In this chapter, we will start with the fundamentals of nuclear radioactivity and the nucleus, moving on to some of the important applications of nuclear physics. Quarks and leptons are beyond the scope of this text. Figure \(\PageIndex{2}\) shows the structures and length scales involved in study of nucleus and elementary particles. In atomic and molecular physics, we were already dealing with structures at nanometer scale, about 500 times shorter than the wavelength of green light. The nuclear scale goes 100,000 times smaller; the size of largest stable nuclei is about 10 femtometers, or 10 − 14 m. All other objects we are studying in nuclear and particle physics are smaller than this, with the most elementary particles, quarks and leptons, having no known size and being considered as "point masses."
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Courses/CSU_San_Bernardino/CHEM_2100%3A_General_Chemistry_I_(Mink)/05%3A_Thermochemistry/5.08%3A_Exercises
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1. A burning match and a bonfire may have the same temperature, yet you would not sit around a burning match on a fall evening to stay warm. Why not? 2. Prepare a table identifying several energy transitions that take place during the typical operation of an automobile. 3. Explain the difference between heat capacity and specific heat of a substance. 4. Calculate the heat capacity, in joules and in calories per degree, of the following: 28.4 g of water 1.00 oz of lead 5. Calculate the heat capacity, in joules and in calories per degree, of the following: 45.8 g of nitrogen gas 1.00 pound of aluminum metal 6. How much heat, in joules and in calories, must be added to a 75.0–g iron block with a specific heat of 0.449 J/g °C to increase its temperature from 25 °C to its melting temperature of 1535 °C? 7. How much heat, in joules and in calories, is required to heat a 28.4-g (1-oz) ice cube from −23.0 °C to −1.0 °C? 8. How much would the temperature of 275 g of water increase if 36.5 kJ of heat were added? 9. If 14.5 kJ of heat were added to 485 g of liquid water, how much would its temperature increase? 10. A piece of unknown substance weighs 44.7 g and requires 2110 J to increase its temperature from 23.2 °C to 89.6 °C. What is the specific heat of the substance? If it is one of the substances found in Table 5.1, what is its likely identity? 11. A piece of unknown solid substance weighs 437.2 g, and requires 8460 J to increase its temperature from 19.3 °C to 68.9 °C. What is the specific heat of the substance? If it is one of the substances found in Table 5.1, what is its likely identity? 12. An aluminum kettle weighs 1.05 kg. What is the heat capacity of the kettle? How much heat is required to increase the temperature of this kettle from 23.0 °C to 99.0 °C? How much heat is required to heat this kettle from 23.0 °C to 99.0 °C if it contains 1.25 L of water (density of 0.997 g/mL and a specific heat of 4.184 J/g °C)? 13. Most people find waterbeds uncomfortable unless the water temperature is maintained at about 85 °F. Unless it is heated, a waterbed that contains 892 L of water cools from 85 °F to 72 °F in 24 hours. Estimate the amount of electrical energy required over 24 hours, in kWh, to keep the bed from cooling. Note that 1 kilowatt-hour (kWh) = 3.6 10 6 J, and assume that the density of water is 1.0 g/mL (independent of temperature). What other assumptions did you make? How did they affect your calculated result (i.e., were they likely to yield “positive” or “negative” errors)? 14. A 500-mL bottle of water at room temperature and a 2-L bottle of water at the same temperature were placed in a refrigerator. After 30 minutes, the 500-mL bottle of water had cooled to the temperature of the refrigerator. An hour later, the 2-L of water had cooled to the same temperature. When asked which sample of water lost the most heat, one student replied that both bottles lost the same amount of heat because they started at the same temperature and finished at the same temperature. A second student thought that the 2-L bottle of water lost more heat because there was more water. A third student believed that the 500-mL bottle of water lost more heat because it cooled more quickly. A fourth student thought that it was not possible to tell because we do not know the initial temperature and the final temperature of the water. Indicate which of these answers is correct and describe the error in each of the other answers. 15. Would the amount of heat measured for the reaction in Example 5.5 be greater, lesser, or remain the same if we used a calorimeter that was a poorer insulator than a coffee cup calorimeter? Explain your answer. 16. Would the amount of heat absorbed by the dissolution in Example 5.6 appear greater, lesser, or remain the same if the experimenter used a calorimeter that was a poorer insulator than a coffee cup calorimeter? Explain your answer. 17. Would the amount of heat absorbed by the dissolution in Example 5.6 appear greater, lesser, or remain the same if the heat capacity of the calorimeter were taken into account? Explain your answer. 18. How many milliliters of water at 23 °C with a density of 1.00 g/mL must be mixed with 180 mL (about 6 oz) of coffee at 95 °C so that the resulting combination will have a temperature of 60 °C? Assume that coffee and water have the same density and the same specific heat. 19. How much will the temperature of a cup (180 g) of coffee at 95 °C be reduced when a 45 g silver spoon (specific heat 0.24 J/g °C) at 25 °C is placed in the coffee and the two are allowed to reach the same temperature? Assume that the coffee has the same density and specific heat as water. 20. A 45-g aluminum spoon (specific heat 0.88 J/g °C) at 24 °C is placed in 180 mL (180 g) of coffee at 85 °C and the temperature of the two become equal. What is the final temperature when the two become equal? Assume that coffee has the same specific heat as water. The first time a student solved this problem she got an answer of 88 °C. Explain why this is clearly an incorrect answer. 21. The temperature of the cooling water as it leaves the hot engine of an automobile is 240 °F. After it passes through the radiator it has a temperature of 175 °F. Calculate the amount of heat transferred from the engine to the surroundings by one gallon of water with a specific heat of 4.184 J/g °C. 22. A 70.0-g piece of metal at 80.0 °C is placed in 100 g of water at 22.0 °C contained in a calorimeter like that shown in Figure 5.12. The metal and water come to the same temperature at 24.6 °C. How much heat did the metal give up to the water? What is the specific heat of the metal? 23. If a reaction produces 1.506 kJ of heat, which is trapped in 30.0 g of water initially at 26.5 °C in a calorimeter like that in Figure 5.12, what is the resulting temperature of the water? 24. A 0.500-g sample of KCl is added to 50.0 g of water in a calorimeter (Figure 5.12). If the temperature decreases by 1.05 °C, what is the approximate amount of heat involved in the dissolution of the KCl, assuming the specific heat of the resulting solution is 4.18 J/g °C? Is the reaction exothermic or endothermic? 25. Dissolving 3.0 g of CaCl 2 ( s ) in 150.0 g of water in a calorimeter (Figure 5.12) at 22.4 °C causes the temperature to rise to 25.8 °C. What is the approximate amount of heat involved in the dissolution, assuming the specific heat of the resulting solution is 4.18 J/g °C? Is the reaction exothermic or endothermic? 26. When 50.0 g of 0.200 M NaCl( aq ) at 24.1 °C is added to 100.0 g of 0.100 M AgNO 3 ( aq ) at 24.1 °C in a calorimeter, the temperature increases to 25.2 °C as AgCl( s ) forms. Assuming the specific heat of the solution and products is 4.20 J/g °C, calculate the approximate amount of heat in joules produced. 27. The addition of 3.15 g of Ba(OH) 2 ·8H 2 O to a solution of 1.52 g of NH 4 SCN in 100 g of water in a calorimeter caused the temperature to fall by 3.1 °C. Assuming the specific heat of the solution and products is 4.20 J/g °C, calculate the approximate amount of heat absorbed by the reaction, which can be represented by the following equation: Ba(OH) 2 ·8H 2 O( s ) + 2NH 4 SCN( aq ) ⟶ Ba(SCN) 2 ( aq ) + 2NH 3 ( aq ) + 10H 2 O( l ) 28. The reaction of 50 mL of acid and 50 mL of base described in Example 5.5 increased the temperature of the solution by 6.9 ºC. How much would the temperature have increased if 100 mL of acid and 100 mL of base had been used in the same calorimeter starting at the same temperature of 22.0 ºC? Explain your answer. 29. If the 3.21 g of NH 4 NO 3 in Example 5.6 were dissolved in 100.0 g of water under the same conditions, how much would the temperature change? Explain your answer. 30. When 1.0 g of fructose, C 6 H 12 O 6 ( s ), a sugar commonly found in fruits, is burned in oxygen in a bomb calorimeter, the temperature of the calorimeter increases by 1.58 °C. If the heat capacity of the calorimeter and its contents is 9.90 kJ/°C, what is q for this combustion? 31. When a 0.740-g sample of trinitrotoluene (TNT), C 7 H 5 N 2 O 6 , is burned in a bomb calorimeter, the temperature increases from 23.4 °C to 26.9 °C. The heat capacity of the calorimeter is 534 J/°C, and it contains 675 mL of water. How much heat was produced by the combustion of the TNT sample? 32. One method of generating electricity is by burning coal to heat water, which produces steam that drives an electric generator. To determine the rate at which coal is to be fed into the burner in this type of plant, the heat of combustion per ton of coal must be determined using a bomb calorimeter. When 1.00 g of coal is burned in a bomb calorimeter (Figure 5.17), the temperature increases by 1.48 °C. If the heat capacity of the calorimeter is 21.6 kJ/°C, determine the heat produced by combustion of a ton of coal (2.000 10 3 pounds). 33. The amount of fat recommended for someone with a daily diet of 2000 Calories is 65 g. What percent of the calories in this diet would be supplied by this amount of fat if the average number of Calories for fat is 9.1 Calories/g? 34. A teaspoon of the carbohydrate sucrose (common sugar) contains 16 Calories (16 kcal). What is the mass of one teaspoon of sucrose if the average number of Calories for carbohydrates is 4.1 Calories/g? 35. What is the maximum mass of carbohydrate in a 6-oz serving of diet soda that contains less than 1 Calorie per can if the average number of Calories for carbohydrates is 4.1 Calories/g? 36. A pint of premium ice cream can contain 1100 Calories. What mass of fat, in grams and pounds, must be produced in the body to store an extra 1.1 10 3 Calories if the average number of Calories for fat is 9.1 Calories/g? 37. A serving of a breakfast cereal contains 3 g of protein, 18 g of carbohydrates, and 6 g of fat. What is the Calorie content of a serving of this cereal if the average number of Calories for fat is 9.1 Calories/g, for carbohydrates is 4.1 Calories/g, and for protein is 4.1 Calories/g? 38. Which is the least expensive source of energy in kilojoules per dollar: a box of breakfast cereal that weighs 32 ounces and costs $4.23, or a liter of isooctane (density, 0.6919 g/mL) that costs $0.45? Compare the nutritional value of the cereal with the heat produced by combustion of the isooctane under standard conditions. A 1.0-ounce serving of the cereal provides 130 Calories. 39. Explain how the heat measured in Example 5.5 differs from the enthalpy change for the exothermic reaction described by the following equation: 40. Using the data in the check your learning section of Example 5.5, calculate Δ H in kJ/mol of AgNO 3 ( aq ) for the reaction: 41. Calculate the enthalpy of solution (Δ H for the dissolution) per mole of NH 4 NO 3 under the conditions described in Example 5.6. 42. Calculate Δ H for the reaction described by the equation. ( Hint : Use the value for the approximate amount of heat absorbed by the reaction that you calculated in a previous exercise.) 43. Calculate the enthalpy of solution (Δ H for the dissolution) per mole of CaCl 2 (refer to Exercise 5.25). 44. Although the gas used in an oxyacetylene torch (Figure 5.7) is essentially pure acetylene, the heat produced by combustion of one mole of acetylene in such a torch is likely not equal to the enthalpy of combustion of acetylene listed in Table 5.2. Considering the conditions for which the tabulated data are reported, suggest an explanation. 45. How much heat is produced by burning 4.00 moles of acetylene under standard state conditions? 46. How much heat is produced by combustion of 125 g of methanol under standard state conditions? 47. How many moles of isooctane must be burned to produce 100 kJ of heat under standard state conditions? 48. What mass of carbon monoxide must be burned to produce 175 kJ of heat under standard state conditions? 49. When 2.50 g of methane burns in oxygen, 125 kJ of heat is produced. What is the enthalpy of combustion per mole of methane under these conditions? 50. How much heat is produced when 100 mL of 0.250 M HCl (density, 1.00 g/mL) and 200 mL of 0.150 M NaOH (density, 1.00 g/mL) are mixed? If both solutions are at the same temperature and the specific heat of the products is 4.19 J/g °C, how much will the temperature increase? What assumption did you make in your calculation? 51. A sample of 0.562 g of carbon is burned in oxygen in a bomb calorimeter, producing carbon dioxide. Assume both the reactants and products are under standard state conditions, and that the heat released is directly proportional to the enthalpy of combustion of graphite. The temperature of the calorimeter increases from 26.74 °C to 27.93 °C. What is the heat capacity of the calorimeter and its contents? 52. Before the introduction of chlorofluorocarbons, sulfur dioxide (enthalpy of vaporization, 6.00 kcal/mol) was used in household refrigerators. What mass of SO 2 must be evaporated to remove as much heat as evaporation of 1.00 kg of CCl 2 F 2 (enthalpy of vaporization is 17.4 kJ/mol)? The vaporization reactions for SO 2 and CCl 2 F 2 are and respectively. 53. Homes may be heated by pumping hot water through radiators. What mass of water will provide the same amount of heat when cooled from 95.0 to 35.0 °C, as the heat provided when 100 g of steam is cooled from 110 °C to 100 °C. 54. Which of the enthalpies of combustion in Table 5.2 the table are also standard enthalpies of formation? 55. Does the standard enthalpy of formation of H 2 O( g ) differ from Δ H ° for the reaction 56. Joseph Priestly prepared oxygen in 1774 by heating red mercury(II) oxide with sunlight focused through a lens. How much heat is required to decompose exactly 1 mole of red HgO( s ) to Hg( l ) and O 2 ( g ) under standard conditions? 57. How many kilojoules of heat will be released when exactly 1 mole of manganese, Mn, is burned to form Mn 3 O 4 ( s ) at standard state conditions? 58. How many kilojoules of heat will be released when exactly 1 mole of iron, Fe, is burned to form Fe 2 O 3 ( s ) at standard state conditions? 59. The following sequence of reactions occurs in the commercial production of aqueous nitric acid: Determine the total energy change for the production of one mole of aqueous nitric acid by this process. 60. Both graphite and diamond burn. For the conversion of graphite to diamond: Which produces more heat, the combustion of graphite or the combustion of diamond? 61. From the molar heats of formation in Appendix G, determine how much heat is required to evaporate one mole of water: 62. Which produces more heat? or for the phase change 63. Calculate for the process from the following information: 64. Calculate for the process from the following information: 65. Calculate Δ H for the process from the following information: 66. Calculate for the process from the following information: 67. Calculate the standard molar enthalpy of formation of NO( g ) from the following data: 68. Using the data in Appendix G, calculate the standard enthalpy change for each of the following reactions: 69. Using the data in Appendix G, calculate the standard enthalpy change for each of the following reactions: 70. The following reactions can be used to prepare samples of metals. Determine the enthalpy change under standard state conditions for each. 71. The decomposition of hydrogen peroxide, H 2 O 2 , has been used to provide thrust in the control jets of various space vehicles. Using the data in Appendix G, determine how much heat is produced by the decomposition of exactly 1 mole of H 2 O 2 under standard conditions. 72. Calculate the enthalpy of combustion of propane, C 3 H 8 ( g ), for the formation of H 2 O( g ) and CO 2 ( g ). The enthalpy of formation of propane is −104 kJ/mol. 73. Calculate the enthalpy of combustion of butane, C 4 H 10 ( g ) for the formation of H 2 O( g ) and CO 2 ( g ). The enthalpy of formation of butane is −126 kJ/mol. 74. Both propane and butane are used as gaseous fuels. Which compound produces more heat per gram when burned? 75. The white pigment TiO 2 is prepared by the reaction of titanium tetrachloride, TiCl 4 , with water vapor in the gas phase: How much heat is evolved in the production of exactly 1 mole of TiO 2 ( s ) under standard state conditions? 76. Water gas, a mixture of H 2 and CO, is an important industrial fuel produced by the reaction of steam with red hot coke, essentially pure carbon: Under the conditions of the reaction, methanol forms as a gas. Calculate for this reaction and for the condensation of gaseous methanol to liquid methanol. (c) Calculate the heat of combustion of 1 mole of liquid methanol to H 2 O( g ) and CO 2 ( g ). 77. In the early days of automobiles, illumination at night was provided by burning acetylene, C 2 H 2 . Though no longer used as auto headlamps, acetylene is still used as a source of light by some cave explorers. The acetylene is (was) prepared in the lamp by the reaction of water with calcium carbide, CaC 2 : Calculate the standard enthalpy of the reaction. The of CaC 2 is −15.14 kcal/mol. 78. From the data in Table 5.2, determine which of the following fuels produces the greatest amount of heat per gram when burned under standard conditions: CO( g ), CH 4 ( g ), or C 2 H 2 ( g ). 79. The enthalpy of combustion of hard coal averages −35 kJ/g, that of gasoline, 1.28 10 5 kJ/gal. How many kilograms of hard coal provide the same amount of heat as is available from 1.0 gallon of gasoline? Assume that the density of gasoline is 0.692 g/mL (the same as the density of isooctane). 80. Ethanol, C 2 H 5 OH, is used as a fuel for motor vehicles, particularly in Brazil. Write the balanced equation for the combustion of ethanol to CO 2 ( g ) and H 2 O( g ), and, using the data in Appendix G, calculate the enthalpy of combustion of 1 mole of ethanol. The density of ethanol is 0.7893 g/mL. Calculate the enthalpy of combustion of exactly 1 L of ethanol. Assuming that an automobile’s mileage is directly proportional to the heat of combustion of the fuel, calculate how much farther an automobile could be expected to travel on 1 L of gasoline than on 1 L of ethanol. Assume that gasoline has the heat of combustion and the density of n–octane, C 8 H 18 density = 0.7025 g/mL). 81. Among the substances that react with oxygen and that have been considered as potential rocket fuels are diborane [B 2 H 6 , produces B 2 O 3 ( s ) and H 2 O( g )], methane [CH 4 , produces CO 2 ( g ) and H 2 O( g )], and hydrazine [N 2 H 4 , produces N 2 ( g ) and H 2 O( g )]. On the basis of the heat released by 1.00 g of each substance in its reaction with oxygen, which of these compounds offers the best possibility as a rocket fuel? The 82. How much heat is produced when 1.25 g of chromium metal reacts with oxygen gas under standard conditions? 83. Ethylene, C 2 H 4 , a byproduct from the fractional distillation of petroleum, is fourth among the 50 chemical compounds produced commercially in the largest quantities. About 80% of synthetic ethanol is manufactured from ethylene by its reaction with water in the presence of a suitable catalyst. Using the data in the table in Appendix G, calculate Δ H ° for the reaction. 84. The oxidation of the sugar glucose, C 6 H 12 O 6 , is described by the following equation: The metabolism of glucose gives the same products, although the glucose reacts with oxygen in a series of steps in the body. How much heat in kilojoules can be produced by the metabolism of 1.0 g of glucose? How many Calories can be produced by the metabolism of 1.0 g of glucose? 85. Propane, C 3 H 8 , is a hydrocarbon that is commonly used as a fuel. Write a balanced equation for the complete combustion of propane gas. Calculate the volume of air at 25 °C and 1.00 atmosphere that is needed to completely combust 25.0 grams of propane. Assume that air is 21.0 percent O 2 by volume. (Hint: We will see how to do this calculation in a later chapter on gases—for now use the information that 1.00 L of air at 25 °C and 1.00 atm contains 0.275 g of O 2 .) The heat of combustion of propane is −2,219.2 kJ/mol. Calculate the heat of formation, of propane given that of H 2 O( l ) = −285.8 kJ/mol and of CO 2 ( g ) = −393.5 kJ/mol. Assuming that all of the heat released in burning 25.0 grams of propane is transferred to 4.00 kilograms of water, calculate the increase in temperature of the water. 86. During a recent winter month in Sheboygan, Wisconsin, it was necessary to obtain 3500 kWh of heat provided by a natural gas furnace with 89% efficiency to keep a small house warm (the efficiency of a gas furnace is the percent of the heat produced by combustion that is transferred into the house). (f) How many kilowatt–hours (1 kWh = 3.6 10 6 J) of electricity would be required to provide the heat necessary to heat the house? Note electricity is 100% efficient in producing heat inside a house. (g) Although electricity is 100% efficient in producing heat inside a house, production and distribution of electricity is not 100% efficient. The efficiency of production and distribution of electricity produced in a coal-fired power plant is about 40%. A certain type of coal provides 2.26 kWh per pound upon combustion. What mass of this coal in kilograms will be required to produce the electrical energy necessary to heat the house if the efficiency of generation and distribution is 40%?
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Courses/University_of_Kentucky/UK%3A_General_Chemistry/17%3A_Electrochemistry/17.3%3A_Standard_Reduction_Potentials
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Skills to Develop Determine standard cell potentials for oxidation-reduction reactions Use standard reduction potentials to determine the better oxidizing or reducing agent from among several possible choices The cell potential results from the difference in the electrical potentials for each electrode. While it is impossible to determine the electrical potential of a single electrode, we can assign an electrode the value of zero and then use it as a reference. The electrode chosen as the zero is shown in Figure 17.4.1 and is called the standard hydrogen electrode (SHE) . The SHE consists of 1 atm of hydrogen gas bubbled through a 1 M HCl solution, usually at room temperature. Platinum, which is chemically inert, is used as the electrode. The reduction half-reaction chosen as the reference is \[\ce{2H+}(aq,\: 1\:M)+\ce{2e-}⇌\ce{H2}(g,\:1\: \ce{atm}) \hspace{20px} E°=\mathrm{0\: V}\] E ° is the standard reduction potential. The superscript “°” on the E denotes standard conditions (1 bar or 1 atm for gases, 1 M for solutes). The voltage is defined as zero for all temperatures. Figure \(\PageIndex{1}\) : Hydrogen gas at 1 atm is bubbled through 1 M HCl solution. Platinum, which is inert to the action of the 1 M HCl, is used as the electrode. Electrons on the surface of the electrode combine with H + in solution to produce hydrogen gas. A galvanic cell consisting of a SHE and Cu 2+ /Cu half-cell can be used to determine the standard reduction potential for Cu 2+ (Figure \(\PageIndex{2}\)). In cell notation, the reaction is \[\ce{Pt}(s)│\ce{H2}(g,\:1\: \ce{atm})│\ce{H+}(aq,\:1\:M)║\ce{Cu^2+}(aq,\:1\:M)│\ce{Cu}(s)\] Electrons flow from the anode to the cathode. The reactions, which are reversible, are \[\begin{align*} &\textrm{Anode (oxidation): }\ce{H2}(g)⟶\ce{2H+}(aq) + \ce{2e-}\\ &\textrm{Cathode (reduction): }\ce{Cu^2+}(aq)+\ce{2e-}⟶\ce{Cu}(s)\\ &\overline{\textrm{Overall: }\ce{Cu^2+}(aq)+\ce{H2}(g)⟶\ce{2H+}(aq)+\ce{Cu}(s)} \end{align*}\] The standard reduction potential can be determined by subtracting the standard reduction potential for the reaction occurring at the anode from the standard reduction potential for the reaction occurring at the cathode. The minus sign is necessary because oxidation is the reverse of reduction. \[E^\circ_\ce{cell}=E^\circ_\ce{cathode}−E^\circ_\ce{anode}\] \[\mathrm{+0.34\: V}=E^\circ_{\ce{Cu^2+/Cu}}−E^\circ_{\ce{H+/H2}}=E^\circ_{\ce{Cu^2+/Cu}}−0=E^\circ_{\ce{Cu^2+/Cu}}\] Figure \(\PageIndex{2}\) : A galvanic cell can be used to determine the standard reduction potential of Cu 2+ . Using the SHE as a reference, other standard reduction potentials can be determined. Consider the cell shown in Figure \(\PageIndex{2}\), where \[\ce{Pt}(s)│\ce{H2}(g,\:1\: \ce{atm})│\ce{H+}(aq,\: 1\:M)║\ce{Ag+}(aq,\: 1\:M)│\ce{Ag}(s)\] Electrons flow from left to right, and the reactions are \[\begin{align*} &\textrm{anode (oxidation): }\ce{H2}(g)⟶\ce{2H+}(aq)+\ce{2e-}\\ &\textrm{cathode (reduction): }\ce{2Ag+}(aq)+\ce{2e-}⟶\ce{2Ag}(s)\\ &\overline{\textrm{overall: }\ce{2Ag+}(aq)+\ce{H2}(g)⟶\ce{2H+}(aq)+\ce{2Ag}(s)} \end{align*}\] The standard reduction potential can be determined by subtracting the standard reduction potential for the reaction occurring at the anode from the standard reduction potential for the reaction occurring at the cathode. The minus sign is needed because oxidation is the reverse of reduction. \[E^\circ_\ce{cell}=E^\circ_\ce{cathode}−E^\circ_\ce{anode}\] \[\mathrm{+0.80\: V}=E^\circ_{\ce{Ag+/Ag}}−E^\circ_{\ce{H+/H2}}=E^\circ_{\ce{Ag+/Ag}}−0=E^\circ_{\ce{Ag+/Ag}}\] It is important to note that the potential is not doubled for the cathode reaction. The SHE is rather dangerous and rarely used in the laboratory. Its main significance is that it established the zero for standard reduction potentials. Once determined, standard reduction potentials can be used to determine the standard cell potential , \(E^\circ_\ce{cell}\), for any cell. For example, for the following cell: \[\ce{Cu}(s)│\ce{Cu^2+}(aq,\:1\:M)║\ce{Ag+}(aq,\:1\:M)│\ce{Ag}(s)\] \[\begin{align*} &\textrm{anode (oxidation): }\ce{Cu}(s)⟶\ce{Cu^2+}(aq)+\ce{2e-}\\ &\textrm{cathode (reduction): }\ce{2Ag+}(aq)+\ce{2e-}⟶\ce{2Ag}(s)\\ &\overline{\textrm{overall: }\ce{Cu}(s)+\ce{2Ag+}(aq)⟶\ce{Cu^2+}(aq)+\ce{2Ag}(s)} \end{align*}\] \[E^\circ_\ce{cell}=E^\circ_\ce{cathode}−E^\circ_\ce{anode}=E^\circ_{\ce{Ag+/Ag}}−E^\circ_{\ce{Cu^2+/Cu}}=\mathrm{0.80\: V−0.34\: V=0.46\: V}\] Again, note that when calculating \(E^\circ_\ce{cell}\), standard reduction potentials always remain the same even when a half-reaction is multiplied by a factor. Standard reduction potentials for selected reduction reactions are shown in Table \(\PageIndex{1}\). A more complete list is provided in Tables P1 or P2 . Figure \(\PageIndex{3}\) : A galvanic cell can be used to determine the standard reduction potential of Ag + . The SHE on the left is the anode and assigned a standard reduction potential of zero. Half-Reaction E° (V) \(\ce{F2}(g)+\ce{2e-}⟶\ce{2F-}(aq)\) +2.866 \(\ce{PbO2}(s)+\ce{SO4^2-}(aq)+\ce{4H+}(aq)+\ce{2e-}⟶\ce{PbSO4}(s)+\ce{2H2O}(l)\) +1.69 \(\ce{MnO4-}(aq)+\ce{8H+}(aq)+\ce{5e-}⟶\ce{Mn^2+}(aq)+\ce{4H2O}(l)\) +1.507 \(\ce{Au^3+}(aq)+\ce{3e-}⟶\ce{Au}(s)\) +1.498 \(\ce{Cl2}(g)+\ce{2e-}⟶\ce{2Cl-}(aq)\) +1.35827 \(\ce{O2}(g)+\ce{4H+}(aq)+\ce{4e-}⟶\ce{2H2O}(l)\) +1.229 \(\ce{Pt^2+}(aq)+\ce{2e-}⟶\ce{Pt}(s)\) +1.20 \(\ce{Br2}(aq)+\ce{2e-}⟶\ce{2Br-}(aq)\) +1.0873 \(\ce{Ag+}(aq)+\ce{e-}⟶\ce{Ag}(s)\) +0.7996 \(\ce{Hg2^2+}(aq)+\ce{2e-}⟶\ce{2Hg}(l)\) +0.7973 \(\ce{Fe^3+}(aq)+\ce{e-}⟶\ce{Fe^2+}(aq)\) +0.771 \(\ce{MnO4-}(aq)+\ce{2H2O}(l)+\ce{3e-}⟶\ce{MnO2}(s)+\ce{4OH-}(aq)\) +0.558 \(\ce{I2}(s)+\ce{2e-}⟶\ce{2I-}(aq)\) +0.5355 \(\ce{NiO2}(s)+\ce{2H2O}(l)+\ce{2e-}⟶\ce{Ni(OH)2}(s)+\ce{2OH-}(aq)\) +0.49 \(\ce{Cu^2+}(aq)+\ce{2e-}⟶\ce{Cu}(s)\) +0.34 \(\ce{Hg2Cl2}(s)+\ce{2e-}⟶\ce{2Hg}(l)+\ce{2Cl-}(aq)\) +0.26808 \(\ce{AgCl}(s)+\ce{e-}⟶\ce{Ag}(s)+\ce{Cl-}(aq)\) +0.22233 \(\ce{Sn^4+}(aq)+\ce{2e-}⟶\ce{Sn^2+}(aq)\) +0.151 \(\ce{2H+}(aq)+\ce{2e-}⟶\ce{H2}(g)\) 0.00 \(\ce{Pb^2+}(aq)+\ce{2e-}⟶\ce{Pb}(s)\) −0.1262 \(\ce{Sn^2+}(aq)+\ce{2e-}⟶\ce{Sn}(s)\) −0.1375 \(\ce{Ni^2+}(aq)+\ce{2e-}⟶\ce{Ni}(s)\) −0.257 \(\ce{Co^2+}(aq)+\ce{2e-}⟶\ce{Co}(s)\) −0.28 \(\ce{PbSO4}(s)+\ce{2e-}⟶\ce{Pb}(s)+\ce{SO4^2-}(aq)\) −0.3505 \(\ce{Cd^2+}(aq)+\ce{2e-}⟶\ce{Cd}(s)\) −0.4030 \(\ce{Fe^2+}(aq)+\ce{2e-}⟶\ce{Fe}(s)\) −0.447 \(\ce{Cr^3+}(aq)+\ce{3e-}⟶\ce{Cr}(s)\) −0.744 \(\ce{Mn^2+}(aq)+\ce{2e-}⟶\ce{Mn}(s)\) −1.185 \(\ce{Zn(OH)2}(s)+\ce{2e-}⟶\ce{Zn}(s)+\ce{2OH-}(aq)\) −1.245 \(\ce{Zn^2+}(aq)+\ce{2e-}⟶\ce{Zn}(s)\) −0.7618 \(\ce{Al^3+}(aq)+\ce{3e-}⟶\ce{Al}(s)\) −1.662 \(\ce{Mg^2}(aq)+\ce{2e-}⟶\ce{Mg}(s)\) −2.372 \(\ce{Na+}(aq)+\ce{e-}⟶\ce{Na}(s)\) −2.71 \(\ce{Ca^2+}(aq)+\ce{2e-}⟶\ce{Ca}(s)\) −2.868 \(\ce{Ba^2+}(aq)+\ce{2e-}⟶\ce{Ba}(s)\) −2.912 \(\ce{K+}(aq)+\ce{e-}⟶\ce{K}(s)\) −2.931 \(\ce{Li+}(aq)+\ce{e-}⟶\ce{Li}(s)\) −3.04 Tables like this make it possible to determine the standard cell potential for many oxidation-reduction reactions. Example \(\PageIndex{1}\): Cell Potentials from Standard Reduction Potentials What is the standard cell potential for a galvanic cell that consists of Au 3+ /Au and Ni 2+ /Ni half-cells? Identify the oxidizing and reducing agents. Solution Using Table \(\PageIndex{1}\), the reactions involved in the galvanic cell, both written as reductions, are \[\ce{Au^3+}(aq)+\ce{3e-}⟶\ce{Au}(s) \hspace{20px} E^\circ_{\ce{Au^3+/Au}}=\mathrm{+1.498\: V}\] \[\ce{Ni^2+}(aq)+\ce{2e-}⟶\ce{Ni}(s) \hspace{20px} E^\circ_{\ce{Ni^2+/Ni}}=\mathrm{−0.257\: V}\] Galvanic cells have positive cell potentials, and all the reduction reactions are reversible. The reaction at the anode will be the half-reaction with the smaller or more negative standard reduction potential. Reversing the reaction at the anode (to show the oxidation) but not its standard reduction potential gives: \[\begin{align*} &\textrm{Anode (oxidation): }\ce{Ni}(s)⟶\ce{Ni^2+}(aq)+\ce{2e-} \hspace{20px} E^\circ_\ce{anode}=E^\circ_{\ce{Ni^2+/Ni}}=\mathrm{−0.257\: V}\\ &\textrm{Cathode (reduction): }\ce{Au^3+}(aq)+\ce{3e-}⟶\ce{Au}(s) \hspace{20px} E^\circ_\ce{cathode}=E^\circ_{\ce{Au^3+/Au}}=\mathrm{+1.498\: V} \end{align*}\] The least common factor is six, so the overall reaction is \(\ce{3Ni}(s)+\ce{2Au^3+}(aq)⟶\ce{3Ni^2+}(aq)+\ce{2Au}(s)\) The reduction potentials are not scaled by the stoichiometric coefficients when calculating the cell potential, and the unmodified standard reduction potentials must be used. \[E^\circ_\ce{cell}=E^\circ_\ce{cathode}−E^\circ_\ce{anode}=\mathrm{1.498\: V−(−0.257\: V)=1.755\: V}\] From the half-reactions, Ni is oxidized, so it is the reducing agent, and Au 3+ is reduced, so it is the oxidizing agent. Exercise \(\PageIndex{1}\) A galvanic cell consists of a Mg electrode in 1 M Mg(NO 3 ) 2 solution and a Ag electrode in 1 M AgNO 3 solution. Calculate the standard cell potential at 25 °C. Answer \[\ce{Mg}(s)+\ce{2Ag+}(aq)⟶\ce{Mg^2+}(aq)+\ce{2Ag}(s) \hspace{20px} E^\circ_\ce{cell}=\mathrm{0.7996\: V−(−2.372\: V)=3.172\: V} \nonumber\] Summary Assigning the potential of the standard hydrogen electrode (SHE) as zero volts allows the determination of standard reduction potentials, E° , for half-reactions in electrochemical cells. As the name implies, standard reduction potentials use standard states (1 bar or 1 atm for gases; 1 M for solutes, often at 298.15 K) and are written as reductions (where electrons appear on the left side of the equation). The reduction reactions are reversible, so standard cell potentials can be calculated by subtracting the standard reduction potential for the reaction at the anode from the standard reduction for the reaction at the cathode. When calculating the standard cell potential, the standard reduction potentials are not scaled by the stoichiometric coefficients in the balanced overall equation. Key Equations \(E^\circ_\ce{cell}=E^\circ_\ce{cathode}−E^\circ_\ce{anode}\) Glossary standard cell potential \( (E^\circ_\ce{cell})\) the cell potential when all reactants and products are in their standard states (1 bar or 1 atm or gases; 1 M for solutes), usually at 298.15 K; can be calculated by subtracting the standard reduction potential for the half-reaction at the anode from the standard reduction potential for the half-reaction occurring at the cathode standard hydrogen electrode (SHE) the electrode consists of hydrogen gas bubbling through hydrochloric acid over an inert platinum electrode whose reduction at standard conditions is assigned a value of 0 V; the reference point for standard reduction potentials standard reduction potential ( E °) the value of the reduction under standard conditions (1 bar or 1 atm for gases; 1 M for solutes) usually at 298.15 K; tabulated values used to calculate standard cell potentials Contributors Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/[email protected] ).
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Courses/Clackamas_Community_College/CH_112%3A_Chemistry_for_Health_Sciences/01%3A_Chemistry_Matter_and_Measurement/1.E%3A_Chemistry_Matter_and_Measurement_(Exercises)
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Does each statement refer to a chemical property or a physical property? Balsa is a very light wood. If held in a flame, magnesium metal burns in air. Mercury has a density of 13.6 g/mL. Human blood is red. Does each statement refer to a chemical property or a physical property? The elements sodium and chlorine can combine to make table salt. The metal tungsten does not melt until its temperature exceeds 3,000°C. The ingestion of ethyl alcohol can lead to disorientation and confusion. The boiling point of isopropyl alcohol, which is used to sterilize cuts and scrapes, is lower than the boiling point of water. Define element . How does it differ from a compound? Define compound . How does it differ from an element? Give two examples of a heterogeneous mixture. Give two examples of a homogeneous mixture. Identify each substance as an element, a compound, a heterogeneous mixture, or a solution. xenon, a substance that cannot be broken down into chemically simpler components blood, a substance composed of several types of cells suspended in a salty solution called plasma water, a substance composed of hydrogen and oxygen Identify each substance as an element, a compound, a heterogeneous mixture, or a solution. sugar, a substance composed of carbon, hydrogen, and oxygen hydrogen, the simplest chemical substance dirt, a combination of rocks and decaying plant matter Identify each substance as an element, a compound, a heterogeneous mixture, or a solution. air, primarily a mixture of nitrogen and oxygen ringer’s lactate, a standard fluid used in medicine that contains salt, potassium, and lactate compounds all dissolved in sterile water tartaric acid, a substance composed of carbon, hydrogen, and oxygen Identify each material as an element, a compound, a heterogeneous mixture, or a solution. equal portions of salt and sand placed in a beaker and shaken up a combination of beeswax dissolved in liquid hexane hydrogen peroxide, a substance composed of hydrogen and oxygen What word describes each phase change? solid to liquid liquid to gas solid to gas What word describes each phase change? liquid to solid gas to liquid gas to solid Answers physical property chemical property physical property physical property 2. chemical property physical property chemical property physical property An element is a substance that cannot be broken down into chemically simpler components. Compounds can be broken down into simpler substances. 4. A compound is composed of two or more elements combined in a fixed ratio. An element is the simplest chemical substance. a salt and pepper mix and a bowl of cereal (answers will vary) 6. vinegar and rubbing alcohol (answers will vary) element heterogeneous mixture compound 8. compound element heterogeneous mixture solution solution compound 10. heterogeneous mixture solution compound 11. melting or fusion boiling or evaporation sublimation 12. freezing condensation deposition
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Courses/Alma_College/Organic_Chemistry_I_(Alma_College)/09%3A_Reactions_of_Alkyl_Halides-_Nucleophilic_Substitutions_and_Eliminations/9.03%3A_The_SN2_Reaction
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Objectives After completing this section, you should be able to write an expression relating reaction rate to the concentration of reagents for a second-order reaction. determine the order of a chemical reaction from experimentally obtained rate data. describe the essential features of the S N 2 mechanism, and draw a generalized transition state for such a reaction. Key Terms Make certain that you can define, and use in context, the key terms below. bimolecular kinetics rate coefficient rate equation reaction rate second-order reaction S N 2 Study Notes Most of the key terms introduced in this section should already be familiar to you from your previous general chemistry course. Reaction rate refers to the change in concentration of a reactant or product per unit of time. Using strict SI units, reaction rates are expressed in mol · L −1 · s −1 , but in some textbooks you will find this value written as M/s. In general, the reaction rate of a given reaction changes with time, as it is dependent on the concentration of one or more of the reactants. An equation which shows the relationship between the reaction rate and the concentrations of the reactants is known as the rate equation. All rate equations contain a proportionality constant, usually given the symbol k , which is known as the rate coefficient. Some textbooks refer to this value as the “rate constant,” but this name is a little misleading as it is not a true constant. The rate coefficient of a given reaction depends on such factors as temperature and the nature of the solvent. S N 2 is short for “bimolecular nucleophilic substitution.” You will encounter abbreviations for other types of reactions later in this chapter. If you are unclear on the point about the inversion of configuration during an S N 2 reaction, construct a molecular model of a chiral alkyl halide, the transition state formed when this substance reacts with a nucleophile in an S N 2 process, and the product obtained from this reaction. Brønsted-Lowry acid-base reactions In many ways, the proton transfer process in a Brønsted-Lowry acid-base reaction can be thought of as simply a special kind of nucleophilic substitution reaction, one in which the electrophile is a hydrogen rather than a carbon. In both reaction types, we are looking at very similar players: an electron-rich species (the nucleophile/base) attacks an electron-poor species (the electrophile/proton), driving off the leaving group/conjugate base. Instead of showing a specific nucleophile like hydroxide, we will simply refer to the nucleophilic reactant as 'Nu'. In a similar fashion, we will call the leaving group 'X'. We will see as we study actual reactions that leaving groups are sometimes negatively charged, sometimes neutral, and sometimes positively charged. We will also see some examples of nucleophiles that are negatively charged and some that are neutral. Therefore, in this general picture we will not include a charge designation on the 'X' or 'Nu' species. We will generalize the three other groups bonded on the electrophilic central carbon as R 1 , R 2 , and R 3 : these symbols could represent hydrogens as well as alkyl groups. Here, then, is the generalized picture of a concerted (single-step) nucleophilic substitution reaction. General S N 2 Reaction Example The S N 2 Mechanism Bimolecular nucleophilic substitution (S N 2) reactions are concerted , meaning they are a one step process . This means that the process whereby the nucleophile attacks and the leaving group leaves is simultaneous. Hence, the bond-making between the nucleophile and the electrophilic carbon occurs at the same time as the bond-breaking between the electrophilic carbon and the halogen. This is called an ' S N 2' mechanism. In the term S N 2, S stands for 'substitution', the subscript N stands for 'nucleophilic', and the number 2 refers to the fact that this is a bimolecular reaction : the overall rate depends on a step in which two separate molecules (the nucleophile and the electrophile) collide. The mechanism starts when lone pair electrons from the nucleophile attacks the electrophilic carbon of the alkyl halide to form a C-Nu sigma bond. Simultaneously, X-C bond is broken when the electrons are pushed onto the leaving group. Overall during this mechanism, a set of lone pair electrons are transferred from the nucleophile to the leaving groups. If you look carefully at the progress of the S N 2 reaction, you will realize something very important about the outcome. The nucleophile, being an electron-rich species, must attack the electrophilic carbon from the back side relative to the location of the leaving group. Approach from the front side simply doesn't work: the leaving group - which is also an electron-rich group - blocks the way. The result of this backside attack is that the stereochemical configuration at the central carbon inverts as the reaction proceeds. In a sense, the molecule is turned inside out. S N 2 reactions that begin with the R enantiomer as the substrate will form the S enantiomer as the product. Those that begin with the S enantiomer as the substrate will form the R enantiomer as the product. This concept also applies to substrates that are cis and substrates that are trans . If the cis configuration is the substrate, the resulting product will be trans . Conversely, if the trans configuration is the substrate, the resulting product will be cis . What this means is that S N 2 reactions whether enzyme catalyzed or not, are inherently stereoselective: when the substitution takes place at a stereocenter, we can confidently predict the stereochemical configuration of the product. Below is an animation illustrating the principles we have just learned, showing the S N 2 reaction between hydroxide ion and methyl iodide. Notice how backside attack by the hydroxide nucleophile results in inversion at the tetrahedral carbon electrophile. Example Bimolecular Nucleophilic Substitution Reactions and Kinetics In the term S N 2, (as previously stated) the number two stands for bimolecular, meaning there are two molecules involved in the rate determining step. The rate of bimolecular nucleophilic substitution reactions depends on the concentration of both the haloalkane and the nucleophile. To understand how the rate depends on the concentrations of both the haloalkane and the nucleophile, let us look at the following example. The hydroxide ion is the nucleophile and methyl iodide is the haloalkane. If we were to double the concentration of either the haloalkane or the nucleophile, we can see that the rate of the reaction would proceed twice as fast as the initial rate. If we were to double the concentration of both the haloalkane and the nucleophile, we can see that the rate of the reaction would proceed four times as fast as the initial rate. The bimolecular nucleophilic substitution reaction follows second-order kinetics; that is, the rate of the reaction depends on the concentration of two first-order reactants. In the case of bimolecular nucleophilic substitution, these two reactants are the haloalkane and the nucleophile. For further clarification on reaction kinetics, the following links may facilitate your understanding of rate laws, rate constants, and second-order kinetics: Definition of a Reaction Rate Rate Laws and Rate Constants The Determination of the Rate Law Second-Order Reactions S N 2 Reactions Are Stereospecific The S N 2 reaction is stereospecific. A stereospecific reaction is one in which different stereoisomers react to give different stereoisomers of the product. For example, if the substrate is an ( R ) enantiomer, a frontside nucleophilic attack results in retention of configuration, and the formation of the ( R ) enantiomer. A backside nucleophilic attack results in inversion of configuration, and the formation of the ( S ) enantiomer. Conversely, if the substrate is an ( S ) enantiomer, a frontside nucleophilic attack results in retention of configuration, and the formation of the ( S ) enantiomer. A backside nucleophilic attack results in inversion of configuration, and the formation of the ( R ) enantiomer. In conclusion, S N 2 reactions that begin with the ( R ) enantiomer as the substrate will form the ( S ) enantiomer as the product. Those that begin with the ( S ) enantiomer as the substrate will form the ( R ) enantiomer as the product. This concept also applies to substrates that are cis and substrates that are trans . If the cis configuration is the substrate, the resulting product will be trans . Conversely, if the trans configuration is the substrate, the resulting product will be cis . Exercise \(\PageIndex{1}\) 1) The reaction below follows the S N 2 mechanism. a) Write the rate law for this reaction. b) Determine the value of the rate coefficient, k, if the initial concentrations are 0.01 M CH 3 Cl, 0.01 M NaOH, and the initial reaction rate is 6 x 10 -10 M/s. c) Calculate the initial reaction rate if the initial reactant concentrations are changed to 0.02 M CH 3 Cl and 0.0005 M NaOH. 2) Predict the product of a nucleophilic substitution of (S)-2-bromopentane reacting with CH 3 CO 2 - , Show stereochemistry. 3) Predict the structure of the product in this S N 2 reaction. Be sure to specify stereochemistry. 4) Since everything is relative in chemistry, one reaction's nucleophile can be another reaction's leaving group. Some functional groups can only react as a nucleophile or electrophile, while other functional groups can react as either a nucleophile or electrophile depending on the reaction conditions. Classify the following compounds as nucleophile, electrophile, or leaving groups. More than one answer may be possible. a) bromoethane b) hydroxide c) water d) chlorocyclohexane e) ethanol f) bromide Answer 1) a) rate = k [CH 3 Cl] [OH - ] b) substitute the data into the rate expression above and apply algebra to solve for k k = 6 x 10 -6 Lmol -1 s -1 c) Using the rate law above, substitute the value for k from the previous question along with the new concentrations to determine the new initial rate. rate = 6 x 10 -10 M/s 2) 3) 4) a) electrophile (Alkyl halides are always electrophiles - one reason they are an o-chem student's best friend.) b) strong nucleophile c) weak nucleophile and good leaving group d) electrophile (Alkyl halides are always electrophiles - one reason they are an o-chem student's best friend.) e) weak nucleophile, a poor electrophile without clever chemistry (stay tuned for future chapters), good leaving group f) good nucleophile and a good leaving group
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Courses/Sewanee%3A_The_University_of_the_South/Organic_Chemistry_Lab_Textbook/05%3A_Filtering_Methods/5.04%3A_Suction_Filtration
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Suction Filtration Overview Suction filtration (vacuum filtration) is the standard technique used for separating a solid-liquid mixture when the goal is to retain the solid (for example in crystallization). Similar to gravity filtration, a solid-liquid mixture is poured onto a filter paper, with the main difference being that the process is aided by suction beneath the funnel (Figures 1.70 + 1.71). The process has advantages and disadvantages in comparison to gravity filtration. Advantages: 1) Suction filtration is much faster than gravity filtration, often taking less than one minute with good seals and a good vacuum source. 2) Suction filtration is more efficient at removing residual liquid, leading to a purer solid. This is especially important in crystallization, as the liquid may contain soluble impurities which could adsorb back onto the solid surface when the solvent evaporates. Disadvantages: The force of suction may draw fine crystals through the filter paper pores, leading to a quantity of material that cannot be recovered from the filter paper, and possibly an additional quantity that is lost in the filtrate. This method therefore works best with large crystals. On small scales, the loss of material to the filter paper and filtrate is significant, and so other methods are recommended for microscale work. Rinsing As the goal of suction filtration is to fully separate a solid from its surrounding liquid, rinsing the solid is necessary if the liquid cannot easily evaporate. In the case of crystallization, the liquid may contain impurities that can reincorporate into the solid if not removed. To rinse a suction-filtered solid, the vacuum is removed and a small portion of cold solvent is poured over the solid (the " filter cake "). In the case of crystallization, the same solvent from the crystallization is used. The solid is then delicately slushed around in the solvent with a glass rod, and the vacuum is reapplied to remove the rinse solvent. To demonstrate the importance of a rinse, Figure 1.72 shows the recovery of a white solid from a yellow liquid using suction filtration. The yellow liquid seemed to be somewhat retained by the solid, as the first crystals collected had a yellow tint (Figure 1.72b). However, rinsing with a few portions of cold solvent were effective at removing the yellow liquid (Figure 1.72d), which could have been reincorporated into the solid without the rinse. Assemble the suction filtration flask Clamp a side-arm Erlenmeyer flask to a ring stand or latticework and attach a thick-walled rubber hose to its sidearm. Connect this thick tubing to the yellow nozzle on the side of the hood. It is best not to bend or strain the tubing as much as is practical, as this may cause poor suction. Place a rubber sleeve (or filter adapter) and Buchner funnel atop the side-arm Erlenmeyer flask (Figure 1.73a). Alternatively, use a Hirsch funnel for small scales (Figure 1.73d). Obtain a filter paper that will fit perfectly into the Buchner or Hirsch funnel. Filter papers are not completely flat and have a subtle arc to their shape (Figure 1.73b). Place the filter paper inside the funnel concave side down (Figure 1.75b+c). The paper should cover all the holes in the funnel, and with the paper arching downward (Figure 1.74a), the solid will be less likely to creep around the edges. Turn on the vacuum using the yellow handle. Wet the filter paper with cold solvent (using the same solvent used in crystallization, if applicable, Figure 1.74b). Suction should drain the liquid and hold the moist filter paper snugly over the holes in the filter. If the solvent does not drain or suction is not occurring, you may need to press down on the funnel (Figure 1.74c) to create a good seal between the glass and rubber sleeve. Lack of suction may also be from a faulty aspirator or a leak in the system: to test for suction remove the tubing from the suction flask and place your finger over the end (Figure 1.74d). Filter and Rinse the Mixture Swirl the mixture to be filtered to dislodge the solid from the sides of the flask. If the solid is thick, use a spatula or stirring rod to free it from the glass (Figure 1.75a). In the context of crystallization, the flask will have previously been in an ice bath. Use a paper towel to dry water residue from the outside of the flask so water does not accidentally pour onto the solid. With a quick motion, swirl and dump the solid into the funnel in portions (Figure 1.75b). If the solid is thick, scoop it out of the flask onto the filter paper (Figure 1.75c). It's best if the solid can be directed toward the middle of the filter paper, as solid near the edges may creep around the filter paper. A small amount of chilled solvent (\(1\)-\(2 \: \text{mL}\) for macroscale work) can be used to help rinse any residual solid from the flask into the funnel (Figure 1.75d). In crystallization, it is not wise to use an excessive amount of solvent as it will decrease the yield by dissolving small amounts of crystals. Again, press on the funnel to create a good seal and efficient drainage if necessary. Rinse the solid on the filter paper to remove contaminants that may remain in the residual liquid. Break the vacuum on the flask by removing the rubber tubing on the filter flask. Add \(1\)-\(2 \: \text{mL}\) of cold solvent (Figure 1.76b). Use a glass stirring rod to break up any solid chunks and distribute the solvent to all portions of the solid (Figure 1.76c), taking care to not rip or dislodge the filter paper. Reapply the vacuum to the flask, and dry the solid with suction for a few minutes. After filtration is complete, again open the flask to the atmosphere by releasing the pinch clamp or opening it elsewhere, and turn off the water connected to the aspirator. Transfer the solid, filter paper and all, to a pre-weighed watch glass using a spatula (Figures 1.77a+b). The filter cake shouldn't be mushy, and if it is, the liquid was not adequately removed (try a different aspirator and repeat suction filtration). Allow the solid to dry overnight if possible before recording a final mass or melting point. The solid will flake off the filter paper more easily when completely dry (Figure 1.77c). If pressed for time, a solid can be quickly dried in the following ways: If the solid is wet with water, it can be placed in a \(110^\text{o} \text{C}\) oven (if the melting point is not below this temperature). If the solid is wet with organic solvent, it should never be placed in an oven as it may ignite. If the solid is wet with organic solvent, it can be pressed between fresh pieces of filter paper (multiple times if needed) to quickly dry them. Inevitably some solid will be lost on the filter paper. Suction Filtration Summary 0 1 2 3 NaN NaN NaN NaN Clamp a side-armed Erlenmeyer flask. Connect thick-walled hosing from the side arm to a vacuum trap and the water aspirator. Place a vacuum sleeve on the Buchner (or Hirsch) funnel, then filter paper on the funnel so it arches downward. Turn on the aspirator. Add a few \(\text{mL}\) of the same solvent used in the flask to wet the filter. The solvent should drain with suction. Swirl the mixture to be filtered to dislodge the solid from the sides of the flask. With a quick motion, pour the slurry into the funnel in portions. In some applications, (e.g. crystallization), rinse with solvent: Open the apparatus to the atmosphere, then turn off the aspirator. Add a few \(\text{mL}\) of cold solvent to the filter paper. Delicately swirl the solid in the solvent with a glass rod. Apply suction again for a few minutes (repeat the rinse step if necessary). Dry the solid on a watch glass along with the filter paper, overnight if possible. The solid will flake off the paper when dried. Table 1.10: Procedural summary for suction filtration.
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Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Topics_in_Thermodynamics_of_Solutions_and_Liquid_Mixtures/00%3A_Front_Matter/02%3A_InfoPage
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This text is disseminated via the Open Education Resource (OER) LibreTexts Project ( https://LibreTexts.org ) and like the hundreds of other texts available within this powerful platform, it is freely available for reading, printing and "consuming." Most, but not all, pages in the library have licenses that may allow individuals to make changes, save, and print this book. Carefully consult the applicable license(s) before pursuing such effects. Instructors can adopt existing LibreTexts texts or Remix them to quickly build course-specific resources to meet the needs of their students. Unlike traditional textbooks, LibreTexts’ web based origins allow powerful integration of advanced features and new technologies to support learning. The LibreTexts mission is to unite students, faculty and scholars in a cooperative effort to develop an easy-to-use online platform for the construction, customization, and dissemination of OER content to reduce the burdens of unreasonable textbook costs to our students and society. The LibreTexts project is a multi-institutional collaborative venture to develop the next generation of open-access texts to improve postsecondary education at all levels of higher learning by developing an Open Access Resource environment. The project currently consists of 14 independently operating and interconnected libraries that are constantly being optimized by students, faculty, and outside experts to supplant conventional paper-based books. These free textbook alternatives are organized within a central environment that is both vertically (from advance to basic level) and horizontally (across different fields) integrated. The LibreTexts libraries are Powered by NICE CXOne and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. This material is based upon work supported by the National Science Foundation under Grant No. 1246120, 1525057, and 1413739. Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not necessarily reflect the views of the National Science Foundation nor the US Department of Education. Have questions or comments? For information about adoptions or adaptions contact [email protected] . More information on our activities can be found via Facebook ( https://facebook.com/Libretexts ), Twitter ( https://twitter.com/libretexts ), or our blog ( http://Blog.Libretexts.org ). This text was compiled on 04/21/2025
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Courses/can/CHEM_231%3A_Organic_Chemistry_I_Textbook/10%3A_Organohalides/10.04%3A_Preparing_Alkyl_Halides_from_Alkenes_-_Allylic_Bromination
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Objectives After completing this section, you should be able to write the equation for the bromination of a symmetrical alkene using N-bromosuccinimide. predict the product formed when a given symmetrical alkene is treated with N-bromosuccinimide. identify the reagent, the symmetrical alkene, or both, needed to produce a given allyl halide by allylic bromination. list the following radicals in order of increasing or decreasing stability: allyl, vinyl, primary alkyl, secondary alkyl, tertiary alkyl, methyl. explain the ease of forming an allyl radical, and the difficulty of forming a vinyl radical, in terms of relative $\ce{\sf{C–H}}$ bond dissociation energies. Key Terms Make certain that you can define, and use in context, the key term below. allylic carbon Study Notes We have discussed the electrophilic addition of X 2 and HX to alkenes as a route to forming alkyl halides (Sections 7.8 and 8.2 ). In this section we introduce bromination at the allyic position with N-bromosuccinimide (NBS). Notice that at the moment we are restricting our studies to the allylic bromination of symmetrical alkenes, such as cyclohexene. When we introduce an element of asymmetry, we find that more than one allyl radical can be formed; therefore, we must assess the relative stability of each radical when trying to predict which product will predominate. The method of doing this assessment is described in the next section. Allylic Bromination Previously, alkyl halides have been produced though reactions with alkenes. Hydrogen halides (HCl, HBr, and HI) react with alkenes in an electrophilic addition reaction discussed in Section 7-8 to yield alkyl halides as products. Also, Bromine (Br 2 ) and chlorine (Cl 2 ) can react with alkenes to provide dihalogenated products as discussed in Section 8-2 . Another method for preparing alkyl halides from alkenes is with N-bromosuccinimide (NBS) in carbon tetrachloride (CCl 4 ) solution with the presence of light. The reaction specifically causes the substitution of bromine with a hydrogen attached to a carbon adjacent to the double bond - the allylic position. General Reaction Mechanism The allylic bromination with NBS is analagous to the alkane halogenation reaction ( Section 10. 2) since it also occurs as a radical chain reaction. NBS is the most commonly used reagent to produce low concentrations of bromine. When suspended in tetrachloride (CCl 4 ), NBS reacts very rapidly with the HBr formed during the reaction mechanism to provide bromine (Br 2 ) which is required for the reaction to continue. Under the correct conditions, NBS provides a constant but very low concentration of Br 2 in the reaction mixture. The low concentration of Br 2 helps to prevent the formation of unwanted side-products. The mechanism starts with the formation of a small amount of bromine radical which then abstracts an allylic hydrogen to form an allylic radical and HBr. The HBr can then react with NBS to form the Br 2 required for the reaction. The allylic radical then abstracts a bromine atom from Br 2 to form the allyl halide product and a bromine radical. The bromine radical produced allows the reaction to continue. The predominance of allylic substitution over other positions is based on bond dissociation energies. An allylic C-H bond has a strength of about 88 kcal/mol which is much weaker than a typical alkyl C-H bond (98 kcal/mol) or vinylic C-H bond (111 kcal/mol). Therefore, an allylic C-H bond is most likely to form a free radical and react. Because a allylic C-H bond requires less energy to undergo homolytic cleavage than even a tertiary C-H bond, it can be inferred that a allylic radical is more stable than a tertiary radical. The ordering of stability in radicals can be expanded to include vinylic and allylic radicals. The enhanced stability of allyl radicals can be attributed to resonance stabilization which will be discussed in the next section.
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Courses/University_of_Georgia/CHEM_3212%3A_Physical_Chemistry_II/08%3A_Gibbs_and_Helmholtz_Energies/8.0S%3A_8.S%3A_Putting_the_Second_Law_to_Work_(Summary)
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Learning Objectives After mastering the material presented in this chapter, one will be able to: Define the free energy function s \(A\) and \(G\), and relate changes in these functions to the spontaneity of a given process an d constant volume and pressure respectively. Use the definitions of entropy and reversible work of expansion to write an equation that combines the first and second laws of thermodynamics. Utilize the combined first and second law relationship to derive Maxwell Relations stemming from the definitions o f \(U\), \(H\), \(A\), and \(G\). Utilize the Maxwell Relations to derive expressions that govern changes in thermodynamic variable as systems move along specified pathways (such as constant temperature, pressure, volume, or adiabatic pathways.) Derive and utilize an expression describing the volume dependenc e of \(A\). Derive and utilize an expression describing the pressure dependenc e of \(G\). Derive and utilize expressions that describe the temperature, dependenc e of \(A\) and \(G\). Derive an expression for, and evaluate the difference betw een \(C_p\) and \(C_V\) for any substance, in terms of \(T\), \(V\), \(\alpha\), and \(\kappa_T\). Vocabulary and Concepts free energy Gibbs Free Energy Gibbs function Gibbs-Helmholtz equation Helmholtz function maximum work Maxwell Relation standard free energy of formation (\(\Delta G_f^Oo\)
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Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/20%3A_Entropy_and_The_Second_Law_of_Thermodynamics/20.06%3A_We_Must_Always_Devise_a_Reversible_Process_to_Calculate_Entropy_Changes
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The second law of thermodynamics can be formulated in many ways, but in one way or another they are all related to the fact that the state function entropy, \(S\), tends to increase over time in isolated systems. For a long time, people have looked at the entire universe as an example of an isolated system and concluded that its entropy must be steadily increasing until \(\delta S_{universe}\) becomes zero. As we will see below, the second law has important consequences for the question of how we can use heat to do useful work. Of late, cosmologists like the late Richard Hawkins have begun to question the assumption that the entropy of the universe is steadily increasing. The key problem is the role that gravity and relativity play in creating black holes. Vacuum Expansion Let's compare two expansions from \(V_1\) to \(V_2\) for an ideal gas, both are isothermal. The first is an irreversible one, where we pull a peg an let the piston move against vacuum: The second one is a reversible isothermal expansion from \(V_1\) to \(V_2\) (and \(P_1\) to \(P_2\)) that we have examined before. In both cases, the is no change in internal energy since \(T\) does not change. During the irreversible expansion, however, there is also no volume work because the piston is expanding against a vacuum and the following integral: \[\int -P_{ext}dV = 0 \nonumber \] integrates to zero. The piston has nothing to perform work against until it slams into the right hand wall. At this point \(V=V_2\) and then \(dV\) becomes zero. This is not true for the reversible isothermal expansion as the external pressure must always equal the internal pressure. No energy and no work means no heat! Clearly the zero heat is irreversible heat (\(q_{irr} = 0\)) and this makes it hard to calculate the entropy of this spontaneous process. But then this process ends in the same final state as the reversible expansion from \(V_1\) to \(V_2\). We know that \(dU\) is still zero, but now \(δw_{rev} = -δq_{rev}\) is nonzero. We calculated its value before: \[q_{rev} = nRT \ln \left(\dfrac{V_2}{V_1} \right) \label{Vacuum} \] The Claussius definition of entropy change can be used to find \(\Delta S\) (under constant temperature). \[\Delta S = \dfrac{q_{rev}}{T} \label{Claussius} \] Substituting Equation \(\ref{Vacuum}\) into Equation \ref{Claussius} results in \[\Delta S = nR \ln \left(\dfrac{V_2}{V_1} \right) \nonumber \] As \(S\) is a state function this equation also holds for the irreversible expansion against vacuum. Always calculate the entropy difference between two points along a reversible path. For the irreversible expansion into vacuum we see that \[\begin{align*}\Delta S_\text{total} &= \Delta S_\text{sys} + \Delta S_\text{surr} \\[4pt] &= nR\ln \left( \dfrac{V_2}{V_1}\right) + 0 \\[4pt] &= nR\ln \left( \dfrac{V_2}{V_1}\right) \end{align*} \nonumber \] For the reversible expansion, heat is transferred to the system while the system does work on the surroundings in order to keep the process isothermal: \[\begin{align*}\Delta S_\text{sys} &= nR\ln \left( \dfrac{V_2}{V_1}\right) \end{align*} \] The entropy change for the surrounding is the opposite of the system: \[\begin{align*}\Delta S_\text{surr} &= -nR\ln \left( \dfrac{V_2}{V_1}\right) \end{align*} \] This is because the amount of heat transferred to the system is the same as the heat transferred from the surroundings and this process is reversible so the system and surroundings are at the same temperature (equilibrium). Heat is related to entropy by the following equation: \[ dS = \frac{\delta q}{T} \nonumber \] Therefore, the total entropy change for the reversible process is zero: \[\begin{align*}\Delta S_\text{total} &= \Delta S_\text{sys} + \Delta S_\text{surr} \\[4pt] &= nR\ln \left( \dfrac{V_2}{V_1}\right)- nR\ln \left( \dfrac{V_2}{V_1}\right) \\[4pt] &=0 \end{align*} \] The Mixing of Two Gases Consider two ideal gases at same pressure separated by a thin wall that is punctured. Both gases behave as if the other one is not there and again we get a spontaneous process, mixing in this case. If the pressure is the same the number of moles of each gas should be proportional to the original volumes, \(V_A\) and \(V_B\), and the total number of moles to the total volume \(V_{tot}\). For gas A we can write: \[\Delta S_A = n_A R \ln \dfrac{V_{tot}}{V_A} = n_A R \ln \dfrac{n_{tot}}{n_A} \nonumber \] and similarly for gas B we can write: \[\Delta S_B = n_B R \ln \dfrac{V_{tot}}{V_B} = n_B R \ln \dfrac{n_{tot}}{n_B} \nonumber \] The total entropy change is therefore the sum of constituent entropy changes: \[ \Delta S = \Delta S_A + \Delta S_B \nonumber \] and the entropy change total per mole of gas is: \[ \dfrac{\Delta S}{n_{tot}} =R \dfrac{\left[n_B \ln \dfrac{n_{tot}}{n_B}+ n_A \ln \dfrac{n_{tot}}{n_A} \right ]}{n_{tot}} \label{EqTot} \] Equation \(\ref{EqTot}\) can be simplified using mole fractions: \[\chi_A = \dfrac{n_A}{n_{tot}} \nonumber \] and the mathematical relationship of logarithms that: \[\ln \left( \dfrac{x}{y} \right)= - \ln \left( \dfrac{y}{x} \right) \nonumber \] to: \[\Delta \bar{S} = -R \left [\chi_A\ln \chi_A +\chi_B \ln \chi_B \right] \label{Molar Entropy} \] In the case of mixing of more than two gases, Equation \(\ref{Molar Entropy}\) can be expressed as: \[\Delta \bar{S} = -R \sum \chi_i\ln \chi_i \label{Sum Entropy} \] This entropy expressed in Equations \(\ref{Molar Entropy}\) and \(\ref{Sum Entropy}\) is known as the entropy of mixing ; its existence is the major reason why there is such a thing as diffusion and mixing when gases, and also solutions (even solid ones), are brought into contact with each other.
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Courses/American_River_College/Chemistry_305_(S21_Zarzana)/Map%3A_Introductory_Chemistry_(Tro)/14%3A_Acids_and_Bases/14.10%3A_AcidBase_Titration
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Learning Objectives Understand the basics of acid-base titrations Understand the use of indicators Perform a titration calculation correctly. The reaction of an acid with a base to make a salt and water is a common reaction in the laboratory, partly because so many compounds can act as acids or bases. Another reason that acid-base reactions are so prevalent is because they are often used to determine quantitative amounts of one or the other. Performing chemical reactions quantitatively to determine the exact amount of a reagent is called a titration . A titration can be performed with almost any chemical reaction for which the balanced chemical equation is known. Here, we will consider titrations that involve acid-base reactions. During an acid-base titration, an acid with a known concentration (a standard solution ) is slowly added to a base with an unknown concentration (or vice versa). A few drops of indicator solution are added to the base. The indicator will signal, by color change, when the base has been neutralized (when [H + ] = [OH - ]). At that point—called the equivalence point, or end point —the titration is stopped. By knowing the volumes of acid and base used, and the concentration of the standard solution, calculations allow us to determine the concentration of the other solution. It is important to accurately measure volumes when doing titrations. The instrument you would use is called a burette (or buret). For example, suppose 25.66 mL (or 0.02566 L) of 0.1078 M HCl was used to titrate an unknown sample of NaOH. What mass of NaOH was in the sample? We can calculate the number of moles of HCl reacted: # mol HCl = (0.02566 L)(0.1078 M) = 0.002766 mol HCl We also have the balanced chemical reaction between HCl and NaOH: \[\ce{HCl + NaOH → NaCl + H2O} \nonumber\] So we can construct a conversion factor to convert to number of moles of NaOH reacted: \[0.002766\cancel{mol\, HCl}\times \frac{1\, mol\, NaOH}{1\cancel{mol\, HCl}}=0.002766\, mol\, NaOH \nonumber\] Then we convert this amount to mass, using the molar mass of NaOH (40.00 g/mol): \[0.002766\cancel{mol\, HCl}\times \frac{40.00\, g\, NaOH}{1\cancel{mol\, HCl}}=0.1106\, g\, NaOH \nonumber\] This is type of calculation is performed as part of a titration. Example \(\PageIndex{1}\): Equivalence Point What mass of Ca(OH) 2 is present in a sample if it is titrated to its equivalence point with 44.02 mL of 0.0885 M HNO 3 ? The balanced chemical equation is as follows: \[\ce{2HNO3 + Ca(OH)2 → Ca(NO3)2 + 2H2O} \nonumber\] Solution In liters, the volume is 0.04402 L. We calculate the number of moles of titrant: # moles HNO 3 = (0.04402 L)(0.0885 M) = 0.00390 mol HNO 3 Using the balanced chemical equation, we can determine the number of moles of Ca(OH) 2 present in the analyte: \[0.00390\cancel{mol\, HNO_{3}}\times \frac{1\, mol\, Ca(OH)_{2}}{2\cancel{mol\, HNO_{3}}}=0.00195\, mol\, Ca(OH)_{2} \nonumber\] Then we convert this to a mass using the molar mass of Ca(OH) 2 : \[0.00195\cancel{mol\, Ca(OH)_{2}}\times \frac{74.1\, g\, Ca(OH)_{2}}{\cancel{mol\, Ca(OH)_{2}}}=0.144\, g\, Ca(OH)_{2} \nonumber\] Exercise \(\PageIndex{1}\) What mass of H 2 C 2 O 4 is present in a sample if it is titrated to its equivalence point with 18.09 mL of 0.2235 M NaOH? The balanced chemical reaction is as follows: \[\ce{H2C2O4 + 2NaOH → Na2C2O4 + 2H2O} \nonumber\] Answer 0.182 g Exercise \(\PageIndex{2}\) If 25.00 mL of HCl solution with a concentration of 0.1234 M is neutralized by 23.45 mL of NaOH, what is the concentration of the base? Answer 0.1316 M NaOH Exercise \(\PageIndex{3}\) A 20.0 mL solution of strontium hydroxide, Sr(OH) 2 , is placed in a flask and a drop of indicator is added. The solution turns color after 25.0 mL of a standard 0.0500 M HCl solution is added. What was the original concentration of the Sr(OH) 2 solution? Answer \(3.12 \times 10^{-2}\;M\) Sr(OH) 2 Indicator Selection for Titrations Which indicator is used depends on which type of titration you are performing. The indicator of choice should change color when enough of one substance (acid or base) has been added to exactly use up the other substance. Only when a strong acid and a strong base are produced will the resulting solution be neutral. The three main types of acid-base titrations, and suggested indicators, are: Titration between . . . Indicator Explanation strong acid and strong base any NaN strong acid and weak base methyl orange changes color in the acidic range (3.2 - 4.4) weak acid and strong base phenolphthalein changes color in the basic range (8.2 - 10.6) Summary A titration is the quantitative reaction of an acid and a base. Indicators are used to show that all the analyte has reacted with the titrant. Contributions & Attributions This page was constructed from content via the following contributor(s) and edited (topically or extensively) by the LibreTexts development team to meet platform style, presentation, and quality: Peggy Lawson (Oxbow Prairie Heights School). Funded by Saskatchewan Educational Technology Consortium. Marisa Alviar-Agnew ( Sacramento City College ) Henry Agnew (UC Davis)
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Courses/University_of_Kansas/KU%3A_CHEM_110_GOB_Chemistry_(Sharpe_Elles)/06%3A_Introduction_to_Chemical_Reactions
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Chemical change is a central concept in chemistry. The goal of chemists is to know how and why a substance changes in the presence of another substance or even by itself. Because there are tens of millions of known substances, there are a huge number of possible chemical reactions. In this chapter, we will find that many of these reactions can be classified into a small number of categories according to certain shared characteristics. 6.0: Prelude to Introduction to Chemical Reactions Although yeast has been used for thousands of years, its true nature has been known only for the last two centuries. Yeasts are single-celled fungi. About 1,000 species are recognized, but the most common species is Saccharomyces cerevisiae, which is used in bread making. Other species are used for the fermentation of alcoholic beverages. Some species can cause infections in humans. 6.1: The Law of Conservation of Matter One scientific law that provides the foundation for understanding in chemistry is the law of conservation of matter. It states that in any given system that is closed to the transfer of matter (in and out), the amount of matter in the system stays constant. A concise way of expressing this law is to say that the amount of matter in a system is conserved. The amount of matter in a closed system is conserved. 6.2: Chemical Equations Chemical reactions are represented by chemical equations that list reactants and products. Proper chemical equations are balanced; the same number of each element’s atoms appears on each side of the equation. 6.3: Quantitative Relationships Based on Chemical Equations A balanced chemical equation not only describes some of the chemical properties of substances—by showing us what substances react with what other substances to make what products—but also shows numerical relationships between the reactants and the products. The study of these numerical relationships is called stoichiometry. A balanced chemical equation gives the ratios in which molecules of substances react and are produced in a chemical reaction. 6.4: Some Types of Chemical Reactions Although there are untold millions of possible chemical reactions, most can be classified into a small number of general reaction types. Classifying reactions has two purposes: it helps us to recognize similarities among them, and it enables us to predict the products of certain reactions. A particular reaction may fall into more than one of the categories that we will define in this book. 6.5: Oxidation-Reduction (Redox) Reactions Chemical reactions in which electrons are transferred are called oxidation-reduction, or redox, reactions. Oxidation is the loss of electrons. Reduction is the gain of electrons. Oxidation and reduction always occur together, even though they can be written as separate chemical equations. 6.6: Redox Reactions in Organic Chemistry and Biochemistry Redox reactions are common in organic and biological chemistry, including the combustion of organic chemicals, respiration, and photosynthesis. 6.E: Introduction to Chemical Reactions (Exercises) 6.S: Introduction to Chemical Reactions (Summary) To ensure that you understand the material in this chapter, you should review the meanings of the following bold terms in the following summary and ask yourself how they relate to the topics in the chapter.
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Bookshelves/Analytical_Chemistry/Instrumental_Analysis_(LibreTexts)/05%3A_Signals_and_Noise_(TBD)/5.03%3A_Signal-to-Noise_Enhancement
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There are two broad approaches we can use to improve the signal-to-noise ratio: hardware and software. Hardware approaches are built into the instrument and include decisions on how the instrument is set-up for making measurements (for example, the choice of a scan rate or a slit width), and how the signal is processed by the instrument (for example, using electronic filters). A few approaches are briefly considered here; others are included with the discussion of individual instruments. Software solutions are computational approaches in which we manipulate the data either while we are collecting it or after data acquisition is complete. Hardware Solutions One way to reduce noise is to focus on the circuitry, or hardware, used to measure the signal. Shielding One way to reduce environmental noise is to prevent it from entering into the instrument's electronic circuitry. One approach is to use a Faraday cage in which the instrument sits within a room or space covered with a conductive material. Electromagnetic radiation from the environment is absorbed by the conductive material and then shunted away to the ground. Rather than encasing the entire instrument in a Faraday cage, particularly sensitive portions of the circuitry can be shielded. Differential Amplifier A difference amplifier (see Chapter 3) is an electrical circuit used to determine the difference between two input voltages or currents and to return that difference as a larger voltage or current. As the magnitude of the noise in the two input signals is generally similar in value—that is, it is in phase—while the signal of interest is not, much of the noise's contribution to the signal is subtracted out. Filtering When the frequency of the noise is quite different from the frequency of the signal, a simple electrical circuit can be used to remove the high frequency noise and pass the low frequency signal; this is called a low-pass filter. See Chapter 2 for details on low-pass filters. Modulation When the signal of interest has a low frequency, the effect of flicker noise becomes significant because a technique that removes low frequency noise will remove the signal as well. Modulation is a process of increasing the frequency of the signal. When complete, a high-pass filter is used to remove the noise. Reversing the modulation returns the original signal, but with much of the noise removed. Software Solutions In this section we will consider three common computational tools for improving the signal-to-noise ratio: signal averaging, digital smoothing, and Fourier filtering. Signal Averaging The most important difference between the signal and the noise is that a signal is determinate (fixed in value) and the noise is indeterminate (random in value). If we measure a pure signal several times, we expect its value to be the same each time; thus, if we add together n scans, we expect that the net signal, \(S_n\), is defined as \[S_n = n S \nonumber \] where \(S\) is the signal for a single scan. Because noise is random, its value varies from one run to the next, sometimes with a value that is larger and sometimes with a value that is smaller, and sometimes with a value that is positive and sometimes with a value that is negative. On average, the standard deviation of the noise increases as we make more scans, but it does so at a slower rate than for the signal \[s_n = \sqrt{n} s \nonumber \] where \(s\) is the standard deviation for a single scan and \(s_n\) is the standard deviation after n scans. Combining these two equations, shows us that the signal-to-noise ratio, \(S/N\), after n scans increases as \[(S/N)_n = \frac{S_n}{s_n} = \frac{nS}{\sqrt{n}s} = \sqrt{n}(S/N)_{n = 1} \nonumber \] where \((S/N)_{n = 1}\) is the signal-to-noise ratio for the initial scan. Thus, when \(n = 4\) the signal-to-noise ratio improves by a factor of 2, and when \(n = 16\) the signal-to-noise ratio increases by a factor of 4. Figure \(\PageIndex{1}\) shows the improvement in the signal-to-noise ratio for 1, 2, 4, and 8 scans. Signal averaging works well when the time it takes to collect a single scan is short and when the analyte's signal is stable with respect to time both because the sample is stable and the instrument is stable; when this is not the case, then we risk a time-dependent change in \(S_\text{analyte}\) and/or \(s_\text{noise}\) Because the equation for \((S/N)_n\) is proportional to the \(\sqrt{n}\), the relative improvement in the signal-to-noise ratio decreases as \(n\) increases; for example, 16 scans gives a \(4 \times\) improvement in the signal-to-noise ratio, but it takes an additional 48 scans (for a total of 64 scans) to achieve a \(8 \times\) improvement in the signal-to-noise ratio. Digital Smoothing Filters One characteristic of noise is that its magnitude fluctuates rapidly in contrast to the underlying signal. We see this, for example, in Figure \(\PageIndex{1}\) where the underlying signal either remains constant or steadily increases or decreases while the noise fluctuates chaotically. Digital smoothing filters take advantage of this by using a mathematical function to average the data for a small range of consecutive data points, replacing the range's middle value with the average signal over that range. Moving Average Filters For a moving average filter, also called a boxcar filter, we replace each point by the average signal for that point and an equal number of points on either side; thus, a moving average filter has a width, \(w\), of 3, 5, 7, ... points. For example, suppose the first five points in a sequence are 0 1 2 3 4 0.8 0.3 0.8 0.2 1.0 then a three-point moving average (\(w = 3)\) returns values of 0 1 2 3 4 NaN 0.63 0.43 0.67 NaN where, for example, 0.63 is the average of 0.80, 0.30, and 0.80. Note that we lose \((w - 1)/2 = (3 - 1)/2 = 1\) points at each end of the data set because we do not have a sufficient number of data points to complete a calculation for the first and the last point. Figure \(\PageIndex{2}\) shows the improvement in the \(S/N\) ratio when using moving average filters with widths of 5, 9, and 13. One limitation to a moving average filter is that it distorts the original data by removing points from both ends, although this is not a serious concern if the points in question are just noise. Of greater concern is the distortion in a signal's height if we use a range that is too wide; for example, Figure \(\PageIndex{3}\), shows how a 23-point moving average filter (shown in blue) applied to the noisy signal in the upper left quadrant of Figure \(\PageIndex{2}\), reduces the height of the original signal (shown in black). Because the filter's width—shown by the red bar—is similar to the peak's width, as the filter passes through the peak it systematically reduces the signal by averaging together values that are mostly smaller than the maximum signal. Savitzky-Golay Filters A moving average filter weights all points equally; that is, points near the edges of the filter contribute to the average as a level equal to points near the filter's center. A Savitzky-Golay filter uses a polynomial model that weights each point differently, placing more weight on points near the center of the filter and less weight on points at the edge of the filter. Specific values depend on the size of the window and the polynomial model; for example, a five-point filter using a second-order polynomial has weights of \[-3/35 \quad \quad 12/35 \quad \quad 17/35 \quad \quad 12/35 \quad \quad -3/35 \nonumber \] For example, suppose the first five points in a sequence are 0 1 2 3 4 0.8 0.3 0.8 0.2 1.0 then this Savitzky-Golay filter returns values of 0 1 2 3 4 NaN NaN 0.41 NaN NaN where, for example, the value for the middle point is \[0.80 \times \frac{-3}{35} + 0.30 \times \frac{12}{35} + 0.80 \times \frac{17}{35} + 0.20 \times \frac{12}{35} + 1.00 \times \frac{-3}{35} = 0.406 \approx 0.41 \nonumber \] Note that we lose \((w - 1)/2 = (5 - 1)/2 = 2\) points at each end of the data set, where w is the filter's range, because we do not have a sufficient number of data points to complete the calculations. For other Savitzky-Golay smoothing filters, see Savitzky, A.; Golay, M. J. E. Anal Chem , 1964 , 36 , 1627-1639 . Figure \(\PageIndex{4}\) shows the improvement in the \(S/N\) ratio when using Savitzky-Golay filters using a second-order polynomial with 5, 9, and 13 points. Because a Savitzky-Golay filter weights points differently than does a moving average smoothing filter, a Savitzky-Golay filter introduces less distortion to the signal, as we see in the following figure. Fourier Filtering This approach to improving the signal-to-noise ratio takes advantage of a mathematical technique called a Fourier transform (FT). The basis of a Fourier transform is that we can express a signal in two separate domains. In the first domain the signal is characterized by one or more peaks, each defined by its position, its width, and its area; this is called the frequency domain. In the second domain, which is called the time domain, the signal consists of a set of oscillations, each defined by its frequency, its amplitude, and its decay rate. The Fourier transform—and the inverse Fourier transform—allow us to move between these two domains. Note The mathematical details behind the Fourier transform are beyond the level of this textbook; for a more in-depth treatment, consult this series of articles from the Journal of Chemical Education : Glasser, L. “Fourier Transforms for Chemists: Part I. Introduction to the Fourier Transform,” J. Chem. Educ. 1987 , 64 , A228–A233. Glasser, L. “Fourier Transforms for Chemists: Part II. Fourier Transforms in Chemistry and Spectroscopy,” J. Chem. Educ. 1987 , 64 , A260–A266. Glasser, L. “Fourier Transforms for Chemists: Part III. Fourier Transforms in Data Treatment,” J. Chem. Educ. 1987 , 64 , A306–A313. Figure \(\PageIndex{6a}\) shows a single peak in the frequency domain and Figure \(\PageIndex{6b}\) shows its equivalent time domain signal. There are correlations between the two domains: the further a peak in the frequency domain is from the origin, the greater it corresponding oscillation frequency in the time domain the broader a peak's width in the frequency domain, the faster its decay rate in the time domain the greater the area under a peak in the frequency domain, the higher its initial intensity in the time domain We can use a Fourier transform to improve the signal-to-noise ratio because the signal is a single broad peak and the noise appears as a multitude of very narrow peaks. As noted above, a broad peak in the frequency domain has a fast decaying signal in the time domain, which means that while the beginning of the time domain signal includes contributions from the signal and the noise, the latter part of the time domain signal includes contributions from noise only. The figure below shows how we can take advantage of this to reduce the noise and improve the signal-to-noise ratio for the noisy signal in Figure \(\PageIndex{7a}\), which has 256 points along the x -axis and has a signal-to-noise ratio of 5.1. First, we use the Fourier transform to convert its original domain into the new domain, the first 128 points of which are shown in Figure \(\PageIndex{7b}\) (note: the first half of the data contains the same information as the second half of the data, so we only need to look at the first half of the data). The points at the beginning are dominated by the signal, which is why there is a systematic decrease in the intensity of the oscillations; the remaining points are dominated by noise, which is why the variation in intensity is random. To filter out the noise we retain the first 24 points as they are and set the intensities of the remaining points to zero (the choice of how many points to retain may require some adjustment). As shown in Figure \(\PageIndex{7c}\), we repeat this for the remaining 128 points, retaining the last 24 points as they are. Finally, we use an inverse Fourier transform to return to our original domain, with the result in Figure \(\PageIndex{7d}\), with the signal-to-noise ratio improving from 5. 1 for the original noisy signal to 11.2 for the filtered signal.
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Courses/Athabasca_University/Chemistry_360%3A_Organic_Chemistry_II/Chapter_28%3A_Biomolecules_-_Nucleic_Acids/28.06_DNA_Sequencing
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Objectives You may omit Section 28.6. You will not be examined on this material. We will discuss one method of reading the sequence of DNA. This method, developed by Sanger won him a second Nobel prize. To sequence a single stranded piece of DNA, the complementary strand is synthesized. Four different reaction mixtures are set up. Each contain all 4 radioactive deoxynucleotides (dATP, dCTP, dGTP, dTTP) required for the reaction and DNA polymerase. In addition, dideoxyATP (ddATP) is added to one reaction tube The dATP and ddATP attach randomly to the growing 3' end of the complementary stranded. If ddATP is added no further nucleotides can be added after since its 3' end has an H and not a OH. That's why they call it dideoxy. The new chain is terminated.. If dATP is added, the chain will continue to grow until another A needs to be added. Hence a whole series of discreet fragments of DNA chains will be made, all terminated when ddATP was added. The same scenario occurs for the other 3 tubes, which contain dCTP and ddCTP, dTTP and ddTTP, and dGTP and ddGTP respectively. All the fragments made in each tube will be placed in separate lanes for electrophoresis, where the fragments will separate by size. Didexoynucleotides Figure: Didexoynucleotides Example 28.6.1 You will pretend to sequence a single stranded piece of DNA as shown below. The new nucleotides are added by the enzyme DNA polymerase to the primer, GACT, in the 5' to 3' direction. You will set up 4 reaction tubes, Each tube contains all the dXTP's. In addition, add ddATP to tube 1, ddTTP to tube 2, ddCTP to tube 3, and ddGTP to tube 4. For each separate reaction mixture, determine all the possible sequences made by writing the possible sequences on one of the unfinished complementary sequences below. Cut the completed sequences from the page, determine the size of the polynucleotide sequences made, and place them as they would migrate (based on size) in the appropriate lane of a imaginary gel which you have drawn on a piece of paper. Lane 1 will contain the nucleotides made in tube 1, etc. Then draw lines under the positions of the cutout nucleotides to represent DNA bands in the gel. Read the sequence of the complementary DNA synthesized. Then write the sequence of the ssDNA that was to be sequenced. 5' T C A A C G A T C T G A 3' (STAND TO SEQUENCE) 3' G A C T 5' (primer) 3' G A C T 5' (primer) 3' G A C T 5' (primer) 3' G A C T 5' (primer) 3' G A C T 5' (primer) 3' G A C T 5' (primer) 3' G A C T 5' (primer) 3' G A C T 5' (primer) Since the DNA fragments have no detectable color, they can not be directly visualized in the gel. Alternative methods are used. In the one described above, radiolabeled ddXTP's where used. Once the sequencing gel is run, it can be dried and the bands visualized by radioautography (also called autoradiography). A place of x-ray film is placed over the dried gel in a dark environment. The radiolabeled bands will emit radiation which will expose the x-ray film directly over the bands. The film can be developed to detect the bands. In a newer technique, the primer can be labeled with a flourescent dye. If a different dye is used for each reaction mixture, all the reaction mixtures can be run in one lane of a gel. (Actually only one reaction mix containing all the ddXTP's together need be performed.) The gel can then be scanned by a laser, which detects fluorescence from the dyes, each at a different wavelength. Figure : DNA sequencing using different fluorescent primers for each ddXTP reaction One recent advance in sequencing allows for real-time determination of a sequence. The four deoxynucleotides are each labeled with a different fluorphore on the 5' phosphate (not the base as above). A tethered DNA polymerase elongates the DNA on a template, releasing the fluorophore into solution (i.e. the fluorophore is not incorporated into the DNA chain). The reaction takes place in a visualization chamber called a zero mode waveguide which is a cylindrical metallic chamber with a width of 70 nm and a volume of 20 zeptoliters (20 x 10-21 L). It sits on a glass support through which laser illumination of the sample is achieved. Given the small volume, non-incorporated fluorescently tagged deoxynucleotides diffuse in and out in the microsecond timescale. When a deoxynucleotide is incorporated into the DNA, its residence time is in the millisecond time scale. This allows for prolonged detection of fluorescence which give a high signal to noise ratio. This method might bring the cost of sequencing the human genome down from the initial billion dollar range to $100. SMART - Single Molecule Real Time Sequencing Pacific Biosystems
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Courses/San_Diego_Miramar_College/Chem_103%3A_Fundamentals_of_GOB_Chemistry_(Garces)/02%3A_Measurements/2.E%3A_Measurement_and_Problem_Solving_(Exercises)
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2.1: Measuring Global Temperatures 2.2: Scientific Notation: Writing Large and Small Numbers 2.3: Significant Figures: Writing Numbers to Reflect Precision Define significant figures . Why are they important? Define the different types of zeros found in a number and explain whether or not they are significant. How many significant figures are in each number? 140 0.009830 15,050 221,560,000 5.67 × 10 3 2.9600 × 10 −5 How many significant figures are in each number? 1.05 9,500 0.0004505 0.00045050 7.210 × 10 6 5.00 × 10 −6 Round each number to three significant figures. 34,705 34,750 34,570 2.4: Significant Figures in Calculations 2.5: The Basic Units of Measurement 2.6: Problem Solving and Unit Conversions 2.7: Solving Multi-step Conversion Problems 2.8: Units Raised to a Power 2.9: Density 2.10: Numerical Problem-Solving Strategies and the Solution Map
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Courses/Ursinus_College/CHEM322%3A_Inorganic_Chemistry/08%3A_Electronic_Structure_of_Coordination_Complexes/8.04%3A_Ligand_Field_Theory/8.4.03%3A_Factors_That_Affect_Ligand_Field_Splitting
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Magnitude of Ligand Field Splitting The magnitude of the ligand field splitting (Δ) dictates whether a complex with four, five, six, or seven d electrons (in an octahedral complex) is high spin or low spin, which affects its magnetic properties, structure, and reactivity. Large values of Δ (i.e., Δ > P) yield a low-spin complex, whereas small values of Δ (i.e., Δ < P) produce a high-spin complex. The magnitude of Δ depends on four factors: the valence of the metal, the principal quantum number of the metal (and thus its location in the periodic table), the geometry, and the nature of the ligand(s). Values of Δ for some representative transition metal complexes are given in Table \(\PageIndex{1}\). Octahedral Complexes Δo (cm−1) Octahedral Complexes.1 Δo (cm−1).1 Tetrahedral Complexes Δt (cm−1) [Ti(H2O)6]3+ 20300 [Fe(CN)6]4− 32800 VCl4 9010 [V(H2O)6]2+ 12600 [Fe(CN)6]3− 35000 [CoCl4]2− 3300 [V(H2O)6]3+ 18900 [CoF6]3− 13000 [CoBr4]2− 2900 [CrCl6]3− 13000 [Co(H2O)6]2+ 9300 [CoI4]2− 2700 [Cr(H2O)6]2+ 13900 [Co(H2O)6]3+ 27000 NaN NaN [Cr(H2O)6]3+ 17400 [Co(NH3)6]3+ 22900 NaN NaN [Cr(NH3)6]3+ 21500 [Co(CN)6]3− 34800 NaN NaN [Cr(CN)6]3− 26600 [Ni(H2O)6]2+ 8500 NaN NaN Cr(CO)6 34150 [Ni(NH3)6]2+ 10800 NaN NaN [MnCl6]4− 7500 [RhCl6]3− 20400 NaN NaN [Mn(H2O)6]2+ 8500 [Rh(H2O)6]3+ 27000 NaN NaN [MnCl6]3− 20000 [Rh(NH3)6]3+ 34000 NaN NaN [Mn(H2O)6]3+ 21000 [Rh(CN)6]3− 45500 NaN NaN [Fe(H2O)6]2+ 10400 [IrCl6]3− 25000 NaN NaN [Fe(H2O)6]3+ 14300 [Ir(NH3)6]3+ 41000 NaN NaN *Energies obtained by spectroscopic measurements are often given in units of wave numbers (cm−1); the wave number is the reciprocal of the wavelength of the corresponding electromagnetic radiation expressed in centimeters: 1 cm−1 = 11.96 J/mol. *Energies obtained by spectroscopic measurements are often given in units of wave numbers (cm−1); the wave number is the reciprocal of the wavelength of the corresponding electromagnetic radiation expressed in centimeters: 1 cm−1 = 11.96 J/mol. *Energies obtained by spectroscopic measurements are often given in units of wave numbers (cm−1); the wave number is the reciprocal of the wavelength of the corresponding electromagnetic radiation expressed in centimeters: 1 cm−1 = 11.96 J/mol. *Energies obtained by spectroscopic measurements are often given in units of wave numbers (cm−1); the wave number is the reciprocal of the wavelength of the corresponding electromagnetic radiation expressed in centimeters: 1 cm−1 = 11.96 J/mol. *Energies obtained by spectroscopic measurements are often given in units of wave numbers (cm−1); the wave number is the reciprocal of the wavelength of the corresponding electromagnetic radiation expressed in centimeters: 1 cm−1 = 11.96 J/mol. *Energies obtained by spectroscopic measurements are often given in units of wave numbers (cm−1); the wave number is the reciprocal of the wavelength of the corresponding electromagnetic radiation expressed in centimeters: 1 cm−1 = 11.96 J/mol. Source of data: Duward F. Shriver, Peter W. Atkins, and Cooper H. Langford, Inorganic Chemistry, 2nd ed. (New York: W. H. Freeman and Company, 1994). Valence of the metal Increasing the valence of a metal ion has two effects: the radius of the metal decreases and ligands are more strongly attracted to it due to Coulombic attraction. Both factors decrease the metal–ligand distance, which in turn causes the ligands to interact more strongly with the d-orbitals. Consequently, the magnitude of Δ o increases as the valence of the metal increases. Typically, Δ o for a M(III) is about 50% greater than for the M(II) of the same metal; for example, for [V(H 2 O) 6 ] 2+ , Δ o = 11,800 cm −1 ; for [V(H 2 O) 6 ] 3+ , Δ o = 17,850 cm −1 . Principal quantum number of the metal For a series of complexes of metals from the same group in the periodic table with the same charge and the same ligands, the magnitude of Δ o increases with increasing principal quantum number: Δ (3d) < Δ (4d) < Δ (5d). The data for hexaammine complexes of the trivalent Group 9 metals illustrate this point: [Co(NH 3 ) 6 ] 3+ : Δ o = 22,900 cm −1 [Rh(NH 3 ) 6 ] 3+ : Δ o = 34,100 cm −1 [Ir(NH 3 ) 6 ] 3+ : Δ o = 40,000 cm −1 The increase in Δ with increasing principal quantum number is due to the larger radial extension of the d orbitals as n increases. In addition, repulsive ligand–ligand interactions are most important for smaller metal ions. Relatively speaking, this results in shorter M–L distances and stronger d orbital–ligand interactions. The increase in Δ going from 3d to 4d and 5d is so large that all 4d and 5d metals will form low spin complexes. Geometry of the complex The number of ligands in a complex as well as how well the ligand geometry overlaps with the d orbitals is also a factor in the magnitude of the ligand field splitting. For example, comparing octahedral and tetrahedral geometries the octahedral geometry has 6 ligands, and those 6 ligands overlap directly with the two d orbitals that lie along the axes, \(d_{x^2-y^2}\) and \(d_{z^2}\). The tetrahedral geometry has fewer ligands than octahedral, and those ligands overlap less ideally with the ligands that lie between the axes, \(d_{xy}\), \(d_{xz}\), and \(d_{yz}\). Because of this the ligand field splitting of a tetrahedral complex is generally less than half (\(~\frac{4}{9}\) than an octahedral complex with the same metal and ligands. For example [FeCl 4 ] - has a Δ t = 5,200 cm -1 and [FeCl 6 ] 3- has a Δ o = 11,600 cm -1 . Nature of the ligands In crystal field theory ligands are all modeled as negative point charges, which means that all ligands should behave identically. Experimentally, it is found that the Δ o observed for a series of complexes of the same metal ion depends strongly on the nature of the ligands. For a series of chemically similar ligands, the magnitude of Δ o decreases as the size of the donor atom increases. For example, Δ o values for halide complexes generally decrease in the order F > Cl > Br > I because smaller, more localized charges, such as we see for F, interact more strongly with the d-orbitals of the metal. In addition, a small neutral ligand with a highly localized lone pair, such as NH 3 , results in significantly larger Δ o values than might be expected. Because the lone pair points directly at the metal ion, the electron density along the M–L axis is greater than for a spherical anion such as F. The experimentally observed order of the crystal field splitting energies produced by different ligands is called the spectrochemical series. Ligands are classified as strong field or weak field based on the spectrochemical series: weak field I - < Br - < Cl - < S CN - < F - < OH - < ox -2 < O NO < H 2 O < SC N - < NH 3 < en < N O 2 < CN - , CO strong field Note that SCN and NO 2 ligands are represented twice in the above spectrochemical series since there are two different Lewis base sites (e.g., free electron pairs to share) on each ligand (e.g., for the SCN ligand, the electron pair on the sulfur or the nitrogen can form the bond to a metal). The specific atom that binds in such ligands is underlined. Ligands on the weak field end of the series (halogens, OH, H 2 O) will tend to form high spin complexes and ligands on the strong field end of the series (CN, CO, NO 2 ) will tend to form low spin complexes. Intermediate ligands in the midle of the series could form high or low spin complexes depending on other factors. How to determine if a complex is high or low spin How many d electrons does the complex have? Only complexes with 4-7 d electrons have high spin and low spin configurations What period is the metal? 3d metals could either be high or low spin 4d and 5d metals will ALWAYS be low spin What is the geometry of the complex? Tetrahedral complexes with 3d metals will almost always be high spin because Δt is generally less than spin pairing energy Square planar complexes will almost always be low spin because they are favored over tetrahedral when crystal field splitting is large Octahedral complexes could be either high or low spin What are the ligands? Strong field ligands will form low spin complexes Weak field ligands will form high spin complexes Intermediate field ligands could form high or low spin complexes. Additional information (like number of unpaired electrons) is needed Adapted by Catherine McCusker (East Tennessee State University)
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Courses/Lumen_Learning/Book%3A_US_History_I_(OS_Collection)_(Lumen)/17%3A_The_Civil_War%2C_1860-1865/17.2%3A_The_Origins_and_Outbreak_of_the_Civil_War
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Learning Objectives By the end of this section, you will be able to: Explain the major events that occurred during the Secession Crisis Describe the creation and founding principles of the Confederate States of America (credit “1865”: modification of work by “Alaskan Dude”/Wikimedia Commons) The 1860 election of Abraham Lincoln was a turning point for the United States. Throughout the tumultuous 1850s, the Fire-Eaters of the southern states had been threatening to leave the Union. With Lincoln’s election, they prepared to make good on their threats. Indeed, the Republican president-elect appeared to be their worst nightmare. The Republican Party committed itself to keeping slavery out of the territories as the country expanded westward, a position that shocked southern sensibilities. Meanwhile, southern leaders suspected that Republican abolitionists would employ the violent tactics of John Brown to deprive southerners of their slave property. The threat posed by the Republican victory in the election of 1860 spurred eleven southern states to leave the Union to form the Confederate States of America, a new republic dedicated to maintaining and expanding slavery. The Union, led by President Lincoln, was unwilling to accept the departure of these states and committed itself to restoring the country. Beginning in 1861 and continuing until 1865, the United States engaged in a brutal Civil War that claimed the lives of over 600,000 soldiers. By 1863, the conflict had become not only a war to save the Union, but also a war to end slavery in the United States. Only after four years of fighting did the North prevail. The Union was preserved, and the institution of slavery had been destroyed in the nation. THE CAUSES OF THE CIVIL WAR Lincoln’s election sparked the southern secession fever into flame, but it did not cause the Civil War. For decades before Lincoln took office, the sectional divisions in the country had been widening. Both the Northern and southern states engaged in inflammatory rhetoric and agitation, and violent emotions ran strong on both sides. Several factors played into the ultimate split between the North and the South. One key irritant was the question of slavery’s expansion westward. The debate over whether new states would be slave or free reached back to the controversy over statehood for Missouri beginning in 1819 and Texas in the 1830s and early 1840s. This question arose again after the Mexican-American War (1846–1848), when the government debated whether slavery would be permitted in the territories taken from Mexico. Efforts in Congress to reach a compromise in 1850 fell back on the principle of popular sovereignty—letting the people in the new territories south of the 1820 Missouri Compromise line decide whether to allow slavery. This same principle came to be applied to the Kansas-Nebraska territories in 1854, a move that added fuel to the fire of sectional conflict by destroying the Missouri Compromise boundary and leading to the birth of the Republican Party. In the end, popular sovereignty proved to be no solution at all. This was especially true in “Bleeding Kansas” in the mid-1850s, as pro- and antislavery forces battled each another in an effort to gain the upper hand. The small but very vocal abolitionist movement further contributed to the escalating tensions between the North and the South. Since the 1830s, abolitionists, led by journalist and reformer William Lloyd Garrison, had cast slavery as a national sin and called for its immediate end. For three decades, the abolitionists—a minority even within the antislavery movement—had had a significant effect on American society by bringing the evils of slavery into the public consciousness. By the 1850s, some of the most radical abolitionists, such as John Brown, had resorted to violence in their efforts to destroy the institution of slavery. The formation of the Liberty Party (1840), the Free-Soil Party (1848), and the Republican Party (1854), all of which strongly opposed the spread of slavery to the West, brought the question solidly into the political arena. Although not all those who opposed the westward expansion of slavery had a strong abolitionist bent, the attempt to limit slaveholders’ control of their human property stiffened the resolve of southern leaders to defend their society at all costs. Prohibiting slavery’s expansion, they argued, ran counter to fundamental American property rights. Across the country, people of all political stripes worried that the nation’s arguments would cause irreparable rifts in the country. Despite the ruptures and tensions, by the 1860s, some hope of healing the nation still existed. Before Lincoln took office, John Crittenden, a senator from Kentucky who had helped form the Constitutional Union Party during the 1860 presidential election, attempted to diffuse the explosive situation by offering six constitutional amendments and a series of resolutions, known as the Crittenden Compromise. Crittenden’s goal was to keep the South from seceding, and his strategy was to transform the Constitution to explicitly protect slavery forever. Specifically, Crittenden proposed an amendment that would restore the 36°30′ line from the Missouri Compromise and extend it all the way to the Pacific Ocean, protecting and ensuring slavery south of the line while prohibiting it north of the line. He further proposed an amendment that would prohibit Congress from abolishing slavery anywhere it already existed or from interfering with the interstate slave trade. Crittenden’s Compromise would protect slavery in all states where it already existed. More importantly, however, it proposed to allow the western expansion of slavery into states below the Missouri Compromise line. Georgia’s Ordinance of Secession and those of the other Deep South states were all based on that of South Carolina, which was drafted just a month after Abraham Lincoln was elected. Republicans, including President-elect Lincoln, rejected Crittenden’s proposals because they ran counter to the party’s goal of keeping slavery out of the territories. The southern states also rejected Crittenden’s attempts at compromise, because it would prevent slaveholders from taking their human chattel north of the 36°30′ line. On December 20, 1860, only a few days after Crittenden’s proposal was introduced in Congress, South Carolina began the march towards war when it seceded from the United States. Three more states of the Deep South—Mississippi, Florida, and Alabama—seceded before the U.S. Senate rejected Crittenden’s proposal on January 16, 1861. Georgia, Louisiana, and Texas joined them in rapid succession on January 19, January 26, and February 1, respectively. In many cases, these secessions occurred after extremely divided conventions and popular votes. A lack of unanimity prevailed in much of the South. Explore the causes, battles, and aftermath of the Civil War at the interactive website offered by the National Parks Service. THE CREATION OF THE CONFEDERATE STATES OF AMERICA The seven Deep South states that seceded quickly formed a new government. In the opinion of many Southern politicians, the federal Constitution that united the states as one nation was a contract by which individual states had agreed to be bound. However, they maintained, the states had not sacrificed their autonomy and could withdraw their consent to be controlled by the federal government. In their eyes, their actions were in keeping with the nature of the Constitution and the social contract theory of government that had influenced the founders of the American Republic. The new nation formed by these men would not be a federal union, but a confederation. In a confederation, individual member states agree to unite under a central government for some purposes, such as defense, but to retain autonomy in other areas of government. In this way, states could protect themselves, and slavery, from interference by what they perceived to be an overbearing central government. The constitution of the Confederate States of America (CSA), or the Confederacy, drafted at a convention in Montgomery, Alabama, in February 1861, closely followed the 1787 Constitution. The only real difference between the two documents centered on slavery. The Confederate Constitution declared that the new nation existed to defend and perpetuate racial slavery, and the leadership of the slaveholding class. Specifically, the constitution protected the interstate slave trade, guaranteed that slavery would exist in any new territory gained by the Confederacy, and, perhaps most importantly, in Article One, Section Nine, declared that “No . . . law impairing or denying the right of property in negro slaves shall be passed.” Beyond its focus on slavery, the Confederate Constitution resembled the 1787 U.S. Constitution. It allowed for a Congress composed of two chambers, a judicial branch, and an executive branch with a president to serve for six years. The convention delegates chose Jefferson Davis of Mississippi to lead the new provisional government as president and Alexander Stephens of Georgia to serve as vice president until elections could be held in the spring and fall of 1861. By that time, four new states—Virginia, Arkansas, Tennessee, and North Carolina—had joined the CSA. As 1861 progressed, the Confederacy claimed Missouri and Kentucky, even though no ordinance of secession had been approved in those states. Southern nationalism ran high, and the Confederacy, buoyed by its sense of purpose, hoped that their new nation would achieve eminence in the world. By the time Lincoln reached Washington, DC, in February 1861, the CSA had already been established. The new president confronted an unprecedented crisis. A conference held that month with delegates from the Southern states failed to secure a promise of peace or to restore the Union. On inauguration day, March 4, 1861, the new president repeated his views on slavery: “I have no purpose, directly or indirectly, to interfere with the institution of slavery in the States where it exists. I believe I have no lawful right to do so, and I have no inclination to do so.” His recognition of slavery in the South did nothing to mollify slaveholders, however, because Lincoln also pledged to keep slavery from expanding into the new western territories. Furthermore, in his inaugural address, Lincoln made clear his commitment to maintaining federal power against the secessionists working to destroy it. Lincoln declared that the Union could not be dissolved by individual state actions, and, therefore, secession was unconstitutional. Read Lincoln’s entire inaugural address at the Yale Avalon project’s website. How would Lincoln’s audience have responded to this speech? FORT SUMTER President Lincoln made it clear to Southern secessionists that he would fight to maintain federal property and to keep the Union intact. Other politicians, however, still hoped to avoid the use of force to resolve the crisis. In February 1861, in an effort to entice the rebellious states to return to the Union without resorting to force, Thomas Corwin, a representative from Ohio, introduced a proposal to amend the Constitution in the House of Representatives. His was but one of several measures proposed in January and February 1861, to head off the impending conflict and save the United States. The proposed amendment would have made it impossible for Congress to pass any law abolishing slavery. The proposal passed the House on February 28, 1861, and the Senate passed the proposal on March 2, 1861. It was then sent to the states to be ratified. Once ratified by three-quarters of state legislatures, it would become law. In his inaugural address, Lincoln stated that he had no objection to the amendment, and his predecessor James Buchanan had supported it. By the time of Lincoln’s inauguration, however, seven states had already left the Union. Of the remaining states, Ohio ratified the amendment in 1861, and Maryland and Illinois did so in 1862. Despite this effort at reconciliation, the Confederate states did not return to the Union. The Confederacy’s attack on Fort Sumter, depicted here in an 1861 lithograph by Currier and Ives, stoked pro-war sentiment on both sides of the conflict. Indeed, by the time of the Corwin amendment’s passage through Congress, Confederate forces in the Deep South had already begun to take over federal forts. The loss of Fort Sumter, in the harbor of Charleston, South Carolina, proved to be the flashpoint in the contest between the new Confederacy and the federal government. A small Union garrison of fewer than one hundred soldiers and officers held the fort, making it a vulnerable target for the Confederacy. Fire-Eaters pressured Jefferson Davis to take Fort Sumter and thereby demonstrate the Confederate government’s resolve. Some also hoped that the Confederacy would gain foreign recognition, especially from Great Britain, by taking the fort in the South’s most important Atlantic port. The situation grew dire as local merchants refused to sell food to the fort’s Union soldiers, and by mid-April, the garrison’s supplies began to run out. President Lincoln let it be known to Confederate leaders that he planned to resupply the Union forces. His strategy was clear: The decision to start the war would rest squarely on the Confederates, not on the Union. On April 12, 1861, Confederate forces in Charleston began a bombardment of Fort Sumter. Two days later, the Union soldiers there surrendered. The attack on Fort Sumter meant war had come, and on April 15, 1861, Lincoln called upon loyal states to supply armed forces to defeat the rebellion and regain Fort Sumter. Faced with the need to choose between the Confederacy and the Union, border states and those of the Upper South, which earlier had been reluctant to dissolve their ties with the United States, were inspired to take action. They quickly voted for secession. A convention in Virginia that had been assembled earlier to consider the question of secession voted to join the Confederacy on April 17, two days after Lincoln called for troops. Arkansas left the Union on May 6 along with Tennessee one day later. North Carolina followed on May 20. Not all residents of the border states and the Upper South wished to join the Confederacy, however. Pro-Union feelings remained strong in Tennessee, especially in the eastern part of the state where slaves were few and consisted largely of house servants owned by the wealthy. The state of Virginia—home of revolutionary leaders and presidents such as George Washington, Thomas Jefferson, James Madison, and James Monroe—literally was split on the issue of secession. Residents in the north and west of the state, where few slaveholders resided, rejected secession. These counties subsequently united to form “West Virginia,” which entered the Union as a free state in 1863. The rest of Virginia, including the historic lands along the Chesapeake Bay that were home to such early American settlements as Jamestown and Williamsburg, joined the Confederacy. The addition of this area gave the Confederacy even greater hope and brought General Robert E. Lee, arguably the best military commander of the day, to their side. In addition, the secession of Virginia brought Washington, DC, perilously close to the Confederacy, and fears that the border state of Maryland would also join the CSA, thus trapping the U.S. capital within Confederate territories, plagued Lincoln. The Confederacy also gained the backing of the Five Civilized Tribes, as they were called, in the Indian Territory. The Five Civilized Tribes comprised the Choctaws, Chickasaws, Creeks, Seminoles, and Cherokees. The tribes supported slavery and many members owned slaves. These Indian slaveholders, who had been forced from their lands in Georgia and elsewhere in the Deep South during the presidency of Andrew Jackson, now found unprecedented common cause with white slaveholders. The CSA even allowed them to send delegates to the Confederate Congress. While most slaveholding states joined the Confederacy, four crucial slave states remained in the Union. Delaware, which was technically a slave state despite its tiny slave population, never voted to secede. Maryland, despite deep divisions, remained in the Union as well. Missouri became the site of vicious fighting and the home of pro-Confederate guerillas but never joined the Confederacy. Kentucky declared itself neutral, although that did little to stop the fighting that occurred within the state. In all, these four states deprived the Confederacy of key resources and soldiers. This map illustrates the southern states that seceded from the Union and formed the Confederacy in 1861, at the outset of the Civil War. Section Summary The election of Abraham Lincoln to the presidency in 1860 proved to be a watershed event. While it did not cause the Civil War, it was the culmination of increasing tensions between the proslavery South and the antislavery North. Before Lincoln had even taken office, seven Deep South states had seceded from the Union to form the CSA, dedicated to maintaining racial slavery and white supremacy. Last-minute efforts to reach a compromise, such as the proposal by Senator Crittenden and the Corwin amendment, went nowhere. The time for compromise had come to an end. With the Confederate attack on Fort Sumter, the Civil War began. A Open Assessments element has been excluded from this version of the text. You can view it online here: http://pb.libretexts.org/ios/?p=312 Review Question Why did the states of the Deep South secede from the Union sooner than the states of the Upper South and the border states? Answer to Review Question Slavery was more deeply entrenched in the Deep South than it was in the Upper South or the border states. The Deep South was home to larger numbers of both slaveholders and slaves. Pro-Union sentiment remained strong in parts of the Upper South and border states, particularly those areas with smaller populations of slaveholders. Glossary Confederacy the new nation formed by the seceding southern states, also known as the Confederate States of America (CSA) Crittenden Compromise a compromise, suggested by Kentucky senator John Crittenden, that would restore the 36°30′ line from the Missouri Compromise and extend it to the Pacific Ocean, allowing slavery to expand into the southwestern territories Fort Sumter a fort in the harbor of Charleston, South Carolina, where the Union garrison came under siege by Confederate forces in an attack on April 12, 1861, beginning the Civil War CC licensed content, Shared previously US History. Authored by : P. Scott Corbett, Volker Janssen, John M. Lund, Todd Pfannestiel, Paul Vickery, and Sylvie Waskiewicz. Provided by : OpenStax College. Located at : http://openstaxcollege.org/textbooks/us-history . License : CC BY: Attribution . License Terms : Download for free at http://cnx.org/content/col11740/latest/
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Courses/Lumen_Learning/Book%3A_Statistics_for_the_Social_Sciences_(Lumen)/12%3A_11-_Chi-Square_Tests/12.04%3A_Assignment-_Using_Technology_with_Data_to_Run_a_Hypothesis_Test
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The purpose of this activity is to give you guided practice in carrying out the two-sample t-test, and to show you how to use software to aid in the process. Background Do undergraduates sleep less than graduate students? A student conducted a study of sleep habits at a large state university. His hypothesis is that undergraduates will party more and sleep less than graduate students. He surveyed random samples of 75 undergraduate students and 50 graduate students. Subjects reported the hours they sleep in a typical night. For this hypothesis test, he defined the population means as follows: μ 1 is the mean number of hours undergraduate students sleep in a typical night. μ 2 is the mean number of hours graduate students sleep in a typical night. Question 1: State the null and alternative hypotheses that are being tested here. Question 2: Explain why we can safely use the two-sample T-test in this case. Comment: Before we move on to carry out the test, it is important to realize that in the two-sample problem, the data can be provided in three possible ways: (i) Sample data in one column, and another column that indicates which sample the observation belongs to. Recall that this is the way the data were given in our leading example (looks vs. personality score and gender): Note that essentially, one column contains the explanatory variable, and one contains the response. (ii) Sample data in different columns—data from each of the two samples appear in a column dedicated to that category. As you’ll see, this is the way the data are provided in this example: (iii) Summarized data—we are not given the actual data, but just the data summaries: sample sizes, sample means and sample standard deviations of both samples. Recall that in our second example, the data were given in this format. Instructions Click on the link corresponding to your statistical package to see instructions for completing the activity, and then answer the questions below. R | StatCrunch | Minitab | Excel 2007 | TI Calculator Question 3: Carry out the test and report the test statistic and P-value. Question 4: Draw your conclusions in context. CC licensed content, Shared previously Concepts in Statistics. Provided by : Open Learning Initiative. Located at : http://oli.cmu.edu . License : CC BY: Attribution
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Courses/CSU_Chico/CSU_Chico%3A_CHEM_451_-_Biochemistry_I/CHEM_451_Test/07%3A_Binding/7.5%3A_New_Methods_in_Drug_Development/E6._DNA_Binding_and_Genomic_Analyses
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Huge numbers (100,000 to 1 million) of different DNA molecules can be covalently attached to silicon or glass chips, as described above for the peptides. These sequences are located at specific x,y coordinates on the chip. DNA probes can then be made from cells (by adding reverse transcriptase to isolated mRNA. forming cDNA), which are then labeled with a fluorescent molecule. Easier yet, mRNA can be isolated from the cells and labeled with a fluorophore. When added to the chip, they will bind through complementary H-bond interactions to specific complementary DNA on the chip. Using this technique, an individual's entire genome could be analyzed in a short amount of time. For example, mutations in certain genes associated with cancer might be detected by fluorescent-DNA probes made from a possible mutant gene to specific DNA molecules in the chip that were designed to bind to mutant probes. In an amazing variation, mRNA can be extracted from two different cells, a control and a tumor cell. The control mRNA can be labeled with a green fluorophore, while the tumor cell mRNA can be labeled with a red fluorphore. They can both be added to the chip containing a library of human genes. If the gene is expressed in both cell types, both types of labeled mRNA will bind and the spot on the chip will appear yellow. Mixing Colors on the Web If the gene is not expressed in either tissue, the spot will appear black. Genes that are only expressed in tumor cells will appear red and in control cells green. In a single experiment, the differential expression of genes in tumor cells can be determined. In this way, tumor-specific proteins can be identified, which could lead to the development of a vaccine against those tumor antigens. A typical microarray analysis for this type of experiment is shown below. (from Nature, 403, 699 (2000). Figure: Detection of Differential Gene Expression in Tumors and Normal Cells Using Microarray Chip Array technologies have continued to evolve. Affymetrix has developed a array chip that contains over 38,000 genes, representing the entire human genome.
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Bookshelves/Analytical_Chemistry/Instrumental_Analysis_(LibreTexts)/35%3A_Appendicies/35.05%3A_Critical_Values_for_Dixon's_Q-Test
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The following table provides critical values for \(Q(\alpha, n)\), where \(\alpha\) is the probability of incorrectly rejecting the suspected outlier and \(n\) is the number of samples in the data set. There are several versions of Dixon’s Q -Test, each of which calculates a value for Q ij where i is the number of suspected outliers on one end of the data set and j is the number of suspected outliers on the opposite end of the data set. The critical values for Q here are for a single outlier, Q 10 , where \[Q_\text{exp} = Q_{10} = \frac {|\text{outlier's value} - \text{nearest value}|} {\text{largest value} - \text{smallest value}} \nonumber \] The suspected outlier is rejected if Q exp is greater than \(Q(\alpha, n)\). For additional information consult Rorabacher, D. B. “Statistical Treatment for Rejection of Deviant Values: Critical Values of Dixon’s ‘ Q ’ Parameter and Related Subrange Ratios at the 95% confidence Level,” Anal. Chem. 1991 , 63 , 139–146. 0 1 2 3 4 5 \(\frac {\alpha \ce{->}} {n \ce{ v }}\) 0.100 0.050 0.040 0.020 0.010 3 0.941 0.970 0.976 0.988 0.994 4 0.765 0.829 0.846 0.889 0.926 5 0.642 0.710 0.729 0.780 0.821 6 0.560 0.625 0.644 0.698 0.740 7 0.507 0.568 0.586 0.637 0.680 8 0.468 0.526 0.543 0.590 0.634 9 0.437 0.493 0.510 0.555 0.598 10 0.412 0.466 0.483 0.527 0.568
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Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/25%3A_Amino_Acids_Peptides_and_Proteins/25.07%3A_Peptides_and_Proteins
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Classification Amino acids are the building blocks of the polyamide structures of peptides and proteins. Each amino acid is linked to another by an amide (or peptide) bond formed between the \(\ce{NH_2}\) group of one and the \(\ce{CO_2H}\) group of the other: In this manner a polymeric structure of repeating amide links is built into a chain or ring. The amide groups are planar and configuration about the \(\ce{C-N}\) bond is usually, but not always, trans ( Section 24-1 ). The pattern of covalent bonds in a peptide or protein is called its primary structure : The distinction between a protein and a peptide is not completely clear. One arbitrary choice is to call proteins only those substances with molecular weights greater than 5000. The distinction might also be made in terms of differences in physical properties, particularly hydration and conformation. Thus proteins, in contrast to peptides, have very long chains that are coiled and folded in particular ways, with water molecules filling the voids in the coils and folds. Hydrogen bonding between the amide groups plays a decisive role in holding the chains in juxtaposition to one another, in what is sometimes called the secondary and tertiary structure .\(^5\) Under the influence of heat, organic solvents, salts, and so on, protein molecules undergo changes, often irreversibly, called denaturation . The conformations of the chains and the degree of hydration are thereby altered, with the result that solubility and ability to crystallize decreases. Most importantly, the physiological properties of the protein usually are destroyed permanently on denaturation. Therefore, if a synthesis of a protein is planned, it would be necessary to duplicate not only the amino-acid sequences but also the exact conformations of the chains and the manner of hydration characteristic of the native protein. With peptides, the chemical and physiological properties of natural and synthetic materials usually are identical, provided the synthesis duplicates all of the structural and configurational elements. What this means is that a peptide automatically assumes the secondary and tertiary structure characteristic of the native peptide on crystallization or dissolution in solvents. Representation of peptide structures of any length with conventional structural formulas is cumbersome. As a result, abbreviations are universally used that employ three-letter symbols for the component amino acids. It is important that you know the conventions for these abbreviations. The two possible dipeptides made up of one glycien and one alanine are Notice that in the conventions used for names and abbreviated formulas the amino acid with the free amino group ( the \(\ce{N}\)-terminal amino acid ) always is written on the left . The amino acid with the free carboxyl group ( the \(\ce{C}\)-terminal amino acid ) always is written on the right . The dash between the three-letter abbreviations for the acids designates that they are linked together by an amide bond. Determination of Amino-Acid Sequences The general procedure for determining the primary structure of a peptide or protein consists of three main steps. First, the number and kind of amino-acid units in the primary structure must be determined. Second, the amino acids at the ends of the chains are identified, and third, the sequence of the component amino acids in the chains is determined. The amino-acid composition usually is obtained by complete acid hydrolysis of the peptide into its component amino acids and analysis of the mixture by ion-exchange chromatography ( Section 25-4C ). This procedure is complicated by the fact that tryptophan is destroyed under acidic conditions. Also, asparagine and glutamine are converted to aspartic and glutamic acids, respectively. Determination of the \(\ce{N}\)-terminal acid in the peptide can be made by treatment of the peptide with 2,4-dinitrofluorobenzene, a substance very reactive in nucleophilic displacements with amines but not amides (see Section 14-6B ). The product is an \(\ce{N}\)-2,4-dinitrophenyl derivative of the peptide which, after hydrolysis of the amide linkages, produces an \(\ce{N}\)-2,4-dinitrophenylamino acid: These amino-acid derivatives can be separated from the ordinary amino acids resulting from hydrolysis of the peptide because the low basicity of the 2,4-dinitrophenyl-substituted nitrogen ( Section 23-7C ) greatly reduces the solubility of the compound in acid solution and alters its chromatographic behavior. The main disadvantage to the method is that the entire peptide must be destroyed in order to identify the one \(\ce{N}\)-terminal acid. A related and more sensitive method makes a sulfonamide of the terminal \(\ce{NH_2}\) group with a reagent called "dansyl chloride". As with 2,4-dinitrofluorobenzene, the peptide must be destroyed by hydrolysis to release the \(\ce{N}\)-sulfonated amino acid, which can be identified spectroscopically in microgram amounts: A powerful method of sequencing a peptide from the \(\ce{N}\)-terminal end is the Edman degradation in which phenyl isothiocyanate, \(\ce{C_6H_5N=C=S}\), reacts selectively with the terminal amino acid under mildly basic conditions. If the reaction mixture is then acidified, the terminal amino acid is cleaved from the peptide as a cyclic thiohydantoin , \(8\): There are simple reagents that react selectively with the carboxyl terminus of a peptide, but they have not proved as generally useful for analysis of the \(\ce{C}\)-terminal amino acids as has the enzyme carboxypeptidase A . This enzyme catalyzes the hydrolysis of the peptide bond connecting the amino acid with the terminal carboxyl groups to the rest of the peptide. Thus the amino acids at the carboxyl end will be removed one by one through the action of the enzyme. Provided that appropriate corrections are made for different rates of hydrolysis of peptide bonds for different amino acids at the carboxyl end of the peptide, the sequence of up to five or six amino acids in the peptide can be deduced from the order of their release by carboxypeptidase. Thus a sequence such as peptide-Ser-Leu-Tyr could be established by observing that carboxypeptidase releases amino acids from the peptide in the order Tyr, Leu, Ser: Determining the amino-acid sequences of large peptides and proteins is very difficult. Although the Edman degradation and even carboxypeptidase can be used to completely sequence small peptides, they cannot be applied successfully to peptide chains with several hundred amino acid units. Success has been obtained with long peptide chains by employing reagents, often enzymes, to selectively cleave certain peptide bonds. In this way the chain can be broken down into several smaller peptides that can be separated and sequenced. The problem then is to determine the sequence of these small peptides in the original structure. To do this, alternative procedures for selective cleavages are carried out that produce different sets of smaller peptides. It is not usually necessary to sequence completely all of the peptide sets. The overall amino-acid composition and the respective end groups of each peptide may suffice to show overlapping sequences from which the complete amino-acid sequence logically can be deduced. The best way to show you how the overlap method of peptide sequencing works is by a specific example. In this example, we will illustrate the use of the two most commonly used enzymes for selective peptide cleavage. One is trypsin , a proteolytic enzyme of the pancreas (MW 24,000) that selectively catalyzes the hydrolysis of the peptide bonds of basic amino acids, lysine and arginine. Cleavage occurs on the carboxyl side of lysine or arginine: Chymotrypsin is a proteolytic enzyme of the pancreas (MW 24,500) that catalyzes the hydrolysis of peptide bonds to the aromatic amino acids, tyrosine, tryptophan, and phenylalanine, more rapidly than to other amino acids. Cleavage occurs on the carboxyl side of the aromatic amino acid: Our example is the sequencing of a peptide (P) derived from partial hydrolysis of a protein which, on complete acid hydrolysis, gave Ala, 3 Gly, Glu, His, 3 Lys, Phe, Tyr, 2 Val, and one molar equivalent of ammonia. 1. Treatment of the peptide (P) with carboxypeptidase released alanine, and with 2,4-dinitrofluorobenzene followed by hydrolysis gave the 2,4-dinitrophenyl derivative of valine. These results establish the \(\ce{N}\)-terminus as valine and the \(\ce{C}\)-terminus as alanine. The known structural elements now are 2. Partial hydrolysis of the peptide (P) with trypsin gave a hexapeptide, a tetrapeptide, a dipeptide, and one molar equivalent of lysine. The peptides, which we will designated respectively as M, N, and O, were sequenced by Edman degradation and found to have structures: \[\begin{array}{ll} \text{Gly}-\text{Ala} & \text{O} \\ \text{Val}-\text{Tyr}-\text{Glu}-\text{Lys} & \text{N} \\ \text{Val}-\text{Gly}-\text{Phe}-\text{Gly}-\text{His}-\text{Lys} & \text{M} \end{array}\] With this information, four possible structures can be written for the original peptide P that are consistent with the known end groups and the fact that trypsin cleaves the peptide P on the carboxyl side of the lysine unit. Thus \[\begin{array}{cc} \text{N}-\text{M}-\text{Lys}-\text{O} & \text{M}-\text{N}-\text{Lys}-\text{O} \\ \text{N}-\text{Lys}-\text{M}-\text{O} & \text{M}-\text{Lys}-\text{N}-\text{O} \end{array}\] 3. Partial hydrolysis of the peptide P using chymotrypsin as catalyst gave three peptides, X, Y, and Z. These were not sequenced, but their amino-acid composition was determined: \[\begin{array}{ll} \text{Gly, Phe, Val} & \text{X} \\ \text{Gly, His, Lys, Tyr, Val} & \text{Y} \\ \text{Ala, Glu, Gly, 2 Lys} & \text{Z} \end{array}\] This information can be used to decide which of the alternative structures deduced above is correct. Chymotrypsin cleaves the peptide on the carboxyl side of the phenylalanine and tyrosine units. Only peptide M contains Phe, and if we compare M with the compositions of X, Y, and Z, we see that only X and Y overlap with M. Peptide Z contains the only Ala unit and must be the \(\ce{C}\)-terminus. If we put together these pieces to get a peptide, P' (which differs from P by not having the nitrogen corresponding to the ammonia formed on complete hydrolysis) then P' must have the structure X-Y-Z: This may not be completely clear, and it will be well to consider the logic in some detail. Peptides M and N both have \(\ce{N}\)-terminal valines, and one of them must be the \(\ce{N}\)-terminal unit. Peptide M overlaps with X and Y, and because X and Y are produced by a cleavage on the carboxyl side of Phe, the X and Y units have to be connected in the order X-Y. Because the other Val is in Y, the \(\ce{N}\)-terminus must be M. This narrows the possibilities to \[\begin{array}{c} \text{M}-\text{N}-\text{Lys}-\text{O} \\ \text{M}-\text{Lys}-\text{N}-\text{O} \end{array}\] There are two Lys units in Z, and this means that only the sequence M-N-Lys-O is consistent with the sequence X-Y-Z, as shown: The final piece of the puzzle is the placement of the mole of ammonia released from the original peptide on acid hydrolysis. The ammonia comes from a primary amide function: \[\ce{R-CONH_2} \overset{\ce{H_3O}^\oplus}{\longrightarrow} \ce{RCO_2H} + \ce{NH_4^-}\] The amide group cannot be at the \(\ce{C}\)-terminus because the peptide would then be inert to carboxypeptidase. The only other possible place is on the side-chain carboxyl of glutamic acid. The complete structure may be written as Using procedures such as those outlined in this section more than 100 proteins have been sequenced. This is an impressive accomplishment considering the complexity and size of many of these molecules (see, for example, Table 25-3). It has been little more than two decades since the first amino acid sequence of a protein was reported by F. Sanger, who determined the primary structure of insulin (1953). This work remains a landmark in the history of chemistry because it established for the first time that proteins have definite primary structures in the same way that other organic molecules do. Up until that time, the concept of definite primary structures for proteins was openly questioned. Sanger developed the method of analysis for \(\ce{N}\)-terminal amino acids using 2,4-dinitrofluorobenzene and received a Nobel Prize in 1958 for his success in determining the amino-acid sequence of insulin. Methods for Forming Peptide Bonds The problems involved in peptide syntheses are of much practical importance and have received considerable attention. The major difficulty in putting together a chain of say 100 amino acids in a particular order is one of overall yield. At least 100 separate synthetic steps would be required, and, if the yield in each step were equal to \(n \times 100\%\), the overall yield would be \(\left( n^{100} \times 100\% \right)\). If the yield in each step were \(90\%\), the overall yield would be only \(0.003\%\). Obviously, a practical laboratory synthesis of a peptide chain must be a highly efficient process. The extraordinary ability of living cells to achieve syntheses of this nature, not of just one but of a wide variety of such substances, is truly impressive. Several methods for the formation of amide bonds have been discussed in Sections 18-7A and 24-3A . The most general reaction is shown below, in which X is some reactive leaving group (see Table 24-1): When applied to coupling two different amino acids, difficulty is to be expected because these same reactions can link two amino acids in a total of four different ways. Thus if we started with a mixture of glycine and alanine, we could generate for dipeptides, Gly-Ala, Ala-Gly, Gly-Gly, and Ala-Ala. To avoid unwanted coupling reactions a protecting group is substituted on the amino function of the acid that is to act as the acylating agent. Furthermore, all of the amino, hydroxyl, and thiol functions that may be acylated to give undesired products usually must be protected. For instance, to synthesize Gly-Ala free of other possible dipeptides, we would have to protect the amino group of glycine and the carboxyl group of alanine: Some methods of protecting amine and hydroxyl functions were discussed previously in Sections 23-13 and 15-9 , respectively. A summary of some commonly used protecting groups for \(\ce{NH_2}\), \(\ce{OH}\), \(\ce{SH}\), and \(\ce{CO_2H}\) functions is in Table 25-2, together with the conditions by which the protecting groups may be removed. The best protecting groups for \(\ce{NH_2}\) functions are phenylmethoxycarbonyl (benzyloxycarbonyl) and tert -butoxycarbonyl. Both groups can be removed by treatment with acid, although the tert -butoxycarbonyl group is more reactive. The phenylmethoxycarbonyl group can be removed by reduction with either hydrogen over a metal catalyst or with sodium in liquid ammonia. This method is most useful when, in the removal step, it is necessary to avoid treatment with acid: Table 25-2: Some Amine and Carboxyl Protecting Groups Used in Peptide Syntheses In most cases, formation of the ethyl ester provides a satisfactory protecting group for the carboxyl function. Conversion of the carboxyl group to a more reactive group and coupling are key steps in peptide synthesis. The coupling reaction must occur readily and quantitatively, and with a minimum of racemization of the chiral centers in the molecule. This last criterion is the Achilles' heel of many possible coupling sequences. The importance of nonracemization can best be appreciated by an example. Consider synthesis of a tripeptide from three protected \(L\)-amino acids, A, B, and C, in two sequential coupling steps, \(\text{C} \overset{\text{B}}{\rightarrow} \text{B}-\text{C} \overset{\text{A}}{\rightarrow} \text{A}-\text{B}-\text{C}\). Suppose that the coupling yield is quantitative, but there is \(20\%\) formation of the \(D\) isomer in the acylating component in each coupling step. Then the tripeptide will consist of a mixture of four diastereomers, only \(64\%\) of which will be the desired \(L\),\(L\),\(L\) diastereomer (Equation 25-7): \(\tag{25-7}\) This is clearly unacceptable, especially for longer-chain peptides. Nine coupling steps with \(20\%\) of the wrong isomer formed in each would give only \(13\%\) of the decapeptide with the correct stereochemistry. The most frequently used carboxyl derivatives in amide coupling are azides, \(\ce{RCO-N_3}\), mixed anhydrides, \(\ce{RCO-O-COR'}\), and esters of moderately acidic phenols, \(\ce{RCO-OAr}\) (see Table 24-1). It also is possible to couple free acid with an amine group using a diimide, \(\ce{R-N=C=N-R}\), most frequently \(\ce{N}\),\(\ce{N'}\)-dicyclohexylcarbodiimide. The diimide reagent may be thought of as a dehydrating agent. The "elements of water" eliminated in the coupling are consumed by the diimide to form a substituted urea. The overall reaction is This reaction takes place because diimides, \(\ce{-N=C=N}-\), have reactive cumulated double-bond systems like those of ketenes, \(\ce{-C=C=O}\); isocyanates, \(\ce{-N=C=O}\); and isothiocyanates, \(\ce{-N=C=S}\); and are susceptible to nucleophilic attack at the central carbon. In the first step of the diimide-coupling reaction, the carboxyl function adds to the imide to give an acyl intermediate, \(9\). This intermediate is an activated carboxyl derivative \(\ce{RCO-X}\) and is much more reactive toward an amino function than is the parent acid. The second step therefore is the aminolysis of \(9\) to give the coupled product and \(\ce{N}\),\(\ce{N'}\)-dicyclohexylurea: After completion of a coupling reaction, and before another amino acid can be added to the \(\ce{N}\)-terminus, it is necessary to remove the protecting group. This must be done by selective reactions that do not destroy the peptide bonds or side-chain protecting groups. This part of peptide synthesis is discussed in Section 23-13 , and some reactions useful for removal of the \(\ce{N}\)-terminal protecting groups are summarized in Table 25-2. In spite of the large number of independent steps involved in the synthesis of even small peptides, each with its attendant problems of yield, racemization, and selectivity, remarkable success has been achieved in the synthesis of large peptides and certain of the smaller peptides. The synthesis of insulin (Figure 25-8) with its 51 amino acid units and 3-disulfide bridges has been achieved by several investigators. Several important hormonal peptides, namely glutathione, oxytocin, vasopressin, and thyrotropic hormone (see Figure 25-9) have been synthesized. A major accomplishment has been the synthesis of an enzyme with ribonuclease activity reported independently by two groups of investigators, led by R. Hirschman (Merck) and R. B. Merrifield (Rockefeller University). This enzyme is one of the simpler proteins, having a linear stricture of 124 amino-acid residues. It is like a peptide, not a protein, in that it assumes the appropriate secondary and tertiary structure without biochemical intervention ( Section 25-7A ). As a specific example of the strategy involved in peptide synthesis, the stepwise synthesis of oxytocin is outlined in Figure 25-10, using the abbreviated notation in common usage. not to \(\ce{C_1}\). Vasopressin (middle left) and oxytocin (middle right) are peptide hormones from the posterior lobe of the pituitary gland. They function primarily to raise blood pressure (vasopressin), as antidiuretic (vasopressin), and to promote contraction of uterus and lactation muscles (oxytocin). The isolation, identification, and synthesis of these hormones was accomplished by Vincent du Vigneaud, for which he was awarded the Nobel Prize in chemistry in 1965. Thyrotropin-releasing hormone (bottom) is one of several small peptide hormones secreted by the anterior lobe of the pituitary gland. These are the "master" hormones that function to stimulate hormone secretion from other endocrine glands. Thyrotropin stimulates the functioning of the thyroid gland. Figure 25-10: Stepwise synthesis of oxytocin by the reactive ester method. In the abbreviations used here \(-\text{Gly}-\ce{NH_2} = \ce{-NHCH_2CONH_2}\) and Solid-Phase Peptide Synthesis The overall yield in a multistep synthesis of a peptide of even modest size is very poor unless each step can be carried out very efficiently. An elegant modification of classical peptide synthesis has been developed by R. B. Merrifield, which offers improved yields by minimizing manipulative losses that normally attend each step of a multistage synthesis. The key innovation is to anchor the \(\ce{C}\)-terminal amino acid to an insoluble support, and then add amino-acid units by the methods used for solution syntheses. After the desired sequence of amino acids has been achieved, the peptide can be cleaved from the support and recovered from solution. All the reactions involved in the synthesis must, of course, be brought to essentially \(100\%\) completion so that a homogeneous product can be obtained. The advantage of having the peptide anchored to a solid support is that laborious purification steps are virtually eliminated; solid material is purified simply by washing and filtering without transferring the material from one container to another. The method has become known as solid-phase peptide synthesis . More of the details of the solid-phase synthesis follow. The nature of the polymer support is of great importance for a successful peptide synthesis. One that is widely used is a cross-linked polystyrene resin of the type employed in ion-exchange chromatography ( Section 25-4C ). It is necessary that the resin be insoluble but have a loose enough structure to absorb organic solvents. Otherwise, the reagents will not be able to penetrate into the spaces between the chains. This is undesirable because the reactions occur on the surface of the resin particles and poor penetration greatly reduces the number of equivalents of reactive sites that can be obtained per gram of resin. Finally, to anchor a peptide chain to the resin, a reactive functional group (usually a chloromethyl group) must be introduced into the resin. This can be done by a Friedel-Crafts chloromethylation reaction, which substitutes the \(\ce{ClCH_2}-\) group in the 4-position of the phenyl groups in the resin: At the start of the peptide synthesis, the \(\ce{C}\)-terminal amino acid is bonded through its carboxyl group to the resin by a nucleophilic attack of the carboxylate ion on the chloromethyl groups. The \(\alpha\)-amino group must be suitably protected, as with tert -butoxycarbonyl, before carrying out this step: Next, the amine protecting group must be removed without cleaving the ester bond to the resin. The coupling step to a second \(\ce{N}\)-protected amino acid follows, with \(\ce{N}\),\(\ce{N'}\)-dicyclohexylcarbodiimide as the coupling reagent of choice: The peptide-bond-forming steps are repeated as many times as needed to build up the desired sequence. Ultimately, the peptide chain is removed from the resin, usually with \(\ce{HBr}\) in anhydrous trifluoroethanoic acid, \(\ce{CF_3CO_2H}\), or with anhydrous \(\ce{HF}\). This treatment also removes the other acid-sensitive protecting groups. The method lends itself beautifully to automatic control, and machines suitably programmed to add reagents and wash the product at appropriate times have been developed. At present, the chain can be extended by six or so amino acid units a day. It is necessary to check the homogeneity of the growing peptide chain at intervals because if any step does not proceed properly, the final product can be seriously contaminated with peptides with the wrong sequence. In the synthesis of the enzyme ribonuclease by the Merrifield method, the 124 amino acids were arranged in the ribonuclease sequence through 369 reactions and some 12,000 individual operations of the automated peptide-synthesis machine without isolation of any intermediates. Separation of Peptides and Proteins In many problems of peptide sequencing and peptide synthesis it is necessary to be able to separate mixtures of peptides and proteins. The principal methods used for this purpose depend on acid-base properties or on molecular sizes and shapes. Ultracentrifugation is widely used for the purification, separation, and molecular-weight determination of proteins. A centrifugal field, up to 500,000 times that of gravity, is applied to the solution, and molecules move downward in the field according to their mass and size. Large molecules also can be separated by gel filtration (or gel chromatography), wherein small molecules are separated from large ones by passing a solution over a gel that has pores of a size that the small molecules can penetrate into and be trapped. Molecules larger than the pore size are carried on with the solvent. This form of chromatographic separation is based on "sieving" rather than on chemical affinity. A wide range of gels with different pore sizes is available, and it is possible to fractionate molecules with molecular weights ranging from 700 to 200,000. The molecular weight of a protein can be estimated by the sizes of the pores that it will, or will not, penetrate. The acid-base properties, and hence ionic character, of peptides and proteins also can be used to achieve separations. Ion-exchange chromatography , similar to that described for amino acids ( Section 25-4C ), is an important separation method. Another method based on acid-base character and molecular size depends on differential rates of migration of the ionized forms of a protein in an electric field ( electrophoresis ). Proteins, like amino acids, have isoelectric points, which are the pH values at which the molecules have no net charge. At all other pH values there will be some degree of net ionic charge. Because different proteins have different ionic properties, they frequently can be separated by electrophoresis in buffered solutions. Another method, which is used for the separation and purification of enzymes, is affinity chromatography , which was described briefly in Section 9-2B . \(^5\) The distinction between secondary and tertiary structure is not sharp. Secondary structure involves consideration of the interactions and spatial relationships of the amino acids in the peptide chains that are close together in the primary structure, whereas tertiary structure is concerned with those that are far apart in the primary structure.
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Courses/De_Anza_College/CHEM_10%3A_Introduction_to_Chemistry_(Parajon_Puenzo)/08%3A_Acids_and_Bases/8.06%3A_The_pH_Scale
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Learning Objectives Define \(pH\). Determine the pH of acidic and basic solutions. Determine the hydrogen (hydronium) ion concentration from pH and vice versa. Molar concentration values of hydrogen \([H^+]\) can be markedly different from one aqueous solution to another. So chemists defined a new scale that succinctly indicates the concentrations of either of these two ions. This is known as the \(pH\) scale. The range of value s from 0 to 14 that describes the acidity or basicity of a solution. You can use \(pH\) to make a quick determination whether a given aqueous solution is acidic, basic, or neutral. pH is a logarithmic scale. A solution that has a pH of 1.0 has 10 times the [H + ] as a solution with a pH of 2.0, which in turn has 10 times the [H + ] as a solution with a pH of 3.0 and so forth. Warning: The pH scale has no limits pH is usually (but not always) between 0 and 14. Knowing the dependence of pH on \([H+]\), we can summarize as follows: If pH < 7, then the solution is acidic. If pH = 7, then the solution is neutral. If pH > 7, then the solution is basic. Figure \(\PageIndex{2}\) illus trates the relationship between pH and the hydrogen ion concentration, along with some examples of various solutions. Because hydrogen ion concentrations are generally less than one (for example \(1.3 \times 10^{-3}\,M\)), th e log of the number will be a negative number. To make pH even easier to work with, pH is defined as the negative log of \([H^+]\) , which will give a positive value for pH. The general formula for determining [H + ] from pH is as follows: [H + ] = 10 −pH Example \(\PageIndex{1}\) Label each solution as acidic, basic, or neutral based only on the stated \(pH\). milk of magnesia, pH = 10.5 pure water, pH = 7 wine, pH = 3.0 Answer With a pH greater than 7, milk of magnesia is basic. (Milk of magnesia is largely Mg(OH) 2 .) Pure water, with a pH of 7, is neutral. With a pH of less than 7, wine is acidic. Exercise \(\PageIndex{1}\) Identify each substance as acidic, basic, or neutral based only on the stated \(pH\). human blood with \(pH\) = 7.4 household ammonia with \(pH\) = 11.0 cherries with \(pH\) = 3.6 Answer a basic Answer b basic Answer c acidic Substance pH stomach acid 1.7 lemon juice 2.2 vinegar 2.9 soda 3.0 wine 3.5 coffee, black 5.0 milk 6.9 pure water 7.0 blood 7.4 seawater 8.5 milk of magnesia 10.5 ammonia solution 12.5 1.0 M NaOH 14.0 *Actual values may vary depending on conditions *Actual values may vary depending on conditions Example \(\PageIndex{2}\): What is the [H + ] for an aqueous solution whose pH is 6? Solution The pH value of 6 denotes that the exponent of 10 is -6. Therefore the answer is [H + ] = 1.0 x 10 −6 M Exercise \(\PageIndex{2}\) What is the [H + ] for an aqueous solution whose pH is 11? Answer [H + ] = 1.0 × 10 −11 M Example \(\PageIndex{3}\): What is the pH of an aqueous solution whose hydrogen ion concentration is 1.0 x 10 −7 M ? Is the solution acidic, basic, or neutral. Solution The hydrogen ion concentration is 1.0 x 10 −7 M. The exponent of 10 is -7, which denoted that pH= -(-7) = 7 . The solution is neutral . Exercise \(\PageIndex{3}\) What is the pH of an aqueous solution whose hydrogen ion concentration is 1.0 x 10 −3 M ? Is the solution acidic, basic, or neutral. Solution The hydrogen ion concentration is 1.0 x 10 −3 M. The exponent of 10 is -3, which denoted that pH= -(-3) = 3 . The solution is acidic . Summary To make pH even easier to work with, pH is defined as the negative log of \([H^+]\) , which will give a positive value for pH. pH is usually (but not always) between 0 and 14. If pH < 7, then the solution is acidic. If pH = 7, then the solution is neutral. If pH > 7, then the solution is basic.
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Courses/Arkansas_Northeastern_College/CH14133%3A_Chemistry_for_General_Education/10%3A_Chemical_Bonding/10.07%3A_Predicting_the_Shapes_of_Molecules
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Learning Objective Determine the shape of simple molecules. Molecules have shapes. There is an abundance of experimental evidence to that effect—from their physical properties to their chemical reactivity. Small molecules—molecules with a single central atom—have shapes that can be easily predicted. The basic idea in molecular shapes is called valence shell electron pair repulsion ( VSEPR ). VSEPR says that electron pairs, being composed of negatively charged particles, repel each other to get as far away from one another as possible. VSEPR makes a distinction between electron group geometry , which expresses how electron groups (bonds and nonbonding electron pairs) are arranged, and molecular geometry , which expresses how the atoms in a molecule are arranged. However, the two geometries are related. There are two types of electron groups : any type of bond—single, double, or triple—and lone electron pairs. When applying VSEPR to simple molecules, the first thing to do is to count the number of electron groups around the central atom. Remember that a multiple bond counts as only one electron group. Any molecule with only two atoms is linear. A molecule whose central atom contains only two electron groups orients those two groups as far apart from each other as possible—180° apart. When the two electron groups are 180° apart, the atoms attached to those electron groups are also 180° apart, so the overall molecular shape is linear. Examples include BeH 2 and CO 2 : The two molecules, shown in the figure below in a "ball and stick" model. A molecule with three electron groups orients the three groups as far apart as possible. They adopt the positions of an equilateral triangle—120° apart and in a plane. The shape of such molecules is trigonal planar . An example is BF 3 : Some substances have a trigonal planar electron group distribution but have atoms bonded to only two of the three electron groups. An example is GeF 2 : From an electron group geometry perspective, GeF 2 has a trigonal planar shape, but its real shape is dictated by the positions of the atoms. This shape is called bent or angular . A molecule with four electron groups about the central atom orients the four groups in the direction of a tetrahedron, as shown in Figure \(\PageIndex{1}\) Tetrahedral Geometry. If there are four atoms attached to these electron groups, then the molecular shape is also tetrahedral . Methane (CH 4 ) is an example. This diagram of CH 4 illustrates the standard convention of displaying a three-dimensional molecule on a two-dimensional surface. The straight lines are in the plane of the page, the solid wedged line is coming out of the plane toward the reader, and the dashed wedged line is going out of the plane away from the reader. NH 3 is an example of a molecule whose central atom has four electron groups, but only three of them are bonded to surrounding atoms. Although the electron groups are oriented in the shape of a tetrahedron, from a molecular geometry perspective, the shape of NH 3 is trigonal pyramidal . H 2 O is an example of a molecule whose central atom has four electron groups, but only two of them are bonded to surrounding atoms. Although the electron groups are oriented in the shape of a tetrahedron, the shape of the molecule is bent or angular . A molecule with four electron groups about the central atom, but only one electron group bonded to another atom, is linear because there are only two atoms in the molecule. Double or triple bonds count as a single electron group. The Lewis electron dot diagram of formaldehyde (CH 2 O) is shown in Figure \(\PageIndex{9}\). The central C atom has three electron groups around it because the double bond counts as one electron group. The three electron groups repel each other to adopt a trigonal planar shape. (The lone electron pairs on the O atom are omitted for clarity.) The molecule will not be a perfect equilateral triangle because the C–O double bond is different from the two C–H bonds, but both planar and triangular describe the appropriate approximate shape of this molecule. Table \(\PageIndex{1}\) summarizes the shapes of molecules based on the number of electron groups and surrounding atoms. Number of Electron Groups on Central Atom Number of Bonding Groups Number of Lone Pairs Electron Geometry Molecular Shape 2 2 0 linear linear 3 3 0 trigonal planar trigonal planar 3 2 1 trigonal planar bent 4 4 0 tetrahedral tetrahedral 4 3 1 tetrahedral trigonal pyramidal 4 2 2 tetrahedral bent Example \(\PageIndex{1}\) What is the approximate shape of each molecule? PCl 3 NOF Solution The first step is to draw the Lewis structure of the molecule. For \(\ce{PCl3}\), the electron dot diagram is as follows: The lone electron pairs on the Cl atoms are omitted for clarity. The P atom has four electron groups with three of them bonded to surrounding atoms, so the molecular shape is trigonal pyramidal. The electron dot diagram for \(\ce{NOF}\) is as follows: The N atom has three electron groups on it, two of which are bonded to other atoms. The molecular shape is bent. Exercise \(\PageIndex{1}\) What is the approximate molecular shape of \(\ce{CH2Cl2}\)? Answer Tetrahedral Exercise \(\PageIndex{2}\) Ethylene (\(\ce{C2H4}\)) has two central atoms. Determine the geometry around each central atom and the shape of the overall molecule. (Hint: hydrogen is a terminal atom.) Answer Trigonal planar about both central C atoms. Summary The approximate shape of a molecule can be predicted from the number of electron groups and the number of surrounding atoms.
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Courses/University_of_California_Davis/Chem_124A%3A_Fundamentals_of_Inorganic_Chemistry/05%3A_Molecular_Orbitals/5.03%3A_Heteronuclear_Diatomic_Molecules
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Diatomic molecules with two non-identical atoms are called heteronuclear diatomic molecules. When atoms are not identical, the molecule forms by combining atomic orbitals of unequal energies. The result is a polar bond in which atomic orbitals contribute unevenly to each molecular orbital . The application of molecular orbital theory to heteronuclear diatomic molecules is similar to the case of homonuclear diatomics, except that the atomic orbitals from each atom have different energies and contribute unequally to molecular orbitals. Recall that atomic orbitals must have compatible symmetry and similar energy to combine into molecular orbitals. In the case where atomic orbitals of like symmetry have different energies, they combine less favorably than orbitals that are closer to one another in energy. As a general rule, orbitals that have energy differences of greater than 10-14 eV do not combine favorably. In the molecular orbital diagram, the closer a molecular orbital is to an atomic orbital, the more that atomic orbital contributes to the molecular orbital. This last point is helpful for back-of-the napkin estimations of what the molecular orbitals "look" like. In this section, you should learn how to generate molecular orbital diagrams of heteronuclear diatomic molecules. To approach such a problem, we must start with a knowledge of the relative energies of electrons in different atomic orbitals. In other words, we need knowledge of the orbital potential energies (or orbital ionization energies). 5.3.1: Orbital ionization energies 5.3.2: Polar bonds The molecular orbital diagram of a heteronuclear diatomic molecule is approached in a similar way to that of homonuclear diatomic molecule. The orbital diagrams may also look similar. A major difference is that the more electronegative atom will have orbitals at a lower energy level. Two examples of heteronuclear diatomic molecules will be explored below as illustrative examples. 5.3.3: Ionic Compounds and Molecular Orbitals Sources: Gray, Harry. Electrons and Chemical Bonding , Benjamin, 1964. Miessler, Gary L, and Donald A. Tarr. Inorganic Chemistry . Upper Saddle River, N.J: Pearson Education, 2014. Print.
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Courses/Lafayette_College/CHEM_440%3A_Structure_Determination/01%3A_Mass_Spectrometry/1.05%3A_The_Mass_Spectra_of_Elements/1.5.02%3A_The_M1_Peak
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This page explains how the M+1 peak in a mass spectrum can be used to estimate the number of carbon atoms in an organic compound. What causes the M+1 peak? If you had a complete (rather than a simplified) mass spectrum, you will find a small line 1 m/z unit to the right of the main molecular ion peak. This small peak is called the M+1 peak. The carbon-13 isotope The M+1 peak is caused by the presence of the 13 C isotope in the molecule. 13 C is a stable isotope of carbon - don't confuse it with the 14 C isotope which is radioactive. Carbon-13 makes up 1.11% of all carbon atoms. If you had a simple compound like methane, CH 4 , approximately 1 in every 100 of these molecules will contain carbon-13 rather than the more common carbon-12. That means that 1 in every 100 of the molecules will have a mass of 17 (13 + 4) rather than 16 (12 + 4). The mass spectrum will therefore have a line corresponding to the molecular ion [ 13 CH 4 ] + as well as [ 12 CH 4 ] + . The line at m/z = 17 will be much smaller than the line at m/z = 16 because the carbon-13 isotope is much less common. Statistically you will have a ratio of approximately 1 of the heavier ions to every 99 of the lighter ones. That's why the M+1 peak is much smaller than the M+ peak. Using the M+1 peak Imagine a compound containing 2 carbon atoms. Either of them has an approximately 1 in 100 chance of being 13 C. There's therefore a 2 in 100 chance of the molecule as a whole containing one 13 C atom rather than a 12 C atom - which leaves a 98 in 100 chance of both atoms being 12 C. That means that the ratio of the height of the M+1 peak to the M+ peak will be approximately 2 : 98. That's pretty close to having an M+1 peak approximately 2% of the height of the M+ peak. Using the relative peak heights to predict the number of carbon atoms If you measure the peak height of the M+1 peak as a percentage of the peak height of the M+ peak, that gives you the number of carbon atoms in the compound. We've just seen that a compound with 2 carbons will have an M+1 peak approximately 2% of the height of the M+ peak. Similarly, you could show that a compound with 3 carbons will have the M+1 peak at about 3% of the height of the M+ peak. The approximations we are making won't hold with more than 2 or 3 carbons. The proportion of carbon atoms which are 13 C isn't 1% - it's 1.11%. And the approximation that a ratio of 2 : 98 is about 2% doesn't hold as the small number increases. Consider a molecule with 5 carbons in it. You could work out that 5.55 (5 x 1.11) molecules will contain 1 13 C to every 94.45 (100 - 5.55) which contain only 12 C atoms. If you convert that to how tall the M+1 peak is as a percentage of the M+ peak, you get an answer of 5.9% (5.55/94.45 x 100). That's close enough to 6% that you might assume wrongly that there are 6 carbon atoms. Above 3 carbon atoms, then, you shouldn't really be making the approximation that the height of the M+1 peak as a percentage of the height of the M+ peak tells you the number of carbons - you will need to do some fiddly sums!
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Bookshelves/Organic_Chemistry/Organic_Chemistry_I_(Morsch_et_al.)/10%3A_Organohalides
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Learning Objectives After you have completed Chapter 10, you should be able to fulfill all of the detailed objectives listed under each individual section. design a multistep synthesis to prepare a given compound from a given starting material using any of the reactions studied up to this point in the course, including those which involve alkyl halides. solve road-map problems requiring a knowledge of any of the reactions or concepts studied up to this point, including those introduced in this chapter. define, and use in context, the key terms introduced. 10.0: Introduction to Organohalides Many organic compounds are closely related to the alkanes. Alkanes react with halogens to produce halogenated hydrocarbons, the simplest of which have a single halogen atom substituted for a hydrogen atom of the alkane. Even more closely related are the cycloalkanes, compounds in which the carbon atoms are joined in a ring, or cyclic fashion. 10.1: Names and Properties of Alkyl Halides Alkyl halides are also known as haloalkanes. This page explains what they are and discusses their physical properties. alkyl halides are compounds in which one or more hydrogen atoms in an alkane have been replaced by halogen atoms (fluorine, chlorine, bromine or iodine). 10.2: Preparing Alkyl Halides from Alkanes - Radical Halogenation Alkanes (the simplest of all organic compounds) undergo very few reactions. One of these reactions is halogenation, or the substitution of a single hydrogen on the alkane for a single halogen to form a haloalkane. This reaction is very important in organic chemistry because it functionalizes alkanes which opens a gateway to further chemical reactions. 10.3: Preparing Alkyl Halides from Alkenes - Allylic Bromination 10.4: Stability of the Allyl Radical - Resonance Revisited 10.5: Preparing Alkyl Halides from Alcohols This page looks at reactions in which the -OH group in an alcohol is replaced by a halogen such as chlorine or bromine. It includes a simple test for an -OH group using phosphorus(V) chloride. 10.6: Reactions of Alkyl Halides - Grignard Reagents The organomagnesium compounds formed by the reaction of an alkyl or aryl halide with magnesium are called Grignard reagents. As you will see throughout the remainder of this course, Grignard reagents can be used to synthesize a wide range of organic compounds and are extremely useful to the organic chemist. 10.7: Organometallic Coupling Reactions 10.8: Oxidation and Reduction in Organic Chemistry In organic chemistry, redox reactions look a little different. Electrons in an organic redox reaction often are transferred in the form of a hydride ion - a proton and two electrons. Because they occur in conjunction with the transfer of a proton, these are commonly referred to as hydrogenation and dehydrogenation reactions: a hydride plus a proton adds up to a hydrogen (H2) molecule. Be careful - do not confuse the terms hydrogenation and dehydrogenation with hydration and dehydration. 10.S: Organohalides (Summary)
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Courses/Grinnell_College/CHM_363%3A_Physical_Chemistry_1_(Grinnell_College)/14%3A_Chemical_Kinetics_I_-_Rate_Laws/14.01%3A_The_Time_Dependence_of_a_Chemical_Reaction_is_Described_by_a_Rate_Law
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The Reaction Rate The rate of a chemical reaction (or the reaction rate ) can be defined by the time needed for a change in concentration to occur. But there is a problem in that this allows for the definition to be made based on concentration changes for either the reactants or the products. Plus, due to stoichiometric concerns, the rates at which the concentrations are generally different! Toward this end, the following convention is used. For a general reaction \[a A + b B \rightarrow c C + d D \nonumber \] the reaction rate can be defined by any of the ratios \[\text{rate} = - \dfrac{1}{a} \dfrac{\Delta [A]}{dt} = - \dfrac{1}{b} \dfrac{\Delta[B]}{dt} = + \dfrac{1}{c} \dfrac{\Delta [C]}{dt} = + \dfrac{1}{d} \dfrac{ \Delta [D]}{dt} \nonumber \] Or for infinitesimal time intervals \[\text{rate} = - \dfrac{1}{a} \dfrac{d[A]}{dt} = - \dfrac{1}{b} \dfrac{d[C]}{dt} = + \dfrac{1}{c} \dfrac{d[C]}{dt} = + \dfrac{1}{d} \dfrac{d[D]}{dt} \nonumber \] Example \(\PageIndex{1}\): Under a certain set of conditions, the rate of the reaction \[\ce{N_2 + 3 H_2 \rightarrow 2 NH_3} \nonumber \] the reaction rate is \(6.0 \times 10^{-4}\, M/s\). Calculate the time-rate of change for the concentrations of N 2 , H 2 , and NH 3 . Solution : Due to the stoichiometry of the reaction, \[\text{rate} = - \dfrac{d[N_2]}{dt} = - \dfrac{1}{3} \dfrac{d[H_2]}{dt} = + \dfrac{1}{2} \dfrac{d[NH_3]}{dt} \nonumber \] so \[\begin{align*} \dfrac{d[N_2]}{dt} &= -6.0 \times 10^{-4} \,M/s \\[4pt] \dfrac{d[H_2]}{dt} &= -2.0 \times 10^{-4} \,M/s \\[4pt] \dfrac{d[NH_3]}{dt} &= 3.0 \times 10^{-4} \,M/s \end{align*} \] Note: The time derivatives for the reactants are negative because the reactant concentrations are decreasing, and those of products are positive since the concentrations of products increase as the reaction progresses. The Rate Law As shown above, the rate of the reaction can be followed experimentally by measuring the rate of the loss of a reactant or the rate of the production of a product. The rate of the reaction is often related to the concentration of some or all of the chemical species present at a given time. An equation called the rate law is used to show this relationship. The rate law cannot be predicted by looking at the balanced chemical reaction but must be determined by experiment. For example, the rate law for the reaction \[\ce{Cl2 (g) + CO (g) → Cl2CO (g) } \nonumber \] was experimentally determined to be \[ \text{rate} = k[Cl_2]^{3/2}[CO] \nonumber \] In this equation, \( k \) is the rate constant, and \( [\ce{Cl_2}] \) and \( [\ce{CO}] \) are the molar concentrations of Cl 2 and of CO. Each exponent is called the order of the given species. Thus, the rate law is second order in Cl 2 and first order in CO. The sum of the individual reactant orders is called the reaction order . This reaction has a reaction order of two and a half. In the next section, we will discuss methods to experimentally determine the rate law.
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Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Nucleic_Acids/DNA/The_Replication_of_DNA
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This page takes a very simplified look at how DNA replicates (copies) itself. It gives only a brief over-view of the process, with no attempt to describe the mechanism. Semi-conservative replication We'll explain exactly what "semi-conservative" means when we have got some diagrams to look at. First imagine what happens if the two individual strands in the DNA double helix start to unzip. The diagram shows this happening in the middle of the DNA double helix - you mustn't assume that the top of the diagram is the end of the chain. It isn't. Further up the double helix, the two strands will still be joined together. In fact, this is happening lots of times along the very long DNA molecule. Lengths of chain become separated to form what are known as "bubbles". If you feel the need to see this in more detail, read the rest of this page, and then have a quick look at the links above. Some of the hydrogen bonds get broken and the two strands become partly separated. The red dotted lines on the diagram just point out the original base pairs. These are not bonds in any form. These base pairs are now much too far apart for any sort of bonding between them. Now suppose that you have a source of nucleotides - phosphate joined to deoxyribose joined to a base, including all the four sorts of bases needed for DNA. The next diagram shows what would happen if a nucleotide containing guanine (G) and one containing cytosine (C) were attracted to the top two bases on the left-hand strand of the unzipped DNA - and then joined together. How did they end up joined together? This is all under the control of a number of enzymes, one of which (DNA polymerase) is responsible for joining up nucleotides along the chain in this way. Now suppose the same sort of thing happened at the top of the right-hand strand. You would end up with . . . Now compare the double strands that you are forming on the left- and right-hand sides. They are exactly the same . . . and if you were to continue this process, they would continue to be the same. And if you compare the patterns of bases in the new DNA being formed with what was in the original DNA before it started to unzip, everything is the same. This is inevitable because of the way the bases pair together. What does semi-conservative replication mean? Let's simplify the last diagram, and assume that the whole copying process is complete. The next diagram focusses on the short bit of the total DNA molecule that we have been looking at. A typical human DNA molecule is around 150 million base pairs long - you will have to imagine the rest of it! You have also got to remember that in reality the whole thing would have coiled into its double helix. Trying to draw that just makes everything look messy and complicated! The original DNA is shown all in blue. The red strands in the daughter DNA are the ones which have been built on the original blue strands during the replication process. You can see that each of the daughter molecules is made of half of the original DNA plus a new strand. That's all "semi-conservative replication" means. Half of the original DNA is conserved (kept) in each of the daughter molecules. The red and blue, of course, have no physical significance apart from as a way of making the diagrams clearer. All three of these DNA molecules will be identical in every way.
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Courses/University_of_Kentucky/UK%3A_General_Chemistry/18%3A_Representative_Metals_Metalloids_and_Nonmetals/18.06%3A_Occurrence_Preparation_and_Properties_of_Carbonates
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Skills to Develop Describe the preparation, properties, and uses of some representative metal carbonates The chemistry of carbon is extensive; however, most of this chemistry is not relevant to this chapter. The other aspects of the chemistry of carbon will appear in the chapter covering organic chemistry. In this chapter, we will focus on the carbonate ion and related substances. The metals of groups 1 and 2, as well as zinc, cadmium, mercury, and lead(II), form ionic carbonates —compounds that contain the carbonate anions, \(\ce{CO3^2-}\). The metals of group 1, magnesium, calcium, strontium, and barium also form hydrogen carbonates —compounds that contain the hydrogen carbonate anion, \(\ce{HCO3-}\), also known as the bicarbonate anion . With the exception of magnesium carbonate, it is possible to prepare carbonates of the metals of groups 1 and 2 by the reaction of carbon dioxide with the respective oxide or hydroxide. Examples of such reactions include: \[ \begin{align} \ce{Na2O}(s)+\ce{CO2}(g) &⟶ \ce{Na2CO3}(s)\\[5pt] \ce{Ca(OH)2}(s)+\ce{CO2}(g) &⟶\ce{CaCO3}(s)+\ce{H2O}(l) \end{align}\] The carbonates of the alkaline earth metals of group 12 and lead(II) are not soluble. These carbonates precipitate upon mixing a solution of soluble alkali metal carbonate with a solution of soluble salts of these metals. Examples of net ionic equations for the reactions are: \[ \begin{align} \ce{Ca^2+}(aq)+\ce{CO3^2-}(aq) &⟶\ce{CaCO3}(s) \\[5pt] \ce{Pb^2+}(aq)+\ce{CO3^2-}(aq) &⟶\ce{PbCO3}(s) \end{align}\] Pearls and the shells of most mollusks are calcium carbonate. Tin(II) or one of the trivalent or tetravalent ions such as Al 3+ or Sn 4+ behave differently in this reaction as carbon dioxide and the corresponding oxide form instead of the carbonate. Alkali metal hydrogen carbonates such as NaHCO 3 and CsHCO 3 form by saturating a solution of the hydroxides with carbon dioxide. The net ionic reaction involves hydroxide ion and carbon dioxide: \[\ce{OH-}(aq)+\ce{CO2}(aq)⟶\ce{HCO3-}(aq)\] It is possible to isolate the solids by evaporation of the water from the solution. Although they are insoluble in pure water, alkaline earth carbonates dissolve readily in water containing carbon dioxide because hydrogen carbonate salts form. For example, caves and sinkholes form in limestone when CaCO 3 dissolves in water containing dissolved carbon dioxide: \[\ce{CaCO3}(s)+\ce{CO2}(aq)+\ce{H2O}(l)⟶\ce{Ca^2+}(aq)+\ce{2HCO3-}(aq)\] Hydrogen carbonates of the alkaline earth metals remain stable only in solution; evaporation of the solution produces the carbonate. Stalactites and stalagmites, like those shown in Figure \(\PageIndex{1}\), form in caves when drops of water containing dissolved calcium hydrogen carbonate evaporate to leave a deposit of calcium carbonate. Figure \(\PageIndex{1}\) : (a) Stalactites and (b) stalagmites are cave formations of calcium carbonate. (credit a: modification of work by Arvind Govindaraj; credit b: modification of work by the National Park Service.) The two carbonates used commercially in the largest quantities are sodium carbonate and calcium carbonate. In the United States, sodium carbonate is extracted from the mineral trona, Na 3 (CO 3 )(HCO 3 )(H 2 O) 2 . Following recrystallization to remove clay and other impurities, heating the recrystallized trona produces Na 2 CO 3 : \[\ce{2Na3(CO3)(HCO3)(H2O)2}(s)⟶\ce{3Na2CO3}(s)+\ce{5H2O}(l)+\ce{CO2}(g)\] Carbonates are moderately strong bases. Aqueous solutions are basic because the carbonate ion accepts hydrogen ion from water in this reversible reaction: \[\ce{CO3^2-}(aq)+\ce{H2O}(l)⇌\ce{HCO3-}(aq)+\ce{OH-}(aq)\] Carbonates react with acids to form salts of the metal, gaseous carbon dioxide, and water. The reaction of calcium carbonate, the active ingredient of the antacid Tums, with hydrochloric acid (stomach acid), as shown in Figure \(\PageIndex{2}\), illustrates the reaction: \[\ce{CaCO3}(s)+\ce{2HCl}(aq)⟶\ce{CaCl2}(aq)+\ce{CO2}(g)+\ce{H2O}(l)\] Figure \(\PageIndex{2}\): The reaction of calcium carbonate with hydrochloric acid is shown. (credit: Mark Ott) Other applications of carbonates include glass making—where carbonate ions serve as a source of oxide ions—and synthesis of oxides. Hydrogen carbonates are amphoteric because they act as both weak acids and weak bases. Hydrogen carbonate ions act as acids and react with solutions of soluble hydroxides to form a carbonate and water: \[\ce{KHCO3}(aq)+\ce{KOH}(aq)⟶\ce{K2CO3}(aq)+\ce{H2O}(l)\] With acids, hydrogen carbonates form a salt, carbon dioxide, and water. Baking soda (bicarbonate of soda or sodium bicarbonate) is sodium hydrogen carbonate. Baking powder contains baking soda and a solid acid such as potassium hydrogen tartrate (cream of tartar), KHC 4 H 4 O 6 . As long as the powder is dry, no reaction occurs; immediately after the addition of water, the acid reacts with the hydrogen carbonate ions to form carbon dioxide: \[\ce{HC4H4O6-}(aq)+\ce{HCO3-}(aq)⟶\ce{C4H4O6^2-}(aq)+\ce{CO2}(g)+\ce{H2O}(l)\] Dough will trap the carbon dioxide, causing it to expand during baking, producing the characteristic texture of baked goods. Summary The usual method for the preparation of the carbonates of the alkali and alkaline earth metals is by reaction of an oxide or hydroxide with carbon dioxide. Other carbonates form by precipitation. Metal carbonates or hydrogen carbonates such as limestone (CaCO 3 ), the antacid Tums (CaCO 3 ), and baking soda (NaHCO 3 ) are common examples. Carbonates and hydrogen carbonates decompose in the presence of acids and most decompose on heating. Glossary bicarbonate anion salt of the hydrogen carbonate ion, \(\ce{HCO3-}\) carbonate salt of the anion \(\ce{CO3^2-}\); often formed by the reaction of carbon dioxide with bases hydrogen carbonate salt of carbonic acid, H 2 CO 3 (containing the anion \(\ce{HCO3-}\)) in which one hydrogen atom has been replaced; an acid carbonate; also known as bicarbonate ion Contributors Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/[email protected] ).
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Courses/Clackamas_Community_College/CH_112%3A_Chemistry_for_Health_Sciences/02%3A_Elements_Atoms_and_the_Periodic_Table/2.04%3A_The_Structure_of_Atoms
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Learning Objectives Describe the three main subatomic particles. State how the subatomic particles are arranged in atoms. There have been several minor but important modifications to Dalton’s atomic theory. For one thing, Dalton considered atoms to be indivisible. We know now that atoms not only can be divided but also are composed of three different kinds of particle, subatomic particles , with their own properties that are different from the chemical properties of atoms. Subatomic Particles The first subatomic particle was identified in 1897 and called the electron. It is an extremely tiny particle, with a mass of about 9.109 × 10 −31 kg. Experiments with magnetic fields showed that the electron has a negative electrical charge. By 1920, experimental evidence indicated the existence of a second particle. A proton has the same amount of charge as an electron, but its charge is positive, not negative. Another major difference between a proton and an electron is mass. Although still incredibly small, the mass of a proton is 1.673 × 10 −27 kg, which is almost 2,000 times greater than the mass of an electron. Because opposite charges attract each other (while like charges repel each other), protons attract electrons (and vice versa). Finally, additional experiments pointed to the existence of a third particle. Evidence produced in 1932 established the existence of the neutron, a particle with about the same mass as a proton but with no electrical charge, it is neutral . We understand now that all atoms can be broken down into subatomic particles: protons, neutrons, and electrons. Table \(\PageIndex{1}\) lists some of their important characteristics and the symbols used to represent each particle. Particle Symbol Mass (kg) Relative Mass (proton = 1) Relative Charge proton p+ 1.673 × 10−27 1.00000 +1 neutron n0 1.675 × 10−27 1.00000 0 electron e− 9.109 × 10−31 0.00055 −1 The Nucleus How are these subatomic particles arranged? Between 1909 and 1911, Ernest Rutherford, a Cambridge physicist, and his associates Hans Geiger and Ernest Marsden performed experiments that provided strong evidence concerning the internal structure of an atom. They took a very thin metal foil, such as gold or platinum, and aimed a beam of positively charged particles (called alpha particles, which are combinations of two protons and two neutrons) from a radioactive source toward the foil. Surrounding the foil was a detector—either a scintillator (a material that glows when hit by such particles) or some unexposed film (which is exposed where the particles hit it). The detector allowed the scientists to determine the distribution of the alpha particles after they interacted with the foil. Figure \(\PageIndex{1}\) shows a diagram of the experimental setup. Rutherford proposed the following model to explain these experimental results. Protons and neutrons are concentrated in a central region he called the nucleus (plural, nuclei ) of the atom. Electrons are outside the nucleus and orbit about it because they are attracted to the positive charge in the nucleus. Most of the mass of an atom is in the nucleus, while the orbiting electrons account for an atom’s size. As a result, an atom consists largely of empty space. Rutherford called his description the “planetary model” of the atom. Figure \(\PageIndex{2}\) shows how this model explains the experimental results. The planetary model of the atom replaced the plum pudding model, which had electrons floating around aimlessly like plums in a “pudding” of positive charge. Rutherford’s model is essentially the same model that we use today to describe atoms but with one important modification. The planetary model suggests that electrons occupy certain specific, circular orbits about the nucleus. We know now that this model is overly simplistic. A better description is that electrons form fuzzy clouds around nuclei. Figure \(\PageIndex{3}\) shows a more modern version of our understanding of atomic structure. Concept Review Exercises What are the charges and the relative masses of the three subatomic particles? Describe the structure of an atom in terms of its protons, neutrons, and electrons. Answers proton: +1, large; neutron: 0, large; electron: −1, small Protons and neutrons are located in a central nucleus, while electrons orbit about the nucleus. Key Takeaways Atoms are composed of three main subatomic particles: protons, neutrons, and electrons. Protons and neutrons are grouped together in the nucleus of an atom, while electrons orbit about the nucleus.
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Courses/Matanuska-Susitna_College/MatSu_College-CHEM_A104_Introduction_to_Organic_and_Biochemistry/14%3A_Organic_Compounds_of_Oxygen/14.06%3A_Reactions_of_Alcohols
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Learning Objectives Give two major types of reactions of alcohols. Describe the result of the oxidation of a primary alcohol. Describe the result of the oxidation of a secondary alcohol. Chemical reactions in alcohols occur mainly at the functional group, but some involve hydrogen atoms attached to the OH -bearing carbon atom or to an adjacent carbon atom. Of the three major kinds of alcohol reactions, which are summarized in Figure \(\PageIndex{1}\), two—dehydration and oxidation—are considered here. The third reaction type— esterification —is covered elsewhere. Dehydration As noted in Figure \(\PageIndex{1}\), an alcohol undergoes dehydration in the presence of a catalyst to form an alkene and water. The reaction removes the OH group from the alcohol carbon atom and a hydrogen atom from an adjacent carbon atom in the same molecule: Under the proper conditions, it is possible for the dehydration to occur between two alcohol molecules. The entire OH group of one molecule and only the hydrogen atom of the OH group of the second molecule are removed. The two ethyl groups attached to an oxygen atom form an ether molecule. ( Ethers are discussed in elsewhere ) Thus, depending on conditions, one can prepare either alkenes or ethers by the dehydration of alcohols. Both dehydration and hydration reactions occur continuously in cellular metabolism, with enzymes serving as catalysts and at a temperature of about 37°C. The following reaction occurs in the "Embden–Meyerhof" pathway Although the participating compounds are complex, the reaction is the same: elimination of water from the starting material. The idea is that if you know the chemistry of a particular functional group, you know the chemistry of hundreds of different compounds. Oxidation Primary and secondary alcohols are readily oxidized. We saw earlier how methanol and ethanol are oxidized by liver enzymes to form aldehydes. Because a variety of oxidizing agents can bring about oxidation, we can indicate an oxidizing agent without specifying a particular one by writing an equation with the symbol [O] above the arrow. For example, we write the oxidation of ethanol—a primary alcohol—to form acetaldehyde—an aldehyde—as follows: We shall see that aldehydes are even more easily oxidized than alcohols and yield carboxylic acids. Secondary alcohols are oxidized to ketones . The oxidation of isopropyl alcohol by potassium dichromate (\(\ce{K2Cr2O7}\)) gives acetone, the simplest ketone: Unlike aldehydes, ketones are relatively resistant to further oxidation, so no special precautions are required to isolate them as they form. Note that in oxidation of both primary (RCH 2 OH) and secondary (R 2 CHOH) alcohols, two hydrogen atoms are removed from the alcohol molecule, one from the OH group and other from the carbon atom that bears the OH group. These reactions can also be carried out in the laboratory with chemical oxidizing agents. One such oxidizing agent is potassium dichromate. The balanced equation (showing only the species involved in the reaction) in this case is as follows: \[\ce{8H^{=} + Cr2O7^{2-} + 3CH3CH2OH -> 3CH3CHO + 2Cr^{3+} + 7H2O} \nonumber \] Alcohol oxidation is important in living organisms. Enzyme-controlled oxidation reactions provide the energy cells need to do useful work. One step in the metabolism of carbohydrates involves the oxidation of the secondary alcohol group in isocitric acid to a ketone group: The overall type of reaction is the same as that in the conversion of isopropyl alcohol to acetone. Tertiary alcohols (R 3 COH) are resistant to oxidation because the carbon atom that carries the OH group does not have a hydrogen atom attached but is instead bonded to other carbon atoms. The oxidation reactions we have described involve the formation of a carbon-to-oxygen double bond. Thus, the carbon atom bearing the OH group must be able to release one of its attached atoms to form the double bond. The carbon-to-hydrogen bonding is easily broken under oxidative conditions, but carbon-to-carbon bonds are not. Therefore tertiary alcohols are not easily oxidized. Example \(\PageIndex{1}\) Write an equation for the oxidation of each alcohol. Use [O] above the arrow to indicate an oxidizing agent. If no reaction occurs, write “no reaction” after the arrow. CH 3 CH 2 CH 2 CH 2 CH 2 OH Solution The first step is to recognize the class of each alcohol as primary, secondary, or tertiary. This alcohol has the OH group on a carbon atom that is attached to only one other carbon atom, so it is a primary alcohol. Oxidation forms first an aldehyde and further oxidation forms a carboxylic acid. This alcohol has the OH group on a carbon atom that is attached to three other carbon atoms, so it is a tertiary alcohol. No reaction occurs. This alcohol has the OH group on a carbon atom that is attached to two other carbon atoms, so it is a secondary alcohol; oxidation gives a ketone. Exercise \(\PageIndex{1}\) Write an equation for the oxidation of each alcohol. Use [O] above the arrow to indicate an oxidizing agent. If no reaction occurs, write “no reaction” after the arrow. Summary Alcohols can be dehydrated to form either alkenes (higher temperature, excess acid) or ethers (lower temperature, excess alcohol). Primary alcohols are oxidized to form aldehydes. Secondary alcohols are oxidized to form ketones. Tertiary alcohols are not readily oxidized.
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Courses/Duke_University/Textbook%3A_Modern_Applications_of_Chemistry_(Cox)/04%3A_Intermolecular_Forces/4.02%3A_Properties_of_Liquids
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Learning Objectives Distinguish between adhesive and cohesive forces Define viscosity, surface tension, and capillary rise Describe the roles of intermolecular attractive forces in each of these properties/phenomena When you pour a glass of water, or fill a car with gasoline, you observe that water and gasoline flow freely. But when you pour syrup on pancakes or add oil to a car engine, you note that syrup and motor oil do not flow as readily. The viscosity of a liquid is a measure of its resistance to flow. Water, gasoline, and other liquids that flow freely have a low viscosity. Honey, syrup, motor oil, and other liquids that do not flow freely, like those shown in Figure \(\PageIndex{1}\), have higher viscosities. We can measure viscosity by measuring the rate at which a metal ball falls through a liquid (the ball falls more slowly through a more viscous liquid) or by measuring the rate at which a liquid flows through a narrow tube (more viscous liquids flow more slowly). The IMFs between the molecules of a liquid, the size and shape of the molecules, and the temperature determine how easily a liquid flows. As Table \(\PageIndex{1}\) shows, the more structurally complex are the molecules in a liquid and the stronger the IMFs between them, the more difficult it is for them to move past each other and the greater is the viscosity of the liquid. As the temperature increases, the molecules move more rapidly and their kinetic energies are better able to overcome the forces that hold them together; thus, the viscosity of the liquid decreases. Substance Formula Viscosity (mPa·s) water H2O 0.890 mercury Hg 1.526 ethanol C2H5OH 1.074 octane C8H18 0.508 ethylene glycol CH2(OH)CH2(OH) 16.1 honey variable ~2,000–10,000 motor oil variable ~50–500 The various IMFs between identical molecules of a substance are examples of cohesive forces . The molecules within a liquid are surrounded by other molecules and are attracted equally in all directions by the cohesive forces within the liquid. However, the molecules on the surface of a liquid are attracted only by about one-half as many molecules. Because of the unbalanced molecular attractions on the surface molecules, liquids contract to form a shape that minimizes the number of molecules on the surface—that is, the shape with the minimum surface area. A small drop of liquid tends to assume a spherical shape, as shown in Figure \(\PageIndex{2}\), because in a sphere, the ratio of surface area to volume is at a minimum. Larger drops are more greatly affected by gravity, air resistance, surface interactions, and so on, and as a result, are less spherical. Surface tension is defined as the energy required to increase the surface area of a liquid, or the force required to increase the length of a liquid surface by a given amount. This property results from the cohesive forces between molecules at the surface of a liquid, and it causes the surface of a liquid to behave like a stretched rubber membrane. Surface tensions of several liquids are presented in Table \(\PageIndex{2}\). Substance Formula Surface Tension (mN/m) water H2O 71.99 mercury Hg 458.48 ethanol C2H5OH 21.97 octane C8H18 21.14 ethylene glycol CH2(OH)CH2(OH) 47.99 Among common liquids, water exhibits a distinctly high surface tension due to strong hydrogen bonding between its molecules. As a result of this high surface tension, the surface of water represents a relatively “tough skin” that can withstand considerable force without breaking. A steel needle carefully placed on water will float. Some insects, like the one shown in Figure \(\PageIndex{3}\), even though they are denser than water, move on its surface because they are supported by the surface tension. The IMFs of attraction between two different molecules are called adhesive forces . Consider what happens when water comes into contact with some surface. If the adhesive forces between water molecules and the molecules of the surface are weak compared to the cohesive forces between the water molecules, the water does not “wet” the surface. For example, water does not wet waxed surfaces or many plastics such as polyethylene. Water forms drops on these surfaces because the cohesive forces within the drops are greater than the adhesive forces between the water and the plastic. Water spreads out on glass because the adhesive force between water and glass is greater than the cohesive forces within the water. When water is confined in a glass tube, its meniscus (surface) has a concave shape because the water wets the glass and creeps up the side of the tube. On the other hand, the cohesive forces between mercury atoms are much greater than the adhesive forces between mercury and glass. Mercury therefore does not wet glass, and it forms a convex meniscus when confined in a tube because the cohesive forces within the mercury tend to draw it into a drop (Figure \(\PageIndex{4}\)). If you place one end of a paper towel in spilled wine, as shown in Figure \(\PageIndex{5}\), the liquid wicks up the paper towel. A similar process occurs in a cloth towel when you use it to dry off after a shower. These are examples of capillary action —when a liquid flows within a porous material due to the attraction of the liquid molecules to the surface of the material and to other liquid molecules. The adhesive forces between the liquid and the porous material, combined with the cohesive forces within the liquid, may be strong enough to move the liquid upward against gravity. Towels soak up liquids like water because the fibers of a towel are made of molecules that are attracted to water molecules. Most cloth towels are made of cotton, and paper towels are generally made from paper pulp. Both consist of long molecules of cellulose that contain many −OH groups. Water molecules are attracted to these −OH groups and form hydrogen bonds with them, which draws the H 2 O molecules up the cellulose molecules. The water molecules are also attracted to each other, so large amounts of water are drawn up the cellulose fibers. Capillary action can also occur when one end of a small diameter tube is immersed in a liquid, as illustrated in Figure \(\PageIndex{6}\). If the liquid molecules are strongly attracted to the tube molecules, the liquid creeps up the inside of the tube until the weight of the liquid and the adhesive forces are in balance. The smaller the diameter of the tube is, the higher the liquid climbs. It is partly by capillary action occurring in plant cells called xylem that water and dissolved nutrients are brought from the soil up through the roots and into a plant. Capillary action is the basis for thin layer chromatography, a laboratory technique commonly used to separate small quantities of mixtures. You depend on a constant supply of tears to keep your eyes lubricated and on capillary action to pump tear fluid away. The height to which a liquid will rise in a capillary tube is determined by several factors as shown in the following equation: \[h=\dfrac{2T\cosθ}{rρg} \label{10.2.1} \] where h is the height of the liquid inside the capillary tube relative to the surface of the liquid outside the tube, T is the surface tension of the liquid, θ is the contact angle between the liquid and the tube, r is the radius of the tube, ρ is the density of the liquid, and g is the acceleration due to gravity, 9.8 m/s 2 . When the tube is made of a material to which the liquid molecules are strongly attracted, they will spread out completely on the surface, which corresponds to a contact angle of 0°. This is the situation for water rising in a glass tube. Example \(\PageIndex{1}\): Capillary Rise At 25 °C, how high will water rise in a glass capillary tube with an inner diameter of 0.25 mm? For water, T = 71.99 mN/m and ρ = 1.0 g/cm 3 . Solution The liquid will rise to a height h given by Equation \(\ref{10.2.1}\) : \[h=\dfrac{2T\cosθ}{rρg} \nonumber \] The Newton is defined as a kg m/s 2 , and so the provided surface tension is equivalent to 0.07199 kg/s 2 . The provided density must be converted into units that will cancel appropriately: ρ = 1000 kg/m 3 . The diameter of the tube in meters is 0.00025 m, so the radius is 0.000125 m. For a glass tube immersed in water, the contact angle is θ = 0°, so cos θ = 1. Finally, acceleration due to gravity on the earth is g = 9.8 m/s 2 . Substituting these values into the equation, and cancelling units, we have: \[h=\mathrm{\dfrac{2(0.07199\:kg/s^2)}{(0.000125\:m)(1000\:kg/m^3)(9.8\:m/s^2)}=0.12\:m=12\: cm} \nonumber \] Exercise \(\PageIndex{1}\) Water rises in a glass capillary tube to a height of 8.4 cm. What is the diameter of the capillary tube? Answer diameter = 0.36 mm Applications: Capillary Action is Used to Draw Blood Many medical tests require drawing a small amount of blood, for example to determine the amount of glucose in someone with diabetes or the hematocrit level in an athlete. This procedure can be easily done because of capillary action, the ability of a liquid to flow up a small tube against gravity, as shown in Figure \(\PageIndex{7}\). When your finger is pricked, a drop of blood forms and holds together due to surface tension—the unbalanced intermolecular attractions at the surface of the drop. Then, when the open end of a narrow-diameter glass tube touches the drop of blood, the adhesive forces between the molecules in the blood and those at the glass surface draw the blood up the tube. How far the blood goes up the tube depends on the diameter of the tube (and the type of fluid). A small tube has a relatively large surface area for a given volume of blood, which results in larger (relative) attractive forces, allowing the blood to be drawn farther up the tube. The liquid itself is held together by its own cohesive forces. When the weight of the liquid in the tube generates a downward force equal to the upward force associated with capillary action, the liquid stops rising. Key Concepts and Summary The intermolecular forces between molecules in the liquid state vary depending upon their chemical identities and result in corresponding variations in various physical properties. Cohesive forces between like molecules are responsible for a liquid’s viscosity (resistance to flow) and surface tension (elasticity of a liquid surface). Adhesive forces between the molecules of a liquid and different molecules composing a surface in contact with the liquid are responsible for phenomena such as surface wetting and capillary rise. Key Equations \(h=\dfrac{2T\cosθ}{rρg}\) Glossary adhesive force force of attraction between molecules of different chemical identities capillary action flow of liquid within a porous material due to the attraction of the liquid molecules to the surface of the material and to other liquid molecules cohesive force force of attraction between identical molecules surface tension energy required to increase the area, or length, of a liquid surface by a given amount viscosity measure of a liquid’s resistance to flow
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Courses/SUNY_Oneonta/Organic_Chemistry_with_a_Biological_Emphasis_(SUNY_Oneonta)/08%3A_Nucleophilic_Substitution_Reactions/8.02%3A_Two_Mechanistic_Models_for_Nucleophilic_Substitution
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As we begin our study of nucleophilic substitution reactions, we will focus at first on simple alkyl halide compounds. While the specific reactions we'll initially consider do not occur in living things, it is nonetheless useful to start with alkyl halides as a model to illustrate some fundamental ideas that we must cover. Later, we will move on to apply what we have earned about alkyl halides to the larger and more complex biomolecules that are undergoing nucleophilic substitution right now in your own cells. The \(S_N2\) mechanism You may recall from our brief introduction to the topic in chapter 6 that there are two mechanistic models for how a nucleophilic substitution reaction can proceed. In one mechanism, the reaction is concerted: it takes place in a single step, and bond-forming and bond-breaking occur simultaneously. This is illustrated by the reaction between chloromethane and hydroxide ion: Recall that the hydroxide ion in this reaction is acting as a nucleophile (an electron-rich, nucleus-loving species), the carbon atom of chloromethane is acting as an electrophile (an electron-poor species which is attracted to electrons), and the chloride ion is the leaving group (where the name is self-evident). Organic chemists refer to this mechanism by the term '\(S_N2\)', where S stands for 'substitution', the subscript N stands for 'nucleophilic', and the number 2 refers to the fact that this is a bimolecular reaction: the overall rate depends on a step in which two separate species collide. A potential energy diagram for this reaction shows the transition state (TS) as the highest point on the pathway from reactants to products. The geometry of an \(S_N2\) reaction is specific: the reaction can only occur when the nucleophile collides with the electrophilic carbon from the opposite side relative to the leaving group. This is referred to as backside attack. Approach from the front side simply doesn't work: the leaving group - which, like the nucleophile is an electron-rich group - blocks the way. 2 The result of backside attack is that the bonding geometry at the electrophilic carbon inverts (turns inside-out) as the reaction proceeds. 3 The transition state of the reaction is illustrated by drawing dotted lines to represent the covalent bonds that are in the process of breaking or forming. Because the formal charge on the oxygen nucleophile changes from negative one to zero as the reaction proceeds, and conversely the charge on the chlorine leaving group changes from zero to negative one, at the transition state both atoms are shown bearing a partial negative charge (the symbol \(\delta \)-). One other drawing convention for transition states is to use brackets, with the double-dagger symbol in subscript. Notice that the transition state for an \(S_N2\) reaction has trigonal bipyramidal geometry: the nucleophile, electrophile, and leaving group form a straight line, and the three substituents on carbon (all hydrogen atoms in this case) are arranged in the same plane at \(120^{\circ}\) angles. What is the measure in degrees) of the \(H-C-O\) angle in the \(S_N2\) transition state illustrated above? Consider what would happen if we were to replace one of the hydrogen atoms in chloromethane with deuterium (the \(^2H\) isotope), and one with tritium (the radioactive \(^3H\) isotope). Now, because it has four different substituents, our carbon electrophile is a chiral center. We'll arbitrarily assume that we start with the \(S\) enantiomer. As the hydroxide nucleophile attacks from the backside and the bonding geometry at carbon inverts, we see that the stereochemistry of the product reflects this inversion: we end up with the \(R\) enantiomer of the chiral product. \(S_N2\) reactions proceed with inversion of stereochemical configuration at the electrophilic carbon. video tutorial/animation: inversion of configuration during SN2 reactions The \(S_N1\) mechanism A second model for the nucleophilic substitution reaction is called the \(S_N1\) mechanism. The '1' in \(S_N1\) indicates that the rate-determining step of the reaction is unimolecular: in other words, the rate-determining step involves a single molecule breaking apart (rather than two molecules colliding as was the case in the \(S_N2\) mechanism.) In an \(S_N1\) mechanism the carbon-leaving group bond breaks first, before the nucleophile approaches, resulting in formation of a carbocation intermediate (step 1): A carbocation is a powerful electrophile: because the carbon lacks a complete octet of valence electrons, it is 'electron-hungry'. In step 2, a lone pair of electrons on the water nucleophile fills the empty p orbital of the carbocation to form a new bond. Notice that this is actually a three-step mechanism, with a final, rapid acid-base step leading to the alcohol product. A potential energy diagram for this \(S_N1\) reaction shows that each of the two positively-charged intermediate stages (\(I_1\) and \(I_2\) in the diagram) can be visualized as a valley in the path of the reaction, higher in energy than both the reactant and product but lower in energy than the transition states. The first, bond-breaking step is the slowest, rate-determining step - notice it has the highest activation energy and leads to the highest-energy species (\(I_1\), the carbocation intermediate). Step 2 is rapid: a new covalent bond forms between a carbocation and and a water nucleophile , and no covalent bonds are broken. Recall from chapter 7 that Bronsted-Lowry proton transfer steps like step 3 are rapid, with low activation energies. The nucleophilic substitution reactions we have seen so far are examples of hydrolysis. This term is one that you will encounter frequently in organic and biological chemistry. Hydrolysis means 'breaking with water': in a hydrolysis reaction, a water molecule (or hydroxide ion) participates in the breaking of a covalent bond. There are many reaction types other than nucleophilic substitution that can accurately be described as hydrolysis, and we will see several examples throughout the remaining chapters of this book. Solvolysis is a more general term, used when a bond in a reagent is broken by a solvent molecule: usually, the solvent in question is water or an alcohol such as methanol or ethanol. Draw a mechanism for the \( S_N1\) solvolysis of tert-butyl chloride in methanol. What new functional group has been formed? We saw that \(S_N2\) reactions result in inversion of stereochemical configuration at the carbon center. What about the stereochemical outcome of \(S_N1\) reactions? Recall that a carbocation is \(sp^2\)-hybridized, with an empty p orbital perpendicular to the plane formed by the three sigma bonds: In the second step of an \(S_N1\) reaction, the nucleophile can attack from either side of the carbocation (the leaving group is already gone, and thus cannot block attack from one side like in an \(S_N2\) reaction). Consider an \(S_N1\) reaction with a chiral tertiary alkyl chloride: Because the nucleophile is free to attack from either side of the carbocation electrophile, the reaction leads to a 50:50 mixture of two stereoisomeric products. In other words: In general nonenzymatic SN1 reaction can occur with either retention or inversion of configuration at the electrophilic carbon, leading to racemization if the carbon is chiral. For an example, consider the hydrolysis of (S)-3-chloro-3-methylhexane. The result of this (nonenzymatic) reaction is a racemic mixture of chiral alcohols. It is important to remember, however, that enzymatic reactions are in almost all cases very specific with regard to stereochemical outcome. A biochemical \(S_N1\) reaction, as we shall see later, can result in either inversion or retention of configuration at the electrophilic carbon, but generally not a mixture of both: the two reactants are bound with specific geometry in the enzyme's active site, so that the nucleophile can approach from one side only. (The following exercises refer to nonbiological reactions) Draw a complete mechanism for the hydrolysis reaction in the previous figure, showing all bond-breaking and bond-forming steps, and all intermediate species. Draw structures representing TS1 and TS2 in the reaction. Use the solid/dash wedge convention to show three dimensions. What is the expected optical rotation of the product mixture? Could the two organic products be separated on a silica column chromatography? Draw the product(s) of the hydrolysis of (R)-3-chloro-3-methyl heptane. What can you predict, if anything, about the optical rotation of the product(s)? Draw the product(s) of the hydrolysis of (3R,5R)-3-chloro-3,5-dimethyl heptane. What can you predict, if anything, about the optical rotation of the product(s)? Before we go on to look at some actual biochemical nucleophilic substitution reactions, we first need to lay the intellectual groundwork by focusing more closely on the characteristics of the three principal partners in the nucleophilic substitution reaction: the nucleophile, the electrophile, and the leaving group. In addition, we need to consider the carbocation intermediate that plays such a key role in the \(S_N1\) mechanism. For the sake of simplicity, we will continue to use simple, non-biological organic molecules and reaction examples as we work through the basic concepts. Video tutorial/animation: SN1 reactions Contributors Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
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Courses/Indiana_Tech/Chemistry_2300_(Budhi)/09%3A_Substitution_and_Elimination_Reactions_of_Alkyl_Halides/9.09%3A_Comparing_the_E2_and_E1_Reactions_of_Alkyl_Halides
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Having discussed the many factors that influence nucleophilic substitution and elimination reactions of alkyl halides, we must now consider the practical problem of predicting the most likely outcome when a given alkyl halide is reacted with a given nucleophile. As we noted earlier, several variables must be considered, the most important being the structure of the alkyl group and the nature of the nucleophilic reactant .In general, in order for an SN1 or E1 reaction to occur, the relevant carbocation intermediate must be relatively stable. Strong nucleophile favor substitution, and strong bases, especially strong hindered bases (such as tert-butoxide) favor elimination. The nature of the halogen substituent on the alkyl halide is usually not very significant if it is Cl, Br or I. In cases where both S N 2 and E2 reactions compete, chlorides generally give more elimination than do iodides, since the greater electronegativity of chlorine increases the acidity of beta-hydrogens. Indeed, although alkyl fluorides are relatively unreactive, when reactions with basic nucleophiles are forced, elimination occurs (note the high electronegativity of fluorine). The following table summarizes the expected outcome of alkyl halide reactions with nucleophiles. It is assumed that the alkyl halides have one or more beta-hydrogens, making elimination possible; and that low dielectric solvents (e.g. acetone, ethanol, tetrahydrofuran & ethyl acetate) are used. When a high dielectric solvent would significantly influence the reaction this is noted in red. Note that halogens bonded to sp 2 or sp hybridized carbon atoms do not normally undergo substitution or elimination reactions with nucleophilic reagents . 0 1 2 3 Nucleophile Anionic Nucleophiles ( Weak Bases: I–, Br–, SCN–, N3–, CH3CO2– , RS–, CN– etc. ) pKa's from -9 to 10 (left to right) Anionic Nucleophiles ( Strong Bases: HO–, RO– ) pKa's > 15 Neutral Nucleophiles ( H2O, ROH, RSH, R3N ) pKa's ranging from -2 to 11 Alkyl Group Anionic Nucleophiles ( Weak Bases: I–, Br–, SCN–, N3–, CH3CO2– , RS–, CN– etc. ) pKa's from -9 to 10 (left to right) Anionic Nucleophiles ( Strong Bases: HO–, RO– ) pKa's > 15 Neutral Nucleophiles ( H2O, ROH, RSH, R3N ) pKa's ranging from -2 to 11 Primary RCH2– Rapid SN2 substitution. The rate may be reduced by substitution of β-carbons, as in the case of neopentyl. Rapid SN2 substitution. E2 elimination may also occur. e.g. ClCH2CH2Cl + KOH ——> CH2=CHCl SN2 substitution. (N ≈ S >>O) Secondary R2CH– SN2 substitution and / or E2 elimination (depending on the basicity of the nucleophile). Bases weaker than acetate (pKa = 4.8) give less elimination. The rate of substitution may be reduced by branching at the β-carbons, and this will increase elimination. E2 elimination will dominate. SN2 substitution. (N ≈ S >>O) In high dielectric ionizing solvents, such as water, dimethyl sulfoxide & acetonitrile, SN1 and E1 products may be formed slowly. Tertiary R3C– E2 elimination will dominate with most nucleophiles (even if they are weak bases). No SN2 substitution due to steric hindrance. In high dielectric ionizing solvents, such as water, dimethyl sulfoxide & acetonitrile, SN1 and E1 products may be expected. E2 elimination will dominate. No SN2 substitution will occur. In high dielectric ionizing solvents SN1 and E1 products may be formed. E2 elimination with nitrogen nucleophiles (they are bases). No SN2 substitution. In high dielectric ionizing solvents SN1 and E1 products may be formed. Allyl H2C=CHCH2– Rapid SN2 substitution for 1º and 2º-halides. For 3º-halides a very slow SN2 substitution or, if the nucleophile is moderately basic, E2 elimination. In high dielectric ionizing solvents, such as water, dimethyl sulfoxide & acetonitrile, SN1 and E1 products may be observed. Rapid SN2 substitution for 1º halides. E2 elimination will compete with substitution in 2º-halides, and dominate in the case of 3º-halides. In high dielectric ionizing solvents SN1 and E1 products may be formed. Nitrogen and sulfur nucleophiles will give SN2 substitution in the case of 1º and 2º-halides. 3º-halides will probably give E2 elimination with nitrogen nucleophiles (they are bases). In high dielectric ionizing solvents SN1 and E1 products may be formed. Water hydrolysis will be favorable for 2º & 3º-halides. Benzyl C6H5CH2– Rapid SN2 substitution for 1º and 2º-halides. For 3º-halides a very slow SN2 substitution or, if the nucleophile is moderately basic, E2 elimination. In high dielectric ionizing solvents, such as water, dimethyl sulfoxide & acetonitrile, SN1 and E1 products may be observed. Rapid SN2 substitution for 1º halides (note there are no β hydrogens). E2 elimination will compete with substitution in 2º-halides, and dominate in the case of 3º-halides. In high dielectric ionizing solvents SN1 and E1 products may be formed. Nitrogen and sulfur nucleophiles will give SN2 substitution in the case of 1º and 2º-halides. 3º-halides will probably give E2 elimination with nitrogen nucleophiles (they are bases). In high dielectric ionizing solvents SN1 and E1 products may be formed. Water hydrolysis will be favorable for 2º & 3º-halides. Contributors William Reusch, Professor Emeritus ( Michigan State U. ), Virtual Textbook of Organic Chemistry
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Courses/Fresno_City_College/Introductory_Chemistry_Atoms_First_for_FCC/10%3A_Chemical_Reactions/10.8%3A_Keeping_Track_of_Redox_Reactions
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There are many important redox reactions, and not all of them can be solved by inspection. For reactions which are more complicated than the simple single replacement reactions introduced in the last section, we need to have ways of keeping track of the movement of electrons. We do this through a process of electron bookkeeping. This process starts with assigning oxidation numbers to each element. Changes in these oxidation numbers are indications of the movement of electrons, and therefore what is oxidized and what is reduced. When balancing a redox reaction, it is important to keep track of the electrons and not just the atoms. Each of these topics will be covered in one of the subsections of this section.
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Courses/Riverland_Community_College/CHEM_1121%3A_General_Organic_and_Biochemistry/17%3A_Proteins_and_Enzymes/17.03%3A_Peptides
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Learning Objectives Explain how a peptide is formed from individual amino acids. Explain why the sequence of amino acids in a protein is important. Two or more amino acids can join together into chains called peptides. Previously , we discussed the reaction between ammonia and a carboxylic acid to form an amide. In a similar reaction, the amino group on one amino acid molecule reacts with the carboxyl group on another, releasing a molecule of water and forming an amide linkage: An amide bond joining two amino acid units is called a peptide bond. Note that the product molecule still has a reactive amino group on the left and a reactive carboxyl group on the right. These can react with additional amino acids to lengthen the peptide. The process can continue until thousands of units have joined, resulting in large proteins. A chain consisting of only two amino acid units is called a dipeptide ; a chain consisting of three is a tripeptide . By convention, peptide and protein structures are depicted with the amino acid whose amino group is free (the N-terminal end) on the left and the amino acid with a free carboxyl group (the C-terminal end) to the right. The general term peptide refers to an amino acid chain of unspecified length. However, chains of about 50 amino acids or more are usually called proteins or polypeptides. In its physiologically active form, a protein may be composed of one or more polypeptide chains. For peptides and proteins to be physiologically active, it is not enough that they incorporate certain amounts of specific amino acids. The order, or sequence , in which the amino acids are connected is also of critical importance. Bradykinin is a nine-amino acid peptide (Figure \(\PageIndex{1}\)) produced in the blood that has the following amino acid sequence: arg-pro-pro-gly-phe-ser-pro-phe-arg This peptide lowers blood pressure, stimulates smooth muscle tissue, increases capillary permeability, and causes pain. When the order of amino acids in bradykinin is reversed, arg-phe-pro-ser-phe-gly-pro-pro-arg the peptide resulting from this synthesis shows none of the activity of bradykinin. Just as millions of different words are spelled with our 26-letter English alphabet, millions of different proteins are made with the 20 common amino acids. However, just as the English alphabet can be used to write gibberish, amino acids can be put together in the wrong sequence to produce nonfunctional proteins. Although the correct sequence is ordinarily of utmost importance, it is not always absolutely required. Just as you can sometimes make sense of incorrectly spelled English words, a protein with a small percentage of “incorrect” amino acids may continue to function. However, it rarely functions as well as a protein having the correct sequence. There are also instances in which seemingly minor errors of sequence have disastrous effects. For example, in some people, every molecule of hemoglobin (a protein in the blood that transports oxygen) has a single incorrect amino acid unit out of about 300 (a single valine replaces a glutamic acid). That “minor” error is responsible for sickle cell anemia, an inherited condition that usually is fatal. Summary The amino group of one amino acid can react with the carboxyl group on another amino acid to form a peptide bond that links the two amino acids together. Additional amino acids can be added on through the formation of addition peptide (amide) bonds. A sequence of amino acids in a peptide or protein is written with the N-terminal amino acid first and the C-terminal amino acid at the end (writing left to right).
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Courses/Lumen_Learning/Book%3A_US_History_II_(OS_Collection)_(Lumen)/09%3A_Age_of_Empire%3A_American_Foreign_Policy%2C_1890-1914/09.10%3A_Assignment%3A_White_Man%E2%80%99s_Burden
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After the United States acquired the Philippines in the wake of the Spanish-American War, the famous British poet Rudyard Kipling (the same man who wrote the Jungle Book ), wrote The White Man’s Burden to the American people. After reading The White Man’s Burden write a paragraph or two that answers the following: What exactly is the burden? Why is it a burden (why does Kipling call it a burden instead of “duty,” “privilege,” or “right”)? What is the tone of the poem? Is it cheerful (“Hey America, welcome to the club, now let’s go exploit some non-whites!”)? Is it a somber warning (“America, do you have any idea what you’ve just got yourself into”)? Is it disappointment (“Oh, great, another country trying to copy Britain and get in on the empire game”)? Or is it something else? Who seems to suffer more, the empire or the colony? Remember to use specific passages and quotes from the poem in support of your answers. It’s a short poem so be thorough in your reading, thinking, and writing. CC licensed content, Original White Man's Burden Assignment. Authored by : Chris Thomas. Provided by : Reynolds Community College. License : CC BY: Attribution
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Courses/Mendocino_College/Introduction_to_Chemistry_(CHM_200)/09%3A_Solutions_and_Aqueous_Mixtures/9.04%3A_Aqueous_Solutions_-_Dissolving_solids_in_water
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Learning Outcomes Define a solution and describe the parts of a solution. Describe how an aqueous solution is formed from both ionic compounds and molecular compounds. Recognize that some compounds are insoluble in water. Describe the differences among strong electrolytes, weak electrolytes, and nonelectrolytes. Forming a Solution When one substance dissolves into another, a solution is formed. A solution is a homogenous mixture consisting of a solute dissolved into a solvent. The solute is the substance that is being dissolved, while the solvent is the dissolving medium . Solutions can be formed with many different types and forms of solutes and solvents. In this chapter, we will focus on solution where the solvent is water. An aqueous solution is water that contains one or more dissolved substance. The dissolved substances in an aqueous solution may be solids, gases, or other liquids. In order to be a true solution, a mixture must be stable. When sugar is fully dissolved into water, it can stand for an indefinite amount of time, and the sugar will not settle out of the solution. Further, if the sugar-water solution is passed through a filter, it will remain with the water. This is because the dissolved particles in a solution are very small, usually less than \(1 \: \text{nm}\) in diameter. Solute particles can be atoms, ions, or molecules, depending on the type of substance that has been dissolved. The Dissolving Process Water typically dissolves most ionic compounds and polar molecules. Nonpolar molecules, such as those found in grease or oil, do not dissolve in water. We will first examine the process that occurs when an ionic compound, such as table salt (sodium chloride), dissolves in water. Water molecules move about continuously due to their kinetic energy. When a crystal of sodium chloride is placed into water, the water's molecules collide with the crystal lattice. Recall that the crystal lattice is composed of alternating positive and negative ions. Water is attracted to the sodium chloride crystal because water is polar; it has both a positive and a negative end. The positively charged sodium ions in the crystal attract the oxygen end of the water molecules because they are partially negative. The negatively charged chloride ions in the crystal attract the hydrogen end of the water molecules because they are partially positive. The action of the polar water molecules takes the crystal lattice apart (see figure below). After coming apart from the crystal, the individual ions are then surrounded by solvent particles in a process called solvation . Note in the figure above that the individual \(\ce{Na^+}\) ions are surrounded by water molecules with the oxygen atom oriented near the positive ion. Likewise, the chloride ions are surrounded by water molecules with the opposite orientation. Hydration is the process of solute particles being surrounded by water molecules arranged in a specific manner . Hydration helps to stabilize aqueous solutions by preventing the positive and negative ions from coming back together and forming a precipitate. Table sugar is made of the molecular compound sucrose \(\left( \ce{C_{12}H_{22}O_{11}} \right)\). Solid sugar consists of individual sugar molecules held together by intermolecular attractive forces. When water dissolves sugar, it separates the individual sugar molecules by disrupting the attractive forces, but it does not break the covalent bonds between the carbon, hydrogen, and oxygen atoms. Dissolved sugar molecules are also hydrated. The hydration shell around a molecule of sucrose is arranged so that its partially negative oxygen atoms are near the partially positive hydrogen atoms in the solvent, and vice versa. Insoluble Compounds Not all compounds dissolve well in water. Some ionic compounds, such as calcium carbonate \(\left( \ce{CaCO_3} \right)\) and silver chloride \(\left( \ce{AgCl} \right)\), are nearly insoluble. This is because the attractions between the ions in the crystal lattice are stronger than the attraction that the water molecules have for the ions. As a result, the crystal remains intact. The solubility of ionic compounds can be predicted using the solubility rules as shown in Table \(\PageIndex{1}\). Table \(\PageIndex{1}\): Solubility rules for ionic compounds in water. Electrolytes and Nonelectrolytes An electrolyte is a compound that conducts an electric current when it is dissolved in water or melted. In order to conduct a current, a substance must contain mobile ions that can move from one electrode to the other. All ionic compounds are electrolytes. When ionic compounds dissolve, they break apart into ions, which are then able to conduct a current. Even insoluble ionic compounds, such as \(\ce{CaCO_3}\), are considered electrolytes because they can conduct a current in the molten (melted) state. A nonelectrolyte is a compound that does not conduct an electric current in either aqueous solution or in the molten state. Many molecular compounds, such a sugar or ethanol, are nonelectrolytes. When these compounds dissolve in water, they do not produce ions. Illustrated below is the difference between an electrolyte and a nonelectrolyte. Dissociation Earlier, you saw how an ionic crystal lattice breaks apart when it is dissolved in water. Dissociation is the separation of ions that occurs when a solid ionic compound dissolves . Simply undo the crisscross method that you learned when writing chemical formulas for ionic compounds, and you are left with the components of an ionic dissociation equation. The subscripts for the ions in the chemical formulas become the coefficients of the respective ions on the product side of the equations. Shown below are dissociation equations for \(\ce{NaCl}\), \(\ce{Ca(NO_3)_2}\), and \(\ce{(NH_4)_3PO_4}\). \[\begin{align} &\ce{NaCl} \left( s \right) \rightarrow \ce{Na^+} \left( aq \right) + \ce{Cl^-} \left( aq \right) \\ &\ce{Ca(NO_3)_2} \left( s \right) \rightarrow \ce{Ca^{2+}} \left( aq \right) + 2 \ce{NO_3^-} \left( aq \right) \\ &\ce{(NH_4)_3PO_4} \left( s \right) \rightarrow 3 \ce{NH_4^+} \left( aq \right) + \ce{PO_4^{3-}} \left( aq \right) \end{align}\] One formula unit of sodium chloride dissociates into one sodium ion and one chloride ion. The calcium nitrate formula unit dissociates into one calcium ion and two nitrate ions, because the \(2+\) charge of each calcium ion requires two nitrate ions (each with a charge of \(1-\)) to form an electrically neutral compound. The ammonium phosphate formula unit dissociates into three ammonium ions and one phosphate ion. Do not confuse the subscripts of the atoms within the polyatomic ion for the subscripts that result from the crisscrossing of the charges that make the original compound neutral. The 3 subscript of the ntirate ion and the 4 subscript of the ammonium ion are part of the polyatomic ion and remain a part of the ionic formula after the compound dissociates. Notice that the compounds are solids \(\left( s \right)\) that become ions when dissolved in water, producing an aqueous solution \(\left( aq \right)\). Nonelectrolytes do not dissociate when forming an aqueous solution. An equation can still be written that simply shows the solid going into solution. For example, the process of dissolving sucrose in water can be written as follows: \[\ce{C_{12}H_{22}O_{11}} \left( s \right) \rightarrow \ce{C_{12}H_{22}O_{11}} \left( aq \right)\] Strong and Weak Electrolytes Some polar molecular compounds are nonelectrolytes when the are in their pure state but become electrolytes when they are dissolved in water. Hydrogen chloride \(\left( \ce{HCl} \right)\) is a gas in its pure molecular state and is a nonelectrolyte. However, when \(\ce{HCl}\) is dissolved in water, it conducts a current well because the \(\ce{HCl}\) molecule ionizes into hydrogen and chloride ions. \[\ce{HCl} \left( g \right) \rightarrow \ce{H^+} \left( aq \right) + \ce{Cl^-} \left( aq \right)\] When \(\ce{HCl}\) is dissolved into water, it is called hydrochloric acid. Ionic compounds and some polar compounds are completely broken apart into ions and thus conduct a current very well. A strong electrolyte is a solution in which almost all of the dissolved solute exists as ions . Some other polar molecular compounds become electrolytes upon being dissolved into water but do not ionize to a very great extent. For example, nitrous acid \(\left( \ce{HNO_2} \right)\) only partially ionizes into hydrogen ions and nitrite ions when dissolved in water. Aqueous nitrous acid is composed of only about \(5\%\) ions and \(95\%\) intact nitrous acid molecules A weak electrolyte is a solution in which only a small fraction of the dissolved solute exists as ions . The equation showing the ionization of a weak electrolyte utilizes an equilibrium arrow, indicating an equilibrium between the reactants and products. \[\ce{HNO_2} \left( aq \right) \rightleftharpoons \ce{H^+} \left( aq \right) + \ce{NO_2^-} \left( aq \right)\]
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Bookshelves/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/V%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_3/08%3A_Photochemical_Reactions/8.04%3A_Photolysis
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We have seen that the absorption of photons (especially in the ultraviolet-visible spectrum) is connected to the excitation of electrons. After excitation, a number of different relaxation pathways lead back to the ground state. Sometimes, absorption of a photon leads to a vastly different outcome. Instead of just relaxing again, the molecules may undergo bond-breaking reactions, instead. An example of this phenomenon is observed in the complex ion [Co(NH 3 )] 6 3+ . Addition of UV light to this complex results in loss of ammonia. In the absence of UV light, however, the complex ion is quite stable. In many cases, loss of a ligand is followed by replacement by a new one. For example, if an aqueous solution of [Co(NH 3 )] 6 3+ is photolysed, an ammonia ligand is easily replaced by water. Exercise \(\PageIndex{1}\) Draw a d orbital splitting diagram for [Co(NH 3 )] 6 3+ . Explain why this complex is normally inert toward substitution. Exercise \(\PageIndex{2}\) Use the d orbital splitting diagram for [Co(NH 3 )] 6 3+ to explain why this complex undergoes substitution upon irradiation with UV light Photolysis is the term used to describe the use of light to initiate bon-breaking events. Photolysis frequently involves the use of high-intensity ultraviolet lamps. The high intensity light is needed in order to provide enough photons to get higher conversion of reactant into a desired product. Two very different things could happen as a result of photon absorption. In one case, the molecule absorbs the photon, then somehow relaxes again, remaining unchanged overall. In the other case, the absorption of the photon results in bond cleavage and the formation of a new product. As a result, for every photon absorbed, there is a certain chance that the molecule will actually undergo a reaction, and a certain chance that the molecule will just relax again. "Quantum yield" is an expression used to define the efficiency of a photolytic reaction. The quantum yield is just the number of molecules of reactant formed per photon absorbed. On a macroscopic level, we might say it is the number of moles of reactant formed per mole of photons absorbed. Quantum yield = number of molecules of product formed / number of photons absorbed The higher the quantum yield, the more efficient the reaction, because it requires less light in order to successfully form the product. Exercise \(\PageIndex{3}\) Calculate the quantum yield in the following cases. 6 mmoles of product results from absorption of 24 mmoles of photons. 54 mmoles of photons are required to produce 3 mmoles of product. 1.2 x 10 -6 moles of product are formed after absorption of 4.2 x 10 -5 moles of photons. In practice, determination of quantum yield is complicated because of the need to calculate exactly how many photons have been absorbed, in addition to how much product has been formed.
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Bookshelves/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_3%3A_The_States_of_Matter/11%3A_Solutions
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Template:HideTOC A solution is a homogeneous mixture composed of two or more substances. In such a mixture, a solute is a substance dissolved in another substance, known as a solvent. 11.1: Composition of Solutions Solutions are homogeneous mixtures of two or more substances whose components are uniformly distributed on a microscopic scale. The component present in the greatest amount is the solvent, and the components present in lesser amounts are the solute(s). The formation of a solution from a solute and a solvent is a physical process, not a chemical one. Substances that are miscible, such as gases, form a single phase in all proportions when mixed. Substances that form separate phases are immiscible. 11.2: Nature of Dissolved Species The solubility of a substance is the maximum amount of a solute that can dissolve in a given quantity of solvent; it depends on the chemical nature of both the solute and the solvent and on the temperature and pressure. When a solution contains the maximum amount of solute that can dissolve under a given set of conditions, it is a saturated solution. Otherwise, it is unsaturated. Supersaturated solutions, which contain more dissolved solute than allowed under particular conditions, are unstable. 11.3: Reaction Stoichiometry in Solutions: Acid-Base Titrations A titration curve is a graph that relates the change in pH of an acidic or basic solution to the volume of added titrant. The characteristics of the titration curve are dependent on the specific solutions being titrated. The pH of the solution at the equivalence point may be greater than, equal to, or less than 7.00. The choice of an indicator for a given titration depends on the expected pH at the equivalence point of the titration, and the range of the color change of the indicator. 11.4: Reaction Stoichiometry in Solutions: Oxidation-Reduction Titrations Redox titration are here the titrant is an oxidizing or reducing agent. In contrast to acid/base titrations, it is convenient for redox titrations to monitor the titration reaction’s potential instead of the concentration of one species. 11.5: Phase Equilibrium in Solutions - Nonvolatile Solutes Colligative properties of a solution depend on only the total number of dissolved particles in solution, not on their chemical identity. Colligative properties include vapor pressure, boiling point, freezing point, and osmotic pressure. The addition of a nonvolatile solute (one without a measurable vapor pressure) decreases the vapor pressure of the solvent. The vapor pressure of the solution is proportional to the mole fraction of solvent in the solution, a relationship known as Raoult’s law. 11.6: Phase Equilibrium in Solutions - Volatile Solutes In contrast, the solubility of gases increases as the partial pressure of the gas above a solution increases. Hence, the effect of increased pressure on the dynamic equilibrium that is established between the dissolved gas molecules in solution and the molecules in the gas phase above the solution. Because the concentration of molecules in the gas phase increases with increasing pressure, the concentration of dissolved gas molecules in 11.7: Colloidal Suspensions A colloid can be classified as a sol, a dispersion of solid particles in a liquid or solid; a gel, a semisolid sol in which all of the liquid phase has been absorbed by the solid particles; an aerosol, a dispersion of solid or liquid particles in a gas; or an emulsion, a dispersion of one liquid phase in another. A colloid can be distinguished from a true solution by its ability to scatter a beam of light, known as the Tyndall effect. 11.E: Solutions (Exercises) These are homework exercises to accompany the Textmap created for "Principles of Modern Chemistry" by Oxtoby et al.
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Courses/Lumen_Learning/Book%3A_World_History_to_1700_(Lumen)/04%3A_Global_Economy/04.2%3A_Chapter_5%3A_The_Greek_World_from_the_Bronze_Age_to_the_Roman_Conquest
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5.1 CHRONOLOGY c. 3300 – 1150 BCE Bronze Age c. 1100 – 700 BCE Dark Ages c. 700 – 480 BCE Archaic Period 480 – 323 BCE Classical Period 431 – 404 BCE The Peloponnesian War 323 – 146 BCE Hellenistic Period 5.2 INTRODUCTION Sometime in the eighth century BCE, an aristocratic resident of the Greek trading colony of Pithekoussai—located on the tiny island of Ischia just off the coast of Naples in Italy—held a symposium at his home. Most of what happened at the party stayed at the party, but what we do know is that it must have been a good one. One of the guests, presumably operating under the influence of his host’s excellent wine, took the liberty of scratching the following ditty onto one of his host’s fine exported ceramic wine cups: “I am the Cup of Nestor, good to drink from. Whoever drinks from this cup, straightaway the desire of beautifully-crowned Aphrodite will seize him.” While party pranks do not commonly make history, this one has: this so-called Cup of Nestor is one of the earliest examples of writing in the Greek alphabet , as well as the earliest known written reference to the Homeric epics. Overall, this cup and the inscription on it exemplify the mobility of the Ancient Greeks and their borrowing of skills and culture from others around the Mediterranean while, at the same time, cultivating a set of values specific to themselves. After all, just like the very residents of Pithekoussai, the cup had originally made the journey all the way from the island of Euboea, off the coast of Athens, to Pithekoussai, on the island of Ischia. Furthermore, the new script, in which the daring guest wrote on the cup, had just recently been borrowed and adapted by the Greeks from the Phoenicians , a seafaring nation based in modern-day Lebanon. Indeed, our clever poet wrote from right to left, just like the Phoenicians. Finally, the poem mentions Nestor, one of the heroes of Homer’s Iliad , an epic about the Trojan War , and a source of common values that all Greeks held dear: military valor, competitive excellence on both the battlefield and in all areas of everyday life, and a sense of brotherhood that manifested itself most obviously in the shared language of all the Greeks. That feeling of kinship facilitated collaboration of all the Greeks in times of crisis from the mythical Trojan War to the Persian Wars , and finally, during the Greeks’ resistance against the Roman conquest. 5.3 QUESTIONS TO GUIDE YOUR READING In what ways did Greek geography and topography impact the history of the ancient Greek world? What are the different periods of Greek history, and what are the chief defining characteristics of each period? What primary sources are available for the study of Greek history, and what are the limitations of these sources? What were the most important developments in the Greek world in the Archaic Period? What was the significance of the Persian Wars for the subsequent history of the Greek World? What were the stages of the Peloponnesian War? How did the outcome of the war impact Greece in the fourth century? What were some of the most important contributions of Classical Athens in the areas of art, government and law, philosophy, and literature? How and why did the Macedonians conquer the Greek world? Why did the empire conquered by Philip and Alexander disintegrate after Alexander’s death? What were some of the strengths and weaknesses of the Hellenistic kingdoms as political entities? Why did they prove to be inherently unstable? What were some of the achievements and legacies of the Hellenistic period? 5.4 KEY TERMS Achaemenid Empire Alexander the Great Alexandria Antigonid dynasty Archaeological evidence Archaic Period Archidamian War Archimedes of Syracuse Aristophanes Aristotle Asia Minor Athenian democracy Athens Battle of Chaeronea Battle of Leuctra Battle of Marathon Battle of Thermopylae Bronze Age Chigi vase Classical Period Cleisthenes Cleopatra VII Crete “Cup of Nestor” Cynic philosophers Cyrus the Great Darius Dark Ages Decelean War Delian League / Athenian Empire Delphi Ekklesia Epicureanism Epidaurus Euripides First Peloponnesian War Gerousia Great Library of Alexandria Greek alphabet Greek colonization Hannukah Hellenistic Period Helots Herodotus Herophilus of Chalcedon Homer Homer’s Iliad Hoplite phalanx Kingdom of Pergamon Kleos Isthmus of Corinth Linear A Linear B Macedonian Wars with Rome / Third Macedonian War Magna Graecia “Mask of Agamemnon” Mauryan Empire Megara Hyblaea Messenia Minoans Mycenaeans Oligarchy Olympic Games Orientalizing style Othismos Pan-Hellenic Peace of Nicias Peisistratus Peloponnese Peloponnesian War Pericles Persian Wars Pharos of Alexandria Philip II of Macedon Phoenicians Plato Polis / poleis Ptolemaic Egypt Ptolemy I Soter Pythia Sarissa Seleucid Empire Septuagint Sicilian Expedition Skepticism Socrates Solon Sophocles Sparta Spartan constitution Stoic philosophy Theban Hegemony Theban Sacred Band Thetes Thucydides Tyranny Tyranny of the Thirty Trojan War Tyrtaeus Wars of the Diadochi Xerxes 5.5 GEOGRAPHY AND TOPOGRAPHY This chapter’s title refers to the Greek World, rather than Greece. While Greece is a unified country today, the territory of the present-day country was not unified under one rule until the rise of the Macedonians in the fourth century BCE. Instead, the basic unit of organization in the period covered in this chapter was the polis , an independent city-state, which consisted of a walled city that controlled and protected the farmland around it. Historians estimate that close to 1,500 of these city-states dotted the ancient Greek landscape. Each of these poleis (plural form of polis ) possessed its own form of government, law-code, army, cults of patron gods, and overall culture that set it apart from the other city-states. While the two most famous poleis , Athens and Sparta , controlled vast territories of farmland, most city-states were quite small, with a population of just a few thousand citizens. Furthermore, the Greek world in antiquity encompassed much more than present-day Greece, extending as far as Italy in the West and the territories of modern-day Turkey and Ukraine in the East. The geography and topography of the Greek mainland and the Mediterranean region surrounding it influenced the history of the Greek people in a number of crucial ways. First, the mountainous nature of mainland Greece, especially in the north, allowed different regions to remain somewhat isolated. The most isolated of all, Thessaly and Macedon, were viewed as uncivilized barbarians by the rest of the Greeks in the Archaic and Classical periods (one oft-mentioned example of their “barbarism” in Greek literature is that they drank their wine undiluted!)and largely kept to themselves until their rise to military prominence in the mid-fourth century BCE. The mountains throughout the northern portion of mainland Greece, in addition to isolating regions from each other and promoting regional culture, also provided tactical defenses in the face of external attacks. Most famously, the Persians learned the hard way about the challenges of navigating the Greek landscape during the second Persian invasion of Greece in 480 BCE. Indeed, the story of the 300 Spartans who fought to the death at the Battle of Thermopylae addresses the challenge of the Persian army trying to cross the mountains to the north of Attica in order to invade Athens by land. The Isthmus , a narrow strip of land controlled by Corinth, played a similar role in separating mainland Greece from the large peninsula of the Peloponnese . An inland city in southern Peloponnese, Sparta conquered Messenia, its surrounding region, early in its history and extended political control over much of the peninsula by early fifth century BCE. Unless the interests of Sparta herself were directly involved, Sparta practiced a policy of isolation and limited military intervention in other city-states affairs and wars, a practice enabled due to Sparta’s far southern location in Peloponnese. No less influential for the history of the Greek city-states than the topographical features were the resources that the land in different regions provided for agriculture and manufacturing. Mainland Greece was notoriously unsuitable for agriculture. Growing the grain staples wheat and barley in the rocky and clay-filled soil of Athens was especially difficult, while the mountainous regions across the entire mainland were optimal for herding, rather than agriculture. One notable exception, however, were olive trees, which grew abundantly. Olive oil, as a result, was ubiquitously used for eating, bathing, and lamps, and even as currency or prize for victors in athletic games. In addition, early on in their history, the inhabitants of Attica and Corinth found a way to profit from the clay in their soil by developing advanced ceramic pot-making and decorating techniques. Remains of Athenian and Corinthian wares have been found at archaeological sites all over the Mediterranean, attesting to their popularity abroad. Finally, precious metals were in short supply in the mainland, but the few that were available had an impact on the history of their regions. Most famously, the discovery of the silver mines at Laurion in Attica contributed to the increased prosperity of Athens in the mid-fifth century BCE. But the topography and geography of mainland Greece and the Peloponnese only tells us a part of the story. The Aegean is filled with islands, some of which remained autonomous, but most came under the control of the Athenian maritime empire in the fifth century BCE. In addition, the Greek colonization movement of the seventh and sixth centuries BCE resulted in the foundation of numerous Greek city-states in Asia Minor (modern-day Turkey), Magna Graecia (southern Italy), Sicily, and the Black Sea littoral. The history of the Greek world from its earliest settlements to the Roman conquest, therefore, is inextricably tied together with the history of the Mediterranean as a whole. And since the Greek areas of influence overlapped with those controlled by the Phoenicians, Persians, and eventually the Romans, interactions, often warlike, were unavoidable as well. As the modern historians Peregrine Horden and Nicholas Purcell noted, the Mediterranean was “the Corrupting Sea” [1] whose inhabitants were like frogs around the pond, watching each other, and borrowing each other’s cultural and technological achievements. As this chapter and the next will show, the Greeks and the Romans were the farthest-leaping frogs of all. 5.6 PERIODS OF GREEK HISTORY Historians today separate Greek history into particular periods, which shared specific features throughout the Greek world: The Bronze Age (c. 3,300 – 1,150 BCE) – a period characterized by the use of bronze tools and weapons. In addition, two particular periods during the Bronze Age are crucial in the development of early Greece: the Minoan Age on the island of Crete (c. 2,000 – 1,450 BCE) and the Mycenean period on mainland Greece (c. 1,600 – 1,100 BCE), both of them characterized by massive palaces, remnants of which still proudly stand today. The Minoan and Mycenaean civilizations had writing (dubbed Linear A and Linear B , respectively), which they used for keeping lists and palace inventories. The Dark Ages (c. 1,100 – 700 BCE) – a period that is “dark” from the archaeological perspective, which means that the monumental palaces of the Mycenean period disappear, and the archaeological record reveals a general poverty and loss of culture throughout the Greek world. For instance, the Linear A and Linear B writing systems disappear. The Greeks do not rediscover writing until the invention of the Greek alphabet at the end of the Dark Ages or the early Archaic Period. Archaic Period (c. 700 – 480 BCE) – the earliest period for which written evidence survives; this is the age of the rise of the Greek city-states, colonization, and the Persian Wars. Classical Period (c. 480 – 323 BCE) – the period from the end of the Persian Wars to the death of Alexander the Great . One of the most important events during this period is the Peloponnesian War (431 – 404 BCE), which pitted Athens against Sparta, and forced all other Greek city-states to choose to join one side or the other. This period ends with the death of Alexander the Great, who had unified the Greek world into a large kingdom with himself at its helm. Hellenistic Period (323 – 146 BCE) – the period from the death of Alexander to the Roman conquest of Greece; this is the age of the Hellenistic monarchies ruling over territories previously conquered by Alexander and his generals. Some historians end this period in 30 BCE, with the death of Cleopatra VII , the last surviving ruler of Egypt who was a descendant of one of Alexander’s generals. The rest of this chapter will be devoted to examining each of these periods in greater detail, covering chief political, military, and cultural developments. 5.7 METHODOLOGY: SOURCES AND PROBLEM Before launching into the story of the early Greek world, it is important to consider the methodology that Greek historians utilize. In other words, how do we know what we know about the Greek world? Modern scholars of ancient history are notoriously obsessed with evaluating their primary sources critically, and with good reason. Studying Greek history, especially in its earliest periods, is like putting together a puzzle, most of whose pieces are missing, and some pieces from another puzzle have also been added in for good measure. Greek history requires careful consideration of a wide range of sources, which fall into two broad categories: literary sources (including both fiction and non-fiction), and material culture. The job of the historian, then, is to reconstruct the story of the Greek people using these very different sources. While historians of the modern world rely on such archival sources as newspapers, magazines, and personal diaries and correspondence of individuals and groups, historians of the ancient world must use every available source to reconstruct the world in which their subject dwelled. Literary sources, such as epics, lyric poetry, and drama, may seem strange for historians to use, as they do not necessarily describe specific historical events. Yet, as in the case of other early civilizations, such sources are a crucial window into the culture and values of the people who produced them. For instance, the Epic of Gilgamesh , discussed in Chapter 2, is a key text for the study of early Mesopotamia. The earliest literary sources for Greek history are the Homeric epics, Iliad and Odyssey . They are, however, one of the most challenging sources to interpret, with one modern historian dubbing them a “historian’s headache.” [2] Composed orally before the existence of the Greek alphabet, the epics were not written down until sometime in the sixth century BCE. The epics most likely do not reflect the society of any particular Greek city-state in any one period, but rather consist of an amalgam of features from the Bronze Age to the early Archaic Period. Their value for historians, as a result, rests more on their impact on subsequent Greek culture, rather than on their providing information about Bronze Age Greeks. More than any other literary source, the Homeric Epics influenced the mentality of the Greeks in thinking about war and what it means to be a hero. Most other literary sources from the Archaic and Classical periods are easier to interpret than the Homeric Epics, as we often can date these later sources more precisely and thus know the period whose values or problems they reflect. There is, however, one important limitation to keep in mind: the overwhelming majority of surviving literature is from Athens, with very few sources from other city-states. Some of this distribution of evidence has to do with the differing values of the city-states themselves. For example, while Greeks of the Classical period considered Sparta to be as great a city as Athens, Spartans valued military valor over all else, so they did not cultivate arts and letters the way Athenians did. As a result, the only literary sources from Sparta are the works of two poets, Alcman and Tyrtaeus. Tyrtaeus’ military elegies, like the Homeric epics, glorify heroic death in battle over life without honor and were likely sung by Spartan warriors as they marched into battle. Several genres of non-fiction survive as well, allowing historians to study specific events and problems in the history of the Greek world, and especially Athens. The works of three major historians survive from Classical Athens. Herodotus , dubbed the Father of History, wrote the Histories about the Persian Wars in mid-fifth century BCE. Thucydides , an Athenian general in the Peloponnesian War, wrote a history of the Peloponnesian War over the course of the war (431 – 404 BCE). Finally, Xenophon wrote a history of the end of the Peloponnesian War, starting with 411 BCE, where Thucydides’ work ended, and into the fourth century. In addition to the works of the historians, philosophical treatises – most notably, those of Plato and Aristotle – provide crucial insight into the political thought, moral values, and perceptions of the world in late fifth and fourth centuries BCE. The approximately 100 surviving courtroom speeches from the same period likewise provide us with a window into the Athenian legal system. Finally, the Hippocratic corpus, a series of medical treatises and physicians’ journals from the Classical period, help us to understand the Greeks’ views of the human body and diseases. But in addition to the geographical restrictions of these sources, which largely document Athens, it is also important to note two other key limitations of the available evidence. First, virtually all of the literary sources were written by men and provide very little evidence of the lives and perspectives of women in the Greek world, except as seen through the eyes of men. Second, most of the authors were wealthy and socially prominent individuals; thus, their perspective does not reflect that of less affluent citizens and slaves. Archaeological evidence thankfully allows historians to fill some of the gaps in the literary evidence, but also comes with problems of its own. One joke that refers to the optimism of archaeologists reflects some of these problems of interpretation: whenever an archaeologist finds three stones that are together, he labels the find as a Minoan palace. Whenever he finds two stones that are together, he thinks he has found a city wall. Whenever he finds one building stone, he thinks he has found a house. Still, archaeological sources provide us with key information about different aspects of everyday life in different city-states. For example, the excavations of the sixth-century BCE colony Megara Hyblaea in Sicily shows that Greek colonists were interested in city planning and in equality of citizens, as demonstrated by the equal size of the lots. Material finds, such as pottery remains, in different sites across the Mediterranean also allow historians to map trading routes – for instance, Figures 5.1 and 5.2 show vases that were made in Euboea and Corinth respectively, but were found in Greek colonies in Italy. In addition, images on pottery provide information about stories and myths that have entered popular culture and that sometimes reflect further aspects of everyday life. For instance, the prevalence of images of women gathering at public fountains on Athenian hydriae (water pots) from the late sixth century BCE shows the importance of the public fountains for the social life of women in Athens in the period. Finally, written archaeological sources, such as inscriptions on stone or pottery shards from all over the Greek world, and papyri from Hellenistic Egypt, are the equivalent of documentary archives from the ancient world. The evidence from epigraphy (inscriptions) includes laws that were written on large stones and set up in public, such as the monumental law-code from Gortyn, Crete, and lists of war-dead, as well as private tomb inscriptions. Papyri, on the other hand, include such private documents as prenuptial agreements (among the strangest are prenuptial documents for brother-sister marriages – legal in Egypt but nowhere else in the Greek world), divorce documents, loans, and village police reports (cattle theft appears to have been a serious problem in the Faiyum in the Hellenistic Period!). Taken together, the literary and archaeological sources allow the historian to complete much more of the puzzle than would have been possible with just one of these types of sources. Still, significant gaps in knowledge remain nevertheless, and are, perhaps, one of the joys of studying ancient history: the historian gets to play the part of a sleuth, attempting to reconstruct the history of events based on just a few available clues. 5.8 FROM MYTHOLOGY TO HISTORY The terms “mythology” and “history” may seem, by modern definitions, to be antithetical. After all, mythology refers to stories that are clearly false, of long-forgotten gods and heroes and their miraculous feats. History, on the other hand, refers to actual events that involved real people. And yet, the idea that the two are opposites would have seemed baffling to a typical resident of the ancient Mediterranean world. Rather, gods and myths were part of the everyday life, and historical events could become subsumed by myths just as easily as myths could become a part of history. For instance, Gilgamesh, the hero of the Sumerian Epic of Gilgamesh , was a real king of Uruk, yet he also became the hero of the epic. Each Greek city-state, in particular, had a foundation myth describing its origins as well as its own patron gods and goddesses. Etiological myths, furthermore, served to explain why certain institutions or practices existed; for instance, the tragic trilogy Oresteia of the Athenian poet Aeschylus tells the etiological myth for the establishment of the Athenian murder courts in the Classical period. Yet, while the Greeks saw mythology and history as related concepts and sometimes as two sides of the same coin, one specific mythical event marked, in the eyes of the earliest known Greek historians, the beginning of the story of Greek-speaking peoples. That event was the Trojan War. 5.8.1 Homer and the Trojan War It is telling that the two earliest Greek historians, Herodotus, writing in the mid-fifth century BCE, and Thucydides, writing in the last third of the fifth century BCE, began their respective histories with the Trojan War, each treating it as a historical event. The Homeric epics Iliad and Odyssey portray the war as an organized attack of a unified Greek army against Troy, a city in Asia Minor (see map 5.2) . The instigating offense? The Trojan prince Paris kidnapped Helen, the most beautiful woman in the world, from her husband Menelaus, king of Sparta. This offense, interpreted as a slight to Menelaus’ honor, prompted Agamemnon, king of Mycenae and Menelaus’ brother, to raise an army from the entire Greek world and sail to Troy. The mythical tradition had it that after a brutal ten- year siege, the Greeks resorted to a trick: they presented the Trojans with a hollow wooden horse, filled with armed soldiers. The Trojans tragically accepted the gift, ostensibly intended as a dedication to the goddess Athena. That same night, the armed contingent emerged from the horse, and the city finally fell to the Greeks. Picking up the story ten years after the end of the Trojan War, the Odyssey then told the story of Odysseus’s struggles to return home after the war and the changes that reverberated throughout the Greek world after the fall of Troy. The Homeric epics were the foundation of Greek education in the Archaic and Classical periods and, as such, are a historian’s best source of pan-Hellenic values. A major theme throughout both epics is personal honor, which Homeric heroes value more than the collective cause. For example, when Agamemnon slights Achilles’ honor in the beginning of the Iliad , Achilles, the best hero of the Greeks, withdraws from battle for much of the epic, even though his action causes the Greeks to start losing battles until he rejoins the fight. A related theme is competitive excellence, with kleos (eternal glory) as its goal: all Greek heroes want to be the best; thus, even while fighting in the same army, they see each other as competition. Ultimately, Achilles has to make a choice: he can live a long life and die unknown, or he can die in battle young and have everlasting glory. Achilles’ selection of the second option made him the inspiration for such historical Greek warriors and generals as Alexander the Great, who brought his scroll copy of the Iliad with him on all campaigns. Finally, the presence of the gods in the background of the Trojan War shows the Greeks’ belief that the gods were everywhere, and acted in the lives of mortals. These gods could be powerful benefactors and patrons of individuals who respected them and sought their favor, or vicious enemies, bent on destruction. Indeed, early in the Iliad , the god Apollo sends a plague on the Greek army at Troy, as punishment for disrespecting his priest. It is important to note that while the Homeric epics influenced Greek values from the Archaic period on, they do not reflect the reality of the Greek world in any one period. Furthermore, they were not composed by a single poet, Homer ; indeed, it is possible that Homer never existed. Because the epics were composed orally by multiple bards over the period of several hundred years, they combine details about technological and other aspects of the Bronze Age with those of the Dark Ages and even the early Archaic Age. For instance, the heroes use bronze weapons side-by-side with iron. Archaeological evidence, however, allows historians to reconstruct to some extent a picture of the Greek world in the Bronze Age and the Dark Ages. 5.8.2 Greece in the Bronze Age, and the Dark Ages While there were people living in mainland Greece already in the Neolithic Period, historians typically begin the study of the Greeks as a unique civilization in the Bronze Age, with the Minoans . The first literate civilization in Europe, the Minoans were a palace civilization that flourished on the island of Crete c. 2,000 – 1,450 BCE. As befits island-dwellers, they were traders and seafarers; indeed, the Greek historian Thucydides credits them with being the first Greeks to sail on ships. Sir Arthur Evans, the archaeologist who first excavated Crete in the early 1900s, dubbed them Minoans, after the mythical Cretan king Minos who was best known for building a labyrinth to house the Minotaur, a monster that was half-man, half-bull. Bulls appear everywhere in surviving Minoan art, suggesting that they indeed held a prominent place in Minoan mythology and religion. Four major palace sites survive on Crete. The most significant of them, Knossos, has been restored and reconstructed for the benefit of modern tourists. Historians hypothesize that the palaces were the homes of local rulers, who ruled and protected the surrounding farmland. The palaces seem to have kept records in two different writing systems, the earliest known in Europe: the Cretan hieroglyphic and Linear scripts. Unfortunately, neither of these systems has been deciphered, but it is likely that these were palace inventories and records pertaining to trade. The palaces had no surrounding walls, suggesting that the Cretans maintained peace with each other and felt safe from outside attacks, since they lived on an island. This sense of security proved to be a mistake as, around 1,450 BCE, the palaces were violently destroyed by invaders, possibly the Mycenaeans who arrived from mainland Greece. Recent discoveries also suggest that at least some of the destruction may have been the result of tsunamis which accompanied the Santorini/ Thera volcanic eruption in the 1600s BCE. The Mycenaeans, similarly to the Minoans, were a palace civilization. Flourishing on mainland Greece c. 1,600 – 1,100 BCE, they received their name from Mycenae, the most elaborate surviving palace and the mythical home of Agamemnon, the commander-in-chief of the Greek army in the Trojan War. The archaeological excavations of graves in Mycenae reveal a prosperous civilization that produced elaborate pottery, bronze weapons and tools, and extravagant jewelry and other objects made of precious metals and gems. One of the most famous finds is the so-called “Mask of Agamemnon ,” a burial mask with which one aristocrat was buried, made of hammered gold. The Mycenaeans also kept palace records in a syllabic script, known as Linear B. Related to the Cretan Linear A script, Linear B, however, has been deciphered, and identified as Greek. Archaeological evidence also shows that sometime in the 1,200s BCE, the Mycenaean palaces suffered a series of attacks and were gradually abandoned over the next century. The period that begins around 1,000 BCE is known as the “Dark Ages” because of the notable decline, in contrast with the preceding period. The Mycenaean Linear B script disappears, and archaeological evidence shows a poorer Greece with a decline in material wealth and life expectancy. Some contact, however, must have remained with the rest of the Mediterranean, as shown by the emergence of the Greek alphabet, adapted from the Phoenician writing system towards the end of the Dark Ages or early in the Archaic Period. 5.9 ARCHAIC GREECE The story of the Greek world in the Dark Ages could mostly be described as a story of fragmentation. With a few exceptions, individual sites had limited contact with each other. The Archaic period, however, appears to have been a time of growing contacts and connections between different parts of mainland Greece. Furthermore, it was a time of expansion, as the establishment of overseas colonies and cities brought the Greeks to Italy and Sicily in the West, and Asia Minor and the Black Sea littoral in the East. Furthermore, while Greeks in the Archaic period saw themselves as citizens of individual city-states, this period also witnessed the rise of a Pan-Hellenic identity, as all Greeks saw themselves connected by virtue of their common language, religion, and Homeric values. This Pan-Hellenic identity was ultimately cemented during the Persian Wars: two invasions of Greece by the Persian Empire at the end of the Archaic period. While warfare in the Iliad consisted largely of duels between individual heroes, the hoplite phalanx was a new mode of fighting that did not rely on the skill of individuals. Rather, it required all soldiers in the line to work together as a whole. Armed in the same way – with a helmet, spear, and the round shield, the hoplon , which gave the hoplites their name – the soldiers were arranged in rows, possibly as much as seven deep. Each soldier carried his shield on his left arm, protecting the left side of his own body and the right side 5.9.1 Rise of the Hoplite Phalanx and the Polis A Corinthian vase, known today as the Chigi Vase , made in the mid-seventh century BCE, presents a tantalizing glimpse of the changing times from the Dark Ages to the Archaic Period. Taking up much of the decorated space on the vase is a battle scene. Two armies of warriors with round shields, helmets, and spears are facing each other and appear to be marching in formation towards each other in preparation for attack. Modern scholars largely consider the vase to be the earliest artistic portrayal of the hoplite phalanx , a new way of fighting that spread around the Greek world in the early Archaic Age and that coincided with the rise of another key institution for subsequent Greek history: the polis , or city-state. From the early Archaic period to the conquest of the Greek world by Philip and Alexander in the late fourth century BCE, the polis was the central unit of organization in the Greek world. While warfare in the Iliad consisted largely of duels between individual heroes, the hoplite phalanx was a new mode of ghting that did not rely on the skill of individuals. Rather, it required all soldiers in the line to work together as a whole. Armed in the same way – with a helmet, spear, and the round shield, the hoplon, which gave the hoplites their name – the soldiers were arranged in rows, possibly as much as seven deep. Each soldier carried his shield on his left arm, protecting the left side of his own body and the right side of his comrade to the left. Working together as one, then, the phalanx would execute the othismos (a mass shove) during battle, with the goal of shoving the enemy phalanx off the battlefield. Historians do not know which came into existence first, the phalanx or the polis , but the two clearly reflect a similar ideology. In fact, the phalanx could be seen as a microcosm of the polis , exemplifying the chief values of the polis on a small scale. Each polis was a fully self-sufficient unit of organization, with its own laws, definition of citizenship, government, army, economy, and local cults. Regardless of the differences between the many poleis in matters of citizenship, government, and law, one key similarity is clear: the survival of the polis depended on the dedication of all its citizens to the collective well-being of the city-state. This dedication included service in the phalanx. As a result, citizenship in most Greek city-states was closely connected to military service, and women were excluded from citizenship. Furthermore, since hoplites had to provide their own armor, these citizen-militias effectively consisted of landowners. This is not to say, though, that the poorer citizens were entirely excluded from serving their city. One example of a way in which they may have participated even in the phalanx appears on the Chigi Vase. Marching between two lines of warriors is an unarmed man, playing a double-reed flute (seen on the right end of the top band in Figure 5.10). Since the success of the phalanx depended on marching together in step, the flute-player’s music would have been essential to ensure that everyone kept the same tempo during the march. 5.9.2 Greek Religion One theory modern scholars have proposed for the rise of the polis connects the locations of the city-states to known cult-sites. The theory argues that the Greeks of the Archaic period built city-states around these precincts of various gods in order to live closer to them and protect them. While impossible to know for sure if this theory or any other regarding the rise of the polis is true, the building of temples in cities during the Archaic period shows the increasing emphasis that the poleis were placing on religion. It is important to note that Greek religion seems to have been, at least to some extent, an element of continuity from the Bronze Age to the Archaic period and beyond. The important role that the gods play in the Homeric epics attests to their prominence in the oral tradition, going back to the Dark Ages. Furthermore, names of the following major gods worshipped in the Archaic period and beyond were found on the deciphered Linear B tablets: Zeus, king of the gods and god of weather, associated with the thunderbolt; Hera, Zeus’ wife and patroness of childbirth; Poseidon, god of the sea; Hermes, messenger god and patron of thieves and merchants; Athena, goddess of war and wisdom and patroness of women’s crafts; Ares, god of war; Dionysus, god of wine; and the twins Apollo, god of the sun and both god of the plague and a healer, and Artemis, goddess of the hunt and the moon. All of these gods continued to be the major divinities in Greek religion for its duration, and many of them were worshipped as patron gods of individual cities, such as Artemis at Sparta, and Athena at Athens. While many local cults of even major gods were truly local in appeal, a few local cults achieved truly Pan-Hellenic appeal. Drawing visitors from all over the Greek world, these Pan-Hellenic cults were seen as belonging equally to all the Greeks. One of the most famous examples is the cult of Asclepius at Epidaurus . Asclepius, son of Apollo, was a healer god, and his shrine at Epidaurus attracted the pilgrims from all over the Greek world. Visitors suffering from illness practiced incubation, that is, spending the night in the temple, in the hopes of receiving a vision in their dreams suggesting a cure. In gratitude for the god’s healing, some pilgrims dedicated casts of their healed body parts. Archaeological findings include a plentitude of ears, noses, arms, and feet. Starting out as local cults, several religious festivals that included athletic competitions as part of the celebration also achieved Pan-Hellenic prominence during the Archaic period. The most influential of these were the Olympic Games . Beginning in 776 BCE, the Olympic Games were held in Olympia every four years in honor of Zeus; they drew competitors from all over the Greek world, and even Persia. The Pan-Hellenic appeal of the Olympics is signified by the impact that these games had on Greek politics: for instance, a truce was in effect throughout the Greek world for the duration of each Olympics. In addition, the Olympics provided a Pan-Hellenic system of dating events by Olympiads or four-year cycles. Finally, perhaps the most politically influential of the Pan-Hellenic cults was the oracle of Apollo at Delphi , established sometime in the eighth century BCE. Available for consultation only nine days a year, the oracle spoke responses to the questions asked by inquirers through a priestess, named the Pythia . The Pythia’s responses came in the form of poetry and were notoriously difficult to interpret. Nevertheless, city-states and major rulers throughout the Greek world considered it essential to consult the oracle before embarking on any major endeavor, such as war or founding a colony. 5.9.3 Maritime Trade and Colonization The historian Herodotus records that sometime c. 630 BCE, the king of the small island of Thera traveled to Delphi to offer a sacrifice and consult the oracle on a few minor points. To his surprise, the oracle’s response had nothing to do with his queries. Instead, the Pythia directed him to found a colony in Libya, in North Africa. Having never heard of Libya, the king ignored the advice. A seven-year drought ensued, and the Therans felt compelled to consult the oracle again. Receiving the same response as before, they finally sent out a group of colonists who eventually founded the city of Cyrene. While this story may sound absurd, it is similar to other foundation stories of Greek colonies and emphasizes the importance of the Delphic oracle. At the same time, though, this story still leaves open the question of motive: why did so many Greek city-states of the Archaic period send out colonies to other parts of the Greek world? Archaeology and foundation legends, such as those recorded by Herodotus, suggest two chief reasons: population pressures along with shortage of productive farmland in the cities on mainland Greece, and increased ease of trade that colonies abroad facilitated. In addition to resolving these two problems, however, the colonies also had the unforeseen impact of increasing interactions of the Greeks with the larger Mediterranean world and the ancient Near East. These interactions are visible, for instance, in the so-called Orientalizing style of art in the Archaic period, a style the Greeks borrowed from the Middle East and Egypt. As section 5.3.5 will show, however, the presence of Greek colonies in Asia Minor also played a major role in bringing about the Greco-Persian Wars. 5.9.4. Aristocracy, Democracy, and Tyranny in Archaic Greece Later Greek historians, including Herodotus and Thucydides, noted a certain trend in the trajectory of the history of most Greek poleis : most city-states started out with a monarchical or quasi-monarchical government. Over time the people gained greater representation, and an assembly of all citizens had at least some degree of political power—although some degree of strife typically materialized between the aristocrats and the poorer elements. Taking advantage of such civic conflicts, tyrants came to power in most city-states for a brief period before the people banded together and drove them out, thenceforth replacing them with a more popular form of government. Many modern historians are skeptical about some of the stories that the Greek historians tell about origins of some polei s; for instance, it is questionable whether the earliest Thebans truly were born from dragon teeth. Similarly, the stories about some of the Archaic tyrants seem to belong more to the realm of legend than history. Nevertheless, the preservation of stories about tyrants in early oral tradition suggests that city-states likely went through periods of turmoil and change in their form of government before developing a more stable constitution. Furthermore, this line of development accurately describes the early history of Athens, the best-documented polis . In the early Archaic period, Athens largely had an aristocratic constitution. Widespread debt-slavery, however, caused significant civic strife in the city and led to the appointment of Solon as lawgiver for the year 594/3 BCE, specifically for the purpose of reforming the laws. Solon created a more democratic constitution and also left poetry documenting justifications for his reforms—and different citizens’ reactions to them. Most controversial of all, Solon instituted a one-time debt-forgiveness, seisachtheia, which literally means “shaking off.” He proceeded to divide all citizens into five classes based on income, assigning a level of political participation and responsibility commensurate with each class. Shortly after Solon’s reforms, a tyrant, Peisistratus , illegally seized control of Athens and remained in power off and on from 561 to 527 BCE. Peisistratus seems to have been a reasonably popular ruler who had the support of a significant portion of the Athenian population. His two sons, Hippias and Hipparchus, however, appear to have been less well-liked. Two men, Harmodius and Aristogeiton, assassinated Hipparchus in 514 BCE; then in 508 BCE, the Athenians, with the help of a Spartan army, permanently drove out Hippias. In subsequent Athenian history, Harmodius and Aristogeiton were considered heroes of the democracy and celebrated as tyrannicides. Immediately following the expulsion of Hippias, Athens underwent a second round of democratic reforms, led by Cleisthenes . The Cleisthenic constitution remained in effect, with few changes, until the Macedonian conquest of Athens in the fourth century BCE and is considered to be the Classical Athenian democracy . Central to the democracy was the participation of all citizens in two types of institutions: the ekklesia , an assembly of all citizens, which functioned as the chief deliberative body of the city; and the law-courts, to which citizens were assigned by lot as jurors. Two chief offices, the generals and the archos, ruled over the city and were appointed for one-year terms. Ten generals were elected annually by the ekklesia for the purpose of leading the Athenian military forces. Finally, the leading political office each year, the nine archons, were appointed by lot from all eligible citizens. While this notion of appointing the top political leaders by lot may seem surprising, it exemplifies the Athenians’ pride in their democracy and their desire to believe that, in theory at least, all Athenian citizens were equally valuable and capable of leading their city-state. Developing in a very different manner from Athens, Sparta was seen by other Greek poleis as a very different sort of city from the rest. Ruled from an early period by two kings – one from each of the two royal houses that ruled jointly – Sparta was a true oligarchy, in which the power rested in its gerousia , a council of thirty elders, whose number included the two kings. While an assembly of all citizens existed as well, its powers were much more limited than were those of the Athenian assembly. Yet because of much more restrictive citizenship rules, Spartan assembly of citizens would have felt as a more selective body, as Figure 5.14 illustrates. A crucial moment in Spartan history was the city’s conquest of the nearby region of Messenia in the eighth century BCE. The Spartans annexed the Messenian territory to their own and made the Messenians helots . While the helots could not be bought or sold, they were permanently tied to the land in a status akin to medieval European serfs. The availability of helot labor allowed the Spartans from that point on to focus their attention on military training. This focus transformed Sparta into the ultimate military state in the Greek world, widely respected by the other Greek poleis for its military prowess. Other Greeks were fascinated by such Spartan practices as the communal bringing up of all children apart from their parents and the requirement that all Spartan girls and women, as well as boys and men, maintain a strict regimen of exercise and training. But while Athens and Sparta sound like each other’s diametrical opposites, the practices of both poleis ultimately derived from the same belief that all city-states held: that, in order to ensure their city’s survival, the citizens must place their city-state’s interests above their own. A democracy simply approached this goal with a different view of the qualifications of its citizens than did an oligarchy. A final note on gender is necessary, in connection with Greek city-states’ definitions of citizenship. Only children of legally married and freeborn citizen parents could be citizens in most city-states. Women had an ambiguous status in the Greek poleis . While not full-fledged citizens themselves, they produced citizens. This view of the primary importance of wives in the city as the mothers of citizens resulted in diametrically opposite laws in Athens and Sparta, showing the different values that the respective cities emphasized. In Athens, if a husband caught his wife with an adulterer in his home, the law allowed the husband to kill said adulterer on the spot. The adultery law was so harsh precisely because adultery put into question the citizenship status of potential children, thereby depriving the city of future citizens. By contrast, Spartan law allowed an unmarried man who wanted offspring to sleep with the wife of another man, with the latter’s consent, specifically for the purpose of producing children. This law reflects the importance that Sparta placed on producing strong future soldiers as well as the communal attitude of the city towards family and citizenship. 5.9.5 The Persian Wars Despite casting their net far and wide in founding colonies, the Greeks seem to have remained in a state of relatively peaceful coexistence with the rest of their Mediterranean neighbors until the sixth century BCE. In the mid-sixth century BCE, Cyrus, an ambitious king of Persia, embarked on a swift program of expansion, ultimately consolidating under his rule the largest empire of the ancient world and earning for himself the title “ Cyrus the Great .” Cyrus’ Achaemenid Empire bordered the area of Asia Minor that had been previously colonized by the Greeks. This expansion of the Persian Empire brought the Persians into direct conflict with the Greeks and became the origin of the Greco-Persian Wars, the greatest military conflict the Greek world had known up until that point. Over the second half of the sixth century, the Persians had taken over the region of Asia Minor, also known as Ionia, installing as rulers of these Greek city-states tyrants loyal to Persia. In 499 BCE, however, the Greek city-states in Asia Minor joined forces to rebel against the Persian rule. Athens and Eretria sent military support for this Ionian Revolt, and the rebelling forces marched on the Persian capital of Sardis and burned it in 498 BCE, before the revolt was finally subdued by the Persians in 493 BCE. Seeking revenge on Athens and Eretria, the Persian king Darius launched an expedition in 490 BCE. Darius’ forces captured Eretria in mid-summer, destroyed the city, and enslaved its inhabitants. Sailing a short distance across the bay, the Persian army then landed at Marathon. The worried Athenians sent a plea for help to Sparta. The Spartans, in the middle of a religious festival, refused to help. So, on September 12, 490 BCE, the Athenians, with only a small force of Plataeans helping, faced the much larger Persian army in the Battle of Marathon . The decisive Athenian victory showed the superiority of the Greek hoplite phalanx and marked the end of the first Persian invasion of Greece. Furthermore, the victory at Marathon, which remained a point of pride for the Athenians for centuries after, demonstrated to the rest of the Greeks that Sparta was not the only great military power in Greece. Darius died in 486 BCE, having never realized his dream of revenge against the Greeks. His son, Xerxes , however, continued his father’s plans and launched in 480 BCE a second invasion of Greece, with an army so large that, as the historian Herodotus claims, it drank entire rivers dry on its march. The Greek world reacted in a much more organized fashion to this second invasion than it did to the first. Led by Athens and Sparta, some seventy Greek poleis formed a sworn alliance to fight together against the Persians. This alliance, the first of its kind, proved to be the key to defeating the Persians as it allowed the allies to split forces strategically in order to guard against Persian attack by both land and sea. The few Greek city-states who declared loyalty to the Persian Empire instead–most notably, Thebes–were seen as traitors for centuries to come by the rest of the Greeks. Marching through mainland Greece from the north, the Persians first confronted the Spartans at the Battle of Thermopylae, a narrow mountain pass that stood in the way of the Persians’ accessing any point south. In this now-legendary battle, 300 Spartans, led by their king Leonidas, successfully defended the pass for two days before being betrayed by a local who showed a roundabout route to the Persians. The Persians then were able to outflank the Spartans and kill them to the last man. This battle, although a loss for the Greeks, bought crucial time for the rest of the Greek forces in preparing to face the Persians. It is also important to note that although the Spartans were considered even in the ancient world to be the heroes of Thermopylae, they were also accompanied by small contingents from several other Greek city-states in this endeavor. The victory at Thermopylae fulfilled the old dream of Darius, as it allowed access to Athens for the Persians. The Athenian statesman Themistocles, however, had ordered a full evacuation of the city in advance of the Persian attack through an unusual interpretation of a Delphic oracle stating that wooden walls will save Athens. Taking the oracle to mean that the wooden walls in question were ships, Themistocles built a massive fleet which he used to send all of the city’s inhabitants to safety. His gamble proved to be successful, and the Persians captured and burned a mostly empty city. The Athenians proceeded to defeat the Persian fleet at the Battle of Salamis, off the coast of Athens, thus shortly before winter turning the tide of the war in favor of the Greeks. Finally, in June of 479 BCE, the Greek forces were able to strike the two final blows, defeating the Persian land and sea forces on the same day in the Battle of Plataea on land and the Battle of Mycale on sea. The victory at Mycale also resulted in a second Ionian revolt, which this time ended in a victory for the Greek city-states in Asia Minor. Xerxes was left to sail home to his diminished empire. It is difficult to overestimate the impact of the Persian Wars on subsequent Greek history. Seen by historians as the end-point of the Archaic Period, the Persian Wars cemented Pan-Hellenic identity, as they saw cooperation on an unprecedented scale among the Greek city-states. In addition, the Persian Wars showed the Greek military superiority over the Persians on both land and sea. Finally, the wars showed Athens in a new light to the rest of the Greeks. As the winners of Marathon in the first invasion and the leaders of the navy during the second invasion, the Athenians emerged from the wars as the rivals of Sparta for military prestige among the Greeks. This last point, in particular, proved to be the most influential for Greek history in the subsequent period. 5.10 THE CLASSICAL PERIOD So far, the story of the Greek world in this chapter has proceeded from a narrative of the fragmented Greek world in the Dark Ages to the emergence and solidification of a Pan-Hellenic identity in the Archaic Period. The story of the Greeks in the Classical Period, by contrast, is best described as the strife for leadership of the Greek world. First, Athens and Sparta spent much of the fifth century BCE battling each other for control of the Greek world. Then, once both were weakened, other states began attempting to fill the power vacuum. Ultimately, the Classical Period will end with the Greek world under the control of a power that was virtually unknown to the Greeks at the beginning of the fifth century BCE: Macedon. 5.10.1 From the Delian League to the Athenian Empire In 478 BCE, barely a year after the end of the Persian Wars, a group of Greek city-states, mainly those located in Ionia and on the island between mainland Greece and Ionia, founded the Delian League , with the aim of continuing to protect the Greeks in Ionia from Persian attacks. Led by Athens, the league first met on the tiny island of Delos. According to Greek mythology, the twin gods Apollo and Artemis were born on Delos. As a result, the island was considered sacred ground and, as such, was a fitting neutral headquarters for the new alliance. The league allowed member states the option of either contributing a tax (an option that most members selected) or contributing ships for the league’s navy. The treasury of the league, where the taxes paid by members were deposited, was housed on Delos. Over the next twenty years, the Delian League gradually transformed from a loose alliance of states led by Athens to a more formal entity. The League’s Athenian leadership, in the meanwhile, grew to be that of an imperial leader. The few members who tried to secede from the League, such as the island of Naxos, quickly learned that doing so was not an option as the revolt was violently subdued. Finally, in 454 BCE, the treasury of the Delian League moved to Athens. That moment marked the transformation of the Delian League into the Athenian Empire . Since the Athenians publicly inscribed each year the one-sixtieth portion of the tribute that they dedicated to Athena, records survive listing the contributing members for a number of years, thereby allowing historians to see the magnitude of the Athenian operation. While only the Athenian side of the story survives, it appears that the Athenians’ allies in the Delian League were not happy with the transformation of the alliance into a full-fledged Athenian Empire. Non-allies were affected a well. The fifth-century BCE Athenian historian Thucydides dramatizes in his history one particularly harsh treatment of a small island, Melos, which effectively refused to join the Athenian cause. To add insult to injury, once the treasury of the Empire had been moved to Athens, the Athenians had used some funds from it for their own building projects, the most famous of these projects being the Parthenon, the great temple to Athena on the Acropolis. The bold decision to move the treasury of the Delian League to Athens was the brainchild of the leading Athenian statesman of the fifth century BCE, Pericles. A member of a prominent aristocratic family, Pericles was a predominant politician for forty years, from the early 460s BCE to his death in 429 BCE, and was instrumental in the development of a more popular democracy in Athens. Under his leadership, an especially vibrant feeling of Athenian patriotic pride seems to have developed, and the decision to move the Delian League treasury to Athens fits into this pattern as well. Shortly after moving the treasury to Athens, Pericles sponsored a Citizenship Decree in 451 BCE that restricted Athenian citizenship from thence onwards only to individuals who had two free- born and legitimately-wed Athenian parents, both of whom were also born of Athenian parents. Then c. 449 BCE, Pericles successfully proposed a decree allowing the Athenians to use Delian League funds for Athenian building projects, and, c. 447 BCE, he sponsored the Athenian Coinage Decree, a decree that imposed Athenian standards of weights and measures on all states that were members of the Delian League. Later in his life, Pericles famously described Athens as “the school of Hellas;” this description would certainly have fit Athens just as much in the mid-fifth century BCE as, in addition to the flourishing of art and architecture, the city was a center of philosophy and drama. The growing wealth and power of Athens in the twenty or so years since the Persian Wars did not escape Sparta and led to increasingly tense relations between the two leading powers in Greece. Sparta had steadily consolidated the Peloponnesian League in this same time-period, but Sparta’s authority over this league was not quite as strict as was the Athenian control over the Delian League. Finally, in the period of 460-445 BCE, the Spartans and the Athenians engaged in a series of battles, to which modern scholars refer as the First Peloponnesian War . In 445 BCE, the two sides swore to a Thirty Years Peace, a treaty that allowed both sides to return to their pre-war holdings, with few exceptions. Still, Spartan unease in this period of Athenian expansion and prosperity, which resulted in the First Peloponnesian War, was merely a sign of much more serious conflict to come. As the Athenian general and historian Thucydides later wrote about the reasons for the Great Peloponnesian War, which erupted in 431 BCE: “But the real cause of the war was one that was formally kept out of sight. The growing power of Athens, and the fear that it inspired in Sparta, made the war inevitable” (Thucydides, I.23). 5.10.2 The Peloponnesian War (431 – 404 BCE) Historians today frown on the use of the term “inevitable” to describe historical events. Still, Thucydides’ point about the inevitability of the Peloponnesian War is perhaps appropriate, as following a conflict that had been bubbling under the surface for fifty years, the war finally broke out over a seemingly minor affair. In 433 BCE, Corcyra, a colony of Corinth that no longer wanted to be under the control of its mother-city, asked Athens for protection against Corinth. The Corinthians claimed that the Athenian support of Corcyra was a violation of the Thirty Years Peace. At a subsequent meeting of the Peloponnesian League in Sparta in 432 BCE, the allies, along with Sparta, voted that the peace had been broken and so declared war against Athens. At the time of the war’s declaration, no one thought that it would last twenty-seven years and would ultimately embroil the entire Greek-speaking world. Rather, the Spartans expected that they would march with an army to Athens, fight a decisive battle, then return home forthwith. The long duration of the war, however, was partly the result of the different strengths of the two leading powers. Athens was a naval empire, with allies scattered all over the Ionian Sea. Sparta, on the other hand, was a land-locked power with supporters chiefly in the Peloponnese and with no navy to speak of at the outset of the war. The Peloponnesian War brought about significant changes in the government of both Athens and Sparta, so that, by the end of the war, neither power looked as it did at its outset. Athens, in particular, became more democratic because of increased need for manpower to row its fleet. The lowest census bracket, the thetes , whose poverty and inability to buy their own armor had previously excluded them from military service, became by the end of the war a full-fledged part of the Athenian forces and required a correspondingly greater degree of political influence. In the case of Sparta, the war had ended the Spartan policy of relative isolationism from the rest of the affairs of the Greek city-states. The length of the war also brought about significant changes to the nature of Greek warfare. While war was previously largely a seasonal affair, with many conflicts being decided with a single battle, the Peloponnesian War forced the Greek city-states to support standing armies. Finally, while sieges of cities and attacks on civilians were previously frowned upon, they became the norm by the end of the Peloponnesian War. In short, Thucydides’s narrative of the war shows that the war had a detrimental effect on human nature, encouraging a previously unprecedented degree of cruelty on both sides. It is important to note, though, that as brutal as sieges could be during the Peloponnesian War, Greek siege warfare during the fifth century BCE was still quite primitive, as no tools existed for ramming or otherwise damaging the city gates or walls. Furthermore, catapults, so useful for targeting a city from the outside, first came into being in 399 BCE, five years after the war had ended. Modern historians divide the Peloponnesian War into three distinct stages, based on the tactics used in each: the Archidamian War , the Peace of Nicias , and the Decelean War . The first stage, the Archidamian War (431 – 421 BCE), is named after the Spartan king Archidamus, who proposed the strategy of annual invasions of Attica at the beginning of the war. Beginning in late spring and early summer of 431 BCE, Archidamus led the Spartan army to invade Attica in order to devastate the agricultural land around the city. The Spartans thereby hoped to provoke the Athenians to a battle. Pericles however, refused to enter into battle against the Spartans, and instead ordered all inhabitants of Attica to retreat within the city. Pericles’ decision was wise, as the Athenians would likely have lost a land battle against the Spartans. His decision, though, had unforeseen repercussions. In 430 BCE, the crowded conditions within Athens resulted in the outbreak of a virulent plague which by some estimates killed as much as twenty-five percent of the city’s population over the following three years. Among the dead was none other than Pericles himself. The plague had significant repercussions for Athens during the first phase of the war because of not only the loss of fighting men to disease and the consequent lowered morale in the city, but also the death of Pericles, the moderate leader. The subsequent leaders who emerged, such as Cleon, were known as war-hawks. Meanwhile, the Spartans continued their annual invasions of Attica until 425 BCE, when luck was finally on the Athenians’ side. In 425 BCE, the Athenian fleet faced a new Spartan fleet in the Battle of Pylos in the Peloponnese. The Athenians won the battle and also managed to trap 420 Spartans on the tiny island of Sphacteria, just off the coast of Pylos. Sending shockwaves through the entire Greek world, the Spartans surrendered. By bringing the hostages to Athens, the Athenians put an end to the annual invasions of Attica. Finally, in 421 BCE, with the death of the most pro-war generals on both sides, the Athenians with their allies signed a peace treaty with Spartans and their allies. Named the “Peace of Nicias” after the Athenian general who brokered this treaty, it was supposed to be a fifty years’ peace; it allowed both sides to return to their pre-war holdings, with a few exceptions. As part of the peace terms, the Spartan hostages from Pylos were finally released. Despite its ambitious casting as a fifty years’ peace, the Peace of Nicias proved to be a short and uneasy time filled with minor battles and skirmishes. One problem with the treaty was that while Athens and all of its allies signed the peace, several key allies of Sparta, including Corinth and Thebes, refused to do so. Furthermore, Athens made the disastrous decision during this stalemate to launch the Sicilian Expedition , a venture that took much of the Athenian fleet to Sicily in 415 BCE. Syracuse, however, proved to be a difficult target, and the expedition ended in 413 BCE with a complete destruction of the Athenian navy. That same year, the Spartans renewed the fighting, launching the third and final phase of the Peloponnesian War. In the third stage of the Peloponnesian war, also known as the Decelean War, the Spartans took the war to Attic soil by occupying Decelea, a village in Attica proper, and transforming it into a military fort. This occupation allowed the Spartans to prevent the Athenians from farming their land and cutting off Athens from most supply routes, effectively crippling the Athenian economy for the remainder of the war. Losing the Sicilian Expedition and the challenge of the Decelean War produced a high level of resentment towards the democratic leaders in Athens. Therefore in 411 BCE, an oligarchic coup briefly replaced the democracy with the rule of the Four Hundred. While this oligarchy was quickly overthrown and the democracy restored, this internal instability highlighted the presence of the aristocratic element in the city as well as the dissatisfaction of at least the aristocratic citizens with the long war. Remarkably, in a testament to the resilience and power of the Athenian state, the Athenians managed to rebuild a navy after the Sicilian Expedition, and even managed to continue to win battles on sea during this final phase of the war. In 405 BCE, however, the Spartan general Lysander defeated Athens in the naval battle of Aegospotami. He proceeded to besiege Athens, and the city finally surrendered in 404 BCE. For the second time in a decade, the Athenian democracy was overthrown, to be replaced this time by the Spartan-sanctioned oligarchy known as the Tyranny of the Thirty . The rule of the Thirty proved to be a much more brutal oligarchy than that of the Four Hundred. A year later, an army formed largely of Athenian democrats in exile marched on the city and overthrew the Thirty. The democracy thus was restored in 403 BCE, and the painful process of recovery from the war and the oligarchic rule could begin. 5.10.3 Athenian Culture during the Peloponnesian War Because it drained Athens of manpower and financial resources, the Peloponnesian War proved to be an utter practical disaster for Athens. Nevertheless, the war period was also the pinnacle of Athenian culture, most notably its tragedy, comedy, and philosophy. Tragedy and comedy in Athens were very much popular entertainment, intended to appeal to all citizens. Thus issues considered in these plays were often ones of paramount concern for the city at the time when the plays were written. As one character in a comedy bitterly joked in an address to the audience, more Athenians attended tragic and comic performances than came to vote at assembly meetings. Not surprisingly, war was a common topic of discussion in the plays. Furthermore, war was not portrayed positively, as the playwrights repeatedly emphasized the costs of war for both winners and losers. Sophocles , one of the two most prominent Athenian tragedians during the Peloponnesian War era, had served his city as a general, albeit at an earlier period; thus, he had direct experience with war. Many of his tragedies that were performed during the war dealt with the darker side of fighting, for both soldiers and generals, and the cities that are affected. By tradition, however, tragedies tackled contemporary issues through integrating them into mythical stories, and the two mythical wars that Sophocles portrayed in his tragedies were the Trojan War, as in Ajax and Philoctetes , and the aftermath of the war of the Seven against Thebes, in which Polynices, the son of Oedipus, led six other heroes to attack Thebes, a city led by his brother Eteocles, as in Oedipus at Colonus . Sophocles’ plays repeatedly showed the emotional and psychological challenges of war for soldiers and civilians alike; they also emphasized the futility of war, as the heroes of his plays, just as in the original myths on which they were based, died tragic, untimely deaths. Sophocles’ younger contemporary, Euripides , had a similar interest in depicting the horrors of war and wrote a number of tragedies on the impact of war on the defeated, such as in Phoenician Women and Hecuba ; both of these plays explored the aftermath of the Trojan War from the perspective of the defeated Trojans. While the tragic playwrights explored the impact of the war on both the fighters and the civilians through narrating mythical events, the comic playwright Aristophanes was far less subtle. The anti-war civilian who saves the day and ends the war was a common hero in the Aristophanic comedies. For instance, in the Acharnians (425 BCE), the main character is a war-weary farmer who, frustrated with the inefficiency of the Athenian leadership in ending the war, brokers his own personal peace with Sparta. Similarly, in Peace (421 BCE), another anti-war farmer fattens up a dung beetle in order to fly to Olympus and beg Zeus to free Peace. Finally, in Lysistrata (411 BCE), the wives of all Greek city-states, missing their husbands who are at war, band together in a plot to end the war by going on a sex-strike until their husbands make peace. By the end of the play, their wish comes true. Undeniably funny, the jokes in these comedies, nevertheless, have a bitter edge, akin to the portrayal of war in the tragedies. The overall impression from the war-era drama is that the playwrights, as well as perhaps the Athenians themselves, spent much of the Peloponnesian War dreaming of peace. While the playwrights were dreaming of the things of this world–most notably war–their contemporary, Socrates , was dreaming of difficult questions. One of the most prominent philosophers of the ancient world, Socrates has not left any writings of his own, but thoughts attributed to him survive in dialogues penned by his student, the fourth-century philosopher Plato . In Plato’s writings, Socrates comes across as someone who loved difficult questions and who was not above confronting any passers-by with such questions as “What is courage?”; “What is moral?”; “What would the ideal city look like?” Using what became known ever since as the “Socratic method,” Socrates continued to probe further every definition and answer that his conversation partners provided, guiding them to delve deeper in their reflections on the topics at hand than they had before. As a result of his love of such debates, Socrates was seen as connected to the Sophists, philosophical debate teachers, who (as Aristophanes joked) could teach anyone to convince others of anything at all, regardless of reality or truth. But Socrates radically differed from the Sophists by not charging fees for his teaching. Instead, as he himself is purported to have said, he was a pest-like gadfly that kept disturbing Athens from growing too content and encouraged all with whom he spoke to keep thinking and questioning. 5.10.4 The Fourth Century BCE In 399 BCE, a seventy-year old Athenian was put on trial for impiety and for corrupting the youth, convicted, and speedily sentenced to death. The trial is especially shocking, since the man in question was none other than Socrates, the philosopher who had spent his life wandering the streets of Athens engaging in endless dialogues regarding the meaning of life. Why did the Athenians suddenly turn against this public teacher and judge him worthy of execution? The answer, most likely, is not the openly-stated causes of the trial, but rather the connections that Socrates previously had to oligarchic leaders. In particular, Socrates had taught Critias, who became one of the Thirty in 404 BCE. Fueled by their hatred of all enemies of the democracy and anyone who had associated with the Thirty, the Athenians condemned Socrates to death. This trial shows how deeply the scars went in the collective psyche and how difficult it was for the Athenians to forget the terrible end of the Peloponnesian War. And while, as usual, more information survives about how the Athenians— more than any other polis —dealt with the aftermath of the war, it is clear that for the rest of the Greek world, their life in the fourth century BCE was very much the result of the Peloponnesian War. The early fourth century saw a power vacuum emerge in the Greek world for the first time since the early Archaic Period. Defeated in the war, Athens was no longer an Empire, while the winner, Sparta, had suffered a catastrophic decline in its population over the course of the Peloponnesian War. At the same time, Thebes had revamped its military, introducing the first two significant changes to the hoplite phalanx way of fighting since its inception: slightly longer spears, and wedge formation. The final key to the Theban military supremacy was the Theban Sacred Band , formed in 378 BCE. An elite core of 300 warriors, the band consisted of 150 couples, based on the assumption that the lovers would fight most bravely in order not to appear to be cowardly to their beloved. In 371 BCE, the Thebans demonstrated the success of their military reforms by defeating the Spartans at the Battle of Leuctra . They continued an aggressive program of military expansion over the next decade, a period known as the Theban Hegemony . Sometime in the 360’s BCE, a young Macedonian prince stayed for several years in Thebes as a hostage. While there, he caught the eye of the military reformer, Epaminondas, who took the prince under his wing. Circa 364 BCE, the prince returned to Macedon, and, in 359 BCE, he ascended to the throne as king Philip II . Up until that point in Greek history, the Macedonians had largely been known for two things: drinking their wine undiluted, which had marked them as complete and utter barbarians in the eyes of the rest of the Greeks, and being excellent horsemen. With Philip at the helm, this estimation was about to change. As soon as he came to the throne, Philip began transforming the Macedonian military into a more successful image of what he had seen at Thebes. Philip further lengthened the already longer spears used by the Thebans, creating the Macedonian sarissa , a spear of about eighteen feet in length, double that of the traditional Greek hoplite spear. He retained the Theban wedge formation but also added heavy cavalry to the line, thus incorporating the Macedonians’ strongest element into the phalanx. The results spoke for themselves, as over the next twenty years, Philip systematically conquered all of mainland Greece, with the exception of Sparta, which he chose to leave alone. Philip’s final great victory, which he shared with his teenage son Alexander, was at the Battle of Chaeronea (338 BCE), in which the Macedonian armies defeated the combined forces of Athens and Thebes. Philip’s conquest of the entire mainland was the end of an era, as for the first time, the entire territory was united under the rule of a king. By all accounts, it appears that Philip was not going to stop at just conquering the Greek world. He did not, however, have this choice. In 336 BCE while on his way to a theatrical performance, Philip was assassinated by one of his own bodyguards. His son Alexander, then twenty years old, succeeded and continued his father’s ambitious program of conquests. Alexander’s first target was the Persian Empire, motivated in part by his love of Homer’s Iliad , and the perception among the Greeks that this new campaign was the continuation of the original, mythical war against Asia. Moving farther and farther East in his campaigns, Alexander conquered the Balkans, Egypt, and the territories of modern-day Lebanon, Syria, and Israel before he achieved a decisive victory over Darius III at the Battle of Gaugamela in 331 BCE. Continuing to move eastwards, Alexander invaded India in 327 BCE, planning to conquer the known world and assuming that he was close to this achievement, since the Greeks of his day were not aware of China’s existence. His war-weary troops, however, rebelled in 326 BCE and demanded to return home (see Chapter 3). It appears that this mutiny was not the first that occurred in Alexander’s army; indeed, over the course of his rule, Alexander had also been the target of a number of failed assassinations. However, this mutiny forced Alexander to give in. Leaving several of his officers behind as satraps, Alexander turned back. In 323 BCE, he and his army reached Babylon, the city that he had hoped to make the new capital of his world empire. There, Alexander fell ill and died at the ripe old age of thirty-three. While Alexander’s rule only lasted thirteen years, his legacy reshaped Greece and the rest of ancient Eurasia for the next several centuries. A charismatic leader, albeit one prone to emotional outbursts, Alexander redefined what it meant to be king and general. His coinage reflects this reinvention. On one coin minted during his lifetime, for instance, appears Alexander dressed as the hero Heracles, while Zeus, whom Alexander alleged to be his real father, appears on the other side. In addition, by conquering territories that were previously not part of the Greek world, Alexander spread Greek culture farther than had anyone else before him. At the same time, by marrying several non-Greek princesses and encouraging such marriages by his troops, Alexander also encouraged the creation of a “melting-pot” empire; he further cemented this creation by founding new cities named after himself all over his new empire. In particular, Alexandria, the city that he founded in Egypt, became a center of Greek civilization—albeit with an Egyptian twist—was seen as a new Athens well into the Roman Empire. Alexander’s brief time in India produced a significant impact as well, as in 321 BCE, Chandragupta Maurya was able to unify India into a single kingdom for the first time, establishing the Mauryan Empire (see Chapter Three). Finally, in the Middle East, North Africa, and the Greek world, Alexander’s generals divided his conquests into several kingdoms that they and their descendants continued to rule until the Romans conquered these respective areas. It appears that Alexander’s melting-pot empire, burning up as a phoenix upon his death, actually allowed several new empires and kingdoms to arise from its ashes. 5.11 HELLENISTIC PERIOD Historians today consider the death of Alexander to be the end point of the Classical Period and the beginning of the Hellenistic Period. That moment, for historians, also marks the end of the polis as the main unit of organization in the Greek world. While city-states continued to exist, the main unit of organization from that point on was the great Hellenistic kingdoms. These kingdoms, encompassing much greater territory than the Greek world had before Alexander, contributed to the thorough Hellenization of the Eastern Mediterranean and the Middle East. The age of the Hellenistic kingdoms also coincided with the rise of Rome as a military power in the West. Ultimately, the Hellenistic kingdoms were conquered and absorbed by Rome. 5.11.1 Hellenistic Kingdoms Although Alexander had several children from his different wives, he did not leave an heir old enough to take power upon his death. Indeed, his only son, Alexander IV, was only born several months after his father’s death. Instead, Alexander’s most talented generals turned against each other in a contest for the control of the empire that they had helped create. These Wars of the Diadochi , as they are known in modern scholarship, ended with a partition of Alexander’s empire into a number of kingdoms, each ruled by dynasties. Of these, the four most influential dynasties which retained power for the remainder of the Hellenistic Age, were the following: Seleucus, who took control of Syria and the surrounding areas, thus creating the Seleucid Empire ; Antigonus Monophthalmos, the One-Eyed, who took over the territory of Asia Minor and northern Syria, establishing the Antigonid Dynasty ; the Attalid Dynasty, which took power over the Kingdom of Pergamon , after the death of its initial ruler, Lysimachus, a general of Alexander; and Ptolemy, Alexander’s most influential general, who took control over Egypt, establishing the Ptolemaic Dynasty . The most imperialistic of Alexander’s successors, Seleucus I Nicator took Syria, swiftly expanding his empire to the east to encompass the entire stretch of territory from Syria to India. At its greatest expanse, this territory’s ethnic diversity was similar to that of Alexander’s original empire, and Seleucus adopted the same policy of ethnic unity as originally practiced by Alexander; some of Seleucus’ later successors, however, attempted to impose Hellenization on some of the peoples under their rule. These successors had difficulties holding on to Seleucus’ conquests. A notable exception, Antiochus III, attempted to expand the Empire into Anatolia and Greece in the early second century BCE but was ultimately defeated by the Romans. The empire’s story for the remainder of its existence is one of almost constant civil wars and increasingly declining territories. The Seleucids seem to have had a particularly antagonistic relationship with their Jewish subjects, going so far as to outlaw Judaism in 168 BCE. The Jewish holiday Hannukah celebrates a miracle that occurred following the historical victory of the Jews, led by Judah Maccabee, over the Seleucids in 165 BCE. Shortly afterwards, the Seleucids had to allow autonomy to the Jewish state; it achieved full independence from Seleucid rule in 129 BCE. In 63 BCE, the Roman general Pompey finally conquered the small remnant of the Seleucid Empire, making it into the Roman province of Syria. Antigonus Monophthalmos, Seleucus’ neighbor, whose holdings included Macedonia, Asia Minor, and the northwestern portion of Syria, harbored ambitious plans that rivaled those of Seleucus. Antigonus’ hopes of reuniting all of Alexander’s original empire under his own rule, however, were never realized as Antigonus died in battle in 301 BCE. The greatest threat to the Antigonids, however, came not from the Seleucid Empire, but from Rome with whom they waged three Macedonian Wars between 214 and 168 BCE. The Roman defeat of king Perseus in 168 BCE at the Battle of Pydna marked the end of the Third Macedonian War , and the end of an era, as control over Greece was now in Roman hands. The smallest and least imperialistic of the successor states, the kingdom of Pergamon, was originally part of a very short-lived empire established by Lysimachus, one of Alexander’s generals. Lysimachus originally held Macedonia and parts of Asia Minor and Thrace but had lost all of these territories by the time of his death in 281 BCE. One of his officers, Philetaerus, however, took over the city of Pergamon, establishing there the Attalid dynasty that transformed Pergamon into a small and successful kingdom. The final Attalid king, Attalus III, left his kingdom to Rome in his will in 133 BCE. Lasting from the death of Alexander in 323 BCE to the death of Cleopatra VII in 30 BCE, the Ptolemaic kingdom proved to be the longest lasting and most successful of the kingdoms carved from Alexander’s initial empire. Its founder, Ptolemy I Soter , was a talented general, as well as an astronomer, philosopher, and historian, who wrote his own histories of Alexander’s campaigns. Aiming to make Alexandria the new Athens of the Mediterranean, Ptolemy spared no expense in building the Museaum, an institution of learning and research that included, most famously, the Great Library, and worked tirelessly to attract scholars and cultured elite to his city. Subsequent Ptolemies continued these works so that Alexandria held its reputation as a cultural capital into Late Antiquity. One example of a particularly impressive scientific discovery is the work of Eratosthenes, the head librarian at the Great Library in the second half of the third century BCE, who accurately calculated the earth’s circumference. But while the Ptolemies brought with them Greek language and culture to Egypt, they were also profoundly influenced by Egyptian customs. Portraying themselves as the new Pharaohs, the Ptolemies even adopted the Egyptian royal custom of brother-sister marriages, a practice that eventually percolated down to the general populace as well. Unfortunately, brother-sister marriages did not prevent strife for power within the royal family. The last of the Ptolemaic rulers, Cleopatra VII, first married and ruled jointly with her brother Ptolemy XIII. After defeating him in a civil war, she then married another brother, Ptolemy XIV, remaining his wife until his death, possibly from sisterly poisoning. Best known for her affairs with Julius Caesar and, after Caesar’s death, with Marcus Antonius, Cleopatra teamed with Marcus Antonius in a bid for the Roman Empire. The last surviving ruler who was descended from one of Alexander’s generals, she was finally defeated by Octavian, the future Roman emperor Augustus, in 30 BCE. The history of the successor states that resulted from the carving of Alexander’s empire shows the imperialistic drive of Greek generals, while also demonstrating the instability of their empires. Historians do not typically engage in counter-factual speculations, but it is very likely that, had he lived longer, Alexander would have seen his empire unravel, as no structure was really in place to hold it together. At the same time, the clash of cultures that Alexander’s empire and the successor states produced resulted in the spread of Greek culture and language further than ever before; simultaneously, it also introduced the Greeks to other peoples, thus bringing foreign customs—such as the brother-sister marriages in Egypt—into the lives of the Greeks living outside the original Greek world. 5.11.2 Hellenistic Culture The Hellenistic kingdoms spread Greek language, culture, and art all over the areas of Alexander’s former conquests. Furthermore, many Hellenistic kings, especially the Ptolemies, were patrons of art and ideas. Thus the Hellenistic era saw the flourishing of art and architecture, philosophy, medical and scientific writing, and even translations of texts of other civilizations into Greek. The undisputed center for these advances was Alexandria. Combining the practical with the ambitious, the Pharos , or Lighthouse, of Alexandria was one of the most famous examples of Hellenistic architecture and has remained a symbol of the city to the present day. Constructed in 280 BCE, it was considered to be one of the Seven Wonders of the Ancient World and was one of the tallest buildings in the world at the time. While its practical purpose was to guide ships into the harbor at night, it also exemplified the bold advances and experimental spirit of Hellenistic architecture. Indeed, it was located on a man-made mole off the coast of the city. The building comprised three layers, the top one of which housed the furnace that produced the light. The structure of the Pharos shows an interest in straight lines and orderly shapes, while its function symbolized the ability of man to subdue the sea, even by night. Similarly, both the scientific and medical texts from the Hellenistic Period reveal a fascination with an ordered universe and an interest in discovering how it worked. Herophilus of Chalcedon , for instance, pioneered dissection in the early third BCE and was especially interested in the human brain and the nervous system. The mathematician Euclid, who lived and worked in Alexandria during the reign of Ptolemy I (323 – 283 BCE), wrote the Elements , an encyclopedic work of mathematics that effectively created the discipline of geometry. Going a step further than Euclid in his research, the third-century BCE scientist and inventor Archimedes of Syracuse specialized in applying mathematical concepts to create such devices as a screw pump and a variety of war machines, including the heat ray. The same fascination with studying the order of the universe appears in Hellenistic philosophy and stems ultimately from the philosophy of Aristotle (384 – 322 BCE), considered to be the last Classical Greek philosopher. Aristotle was a prolific polymath, who wrote on political theory, poetry, music, and a variety of sciences, to list just some of his interests. Engrossed in seeing all disciplines as part of a larger world order, Aristotle specifically argued for empiricism, that is, the belief that knowledge is acquired from sensory experiences rather than from intuition. In the sciences, for instance, this approach required experiments and the careful gathering of data. While Aristotle’s influence on the Hellenistic philosophers is undeniable, the alternate theories that some of the philosophers developed regarding the structure of the universe and the place of humanity in it differs drastically from Aristotle’s view. For instance, Skepticism , especially as formulated by Pyrrho in the third century BCE, argued that it was impossible to reach any accurate conclusions about the world and the key to happiness was to stop trying. Cynic philosophers , starting in the fourth century BCE, advocated the ascetic life of simplicity and freedom from possessions. A related philosophy, Stoicism , argued for letting go of all emotions and developing a self-control that would allow one to live in accordance to nature. On the other hand, the third-century philosophy of Epicureanism argued for the absence of pain as the ultimate goal in life and saw the universe as ruled by random chance, separate from the intervention of the gods. All of these philosophies, and many others that co-existed with them, aimed to provide a coherent system that made sense of the world and provided a purpose for human life. Finally, in a testament to the deep influence of the Hellenistic language culture on the conquered regions, the Hellenistic Period saw the translation of texts of other civilizations into Greek. One particularly influential example was the translation of the Hebrew Old Testament into Greek. Jews formed a significant minority of the population of Alexandria , the capital of Ptolemaic Egypt, as well as other major cities around the Mediterranean, such as Antioch. By the third century BCE, these Jews appear to have largely lost the knowledge of Hebrew; thus, a translation of the sacred texts into Greek was necessary. In addition, as later legend has it, Ptolemy II Philadelphus allegedly commissioned seventy-two scholars to translate the Old Testament into Greek for his Royal Library. Whether indeed solicited by Ptolemy II or not, the translation was likely completed over the course of the third throughfirst centuries BCE. Named after the legendary seventy-two (or, in some versions, seventy) translators, the text was titled the Septuagint . The completion of this translation showed the thorough Hellenization of even the Jews, who had largely kept themselves apart from mainstream culture of the cities in which they lived. 5.12 CONCLUSION “Captive Greece has conquered her rude conqueror,” the Roman poet Horace famously wrote in the late first century BCE. This comment about the deep influence of Greek culture on the Roman world, even after the Roman conquest of Greece was complete, continued to be the case well after the days of Horace. Ultimately, the impact of the Hellenization of the Mediterranean and the Middle East, which started with Alexander’s conquests, lasted far beyond the Hellenistic kingdoms, as the Greek language continued to be the language of the Eastern Roman Empire and, subsequently, the Byzantine Empire up until the conquest of that territory by the Ottomans in 1453 CE. In some respects, this spread of the Greeks and their civilization ultimately changed what it meant to be Greek–or, rather, it created a more universal Greek identity, which largely replaced the polis -specific view of citizenship and identity that existed before Philip’s conquest of Greece. And yet, certain cultural constants persisted. The first of these was Homer, whose epics continued to be as great an inspiration to the Greeks of the Roman world as they were to their Archaic Age counterparts. For instance, the Homeric values were likely the reason for the minimal advances in military technology in the Greek world, as honor was more important than military success at all cost. The second cultural constant was the work of the philosophers Plato and Aristotle, in whose shadows all subsequent philosophers of the Greco-Roman world labored. Even as the Greek-speaking portions of the Roman Empire turned to Christianity, they could not abandon their philosophical roots, resulting, for instance, in the Gnostic heresies. Horace’s cheeky comment thus proved to be true far longer than he could have expected. 5.13 WORKS CONSULTED AND FURTHER READING Burkert, Walter. The Orientalizing Revolution: Near Eastern Influence on Greek Culture in the Early Archaic Age . Cambridge: Harvard University Press, 1998. Bury, J. B. and Russell Meiggs. A History of Greece to the Death of Alexander the Great . New York: St. Martin’s Press. Cartledge, Paul. Sparta and Laconia: A Regional History 1300 – 362 BC . New York: Routledge, 2002. Dickinson, Oliver. The Aegean Bronze Age . Cambridge: Cambridge University Press, 1994. Gruen, Erich. Diaspora: Jews amidst Greeks and Romans . Cambridge: Harvard University Press, 2004. Hanson, Victor Davis. A War Like No Other: How the Athenians and Spartans Fought the Peloponnesian War . New York: Random House Trade Paperbacks, 2006. Horden, Peregrine and Nicholas Purcell, The Corrupting Sea: A Study of Mediterranean History . Oxford: Wiley-Blackwell, 2000. Kagan, Donald. The Archidamian War . Ithaca: Cornell University Press, 1990. ———. The Fall of the Athenian Empire. Ithaca: Cornell University Press, 1991. ———. The Outbreak of the Peloponnesian War . Ithaca: Cornell University Press, 1989. ———. The Peace of Nicias and the Sicilian Expedition . Ithaca: Cornell University Press, 1991. Lane Fox, Robin. Alexander the Great . New York: Penguin, 2004. Lendon, J. E. Soldiers and Ghosts: A History of Battle in Classical Antiquity . New Haven: Yale University Press, 2006. Lewis, Naphthali. Greeks in Ptolemaic Egypt . Durham: American Society of Papyrologists, 2001. Long, A. A. Hellenistic Philosophy: Stoics, Epicureans, Sceptics . Berkeley: University of California Press, 1986. Meiggs, Russell. The Athenian Empire . Oxford: Oxford University Press, 1979. Morkot, Robert. The Penguin Historical Atlas of Ancient Greece . London: Penguin Books, 1997. Murray, Oswyn. Early Greece . Cambridge: Harvard University Press, 1993. Ober, Josiah. Mass and Elite in Democratic Athens: Rhetoric, Ideology, and the Power of the People . Princeton: Princeton University Press, 1991. de Polignac, François. Cults, Territory, and the Origins of the Greek City-State . Chicago: University of Chicago Press, 1995. Pomeroy, Sarah, Stanley Burstein, Walter Donlan, Jennifer Tolbert Roberts, and David Tandy. Ancient Greece: A Political, Social, and Cultural History . Oxford: Oxford University Press, 2011. Walbank, F. W. The Hellenistic World . Cambridge: Harvard University Press, 1993. 5.14 LINKS TO PRIMARY SOURCES Ancient History Sourcebook at Fordham University (collection of materials from all periods of Greek history) http://legacy.fordham.edu/Halsall/ancient/asbook07.asp Archimedes’ Inventions http://archimedesjack.weebly.com/discoveries-and inventions.html Aristophanes, excerpt from Clouds , making fun of Socrates http://www.thenagain.info/Classes/Sources/Aristophanes-Clouds.htm Aristophanes, Lysistrata http://drama.eserver.org/plays/classical/aristophanes/lysistrata.txt Aristotle on Spartan Women http://legacy.fordham.edu/Halsall/ancient/aristotle-spartanwomen.asp Athens, photographic archive of archaeological materials http://web.archive.org/web/20040607030018/www.stoa.org/athens/ Cyrene, primary sources on founding of the colony http://legacy.fordham.edu/Halsall/ancient/630cyrene.asp Euripides, The Phoenician Women http://www.users.globalnet.co.uk/~loxias/phoenissae.htm The Gortyn Code, selections https://legacy.fordham.edu/halsall/ancient/450-gortyn.asp Herodotus, Histories , excerpt on the Battle of Marathon http://www.thenagain.info/Classes/Sources/Herodotus-Marathon.html History of Greek Philosophy http://faculty.washington.edu/smcohen/320/index.html http://www.uh.edu/~cfreelan/courses/riceanc.html Justin, on the beginning of the reign of Philip II of Macedon http://legacy.fordham.edu/Halsall/ancient/justin-philip.asp Linear B Writing http://www.ancientscripts.com/linearb.html Plutarch, Lives http://classics.mit.edu/Browse/browse-Plutarch.html Solon, selected fragments from political poems http://homepage.usask.ca/~jrp638/DeptTransls/Solon.html Sophocles, Antigone http://www.stoa.org/diotima/anthology/ant/antigstruct.htm Thucydides, Pericles’ “Funeral Oration” http://legacy.fordham.edu/Halsall/ancient/pericles-funeralspeech.asp Thucydides, “Melian Dialogue” from the History of the Peloponnesian War https://www.mtholyoke.edu/acad/intrel/melian.htm Tyrants, primary sources documents https://legacy.fordham.edu/Halsall/ancient/650tyranny.asp Xenophon on the Battle of Leuctra http://legacy.fordham.edu/Halsall/ancient/371leuctra.asp Xenophon on the Spartan Constitution http://legacy.fordham.edu/Halsall/ancient/xeno-sparta1.asp Peregrine Horden and Nicholas Purcell, The Corrupting Sea: A Study of Mediterranean History (Oxford: Wiley-Blackwell, 2000). ↵ Kurt Raaflaub, “A Historian’s Deadache: How to Read ‘Homeric Society’?” in N. Fisher and H. Van wees eds., Archaic Greece: New Approaches and New Evidence (London: Duckworth: 1998). ↵ CC licensed content, Shared previously Authored by : Nadejda Williams. Provided by : World History - Cultures, States, and Societies to 1500. Located at : http://oer.galileo.usg.edu/history-textbooks/2/?utm_source=oer.galileo.usg.edu%2Fhistory-textbooks%2F2&utm_medium=PDF&utm_campaign=PDFCoverPages . License : CC BY-SA: Attribution-ShareAlike
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Courses/Case_Western_Reserve_University/CHEM_121%3A_Concepts_for_a_Molecular_View_of_Biology_II_(Cunningham)/1%3A_Organic_Chemistry_Basics/1.13%3A_Chemical_Properties_of_Alkenes
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Learning Objectives To write equations for the addition reactions of alkenes with hydrogen, halogens, and water Alkenes are valued mainly for addition reactions, in which one of the bonds in the double bond is broken. Each of the carbon atoms in the bond can then attach another atom or group while remaining joined to each other by a single bond. Perhaps the simplest addition reaction ishydrogenation—a reaction with hydrogen (H 2 ) in the presence of a catalyst such as nickel (Ni) or platinum (Pt). The product is an alkane having the same carbon skeleton as the alkene. Alkenes also readily undergo halogenation —the addition of halogens. Indeed, the reaction with bromine (Br 2 ) can be used to test for alkenes. Bromine solutions are brownish red. When we add a Br 2 solution to an alkene, the color of the solution disappears because the alkene reacts with the bromine: Another important addition reaction is that between an alkene and water to form an alcohol. This reaction, called hydration , requires a catalyst—usually a strong acid, such as sulfuric acid (H 2 SO 4 ): The hydration reaction is discussed later, where we deal with this reaction in the synthesis of alcohols. Example \(\PageIndex{1}\) Write the equation for the reaction between CH 3 CH=CHCH 3 and each substance. H 2 (Ni catalyst) Br 2 H 2 O (H 2 SO 4 catalyst) SOLUTION In each reaction, the reagent adds across the double bond. Exercise \(\PageIndex{1}\) Write the equation for each reaction. CH 3 CH 2 CH=CH 2 with H 2 (Ni catalyst) CH 3 CH=CH 2 with Cl 2 CH 3 CH 2 CH=CHCH 2 CH 3 with H 2 O (H 2 SO 4 catalyst) Concept Review Exercises What is the principal difference in properties between alkenes and alkanes? How are they alike? If C 12 H 24 reacts with HBr in an addition reaction, what is the molecular formula of the product? Answers Alkenes undergo addition reactions; alkanes do not. Both burn. C 12 H 24 Br 2 Key Takeaway Alkenes undergo addition reactions, adding such substances as hydrogen, bromine, and water across the carbon-to-carbon double bond. Exercises Complete each equation. (CH 3 ) 2 C=CH 2 + Br 2 → \(\mathrm{CH_2\textrm{=C}(CH_3)CH_2CH_3 + H_2 \xrightarrow{Ni}}\) Complete each equation. \(\mathrm{CH_2\textrm{=CHCH=C}H_2 + 2H_2\xrightarrow{Ni}}\) \(\mathrm{(CH_3)_2\textrm{C=C}(CH_3)_2 + H_2O \xrightarrow{H_2SO_4}}\) Answer (CH 3 ) 2 CBrCH 2 Br CH 3 CH(CH 3 )CH 2 CH 3
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Courses/Athabasca_University/Chemistry_350%3A_Organic_Chemistry_I/07%3A_Alkenes-_Structure_and_Reactivity/7.10%3A_The_Hammond_Postulate
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Objective After completing this section, you should be able to use the Hammond postulate to explain the formation of the most stable carbocation during the addition of a protic acid, HX, to an alkene. Key Terms Make certain that you can define, and use in context, the key term below. Hammond postulate Now, back to transition states. Chemists are often very interested in trying to learn about what the transition state for a given reaction looks like, but addressing this question requires an indirect approach because the transition state itself cannot be observed. In order to gain some insight into what a particular transition state looks like, chemists often invoke the Hammond postulate , which states that a transition state resembles the structure of the nearest stable species . For an exergonic reaction, therefore, the transition state resembles the reactants more than it does the products. If we consider a hypothetical exergonic reaction between compounds A and B to form AB, the distance between A and B would be relatively large at the transition state, resembling the starting state where A and B are two isolated species. In the hypothetical endergonic reaction between C and D to form CD, however, the bond formation process would be much further along at the TS point, resembling the product. The Hammond Postulate is a very simplistic idea, which relies on an assumption that potential energy surfaces are parabolic. Although such an assumption is not rigorously true, it is fairly reliable and allows chemists to make energetic arguments about transition states by employing arguments about the stability of a related species. Since the formation of a reactive intermediate is very reliably endergonic , arguments about the stability of reactive intermediates can serve as proxy arguments about transition state stability. The Hammond Postulate and the S N 1 Reaction the Hammond postulate suggests that the activation energy of the rate-determining first step will be inversely proportional to the stability of the carbocation intermediate. The stability of carbocations was discussed earlier, and a qualitative relationship is given below: 0 1 2 3 4 5 6 7 8 9 10 11 Carbocation Stability CH3(+) < CH3CH2(+) < (CH3)2CH(+) ≈ CH2=CH-CH2(+) < C6H5CH2(+) ≈ (CH3)3C(+) Consequently, we expect that 3º-alkyl halides will be more reactive than their 2º and 1º-counterparts in reactions that follow an S N 1 mechanism. This is opposite to the reactivity order observed for the S N 2 mechanism. Allylic and benzylic halides are exceptionally reactive by either mechanism. Exercise Exercise \(\PageIndex{1}\) Consider the second step in the electrophilic addition of HBr to an alkene. Is this step exergonic or endergonic and does the transition state resemble the product or the reactant (cation)? Draw out an energy diagram of this step reaction. Answer As shown, the second step going from intermediate cation (A) to final product (C) it is exergonic. The transition state (B) resembles more the reactant (intermediate cation - A).
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Courses/Pasadena_City_College/CHEM_001A%3A_General_Chemistry_and_Chemical_Analysis/13%3A_Solids_and_Modern_Materials/13.07%3A_Ceramics_Cement_and_Glass
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0 1 NaN Howard University General Chemistry: An Atoms First Approach Unit 1: Atomic Theory Unit 2: Molecular Structure Unit 3: Stoichiometry Unit 4: Thermochem & Gases Unit 5: States of Matter Unit 6: Kinetics & Equilibria Unit 7: Electro & Thermo Chemistry Unit 8: Materials Unit 1: Atomic Theory Unit 2: Molecular Structure Unit 3: Stoichiometry Unit 4: Thermochem & Gases Unit 5: States of Matter Unit 6: Kinetics & Equilibria Unit 7: Electro & Thermo Chemistry Unit 8: Materials Some liquids become extremely viscous as the temperature falls toward their freezing points, often because they consist of macromolecules. An example is quartz, SiO 2 , seen . When quartz melts (at 1610°C), a few Si—O bonds break, but most remain intact. The liquid contains large covalently bonded fragments of the original structure and is highly viscous. When the liquid is cooled, the macromolecular fragments cannot readily slide past one another to attain the regular solid structure of quartz. Instead, a collection of interconnected, randomly oriented tetrahedrons of oxygen atoms surrounding silicon atoms is formed, as shown in the figure below. The material having this structure is known as fused silica . Fused silica is an example of an amorphous material or glass . It is highly rigid at room temperature, but it does not have the long-range microscopic regularity of a solid crystal lattice. Consequently it cannot be made to cleave along a plane. Instead, like ordinary window glass, it shatters into irregular fragments when struck sharply. (Window glass is primarily silica, but oxides of sodium and calcium are added to lower the melting point.) Since the microscopic structure of a glass is random, like that of a liquid, scientific purists describe glasses as highly viscous liquids, not as solids. To the left is an example of fused silica , which has many industry applications due to its high purity, high melting temperature, and high radiation resistance.
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Courses/can/CHEM_210_General_Chemistry_I_(Puenzo)/02%3A_Atoms_and_Elements/2.09%3A_Molar_Mass_-_Counting_Atoms_by_Weighing_Them
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Learning Objectives Make sure you thoroughly understand the following essential ideas: Define Avogadro's number and explain why it is important to know. Define the mole . Be able to calculate the number of moles in a given mass of a substance, or the mass corresponding to a given number of moles. Define molecular weight , formula weight , and molar mass ; explain how the latter differs from the first two. Be able to find the number of atoms or molecules in a given weight of a substance. Find the molar volume of a solid or liquid, given its density and molar mass. Explain how the molar volume of a metallic solid can lead to an estimate of atomic diameter. The chemical changes we observe always involve discrete numbers of atoms that rearrange themselves into new configurations. These numbers are HUGE— far too large in magnitude for us to count or even visualize, but they are still numbers , and we need to have a way to deal with them. We also need a bridge between these numbers, which we are unable to measure directly, and the weights of substances, which we do measure and observe. The mole concept provides this bridge, and is central to all of quantitative chemistry. Counting Atoms: Avogadro's Number Owing to their tiny size, atoms and molecules cannot be counted by direct observation. But much as we do when "counting" beans in a jar, we can estimate the number of particles in a sample of an element or compound if we have some idea of the volume occupied by each particle and the volume of the container. Once this has been done, we know the number of formula units (to use the most general term for any combination of atoms we wish to define) in any arbitrary weight of the substance. The number will of course depend both on the formula of the substance and on the weight of the sample. However, if we consider a weight of substance that is the same as its formula (molecular) weight expressed in grams, we have only one number to know: Avogadro's number . Avogadro's number Avogadro's number is known to ten significant digits: \[N_A = 6.022141527 \times 10^{23}.\] However, you only need to know it to three significant figures: \[N_A \approx 6.02 \times 10^{23}. \label{3.2.1}\] So \(6.02 \times 10^{23}\) of what ? Well, of anything you like: apples, stars in the sky, burritos. However, the only practical use for \(N_A\) is to have a more convenient way of expressing the huge numbers of the tiny particles such as atoms or molecules that we deal with in chemistry. Avogadro's number is a collective number , just like a dozen. Students can think of \(6.02 \times 10^{23}\) as the "chemist's dozen". Before getting into the use of Avogadro's number in problems, take a moment to convince yourself of the reasoning embodied in the following examples. Example \(\PageIndex{1}\): Mass ratio from atomic weights The atomic weights of oxygen and carbon are 16.0 and 12.0 atomic mass units (\(u\)), respectively. How much heavier is the oxygen atom in relation to carbon? Solution Atomic weights represent the relative masses of different kinds of atoms. This means that the atom of oxygen has a mass that is \[\dfrac{16\, \cancel{u}}{12\, \cancel{u}} = \dfrac{4}{3} ≈ 1.33 \nonumber\] as great as the mass of a carbon atom. Example \(\PageIndex{2}\): Mass of a single atom The absolute mass of a carbon atom is 12.0 unified atomic mass units (\(u\)). How many grams will a single oxygen atom weigh? Solution The absolute mass of a carbon atom is 12.0 \(u\) or \[12\,\cancel{u} \times \dfrac{1.6605 \times 10^{–24}\, g}{1 \,\cancel{u}} = 1.99 \times 10^{–23} \, g \text{ (per carbon atom)} \nonumber\] The mass of the oxygen atom will be 4/3 greater (from Example \(\PageIndex{1}\)): \[ \left( \dfrac{4}{3} \right) 1.99 \times 10^{–23} \, g = 2.66 \times 10^{–23} \, g \text{ (per oxygen atom)} \nonumber\] Alternatively we can do the calculation directly like with carbon: \[16\,\cancel{u} \times \dfrac{1.6605 \times 10^{–24}\, g}{1 \,\cancel{u}} = 2.66 \times 10^{–23} \, g \text{ (per oxygen atom)} \nonumber\] Example \(\PageIndex{3}\): Relative masses from atomic weights Suppose that we have \(N\) carbon atoms, where \(N\) is a number large enough to give us a pile of carbon atoms whose mass is 12.0 grams. How much would the same number, \(N\), of oxygen atoms weigh? Solution We use the results from Example \(\PageIndex{1}\) again. The collection of \(N\) oxygen atoms would have a mass of \[\dfrac{4}{3} \times 12\, g = 16.0\, g. \nonumber\] Exercise \(\PageIndex{1}\) What is the numerical value of \(N\) in Example \(\PageIndex{3}\)? Answer Using the results of Examples \(\PageIndex{2}\) and \(\PageIndex{3}\). \[N \times 1.99 \times 10^{–23} \, g \text{ (per carbon atom)} = 12\, g \nonumber\] or \[N = \dfrac{12\, \cancel{g}}{1.99 \times 10^{–23} \, \cancel{g} \text{ (per carbon atom)}} = 6.03 \times 10^{23} \text{atoms} \nonumber \] There are a lot of atoms in 12 g of carbon. Things to understand about Avogadro's number It is a number , just as is "dozen", and thus is dimensionless . It is a huge number, far greater in magnitude than we can visualize Its practical use is limited to counting tiny things like atoms, molecules, "formula units", electrons, or photons. The value of N A can be known only to the precision that the number of atoms in a measurable weight of a substance can be estimated. Because large numbers of atoms cannot be counted directly, a variety of ingenious indirect measurements have been made involving such things as Brownian motion and X-ray scattering . The current value was determined by measuring the distances between the atoms of silicon in an ultrapure crystal of this element that was shaped into a perfect sphere. (The measurement was made by X-ray scattering.) When combined with the measured mass of this sphere, it yields Avogadro's number. However, there are two problems with this: The silicon sphere is an artifact, rather than being something that occurs in nature, and thus may not be perfectly reproducible. The standard of mass, the kilogram, is not precisely known, and its value appears to be changing. For these reasons, there are proposals to revise the definitions of both N A and the kilogram. The Mole The identity of a substance is defined not only by the types of atoms or ions it contains, but by the quantity of each type of atom or ion. For example, water, H2OH2O, and hydrogen peroxide, H2O2H2O2, are alike in that their respective molecules are composed of hydrogen and oxygen atoms. However, because a hydrogen peroxide molecule contains two oxygen atoms, as opposed to the water molecule, which has only one, the two substances exhibit very different properties. Today, we possess sophisticated instruments that allow the direct measurement of these defining microscopic traits; however, the same traits were originally derived from the measurement of macroscopic properties (the masses and volumes of bulk quantities of matter) using relatively simple tools (balances and volumetric glassware). This experimental approach required the introduction of a new unit for amount of substances, the mole , which remains indispensable in modern chemical science. The mole is an amount unit similar to familiar units like pair, dozen, gross, etc. It provides a specific measure of the number of atoms or molecules in a bulk sample of matter. A mole is defined as the amount of substance containing the same number of discrete entities (such as atoms, molecules, and ions) as the number of atoms in a sample of pure 12 C weighing exactly 12 g. One Latin connotation for the word “mole” is “large mass” or “bulk,” which is consistent with its use as the name for this unit. The mole provides a link between an easily measured macroscopic property, bulk mass, and an extremely important fundamental property, number of atoms, molecules, and so forth. The number of entities composing a mole has been experimentally determined to be 6.02214179×10236.02214179×1023, a fundamental constant named Avogadro’s number (NANA) or the Avogadro constant in honor of Italian scientist Amedeo Avogadro. This constant is properly reported with an explicit unit of “per mole,” a conveniently rounded version being 6.022×1023/mol6.022×1023/mol. Consistent with its definition as an amount unit, 1 mole of any element contains the same number of atoms as 1 mole of any other element. The masses of 1 mole of different elements, however, are different, since the masses of the individual atoms are drastically different. The molar mass of an element (or compound) is the mass in grams of 1 mole of that substance, a property expressed in units of grams per mole (g/mol) (Figure 3.9.43.9.4). Because the definitions of both the mole and the atomic mass unit are based on the same reference substance, 12 C, the molar mass of any substance is numerically equivalent to its atomic or formula weight in amu. Per the amu definition, a single 12 C atom weighs 12 amu (its atomic mass is 12 amu). According to the definition of the mole, 12 g of 12 C contains 1 mole of 12 C atoms (its molar mass is 12 g/mol). This relationship holds for all elements, since their atomic masses are measured relative to that of the amu-reference substance, 12 C. Extending this principle, the molar mass of a compound in grams is likewise numerically equivalent to its formula mass in amu (Figure 3.9.53.9.5). Moles and their Uses The mole (abbreviated mol) is the the SI measure of quantity of a "chemical entity" , which can be an atom, molecule, formula unit, electron or photon. One mole of anything is just Avogadro's number of that something. Or, if you think like a lawyer, you might prefer the official SI definition: Definition: The Mole The mole is the amount of substance of a system which contains as many elementary entities as there are atoms in 0.012 kilogram of carbon 12 Avogadro's number (Equation \ref{3.2.1}) like any pure number, is dimensionless. However, it also defines the mole, so we can also express N A as 6.02 × 10 23 mol –1 ; in this form, it is properly known as Avogadro's constant . This construction emphasizes the role of Avogadro's number as a conversion factor between number of moles and number of "entities". Example \(\PageIndex{4}\): number of moles in N particles How many moles of nickel atoms are there in 80 nickel atoms? Solution \[\dfrac{80 \;atoms}{6.02 \times 10^{23} \; atoms\; mol^{-1}} = 1.33 \times 10^{-22} mol \nonumber\] Is this answer reasonable? Yes, because 80 is an extremely small fraction of \(N_A\). Molar Mass The atomic weight, molecular weight, or formula weight of one mole of the fundamental units (atoms, molecules, or groups of atoms that correspond to the formula of a pure substance) is the ratio of its mass to 1/12 the mass of one mole of C 12 atoms, and being a ratio, is dimensionless. But at the same time, this molar mass (as many now prefer to call it) is also the observable mass of one mole ( N A ) of the substance, so we frequently emphasize this by stating it explicitly as so many grams (or kilograms) per mole: g mol –1 . It is important always to bear in mind that the mole is a number and not a mass . But each individual particle has a mass of its own, so a mole of any specific substance will always correspond to a certain mass of that substance. Example \(\PageIndex{5}\): Boron content of borax Borax is the common name of sodium tetraborate, \(\ce{Na2B4O7}\). how many moles of boron are present in 20.0 g of borax? how many grams of boron are present in 20.0 g of borax? Solution The formula weight of \(\ce{Na2B4O7}\) so the molecular weight is: \[(2 \times 23.0) + (4 \times 10.8) + (7 \times 16.0) = 201.2 \nonumber\] 20 g of borax contains (20.0 g) ÷ (201 g mol –1 ) = 0.10 mol of borax, and thus 0.40 mol of B. 0.40 mol of boron has a mass of (0.40 mol) × (10.8 g mol –1 ) = 4.3 g . Example \(\PageIndex{6}\): Magnesium in chlorophyll The plant photosynthetic pigment chlorophyll contains 2.68 percent magnesium by weight. How many atoms of Mg will there be in 1.00 g of chlorophyll? Solution Each gram of chlorophyll contains 0.0268 g of Mg, atomic weight 24.3. Number of moles in this weight of Mg: (.0268 g) / (24.2 g mol –1 ) = 0.00110 mol Number of atoms: (0.00110 mol) × (6.02E23 mol –1 ) = \(6.64 \times 10^{20}\) Is this answer reasonable? (Always be suspicious of huge-number answers!) Yes, because we would expect to have huge numbers of atoms in any observable quantity of a substance. Molar Volume This is the volume occupied by one mole of a pure substance. Molar volume depends on the density of a substance and, like density, varies with temperature owing to thermal expansion, and also with the pressure. For solids and liquids, these variables ordinarily have little practical effect, so the values quoted for 1 atm pressure and 25°C are generally useful over a fairly wide range of conditions. This is definitely not the case with gases, whose molar volumes must be calculated for a specific temperature and pressure. Example \(\PageIndex{7}\): Molar Volume of a Liquid Methanol, CH 3 OH, is a liquid having a density of 0.79 g per milliliter. Calculate the molar volume of methanol. Solution The molar volume will be the volume occupied by one molar mass (32 g) of the liquid. Expressing the density in liters instead of mL, we have \[V_M = \dfrac{32\; g\; mol^{–1}}{790\; g\; L^{–1}}= 0.0405 \;L \;mol^{–1} \nonumber\] The molar volume of a metallic element allows one to estimate the size of the atom. The idea is to mentally divide a piece of the metal into as many little cubic boxes as there are atoms, and then calculate the length of each box. Assuming that an atom sits in the center of each box and that each atom is in direct contact with its six neighbors (two along each dimension), this gives the diameter of the atom. The manner in which atoms pack together in actual metallic crystals is usually more complicated than this and it varies from metal to metal, so this calculation only provides an approximate value. Example \(\PageIndex{8}\): Radius of a Strontium Atom The density of metallic strontium is 2.60 g cm –3 . Use this value to estimate the radius of the atom of Sr, whose atomic weight is 87.6. Solution The molar volume of Sr is: \[\dfrac{87.6 \; g \; mol^{-1}}{2.60\; g\; cm^{-3}} = 33.7\; cm^3\; mol^{–1}\] The volume of each "box" is" \[\dfrac{33.7\; cm^3 mol^{–1}} {6.02 \times 10^{23}\; mol^{–1}} = 5.48 \times 10^{-23}\; cm^3\] The side length of each box will be the cube root of this value, \(3.79 \times 10^{–8}\; cm\). The atomic radius will be half this value, or \[1.9 \times 10^{–8}\; cm = 1.9 \times 10^{–10}\; m = 190 pm\] Note : Your calculator probably has no cube root button, but you are expected to be able to find cube roots; you can usually use the x y button with y =0.333. You should also be able estimate the magnitude of this value for checking. The easiest way is to express the number so that the exponent is a multiple of 3. Take \(54 \times 10^{-24}\), for example. Since 3 3 =27 and 4 3 = 64, you know that the cube root of 55 will be between 3 and 4, so the cube root should be a bit less than 4 × 10 –8 . So how good is our atomic radius? Standard tables give the atomic radius of strontium is in the range 192-220 pm.
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Bookshelves/Analytical_Chemistry/Analytical_Chemistry_2.1_(Harvey)/15%3A_Quality_Assurance/15.03%3A_Quality_Assessment
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The written directives of a quality control program are a necessary, but not a sufficient condition for obtaining and maintaining a state of statistical control. Although quality control directives explain how to conduct an analysis, they do not indicate whether the system is under statistical control. This is the role of quality assessment , the second component of a quality assurance program. The goals of quality assessment are to determine when an analysis has reached a state of statistical control, to detect when an analysis falls out of statistical control, and to suggest possible reasons for this loss of statistical control. For convenience, we divide quality assessment into two categories: internal methods coordinated within the laboratory, and external methods organized and maintained by an outside agency. Internal Methods of Quality Assessment The most useful methods for quality assessment are those coordinated by the laboratory, which provide immediate feedback about the analytical method’s state of statistical control. Internal methods of quality assessment include the analysis of duplicate samples, the analysis of blanks, the analysis of standard samples, and spike recoveries. Analysis of Duplicate Samples An effective method for determining the precision of an analysis is to analyze duplicate samples . Duplicate samples are obtained by dividing a single gross sample into two parts, although in some cases the duplicate samples are independently collected gross samples. We report the results for the duplicate samples, X 1 and X 2 , by determining the difference, d , or the relative difference, ( d ) r , between the two samples \[d = X_1 - X_2 \nonumber\] \[(d)_r = \frac {d} {(X_1 + X_2)/2} \times 100 \nonumber\] and comparing to an accepted value, such as those in Table 15.3.1 for the analysis of waters and wastewaters. Alternatively, we can estimate the standard deviation using the results for a set of n duplicates \[s = \sqrt{\frac {\sum_{i = 1}^n d_i^2} {2n}} \nonumber\] where d i is the difference between the i th pair of duplicates. The degrees of freedom for the standard deviation is the same as the number of duplicate samples. If we combine duplicate samples from several sources, then the precision of the measurement process must be approximately the same for each. analyte \((d)_r\): [analyte] < 20 \(\times\) MDL (\(\pm\)%) \((d)_r\): [analyte] > 20 \(\times\) MDL (\(\pm\)%) spike recovery limit (%) acids 40 20 \(\times\) MDL (\(\pm\)%)">20 60–140 anions 25 20 \(\times\) MDL (\(\pm\)%)">10 80–120 bases or neutrals 40 20 \(\times\) MDL (\(\pm\)%)">20 70–130 carbamate pesticides 40 20 \(\times\) MDL (\(\pm\)%)">20 50–150 herbicides 40 20 \(\times\) MDL (\(\pm\)%)">20 40–160 metals 25 20 \(\times\) MDL (\(\pm\)%)">10 80–120 other inorganics 25 20 \(\times\) MDL (\(\pm\)%)">10 80–120 volatile organics 40 20 \(\times\) MDL (\(\pm\)%)">20 70–130 Abbreviation: MDL = method's detection limit Source: Table 1020.1 in Standard Methods for the Analysis of Water and Wastewater, American Public Health Association: Washington, D. C., 18th Ed. 1992. Abbreviation: MDL = method's detection limit Source: Table 1020.1 in Standard Methods for the Analysis of Water and Wastewater, American Public Health Association: Washington, D. C., 18th Ed. 1992. Abbreviation: MDL = method's detection limit Source: Table 1020.1 in Standard Methods for the Analysis of Water and Wastewater, American Public Health Association: Washington, D. C., 18th Ed. 1992. Abbreviation: MDL = method's detection limit Source: Table 1020.1 in Standard Methods for the Analysis of Water and Wastewater, American Public Health Association: Washington, D. C., 18th Ed. 1992. To evaluate the precision for the determination of potassium in blood serum, duplicate analyses were performed on six samples, yielding the following results in mg K/L. duplicate \(X_1\) \(X_2\) 1 160 147 2 196 202 3 207 196 4 185 193 5 172 188 6 133 119 Estimate the standard deviation for the analysis. Solution To estimate the standard deviation we first calculate the difference, \(d\), and the squared difference, \(d^{2}\), for each duplicate. The results of these calculations are summarized in the following table. duplicate \(d=X_{1}-X_{2}\) \(d^{2}\) 1 13 169 2 –6 36 3 11 121 4 –8 64 5 –16 256 6 14 196 Finally, we calculate the standard deviation \[s=\sqrt{\frac{169+36+121+64+256+196}{2 \times 6}}=8.4 \nonumber\] To evaluate the precision of a glucometer—a device a patient uses at home to monitor his or her blood glucose level—duplicate analyses are performed on samples drawn from five individuals, yielding the following results in mg glucose/100 mL. duplicate \(X_1\) \(X_2\) 1 148.5 149.1 2 96.5 98.8 3 174.9 174.5 4 118.1 118.9 5 72.7 70.4 Estimate the standard deviation for the analysis. Answer To estimate the standard deviation we first calculate the difference, d , and the squared difference, \(d^{2}\), for each duplicate. The results of these calculations are summarized in the following table. duplicate \(d=X_{1}-X_{2}\) \(d^{2}\) 1 –0.6 0.36 2 –2.3 5.29 3 0.4 0.16 4 –0.8 0.64 5 2.3 5.29 Finally, we calculate the standard deviation. \[s=\sqrt{\frac{0.36+5.29+0.16+0.64+5.29}{2 \times 5}}=1.08 \nonumber\] Analysis of Blanks We introduced the use of a blank in Chapter 3 as a way to correct the signal for contributions from sources other than the analyte. The most common blank is a method blank in which we take an analyte free sample through the analysis using the same reagents, glassware, and instrumentation. A method blank allows us to identify and to correct systematic errors due to impurities in the reagents, contaminated glassware, and poorly calibrated instrumentation. At a minimum, a new method blank is analyzed whenever we prepare a new reagent, or after we analyze a sample with a high concentration of analyte as residual carryover of analyte may produce a positive determinate error. When we collect samples in the field, additional blanks are needed to correct for potential sampling errors [Keith, L. H. Environmental Sampling and Analysis: A Practical Guide , Lewis Publishers: Chelsea, MI, 1991]. A field blank is an analyte-free sample carried from the laboratory to the sampling site. At the sampling site the blank is transferred to a clean sample container, which exposes it to the local environment. The field blank is then preserved and transported back to the laboratory for analysis. A field blank helps identify systematic errors due to sampling, transport, and analysis. A trip blank is an analyte-free sample carried from the laboratory to the sampling site and back to the laboratory without being opened. A trip blank helps to identify systematic errors due to cross-contamination of volatile organic compounds during transport, handling, storage, and analysis. A method blank also is called a reagent blank. The contamination of reagents over time is a significant concern. The regular use of a method blank compensates for this contamination. Analysis of Standards Another tool for monitoring an analytical method’s state of statistical control is to analyze a standard that contains a known concentration of analyte. A standard reference material (SRM) is the ideal choice, provided that the SRM’s matrix is similar to that of our samples. A variety of SRMs are available from the National Institute of Standards and Technology (NIST). If a suitable SRM is not available, then we can use an independently prepared synthetic sample if it is prepared from reagents of known purity. In all cases, the analyte’s experimentally determined concentration in the standard must fall within predetermined limits before the analysis is considered under statistical control. Table 4.2.6 in Chapter 4 provides a summary of SRM 2346, a standard sample of Gingko biloba leaves with certified values for the concentrations of flavonoids, terpene ketones, and toxic elements, such as mercury and lead. Spike Recoveries One of the most important quality assessment tools is the recovery of a known addition, or spike, of analyte to a method blank, a field blank, or a sample. To determine a spike recovery , the blank or sample is split into two portions and a known amount of a standard solution of analyte is added to one portion. The analyte’s concentration is determined for both the spiked, F , and unspiked portions, I , and the percent recovery, % R , is calculated as \[\% R=\frac{F-I}{A} \times 100 \nonumber\] where A is the concentration of analyte added to the spiked portion. A spike recovery for the analysis of chloride in well water was performed by adding 5.00 mL of a 250.0 ppm solution of Cl – to a 50-mL volumetric flask and diluting to volume with the sample. An unspiked sample was prepared by adding 5.00 mL of distilled water to a separate 50-mL volumetric flask and diluting to volume with the sample. Analysis of the sample and the spiked sample return chloride concentrations of 18.3 ppm and 40.9 ppm, respectively. Determine the spike recovery. Solution To calculate the concentration of the analyte added in the spike, we take into account the effect of dilution. \[A=250.0 \mathrm{ppm} \times \frac{5.00 \mathrm{mL}}{50.0 \mathrm{mL}}=25.0 \mathrm{ppm} \nonumber\] Thus, the spike recovery is \[\% R=\frac{40.9-18.3}{25.0} \times 100=90.4 \% \nonumber\] To test a glucometer, a spike recovery is carried out by measuring the amount of glucose in a sample of a patient’s blood before and after spiking it with a standard solution of glucose. Before spiking the sample the glucose level is 86.7 mg/100 mL and after spiking the sample it is 110.3 mg/100 mL. The spike is prepared by adding 10.0 μL of a 25 000 mg/100mL standard to a 10.0-mL portion of the blood. What is the spike recovery for this sample. Answer Adding a 10.0-μL spike to a 10.0-mL sample is a 1000-fold dilution; thus, the concentration of added glucose is 25.0 mg/100 mL and the spike recovery is \[\% R=\frac{110.3-86.7}{25.0} \times 100=94.4 \% \nonumber\] We can use a spike recovery on a method blank and a field blank to evaluate the general performance of an analytical procedure. A known concentration of analyte is added to each blank at a concentration that is 5 to 50 times the method’s detection limit. A systematic error during sampling and transport will result in an unacceptable recovery for the field blank, but not for the method blank. A systematic error in the laboratory, however, affects the recoveries for both the field blank and the method blank. Spike recoveries on a sample are used to detect systematic errors due to the sample’s matrix, or to evaluate the stability of a sample after its collection. Ideally, samples are spiked in the field at a concentration that is 1 to 10 times the analyte’s expected concentration or 5 to 50 times the method’s detection limit, whichever is larger. If the recovery for a field spike is unacceptable, then a duplicate sample is spiked in the laboratory and analyzed immediately. If the laboratory spike’s recovery is acceptable, then the poor recovery for the field spike likely is the result of the sample’s deterioration during storage. If the recovery for the laboratory spike also is unacceptable, the most probable cause is a matrix-dependent relationship between the analytical signal and the analyte’s concentration. In this case the sample is analyzed by the method of standard additions. Typical limits for spike recoveries for the analysis of waters and wastewaters are shown in Table 15.3.1 . Figure 15.4.1 , which we will discuss in the next section, illustrates the use of spike recoveries as part of a quality assessment program. External Methods of Quality Assessment Internal methods of quality assessment always carry some level of suspicion because there is a potential for bias in their execution and interpretation. For this reason, external methods of quality assessment also play an important role in a quality assurance program. One external method of quality assessment is the certification of a laboratory by a sponsoring agency. Certification of a lab is based on its successful analysis of a set of proficiency standards prepared by the sponsoring agency. For example, laboratories involved in environmental analyses may be required to analyze standard samples prepared by the Environmental Protection Agency. A second example of an external method of quality assessment is a laboratory’s voluntary participation in a collaborative test sponsored by a professional organization, such as the Association of Official Analytical Chemists. Finally, an individual contracting with a laboratory can perform his or her own external quality assessment by submitting blind duplicate samples and blind standards to the laboratory for analysis. If the results for the quality assessment samples are unacceptable, then there is good reason to question the laboratory’s results for other samples. See Chapter 14 for a more detailed description of collaborative testing.
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Courses/Los_Angeles_Trade_Technical_College/Chem_51/1.6%3A_Chemical_Reactions_and_Equations/6.02%3A_The_Chemical_Equation_-_Writing_and_Balancing
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Learning Objectives Define chemical equation . Identify the parts of a chemical equation. A chemical reaction expresses a chemical change. For example, one chemical property of hydrogen is that it will react with oxygen to make water. We can write that as follows: \[\text{hydrogen reacts with oxygen to make water}\] We can represent this chemical change more succinctly as \[\text{hydrogen} + \text{oxygen} → \text{water}\] where the + sign means that the two substances interact chemically with each other and the → symbol implies that a chemical reaction takes place. But substances can also be represented by chemical formulas. Remembering that hydrogen and oxygen both exist as diatomic molecules, we can rewrite our chemical change as \[H_2 + O_2 → H_2O\] This is an example of a chemical equation, which is a concise way of representing a chemical reaction . The initial substances are called reactants and the final substances are called products . Unfortunately, it is also an incomplete chemical equation. The law of conservation of matter says that matter cannot be created or destroyed. In chemical equations, the number of atoms of each element in the reactants must be the same as the number of atoms of each element in the products. If we count the number of hydrogen atoms in the reactants and products, we find two hydrogen atoms. But if we count the number of oxygen atoms in the reactants and products, we find that there are two oxygen atoms in the reactants but only one oxygen atom in the products. What can we do? Can we change the subscripts in the formula for water so that it has two oxygen atoms in it? No; you cannot change the formulas of individual substances because the chemical formula for a given substance is characteristic of that substance. What you can do, however, is to change the number of molecules that react or are produced. We do this one element at a time, going from one side of the reaction to the other, changing the number of molecules of a substance until all elements have the same number of atoms on each side. To accommodate the two oxygen atoms as reactants, let us assume that we have two water molecules as products: \[H_2 + O_2 → 2H_2O\] The 2 in front of the formula for water is called a coefficient . Now there is the same number of oxygen atoms in the reactants as there are in the product. But in satisfying the need for the same number of oxygen atoms on both sides of the reaction, we have also changed the number of hydrogen atoms on the product side, so the number of hydrogen atoms is no longer equal. No problem-simply go back to the reactant side of the equation and add a coefficient in front of the H 2 . The coefficient that works is 2: \[2H_2 + O_2 → 2H_2O\] There are now four hydrogen atoms in the reactants and also four atoms of hydrogen in the product. There are two oxygen atoms in the reactants and two atoms of oxygen in the product. The law of conservation of matter has been satisfied. When the reactants and products of a chemical equation have the same number of atoms of all elements present, we say that an equation is balanced . All proper chemical equations are balanced. If a substance does not have a coefficient written in front of it, it is assumed to be 1. Also, the convention is to use all whole numbers when balancing chemical equations. This sometimes makes us do a bit more "back and forth" work when balancing a chemical equation. Example \(\PageIndex{1}\): Write and balance the chemical equation for each given chemical reaction. Hydrogen and chlorine react to make HCl. Ethane, C 2 H 6 , reacts with oxygen to make carbon dioxide and water. Solution Let us start by simply writing a chemical equation in terms of the formulas of the substances, remembering that both elemental hydrogen and chlorine are diatomic: H 2 + Cl 2 → HCl There are two hydrogen atoms and two chlorine atoms in the reactants and one of each atom in the product. We can fix this by including the coefficient 2 on the product side: H 2 + Cl 2 → 2HCl Now there are two hydrogen atoms and two chlorine atoms on both sides of the chemical equation, so it is balanced. Start by writing the chemical equation in terms of the substances involved: C 2 H 6 + O 2 → CO 2 + H 2 O We have two carbon atoms on the left, so we need two carbon dioxide molecules on the product side, so that each side has two carbon atoms; that element is balanced. We have six hydrogen atoms in the reactants, so we need six hydrogen atoms in the products. We can get this by having three water molecules: C 2 H 6 + O 2 → 2CO 2 + 3H 2 O Now we have seven oxygen atoms in the products (four from the CO 2 and three from the H 2 O). That means we need seven oxygen atoms in the reactants. However, because oxygen is a diatomic molecule, we can only get an even number of oxygen atoms at a time. We can achieve this by multiplying the other coefficients by 2: 2C 2 H 6 + O 2 → 4CO 2 + 6H 2 O By multiplying everything else by 2, we do not unbalance the other elements, and we now get an even number of oxygen atoms in the product-14. We can get 14 oxygen atoms on the reactant side by having 7 oxygen molecules: 2C 2 H 6 + 7O 2 → 4CO 2 + 6H 2 O As a check, recount everything to determine that each side has the same number of atoms of each element. This chemical equation is now balanced. Exercise \(\PageIndex{1}\) Write and balance the chemical equation that represents nitrogen and hydrogen reacting to produce ammonia, NH 3 . Answer: N 2 + 3H 2 → 2NH 3 Many chemical equations also include phase labels for the substances: (s) for solid, (ℓ) for liquid, (g) for gas, and (aq) for aqueous (i.e., dissolved in water). Special conditions, such as temperature, may also be listed above the arrow. For example, \[2NaHCO_{3}(s)\overset{200^{\circ}C}{\rightarrow}Na_{2}CO_{3}(s)+CO_{2}(g)+H_{2}O(l)\] Key Takeaways A chemical equation is a concise description of a chemical reaction. Proper chemical equations are balanced.
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Courses/Bellingham_Technical_College/CHEM_110%3A_Bellingham_Technical_College/04%3A_Module_4/4.11%3A_Phase_Transitions_-_Melting_Boiling_and_Subliming
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Learning Objective Describe what happens during a phase change. Calculate the energy change needed for a phase change. Substances can change phase—often because of a temperature change. At low temperatures, most substances are solid; as the temperature increases, they become liquid; at higher temperatures still, they become gaseous. The process of a solid becoming a liquid is called melting (an older term that you may see sometimes is fusion ). The opposite process, a liquid becoming a solid, is called solidification . For any pure substance, the temperature at which melting occurs—known as the melting point —is a characteristic of that substance. It requires energy for a solid to melt into a liquid. Every pure substance has a certain amount of energy it needs to change from a solid to a liquid. This amount is called the enthalpy of fusion (or heat of fusion) of the substance, represented as Δ H fus . Some Δ H fus values are listed in Table \(\PageIndex{1}\); it is assumed that these values are for the melting point of the substance. Note that the unit of Δ H fus is kilojoules per mole, so we need to know the quantity of material to know how much energy is involved. The Δ H fus is always tabulated as a positive number. However, it can be used for both the melting and the solidification processes, minding that melting is always endothermic (so Δ H will be positive), while solidification is always exothermic (so Δ H will be negative). Substance (Melting Point) ΔHfus (kJ/mol) Water (0°C) 6.01 Aluminum (660°C) 10.70 Benzene (5.5°C) 9.95 Ethanol (−114.3°C) 5.02 Mercury (−38.8°C) 2.29 Example \(\PageIndex{1}\) What is the energy change when 45.7 g of \(\ce{H2O}\) melt at 0°C? Solution The \(ΔH_{fus}\) of \(\ce{H2O}\) is 6.01 kJ/mol. However, our quantity is given in units of grams, not moles, so the first step is to convert grams to moles using the molar mass of \(\ce{H_2O}\), which is 18.0 g/mol. Then we can use \(ΔH_{fus}\) as a conversion factor. Because the substance is melting, the process is endothermic, so the energy change will have a positive sign. \[45.7\cancel{g\: H_{2}O}\times \frac{1\cancel{mol\: H_{2}O}}{18.0\cancel{g}}\times \frac{6.01kJ}{\cancel{mol}}=15.3\,kJ \nonumber \nonumber \] Without a sign, the number is assumed to be positive. Exercise \(\PageIndex{1}\) What is the energy change when 108 g of \(\ce{C6H6}\) freeze at 5.5°C? Answer −13.8 kJ During melting, energy goes exclusively to changing the phase of a substance; it does not go into changing the temperature of a substance. Hence melting is an isothermal process because a substance stays at the same temperature. Only when all of a substance is melted does additional energy go to changing its temperature. What happens when a solid becomes a liquid? In a solid, individual particles are stuck in place because the intermolecular forces cannot be overcome by the energy of the particles. When more energy is supplied (e.g., by raising the temperature), there comes a point at which the particles have enough energy to move around, but not enough energy to separate. This is the liquid phase: particles are still in contact, but are able to move around each other. This explains why liquids can assume the shape of their containers: the particles move around and, under the influence of gravity, fill the lowest volume possible (unless the liquid is in a zero-gravity environment—see Figure \(\PageIndex{1}\). The phase change between a liquid and a gas has some similarities to the phase change between a solid and a liquid. At a certain temperature, the particles in a liquid have enough energy to become a gas. The process of a liquid becoming a gas is called boiling (or vaporization) , while the process of a gas becoming a liquid is called condensation . However, unlike the solid/liquid conversion process, the liquid/gas conversion process is noticeably affected by the surrounding pressure on the liquid because gases are strongly affected by pressure. This means that the temperature at which a liquid becomes a gas, the boiling point , can change with surrounding pressure. Therefore, we define the normal boiling point as the temperature at which a liquid changes to a gas when the surrounding pressure is exactly 1 atm, or 760 torr. Unless otherwise specified, it is assumed that a boiling point is for 1 atm of pressure. Like the solid/liquid phase change, the liquid/gas phase change involves energy. The amount of energy required to convert a liquid to a gas is called the enthalpy of vaporization (or heat of vaporization), represented as Δ H vap . Some Δ H vap values are listed in Table \(\PageIndex{2}\); it is assumed that these values are for the normal boiling point temperature of the substance, which is also given in the table. The unit for Δ H vap is also kilojoules per mole, so we need to know the quantity of material to know how much energy is involved. The Δ H vap is also always tabulated as a positive number. It can be used for both the boiling and the condensation processes as long as you keep in mind that boiling is always endothermic (so Δ H will be positive), while condensation is always exothermic (so Δ H will be negative). Substance (Normal Boiling Point) ΔHvap (kJ/mol) Water (100°C) 40.68 Bromine (59.5°C) 15.40 Benzene (80.1°C) 30.80 Ethanol (78.3°C) 38.60 Mercury (357°C) 59.23 Example \(\PageIndex{2}\) What is the energy change when 66.7 g of Br 2 (g) condense to a liquid at 59.5°C? Solution The Δ H vap of Br 2 is 15.4 kJ/mol. Even though this is a condensation process, we can still use the numerical value of Δ H vap as long as we realize that we must take energy out, so the Δ H value will be negative. To determine the magnitude of the energy change, we must first convert the amount of Br 2 to moles. Then we can use Δ H vap as a conversion factor. \[66.7\cancel{g\: Br_{2}}\times \frac{1\cancel{mol\: Br_{2}}}{159.8\cancel{g}}\times \frac{15.4kJ}{\cancel{mol}}=6.43\,kJ \nonumber \nonumber \] Because the process is exothermic, the actual value will be negative: Δ H = −6.43 kJ. Exercise \(\PageIndex{2}\) What is the energy change when 822 g of \(\ce{C2H5OH(ℓ)}\) boil at its normal boiling point of 78.3°C? Answer 689 kJ As with melting, the energy in boiling goes exclusively to changing the phase of a substance; it does not go into changing the temperature of a substance. So boiling is also an isothermal process. Only when all of a substance has boiled does any additional energy go to changing its temperature. What happens when a liquid becomes a gas? We have already established that a liquid is composed of particles in contact with each other. When a liquid becomes a gas, the particles separate from each other, with each particle going its own way in space. This is how gases tend to fill their containers. Indeed, in the gas phase most of the volume is empty space; only about 1/1,000th of the volume is actually taken up by matter (Figure \(\PageIndex{1}\)). It is this property of gases that explains why they can be compressed, a fact that is considered in Chapter 6. Under some circumstances, the solid phase can transition directly to the gas phase without going through a liquid phase, and a gas can directly become a solid. The solid-to-gas change is called sublimation , while the reverse process is called deposition. Sublimation is isothermal, like the other phase changes. There is a measurable energy change during sublimation—this energy change is called the enthalpy of sublimation , represented as Δ H sub . The relationship between the Δ H sub and the other enthalpy changes is as follows: \[ΔH_{sub} = ΔH_{fus} + ΔH_{vap}\nonumber \] As such, Δ H sub is not always tabulated because it can be simply calculated from Δ H fus and Δ H vap . There are several common examples of sublimation. A well-known product, dry ice, is actually solid CO 2 . Dry ice is dry because it sublimes, with the solid bypassing the liquid phase and going straight to the gas phase. The sublimation occurs at temperature of −77°C, so it must be handled with caution. If you have ever noticed that ice cubes in a freezer tend to get smaller over time, it is because the solid water is very slowly subliming. "Freezer burn" isn't actually a burn; it occurs when certain foods, such as meats, slowly lose solid water content because of sublimation. The food is still good, but looks unappetizing. Reducing the temperature of a freezer will slow the sublimation of solid water. Chemical equations can be used to represent a phase change. In such cases, it is crucial to use phase labels on the substances. For example, the chemical equation for the melting of ice to make liquid water is as follows: \[H_2O(s) → H_2O(ℓ)\nonumber \] No chemical change is taking place; however, a physical change is taking place. Summary Phase changes can occur between any two phases of matter. All phase changes occur with a simultaneous change in energy. All phase changes are isothermal.
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Courses/Harper_College/CHM_110%3A_Fundamentals_of_Chemistry/04%3A_Water/4.18%3A_The_pH_Scale
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Learning Objectives Define pH . Determine the pH of acidic and basic solutions. As we have seen, [H + ] and [OH − ] values can be markedly different from one aqueous solution to another. So chemists defined a new scale that succinctly indicates the concentrations of either of these two ions. pH is commonly defined as a logarithmic function of [H + ]: \[\text{pH} = −\log [\ce{H^{+}}] \label{1} \] pH is usually (but not always) between 0 and 14. Knowing the dependence of pH on [H + ], we can summarize as follows: If pH < 7, then the solution is acidic. If pH = 7, then the solution is neutral. If pH > 7, then the solution is basic. This is known as the pH scale and is the range of values from 0 to 14 that describes the acidity or basicity of a solution. You can use pH to quickly determine whether a given aqueous solution is acidic, basic, or neutral. Example \(\PageIndex{1}\) Label each solution as acidic, basic, or neutral based only on the stated pH. milk of magnesia, pH = 10.5 pure water, pH = 7 wine, pH = 3.0 Solution With a pH greater than 7, milk of magnesia is basic. (Milk of magnesia is largely Mg(OH) 2 .) Pure water, with a pH of 7, is neutral. With a pH of less than 7, wine is acidic. Exercise \(\PageIndex{1}\) Identify each substance as acidic, basic, or neutral based only on the stated pH. human blood, pH = 7.4 household ammonia, pH = 11.0 cherries, pH = 3.6 Answers basic basic acidic Table \(\PageIndex{1}\) gives the typical pH values of some common substances. Note that several food items are on the list, and most of them are acidic. Substance pH stomach acid 1.7 lemon juice 2.2 vinegar 2.9 soda 3.0 wine 3.5 coffee, black 5.0 milk 6.9 pure water 7.0 blood 7.4 seawater 8.5 milk of magnesia 10.5 ammonia solution 12.5 1.0 M NaOH 14.0 *Actual values may vary depending on conditions *Actual values may vary depending on conditions pH is a logarithmic scale. A solution that has a pH of 1.0 has 10 times the [H + ] as a solution with a pH of 2.0, which in turn has 10 times the [H + ] as a solution with a pH of 3.0 and so forth. Using the definition of pH (Equation \ref{1}), it is also possible to calculate [H + ] (and [OH − ]) from pH and vice versa. The general formula for determining [H + ] from pH is as follows: \[[\ce{H^{+}}] = 10^{−\text{pH}} \nonumber\] Key Takeaways pH is a logarithmic function of [H + ]. [H + ] can be calculated directly from pH. pOH is related to pH and can be easily calculated from pH.
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Bookshelves/Organic_Chemistry/Organic_Chemistry_I_(Liu)/06%3A_Structural_Identification_of_Organic_Compounds-_IR_and_NMR_Spectroscopy/6.08%3A_C_NMR_Spectroscopy
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For carbon element, the most abundant isotope 12 C (with ~99% natural abundance) does not have a nuclear magnetic moment, and thus is NMR-inactive. The C NMR is therefore based on the 13 C isotope, that accounts for about 1% of carbon atoms in nature and has a magnetic dipole moment just like a proton. The theories we have learned about 1 H NMR spectroscopy also applies to 13 C NMR, however with several important differences about the spectrum. The magnetic moment of a 13 C nucleus is much weaker than that of a proton, meaning that 13 C NMR signals are inherently much weaker than proton signals. This, combined with the low natural abundance of 13 C, means that it is much more difficult to observe carbon signals. Usually, sample with high concentration and large number of scans (thousands or more) are required in order to bring the signal-to-noise ratio down to acceptable levels for 13 C NMR spectra. Chemical Equivalent For carbons that are chemical equivalent, they only show one signal in 13 C NMR as like protons for 1 HNMR. So it is very important to be able to identify equivalent carbons in the structure, in order to interpret 13 C NMR spectrum correctly. Taking toluene as an example, there are five sets of different carbon atoms (shown in different colors), so there are five signals in the 13 C NMR spectrum of toluene. Chemical Shift 13 C nuclei has different value of g (the magnetogyric ratio) comparing to 1 H nuclei, so the resonance frequencies of 13 C nuclei is different to those of protons in the same applied field (referring to formula. 6.4 , in section 6.5 ). In an instrument with a 7.05 Tesla magnet, protons resonate at about 300 MHz, while carbons resonate at about 75 MHz. This allows us to look at 13 C signals using a completely separate ‘window’ of radio frequencies. Just like in 1 H NMR, tetramethylsilane (TMS) is also used as the standard compound in 13 C NMR experiments to define the 0 ppm, however it is the signal from the four equivalent carbon atoms in TMS that serves as the standard. Chemical shifts for 13 C nuclei in organic molecules are spread out over a much wider range of about 220 ppm ( see Table 6.3 ) . The chemical shift of a 13 C nucleus is influenced by essentially the same factors that influence the chemical shift a proton: the deshielding effect of electronegative atoms and anisotropy effects tend to shift signals downfield (higher resonance frequency, with higher chemical shifts). In addition, sp 2 hybridization results in a large downfield shift. The 13 C NMR signals for carbonyl carbons are generally the furthest downfield (170-220 ppm), due to both sp 2 hybridization and to the double bond to oxygen. Integration and Coupling in 13 C NMR Unlike 1 H NMR, the area under a 13 C NMR signal cannot easily be used to determine the number of carbons to which it corresponds. The signals for some types of carbons are inherently weaker than for other types, for example peaks corresponding to carbonyl carbons are much smaller than those for methyl or methylene (CH 2 ) peaks. For this reason, signal integration is generally not useful in 13 C NMR spectroscopy. Because of the low natural abundance of 13 C nuclei, the spin-spin coupling between two nonequivalent 13 C atoms is negligible. 13 C nuclei are coupled to nearby protons, however, which results in complicated spectra. For clarity, chemists generally use a technique called broadband decoupling , which essentially ‘turns off’ C-H coupling, resulting in a spectrum in which all carbon signals are singlets . Below is the proton-decoupled 13 C NMR spectrum of ethyl acetate in CDCl 3 ( Fig. 6.8a ), showing the expected four signals, one for each of the carbons. For our class purpose, 13 C NMR spectra are usually used as supporting information to confirm the structure of a compound. Below are 13 C NMR spectra for methylbenzene (common name toluene) and methyl methacrylate. Refer to Table 6.3 to match the spectra to the correct structure. Answers to Practice Questions Chapter 6
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Courses/De_Anza_College/CHEM_10%3A_Introduction_to_Chemistry_(Parajon_Puenzo)/04%3A_Nuclear_Chemistry/4.07%3A_Energy_from_the_Nucleus
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Learning Objectives Explain where nuclear energy comes from. Describe the difference between fission and fusion. Know key examples of nuclear fission and nuclear fusion. A combination of radiochemistry and radiation chemistry is used to study nuclear reactions such as fission and fusion. Some early evidence for nuclear fission was the formation of a short-lived radioisotope of barium which was isolated from neutron irradiated uranium ( 139 Ba, with a half-life of 83 minutes and 140 Ba, with a half-life of 12.8 days, are major fission products of uranium). Nuclear fission is the splitting of an atomic nucleus. In nuclear weapons and reactors, neutrons hit unstable nuclei to form smaller atoms. Nuclear Fusion is the bringing together of two atomic nuclei to form a larger atom. Figure \(\PageIndex{1}\) illustrates the difference between nuclear fission and nuclear fusion. Einstein and the Equivalence of Mass and Energy Nuclear changes occur with a simultaneous release of energy. Where does this energy come from? If we could precisely measure the masses of the reactants and products of a nuclear reaction, we would notice that the amount of mass drops slightly in the conversion from reactants to products. Consider the following nuclear equation, in which the molar mass of each species is indicated to four decimal places: \[_{235.0439}^{235}\textrm{U}\rightarrow _{138.9088}^{139}\textrm{Ba}+_{93.9343}^{94}\textrm{Kr}+_{2\times 1.0087}^{2^{1}}\textrm{n} \nonumber \] If we compare the mass of the reactant (235.0439) to the masses of the products (sum=234.8605) we notice a mass difference of -0.1834 g or -0.0001834 kg. Where did this mass go? According to Albert Einstein's theory of relativity, energy ( E ) and mass ( m ) are related by the following equation: \[E = mc^2 \nonumber \] where c is the speed of light, or \[c=3.00\times 10^{8}\, m/s \nonumber \] In the course of the chemical reaction for uranium, the mass difference is converted to energy, which is given off by the reaction: \[E=(-0.0001834\, kg)(3.00\times 10^{8}\, m/s)^{2}=-1.65\times 10^{13}J=-1.65\times 10^{10}kJ \nonumber \] (For the units to work out, mass must be expressed in units of kilograms.) That is, 16.5 billion kJ of energy is given off every time 1 mol of uranium-235 undergoes this nuclear reaction. This is an extraordinary amount of energy. Compare it to combustion reactions of hydrocarbons, which give off about 650kJ/mol of energy for every CH 2 unit in the hydrocarbon-on the order of hundreds of kilojoules per mole. Nuclear reactions give off billions of kilojoules per mole. If this energy could be properly harvested, it would be a significant source of energy for our society. Nuclear energy involves the controlled harvesting of energy from fission reactions. The reaction can be controlled because the fission of uranium-235 (and a few other isotopes, such as plutonium-239) can be artificially initiated by injecting a neutron into a uranium nucleus. The overall nuclear equation, with energy included as a product, is then as follows: \[_{}^{235}\textrm{U}\: +\: _{ }^{1}\textrm{n}\rightarrow \: _{ }^{139}\textrm{Ba}\: +\: _{ }^{94}\textrm{Kr}\: +\: 3_{ }^{1}\textrm{n} \nonumber \] Thus by the careful addition of extra neutrons into a sample of uranium, we can control the fission process and obtain energy that can be used for other purposes. (Artificial or induced radioactivity, in which neutrons are injected into a sample of matter that subsequently cause fission, was first demonstrated in 1934 by Irène Joliot-Curie and Frédéric Joliot, the daughter and son-in-law of Marie Curie.) Binding Energy The forces that bind nucleons together in an atomic nucleus are much greater than those that bind an electron to an atom through electrostatic attraction. This is evident by the relative sizes of the atomic nucleus and the atom (\(10^{-15}\) and \(10^{-10}\)m, respectively). The energy required to pry a nucleon from the nucleus is therefore much larger than that required to remove (or ionize) an electron in an atom. In general, all nuclear changes involve large amounts of energy per particle undergoing the reaction. This has numerous practical applications. As shown in figure \(\PageIndex{2}\), energy is put into the system to break apart the nucleus. The amount of energy required is called the total binding energy (BE), \(E_b\). The binding energy is equal to the amount of energy released in forming the nucleus, and is therefore given by \[E_b = (\Delta m)c^2. \label{BE} \] In nuclear physics, one of the most important experimental quantities is the binding energy per nucleon (BEN) , which is defined by \[BEN = \dfrac{E_b}{A} \label{BEN} \] This quantity is the average energy required to remove an individual nucleon (proton or neutron) from a nucleus—analogous to the ionization energy of an electron in an atom. If the BEN is relatively large, the nucleus is relatively stable. BEN values are estimated from nuclear scattering experiments. Figure \(\PageIndex{3}\) shows the relative binding energies for various isotopes. Of these elements, fission requires heavy, unstable nuclei. This means selected atoms would have low binding energies and would have large atomic masses. Nuclei that are larger than Fe-56 may undergo fission. Uranium-238 and uranium-235 both have lower binding energies with heavy masses-. These two isotopes would be suitable for splitting based on these requirements. Of these two isotopes, only uranium-235 is readily fissionable. This is due to odd neutron count contributing to additional instability. Another fissionable isotope not shown in Figure \(\PageIndex{3}\) is plutonium-239. This is a synthetic isotope produced by transmutation and decay reactions. Like uranium-235, it has low binding energy, high mass, and an odd number of neutrons. U-235 and Pu-239 are used in atomic bombs (fission based) and nuclear reactors. Positively charged centers of atoms make nuclear fusion extremely difficult. For this reason, smaller atoms are suitable for fusion reactions. In addition, these particular isotopes need to have low binding energies in order to undergo fusion. Atoms that possess these two qualities are H-2 and H-3. For fusion to occur, extreme temperatures are required to fuse these deuterium and tritium together. As we will see, the BEN-versus- A graph implies that nuclei divided or combined release an enormous amount of energy. This is the basis for a wide range of phenomena, from the production of electricity at a nuclear power plant to sunlight. Nuclear Fission In both fission and fusion, large amounts of energy are given off in the form of heat, light, and gamma radiation. Italian physicist, Enrico Fermi, performed the first fission reaction in 1934. He was unaware that he had split a uranium atom into two smaller nuclei. Nuclear fission of heavy elements was discovered on December 17, 1938 by German Otto Hahn and his assistant Fritz Strassmann, and explained theoretically in January 1939 by Lise Meitner (Figure \(\PageIndex{4}\)) and her nephew Otto Robert Frisch. Frisch named the process "fission" by analogy with biological fission of living cells. Image take from: https://upload.wikimedia.org/wikiped...ner12crop2.J Since then, fission has been observed in many other isotopes, including most actinide isotopes that have an odd number of neutrons. A typical nuclear fission reaction is shown in Figure \(\PageIndex{5}\). Among the products of Meitner, Hahn, and Strassman’s fission reaction were barium, krypton, lanthanum, and cerium, all of which have nuclei that are more stable than uranium-235. Since then, hundreds of different isotopes have been observed among the products of fissionable substances. Fission has been used in nuclear weapons and powers all nuclear reactors. Approximately fifty-five countries worldwide possess fission technology in the form of research or energy reactors. Less than ten of these countries have fission weapons. A tremendous amount of energy is produced by the fission of heavy elements. For instance, when one mole of U-235 undergoes fission, the products weigh about 0.2 grams less than the reactants; this “lost” mass is converted into a very large amount of energy, about 1.8 × 10 10 kJ per mole of U-235. Nuclear fission reactions produce incredibly large amounts of energy compared to chemical reactions. The fission of 1 kilogram of uranium-235, for example, produces about 2.5 million times as much energy as is produced by burning 1 kilogram of coal. Nuclear Chain Reaction As described earlier, when undergoing fission U-235 produces two “medium-sized” nuclei, and two or three neutrons. These neutrons may then cause the fission of other uranium-235 atoms, which in turn provide more neutrons that can cause fission of even more nuclei, and so on. If this occurs, we have a nuclear chain reaction (Figure \(\PageIndex{6}\)). On the other hand, if too many neutrons escape the bulk material without interacting with a nucleus, then no chain reaction will occur. Material that can sustain a nuclear fission chain reaction is said to be fissile or fissionable . (Technically, fissile material can undergo fission with neutrons of any energy, whereas fissionable material requires high-energy neutrons.) Nuclear fission becomes self-sustaining when the number of neutrons produced by fission equals or exceeds the number of neutrons absorbed by splitting nuclei plus the number that escape into the surroundings. The amount of a fissionable material that will support a self-sustaining chain reaction is a critical mass . An amount of fissionable material that cannot sustain a chain reaction is a subcritical mass . An amount of material in which there is an increasing rate of fission is known as a supercritical mass . The critical mass depends on the type of material: its purity, the temperature, the shape of the sample, and how the neutron reactions are controlled (Figure \(\PageIndex{7}\)). Thermonuclear Reactions Thermonuclear fusion is a way to achieve nuclear fusion by using extremely high temperatures. There are two forms of thermonuclear fusion: uncontrolled , in which the resulting energy is released in an uncontrolled manner, as it is in thermonuclear weapons ("hydrogen bombs") and in most stars; and controlled , where the fusion reactions take place in an environment allowing some or all of the energy released to be harnessed for constructive purposes. The discussion below is limited to the fusion process that powers the sun and the stars. The prime energy producer in the sun is the fusion of hydrogen to form helium, which occurs at a solar-core temperature of 14 million kelvin. The net result is the fusion of four protons into one alpha particle, with the release of two positrons, two neutrinos (which changes two of the protons into neutrons), and energy (Figure \(\PageIndex{8}\)). This is a net reaction of a more complicated series of events: \[\ce{4^1_1H ⟶ ^4_2He + 2 ^1_{0}n} \nonumber \] A helium nucleus has a mass that is 0.7% less than that of four hydrogen nuclei; this lost mass is converted into energy during the fusion. This reaction produces about 3.6 × 10 11 kJ of energy per mole of \(\ce{^4_2He}\) produced. This is somewhat larger than the energy produced by the nuclear fission of one mole of U-235 (1.8 × 10 10 kJ), and over 3 million times larger than the energy produced by the (chemical) combustion of one mole of octane (5471 kJ). Summary Nuclear energy comes from tiny mass changes in nuclei as radioactive processes occur. In fission, large nuclei break apart and release energy; in fusion, small nuclei merge together and release energy. The continues process whereby neutrons produced from the initial fission of a large nucleus like U-235 cause further fission of another nucleus. In a critical mass, a large enough number of neutrons in the fissile material induce fission to create a chain reaction. The most important fusion process in nature is the one that powers the stars. The net result is the fusion of four protons into one alpha particle, with the release of two positrons, two neutrinos (which changes two of the protons into neutrons), and energy
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Bookshelves/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/08%3A_Basic_Concepts_of_Chemical_Bonding/8.07%3A_Exceptions_to_the_Octet_Rule
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Learning Objectives To assign a Lewis dot symbol to elements not having an octet of electrons in their compounds. Three cases can be constructed that do not follow the octet rule, and as such, they are known as the exceptions to the octet rule. Following the Octet Rule for Lewis Dot Structures leads to the most accurate depictions of stable molecular and atomic structures and because of this we always want to use the octet rule when drawing Lewis Dot Structures. However, it is hard to imagine that one rule could be followed by all molecules. There is always an exception, and in this case, three exceptions: When there are an odd number of valence electrons When there are too few valence electrons When there are too many valence electrons Exception 1: Species with Odd Numbers of Electrons The first exception to the Octet Rule is when there are an odd number of valence electrons. An example of this would be Nitrogen (II) Oxide also called nitric oxide (\(\ce{NO}\). Nitrogen has 5 valence electrons while Oxygen has 6. The total would be 11 valence electrons to be used. The Octet Rule for this molecule is fulfilled in the above example, however that is with 10 valence electrons. The last one does not know where to go. The lone electron is called an unpaired electron. But where should the unpaired electron go? The unpaired electron is usually placed in the Lewis Dot Structure so that each element in the structure will have the lowest formal charge possible. The formal charge is the perceived charge on an individual atom in a molecule when atoms do not contribute equal numbers of electrons to the bonds they participate in . No formal charge at all is the most ideal situation. An example of a stable molecule with an odd number of valence electrons would be nitric oxide. nitric oxide has 11 valence electrons. If you need more information about formal charges, see Lewis Structures. If we were to imagine nitric oxide had ten valence electrons we would come up with the Lewis Structure (Figure \(\PageIndex{1}\)): Let's look at the formal charges of Figure \(\PageIndex{2}\) based on this Lewis structure. Nitrogen normally has five valence electrons. In Figure \(\PageIndex{1}\), it has two lone pair electrons and it participates in two bonds (a double bond) with oxygen. This results in nitrogen having a formal charge of +1. Oxygen normally has six valence electrons. In Figure \(\PageIndex{1}\), oxygen has four lone pair electrons and it participates in two bonds with nitrogen. Oxygen therefore has a formal charge of 0. The overall molecule here has a formal charge of +1 (+1 for nitrogen, 0 for oxygen. +1 + 0 = +1). However, if we add the eleventh electron to nitrogen (because we want the molecule to have the lowest total formal charge), it will bring both the nitrogen and the molecule's overall charges to zero, the most ideal formal charge situation. That is exactly what is done to get the correct Lewis structure for nitric oxide (Figure \(\PageIndex{2}\)): Free Radicals There are actually very few stable molecules with odd numbers of electrons that exist, since that unpaired electron is willing to react with other unpaired electrons. Most odd electron species are highly reactive, which we call Free Radicals. Because of their instability, free radicals bond to atoms in which they can take an electron from in order to become stable, making them very chemically reactive. Radicals are found as both reactants and products, but generally react to form more stable molecules as soon as they can. In order to emphasize the existence of the unpaired electron, radicals are denoted with a dot in front of their chemical symbol as with \(\cdot OH\), the hydroxyl radical. An example of a radical you may by familiar with already is the gaseous chlorine atom, denoted \(\cdot Cl\). Interestingly, an odd Number of Valence Electrons will result in the molecule being paramagnetic. Exception 2: Incomplete Octets The second exception to the Octet Rule is when there are too few valence electrons that results in an incomplete Octet. There are even more occasions where the octet rule does not give the most correct depiction of a molecule or ion. This is also the case with incomplete octets. Species with incomplete octets are pretty rare and generally are only found in some beryllium, aluminum, and boron compounds including the boron hydrides. Let's take a look at one such hydride, BH 3 (Borane). If one were to make a Lewis structure for BH 3 following the basic strategies for drawing Lewis structures, one would probably come up with this structure (Figure \(\PageIndex{2}\)): The problem with this structure is that boron has an incomplete octet; it only has six electrons around it. Hydrogen atoms can naturally only have only 2 electrons in their outermost shell (their version of an octet), and as such there are no spare electrons to form a double bond with boron. One might surmise that the failure of this structure to form complete octets must mean that this bond should be ionic instead of covalent. However, boron has an electronegativity that is very similar to hydrogen, meaning there is likely very little ionic character in the hydrogen to boron bonds, and as such this Lewis structure, though it does not fulfill the octet rule, is likely the best structure possible for depicting BH 3 with Lewis theory. One of the things that may account for BH 3 's incomplete octet is that it is commonly a transitory species, formed temporarily in reactions that involve multiple steps. Let's take a look at another incomplete octet situation dealing with boron, BF 3 (Boron trifluorine). Like with BH 3 , the initial drawing of a Lewis structure of BF 3 will form a structure where boron has only six electrons around it (Figure \(\PageIndex{4}\)). If you look Figure \(\PageIndex{4}\), you can see that the fluorine atoms possess extra lone pairs that they can use to make additional bonds with boron, and you might think that all you have to do is make one lone pair into a bond and the structure will be correct. If we add one double bond between boron and one of the fluorines we get the following Lewis Structure (Figure \(\PageIndex{5}\)): Each fluorine has eight electrons, and the boron atom has eight as well! Each atom has a perfect octet, right? Not so fast. We must examine the formal charges of this structure. The fluorine that shares a double bond with boron has six electrons around it (four from its two lone pairs of electrons and one each from its two bonds with boron). This is one less electron than the number of valence electrons it would have naturally (Group Seven elements have seven valence electrons), so it has a formal charge of +1. The two flourines that share single bonds with boron have seven electrons around them (six from their three lone pairs and one from their single bonds with boron). This is the same amount as the number of valence electrons they would have on their own, so they both have a formal charge of zero. Finally, boron has four electrons around it (one from each of its four bonds shared with fluorine). This is one more electron than the number of valence electrons that boron would have on its own, and as such boron has a formal charge of -1. This structure is supported by the fact that the experimentally determined bond length of the boron to fluorine bonds in BF 3 is less than what would be typical for a single bond (see Bond Order and Lengths). However, this structure contradicts one of the major rules of formal charges: Negative formal charges are supposed to be found on the more electronegative atom(s) in a bond, but in the structure depicted in Figure \(\PageIndex{5}\), a positive formal charge is found on fluorine, which not only is the most electronegative element in the structure, but the most electronegative element in the entire periodic table (\(\chi=4.0\)). Boron on the other hand, with the much lower electronegativity of 2.0, has the negative formal charge in this structure. This formal charge-electronegativity disagreement makes this double-bonded structure impossible. However the large electronegativity difference here, as opposed to in BH 3 , signifies significant polar bonds between boron and fluorine, which means there is a high ionic character to this molecule. This suggests the possibility of a semi-ionic structure such as seen in Figure \(\PageIndex{6}\): None of these three structures is the "correct" structure in this instance. The most "correct" structure is most likely a resonance of all three structures: the one with the incomplete octet (Figure \(\PageIndex{4}\)), the one with the double bond (Figure \(\PageIndex{5}\)), and the one with the ionic bond (Figure \(\PageIndex{6}\)). The most contributing structure is probably the incomplete octet structure (due to Figure \(\PageIndex{5}\) being basically impossible and Figure \(\PageIndex{6}\) not matching up with the behavior and properties of BF 3 ). As you can see even when other possibilities exist, incomplete octets may best portray a molecular structure. As a side note, it is important to note that BF 3 frequently bonds with a F - ion in order to form BF 4 - rather than staying as BF 3 . This structure completes boron's octet and it is more common in nature. This exemplifies the fact that incomplete octets are rare, and other configurations are typically more favorable, including bonding with additional ions as in the case of BF 3 . Example \(\PageIndex{1}\): \(NF_3\) Draw the Lewis structure for boron trifluoride (BF 3 ). Solution 1. Add electrons (3*7) + 3 = 24 2. Draw connectivities: 3. Add octets to outer atoms: 4. Add extra electrons (24-24=0) to central atom: 5. Does central electron have octet? NO. It has 6 electrons Add a multiple bond (double bond) to see if central atom can achieve an octet: 6. The central Boron now has an octet (there would be three resonance Lewis structures) However... In this structure with a double bond the fluorine atom is sharing extra electrons with the boron. The fluorine would have a '+' partial charge, and the boron a '-' partial charge, this is inconsistent with the electronegativities of fluorine and boron. Thus, the structure of BF 3 , with single bonds, and 6 valence electrons around the central boron is the most likely structure BF 3 reacts strongly with compounds which have an unshared pair of electrons which can be used to form a bond with the boron: Exception 3: Expanded Valence Shells More common than incomplete octets are expanded octets where the central atom in a Lewis structure has more than eight electrons in its valence shell. In expanded octets, the central atom can have ten electrons, or even twelve. Molecules with expanded octets involve highly electronegative terminal atoms, and a nonmetal central atom found in the third period or below , which those terminal atoms bond to. For example, \(PCl_5\) is a legitimate compound (whereas \(NCl_5\)) is not: Expanded valence shells are observed only for elements in period 3 (i.e. n=3) and beyond The 'octet' rule is based upon available n s and n p orbitals for valence electrons (2 electrons in the s orbitals, and 6 in the p orbitals). Beginning with the n=3 principle quantum number, the d orbitals become available ( l =2). The orbital diagram for the valence shell of phosphorous is: Hence, the third period elements occasionally exceed the octet rule by using their empty d orbitals to accommodate additional electrons. Size is also an important consideration: The larger the central atom, the larger the number of electrons which can surround it Expanded valence shells occur most often when the central atom is bonded to small electronegative atoms, such as F, Cl and O. There is currently much scientific exploration and inquiry into the reason why expanded valence shells are found. The top area of interest is figuring out where the extra pair(s) of electrons are found. Many chemists think that there is not a very large energy difference between the 3p and 3d orbitals, and as such it is plausible for extra electrons to easily fill the 3d orbital when an expanded octet is more favorable than having a complete octet. This matter is still under hot debate, however and there is even debate as to what makes an expanded octet more favorable than a configuration that follows the octet rule. One of the situations where expanded octet structures are treated as more favorable than Lewis structures that follow the octet rule is when the formal charges in the expanded octet structure are smaller than in a structure that adheres to the octet rule, or when there are less formal charges in the expanded octet than in the structure a structure that adheres to the octet rule. Example \(\PageIndex{2}\): The \(SO_4^{-2}\) ion Such is the case for the sulfate ion, SO 4 -2 . A strict adherence to the octet rule forms the following Lewis structure: If we look at the formal charges on this molecule, we can see that all of the oxygen atoms have seven electrons around them (six from the three lone pairs and one from the bond with sulfur). This is one more electron than the number of valence electrons then they would have normally, and as such each of the oxygens in this structure has a formal charge of -1. Sulfur has four electrons around it in this structure (one from each of its four bonds) which is two electrons more than the number of valence electrons it would have normally, and as such it carries a formal charge of +2. If instead we made a structure for the sulfate ion with an expanded octet, it would look like this: Looking at the formal charges for this structure, the sulfur ion has six electrons around it (one from each of its bonds). This is the same amount as the number of valence electrons it would have naturally. This leaves sulfur with a formal charge of zero. The two oxygens that have double bonds to sulfur have six electrons each around them (four from the two lone pairs and one each from the two bonds with sulfur). This is the same amount of electrons as the number of valence electrons that oxygen atoms have on their own, and as such both of these oxygen atoms have a formal charge of zero. The two oxygens with the single bonds to sulfur have seven electrons around them in this structure (six from the three lone pairs and one from the bond to sulfur). That is one electron more than the number of valence electrons that oxygen would have on its own, and as such those two oxygens carry a formal charge of -1. Remember that with formal charges, the goal is to keep the formal charges (or the difference between the formal charges of each atom) as small as possible. The number of and values of the formal charges on this structure (-1 and 0 (difference of 1) in Figure \(\PageIndex{12}\), as opposed to +2 and -1 (difference of 3) in Figure \(\PageIndex{12}\)) is significantly lower than on the structure that follows the octet rule, and as such an expanded octet is plausible, and even preferred to a normal octet, in this case. Example \(\PageIndex{3}\): The \(ICl_4^-\) Ion Draw the Lewis structure for \(ICl_4^-\) ion. Solution 1. Count up the valence electrons: 7+(4*7)+1 = 36 electrons 2. Draw the connectivities: 3. Add octet of electrons to outer atoms: 4. Add extra electrons (36-32= 4 ) to central atom: 5. The ICl 4 - ion thus has 12 valence electrons around the central Iodine (in the 5 d orbitals) Expanded Lewis structures are also plausible depictions of molecules when experimentally determined bond lengths suggest partial double bond characters even when single bonds would already fully fill the octet of the central atom. Despite the cases for expanded octets, as mentioned for incomplete octets, it is important to keep in mind that, in general, the octet rule applies. Summary Following the Octet Rule for Lewis Dot Structures leads to the most accurate depictions of stable molecular and atomic structures and because of this we always want to use the octet rule when drawing Lewis Dot Structures. There are three exceptions: (1) When there are an odd number of valence electrons, (2) When there are too few valence electrons, and (3) when there are too many valence electrons References Petrucci, Ralph H.; Harwood, William S.; Herring, F. G.; Madura, Jeffrey D. General Chemistry: Principles & Modern Applications . 9th Ed. New Jersey. Pearson Education, Inc. 2007. Moore, John W.; Stanitski, Conrad L. ; Jurs, Peter C. Chemistry; The Molecular Science . 2nd Ed. 2004.
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Courses/University_of_Kentucky/UK%3A_General_Chemistry/11%3A_Solutions_and_Colloids/11.1%3A_The_Dissolution_Process
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Skills to Develop Describe the basic properties of solutions and how they form Predict whether a given mixture will yield a solution based on molecular properties of its components Explain why some solutions either produce or absorb heat when they form An earlier chapter of this text introduced solutions , defined as homogeneous mixtures of two or more substances. Often, one component of a solution is present at a significantly greater concentration, in which case it is called the solvent. The other components of the solution present in relatively lesser concentrations are called solutes. Sugar is a covalent solid composed of sucrose molecules, \(\mathrm{C_{12}H_{22}O_{11}}\). When this compound dissolves in water, its molecules become uniformly distributed among the molecules of water: \[\mathrm{C_{12}H_{22}O}_{11(s)}⟶\mathrm{C_{12}H_{22}O}_{11(aq)} \label{Eq1}\] The subscript “aq” in the equation signifies that the sucrose molecules are solutes and are therefore individually dispersed throughout the aqueous solution (water is the solvent). Although sucrose molecules are heavier than water molecules, they remain dispersed throughout the solution; gravity does not cause them to “settle out” over time. Potassium dichromate, \(\mathrm{K_2Cr_2O_7}\), is an ionic compound composed of colorless potassium ions, \(\mathrm{K^+}\), and orange dichromate ions, \(\mathrm{Cr_2O_7^{2−}}\) . When a small amount of solid potassium dichromate is added to water, the compound dissolves and dissociates to yield potassium ions and dichromate ions uniformly distributed throughout the mixture (Figure \(\PageIndex{1}\)), as indicated in this equation: \[\mathrm{K_2Cr_2O}_{7(s)}⟶\mathrm{2K^+}_{(aq)}+\mathrm{Cr_2O_7^{2-}}_{(aq)} \label{Eq2}\] As with the mixture of sugar and water, this mixture is also an aqueous solution. Its solutes, potassium and dichromate ions, remain individually dispersed among the solvent (water) molecules. Figure \(\PageIndex{1}\): When potassium dichromate (\(K_2Cr_2O_7\)) is mixed with water, it forms a homogeneous orange solution. (credit: modification of work by Mark Ott). Water is used so often as a solvent that the word solution has come to imply an aqueous solution to many people. However, almost any gas, liquid, or solid can act as a solvent. Many alloys are solid solutions of one metal dissolved in another; for example, US five-cent coins contain nickel dissolved in copper. Air is a gaseous solution, a homogeneous mixture of nitrogen, oxygen, and several other gases. Oxygen (a gas), alcohol (a liquid), and sugar (a solid) all dissolve in water (a liquid) to form liquid solutions. Table \(\PageIndex{1}\) gives examples of several different solutions and the phases of the solutes and solvents. Solution Solute Solvent air O2(g) N2(g) soft drinks CO2(g) H2O(l) hydrogen in palladium H2(g) Pd(s) rubbing alcohol H2O(l) C3H8O(l) (2-propanol) saltwater NaCl(s) H2O(l) brass Zn(s) Cu(s) Solutions exhibit these defining traits: They are homogeneous; that is, after a solution is mixed, it has the same composition at all points throughout (its composition is uniform). The physical state of a solution—solid, liquid, or gas—is typically the same as that of the solvent, as demonstrated by the examples in Table \(\PageIndex{1}\). The components of a solution are dispersed on a molecular scale; that is, they consist of a mixture of separated molecules, atoms, and/or ions. The dissolved solute in a solution will not settle out or separate from the solvent. The composition of a solution, or the concentrations of its components, can be varied continuously, within limits. The Formation of Solutions The formation of a solution is an example of a spontaneous process , a process that occurs under specified conditions without the requirement of energy from some external source. Sometimes we stir a mixture to speed up the dissolution process, but this is not necessary; a homogeneous solution would form if we waited long enough. The topic of spontaneity is critically important to the study of chemical thermodynamics and is treated more thoroughly in a later chapter of this text. For purposes of this chapter’s discussion, it will suffice to consider two criteria that favor , but do not guarantee, the spontaneous formation of a solution: a decrease in the internal energy of the system (an exothermic change, as discussed in the previous chapter on thermochemistry) an increase in the disorder in the system (which indicates an increase in the entropy of the system, as you will learn about in the later chapter on thermodynamics) In the process of dissolution, an internal energy change often, but not always, occurs as heat is absorbed or evolved. An increase in disorder always results when a solution forms. When the strengths of the intermolecular forces of attraction between solute and solvent species in a solution are no different than those present in the separated components, the solution is formed with no accompanying energy change. Such a solution is called an ideal solution . A mixture of ideal gases (or gases such as helium and argon, which closely approach ideal behavior) is an example of an ideal solution, since the entities comprising these gases experience no significant intermolecular attractions. When containers of helium and argon are connected, the gases spontaneously mix due to diffusion and form a solution (Figure \(\PageIndex{2}\)). The formation of this solution clearly involves an increase in disorder, since the helium and argon atoms occupy a volume twice as large as that which each occupied before mixing. Figure \(\PageIndex{2}\): Samples of helium and argon spontaneously mix to give a solution in which the disorder of the atoms of the two gases is increased. Ideal solutions may also form when structurally similar liquids are mixed. For example, mixtures of the alcohols methanol (CH 3 OH) and ethanol (C 2 H 5 OH) form ideal solutions, as do mixtures of the hydrocarbons pentane, C 5 H 12 , and hexane, C 6 H 14 . Placing methanol and ethanol, or pentane and hexane, in the bulbs shown in Figure \(\PageIndex{2}\) will result in the same diffusion and subsequent mixing of these liquids as is observed for the He and Ar gases (although at a much slower rate), yielding solutions with no significant change in energy. Unlike a mixture of gases, however, the components of these liquid-liquid solutions do, indeed, experience intermolecular attractive forces. But since the molecules of the two substances being mixed are structurally very similar, the intermolecular attractive forces between like and unlike molecules are essentially the same, and the dissolution process, therefore, does not entail any appreciable increase or decrease in energy. These examples illustrate how diffusion alone can provide the driving force required to cause the spontaneous formation of a solution. In some cases, however, the relative magnitudes of intermolecular forces of attraction between solute and solvent species may prevent dissolution. Figure \(\PageIndex{3}\): This schematic representation of dissolution shows a stepwise process involving the endothermic separation of solute and solvent species (Steps 1 and 2) and exothermic solvation (Step 3). Three types of intermolecular attractive forces are relevant to the dissolution process: solute-solute, solvent-solvent, and solute-solvent. As illustrated in Figure \(\PageIndex{3}\), the formation of a solution may be viewed as a stepwise process in which energy is consumed to overcome solute-solute and solvent-solvent attractions (endothermic processes) and released when solute-solvent attractions are established (an exothermic process referred to as solvation ). The relative magnitudes of the energy changes associated with these stepwise processes determine whether the dissolution process overall will release or absorb energy. In some cases, solutions do not form because the energy required to separate solute and solvent species is so much greater than the energy released by solvation. Figure \(\PageIndex{3}\) : A mixture of nonpolar cooking oil and polar water does not yield a solution. (credit: Gautam Dogra). For example, cooking oils and water will not mix to any appreciable extent to yield solutions (Figure \(\PageIndex{3}\)). Hydrogen bonding is the dominant intermolecular attractive force present in liquid water; the nonpolar hydrocarbon molecules of cooking oils are not capable of hydrogen bonding, instead being held together by dispersion forces. Forming an oil-water solution would require overcoming the very strong hydrogen bonding in water, as well as the significantly strong dispersion forces between the relatively large oil molecules. And, since the polar water molecules and nonpolar oil molecules would not experience very strong intermolecular attraction, very little energy would be released by solvation. On the other hand, a mixture of ethanol and water will mix in any proportions to yield a solution. In this case, both substances are capable of hydrogen bonding, and so the solvation process is sufficiently exothermic to compensate for the endothermic separations of solute and solvent molecules. Figure \(\PageIndex{4}\): An instant cold pack gets cold when certain salts, such as ammonium nitrate, dissolve in water—an endothermic process. As noted at the beginning of this module, spontaneous solution formation is favored, but not guaranteed, by exothermic dissolution processes. While many soluble compounds do, indeed, dissolve with the release of heat, some dissolve endothermically. Ammonium nitrate (NH 4 NO 3 ) is one such example and is used to make instant cold packs for treating injuries like the one pictured in Figure \(\PageIndex{4}\). A thin-walled plastic bag of water is sealed inside a larger bag with solid NH 4 NO 3 . When the smaller bag is broken, a solution of NH 4 NO 3 forms, absorbing heat from the surroundings (the injured area to which the pack is applied) and providing a cold compress that decreases swelling. Endothermic dissolutions such as this one require a greater energy input to separate the solute species than is recovered when the solutes are solvated, but they are spontaneous nonetheless due to the increase in disorder that accompanies formation of the solution. Video \(\PageIndex{1}\) : Watch this brief video illustrating endothermic and exothermic dissolution processes. Summary A solution forms when two or more substances combine physically to yield a mixture that is homogeneous at the molecular level. The solvent is the most concentrated component and determines the physical state of the solution. The solutes are the other components typically present at concentrations less than that of the solvent. Solutions may form endothermically or exothermically, depending upon the relative magnitudes of solute and solvent intermolecular attractive forces. Ideal solutions form with no appreciable change in energy. Footnotes 1 If bubbles of gas are observed within the liquid, the mixture is not homogeneous and, thus, not a solution. Glossary alloy solid mixture of a metallic element and one or more additional elements ideal solution solution that forms with no accompanying energy change solvation exothermic process in which intermolecular attractive forces between the solute and solvent in a solution are established spontaneous process physical or chemical change that occurs without the addition of energy from an external source Contributors Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/[email protected] ).
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Bookshelves/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/IV%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_2/05%3A_Oxidative_Addition_and_Reductive_Elimination/5.05%3A_Oxidative_Addition_in_Action-__Catalytic_Hydrogenation
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Catalytic hydrogenation is a tremendously important reaction. It is an essential step in the synthesis of many fine chemicals as well as bulk commodities. In catalytic hydrogenation, a pair of hydrogen atoms are added across a double bond, turning an alkene into an alkane. In general, the reaction requires a large excess of hydrogen gas, often under high pressure. The reaction can be performed under either homogeneous or heterogeneous conditions. Homogeneous reactions employ a soluble catalyst. Soluble catalysts are those that dissolve under the reaction conditions; they often provide superior control over the reaction. Heterogeneous catalysts do not dissolve; they are solids that sit on the bottom of the reaction, like sand in a lakebed. One of the advantages of heterogeneous catalysts is that they can easily be filtered away from the rest of the reaction, making purification of the product much more straightforward. Exercise \(\PageIndex{1}\) Indicate whether the following mixtures are homogeneous or heterogeneous. a) Kool-aid b) a glass of pop with ice c) orange juice d) cranberry juice Answer a homogeneous Answer b heterogeneous Answer c heterogeneous Answer d homogeneous Because of the importance of hydrogenation, a number of catalysts have been developed over the years that are capable of performing the reaction. There are a number of motivations for working on catalyst development. One reason is speed: the faster the catalyst, the more product can be made and the more economical the process. Another reason is selectivity. Suppose there are two double bonds present in a molecule. Maybe you only want to hydrogenate one of these double bonds. By choosing the proper catalyst, you may be able to do that. Let's take a look at a few different examples of catalysts with different selectivities. Perhaps the most commonly used catalyst is palladium metal, a heterogeneous catalyst. Very often, expensive metals like palladium are not used in their pure state. For example, palladium is often dispersed on a "solid support", such as carbon. There are a couple of benefits of doing that. First, the expensive palladium metal is stretched a little further by mixing it with carbon, which is much cheaper. Usually, this mixture is about 5% palladium and 95% carbon, although different compositions can be used. In addition, use of a solid support helps to spread the metal particles out spatially. When the metal isn't all clumped together, it has an increased surface area. That means there are more places available for hydrogen and alkenes to bind and undergo the hydrogenation reaction. Finally, a solid support often tunes the reactivity of the metal that is stuck to it. The solid support might change the rate of the reaction or alter the selectivity because of interactions between the metal and the support. Palladium on carbon, or Pd/C, provides an example of what we mean by selectivity. It is very good at adding hydrogen to alkenes. It can hydrogenate alkynes, too. However, it is not very good at hydrogenating more stable double bonds, such as those in conjugated dienes, or in benzene or other aromatics. In contrast, platinum oxide is much more general, hydrogenating regular alkenes and also conjugated ones. Under the right conditions, platinum oxide can even be used to hydrogenate benzene. That usually means very high pressure of hydrogen gas. Also, hydrogenation with palladium doesn't work very well with carbonyls. It usually won't reduce aldehydes or ketones. You may remember that other reducing agents (compounds that add hydrogen to carbon atoms) such as LiAlH 4 can react with carbonyls quite easily, so palladium with hydrogen is very complementary to those reagents. Even PtO 2 can't induce hydrogen to add across a carbonyl, although another heterogeneous catalyst, a ferocious one called Raney Nickel, can do the job. However, not even Raney Nickel does very well at hydrogenating more stable carbonyls, such as amides, esters and carboxylic acids. Those are the ones at the bottom of the energetic "ski hill", so they are the least reactive carbonyls. They are difficult to hydrogenate, and are usually left alone. On the other hand, palladium does just fine with some seemingly related compounds, containing imines and nitro groups. Although these groups contain multiple bonds and nitrogen, they do not have the same stability of amides. Imines and nitro groups behave a little more like simple carbonyl compounds when it comes to hydrogenation. Exercise \(\PageIndex{2}\) Provide products for the following reactions. We can make the palladium catalyst even more selective by preparing it in a different way. Lindlar's catalyst is a very dramatic example of how reactivity can be tuned by using different compositions. To make Lindlar's catalyst, palladium is supported on calcium carbonate rather than carbon, together with other components, such as lead acetate and quinoline. That last component turns out to be the key to Lindlar's catalyst. It tunes the reactivity so that the catalyst can react with alkynes but not with alkenes. As a result, if a compound is hydrogenated with a palladium catalyst in the presence of quinoline, an alkene is produced. Without the quinoline, you would get an alkane. Normally, if a catalyst hydrogenates an alkyne, no alkene is observed. That's because the alkene also reacts under the same conditions and is quickly converted to an alkane. Not so fast, you say. If we just add one equivalent of hydrogen (that is, one molecule of hydrogen for every molecule of alkyne) then the reaction will stop after forming an alkene. That's very clever of you. However, you've missed a couple of important concepts. First of all, we are never dealing with individual molecules when we run a reaction; instead, we are dealing with vast numbers of molecules at a time. That means we will deal with statistical distributions. Maybe some molecules of hydrogen react with alkyne to produce alkene. Maybe some molecules go ahead and react with that alkene to produce alkane. Now, if we only added enough hydrogen for every alkyne to react with one H 2 , and some of them have already reacted with two, then somebody will be left out. There will be some leftover alkyne, too. That means we have made a mixture of alkyne, alkene and alkane. The other missing concept is that hydrogenation reactions usually run under a high pressure of hydrogen gas. That means many equivalents of hydrogen are needed in order to push the reaction forward. How does the presence of quinoline change the catalyst so dramatically? It seems that the answer is based on steric crowdedness. Although alkenes are flat, and don't seem very crowded, they may be crowded compared to an alkyne, which is linear. That difference makes alkynes even more reactive than alkenes with respect to hydrogenation. When these compounds bind to the surface of the catalyst, the alkyne takes up less space than the alkenes. It mostly lies along one dimension, whereas the alkenes are spread out into two dimensions. The quinoline seems to simply take up space when it binds to the surface of the catalyst. As a result, the wider alkene can't bind as well as the narrow alkyne. So it turns out that Lindlar's catalyst is a big deal. It provides a very selective reaction: production of an alkene from an alkyne. Furthermore, it doesn't just make any alkene. It only makes cis -alkenes. That's because the hydrogen atoms are both delivered from the surface of the metal. The alkyne binds to the surface of the metal and accepts the hydrogen atoms from that surface. As a result, both hydrogens end up on the same side of the new double bond. The alkene formed is then a cis -alkene. The addition of two hydrogen atoms to the same side of the molecule is not limited to Lindlar's catalyst. It's a general feature of catalytic hydrogenation. As a result, catalytic hydrogenations are often diastereoselective; they result in the formation of one diastereomer, but not the other. Exercise \(\PageIndex{3}\) Provide the missing reagents or products for the following reactions. What about homogeneous catalysts? The most common one is Wilkinson's catalyst, (Ph 3 P) 3 RhCl. Wilkinson's catalyst, like Pd/C, is good at reacting with alkenes but leaving polar bonds alone. It is also highly selective, reacting only with the least sterically crowded alkenes. It only reacts with monosubstituted and disubstituted alkenes. If an alkene has fewer than two hydrogens attached to the double bond, Wilkinson's catalyst leaves it alone. In contrast, Crabtree's catalyst, [(COD)(PCy 3 )(py)Ir]PF 6 is a much more reactive catalyst. In part, that's because it is a more electrophilic, cationic catalyst; the PF 6 is a non-reactive counterion. In addition, Crabtree's catalyst contains a sacrificial alkene ligand. COD is cyclooctadiene, a bidentate ligand that contains two double bonds. What happens to that ligand when the catalyst is exposed to hydrogen? It gets hydrogenated, of course. Without double bonds, it can no longer be a ligand. That leaves the catalyst with two open coordination sites, although really these sites are occupied by solvent molecules. Nevertheless, the solvent molecules bind only loosely, and can easily leave to make room for an alkene. As a result, Crabtree's catalyst is much less sensitive to steric crowding. Unlike Wilkinson's catalyst, it is perfectly capable of hydrogenating trisubstituted or even tetrasubstituted alkenes. Exercise \(\PageIndex{4}\) Provide the missing reagents or products in the following reactions. Exercise \(\PageIndex{5}\) Provide products for the following reactions. Exercise \(\PageIndex{6}\) Provide products for the following reactions. Exercise \(\PageIndex{7}\) Provide reagents for the following reactions.
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Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/An_Introduction_to_the_Electronic_Structure_of_Atoms_and_Molecules_(Bader)/09%3A_Preface_to_Calculations
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The beginning student of chemistry must have a knowledge of the theory which forms the basis for our understanding of chemistry and he must acquire this knowledge before he has the mathematical background required for a rigorous course of study in quantum mechanics. The present approach is designed to meet this need by stressing the physical or observable aspects of the theory through an extensive use of the electronic charge density. The manner in which the negative charge of an atom or a molecule is arranged in three-dimensional space is determined by the electronic charge density distribution. Thus, it determines directly the sizes and shapes of molecules, their electrical moments and, indeed, all of their chemical and physical properties. Since the charge density describes the distribution of negative charge in real space, it is a physically measurable quantity. Consequently, when used as a basis for the discussion of chemistry, the charge density allows for a direct physical picture and interpretation. In particular, the forces exerted on a nucleus in a molecule by the other nuclei and by the electronic charge density may be rigorously calculated and interpreted in terms of classical electrostatics. Thus, given the molecular charge distribution, the stability of a chemical bond may be discussed in terms of the electrostatic requirement of achieving a zero force on the nuclei in the molecule. A chemical bond is the result of the accumulation of negative charge density in the region between any pair of nuclei to an extent sufficient to balance the forces of repulsion. This is true of any chemical bond, ionic or covalent, and even of the shallow minimum in the potential curves arising from van der Waals' forces. In this treatment, the classifications of bonding, ionic or covalent, are retained, but they are given physical definitions in terms of the actual distribution of charge within the molecule. In covalent bonding the valence charge density is distributed over the whole molecule and the attractive forces responsible for binding the nuclei are exerted by the charge density equally shared between them in the internuclear region. In ionic bonding, the valence charge density is localized in the region of a single nucleus and in this extreme of binding the charge density localized on a single nucleus exerts the attractive force which binds both nuclei. This web page begins with a discussion of the need for a new mechanics to describe the events at the atomic level. This is illustrated through a discussion of experiments with electrons and light, which are found to be inexplicable in terms of the mechanics of Newton. The basic concepts of the quantum description of a bound electron, such as quantization, degeneracy and its probabilistic aspect, are introduced by contrasting the quantum and classical results for similar one-dimensional systems. The atomic orbital description of the many-electron atom and the Pauli exclusion principle are considered in some detail, and the experimental consequences of their predictions regarding the energy, angular momentum and magnetic properties of atoms are illustrated. The alternative interpretation of the probability distribution (for a stationary state of an atom) as a representation of a static distribution of electronic charge in real space is stressed, in preparation for the discussion of the chemical bond. Chemical binding is discussed in terms of the molecular charge distribution and the forces which it exerts on the nuclei, an approach which may be rigorously presented using electrostatic concepts. The discussion is enhanced through the extensive use of diagrams to illustrate both the molecular charge distributions and the changes in the atomic charge distributions accompanying the formation of a chemical bond. The above topics are covered in the first seven sections of this web page. The final section is for the reader interested in the extension of the orbital concept to the molecular cases. An elementary account of the use of symmetry in predicting the electronic structure of molecules is given in this section. Hamilton, 1970 Acknowledgement The physical picture of chemical binding afforded by the study of molecular charge distributions has been forced to await the availability of molecular wave functions of considerable quality. The author, together with the reader who finds the approach presented in this volume helpful in his understanding of chemistry, is indebted to the people who overcame the formidable mathematical obstacles encountered in obtaining these wave functions. This webpage is dedicated to Pamela, Carolyn, Kimberly and Suzanne.
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Courses/Thompson_Rivers_University/TRU%3A_Fundamentals_and_Principles_of_Chemistry_(CHEM_1510_and_CHEM_1520)/02%3A_Gases/2.07%3A_Non-Ideal_Gas_Behavior
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Learning Objectives Describe the physical factors that lead to deviations from ideal gas behavior Explain how these factors are represented in the van der Waals equation Define compressibility (Z) and describe how its variation with pressure reflects non-ideal behavior Quantify non-ideal behavior by comparing computations of gas properties using the ideal gas law and the van der Waals equation Thus far, the ideal gas law, PV = nRT , has been applied to a variety of different types of problems, ranging from reaction stoichiometry and empirical and molecular formula problems to determining the density and molar mass of a gas. As mentioned in the previous modules of this chapter, however, the behavior of a gas is often non-ideal, meaning that the observed relationships between its pressure, volume, and temperature are not accurately described by the gas laws. In this section, the reasons for these deviations from ideal gas behavior are considered. One way in which the accuracy of PV = nRT can be judged is by comparing the actual volume of 1 mole of gas (its molar volume, Vm) to the molar volume of an ideal gas at the same temperature and pressure. This ratio is called the compressibility factor (Z) with: \[\mathrm{Z=\dfrac{molar\: volume\: of\: gas\: at\: same\:\mathit{T}\:and\:\mathit{P}}{molar\: volume\: of\: ideal\: gas\: at\: same\:\mathit{T}\:and\:\mathit{P}}}=\left(\dfrac{PV_m}{RT}\right)_\ce{measured} \nonumber \] Ideal gas behavior is therefore indicated when this ratio is equal to 1, and any deviation from 1 is an indication of non-ideal behavior. Figure \(\PageIndex{1}\) shows plots of Z over a large pressure range for several common gases. As is apparent from Figure \(\PageIndex{1}\), the ideal gas law does not describe gas behavior well at relatively high pressures. To determine why this is, consider the differences between real gas properties and what is expected of a hypothetical ideal gas. Particles of a hypothetical ideal gas have no significant volume and do not attract or repel each other . In general, real gases approximate this behavior at relatively low pressures and high temperatures. However, at high pressures, the molecules of a gas are crowded closer together, and the amount of empty space between the molecules is reduced. At these higher pressures, the volume of the gas molecules themselves becomes appreciable relative to the total volume occupied by the gas (Figure \(\PageIndex{2}\)). The gas therefore becomes less compressible at these high pressures, and although its volume continues to decrease with increasing pressure, this decrease is not proportional as predicted by Boyle’s law. At relatively low pressures, gas molecules have practically no attraction for one another because they are (on average) so far apart, and they behave almost like particles of an ideal gas. At higher pressures, however, the force of attraction is also no longer insignificant. This force pulls the molecules a little closer together, slightly decreasing the pressure (if the volume is constant) or decreasing the volume (at constant pressure) (Figure \(\PageIndex{3}\)). This change is more pronounced at low temperatures because the molecules have lower KE relative to the attractive forces, and so they are less effective in overcoming these attractions after colliding with one another. There are several different equations that better approximate gas behavior than does the ideal gas law. The first, and simplest, of these was developed by the Dutch scientist Johannes van der Waals in 1879. The van der Waals equation improves upon the ideal gas law by adding two terms: one to account for the volume of the gas molecules and another for the attractive forces between them. The constant a corresponds to the strength of the attraction between molecules of a particular gas, and the constant b corresponds to the size of the molecules of a particular gas. The “correction” to the pressure term in the ideal gas law is \(\dfrac{n^2a}{V^2}\), and the “correction” to the volume is nb . Note that when V is relatively large and n is relatively small, both of these correction terms become negligible, and the van der Waals equation reduces to the ideal gas law, PV = nRT . Such a condition corresponds to a gas in which a relatively low number of molecules is occupying a relatively large volume, that is, a gas at a relatively low pressure. Experimental values for the van der Waals constants of some common gases are given in Table \(\PageIndex{1}\). Gas a (L2 atm/mol2) b (L/mol) N2 1.3900 0.0391 O2 1.3600 0.0318 CO2 3.5900 0.0427 H2O 5.4600 0.0305 He 0.0342 0.0237 CCl4 20.4000 0.1383 At low pressures, the correction for intermolecular attraction, a , is more important than the one for molecular volume, b . At high pressures and small volumes, the correction for the volume of the molecules becomes important because the molecules themselves are incompressible and constitute an appreciable fraction of the total volume. At some intermediate pressure, the two corrections have opposing influences and the gas appears to follow the relationship given by PV = nRT over a small range of pressures. This behavior is reflected by the “dips” in several of the compressibility curves shown in Figure \(\PageIndex{1}\). The attractive force between molecules initially makes the gas more compressible than an ideal gas, as pressure is raised (Z decreases with increasing P ). At very high pressures, the gas becomes less compressible (Z increases with P ), as the gas molecules begin to occupy an increasingly significant fraction of the total gas volume. Strictly speaking, the ideal gas equation functions well when intermolecular attractions between gas molecules are negligible and the gas molecules themselves do not occupy an appreciable part of the whole volume. These criteria are satisfied under conditions of low pressure and high temperature . Under such conditions, the gas is said to behave ideally, and deviations from the gas laws are small enough that they may be disregarded—this is, however, very often not the case. Example \(\PageIndex{1}\): Comparison of Ideal Gas Law and van der Waals Equation A 4.25-L flask contains 3.46 mol CO 2 at 229 °C. Calculate the pressure of this sample of CO 2 : from the ideal gas law from the van der Waals equation Explain the reason(s) for the difference. Solution (a) From the ideal gas law: \(P=\dfrac{nRT}{V}=\mathrm{\dfrac{3.46\cancel{mol}×0.08206\cancel{L}atm\cancel{mol^{−1}}\cancel{K^{−1}}×502\cancel{K}}{4.25\cancel{L}}=33.5\:atm}\) (b) From the van der Waals equation: \(\left(P+\dfrac{n^2a}{V^2}\right)×(V−nb)=nRT⟶P=\dfrac{nRT}{(V−nb)}−\dfrac{n^2a}{V^2}\) \(P=\mathrm{\dfrac{3.46\:mol×0.08206\:L\:atm\:mol^{−1}\:K^{−1}×502\: K}{(4.25\:L−3.46\:mol×0.0427\:L\:mol^{−1})}−\dfrac{(3.46\:mol)^2×3.59\:L^2\:atm\:mol^2}{(4.25\:L)^2}}\) This finally yields P = 32.4 atm. (c) This is not very different from the value from the ideal gas law because the pressure is not very high and the temperature is not very low. The value is somewhat different because CO 2 molecules do have some volume and attractions between molecules, and the ideal gas law assumes they do not have volume or attractions. Exercise \(\PageIndex{1}\) A 560-mL flask contains 21.3 g N 2 at 145 °C. Calculate the pressure of N 2 : from the ideal gas law from the van der Waals equation Explain the reason(s) for the difference. Answer a 46.562 atm Answer b 46.594 atm Answer c The van der Waals equation takes into account the volume of the gas molecules themselves as well as intermolecular attractions. Summary Gas molecules possess a finite volume and experience forces of attraction for one another. Consequently, gas behavior is not necessarily described well by the ideal gas law. Under conditions of low pressure and high temperature, these factors are negligible, the ideal gas equation is an accurate description of gas behavior, and the gas is said to exhibit ideal behavior. However, at lower temperatures and higher pressures, corrections for molecular volume and molecular attractions are required to account for finite molecular size and attractive forces. The van der Waals equation is a modified version of the ideal gas law that can be used to account for the non-ideal behavior of gases under these conditions. Key Equations \(\mathrm{Z=\dfrac{molar\:volume\: of\: gas\: at\: same\:\mathit{T}\:and\:\mathit{P}}{molar\: volume\: of\: ideal\: gas\: at\: same\:\mathit{T}\:and\:\mathit{P}}}=\left(\dfrac{P×V_m}{R×T}\right)_\ce{measured}\) \(\left(P+\dfrac{n^2a}{V^2}\right)×(V−nb)=nRT\) Glossary compressibility factor (Z) ratio of the experimentally measured molar volume for a gas to its molar volume as computed from the ideal gas equation van der Waals equation modified version of the ideal gas equation containing additional terms to account for non-ideal gas behavior
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Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Acid_Halides/Properties_of_Acyl_Halides
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This page defines acyl halides and discusses their simple physical properties, introducing chemical reactivity in a general way. Acyl halides as "acid derivatives" A carboxylic acid such as ethanoic acid has the following structure: There are a number of related compounds in which the -OH group in the acid is replaced by somet hing else that leaves the acyl carbon in a +3 oxidation state. Compounds like this are described as acid derivatives. Acyl halides (also known as acid halides) are one example of an acid derivative. In this example, the -OH group has been replaced by a chlorine atom; chlorine is the most commonly used acid halide. The acyl group The acyl group is a hydrocarbon group attached to a carbon-oxygen double bond: The "R" group is normally restricted to an alkyl group. It could, however, be a group based on a benzene ring. Naming acyl halides The easiest way to name an acyl halide is to consider the relationship with the corresponding carboxylic acid: carboxylic acid name acyl halide name acyl halide formula ethanoic acid ethanoyl chloride CH3COCl propanoic acid propanoyl chloride CH3CH2COCl butanoic acid butanoyl chloride CH3CH2CH2COCl butanoic acid butanoyl bromide CH3CH2CH2COBr butanoic acid butanoyl iodide CH3CH2CH2COI If something is substituted into the hydrocarbon chain, the carbon in the -COX (X = halide) group is the number 1 carbon. For example, 2-methylbutanoyl chloride is named as follows: Physical properties of acyl halides An acyl halide such as ethanoyl chloride is a colorless, fuming liquid. The strong smell of ethanoyl chloride is a mixture of the smell of vinegar ( ethanoic acid) and the acrid smell of hydrogen chloride gas. The smell and the fumes originate from the reactions between ethanoyl chloride and water vapor in the air. Solubility in water Acyl halides do not dissolve in water because they react (often violently) with it to produce carboxylic acids and hydrogen halides (e.g. HCl). The strong reaction makes it impossible to obtain a simple aqueous solution of an acyl halide. Boiling points Taking ethanoyl chloride as a typical example: ethanoyl chloride boils at 51°C and is a polar molecule. Therefore, it has dipole-dipole attractions between its molecules as well as van der Waals dispersion forces. However, it does not form hydrogen bonds . Its boiling point is, therefore, higher than an alkane of similar size (which has no permanent dipoles) such as ethane, which boils at -88.5 °C , but not as high as a similarly sized alcohol (which forms hydrogen bonds in addition to everything else) such as ethanol, which boils at 78.37 °C . Reactivity Acyl halides are extremely reactive, and in each of their reactions the halogen atom is replaced by another functional group. In each case, in the first step, hydrogen halide (normally hydrogen chloride) gas is produced as steamy acidic fumes. However, in some cases the hydrogen halide goes on to react with one of the substances in the reaction mixture. Taking ethanoyl chloride as typical, the initial reaction is the following: These reactions involve water, alcohols and phenols, or ammonia and amines as nucleophiles. Each of these particular cases contains a very electronegative element with an active lone pair of electrons—either oxygen or nitrogen. The most commonly performed reaction with acyl halides and the example reaction is known as nucleophilic acyl substitution . Contributors Jim Clark ( Chemguide.co.uk ) Mark Tye (Diablo Valley College)
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Courses/Sacramento_City_College/SCC%3A_CHEM_300_-_Beginning_Chemistry/SCC%3A_CHEM_300_-_Beginning_Chemistry_(Faculty)/05%3A_Molecules_and_Compounds/5.01%3A_Sugar_and_Salt
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Sodium chloride, also known as table salt, is an ionic compound with the chemical formula \(\ce{NaCl}\), representing a 1:1 ratio of sodium and chloride ions. It is commonly used as a condiment and food preservative. Salt can be created by adding two very reactive elements together: sodium (\(\ce{Na (s)}\) metal and chlorine (\(\ce{Cl2 (g)}\) gas. \[ \ce{2Na (s) + Cl2(g) \rightarrow 2NaCl (s)} \label{eq1} \] The element sodium (Figure \(\PageIndex{1a}\)) is a very reactive metal; given the opportunity, it will react with the sweat on your hands and form sodium hydroxide, which is a very corrosive substance. The element chlorine (Figure \(\PageIndex{1b}\)) is a pale yellow, corrosive gas that should not be inhaled due to its poisonous nature. Bring these two hazardous substances together, however, and they react to make the ionic compound sodium chloride (Figure \(\PageIndex{1c}\)), known simply as salt. Salt is necessary for life. \(\ce{Na^{+}}\) ions are one of the main ions in the human body and are necessary to regulate the fluid balance in the body. \(\ce{Cl^{−}}\) ions are necessary for proper nerve function and respiration. Both of these ions are supplied by salt. The taste of salt is one of the fundamental tastes; salt is probably the most ancient flavoring known, and one of the few rocks we eat. Clearly when the elemental sodium and chlorine combine (Equation \ref{eq1}), the resulting salt product has radically different properties (both physical and chemical). This reaction is spectacular to observe (Video \(\PageIndex{1}\)). Video \(\PageIndex{1}\): Making Table Salt using Sodium Metal and Chlorine gas Another compound is sugar, which is the generic name for sweet, soluble carbohydrates, many of which are used in food. Sugar has the chemical formulate \(\ce{C12H22O11}\) and is constructed from different elements than salt: carbon , hydrogen and oxygen . While sugar qualitatively resembles table salt (often confused in the kitchen), the two have distinctly different physical and chemical properties. There are various types of sugar derived from different sources. While sugar is made with carbon, hydrogen, and oxygen, it is considerably harder to synthesize from its constituent elements than table salt is (Equation \ref{eq1}). However, the thermal decomposition is considerably easier and can be represented as a dehydration of sucrose to pure carbon and water vapor in Equation \ref{eq2}, and demonstrated in Video \(\PageIndex{2}\). \[\ce{C12H22O11 (s) + heat → 12C (s) + 11H2O (g)} \label{eq2} \] Video \(\PageIndex{2}\): A science experiment in the kitchen shows what happens to sugar molecules when they are heated. The experiment did not disappoint! As with salt, sugar has radically different properties (both physical and chemical) than its constituent elements. This difference in properties, of constituent elements and compounds, is a central feature of chemical reactions. Contributions & Attributions Wikipedia
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Courses/BethuneCookman_University/B-CU%3A_CH-141_General_Chemistry_1/Map%3A_Chemistry_-_Atoms_First_(OpenSTAX)/10%3A_Liquids_and_Solids/10.0%3A_Prelude_to_Liquids_and_Solids
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The great distances between atoms and molecules in a gaseous phase, and the corresponding absence of any significant interactions between them, allows for simple descriptions of many physical properties that are the same for all gases, regardless of their chemical identities. As described in the final module of the chapter on gases, this situation changes at high pressures and low temperatures—conditions that permit the atoms and molecules to interact to a much greater extent. In the liquid and solid states, these interactions are of considerable strength and play an important role in determining a number of physical properties that do depend on the chemical identity of the substance. In this chapter, the nature of these interactions and their effects on various physical properties of liquid and solid phases will be examined.
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Courses/Saint_Francis_University/CHEM_113%3A_Human_Chemistry_I_(Zovinka)/05%3A_Classification_and_Balancing_of_Chemical_Reactions/5.03%3A_Precipitation_Reactions_and_Solubility_Guidelines
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Learning Objectives Use solubility rules to predict whether a precipitate will form. Write and balance chemical equations for precipitation reactions. There are many types of chemical reactions that you will learn about in chemistry. In this chapter, we will focus on some reactions of ionic compounds: precipitation reactions (a type of double-replacement reaction ), acid-base neutralization reactions , and oxidation-reduction reactions . The practice of barter (trading one thing for another) has been in existence since the beginning of time. In the past, for example, items like chickens were bartered for newspapers. Person A had something that person B wanted, and vice versa. So, when person A and person B traded items, they each had something new. Some chemical reactions are like that—compounds swap parts, and the products are new materials. A double-replacement reaction is a reaction in which the positive and negative ions of two ionic compounds exchange places to form two new compounds. The general form of a double-replacement (also called double-displacement) reaction is: \[\ce{AB} + \ce{CD} \rightarrow \ce{AD} + \ce{CB}\] In this reaction, \(\ce{A}\) and \(\ce{C}\) are positively-charged cations, while \(\ce{B}\) and \(\ce{D}\) are negatively-charged anions. Double-replacement reactions generally occur between substances in aqueous solution. In order for a reaction to occur, one of the products is usually a solid precipitate, a gas, or a molecular compound such as water. Formation of a Precipitate A precipitate forms in a double-replacement reaction when the cations from one of the reactants combine with the anions from the other reactant to form an insoluble ionic compound. When aqueous solutions of potassium iodide and lead (II) nitrate are mixed, the following reaction occurs: \[2 \ce{KI} \left( aq \right) + \ce{Pb(NO_3)_2} \left( aq \right) \rightarrow 2 \ce{KNO_3} \left( aq \right) + \ce{PbI_2} \left( s \right)\] There are very strong attractive forces that occur between \(\ce{Pb^{2+}}\) and \(\ce{I^-}\) ions and the result is a brilliant yellow precipitate (see Figure \(\PageIndex{1}\) below). The other product of the reaction, potassium nitrate, remains soluble. To judge whether double-replacement reactions will occur, we need to know what kinds of ionic compounds form precipitates. For this, we use solubility rules , which are general statements that predict which ionic compounds dissolve (are soluble) and which do not (are not soluble, or insoluble). Table \(\PageIndex{1}\) lists some general solubility rules. We need to consider each ionic compound (both the reactants and the possible products) in light of the solubility rules. If a compound is soluble, we use the ( aq ) label with it, indicating that it dissolves. If a compound is not soluble, we use the ( s ) label with it and assume that it will precipitate out of solution. If everything is soluble, then no reaction will be expected. These compounds generally dissolve in water (are soluble): Exceptions: All compounds of Li+, Na+, K+, Rb+, Cs+, and NH4+ NaN All compounds of NO3− and C2H3O2− NaN Compounds of Cl−, Br−, I− Ag+, Hg22+, Pb2+ Compounds of SO42 Hg22+, Pb2+, Sr2+, Ba2+ These compounds generally do not dissolve in water (are insoluble): Exceptions: Compounds of CO32− and PO43− Compounds of Li+, Na+, K+, Rb+, Cs+, and NH4+ Compounds of OH− Compounds of Li+, Na+, K+, Rb+, Cs+, NH4+, Sr2+, and Ba2+ For example, consider the possible double-replacement reaction between Na 2 SO 4 and SrCl 2 . The solubility rules say that all ionic sodium compounds are soluble and all ionic chloride compounds are soluble, except for Ag + , Hg 2 2 + , and Pb 2 + , which are not being considered here. Therefore, Na 2 SO 4 and SrCl 2 are both soluble. The possible double-replacement reaction products are NaCl and SrSO 4 . Are these soluble? NaCl is (by the same rule we just quoted), but what about SrSO 4 ? Compounds of the sulfate ion are generally soluble, but Sr 2 + is an exception: we expect it to be insoluble—a precipitate. Therefore, we expect a reaction to occur, and the balanced chemical equation would be: \[\ce{Na2SO4(aq) + SrCl2(aq) → 2NaCl(aq) + SrSO4(s)}\nonumber \] You would expect to see a visual change corresponding to SrSO 4 precipitating out of solution (Figure \(\PageIndex{2}\)). Example \(\PageIndex{4}\): Will a precipitation reaction occur? If so, identify the products. Ca(NO 3 ) 2 + KBr → ? NaOH + FeCl 2 → ? Solution According to the solubility rules, both Ca(NO 3 ) 2 and KBr are soluble. Now we consider what the double-replacement products would be by switching the cations (or the anions)—namely, CaBr 2 and KNO 3 . However, the solubility rules predict that these two substances would also be soluble, so no precipitate would form. Thus, we predict no reaction in this case. According to the solubility rules, both NaOH and FeCl 2 are expected to be soluble. If we assume that a double-replacement reaction may occur, we need to consider the possible products, which would be NaCl and Fe(OH) 2 . NaCl is soluble, but, according to the solubility rules, Fe(OH) 2 is not. Therefore, a reaction would occur, and Fe(OH) 2 (s) would precipitate out of solution. The balanced chemical equation is \[\ce{2NaOH(aq) + FeCl2(aq) → 2NaCl(aq) + Fe(OH)2(s)}\nonumber \] Exercise \(\PageIndex{4}\) \[\ce{Sr(NO3)2 + KCl → }\nonumber \] Answer No reaction; all possible products are soluble. Key Takeaways A single-replacement reaction replaces one element for another in a compound. The periodic table or an activity series can help predict whether single-replacement reactions occur. A double-replacement reaction exchanges the cations (or the anions) of two ionic compounds. A precipitation reaction is a double-replacement reaction in which one product is a solid precipitate. Solubility rules are used to predict whether some double-replacement reactions will occur.
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Courses/Prince_Georges_Community_College/CHEM_1010%3A_General_Chemistry_(Miller)/09%3A_Chemical_Bonding_I-_Lewis_Structures_and_Determining_Molecular_Shapes/9.02%3A_Representing_Valance_Electrons_with_Dots
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Learning Objectives To use Lewis electron dot symbols to predict the number of bonds an element will form. Why are some substances chemically bonded molecules and others are an association of ions? The answer to this question depends upon the electronic structures of the atoms and nature of the chemical forces within the compounds. Although there are no sharply defined boundaries, chemical bonds are typically classified into three main types: ionic bonds, covalent bonds, and metallic bonds. In this chapter, each type of bond wil be discussed and the general properties found in typical substances in which the bond type occurs Ionic bonds results from electrostatic forces that exist between ions of opposite charge . These bonds typically involves a metal with a nonmetal Covalent bonds result from the sharing of electrons between two atoms . The bonds typically involves one nonmetallic element with another Metallic bonds These bonds are found in solid metals (copper, iron, aluminum) with each metal bonded to several neighboring groups and bonding electrons free to move throughout the 3-dimensional structure. Each bond classification is discussed in detail in subsequent sections of the chapter. Let's look at the preferred arrangements of electrons in atoms when they form chemical compounds. Lewis Symbols At the beginning of the 20 th century, the American chemist G. N. Lewis (1875–1946) devised a system of symbols—now called Lewis electron dot symbols (often shortened to Lewis dot symbols ) that can be used for predicting the number of bonds formed by most elements in their compounds. Each Lewis dot symbol consists of the chemical symbol for an element surrounded by dots that represent its valence electrons. Lewis Dot symbols: convenient representation of valence electrons allows you to keep track of valence electrons during bond formation consists of the chemical symbol for the element plus a dot for each valence electron To write an element’s Lewis dot symbol, we place dots representing its valence electrons, one at a time, around the element’s chemical symbol. Up to four dots are placed above, below, to the left, and to the right of the symbol (in any order, as long as elements with four or fewer valence electrons have no more than one dot in each position). The next dots, for elements with more than four valence electrons, are again distributed one at a time, each paired with one of the first four. For example, the electron configuration for atomic sulfur is [Ne]3s 2 3p 4 , thus there are six valence electrons. Its Lewis symbol would therefore be: Fluorine, for example, with the electron configuration [He]2 s 2 2 p 5 , has seven valence electrons, so its Lewis dot symbol is constructed as follows: Lewis used the unpaired dots to predict the number of bonds that an element will form in a compound. Consider the symbol for nitrogen in Figure 8.1.2. The Lewis dot symbol explains why nitrogen, with three unpaired valence electrons, tends to form compounds in which it shares the unpaired electrons to form three bonds. Boron, which also has three unpaired valence electrons in its Lewis dot symbol, also tends to form compounds with three bonds, whereas carbon, with four unpaired valence electrons in its Lewis dot symbol, tends to share all of its unpaired valence electrons by forming compounds in which it has four bonds. The Octet Rule In 1904, Richard Abegg formulated what is now known as Abegg's rule , which states that the difference between the maximum positive and negative valences of an element is frequently eight. This rule was used later in 1916 when Gilbert N. Lewis formulated the "octet rule" in his cubical atom theory. The octet rule refers to the tendency of atoms to prefer to have eight electrons in the valence shell . When atoms have fewer than eight electrons, they tend to react and form more stable compounds. Atoms will react to get in the most stable state possible. A complete octet is very stable because all orbitals will be full. Atoms with greater stability have less energy, so a reaction that increases the stability of the atoms will release energy in the form of heat or light ;reactions that decrease stability must absorb energy, getting colder. When discussing the octet rule, we do not consider d or f electrons. Only the s and p electrons are involved in the octet rule, making it a useful rule for the main group elements (elements not in the transition metal or inner-transition metal blocks); an octet in these atoms corresponds to an electron configurations ending with s 2 p 6 . Definition: Octet Rule A stable arrangement is attended when the atom is surrounded by eight electrons. This octet can be made up by own electrons and some electrons which are shared. Thus, an atom continues to form bonds until an octet of electrons is made. This is known as octet rule by Lewis. Normally two electrons pairs up and forms a bond, e.g., \(\ce{H_2}\) For most atoms there will be a maximum of eight electrons in the valence shell (octet structure), e.g., \(\ce{CH_4}\) The other tendency of atoms is to maintain a neutral charge. Only the noble gases (the elements on the right-most column of the periodic table) have zero charge with filled valence octets. All of the other elements have a charge when they have eight electrons all to themselves. The result of these two guiding principles is the explanation for much of the reactivity and bonding that is observed within atoms: atoms seek to share electrons in a way that minimizes charge while fulfilling an octet in the valence shell. The noble gases rarely form compounds. They have the most stable configuration (full octet, no charge), so they have no reason to react and change their configuration. All other elements attempt to gain, lose, or share electrons to achieve a noble gas configuration. Example \(\PageIndex{1}\): Salt The formula for table salt is NaCl. It is the result of Na + ions and Cl - ions bonding together. If sodium metal and chlorine gas mix under the right conditions, they will form salt. The sodium loses an electron, and the chlorine gains that electron. In the process, a great amount of light and heat is released. The resulting salt is mostly unreactive — it is stable. It will not undergo any explosive reactions, unlike the sodium and chlorine that it is made of. Why? Solution Referring to the octet rule, atoms attempt to get a noble gas electron configuration, which is eight valence electrons. Sodium has one valence electron, so giving it up would result in the same electron configuration as neon. Chlorine has seven valence electrons, so if it takes one it will have eight (an octet). Chlorine has the electron configuration of argon when it gains an electron. The octet rule could have been satisfied if chlorine gave up all seven of its valence electrons and sodium took them. In that case, both would have the electron configurations of noble gasses, with a full valence shell. However, their charges would be much higher. It would be Na 7 - and Cl 7 + , which is much less stable than Na + and Cl - . Atoms are more stable when they have no charge, or a small charge. Lewis dot symbols can also be used to represent the ions in ionic compounds. The reaction of cesium with fluorine, for example, to produce the ionic compound CsF can be written as follows: No dots are shown on Cs + in the product because cesium has lost its single valence electron to fluorine. The transfer of this electron produces the Cs + ion, which has the valence electron configuration of Xe, and the F − ion, which has a total of eight valence electrons (an octet) and the Ne electron configuration. This description is consistent with the statement that among the main group elements, ions in simple binary ionic compounds generally have the electron configurations of the nearest noble gas. The charge of each ion is written in the product, and the anion and its electrons are enclosed in brackets. This notation emphasizes that the ions are associated electrostatically; no electrons are shared between the two elements. Atoms often gain, lose, or share electrons to achieve the same number of electrons as the noble gas closest to them in the periodic table. As you might expect for such a qualitative approach to bonding, there are exceptions to the octet rule, which we describe elsewhere. These include molecules in which one or more atoms contain fewer or more than eight electrons. Summary Lewis dot symbols can be used to predict the number of bonds formed by most elements in their compounds. One convenient way to predict the number and basic arrangement of bonds in compounds is by using Lewis electron dot symbols , which consist of the chemical symbol for an element surrounded by dots that represent its valence electrons, grouped into pairs often placed above, below, and to the left and right of the symbol. The structures reflect the fact that the elements in period 2 and beyond tend to gain, lose, or share electrons to reach a total of eight valence electrons in their compounds, the so-called octet rule . Hydrogen, with only two valence electrons, does not obey the octet rule.
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Courses/BethuneCookman_University/BCU%3A_CH_332_Physical_Chemistry_II/Text/00%3A_Front_Matter/02%3A_InfoPage
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Courses/Saint_Francis_University/CHEM_113%3A_Human_Chemistry_I_(Zovinka)/06%3A_Chemical_Reactions_-_Mole_and_Mass_Relationships/6.01%3A_The_Mole_and_Avogadros_Number
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Learning Objectives Calculate formula masses for covalent and ionic compounds. Define the amount unit mole and the related quantity Avogadro’s number. Calculate molar mass of a compound from the molecular formula. We can argue that modern chemical science began when scientists started exploring the quantitative as well as the qualitative aspects of chemistry. For example, Dalton’s atomic theory was an attempt to explain the results of measurements that allowed him to calculate the relative masses of elements combined in various compounds. Understanding the relationship between the masses of atoms and the chemical formulas of compounds allows us to quantitatively describe the composition of substances. Formula Mass In an earlier chapter, we described the development of the atomic mass unit, the concept of average atomic masses, and the use of chemical formulas to represent the elemental makeup of substances. These ideas can be extended to calculate the formula mass of a substance, which is equal to the sum of the atomic masses for all the atoms represented in the substance’s formula. Formula Mass for Covalent Substances For covalent substances, the formula represents the numbers and types of atoms composing a single molecule of the substance; therefore, the formula mass may be correctly referred to as a molecular mass. Consider chloroform (CHCl 3 ), a covalent compound once used as a surgical anesthetic and now primarily used in the production of tetrafluoroethylene, the building block for the “anti-stick” polymer, Teflon. The molecular formula of chloroform indicates that a single molecule contains one carbon atom, one hydrogen atom, and three chlorine atoms. The average molecular mass of a chloroform molecule is therefore equal to the sum of the average atomic masses of these atoms. Figure \(\PageIndex{1}\) outlines the calculations used to derive the molecular mass of chloroform, which is 119.37 amu. Likewise, the molecular mass of an aspirin molecule, C 9 H 8 O 4 , is the sum of the atomic masses of nine carbon atoms, eight hydrogen atoms, and four oxygen atoms, which amounts to 180.15 amu (Figure \(\PageIndex{2}\)). Example \(\PageIndex{1}\) : Computing Molecular Mass for a Covalent Compound Ibuprofen, C 13 H 18 O 2 , is a covalent compound and the active ingredient in several popular nonprescription pain medications, such as Advil and Motrin. What is the molecular mass (amu) for this compound? Solution Molecules of this compound are comprised of 13 carbon atoms, 18 hydrogen atoms, and 2 oxygen atoms. Following the approach described above, the average molecular mass for this compound is therefore: Exercise \(\PageIndex{1}\) Acetaminophen, C 8 H 9 NO 2 , is a covalent compound and the active ingredient in several popular nonprescription pain medications, such as Tylenol. What is the molecular mass (amu) for this compound? Answer 151.16 amu Formula Mass for Ionic Compounds Ionic compounds are composed of discrete cations and anions combined in ratios to yield electrically neutral bulk matter. The formula mass for an ionic compound is calculated in the same way as the formula mass for covalent compounds: by summing the average atomic masses of all the atoms in the compound’s formula. Keep in mind, however, that the formula for an ionic compound does not represent the composition of a discrete molecule, so it may not correctly be referred to as the “molecular mass.” As an example, consider sodium chloride, NaCl, the chemical name for common table salt. Sodium chloride is an ionic compound composed of sodium cations, Na + , and chloride anions, Cl − , combined in a 1:1 ratio. The formula mass for this compound is computed as 58.44 amu (Figure \(\PageIndex{3}\)). Note that the average masses of neutral sodium and chlorine atoms were used in this computation, rather than the masses for sodium cations and chlorine anions. This approach is perfectly acceptable when computing the formula mass of an ionic compound. Even though a sodium cation has a slightly smaller mass than a sodium atom (since it is missing an electron), this difference will be offset by the fact that a chloride anion is slightly more massive than a chloride atom (due to the extra electron). Moreover, the mass of an electron is negligibly small with respect to the mass of a typical atom. Even when calculating the mass of an isolated ion, the missing or additional electrons can generally be ignored, since their contribution to the overall mass is negligible, reflected only in the nonsignificant digits that will be lost when the computed mass is properly rounded. The few exceptions to this guideline are very light ions derived from elements with precisely known atomic masses. Example \(\PageIndex{2}\): Computing Formula Mass for an Ionic Compound Aluminum sulfate, Al 2 (SO 4 ) 3 , is an ionic compound that is used in the manufacture of paper and in various water purification processes. What is the formula mass (amu) of this compound? Solution The formula for this compound indicates it contains Al 3 + and SO 4 2 − ions combined in a 2:3 ratio. For purposes of computing a formula mass, it is helpful to rewrite the formula in the simpler format, Al 2 S 3 O 12 . Following the approach outlined above, the formula mass for this compound is calculated as follows: Exercise \(\PageIndex{2}\) Calcium phosphate, \(\ce{Ca3(PO4)2}\), is an ionic compound and a common anti-caking agent added to food products. What is the formula mass (amu) of calcium phosphate? Answer 310.18 amu The Mole So far, we have been talking about chemical substances in terms of individual atoms and molecules. Yet we do not typically deal with substances one atom or molecule at a time; we work with millions, billions, and trillions of atoms and molecules at a time. We need a way to deal with macroscopic, rather than microscopic, amounts of matter. We need a unit of amount that relates quantities of substances on a scale that we can interact with. Chemistry uses a unit called mole . The mole (mol) is an counting term similar to familiar units like pair, dozen, gross, etc. It provides a specific measure of the number of atoms or molecules in a bulk sample of matter. A mole is defined as the amount of substance containing the same number of discrete entities (such as atoms, molecules, and ions) as the number of atoms in a sample of pure 12 C weighing exactly 12 g. One Latin connotation for the word “mole” is “large mass” or “bulk,” which is consistent with its use as the name for this unit. The mole provides a link between an easily measured macroscopic property, bulk mass, and an extremely important fundamental property, number of atoms, molecules, and so forth. The number of entities composing a mole has been experimentally determined to be \(6.02214179 \times 10^{23}\), a fundamental constant named Avogadro’s number (\(N_A\)) or the Avogadro constant in honor of Italian scientist Amedeo Avogadro. This constant is properly reported with an explicit unit of “per mole,” a conveniently rounded version being \(6.022 \times 10^{23}/\ce{mol}\). How big is a mole? It is very large. Suppose you had a mole of dollar bills that need to be counted. If everyone on earth (about 6 billion people) counted one bill per second, it would take about 3.2 million years to count all the bills. A mole of sand would fill a cube about 32 km on a side. A mole of pennies stacked on top of each other would have about the same diameter as our galaxy, the Milky Way. Atoms and molecules are very tiny, so one mole of carbon atoms would make a cube that is 1.74 cm on a side, small enough to carry in your pocket. One mole of water molecules is approximately 18 mL or just under 4 teaspoons of water. Example \(\PageIndex{3}\) How many molecules are present in 2.76 mol of H 2 O? How many atoms is this? Solution The definition of a mole is an equality that can be used to construct a conversion factor. Also, because we know that there are three atoms in each molecule of H 2 O, we can also determine the number of atoms in the sample. \[2.76\, \cancel{mol\, H_{2}O}\times \frac{6.022\times 10^{23}molecules\, H_{2}O}{\cancel{mol\, H_{2}O}}=1.66\times 10^{24}molecules\, H_{2}O \nonumber\nonumber \] To determine the total number of atoms, we have \[1.66\times 10^{24}\cancel{molecules\, H_{2}O}\times \frac{3\, atoms}{1\, molecule}=4.99\times 10^{24}\, atoms \nonumber\nonumber \] Exercise \(\PageIndex{3}\) How many molecules are present in 4.61 × 10 −2 mol of \(\ce{O2}\)? Answer 2.78 × 10 22 molecules Molar Mass Why is the mole unit so important? It represents the link between the microscopic and the macroscopic, especially in terms of mass. A mole of a substance has the same mass in grams as one unit (atom or molecules) has in atomic mass units . The mole unit allows us to express amounts of atoms and molecules in visible amounts that we can understand. For example, we already know that, by definition, a mole of carbon has a mass of exactly 12 g. This means that exactly 12 g of C has 6.022 × 10 23 atoms: 12 g C = 6.022 × 10 23 atoms C We can use this equality as a conversion factor between the number of atoms of carbon and the number of grams of carbon. How many grams are there, say, in 1.50 × 10 25 atoms of carbon? This is a one-step conversion: \[1.50\times 10^{25}\cancel{atoms\, C}\times \frac{12.0000\, g\, C}{6.022\times 10^{23}\cancel{atoms\, C}}=299\, g\, C\nonumber \] But it also goes beyond carbon. Previously we defined atomic and molecular masses as the number of atomic mass units per atom or molecule. Now we can do so in terms of grams. The atomic mass of an element is the number of grams in 1 mol of atoms of that element, while the molecular mass of a compound is the number of grams in 1 mol of molecules of that compound. Sometimes these masses are called molar masses to emphasize the fact that they are the mass for 1 mol of things. (The term molar is the adjective form of mole and has nothing to do with teeth.) Consistent with its definition as an amount unit, 1 mole of any element contains the same number of atoms as 1 mole of any other element. The masses of 1 mole of different elements, however, are different, since the masses of the individual atoms are drastically different. The molar mass of an element (or compound) is the mass in grams of 1 mole of that substance, a property expressed in units of grams per mole (g/mol) (Figure \(\PageIndex{1}\)). Because the definitions of both the mole and the atomic mass unit are based on the same reference substance, 12 C, the molar mass of any substance is numerically equivalent to its atomic or formula weight in amu. Per the amu definition, a single 12 C atom weighs 12 amu (its atomic mass is 12 amu). According to the definition of the mole, 12 g of 12 C contains 1 mole of 12 C atoms (its molar mass is 12 g/mol). This relationship holds for all elements, since their atomic masses are measured relative to that of the amu-reference substance, 12 C. Element Average Atomic Mass (amu) Molar Mass (g/mol) Atoms/Mole C 12.010 12.010 \(6.022 \times 10^{23}\) H 1.008 1.008 \(6.022 \times 10^{23}\) O 16.000 16.000 \(6.022 \times 10^{23}\) Na 22.990 22.990 \(6.022 \times 10^{23}\) Cl 33.450 35.450 \(6.022 \times 10^{23}\) While atomic mass and molar mass are numerically equivalent, keep in mind that they are vastly different in terms of scale, as represented by the vast difference in the magnitudes of their respective units (amu versus g). To appreciate the enormity of the mole, consider a small drop of water after a rainfall. Although this represents just a tiny fraction of 1 mole of water (~18 g), it contains more water molecules than can be clearly imagined. If the molecules were distributed equally among the roughly seven billion people on earth, each person would receive more than 100 billion molecules. The relationships between formula mass, the mole, and Avogadro’s number can be applied to compute various quantities that describe the composition of substances and compounds. For example, if we know the mass and chemical composition of a substance, we can determine the number of moles and calculate number of atoms or molecules in the sample. Likewise, if we know the number of moles of a substance, we can derive the number of atoms or molecules and calculate the substance’s mass. Here are some examples. The mass of 1 hydrogen atom is 1.0079 u; the mass of 1 mol of hydrogen atoms is 1.0079 g. Elemental hydrogen exists as a diatomic molecule, H 2 . One molecule has a mass of 1.0079 u + 1.0079 u = 2.0158 u, while 1 mol of H 2 has a mass of 1.0079 g + 1.0079 g = 2.0158 g. One molecule of H 2 O has a mass of about 18.01 u; 1 mol H 2 O has a mass of 18.01 g. A single unit of NaCl has a mass of 58.45 u; NaCl has a molar mass of 58.45 g. In each of these moles of substances, there are 6.022 × 10 23 units: 6.022 × 10 23 atoms of H, 6.022 × 10 23 molecules of H 2 and H 2 O, 6.022 × 10 23 units of NaCl ions. These relationships give us plenty of opportunities to construct conversion factors for simple calculations. Example \(\PageIndex{4}\): Sugar What is the molar mass of sugar (\(\ce{C6H12O6}\))? Solution To determine the molar mass, we simply add the atomic masses of the atoms in the molecular formula; but express the total in grams per mole, not atomic mass units. The masses of the atoms can be taken from the periodic table. 0 1 6 C = 6 × 12.011 = 72.066 12 H = 12 × 1.0079 = 12.0948 6 O = 6 × 15.999 = 95.994 TOTAL = 180.155 g/mol Per convention, the unit grams per mole is written as a fraction. Exercise \(\PageIndex{4}\) What is the molar mass of \(\ce{AgNO3}\)? Answer 169.87 g/mol Summary The mole is a key unit in chemistry. The molar mass of a substance, in grams, is numerically equal to one atom's or molecule's mass in atomic mass units.
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Bookshelves/Introductory_Chemistry/Beginning_Chemistry_(Ball)/02%3A_Measurements/2.E%3A_Measurements_(Exercises)
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Exercises (Expressing Numbers) 1. Express these numbers in scientific notation. 56.9 563,100 0.0804 0.00000667 2. Express these numbers in scientific notation. −890,000 602,000,000,000 0.0000004099 0.000000000000011 3. Express these numbers in scientific notation. 0.00656 65,600 4,567,000 0.000005507 4. Express these numbers in scientific notation. 65 −321.09 0.000077099 0.000000000218 5. Express these numbers in standard notation. 1.381 × 10 5 5.22 × 10 −7 9.998 × 10 4 6. Express these numbers in standard notation. 7.11 × 10 −2 9.18 × 10 2 3.09 × 10 −10 7. Express these numbers in standard notation. 8.09 × 10 0 3.088 × 10 −5 −4.239 × 10 2 8. Express these numbers in standard notation. 2.87 × 10 −8 1.78 × 10 11 1.381 × 10 −23 9. These numbers are not written in proper scientific notation. Rewrite them so that they are in proper scientific notation. 72.44 × 10 3 9,943 × 10 −5 588,399 × 10 2 10. These numbers are not written in proper scientific notation. Rewrite them so that they are in proper scientific notation. 0.000077 × 10 −7 0.000111 × 10 8 602,000 × 10 18 11. These numbers are not written in proper scientific notation. Rewrite them so that they are in proper scientific notation. 345.1 × 10 2 0.234 × 10 −3 1,800 × 10 −2 12. These numbers are not written in proper scientific notation. Rewrite them so that they are in proper scientific notation. 8,099 × 10 −8 34.5 × 10 0 0.000332 × 10 4 13. Write these numbers in scientific notation by counting the number of places the decimal point is moved. 123,456.78 98,490 0.000000445 14. Write these numbers in scientific notation by counting the number of places the decimal point is moved. 0.000552 1,987 0.00000000887 15. Use your calculator to evaluate these expressions. Express the final answer in proper scientific notation. 456 × (7.4 × 10 8 ) = ? (3.02 × 10 5 ) ÷ (9.04 × 10 15 ) = ? 0.0044 × 0.000833 = ? 16. Use your calculator to evaluate these expressions. Express the final answer in proper scientific notation. 98,000 × 23,000 = ? 98,000 ÷ 23,000 = ? (4.6 × 10 −5 ) × (2.09 × 10 3 ) = ? 17. Use your calculator to evaluate these expressions. Express the final answer in proper scientific notation. 45 × 132 ÷ 882 = ? [(6.37 × 10 4 ) × (8.44 × 10 −4 )] ÷ (3.2209 × 10 15 ) = ? 18. Use your calculator to evaluate these expressions. Express the final answer in proper scientific notation. (9.09 × 10 8 ) ÷ [(6.33 × 10 9 ) × (4.066 × 10 −7 )] = ? 9,345 × 34.866 ÷ 0.00665 = ? Answers 5.69 × 10 1 5.631 × 10 5 8.04 × 10 −2 6.67 × 10 −6 6.56 × 10 −3 6.56 × 10 4 4.567 × 10 6 5.507 × 10 −6 138,100 0.000000522 99,980 8.09 0.00003088 −423.9 7.244 × 10 4 9.943 × 10 −2 5.88399 × 10 7 3.451 × 10 4 2.34 × 10 −4 1.8 × 10 1 1.2345678 × 10 5 9.849 × 10 4 4.45 × 10 −7 3.3744 × 10 11 3.3407 × 10 −11 3.665 × 10 −6 6.7346 × 10 0 1.6691 × 10 −14 Exercises (Expressing Units) 1. Identify the unit in each quantity. 2 boxes of crayons 3.5 grams of gold 2. Identify the unit in each quantity. 32 oz of cheddar cheese 0.045 cm 3 of water 3. Identify the unit in each quantity. 9.58 s (the current world record in the 100 m dash) 6.14 m (the current world record in the pole vault) 4. Identify the unit in each quantity. 2 dozen eggs 2.4 km/s (the escape velocity of the moon, which is the velocity you need at the surface to escape the moon's gravity) 5. Indicate what multiplier each prefix represents. k m M 6. Indicate what multiplier each prefix represents. c G μ 7. Give the prefix that represents each multiplier. 1/1,000th × 1,000 × 1,000,000,000 × 8. Give the prefix that represents each multiplier. 1/1,000,000,000th × 1/100th × 1,000,000 × 9. Complete the following table with the missing information. Unit Abbreviation kilosecond NaN NaN mL NaN Mg centimeter NaN 10. Complete the following table with the missing information. Unit Abbreviation kilometer per second NaN second NaN NaN cm3 NaN μL nanosecond NaN 11. Express each quantity in a more appropriate unit. There may be more than one acceptable answer. 3.44 × 10 −6 s 3,500 L 0.045 m 12. Express each quantity in a more appropriate unit. There may be more than one acceptable answer. 0.000066 m/s (Hint: you need consider only the unit in the numerator.) 4.66 × 10 6 s 7,654 L 13. Express each quantity in a more appropriate unit. There may be more than one acceptable answer. 43,600 mL 0.0000044 m 1,438 ms 14. Express each quantity in a more appropriate unit. There may be more than one acceptable answer. 0.000000345 m 3 47,000,000 mm 3 0.00665 L 15. Multiplicative prefixes are used for other units as well, such as computer memory. The basic unit of computer memory is the byte (b). What is the unit for one million bytes? 16. You may have heard the terms microscale or nanoscale to represent the sizes of small objects. What units of length do you think are useful at these scales? What fractions of the fundamental unit of length are these units? 17. Acceleration is defined as a change in velocity per time. Propose a unit for acceleration in terms of the fundamental SI units. 18. Density is defined as the mass of an object divided by its volume. Propose a unit of density in terms of the fundamental SI units. Answers 1. boxes of crayons grams of gold 3. seconds meters 5. 1,000 x 1/1,000 x 1,000,000 x 7. milli- kilo- giga- 9. Unit Abbreviation kilosecond ks milliliter mL megagram Mg centimeter cm 11. 3.44 μs 3.5 kL 4.5 cm 13. 43.6 mL 4.4 μm 1.438 s 15. megabytes (Mb) 17. meters/second 2 Exercises (Significant Figures) 1. Express each measurement to the correct number of significant figures. 2. Express each measurement to the correct number of significant figures. 3. How many significant figures do these numbers have? 23 23.0 0.00023 0.0002302 4. How many significant figures do these numbers have? 5.44 × 10 8 1.008 × 10 −5 43.09 0.0000001381 5. How many significant figures do these numbers have? 765,890 765,890.0 1.2000 × 10 5 0.0005060 6. How many significant figures do these numbers have? 0.009 0.0000009 65,444 65,040 7. Compute and express each answer with the proper number of significant figures, rounding as necessary. 56.0 + 3.44 = ? 0.00665 + 1.004 = ? 45.99 − 32.8 = ? 45.99 − 32.8 + 75.02 = ? 8. Compute and express each answer with the proper number of significant figures, rounding as necessary. 1.005 + 17.88 = ? 56,700 − 324 = ? 405,007 − 123.3 = ? 55.5 + 66.66 − 77.777 = ? 9. Compute and express each answer with the proper number of significant figures, rounding as necessary. 56.7 × 66.99 = ? 1.0÷ 77 = ? 1.00÷ 77.0 = ? 6.022 × 1.89 = ? 10. Compute and express each answer with the proper number of significant figures, rounding as necessary. 0.000440 × 17.22 = ? 203,000 ÷ 0.044 = ? 67 × 85.0 × 0.0028 = ? 999,999 ÷ 3,310 = ? 11. a. Write the number 87,449 in scientific notation with four significant figures. b. Write the number 0.000066600 in scientific notation with five significant figures. 12. a. Write the number 306,000,000 in scientific notation to the proper number of significant figures. b. Write the number 0.0000558 in scientific notation with two significant figures 13. Perform each calculation and limit each answer to three significant figures. 67,883 × 0.004321 = ? (9.67 × 10 3 ) × 0.0055087 = ? 14. Perform each calculation and limit each answer to four significant figures. 18,900 × 76.33 ÷ 0.00336 = ? 0.77604 ÷ 76,003 × 8.888 = ? Answers 1. 375 psi 1.30 cm 3. two three two four 5. five seven five four 7. 59.4 1.011 13.2 88.2 9. 3.80 × 10 3 0.013 0.0130 11.4 11. a. \(8.745 \times 10^4\) b. \(6.6600 \times 10^{-5}\) 13. a. 293 b. 53.3 Exercises (Converting Units) 1. Write the two conversion factors that exist between the two given units. milliliters and liters microseconds and seconds kilometers and meters 2. Write the two conversion factors that exist between the two given units. kilograms and grams milliseconds and seconds centimeters and meters 3. Perform the following conversions. 5.4 km to meters 0.665 m to millimeters 0.665 m to kilometers 4. Perform the following conversions. 90.6 mL to liters 0.00066 ML to liters 750 L to kiloliters 5. Perform the following conversions. 17.8 μg to grams 7.22 × 10 2 kg to grams 0.00118 g to nanograms 6. Perform the following conversions. 833 ns to seconds 5.809 s to milliseconds 2.77 × 10 6 s to megaseconds 7. Perform the following conversions. 9.44 m 2 to square centimeters 3.44 × 10 8 mm 3 to cubic meters 8. Perform the following conversions. 0.00444 cm 3 to cubic meters 8.11 × 10 2 m 2 to square nanometers 9. Why would it be inappropriate to convert square centimeters to cubic meters? 10. Why would it be inappropriate to convert from cubic meters to cubic seconds? 11. Perform the following conversions. 45.0 m/min to meters/second 0.000444 m/s to micrometers/second 60.0 km/h to kilometers/second 12. Perform the following conversions. 3.4 × 10 2 cm/s to centimeters/minute 26.6 mm/s to millimeters/hour 13.7 kg/L to kilograms/milliliters 13. Perform the following conversions. 0.674 kL to milliliters 2.81 × 10 12 mm to kilometers 94.5 kg to milligrams 14. Perform the following conversions. 6.79 × 10 −6 kg to micrograms 1.22 mL to kiloliters 9.508 × 10 −9 ks to milliseconds 15. Perform the following conversions. 6.77 × 10 14 ms to kiloseconds 34,550,000 cm to kilometers 16. Perform the following conversions. 4.701 × 10 15 mL to kiloliters 8.022 × 10 −11 ks to microseconds 17. Perform the following conversions. Note that you will have to convert units in both the numerator and the denominator. 88 ft/s to miles/hour (Hint: use 5,280 ft = 1 mi.) 0.00667 km/h to meters/second 18. Perform the following conversions. Note that you will have to convert units in both the numerator and the denominator. 3.88 × 10 2 mm/s to kilometers/hour 1.004 kg/L to grams/milliliter 19. What is the area in square millimeters of a rectangle whose sides are 2.44 cm × 6.077 cm? Express the answer to the proper number of significant figures. 20. What is the volume in cubic centimeters of a cube with sides of 0.774 m? Express the answer to the proper number of significant figures. 21. The formula for the area of a triangle is 1/2 × base × height. What is the area of a triangle in square centimeters if its base is 1.007 m and its height is 0.665 m? Express the answer to the proper number of significant figures. 22. The formula for the area of a triangle is 1/2 × base × height. What is the area of a triangle in square meters if its base is 166 mm and its height is 930.0 mm? Express the answer to the proper number of significant figures. Answers 1. \(\dfrac{1000mL}{1L} \;and\; \dfrac{1L}{1000mL}\nonumber \) \(\dfrac{1000000\mu s}{1s}\;and\; \dfrac{1s}{1000000\mu s}\nonumber \) \(\dfrac{1000m}{1km}\;and\; \dfrac{1km}{1000m}\nonumber \) 3. 5,400 m 665 mm 6.65 × 10 −4 km 5. 1.78 × 10 −5 g 7.22 × 10 5 g 1.18 × 10 6 ng 7. 94,400 cm 2 0.344 m 3 9. One is a unit of area, and the other is a unit of volume. 11. 0.75 m/s 444 µm/s 1.666 × 10 −2 km/s 13. 674,000 mL 2.81 × 10 6 km 9.45 × 10 7 mg 15. 6.77 × 10 8 ks 345.5 km 17. 6.0 × 10 1 mi/h 0.00185 m/s 19. 1.48 × 10 3 mm 2 21. 3.35 × 10 3 cm 2 Exercises (Other Units - Temperature and Density) 1. Perform the following conversions. 255°F to degrees Celsius −255°F to degrees Celsius 50.0°C to degrees Fahrenheit −50.0°C to degrees Fahrenheit 2. Perform the following conversions. 1,065°C to degrees Fahrenheit −222°C to degrees Fahrenheit 400.0°F to degrees Celsius 200.0°F to degrees Celsius 3. Perform the following conversions. 100.0°C to kelvins −100.0°C to kelvins 100 K to degrees Celsius 300 K to degrees Celsius 4. Perform the following conversions. 1,000.0 K to degrees Celsius 50.0 K to degrees Celsius 37.0°C to kelvins −37.0°C to kelvins Convert 0 K to degrees Celsius. What is the significance of the temperature in degrees Celsius? Convert 0 K to degrees Fahrenheit. What is the significance of the temperature in degrees Fahrenheit? The hottest temperature ever recorded on the surface of the earth was 136°F in Libya in 1922. What is the temperature in degrees Celsius and in kelvins? The coldest temperature ever recorded on the surface of the earth was −128.6°F in Vostok, Antarctica, in 1983. What is the temperature in degrees Celsius and in kelvins? Give at least three possible units for density. What are the units when density is inverted? Give three examples. A sample of iron has a volume of 48.2 cm 3 . What is its mass? A sample of air has a volume of 1,015 mL. What is its mass? The volume of hydrogen used by the Hindenburg , the German airship that exploded in New Jersey in 1937, was 2.000 × 10 8 L. If hydrogen gas has a density of 0.0899 g/L, what mass of hydrogen was used by the airship? The volume of an Olympic-sized swimming pool is 2.50 × 10 9 cm 3 . If the pool is filled with alcohol ( d = 0.789 g/cm 3 ), what mass of alcohol is in the pool? A typical engagement ring has 0.77 cm 3 of gold. What mass of gold is present? A typical mercury thermometer has 0.039 mL of mercury in it. What mass of mercury is in the thermometer? What is the volume of 100.0 g of lead if lead has a density of 11.34 g/cm 3 ? What is the volume of 255.0 g of uranium if uranium has a density of 19.05 g/cm 3 ? What is the volume in liters of 222 g of neon if neon has a density of 0.900 g/L? What is the volume in liters of 20.5 g of sulfur hexafluoride if sulfur hexafluoride has a density of 6.164 g/L? Which has the greater volume, 100.0 g of iron ( d = 7.87 g/cm 3 ) or 75.0 g of gold ( d = 19.3 g/cm 3 )? Which has the greater volume, 100.0 g of hydrogen gas ( d = 0.0000899 g/cm 3 ) or 25.0 g of argon gas ( d = 0.00178 g/cm 3 )? Answers 1. 124°C −159°C 122°F −58°F 3. 373 K 173 K −173°C 27°C 5. −273°C. This is the lowest possible temperature in degrees Celsius. 7. 57.8°C; 331 K 9. g/mL, g/L, and kg/L (answers will vary) 11. 379 g 13. 1.80 × 10 7 g 15. 15 g 17. 8.818 cm 3 19. 247 L 21. The 100.0 g of iron has the greater volume Additional Exercises Evaluate 0.00000000552 × 0.0000000006188 and express the answer in scientific notation. You may have to rewrite the original numbers in scientific notation first. Evaluate 333,999,500,000 ÷ 0.00000000003396 and express the answer in scientific notation. You may need to rewrite the original numbers in scientific notation first. Express the number 6.022 × 10 23 in standard notation. Express the number 6.626 × 10 −34 in standard notation. When powers of 10 are multiplied together, the powers are added together. For example, 10 2 × 10 3 = 10 2+3 = 10 5 . With this in mind, can you evaluate (4.506 × 10 4 ) × (1.003 × 10 2 ) without entering scientific notation into your calculator? When powers of 10 are divided into each other, the bottom exponent is subtracted from the top exponent. For example, 10 5 /10 3 = 10 5−3 = 10 2 . With this in mind, can you evaluate (8.552 × 10 6 ) ÷ (3.129 × 10 3 ) without entering scientific notation into your calculator? Consider the quantity two dozen eggs. Is the number in this quantity "two" or "two dozen"? Justify your choice. Consider the quantity two dozen eggs. Is the unit in this quantity "eggs" or "dozen eggs"? Justify your choice. Fill in the blank: 1 km = ______________ μm. Fill in the blank: 1 Ms = ______________ ns. Fill in the blank: 1 cL = ______________ ML. Fill in the blank: 1 mg = ______________ kg. Express 67.3 km/h in meters/second. Express 0.00444 m/s in kilometers/hour. Using the idea that 1.602 km = 1.000 mi, convert a speed of 60.0 mi/h into kilometers/hour. Using the idea that 1.602 km = 1.000 mi, convert a speed of 60.0 km/h into miles/hour. Convert 52.09 km/h into meters/second. Convert 2.155 m/s into kilometers/hour. Use the formulas for converting degrees Fahrenheit into degrees Celsius to determine the relative size of the Fahrenheit degree over the Celsius degree. Use the formulas for converting degrees Celsius into kelvins to determine the relative size of the Celsius degree over kelvins. What is the mass of 12.67 L of mercury? What is the mass of 0.663 m 3 of air? What is the volume of 2.884 kg of gold? What is the volume of 40.99 kg of cork? Assume a density of 0.22 g/cm 3 . Answers 1. 3.42 × 10 −18 3. 602,200,000,000,000,000,000,000 5. 4.520 × 10 6 7. The quantity is two; dozen is the unit. 9.1,000,000,000 11. 1/100,000,000 13. 18.7 m/s 15. 96.1 km/h 17. 14.47 m/s 19. One Fahrenheit degree is nine-fifths the size of a Celsius degree. 21. 1.72 × 10 5 g 23. 149 mL
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Courses/University_of_British_Columbia/UBC_Introductory_Chemistry/02%3A_Naming_and_Composition/2.04%3A_Naming_Ionic_Compounds
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Learning Objectives To use the rules for naming ionic compounds. After learning a few more details about the names of individual ions, you will be one step away from knowing how to name ionic compounds. This section begins the formal study of nomenclature, the systematic naming of chemical compounds. Naming Ions The name of a monatomic cation is simply the name of the element followed by the word ion . Thus, Na + is the sodium ion, Al 3 + is the aluminum ion, Ca 2 + is the calcium ion, and so forth. We have seen that some elements lose different numbers of electrons, producing ions of different charges (Figure 3.3). Iron, for example, can form two cations, each of which, when combined with the same anion, makes a different compound with unique physical and chemical properties. Thus, we need a different name for each iron ion to distinguish Fe 2 + from Fe 3 + . The same issue arises for other ions with more than one possible charge. There are two ways to make this distinction. In the simpler, more modern approach, called the Stock system , an ion’s positive charge is indicated by a roman numeral in parentheses after the element name, followed by the word ion . Thus, Fe 2 + is called the iron(II) ion, while Fe 3 + is called the iron(III) ion. This system is used only for elements that form more than one common positive ion. We do not call the Na + ion the sodium(I) ion because (I) is unnecessary. Sodium forms only a 1+ ion, so there is no ambiguity about the name sodium ion . Element Stem Charge Modern Name Common Name iron ferr- 2+ iron(II) ion ferrous ion iron ferr- 3+ iron(III) ion ferric ion copper cupr- 1+ copper(I) ion cuprous ion copper cupr- 2+ copper(II) ion cupric ion tin stann- 2+ tin(II) ion stannous ion tin stann- 4+ tin(IV) ion stannic ion lead plumb- 2+ lead(II) ion plumbous ion lead plumb- 4+ lead(IV) ion plumbic ion chromium chrom- 2+ chromium(II) ion chromous ion chromium chrom- 3+ chromium(III) ion chromic ion gold aur- 1+ gold(I) ion aurous ion gold aur- 3+ gold(III) ion auric ion The second system, called the common system , is not conventional but is still prevalent and used in the health sciences. This system recognizes that many metals have two common cations. The common system uses two suffixes (- ic and - ous ) that are appended to the stem of the element name. The -ic suffix represents the greater of the two cation charges, and the -ous suffix represents the lower one. In many cases, the stem of the element name comes from the Latin name of the element. Table \(\PageIndex{1}\) lists the elements that use the common system, along with their respective cation names. Ion Name F− fluoride ion Cl− chloride ion Br− bromide ion I− iodide ion O2− oxide ion S2− sulfide ion P3− phosphide ion N3− nitride ion The name of a monatomic anion consists of the stem of the element name, the suffix - ide , and then the word ion . Thus, as we have already seen, Cl − is “chlor-” + “-ide ion,” or the chloride ion. Similarly, O 2− is the oxide ion, Se 2 − is the selenide ion, and so forth. Table \(\PageIndex{2}\) lists the names of some common monatomic ions. The polyatomic ions have their own characteristic names, as discussed earlier. Example \(\PageIndex{1}\) Name each ion. Ca 2 + S 2− SO 3 2 − NH 4 + Cu + Solution the calcium ion the sulfide ion the sulfite ion the ammonium ion the copper(I) ion or the cuprous ion Exercise \(\PageIndex{1}\) Name each ion. Fe 2 + Fe 3 + SO 4 2 − Ba 2 + HCO 3 − Answer a: iron(II) ion Answer b: iron(III) ion Answer c: sulfate ion Answer d: barium ion Answer e: hydrogen carbonate ion or bicarbonate ion Example \(\PageIndex{2}\) Write the formula for each ion. the bromide ion the phosphate ion the cupric ion the magnesium ion Solution Br − PO 4 3 − Cu 2 + Mg 2 + Exercise \(\PageIndex{2}\) Write the formula for each ion. the fluoride ion the carbonate ion the stannous ion the potassium ion Answer a: F - Answer b: CO 3 2 - Answer c: Sn 2+ Answer d: K + Naming Binary Ionic Compounds with a Metal that Forms Only One Type of Cation A binary ionic compound is a compound composed of a monatomic metal cation and a monatomic nonmetal anion . The metal cation is named first, followed by the nonmetal anion as illustrated in Figure \(\PageIndex{1}\) for the compound BaCl 2 . The word ion is dropped from both parts. Subscripts in the formula do not affect the name. Example \(\PageIndex{3}\): Naming Ionic Compounds Name each ionic compound. CaCl 2 AlF 3 KCl Solution Using the names of the ions, this ionic compound is named calcium chloride. The name of this ionic compound is aluminum fluoride. The name of this ionic compound is potassium chloride Exercise \(\PageIndex{3}\) Name each ionic compound. AgI MgO Ca 3 P 2 Answer a: silver iodide Answer b: magnesium oxide Answer c: calcium phosphide Naming Binary Ionic Compounds with a Metal That Forms More Than One Type of Cation If you are given a formula for an ionic compound whose cation can have more than one possible charge, you must first determine the charge on the cation before identifying its correct name. For example, consider FeCl 2 and FeCl 3 . In the first compound, the iron ion has a 2+ charge because there are two Cl − ions in the formula (1− charge on each chloride ion). In the second compound, the iron ion has a 3+ charge, as indicated by the three Cl − ions in the formula. These are two different compounds that need two different names. By the Stock system, the names are iron(II) chloride and iron(III) chloride (Figure \(\PageIndex{2}\)). Name of cation (metal) + (Roman Numeral in parenthesis) + Base name of anion (nonmetal) and -ide Name of cation (metal) + (Roman Numeral in parenthesis) + Base name of anion (nonmetal) and -ide.1 NaN NaN If we were to use the stems and suffixes of the common system, the names would be ferrous chloride and ferric chloride, respectively (Figure \(\PageIndex{3}\)) . "Old" base name of cation (metal) and -ic or -ous + Base name of anion (nonmetal) and -ide "Old" base name of cation (metal) and -ic or -ous + Base name of anion (nonmetal) and -ide.1 -ous (for ions with lower charge) -ic (for ions with higher charge) Example \(\PageIndex{4}\): Name each ionic compound. Co 2 O 3 FeCl 2 Solution Unnamed: 0 Explanation Answer a We know that cobalt can have more than one possible charge; we just need to determine what it is. Oxide always has a 2− charge, so with three oxide ions, we have a total negative charge of 6−. This means that the two cobalt ions have to contribute 6+, which for two cobalt ions means that each one is 3+. Therefore, the proper name for this ionic compound is cobalt(III) oxide. cobalt(III) oxide b Iron can also have more than one possible charge. Chloride always has a 1− charge, so with two chloride ions, we have a total negative charge of 2−. This means that the one iron ion must have a 2+ charge. Therefore, the proper name for this ionic compound is iron(II) chloride. iron(II) chloride Exercise \(\PageIndex{4}\) Name each ionic compound. AuCl 3 PbO 2 CuO Answer a: gold(III) chloride Answer b: lead(IV) oxide Answer c: copper(II) oxide Naming Ionic Compounds with Polyatomic Ions The process of naming ionic compounds with polyatomic ions is the same as naming binary ionic compounds. The cation is named first, followed by the anion. One example is the ammonium sulfate compound in Figure \(\PageIndex{6}\). Example \(\PageIndex{5}\): Naming Ionic Compounds Write the proper name for each ionic compound. (NH 4 ) 2 S AlPO 4 , Fe 3 (PO 4 ) 2 Solution Explanation Answer a. The ammonium ion has a 1+ charge and the sulfide ion has a 2− charge. Two ammonium ions need to balance the charge on a single sulfide ion. The compound’s name is ammonium sulfide. ammonium sulfide b. The ions have the same magnitude of charge, one of each (ion) is needed to balance the charges. The name of the compound is aluminum phosphate. aluminum phosphate c. Neither charge is an exact multiple of the other, so we have to go to the least common multiple of 6. To get 6+, three iron(II) ions are needed, and to get 6−, two phosphate ions are needed . The compound’s name is iron(II) phosphate. iron(II) phosphate Exercise \(\PageIndex{5A}\) Write the proper name for each ionic compound. (NH 4 ) 3 PO 4 Co(NO 2 ) 3 Answer a: ammonium phosphate Answer b: cobalt(III) nitrite Figure \(\PageIndex{1}\) is a synopsis of how to name simple ionic compounds. Exercise \(\PageIndex{5B}\) Name each ionic compound. ZnBr 2 Al 2 O 3 (NH 4 ) 3 PO 4 AuF 3 AgF Answer a: zinc bromide Answer b: aluminum oxide Answer c: ammonium phosphate Answer d: gold(III) fluoride or auric fluoride Answer e: silver fluoride Summary Ionic compounds are named by stating the cation first, followed by the anion. Positive and negative charges must balance. Some anions have multiple forms and are named accordingly with the use of roman numerals in parentheses. Ternary compounds are composed of three or more elements.
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Courses/Sacramento_City_College/SCC%3A_CHEM_300_-_Beginning_Chemistry/SCC%3A_CHEM_300_-_Beginning_Chemistry_(Alviar-Agnew)/09%3A_Electrons_in_Atoms_and_the_Periodic_Table/9.E%3A_Homework_Chapter_9
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Electromagnetic Radiation 1. Which of the following types of electromagnetic radiation has the shortest wavelength? a) gamma rays b) microwave c) visible d) radio waves 2. Which of the following types of electromagnetic radiation has the longest wavelength? a) ultraviolet b) radio waves c) x-ray d) infrared 3. Rank the following types of electromagnetic radiation in order of increasing wavelength: Infrared, Visible Light, X-rays, Radio Waves, Gamma Rays, Microwaves, Ultraviolet. The Quantum Mechanical Model 4. Sketch a 3D view (including x, y and z axes) of the 1s, 2s, and 3s orbitals respectively. 5. Sketch three different diagrams, each with a 3D view of the 2p orbitals along a different axis. 6. Sketch as many diagrams as possible, each with a 3D view of the 3d orbitals. Electron Configurations 7. Write full electron configurations for each of the following elements: a) Li: b) N: c) Cl: d) Ga: 8. Write full electron configurations for each of the following elements: a) Ca: b) Ar: c) Mn: d) P: 9. Write full electron configurations for each of the following elements: a) Xe: b) Ag: c) Cs: d) Hg: 10. Write full orbital diagrams and indicate the number of unpaired electrons for each element. a) Ne: b) O: c) Na: d) Si: 11. Write full orbital diagrams and indicate the number of unpaired electrons for each element. a) P: b) K: c) He: d) Al: 12. Write full orbital diagrams and indicate the number of unpaired electrons for each element. a) F: b) Be: c) B: d) Cl: 13. Write the noble gas electron configuration for the following elements. a) Nb: b) Se: c) Ca: d) Zn: 14. Write the noble gas electron configuration for the following elements. a) Ta: b) Fr: c) Ag: d) Al: 15. Write the noble gas electron configuration for the following elements. a) In: b) Ba: c) V: d) Y: Valence Electrons 16. Write orbital diagrams for the valence electrons of the following elements and indicate the number of unpaired electrons. a) Ar: b) F: c) As: d) Zn: 17. Write orbital diagrams for the valence electrons of the following elements and indicate the number of unpaired electrons. a) Cd: b) Na: c) I: d) Sn: 18. Write orbital diagrams for the valence electrons of the following elements and indicate the number of unpaired electrons. a) Sc: b) He: c) Hf: d) O: Electron Configurations and the Period Table 19. Write out the outer electron configuration for the following columns on the periodic table. a) 1A: b) 5A: c) 6A: 20. Write out the outer electron configuration for the following columns on the periodic table. a) 3A: b) 7A: c) 4A: 21. Write out the outer electron configuration for the following columns on the periodic table. a) 8A: b) 2A: 22. Write the noble gas electron configuration for the following elements. a) Au: b) Ni: c) S: d) Sr: 23. Write the noble gas electron configuration for the following elements. a) Ra: b) Pd: c) Br: d) Po: 24. Write the noble gas electron configuration for the following elements. a) Ti: b) Ru: c) B: d) Kr: 25. Write the noble gas electron configuration for the following elements. a) At: b) Pb: c) Ba: d) Ce: 26. Write the noble gas electron configuration for the following elements. a) Er: b) Pd: c) Ge: d) F: 27. Write the noble gas electron configuration for the following elements. a) Re: b) Al: c) Lr: d) Os: 28. How many 4d electrons are in an atom of each of the following elements? a) Y: b) Ag: c) Cs: d) Tc: 29. How many 3s electrons are in an atom of each of the following elements? a) Mg: b) K: c) Na: d) Li: 30. How many 5p electrons are in an atom of each of the following elements? a) Te: b) Xe: c) Ir: 31. Name an element in the 5 th period of the periodic table with: a) four 5p electrons: b) two valence electrons: c) two 5s electrons and three 4d electrons: d) one 5p electron: 32. Name an element in the 3 rd period of the periodic table with: a) three 3p electrons: b) one 3s electron: c) six 3p electrons: d) three valence electrons: 33. Name an element in the 4 th period of the periodic table with: a) five valence electrons: b) two 4p electrons: c) nine 3d orbital: d) one 4s electron: 34. Identify the element with the following noble gas electron configuration: a) [Ne] 3s 2 : b) [Xe] 6s 2 4f 7 : c) [Ar] 4s 2 3d 9 : d) [Ar] 4s 2 3d 10 4p 1 : 35. Identify the element with the following noble gas electron configuration: a) [Rn] 7s 2 5f 13 : b) [He] 2s 2 2p 2 : c) [Ar] 4s 2 3d 10 4p 6 : d) [Xe] 6s 2 4f 14 5d 4 : 36. Identify the element with the following noble gas electron configuration: a) [Ar] 4s 1 : b) [Ar] 4s 2 3d 10 4p 5 : c) [Kr] 5s 2 4d 2 : d) [Ne] 3s 2 3p 2 : Periodic Trends 37. Circle the element with the larger atoms in each given pair. a) Rb or Na b) F or Te c) Kr or Ge d) Pb or H 38. Circle the element with the smaller atoms in each given pair. a) He or Mg b) Ba or Se c) Rn or Cs d) Ne or Ca 39. Circle the element with more metallic characteristics in each given pair. a) Y or Cl b) Zn or K c) Tc or B d) H or Cs 40. Arrange the following elements in order of increasing atomic size: Rb, Cu, Cl, Cs, Ne 41. Arrange the following elements in order of increasing atomic size: O, Be, Sr, K, Kr 42. Arrange the following elements in order of decreasing metallic characteristics: Ti, Ne, O, Na, Cs Cumulative Challenge Problems 43. How many electrons can occupy the following quantum shells? a) n=1 b) n=2 c) n=3 d) n=4 44. Write full electron configurations for the following ions: a) Mg 2 + : b) Zn 2 + : c) O 2- : d) Cl - : 45. Write noble gas electron configurations for the following ions: a) Mn 2 + : b) Al 3 + : c) I - : d) Li + : 46. Based on your answers in problems 2 and 3 above, what conclusion can you draw regarding the difference between transition metal ions and other ions? 47. Come up with five different ions that have the same electron configuration of Argon. 48. Correct and rewrite the following incorrect electron configurations based on how many electrons they have. a) 1s 6 2p 3 2s 6 b) 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 4p 6 5s 2 5p 6 c) 1s 1 2s 2 3s 3 4s 4 5s 5 6s 6 d) 1s 2 1p 6 1d 10 2s 2 2p 6 2d 10 3s 2 3p 6 3d 10 4s 2 (for this one, provide your answer in the noble gas notation)
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Courses/Martin_Luther_College/Organic_Chemistry_-_MLC/02%3A_Tools_of_Organic_Chemistry/2.02%3A_Column_Chromatography/2.2.04%3A_Crystallization/2.2.4.03%3A_Choice_of_Solvent/2.2.4.3D%3A_Using_Solubility_Data
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If you are not following a procedure where a crystallization solvent has been specified, it may be helpful to consult solubility data for the desired compound. Qualitative (and sometimes quantitative) solubility data can be found for many compounds in the CRC\(^5\) and Merck Index.\(^6\) If the goal is to use just one solvent for crystallization, it's best to look for a solvent in which the desired compound is slightly soluble. Although the percent recovery will be less than \(100\%\) (a portion will remain in the mother liquid), there is a good chance that a slightly soluble solid will dissolve when heated. It is possible that solvents listed as insoluble will not dissolve the compound at any temperature. For example, below are entries in the CRC and Merck Index for the compound 1,4-dinitrobenzene: CRC Handbook: "i \(\ce{H_2O}\); sl \(\ce{EtOH}\), chl; s ace, bz, tol" Translation: The compound is insoluble in water; slightly soluble in ethanol and chloroform; and soluble in acetone, benzene, and toluene. Merck Index: "White crystals. . . one gram dissolves in \(12,500 \: \text{mL}\) cold water, \(555 \: \text{mL}\) boiling water, \(300 \: \text{mL}\) alcohol; sparingly soluble in benzene, chloroform, ethyl acetate." Translation: The compound is very insoluble in cold water, but much more soluble in hot water. It is slightly soluble in ethanol, and basically insoluble in benzene, chloroform, and ethyl acetate. 1,4-dinitrobenzene is slightly soluble in ethanol, making it a good "first guess" as a crystallization solvent. Indeed Perrin's Purification of Organic Chemicals \(^7\) recommends crystallization of 1,4-dinitrobenzene using ethanol or ethyl acetate. \(^5\) Handbook of Chemistry and Physics , CRC Press, 84^\text{th}\) edition, 2003-2004 . \(^6\) The Merck Index , Merck Research Laboratories, 12\(^\text{th}\) edition, 1996 . \(^7\)D.D. Perrin, W.L.F. Armarego, Purification of Organic Chemicals , Pergamon Press, 3\(^\text{rd}\) edition, 1988 .
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Courses/Mount_Aloysius_College/CHEM_100%3A_General_Chemistry_(O'Connor)/01%3A_Chemistry_Matter_and_Measurement
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The study of chemistry will open your eyes to a fascinating world. Chemical processes are continuously at work all around us. They happen as you cook and eat food, strike a match, shampoo your hair, and even read this page. Chemistry is called the central science because a knowledge of chemical principles is essential for other sciences. You might be surprised at the extent to which chemistry pervades your life. 1.1: Prelude to Chemistry, Matter, and Measurement Quantities and measurements are as important in our everyday lives as they are in medicine. The posted speed limits on roads and highways, such as 55 miles per hour (mph), are quantities we might encounter all the time. Both parts of a quantity, the amount (55) and the unit (mph), must be properly communicated to prevent potential problems. In chemistry, as in any technical endeavor, the proper expression of quantities is a necessary fundamental skill. 1.2: The Classification of Matter Matter can be described with both physical properties and chemical properties. Matter can be identified as an element, a compound, or a mixture. 1.3: Measurements Chemists measure the properties of matter and express these measurements as quantities. A quantity is an amount of something and consists of a number and a unit. The number tells us how many (or how much), and the unit tells us what the scale of measurement is. For example, when a distance is reported as “5 kilometers,” we know that the quantity has been expressed in units of kilometers and that the number of kilometers is 5. 1.4: Expressing Numbers - Scientific Notation Scientific notation is a system for expressing very large or very small numbers in a compact manner. It uses the idea that such numbers can be rewritten as a simple number multiplied by 10 raised to a certain exponent, or power. Scientific notation expressed numbers using powers of 10. 1.5: Expressing Numbers - Significant Figures Significant figures properly report the number of measured and estimated digits in a measurement. There are rules for applying significant figures in calculations. 1.6: The International System of Units Recognize the SI base units. Combining prefixes with base units creates new units of larger or smaller sizes. 1.7: Converting Units The ability to convert from one unit to another is an important skill. A unit can be converted to another unit of the same type with a conversion factor. 1.8: 1.8 Dosage Calculations Conversion factors are important in calculating dosages. 1.E: Chemistry, Matter, and Measurement (Exercises) These are homework exercises to accompany Chapter 1 of the Ball et al. "The Basics of GOB Chemistry" Textmap. 1.S: Chemistry, Matter, and Measurement (Summary) To ensure that you understand the material in this chapter, you should review the meanings of the bold terms in the following summary and ask yourself how they relate to the topics in the chapter. Template:HideTOC Thumbnail: Two small test tubes held in spring clamps. (CC BY-SA 3.0; mitchell125 ).
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Bookshelves/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/07%3A_The_Quantum-Mechanical_Model_of_the_Atom/7.01%3A_Schrodinger's_Cat
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In the early 1930's Erwin Schrödinger published a way of thinking about the circumstance of radioactive decay that is still useful. We imagine an apparatus containing just one Nitrogen-13 atom and a detector that will respond when the atom decays. Connected to the detector is a relay connected to a hammer, and when the atom decays the relay releases the hammer which then falls on a glass vial containing poison gas. We take the entire apparatus and put it in a box. We also place a cat in the box, close the lid, and wait 10 minutes. We then ask: Is the cat alive or dead? The answer according to quantum mechanics is that it is 50% dead and 50% alive. Quantum Mechanics describes the world in terms of a wave function . DeWitt wrote about the cat that "at the end of [one half-life] the total wave function for the system will have a form in which the living cat and dead cat are mixed in equal portions." (Reference: B.S. DeWitt and N. Graham, eds., The Many-Worlds Interpretation of Quantum Mechanics (Princeton, 1973), pg. 156.) When we open the box, we "collapse the wave function" or "collapse the state" and have either a live cat or a dead cat. Of course, this is just a thought experiment. So far as I know nobody has actually every done this experiment. In a sense the cat is a "red herring" [sorry!]. The paradox is just an illuminating way of thinking about the consequences of radioactive decay being totally random. Imagine we have a friend waiting outside when we open the box. For us the wave function collapses and we have, say, a live cat. But our friend's wave function does not collapse until he comes into the room. This leads to a strong solipsism, since our friend can they say that we owe our objective existence to his kind intervention in coming into the room and collapsing our state. As Heisenberg said, then, "The wave function represents partly a fact and partly our knowledge of a fact." Our friend needn't have come into the room to collapse his wave function: if we have a cell phone we can call him and tell him the result of the experiment. Of course, this assumes that we don't lie to him and tell him the cat is dead when it is alive. Unexplained but apparently true is the fact that when a state collapses, it collapses into the same state for everybody. If we see a live cat everybody sees a live cat (unless they or us are hallucinating).
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Bookshelves/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_4%3A_Equilibrium_in_Chemical_Reactions/14%3A_Chemical_Equilibrium/14.1%3A_The_Nature_of_Chemical_Equilibrium
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Learning Objectives To understand what is meant by chemical equilibrium. In the last chapter , we discussed the principles of chemical kinetics, which deal with the rate of change , or how quickly a given chemical reaction occurs. We now turn our attention to the extent to which a reaction occurs and how reaction conditions affect the final concentrations of reactants and products. For most of the reactions that we have discussed so far, you may have assumed that once reactants are converted to products, they are likely to remain that way. In fact, however, virtually all chemical reactions are reversible to some extent. That is, an opposing reaction occurs in which the products react, to a greater or lesser degree, to re-form the reactants. Eventually, the forward and reverse reaction rates become the same, and the system reaches chemical equilibrium, the point at which the composition of the system no longer changes with time. Chemical equilibrium is a dynamic process that consists of a forward reaction, in which reactants are converted to products, and a reverse reaction, in which products are converted to reactants. At equilibrium, the forward and reverse reactions proceed at equal rates. Consider, for example, a simple system that contains only one reactant and one product, the reversible dissociation of dinitrogen tetroxide (\(\ce{N_2O_4}\)) to nitrogen dioxide (\(\ce{NO_2}\)). You may recall that \(\ce{NO_2}\) is responsible for the brown color we associate with smog. When a sealed tube containing solid \(\ce{N_2O_4}\) (mp = −9.3°C; bp = 21.2°C) is heated from −78.4°C to 25°C, the red-brown color of \(\ce{NO_2}\) appears (Figure \(\PageIndex{1}\)). The reaction can be followed visually because the product (\(\ce{NO_2}\)) is colored, whereas the reactant (\(\ce{N_2O_4}\)) is colorless: \[\underset{colorless }{\ce{N2O4 (g)}} \ce{ <=>[k_f][k_r] } \underset{red-brown }{\ce{2NO2(g)}}\label{Eq1} \] The double arrow indicates that both the forward reaction \[\ce{N2O4 (g) ->[k_f] 2NO2(g)} \label{eq1B} \] and reverse reaction \[\ce{2NO2(g) ->[k_r] N2O4 (g) } \label{eq1C} \] occurring simultaneously (i.e, the reaction is reversible). However, this does not necessarily mean the system is equilibrium as the following chapter demonstrates. Figure \(\PageIndex{2}\) shows how the composition of this system would vary as a function of time at a constant temperature. If the initial concentration of \(\ce{NO_2}\) were zero, then it increases as the concentration of \(\ce{N_2O_4}\) decreases. Eventually the composition of the system stops changing with time, and chemical equilibrium is achieved. Conversely, if we start with a sample that contains no \(\ce{N_2O_4}\) but an initial \(\ce{NO_2}\) concentration twice the initial concentration of \(\ce{N_2O_4}\) (Figure \(\PageIndex{2a}\)), in accordance with the stoichiometry of the reaction, we reach exactly the same equilibrium composition (Figure \(\PageIndex{2b}\)). Thus equilibrium can be approached from either direction in a chemical reaction. Figure \(\PageIndex{3}\) shows the forward and reverse reaction rates for a sample that initially contains pure \(\ce{NO_2}\). Because the initial concentration of \(\ce{N_2O_4}\) is zero, the forward reaction rate (dissociation of \(\ce{N_2O_4}\)) is initially zero as well. In contrast, the reverse reaction rate (dimerization of \(\ce{NO_2}\)) is initially very high (\(2.0 \times 10^6\, M/s\)), but it decreases rapidly as the concentration of \(\ce{NO_2}\) decreases. As the concentration of \(\ce{N_2O_4}\) increases, the rate of dissociation of \(\ce{N_2O_4}\) increases—but more slowly than the dimerization of \(\ce{NO_2}\)—because the reaction is only first order in \(\ce{N_2O_4}\) (rate = \(k_f[N_2O_4]\), where \(k_f\) is the rate constant for the forward reaction in Equations \(\ref{Eq1}\) and \(\ref{eq1B}\)). Eventually, the forward and reverse reaction rates become identical, \(k_f = k_r\), and the system has reached chemical equilibrium. If the forward and reverse reactions occur at different rates, then the system is not at equilibrium. The rate of dimerization of \(\ce{NO_2}\) (reverse reaction) decreases rapidly with time, as expected for a second-order reaction. Because the initial concentration of \(\ce{N_2O_4}\) is zero, the rate of the dissociation reaction (forward reaction) at \(t = 0\) is also zero. As the dimerization reaction proceeds, the \(\ce{N_2O_4}\) concentration increases, and its rate of dissociation also increases. Eventually the rates of the two reactions are equal: chemical equilibrium has been reached, and the concentrations of \(\ce{N_2O_4}\) and \(\ce{NO_2}\) no longer change. At equilibrium, the forward reaction rate is equal to the reverse reaction rate. Example \(\PageIndex{1}\) The three reaction systems (1, 2, and 3) depicted in the accompanying illustration can all be described by the equation: \[2A \rightleftharpoons B \nonumber \] where the blue circles are \(A\) and the purple ovals are \(B\). Each set of panels shows the changing composition of one of the three reaction mixtures as a function of time. Which system took the longest to reach chemical equilibrium? Given : three reaction systems Asked for: relative time to reach chemical equilibrium Strategy : Compare the concentrations of A and B at different times. The system whose composition takes the longest to stabilize took the longest to reach chemical equilibrium. Solution : In systems 1 and 3, the concentration of A decreases from \(t_0\) through \(t_2\) but is the same at both \(t_2\) and \(t_3\). Thus systems 1 and 3 are at equilibrium by \(t_3\). In system 2, the concentrations of A and B are still changing between \(t_2\) and \(t_3\), so system 2 may not yet have reached equilibrium by \(t_3\). Thus system 2 took the longest to reach chemical equilibrium. Exercise \(\PageIndex{1}\) In the following illustration, A is represented by blue circles, B by purple squares, and C by orange ovals; the equation for the reaction is A + B ⇌ C. The sets of panels represent the compositions of three reaction mixtures as a function of time. Which, if any, of the systems shown has reached equilibrium? Answer system 2 A Video Introduction to Dynamic Equilibrium: Introduction to Dynamic Equilibrium(opens in new window) [youtu.be] Summary At equilibrium, the forward and reverse reactions of a system proceed at equal rates. Chemical equilibrium is a dynamic process consisting of forward and reverse reactions that proceed at equal rates. At equilibrium, the composition of the system no longer changes with time. The composition of an equilibrium mixture is independent of the direction from which equilibrium is approached.
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Courses/Stanford_Online_High_School/TEN2C-Carbon/4%3A_Chirality/4.03%3A_Naming_Enantiomers_by_the_RS_System
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To name the enantiomers of a compound unambiguously, their names must include the "handedness" of the molecule. The method for this is formally known as R/S nomenclature. Introduction The method of unambiguously assigning the handedness of molecules was originated by three chemists: R.S. Cahn, C. Ingold, and V. Prelog and is also often called the Cahn-Ingold-Prelog rules . In addition to the Cahn-Ingold system, there are two ways of experimentally determining the absolute configuration of an enantiomer: X-ray diffraction analysis. Note that there is no correlation between the sign of rotation and the structure of a particular enantiomer. Chemical correlation with a molecule whose structure has already been determined via X-ray diffraction. However, for non-laboratory purposes, it is beneficial to focus on the R/S system. The sign of optical rotation , although different for the two enantiomers of a chiral molecule, at the same temperature, cannot be used to establish the absolute configuration of an enantiomer; this is because the sign of optical rotation for a particular enantiomer may change when the temperature changes. Stereocenters are labeled R or S The "right hand" and "left hand" nomenclature is used to name the enantiomers of a chiral compound. The stereocenters are labeled as R or S. Consider the first picture: a curved arrow is drawn from the highest priority ( 1 ) substituent to the lowest priority (4 ) substituent. If the arrow points in a counterclockwise direction ( left when leaving the 12 o' clock position), the configuration at stereocenter is considered S ("Sinister" → Latin= "left"). If, however, the arrow points clockwise,( Right when leaving the 12 o' clock position) then the stereocenter is labeled R ("Rectus" → Latin= "right"). The R or S is then added as a prefix, in parenthesis, to the name of the enantiomer of interest. For example: ( R )-2-Bromobutane and ( S )-2,3- Dihydroxypropanal. Sequence rules to assign priorities to substituents Before applying the R and S nomenclature to a stereocenter, the substituents must be prioritized according to the following rules: Rule 1 First, examine at the atoms directly attached to the stereocenter of the compound. A substituent with a higher atomic number takes precedence over a substituent with a lower atomic number. Hydrogen is the lowest possible priority substituent, because it has the lowest atomic number. When dealing with isotopes, the atom with the higher atomic mass receives higher priority. When visualizing the molecule, the lowest priority substituent should always point away from the viewer (a dashed line indicates this). To understand how this works or looks, imagine that a clock and a pole. Attach the pole to the back of the clock, so that when when looking at the face of the clock the pole points away from the viewer in the same way the lowest priority substituent should point away. Then, draw an arrow from the highest priority atom to the 2nd highest priority atom to the 3rd highest priority atom. Because the 4th highest priority atom is placed in the back, the arrow should appear like it is going across the face of a clock. If it is going clockwise, then it is an R-enantiomer; If it is going counterclockwise, it is an S-enantiomer. When looking at a problem with wedges and dashes, if the lowest priority atom is not on the dashed line pointing away, the molecule must be rotated. Remember that Wedges indicate coming towards the viewer. Dashes indicate pointing away from the viewer. Rule 2 If there are two substituents with equal rank, proceed along the two substituent chains until there is a point of difference. First, determine which of the chains has the first connection to an atom with the highest priority (the highest atomic number). That chain has the higher priority. If the chains are similar, proceed down the chain, until a point of difference. For example : an ethyl substituent takes priority over a methyl substituent. At the connectivity of the stereocenter, both have a carbon atom, which are equal in rank. Going down the chains, a methyl has only has hydrogen atoms attached to it, whereas the ethyl has another carbon atom. The carbon atom on the ethyl is the first point of difference and has a higher atomic number than hydrogen; therefore the ethyl takes priority over the methyl. Rule 3 If a chain is connected to the same kind of atom twice or three times, check to see if the atom it is connected to has a greater atomic number than any of the atoms that the competing chain is connected to. If none of the atoms connected to the competing chain(s) at the same point has a greater atomic number: the chain bonded to the same atom multiple times has the greater priority If however, one of the atoms connected to the competing chain has a higher atomic number: that chain has the higher priority. Example 2 A 1-methylethyl substituent takes precedence over an ethyl substituent. Connected to the first carbon atom, ethyl only has one other carbon, whereas the 1-methylethyl has two carbon atoms attached to the first; this is the first point of difference. Therefore, 1-methylethyl ranks higher in priority than ethyl, as shown below: However: Remember that being double or triple bonded to an atom means that the atom is connected to the same atom twice. In such a case, follow the same method as above. Caution!! Keep in mind that priority is determined by the first point of difference along the two similar substituent chains. After the first point of difference, the rest of the chain is irrelevant. When looking for the first point of difference on similar substituent chains, one may encounter branching. If there is branching, choose the branch that is higher in priority. If the two substituents have similar branches, rank the elements within the branches until a point of difference. After all your substituents have been prioritized in the correct manner, you can now name/label the molecule R or S. Put the lowest priority substituent in the back (dashed line). Proceed from 1 to 2 to 3. (it is helpful to draw or imagine an arcing arrow that goes from 1--> 2-->3) Determine if the direction from 1 to 2 to 3 clockwise or counterclockwise. i) If it is clockwise it is R. ii) if it is counterclockwise it is S . USE YOUR MODELING KIT: Models assist in visualizing the structure. When using a model, make sure the lowest priority is pointing away from you. Then determine the direction from the highest priority substituent to the lowest: clockwise (R) or counterclockwise (S). IF YOU DO NOT HAVE A MODELING KIT : remember that the dashes mean the bond is going into the screen and the wedges means that bond is coming out of the screen. If the lowest priority bond is not pointing to the back, mentally rotate it so that it is. However, it is very useful when learning organic chemistry to use models. If you have a modeling kit use it to help you solve the following practice problems. Exercise \(\PageIndex{1}\) Are the following R or S? Answer S: I > Br > F > H. The lowest priority substituent, H, is already going towards the back. It turns left going from I to Br to F, so it's a S. R: Br > Cl > CH 3 > H. You have to switch the H and Br in order to place the H, the lowest priority, in the back. Then, going from Br to Cl, CH 3 is turning to the right, giving you a R. Neither R or S : This molecule is achiral. Only chiral molecules can be named R or S. R: OH > CN > CH 2 NH 2 > H. The H, the lowest priority, has to be switched to the back. Then, going from OH to CN to CH 2 NH 2 , you are turning right, giving you a R. (5) S: \(\ce{-COOH}\) > \(\ce{-CH_2OH}\) > \(\ce{C#CH}\) > \(\ce{H}\). Then, going from \(\ce{-COOH}\) to \(\ce{-CH_2OH}\) to \(\ce{-C#CH}\) you are turning left, giving you a S configuration. References Schore and Vollhardt. Organic Chemistry Structure and Function. New York:W.H. Freeman and Company, 2007. McMurry, John and Simanek, Eric. Fundamentals of Organic Chemistry . 6th Ed. Brooks Cole, 2006.
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Bookshelves/Introductory_Chemistry/Introductory_Chemistry_(LibreTexts)/11%3A_Gases/11.05%3A_Charless_Law-_Volume_and_Temperature
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Learning Objectives Learn and apply Charles's Law. Everybody enjoys the smell and taste of freshly-baked bread. It is light and fluffy as a result of the action of yeast on sugar. The yeast converts the sugar to carbon dioxide, which at high temperatures causes the dough to expand. The end result is an enjoyable treat, especially when covered with melted butter. Charles's Law French physicist Jacques Charles (1746-1823) studied the effect of temperature on the volume of a gas at constant pressure. Charles's Law states that the volume of a given mass of gas varies directly with the absolute temperature of the gas when pressure is kept constant. The absolute temperature is temperature measured with the Kelvin scale. The Kelvin scale must be used because zero on the Kelvin scale corresponds to a complete stop of molecular motion. Mathematically, the direct relationship of Charles's Law can be represented by the following equation: \[\dfrac{V}{T} = k \nonumber \] As with Boyle's Law, \(k\) is constant only for a given gas sample. The table below shows temperature and volume data for a set amount of gas at a constant pressure. The third column is the constant for this particular data set and is always equal to the volume divided by the Kelvin temperature. Temperature \(\left( \text{K} \right)\) Volume \(\left( \text{mL} \right)\) \(\dfrac{V}{T} = k\) \(\left( \dfrac{\text{mL}}{\text{K}} \right)\) 50 20 0.4 100 40 0.4 150 60 0.4 200 80 0.4 300 120 0.4 500 200 0.4 1000 400 0.4 When this data is graphed, the result is a straight line, indicative of a direct relationship, shown in the figure below. Notice that the line goes exactly toward the origin, meaning that as the absolute temperature of the gas approaches zero, its volume approaches zero. However, when a gas is brought to extremely cold temperatures, its molecules would eventually condense into the liquid state before reaching absolute zero. The temperature at which this change into the liquid state occurs varies for different gases. Charles's Law can also be used to compare changing conditions for a gas. Now we use \(V_1\) and \(T_1\) to stand for the initial volume and temperature of a gas, while \(V_2\) and \(T_2\) stand for the final volume and temperature. The mathematical relationship of Charles's Law becomes: \[\dfrac{V_1}{T_1} = \dfrac{V_2}{T_2} \nonumber \] This equation can be used to calculate any one of the four quantities if the other three are known. The direct relationship will only hold if the temperatures are expressed in Kelvin. Temperatures in Celsius will not work. Recall the relationship that \(\text{K} = \: ^\text{o} \text{C} + 273\). Example \(\PageIndex{1}\): A balloon is filled to a volume of \(2.20 \: \text{L}\) at a temperature of \(22^\text{o} \text{C}\). The balloon is then heated to a temperature of \(71^\text{o} \text{C}\). Find the new volume of the balloon. Solution Steps for Problem Solving Unnamed: 1 Identify the "given" information and what the problem is asking you to "find." Given: \(V_1 = 2.20 \: \text{L}\) and \(T_1 = 22^\text{o} \text{C} = 295 \: \text{K}\) \(T_2 = 71^\text{o} \text{C} = 344 \: \text{K}\) Find: V2 = ? L List other known quantities. The temperatures have first been converted to Kelvin. Plan the problem. First, rearrange the equation algebraically to solve for \(V_2\). \[V_2 = \dfrac{V_1 \times T_2}{T_1} \nonumber \] Cancel units and calculate. Now substitute the known quantities into the equation and solve. \[V_2 = \dfrac{2.20 \: \text{L} \times 344 \: \cancel{\text{K}}}{295 \: \cancel{\text{K}}} = 2.57 \: \text{L} \nonumber \] Think about your result. The volume increases as the temperature increases. The result has three significant figures. Exercise \(\PageIndex{1}\) If V 1 = 3.77 L and T 1 = 255 K, what is V 2 if T 2 = 123 K? Answer 1.82 L Example \(\PageIndex{2}\): A sample of a gas has an initial volume of 34.8 L and an initial temperature of −67°C. What must be the temperature of the gas for its volume to be 25.0 L? Solution Steps for Problem Solving Unnamed: 1 Identify the "given" information and what the problem is asking you to "find." Given: Given:T1 = -27oC and V1 = 34.8 L V2 = 25.0 L Find: T2 = ? K List other known quantities. K = -27oC + 273 Plan the problem. 1. Convert the initial temperature to Kelvin 2. Rearrange the equation algebraically to solve for \(T_2\). \[T_2 = \dfrac{V_2 \times T_1}{V_1} \nonumber \] Cancel units and calculate. 1. −67°C + 273 = 206 K 2. Substitute the known quantities into the equation and solve. \[T_2 = \dfrac{25.0 \: \cancel{\text{L}} \times 206 \: \text{K}}{34.8 \: \cancel{\text{L}}} = 148 \: \text{K} \nonumber \] Think about your result. This is also equal to −125°C. As temperature decreases, volume decreases—which it does in this example. Exercise \(\PageIndex{2}\) If V 1 = 623 mL, T 1 = 255°C, and V 2 = 277 mL, what is T 2 ? Answer 235 K, or −38°C Summary Charles’s Law relates the volume and temperature of a gas at constant pressure and amount.
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Courses/San_Diego_Miramar_College/Chemistry_201%3A_General_Chemistry_II_(Garces)/01%3A_Liquids_Solids_and_Intermolecular_Forces/1.03%3A_Intermolecular_Forces-_The_Forces_that_Hold_Condensed_Phases_Together
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Learning Objectives To describe the intermolecular forces in liquids. The properties of liquids are intermediate between those of gases and solids, but are more similar to solids. In contrast to intra molecular forces, such as the covalent bonds that hold atoms together in molecules and polyatomic ions, inter molecular forces hold molecules together in a liquid or solid. Intermolecular forces are generally much weaker than covalent bonds. For example, it requires 927 kJ to overcome the intramolecular forces and break both O–H bonds in 1 mol of water, but it takes only about 41 kJ to overcome the intermolecular attractions and convert 1 mol of liquid water to water vapor at 100°C. (Despite this seemingly low value, the intermolecular forces in liquid water are among the strongest such forces known!) Given the large difference in the strengths of intra- and intermolecular forces, changes between the solid, liquid, and gaseous states almost invariably occur for molecular substances without breaking covalent bonds . The properties of liquids are intermediate between those of gases and solids, but are more similar to solids. Intermolecular forces determine bulk properties, such as the melting points of solids and the boiling points of liquids. Liquids boil when the molecules have enough thermal energy to overcome the intermolecular attractive forces that hold them together, thereby forming bubbles of vapor within the liquid. Similarly, solids melt when the molecules acquire enough thermal energy to overcome the intermolecular forces that lock them into place in the solid. Intermolecular forces are electrostatic in nature; that is, they arise from the interaction between positively and negatively charged species. Like covalent and ionic bonds, intermolecular interactions are the sum of both attractive and repulsive components. Because electrostatic interactions fall off rapidly with increasing distance between molecules, intermolecular interactions are most important for solids and liquids, where the molecules are close together. These interactions become important for gases only at very high pressures, where they are responsible for the observed deviations from the ideal gas law at high pressures. In this section, we explicitly consider three kinds of intermolecular interactions. There are two additional types of electrostatic interaction that you are already familiar with: the ion–ion interactions that are responsible for ionic bonding, and the ion–dipole interactions that occur when ionic substances dissolve in a polar substance such as water. The first two are often described collectively as van der Waals forces. Dipole–Dipole Interactions Polar covalent bonds behave as if the bonded atoms have localized fractional charges that are equal but opposite (i.e., the two bonded atoms generate a dipole ). If the structure of a molecule is such that the individual bond dipoles do not cancel one another, then the molecule has a net dipole moment. Molecules with net dipole moments tend to align themselves so that the positive end of one dipole is near the negative end of another and vice versa, as shown in Figure \(\PageIndex{1a}\). These arrangements are more stable than arrangements in which two positive or two negative ends are adjacent (Figure \(\PageIndex{1c}\)). Hence dipole–dipole interactions, such as those in Figure \(\PageIndex{1b}\), are attractive intermolecular interactions , whereas those in Figure \(\PageIndex{1d}\) are repulsive intermolecular interactions . Because molecules in a liquid move freely and continuously, molecules always experience both attractive and repulsive dipole–dipole interactions simultaneously, as shown in Figure \(\PageIndex{2}\). On average, however, the attractive interactions dominate. Because each end of a dipole possesses only a fraction of the charge of an electron, dipole–dipole interactions are substantially weaker than the interactions between two ions, each of which has a charge of at least ±1, or between a dipole and an ion, in which one of the species has at least a full positive or negative charge. In addition, the attractive interaction between dipoles falls off much more rapidly with increasing distance than do the ion–ion interactions. Recall that the attractive energy between two ions is proportional to 1/ r , where r is the distance between the ions. Doubling the distance ( r → 2 r ) decreases the attractive energy by one-half. In contrast, the energy of the interaction of two dipoles is proportional to 1/ r 3 , so doubling the distance between the dipoles decreases the strength of the interaction by 2 3 , or 8-fold. Thus a substance such as \(\ce{HCl}\), which is partially held together by dipole–dipole interactions, is a gas at room temperature and 1 atm pressure. Conversely, \(\ce{NaCl}\), which is held together by interionic interactions, is a high-melting-point solid. Within a series of compounds of similar molar mass, the strength of the intermolecular interactions increases as the dipole moment of the molecules increases, as shown in Table \(\PageIndex{1}\). Compound Molar Mass (g/mol) Dipole Moment (D) Boiling Point (K) C3H6 (cyclopropane) 42 0.0 240 CH3OCH3 (dimethyl ether) 46 1.3 248 CH3CN (acetonitrile) 41 3.9 355 The attractive energy between two ions is proportional to 1/r, whereas the attractive energy between two dipoles is proportional to 1/r6. Video Discussing Dipole Intermolecular Forces. Source: Dipole Intermolecular Force, YouTube(opens in new window) [youtu.be] Example \(\PageIndex{1}\) Arrange ethyl methyl ether (CH 3 OCH 2 CH 3 ), 2-methylpropane [isobutane, (CH 3 ) 2 CHCH 3 ], and acetone (CH 3 COCH 3 ) in order of increasing boiling points. Their structures are as follows: Given: compounds. Asked for: order of increasing boiling points. Strategy: Compare the molar masses and the polarities of the compounds. Compounds with higher molar masses and that are polar will have the highest boiling points. Solution: The three compounds have essentially the same molar mass (58–60 g/mol), so we must look at differences in polarity to predict the strength of the intermolecular dipole–dipole interactions and thus the boiling points of the compounds. The first compound, 2-methylpropane, contains only C–H bonds, which are not very polar because C and H have similar electronegativities. It should therefore have a very small (but nonzero) dipole moment and a very low boiling point. Ethyl methyl ether has a structure similar to H 2 O; it contains two polar C–O single bonds oriented at about a 109° angle to each other, in addition to relatively nonpolar C–H bonds. As a result, the C–O bond dipoles partially reinforce one another and generate a significant dipole moment that should give a moderately high boiling point. Acetone contains a polar C=O double bond oriented at about 120° to two methyl groups with nonpolar C–H bonds. The C–O bond dipole therefore corresponds to the molecular dipole, which should result in both a rather large dipole moment and a high boiling point. Thus we predict the following order of boiling points: 2-methylpropane < ethyl methyl ether < acetone This result is in good agreement with the actual data: 2-methylpropane, boiling point = −11.7°C, and the dipole moment (μ) = 0.13 D; methyl ethyl ether, boiling point = 7.4°C and μ = 1.17 D; acetone, boiling point = 56.1°C and μ = 2.88 D. Exercise \(\PageIndex{1}\) Arrange carbon tetrafluoride (CF 4 ), ethyl methyl sulfide (CH 3 SC 2 H 5 ), dimethyl sulfoxide [(CH 3 ) 2 S=O], and 2-methylbutane [isopentane, (CH 3 ) 2 CHCH 2 CH 3 ] in order of decreasing boiling points. Answer dimethyl sulfoxide (boiling point = 189.9°C) > ethyl methyl sulfide (boiling point = 67°C) > 2-methylbutane (boiling point = 27.8°C) > carbon tetrafluoride (boiling point = −128°C) London Dispersion Forces Thus far, we have considered only interactions between polar molecules. Other factors must be considered to explain why many nonpolar molecules, such as bromine, benzene, and hexane, are liquids at room temperature; why others, such as iodine and naphthalene, are solids. Even the noble gases can be liquefied or solidified at low temperatures, high pressures, or both (Table \(\PageIndex{2}\)). What kind of attractive forces can exist between nonpolar molecules or atoms? This question was answered by Fritz London (1900–1954), a German physicist who later worked in the United States. In 1930, London proposed that temporary fluctuations in the electron distributions within atoms and nonpolar molecules could result in the formation of short-lived instantaneous dipole moments, which produce attractive forces called London dispersion forces between otherwise nonpolar substances. Substance Molar Mass (g/mol) Melting Point (°C) Boiling Point (°C) Ar 40 −189.4 −185.9 Xe 131 −111.8 −108.1 N2 28 −210 −195.8 O2 32 −218.8 −183.0 F2 38 −219.7 −188.1 I2 254 113.7 184.4 CH4 16 −182.5 −161.5 Consider a pair of adjacent He atoms, for example. On average, the two electrons in each He atom are uniformly distributed around the nucleus. Because the electrons are in constant motion, however, their distribution in one atom is likely to be asymmetrical at any given instant, resulting in an instantaneous dipole moment. As shown in part (a) in Figure \(\PageIndex{3}\), the instantaneous dipole moment on one atom can interact with the electrons in an adjacent atom, pulling them toward the positive end of the instantaneous dipole or repelling them from the negative end. The net effect is that the first atom causes the temporary formation of a dipole, called an induced dipole, in the second. Interactions between these temporary dipoles cause atoms to be attracted to one another. These attractive interactions are weak and fall off rapidly with increasing distance. London was able to show with quantum mechanics that the attractive energy between molecules due to temporary dipole–induced dipole interactions falls off as 1/ r 6 . Doubling the distance therefore decreases the attractive energy by 2 6 , or 64-fold. Instantaneous dipole–induced dipole interactions between nonpolar molecules can produce intermolecular attractions just as they produce interatomic attractions in monatomic substances like Xe. This effect, illustrated for two H 2 molecules in part (b) in Figure \(\PageIndex{3}\), tends to become more pronounced as atomic and molecular masses increase (Table \(\PageIndex{2}\)). For example, Xe boils at −108.1°C, whereas He boils at −269°C. The reason for this trend is that the strength of London dispersion forces is related to the ease with which the electron distribution in a given atom can be perturbed. In small atoms such as He, the two 1 s electrons are held close to the nucleus in a very small volume, and electron–electron repulsions are strong enough to prevent significant asymmetry in their distribution. In larger atoms such as Xe, however, the outer electrons are much less strongly attracted to the nucleus because of filled intervening shells. As a result, it is relatively easy to temporarily deform the electron distribution to generate an instantaneous or induced dipole. The ease of deformation of the electron distribution in an atom or molecule is called its polarizability. Because the electron distribution is more easily perturbed in large, heavy species than in small, light species, we say that heavier substances tend to be much more polarizable than lighter ones. For similar substances, London dispersion forces get stronger with increasing molecular size. The polarizability of a substance also determines how it interacts with ions and species that possess permanent dipoles. Thus, London dispersion forces are responsible for the general trend toward higher boiling points with increased molecular mass and greater surface area in a homologous series of compounds, such as the alkanes (part (a) in Figure \(\PageIndex{4}\)). The strengths of London dispersion forces also depend significantly on molecular shape because shape determines how much of one molecule can interact with its neighboring molecules at any given time. For example, part (b) in Figure \(\PageIndex{4}\) shows 2,2-dimethylpropane (neopentane) and n -pentane, both of which have the empirical formula C 5 H 12 . Neopentane is almost spherical, with a small surface area for intermolecular interactions, whereas n -pentane has an extended conformation that enables it to come into close contact with other n -pentane molecules. As a result, the boiling point of neopentane (9.5°C) is more than 25°C lower than the boiling point of n -pentane (36.1°C). All molecules, whether polar or nonpolar, are attracted to one another by London dispersion forces in addition to any other attractive forces that may be present. In general, however, dipole–dipole interactions in small polar molecules are significantly stronger than London dispersion forces, so the former predominate. Video Discussing London/Dispersion Intermolecular Forces. Source: Dispersion Intermolecular Force, YouTube(opens in new window) [youtu.be] Example \(\PageIndex{2}\) Arrange n -butane, propane, 2-methylpropane [isobutene, (CH 3 ) 2 CHCH 3 ], and n -pentane in order of increasing boiling points. Given: compounds Asked for: order of increasing boiling points Strategy: Determine the intermolecular forces in the compounds, and then arrange the compounds according to the strength of those forces. The substance with the weakest forces will have the lowest boiling point. Solution: The four compounds are alkanes and nonpolar, so London dispersion forces are the only important intermolecular forces. These forces are generally stronger with increasing molecular mass, so propane should have the lowest boiling point and n -pentane should have the highest, with the two butane isomers falling in between. Of the two butane isomers, 2-methylpropane is more compact, and n -butane has the more extended shape. Consequently, we expect intermolecular interactions for n -butane to be stronger due to its larger surface area, resulting in a higher boiling point. The overall order is thus as follows, with actual boiling points in parentheses: propane (−42.1°C) < 2-methylpropane (−11.7°C) < n -butane (−0.5°C) < n -pentane (36.1°C). Exercise \(\PageIndex{2}\) Arrange GeH 4 , SiCl 4 , SiH 4 , CH 4 , and GeCl 4 in order of decreasing boiling points. Answer GeCl 4 (87°C) > SiCl 4 (57.6°C) > GeH 4 (−88.5°C) > SiH 4 (−111.8°C) > CH 4 (−161°C) Hydrogen Bonds Molecules with hydrogen atoms bonded to electronegative atoms such as O, N, and F (and to a much lesser extent, Cl and S) tend to exhibit unusually strong intermolecular interactions. These result in much higher boiling points than are observed for substances in which London dispersion forces dominate, as illustrated for the covalent hydrides of elements of groups 14–17 in Figure \(\PageIndex{5}\). Methane and its heavier congeners in group 14 form a series whose boiling points increase smoothly with increasing molar mass. This is the expected trend in nonpolar molecules, for which London dispersion forces are the exclusive intermolecular forces. In contrast, the hydrides of the lightest members of groups 15–17 have boiling points that are more than 100°C greater than predicted on the basis of their molar masses. The effect is most dramatic for water: if we extend the straight line connecting the points for H 2 Te and H 2 Se to the line for period 2, we obtain an estimated boiling point of −130°C for water! Imagine the implications for life on Earth if water boiled at −130°C rather than 100°C. Why do strong intermolecular forces produce such anomalously high boiling points and other unusual properties, such as high enthalpies of vaporization and high melting points? The answer lies in the highly polar nature of the bonds between hydrogen and very electronegative elements such as O, N, and F. The large difference in electronegativity results in a large partial positive charge on hydrogen and a correspondingly large partial negative charge on the O, N, or F atom. Consequently, H–O, H–N, and H–F bonds have very large bond dipoles that can interact strongly with one another. Because a hydrogen atom is so small, these dipoles can also approach one another more closely than most other dipoles. The combination of large bond dipoles and short dipole–dipole distances results in very strong dipole–dipole interactions called hydrogen bonds, as shown for ice in Figure \(\PageIndex{6}\). A hydrogen bond is usually indicated by a dotted line between the hydrogen atom attached to O, N, or F (the hydrogen bond donor ) and the atom that has the lone pair of electrons (the hydrogen bond acceptor ). Because each water molecule contains two hydrogen atoms and two lone pairs, a tetrahedral arrangement maximizes the number of hydrogen bonds that can be formed. In the structure of ice, each oxygen atom is surrounded by a distorted tetrahedron of hydrogen atoms that form bridges to the oxygen atoms of adjacent water molecules. The bridging hydrogen atoms are not equidistant from the two oxygen atoms they connect, however. Instead, each hydrogen atom is 101 pm from one oxygen and 174 pm from the other. In contrast, each oxygen atom is bonded to two H atoms at the shorter distance and two at the longer distance, corresponding to two O–H covalent bonds and two O⋅⋅⋅H hydrogen bonds from adjacent water molecules, respectively. The resulting open, cagelike structure of ice means that the solid is actually slightly less dense than the liquid, which explains why ice floats on water, rather than sinks. Each water molecule accepts two hydrogen bonds from two other water molecules and donates two hydrogen atoms to form hydrogen bonds with two more water molecules, producing an open, cage like structure. The structure of liquid water is very similar, but in the liquid, the hydrogen bonds are continually broken and formed because of rapid molecular motion. Hydrogen bond formation requires both a hydrogen bond donor and a hydrogen bond acceptor. Because ice is less dense than liquid water, rivers, lakes, and oceans freeze from the top down. In fact, the ice forms a protective surface layer that insulates the rest of the water, allowing fish and other organisms to survive in the lower levels of a frozen lake or sea. If ice were denser than the liquid, the ice formed at the surface in cold weather would sink as fast as it formed. Bodies of water would freeze from the bottom up, which would be lethal for most aquatic creatures. The expansion of water when freezing also explains why automobile or boat engines must be protected by “antifreeze” and why unprotected pipes in houses break if they are allowed to freeze. Video Discussing Hydrogen Bonding Intermolecular Forces. Source: Hydrogen Bonding Intermolecular Force, YouTube(opens in new window) [youtu.be] Example \(\PageIndex{3}\) Considering CH 3 OH , C 2 H 6 , Xe, and (CH 3 ) 3 N, which can form hydrogen bonds with themselves? Draw the hydrogen-bonded structures. Given: compounds Asked for: formation of hydrogen bonds and structure Strategy: Identify the compounds with a hydrogen atom attached to O, N, or F. These are likely to be able to act as hydrogen bond donors. Of the compounds that can act as hydrogen bond donors, identify those that also contain lone pairs of electrons, which allow them to be hydrogen bond acceptors. If a substance is both a hydrogen donor and a hydrogen bond acceptor, draw a structure showing the hydrogen bonding. Solution: A. Of the species listed, xenon (Xe), ethane (C 2 H 6 ), and trimethylamine [(CH 3 ) 3 N] do not contain a hydrogen atom attached to O, N, or F; hence they cannot act as hydrogen bond donors. B. The one compound that can act as a hydrogen bond donor, methanol (CH 3 OH), contains both a hydrogen atom attached to O (making it a hydrogen bond donor) and two lone pairs of electrons on O (making it a hydrogen bond acceptor); methanol can thus form hydrogen bonds by acting as either a hydrogen bond donor or a hydrogen bond acceptor. The hydrogen-bonded structure of methanol is as follows: Exercise \(\PageIndex{3}\) Considering CH 3 CO 2 H, (CH 3 ) 3 N, NH 3 , and CH 3 F, which can form hydrogen bonds with themselves? Draw the hydrogen-bonded structures. Answer CH 3 CO 2 H and NH 3 ; Although hydrogen bonds are significantly weaker than covalent bonds, with typical dissociation energies of only 15–25 kJ/mol, they have a significant influence on the physical properties of a compound. Compounds such as HF can form only two hydrogen bonds at a time as can, on average, pure liquid NH 3 . Consequently, even though their molecular masses are similar to that of water, their boiling points are significantly lower than the boiling point of water, which forms four hydrogen bonds at a time. Example \(\PageIndex{4}\): Buckyballs Arrange C 60 (buckminsterfullerene, which has a cage structure), NaCl, He, Ar, and N 2 O in order of increasing boiling points. Given: compounds. Asked for: order of increasing boiling points. Strategy: Identify the intermolecular forces in each compound and then arrange the compounds according to the strength of those forces. The substance with the weakest forces will have the lowest boiling point. Solution Electrostatic interactions are strongest for an ionic compound, so we expect NaCl to have the highest boiling point. To predict the relative boiling points of the other compounds, we must consider their polarity (for dipole–dipole interactions), their ability to form hydrogen bonds, and their molar mass (for London dispersion forces). Helium is nonpolar and by far the lightest, so it should have the lowest boiling point. Argon and N 2 O have very similar molar masses (40 and 44 g/mol, respectively), but N 2 O is polar while Ar is not. Consequently, N 2 O should have a higher boiling point. A C 60 molecule is nonpolar, but its molar mass is 720 g/mol, much greater than that of Ar or N 2 O. Because the boiling points of nonpolar substances increase rapidly with molecular mass, C 60 should boil at a higher temperature than the other nonionic substances. The predicted order is thus as follows, with actual boiling points in parentheses: He (−269°C) < Ar (−185.7°C) < N 2 O (−88.5°C) < C 60 (>280°C) < NaCl (1465°C). Exercise \(\PageIndex{4}\) Arrange 2,4-dimethylheptane, Ne, CS 2 , Cl 2 , and KBr in order of decreasing boiling points. Answer KBr (1435°C) > 2,4-dimethylheptane (132.9°C) > CS 2 (46.6°C) > Cl 2 (−34.6°C) > Ne (−246°C) Example \(\PageIndex{5}\) Identify the most significant intermolecular force in each substance. C 3 H 8 CH 3 OH H 2 S Solution Although C–H bonds are polar, they are only minimally polar. The most significant intermolecular force for this substance would be dispersion forces. This molecule has an H atom bonded to an O atom, so it will experience hydrogen bonding. Although this molecule does not experience hydrogen bonding, the Lewis electron dot diagram and VSEPR indicate that it is bent, so it has a permanent dipole. The most significant force in this substance is dipole-dipole interaction. Exercise \(\PageIndex{6}\) Identify the most significant intermolecular force in each substance. HF HCl Answer a hydrogen bonding Answer b dipole-dipole interactions Summary Intermolecular forces are electrostatic in nature and include van der Waals forces and hydrogen bonds. Molecules in liquids are held to other molecules by intermolecular interactions, which are weaker than the intramolecular interactions that hold the atoms together within molecules and polyatomic ions. Transitions between the solid and liquid, or the liquid and gas phases, are due to changes in intermolecular interactions, but do not affect intramolecular interactions. The three major types of intermolecular interactions are dipole–dipole interactions, London dispersion forces (these two are often referred to collectively as van der Waals forces ), and hydrogen bonds. Dipole–dipole interactions arise from the electrostatic interactions of the positive and negative ends of molecules with permanent dipole moments; their strength is proportional to the magnitude of the dipole moment and to 1/ r 3 , where r is the distance between dipoles. London dispersion forces are due to the formation of instantaneous dipole moments in polar or nonpolar molecules as a result of short-lived fluctuations of electron charge distribution, which in turn cause the temporary formation of an induced dipole in adjacent molecules; their energy falls off as 1/ r 6 . Larger atoms tend to be more polarizable than smaller ones, because their outer electrons are less tightly bound and are therefore more easily perturbed. Hydrogen bonds are especially strong dipole–dipole interactions between molecules that have hydrogen bonded to a highly electronegative atom, such as O, N, or F. The resulting partially positively charged H atom on one molecule (the hydrogen bond donor ) can interact strongly with a lone pair of electrons of a partially negatively charged O, N, or F atom on adjacent molecules (the hydrogen bond acceptor ). Because of strong O⋅⋅⋅H hydrogen bonding between water molecules, water has an unusually high boiling point, and ice has an open, cage like structure that is less dense than liquid water.
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Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/32%3A_Math_Chapters/32.10%3A_Fourier_Analysis/32.10.01%3A_Fourier_Analysis_in_Matlab
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Fourier analysis encompasses a vast spectrum of mathematics with parts that, at first glance, may appear quite different. In the sciences and engineering the process of decomposing a function into simpler pieces is often called an analysis. The corresponding operation of rebuilding the function from these pieces is known as synthesis. In this context the term Fourier synthesis describes the act of rebuilding and the term Fourier analysis describes the process of breaking the function into a sum of simpler pieces. In mathematics, the term Fourier analysis often refers to the study of both operations. Introduction In Fourier analysis, the term Fourier transform often refers to the process that decomposes a given function into the basic pieces. This process results in another function that describes how much of each basic piece are in the original function. It is common practice to also use the term Fourier transform to refer to this function. However, the transform is often given a more specific name depending upon the domain and other properties of the function being transformed, as elaborated below. Moreover, the original concept of Fourier analysis has been extended over time to apply to more and more abstract and general situations and the general field is often known as harmonic analysis. (Continuous) Fourier transform Most often, the unqualified term Fourier transform refers to the transform of functions of a continuous real argument, such as time (t). In this case the Fourier transform describes a function ƒ(t) in terms of basic complex exponentials of various frequencies. In terms of ordinary frequency ν, the Fourier transform is given by the complex number. Evaluating this quantity for all values of ν produces the frequency-domain function. Matlab and the FFT Matlab's FFT function is an effective tool for computing the discrete Fourier transform of a signal.The following code examples will help you to understand the details of using the FFT function. Example 1 The typical syntax for computing the FFT of a signal is FFT(x,N) where x is the signal, x[n], you wish to transform, and N is the number of points in the FFT. N must be at least as large as the number of samples in x[n]. To demonstrate the effect of changing the value of N, synthesize a cosine with 30 samples at 10 samples per period. n = [0:29]; x = cos(2*pi*n/10); Define 3 different values for N. Then take the transform of x[n] for each of the 3 values that were defined. The abs function find the magnitude of the transform, as we are not concerned with distinguishing between real and imaginary components. N1 = 64; N2 = 128; N3 = 256; X1 = abs(fft(x,N1)); X2 = abs(fft(x,N2)); X3 = abs(fft(x,N3)); The frequency scale begins at 0 and extends to N - 1 for an N-point FFT. We then normalize the scale so that it extends from 0 to 1 - 1/N F1 = [0 : N1 - 1]/N1; F2 = [0 : N2 - 1]/N2; F3 = [0 : N3 - 1]/N3; Plot each of the transforms one above the other subplot(3,1,1) plot(F1,X1,'-x'),title('N = 64'),axis([0 1 0 20]) subplot(3,1,2) plot(F2,X2,'-x'),title('N = 128'),axis([0 1 0 20]) subplot(3,1,3) plot(F3,X3,'-x'),title('N = 256'),axis([0 1 0 20]) Upon examining the plot one can see that each of the transforms adheres to the same shape, differing only in the number of samples used to approximate that shape. What happens if N is the same as the number of samples in x[n]? To find out, set N1 = 30. What does the resulting plot look like? Why does it look like this? Example 2 In the last example the length of x[n] was limited to 3 periods in length. Now, let's choose a large value for N (for a transform with many points), and vary the number of repetitions of the fundamental period. n = [0:29]; x1 = cos(2*pi*n/10); % 3 periods x2 = [x1 x1]; % 6 periods x3 = [x1 x1 x1]; % 9 periods N = 2048; X1 = abs(fft(x1,N)); X2 = abs(fft(x2,N)); X3 = abs(fft(x3,N)); F = [0:N-1]/N; subplot(3,1,1) plot(F,X1),title('3 periods'),axis([0 1 0 50]) subplot(3,1,2) plot(F,X2),title('6 periods'),axis([0 1 0 50]) subplot(3,1,3) plot(F,X3),title('9 periods'),axis([0 1 0 50]) The previous code will produce three plots. The first plot, the transform of 3 periods of a cosine, looks like the magnitude of 2 sincs with the center of the first sinc at 0.1fs and the second at 0.9fs. The second plot also has a sinc-like appearance, but its frequency is higher and it has a larger magnitude at 0:1fs and 0:9fs. Similarly, the third plot has a larger sinc frequency and magnitude. As x[n] is extended to a large number of periods, the sincs will begin to look more and more like impulses. But I thought a sinusoid transformed to an impulse, why do we have sincs in the frequency domain? When the FFT is computed with an N larger than the number of samples in x[n], it fills in the samples after x[n] with zeros. Example 2 had an x[n] that was 30 samples long, but the FFT had an N = 2048. When Matlab computes the FFT, it automatically fills the spaces from n = 30 to n = 2047 with zeros. This is like taking a sinusoid and multiplying it with a rectangular box of length 30. A multiplication of a box and a sinusoid in the time domain should result in the convolution of a sinc with impulses in the frequency domain. Furthermore, increasing the width of the box in the time domain should increase the frequency of the sinc in the frequency domain. The previous Matlab experiment supports this conclusion. Identifying Signals in noise with FFT A common use of Fourier transforms is to find the frequency components of a signal buried in a noisy time domain signal. Consider data sampled at 1000 Hz. Form a signal containing a 50 Hz sinusoid of amplitude 0.7 and 120 Hz sinusoid of amplitude 1 and corrupt it with some zero-mean random noise: Fs = 1000; % Sampling frequency T = 1/Fs; % Sample time L = 1000; % Length of signal t = (0:L-1)*T; % Time vector x = 0.7*sin(2*pi*50*t) + sin(2*pi*120*t); % Sum of a 50 Hz sinusoid and a 120 Hz sinusoid y = x + 2*randn(size(t)); % Sinusoids plus noise plot(Fs*t(1:50),y(1:50)) title('Signal Corrupted with Zero-Mean Random Noise') xlabel('time (milliseconds)') It is difficult to identify the frequency components by looking at the original signal. Converting to the frequency domain, the discrete Fourier transform of the noisy signal y is found by taking the fast Fourier transform (FFT): NFFT = 2^nextpow2(L); % Next power of 2 from length of y Y = fft(y,NFFT)/L; f = Fs/2*linspace(0,1,NFFT/2+1); % Plot single-sided amplitude spectrum. plot(f,2*abs(Y(1:NFFT/2+1))) title('Single-Sided Amplitude Spectrum of y(t)') xlabel('Frequency (Hz)') ylabel('|Y(f)|') The main reason the amplitudes are not exactly at 0.7 and 1 is because of the noise. Several executions of this code (including recomputation of y) will produce different approximations to 0.7 and 1. The other reason is that you have a finite length signal. Increasing L from 1000 to 10000 in the example above will produce much better approximations on average. Outside Links http://www.chem.uoa.gr/applets/Apple...FourAnal2.html
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Courses/Modesto_Junior_College/Chemistry_150_-_Bunag/Textbook_for_Chemistry_150/04%3A_Valence_Electrons_and_Bonding/4.11%3A_Applications_and_Solubility_of_Covalent_Compounds
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Learning Objectives Know the basic difference between the terms polar and nonpolar. Apply the like dissolves like solubility rule for covalent compounds Understand why molecular shapes are important to pharmaceutical chemists and biologists. Appreciate how medications/vitamins need to be soluble in the body. Pick a vitamin and memorize highlighted information Realize that molecules are three dimensional in nature. Know the applications of covalent compounds in this section The Solubility of Covalent Compounds Unlike ionic solubility, covalent compound solubility cannot be determined by a table. Instead, structures and three-dimensional shapes must be drawn. Once a correct geometry has been determined, the compound would be classified as being polar on nonpolar. Polar species are soluble in water, while nonpolar species are soluble in oils and fats. Covalent solubility uses the like dissolves like rule. This means that substances with the same type of polarity will be soluble in one another. Moreover, compounds with differing polarities will be insoluble in one another. Oil and water form a heterogeneous mixture due to their differing polarities; these substances are immiscible (not mixable) (Figure \(\PageIndex{1a}\)). In contrast, alcohol dissolves in water to form a homogeneous mixture (Figure \(\PageIndex{1b}\)). In this class, we will not explore molecular geometries that are used to determine polarity. Instead, the polarity of a substance will be provided. It is important to remember that water is polar and oil/fat is nonpolar. If the polarity of a substance is given, you should be able to classify it as being water or oil/fat soluble. Phet Simulation: Molecular Shapes Template:HideTOC In determining polarity, chemists look to the power of atom's' nucleus. The protons from an atom's nucleus are capable of attracting another atom's electrons. Within a covalent bond, valence electrons are pulled toward's an atom that has a more powerful nucleus. This pull is called electronegativity. If different atoms are connected in a bond, then one tends to be more electronegative than the other. Molecules that have an overall pull in one direction are labeled as being polar species. Look at the structure of water that is shown below. This structure of this molecule shows the bonding electrons being pulled towards oxygen. Therefore, oxygen has a more powerful nucleus than hydrogen. Water's bent molecular shape does not cancel out the individual dipole pulls. As a result, water has an overall pull and is classified as being polar. Carbon dioxide also has local dipoles (arrows) that pull in opposing directions. This molecule does not have an overall pull in one direction and is classified as being nonpolar. Some other molecules are shown in the figure below. The top three assymetrical molecules are all polar. They all have dipoles (pulls) that do not cancel. All of these molecules would be soluble in water. The bottom three molecules are nonpolar. These are symmetrical molecules that have dipoles that cancel. Both of these molecules would be oil or fat soluble. Covalent solubility is important in the pharmaceutical industry. If a medication is not water soluble, then it will not dissolve in the bloodstream and react in the active site of the body in a timely and potent fashion. Watch the video below to obtain a basic understanding of how ibuprofen travels through the body to reduce pain and/or inflammation. The polarity of vitamins can affect how long they remain in the body. Vitamins B and C are both water soluble and remain in the body for a short period of time. Our bodies must intake these vitamins more frequently. On the other hand, vitamins A, D, E, and K are all fat or oil soluble. Nutrients that are fat soluble remain in the body longer. These vitamins accumulate easily and can be toxic if recommended doses are surpassed. Exercise Pick one vitamin from the table below and memorize the common/chemical names, one source, and one function. In addition, note the solubility (water or fat) of your selected vitamin. Vitamins with their sources and functions. Image taken from: https://upload.wikimedia.org/wikiped...mins_Table.png Applications of Covalent Compounds Carbon dioxide Sulfur dioxide/Sulfur trioxide Dihydrogen Dioxide Dihydrogen monosulfide Carbon monoxide Carbon tetrahydride (methane) Contributors Elizabeth R. Gordon (Furman University) Isabella Quiros (Furman University)
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Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Thermodynamics_and_Chemical_Equilibrium_(Ellgen)/15%3A_Chemical_Potential_Fugacity_Activity_and_Equilibrium/15.09%3A_Problems
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Problems Use data from the table below to find the thermodynamic properties requested in problems 1 to 7. Properties \(\boldsymbol{C}{\boldsymbol{H}}_{\boldsymbol{3}}\boldsymbol{OH}\) \(\boldsymbol{C}{\boldsymbol{H}}_{\boldsymbol{3}}\boldsymbol{C}{\boldsymbol{H}}_{\boldsymbol{2}}\boldsymbol{OH}\) Density, \(\mathrm{g}\ {\mathrm{cm}}^{-3}\) at 20 C 0.7914 0.7893 Mol mass, \(\mathrm{\ g}\ {\mathrm{mol}}^{-1}\) 32.04 46.07 bp, C 64.6 78.2 \({\Delta }_fG^o\left(300\mathrm{K},\ {HIG}^o\right)\)\(\mathrm{kJ}\ {\mathrm{mol}}^{-1}\) –159.436 –162.934 Vapor pressure at320 K, bar 0.5063 0.2764 Virial coefficient, \(B\), \(\mathrm{\ }{\mathrm{m}}^3\ {\mathrm{mol}}^{-1}\) \(-1.421\times {10}^{-3}\) \(-2.710\times {10}^{-3}\) 1. Find the chemical potentials of the pure gases, assuming that they are ideal, taking the hypothetical ideal gas standard state as the standard state for activity (\(f=\tilde{a}=P\)). 2. Find the chemical potentials of the mixed gases, assuming that they are ideal, taking the hypothetical ideal gas standard state as the standard state for activity (\(f={\tilde{a}}_A=x_AP)\). 3. Find the chemical potentials of the mixed gases, assuming them to obey the Virial equation, \[{PV}/{RT=1+\left({BP}/{RT}\right)}, \nonumber \] assuming that the partial molar volumes in the mixture are equal to the partial molar volumes of the pure gases at the same pressure, and taking the hypothetical ideal gas standard state as the standard state for activity. 4. Find the standard chemical potentials of the pure liquids at 320 K, assuming that the gases behave ideally. 5. Find the standard chemical potentials of the pure liquids at 320 K, assuming that the gases obey the Virial equation. 6. Find the chemical potential of the pure liquids as a function of pressure, assuming that the partial molar volumes of the pure liquids are constant and that the gases obey the Virial equation. 7. Find the activity and chemical potential of the pure liquids at 101 bar, taking the pure liquids at 1 bar as the standard state for activity. 8. A system is created by mixing one mole of gas \(A\) with one mole of gas \(C\). Reaction occurs according to the stoichiometry \(A+C\ \ \rightleftharpoons \ D\). Assume that the behaviors of these gases in their equilibrium mixture are adequately approximated by the Virial equations \({P{\overline{V}}_A}/{RT=1+\left({B_AP}/{RT}\right)}\), etc. (a) Show that the fugacity of gas A is given by \[{ \ln f_A={ \ln \left(\frac{x_AP}{P^o}\right)\ }\ }+\frac{B_AP}{RT} \nonumber \] (b) Write an equation for \({\mu }_A={\mu }_A\left(x_A,P\right)\) at constant temperature, \(T\). (c) Write an equation for \({\Delta }_r\mu\). (d) Assume that \({\Delta }_r{\mu }^o={\mu }^o_D-{\mu }^o_A-{\mu }^o_C=1000\ \mathrm{J}\ {\mathrm{mol}}^{-1}\). If all three gases behave ideally (\(B_A=B_C=B_D=0\)), what is \(K_P\) for this reaction at 300 K? At equilibrium at 300 K and 1 bar, what are the mole fractions of \(A\), \(C\), and \(D\)? (e) Under the assumptions in part (d), what are the equilibrium mole fractions of \(A\), \(C\), and \(D\) at 300 K and 10 bar? (f) Suppose that, contrary to the assumptions in (d) and (e), the Virial coefficients are not zero and that \(B_A=B_C=B_D=-{10}^{-3}\ {\mathrm{m}}^3\ {\mathrm{mol}}^{-1}\). At equilibrium at 300 K and 1 bar, what are the mole fractions of \(A\), \(C\), and \(D\)? (g) Under the assumptions in part (f), what are the equilibrium mole fractions of \(A\), \(C\), and \(D\) at 300 K and 10 bar? 9. Suppose that the reaction \(A+C\ \ \rightleftharpoons \ D\) occurs in an inert solvent and that it is convenient to express concentrations as molarities. A frequently convenient choice of activity standard state for a solute is a hypothetical one-molar solution in which the chemical potential of the solute is equal to the chemical potential of the solute in a very (“infinitely”) dilute solution in the same solvent. Then \({\tilde{a}}_A=\left[A\right]{\gamma }_A\), and \({\gamma }_A\to 1\) as \(\left[A\right]\to 0\). The thermodynamic equilibrium constant becomes \[K_a=\frac{{\tilde{a}}_D}{{\tilde{a}}_A{\tilde{a}}_C}=\frac{\left[D\right]}{\left[A\right]\left[C\right]}\frac{{\gamma }^d_D}{{\gamma }^a_A{\gamma }^c_C}=K_cK_{\gamma } \nonumber \] where we introduce \[K_c=\frac{\left[D\right]}{\left[A\right]\left[C\right]} \nonumber \] and \[K_{\gamma }=\frac{{\gamma }^d_D}{{\gamma }^a_A{\gamma }^c_C} \nonumber \] In a very dilute solution, \(K_{\gamma }\to 1\) and \(K_c=K_a\). Therefore, we can estimate \(K_a\) by finding the limiting value of \(K_c\) as all of the concentrations become very small. From values of \(K_c\) at higher concentrations, we can develop an empirical equation for \(K_{\gamma }\). The form of this equation can be anything that can adequately represent the experimental data. Note, however, that finding an empirical model for \(K_{\gamma }\) does not solve the problem of finding empirical models for \({\gamma }_A\), \({\gamma }_C\), and \({\gamma }_D\) individually. (a) Given that the hypothetical one-molar solution is chosen to be the activity standard state for all three species, what is the physical significance of \({\Delta }_r{\widetilde{\mu }}^o\)? (b) A simple function that has the properties required of \({\gamma }_A\) is \({\gamma }_A={\alpha }^{\left[A\right]}\), where \(\alpha\) is a constant. Represent , \({\gamma }_C\) and , \({\gamma }_D\) by similar functions and show that this leads to \({ \ln K_{\gamma }={\beta }_A\ }\left[A\right]+{\beta }_C\left[C\right]+{\beta }_D\left[D\right]\), where \({\beta }_A\), \({\beta }_C\), and \({\beta }_D\) are constants. (c) A series of solutions is prepared. The equilibrium concentrations of \(A\), \(B\), and \(D\) in these solutions are given below. Calculate \(K_c\) for each solution. Estimate \(K_a\) and the parameters \({\beta }_A\), \({\beta }_C\), and \({\beta }_D\) in the equation of part (b). \(\left[\boldsymbol{A}\right]\) [B] [C] 1.96 x 10\({}^{-3}\) 1.96 x 10\({}^{-3}\) 3.84 x 10\({}^{-5}\) 7.85 x 10\({}^{-3}\) 1.85 x 10\({}^{-3}\) 1.45 x 10\({}^{-4}\) 3.94 x 10\({}^{-2}\) 1.44 x 10\({}^{-3}\) 5.57 x 10\({}^{-4}\) 1.99 x 10\({}^{-1}\) 7.15 x 10\({}^{-4}\) 1.29 x 10\({}^{-3}\) 9.98 x 10\({}^{-1}\) 2.84 x 10\({}^{-4}\) 1.72 x 10\({}^{-3}\) 1.85 x 10\({}^{-3}\) 7.85 x 10\({}^{-3}\) 1.45 x 10\({}^{-4}\) 1.44 x 10\({}^{-3}\) 3.94 x 10\({}^{-2}\) 5.60 x 10\({}^{-4}\) 6.97 x 10\({}^{-4}\) 1.99 x 10\({}^{-1}\) 1.30 x 10\({}^{-3}\) 2.38 x 10\({}^{-4}\) 9.98 x 10\({}^{-1}\) 1.76 x 10\({}^{-3}\) 9.32 x 10\({}^{-3}\) 7.32 x 10\({}^{-3}\) 6.78 x 10\({}^{-4}\) 3.24 x 10\({}^{-2}\) 3.04 x 10\({}^{-2}\) 9.57 x 10\({}^{-3}\) 1.06 x 10\({}^{-1}\) 1.04 x 10\({}^{-1}\) 9.59 x 10\({}^{-2}\) (d) Using the values you find in part (c), estimate the equilibrium concentrations of \(A\), \(B\), and \(D\) when a solution is prepared by mixing one mole of \(A\) with one mole of \(C\) and sufficient solvent to make 1 L of solution at equilibrium. 10. An ester, \(RCO_2R'\), undergoes hydrolysis in an ether solvent: \[\ce{RCO2R^{'} + H2O <=> RCO2H + HOR^{'}} \nonumber \] We can express the activity of any of these species as the product of a concentration (in any convenient units) and an activity coefficient. When all of the reactants and products are present at low concentrations, the activity coefficients are approximately unity. The standard state for each species becomes a hypothetical solution of unit concentration in which the chemical potential (per mole) of that species is the same as its chemical potential in an arbitrarily (infinitely) dilute solution. A solution is prepared by mixing \(2\times {10}^{-3}\) mole of the ester and \({10}^{-1}\) mole water in sufficient ether to make 1 L of solution. When equilibrium is reached, the acid and alcohol concentrations are \(9.66\times {10}^{-3}\) molar. (a) What is the equilibrium constant for this reaction? (b) A solution is prepared by mixing \(2\times {10}^{-2}\) mole of the acid, 3\(\times {10}^{-2}\) mole of the alcohol, and \({10}^{-1}\) mole of water in sufficient ether to make 1 L of solution. When equilibrium is reached, what is the concentration of the ester? (c) With this choice of the standard states, what physical process does the standard chemical-potential change, \(\Delta {\widetilde{\mu }}^o\), describe?
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Courses/Lumen_Learning/Book%3A_US_History_II_(Lumen)/15%3A_Age_of_Empire-_American_Foreign_Policy_1890-1914/15.05%3A_Roosevelts_Big_Stick_Foreign_Policy
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Learning Objectives By the end of this section, you will be able to: Explain the meaning of “big stick” foreign policy Describe Theodore Roosevelt’s use of the “big stick” to construct the Panama Canal Explain the role of the United States in ending the Russo-Japanese War While President McKinley ushered in the era of the American empire through military strength and economic coercion, his successor, Theodore Roosevelt, established a new foreign policy approach, allegedly based on a favorite African proverb, “speak softly, and carry a big stick, and you will go far.” At the crux of his foreign policy was a thinly veiled threat. Roosevelt believed that in light of the country’s recent military successes, it was unnecessary to use force to achieve foreign policy goals, so long as the military could threaten force. This rationale also rested on the young president’s philosophy, which he termed the “strenuous life,” and that prized challenges overseas as opportunities to instill American men with the resolve and vigor they allegedly had once acquired in the Trans-Mississippi West. Roosevelt was often depicted in cartoons wielding his “big stick” and pushing the U.S. foreign agenda, often through the power of the U.S. Navy. Roosevelt believed that while the coercive power wielded by the United States could be harmful in the wrong hands, the Western Hemisphere’s best interests were also the best interests of the United States. He felt, in short, that the United States had the right and the obligation to be the policeman of the hemisphere. This belief, and his strategy of “speaking softly and carrying a big stick,” shaped much of Roosevelt’s foreign policy. THE CONSTRUCTION OF THE PANAMA CANAL As early as the mid-sixteenth century, interest in a canal across the Central American isthmus began to take root, primarily out of trade interests. The subsequent discovery of gold in California in 1848 further spurred interest in connecting the Atlantic and Pacific Oceans, and led to the construction of the Panama Railway, which began operations in 1855. Several attempts by France to construct a canal between 1881 and 1894 failed due to a combination of financial crises and health hazards, including malaria and yellow fever , which led to the deaths of thousands of French workers. Upon becoming president in 1901, Roosevelt was determined to succeed where others had failed. Following the advice that Mahan set forth in his book The Influence of Seapower upon History , he sought to achieve the construction of a canal across Central America, primarily for military reasons associated with empire, but also for international trade considerations. The most strategic point for the construction was across the fifty-mile isthmus of Panama, which, at the turn of the century, was part of the nation of Colombia. Roosevelt negotiated with the government of Colombia, sometimes threatening to take the project away and build through Nicaragua, until Colombia agreed to a treaty that would grant the United States a lease on the land across Panama in exchange for a payment of $10 million and an additional $250,000 annual rental fee. The matter was far from settled, however. The Colombian people were outraged over the loss of their land to the United States, and saw the payment as far too low. Influenced by the public outcry, the Colombian Senate rejected the treaty and informed Roosevelt there would be no canal. Undaunted, Roosevelt chose to now wield the “big stick.” In comments to journalists, he made it clear that the United States would strongly support the Panamanian people should they choose to revolt against Colombia and form their own nation. In November 1903, he even sent American battleships to the coast of Colombia, ostensibly for practice maneuvers, as the Panamanian revolution unfolded. The warships effectively blocked Colombia from moving additional troops into the region to quell the growing Panamanian uprising. Within a week, Roosevelt immediately recognized the new country of Panama, welcoming them to the world community and offering them the same terms—$10 million plus the annual $250,000 rental fee—he had previously offered Colombia. Following the successful revolution, Panama became an American protectorate, and remained so until 1939. Once the Panamanian victory was secured, with American support, construction on the canal began in May 1904. For the first year of operations, the United States worked primarily to build adequate housing, cafeterias, warehouses, machine shops, and other elements of infrastructure that previous French efforts had failed to consider. Most importantly, the introduction of fumigation systems and mosquito nets following Dr. Walter Reed’s discovery of the role of mosquitoes in the spread of malaria and yellow fever reduced the death rate and restored the fledgling morale among workers and American-born supervisors. At the same time, a new wave of American engineers planned for the construction of the canal. Even though they decided to build a lock-system rather than a sea-level canal, workers still had to excavate over 170 million cubic yards of earth with the use of over one hundred new rail-mounted steam shovels. Excited by the work, Roosevelt became the first sitting U.S. president to leave the country while in office. He traveled to Panama where he visited the construction site, taking a turn at the steam shovel and removing dirt. The canal opened in 1914, permanently changing world trade and military defense patterns. Recurring landslides made the excavation of the Culebra Cut one of the most technically challenging elements in the construction of the Panama Canal. This timeline of the Panama Canal illustrates the efforts involved in both the French and U.S. canal projects. THE ROOSEVELT COROLLARY With the construction of the canal now underway, Roosevelt next wanted to send a clear message to the rest of the world—and in particular to his European counterparts—that the colonization of the Western Hemisphere had now ended, and their interference in the countries there would no longer be tolerated. At the same time, he sent a message to his counterparts in Central and South America, should the United States see problems erupt in the region, that it would intervene in order to maintain peace and stability throughout the hemisphere. Roosevelt articulated this seeming double standard in a 1904 address before Congress, in a speech that became known as the Roosevelt Corollary. The Roosevelt Corollary was based on the original Monroe Doctrine of the early nineteenth century, which warned European nations of the consequences of their interference in the Caribbean. In this addition, Roosevelt states that the United States would use military force “as an international police power” to correct any “chronic wrongdoing” by any Latin American nation that might threaten stability in the region. Unlike the Monroe Doctrine, which proclaimed an American policy of noninterference with its neighbors’ affairs, the Roosevelt Corollary loudly proclaimed the right and obligation of the United States to involve itself whenever necessary. Roosevelt immediately began to put the new corollary to work. He used it to establish protectorates over Cuba and Panama, as well as to direct the United States to manage the Dominican Republic’s custom service revenues. Despite growing resentment from neighboring countries over American intervention in their internal affairs, as well as European concerns from afar, knowledge of Roosevelt’s previous actions in Colombia concerning acquisition of land upon which to build the Panama Canal left many fearful of American reprisals should they resist. Eventually, Presidents Herbert Hoover and Franklin Roosevelt softened American rhetoric regarding U.S. domination of the Western Hemisphere, with the latter proclaiming a new “Good Neighbor Policy” that renounced American intervention in other nations’ affairs. However, subsequent presidents would continue to reference aspects of the Roosevelt Corollary to justify American involvement in Haiti, Nicaragua, and other nations throughout the twentieth century. The map below shows the widespread effects of Roosevelt’s policies throughout Latin America. From underwriting a revolution in Panama with the goal of building a canal to putting troops in Cuba, Roosevelt vastly increased the U.S. impact in Latin America. The Roosevelt Corollary and Its Impact In 1904, Roosevelt put the United States in the role of the “police power” of the Western Hemisphere and set a course for the U.S. relationship with Central and Latin America that played out over the next several decades. He did so with the Roosevelt Corollary, in which he stated: It is not true that the United States feels any land hunger or entertains any projects as regards the other nations of the Western Hemisphere save as such are for their welfare. All that this country desires is to see the neighboring countries stable, orderly, and prosperous. Any country whose people conduct themselves well can count upon our hearty friendship. . . . Chronic wrongdoing, or an impotence which results in a general loosening of the ties of civilized society, may in America, as elsewhere, require intervention by some civilized nation, and in the Western Hemisphere the adherence of the United States to the Monroe Doctrine may force the United States, however, reluctantly, in flagrant cases of such wrongdoing or impotence, to the exercise of an international police power.” In the twenty years after he made this statement, the United States would use military force in Latin America over a dozen times. The Roosevelt Corollary was used as a rationale for American involvement in the Dominican Republic, Nicaragua, Haiti, and other Latin American countries, straining relations between Central America and its dominant neighbor to the north throughout the twentieth century. AMERICAN INTERVENTION IN THE RUSSO-JAPANESE WAR Although he supported the Open Door notes as an excellent economic policy in China, Roosevelt lamented the fact that the United States had no strong military presence in the region to enforce it. Clearly, without a military presence there, he could not as easily use his “big stick” threat credibly to achieve his foreign policy goals. As a result, when conflicts did arise on the other side of the Pacific, Roosevelt adopted a policy of maintaining a balance of power among the nations there. This was particularly evident when the Russo-Japanese War erupted in 1904. Japan’s defense against Russia was supported by President Roosevelt, but when Japan’s ongoing victories put the United States’ own Asian interests at risk, he stepped in. In 1904, angered by the massing of Russian troops along the Manchurian border, and the threat it represented to the region, Japan launched a surprise naval attack upon the Russian fleet. Initially, Roosevelt supported the Japanese position. However, when the Japanese fleet quickly achieved victory after victory, Roosevelt grew concerned over the growth of Japanese influence in the region and the continued threat that it represented to China and American access to those markets. Wishing to maintain the aforementioned balance of power, in 1905, Roosevelt arranged for diplomats from both nations to attend a secret peace conference in Portsmouth, New Hampshire. The resultant negotiations secured peace in the region, with Japan gaining control over Korea, several former Russian bases in Manchuria, and the southern half of Sakhalin Island. These negotiations also garnered the Nobel Peace Prize for Roosevelt, the first American to receive the award. When Japan later exercised its authority over its gains by forcing American business interests out of Manchuria in 1906–1907, Roosevelt felt he needed to invoke his “big stick” foreign policy, even though the distance was great. He did so by sending the U.S. Great White Fleet on maneuvers in the western Pacific Ocean as a show of force from December 1907 through February 1909. Publicly described as a goodwill tour, the message to the Japanese government regarding American interests was equally clear. Subsequent negotiations reinforced the Open Door policy throughout China and the rest of Asia. Roosevelt had, by both the judicious use of the “big stick” and his strategy of maintaining a balance of power, kept U.S. interests in Asia well protected. Browse the Smithsonian National Portrait Gallery to follow Theodore Roosevelt from Rough Rider to president and beyond. Section Summary When Roosevelt succeeded McKinley as president, he implemented a key strategy for building an American empire: the threat, rather than the outright use, of military force. McKinley had engaged the U.S. military in several successful skirmishes and then used the country’s superior industrial power to negotiate beneficial foreign trade agreements. Roosevelt, with his “big stick” policy, was able to keep the United States out of military conflicts by employing the legitimate threat of force. Nonetheless, as negotiations with Japan illustrated, the maintenance of an empire was fraught with complexity. Changing alliances, shifting economic needs, and power politics all meant that the United States would need to tread carefully to maintain its status as a world power. A Open Assessments element has been excluded from this version of the text. You can view it online here: http://pb.libretexts.org/u2/?p=216 Review Question Compare Roosevelt’s foreign policy in Latin America and Asia. Why did he employ these different methods? Review Question Roosevelt’s strategy of “speaking softly and carrying a big stick” worked well in Latin America, where the United States had a strong military presence and could quickly and easily act on any threat of military action. Roosevelt’s threat of force was therefore credible in that region, and he was able to wield it effectively. In Asia, however, the United States had less of a military presence. Instead, Roosevelt sought to maintain a balance of power, wherein the various Asian countries kept each other in check and no single player grew too powerful. When the power balance tipped, Roosevelt acted to broker a peace deal between Russia and Japan as a means of restoring balance. Glossary Roosevelt Corollary a statement by Theodore Roosevelt that the United States would use military force to act as an international police power and correct any chronic wrongdoing by any Latin American nation threatening the stability of the region CC licensed content, Original Revision and Adaptation. Authored by : Kimlisa Duchicela. License : CC BY: Attribution CC licensed content, Shared previously US History. Authored by : P. Scott Corbett, Volker Janssen, John M. Lund, Todd Pfannestiel, Paul Vickery, and Sylvie Waskiewicz. Provided by : OpenStax College. Located at : http://openstaxcollege.org/textbooks/us-history . License : CC BY: Attribution . License Terms : Download for free at http://cnx.org/content/col11740/latest/
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Courses/Riverland_Community_College/CHEM_1121%3A_General_Organic_and_Biochemistry/05%3A_Covalent_Bonding_and_Simple_Molecular_Compounds/5.01%3A_Covalent_Bonds
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Learning Objectives To describe how a covalent bond forms. To apply the octet rule to covalent compounds You have already seen examples of substances that contain covalent bonds. One substance mentioned previously was water (\(\ce{H2O}\)). You can tell from its formula that it is not an ionic compound; it is not composed of a metal and a nonmetal. Consequently, its properties are different from those of ionic compounds. Electron Sharing Previously, we discussed ionic bonding where electrons can be transferred from one atom to another so that both atoms have an energy-stable outer electron shell. Because most filled electron shells have eight electrons in them, chemists called this tendency the octet rule. However, there is another way an atom can achieve a full valence shell: atoms can share electrons. This concept can be illustrated by using two hydrogen atoms, each of which has a single electron in its valence shell. (For small atoms such as hydrogen atoms, the valence shell will be the first shell, which holds only two electrons.) We can represent the two individual hydrogen atoms as follows: In contrast, when two hydrogen atoms get close enough together to share their electrons, they can be represented as follows: By sharing their valence electrons, both hydrogen atoms now have two electrons in their respective valence shells. Because each valence shell is now filled, this arrangement is more stable than when the two atoms are separate. The sharing of electrons between atoms is called a covalent bond, and the two electrons that join atoms in a covalent bond are called a bonding pair of electrons. A discrete group of atoms connected by covalent bonds is called a molecule—the smallest part of a compound that retains the chemical identity of that compound. Chemists frequently use Lewis diagrams to represent covalent bonding in molecular substances. For example, the Lewis diagrams of two separate hydrogen atoms are as follows: The Lewis diagram of two hydrogen atoms sharing electrons looks like this: This depiction of molecules is simplified further by using a dash to represent a covalent bond. The hydrogen molecule is then represented as follows: Remember that the dash, also referred to as a single bond, represents a pair of electrons. The bond in a hydrogen molecule, measured as the distance between the two nuclei, is about 7.4 × 10 −11 m, or 74 picometers (pm; 1 pm = 1 × 10 −12 m). This particular bond length represents a balance between several forces: the attractions between oppositely charged electrons and nuclei, the repulsion between two negatively charged electrons, and the repulsion between two positively charged nuclei. If the nuclei were closer together, they would repel each other more strongly; if the nuclei were farther apart, there would be less attraction between the positive and negative particles. Fluorine is another element whose atoms bond together in pairs to form diatomic (two-atom) molecules. Two separate fluorine atoms have the following electron dot diagrams: Each fluorine atom contributes one valence electron, making a single bond and giving each atom a complete valence shell, which fulfills the octet rule: The circles show that each fluorine atom has eight electrons around it. As with hydrogen, we can represent the fluorine molecule with a dash in place of the bonding electrons: Each fluorine atom has six electrons, or three pairs of electrons, that are not participating in the covalent bond. Rather than being shared, they are considered to belong to a single atom. These are called nonbonding pairs (or lone pairs) of electrons. Covalent Bonds between Different Atoms Now that we have looked at electron sharing between atoms of the same element, let us look at covalent bond formation between atoms of different elements. Consider a molecule composed of one hydrogen atom and one fluorine atom: Each atom needs one additional electron to complete its valence shell. By each contributing one electron, they make the following molecule: In this molecule, the hydrogen atom does not have nonbonding electrons, while the fluorine atom has six nonbonding electrons (three lone electron pairs). The circles show how the valence electron shells are filled for both atoms. Example \(\PageIndex{1}\) Use Lewis diagrams to indicate the formation of the following: Cl 2 HBr Solution a. When two chlorine atoms form a chlorine molecule, they share one pair of electrons. In Cl 2 molecule, each chlorine atom is surrounded by an octet number of electrons. The Lewis diagram for a Cl 2 molecule is similar to the one for F 2 (shown above). b. When a hydrogen atom and a bromine atom form HBr, they share one pair of electrons. In the HBr molecule, H achieves a full valence of two electrons ( duet ) while Br achieves an octet . The Lewis diagram for HBr is similar to that for HF shown above. Exercise \(\PageIndex{1}\) Draw the Lewis diagram for each compound. a molecule composed of one chlorine atom and one fluorine atom a molecule composed of one hydrogen atom and one iodine atom Answer a: Answer b: Covalent Bonds in Larger Molecules The formation of a water molecule from two hydrogen atoms and an oxygen atom can be illustrated using Lewis dot symbols (shown below). The structure on the right is the Lewis electron structure , or Lewis structure , for \(\ce{H2O}\). With two bonding pairs and two lone pairs, the oxygen atom has now completed its octet . Moreover, by sharing a bonding pair with oxygen, each hydrogen atom now has a full valence shell of two electrons. Chemists usually indicate a bonding pair by a single line, as shown (below). Other large molecules are constructed in a similar fashion, with some atoms participating in more than one covalent bond. For example, methane (\(\ce{CH4}\)), the central carbon atom bonded to four hydrogen atoms, can be represented using either of the Lewis structures below. Again, sharing electrons between C and H atoms results in C achieving and octet while H achieving a duet number of electrons. How Many Covalent Bonds Are Formed? The number of bonds that an atom can form can often be predicted from the number of electrons needed to reach an octet (eight valence electrons). In the Lewis structure, the number of bonds formed by an element in a neutral compound is the same as the number of unpaired electrons it must share with other atoms to complete its octet of electrons. For example, each atom of a group 4A (14) element has four electrons in its outermost shell and therefore requires four more electrons to reach an octet. These four electrons can be gained by forming four covalent bonds , as illustrated here for carbon in CH 4 (methane). Group 5A (15) elements such as nitrogen have five valence electrons in the atomic Lewis symbol: one lone pair and three unpaired electrons. To obtain an octet, these atoms form three covalent bonds , as in NH 3 (ammonia). Oxygen and other atoms in group 6A (16) obtain an octet by forming two covalent bonds . Fluorine and the other halogens in group 7A (17) have seven valence electrons and can obtain an octet by forming one covalent bond . Typically, the atoms of group 4A form 4 covalent bonds; group 5A form 3 bonds; group 6A form 2 bonds; and group 7A form one bond. The number of electrons required to obtain an octet determines the number of covalent bonds an atom can form. This is summarized in the table below. In each case, the sum of the number of bonds and the number of lone pairs is 4, which is equivalent to eight (octet) electrons. Atom (Group number) Number of Bonds Number of Lone Pairs Carbon (Group 14 or 4A) 4 0 Nitrogen (Group 15 or 5A) 3 1 Oxygen (Group 16 or 6A) 2 2 Fluorine (Group 17 or 7A) 1 3 Because hydrogen only needs two electrons to fill its valence shell, it follows the duet rule. It is an exception to the octet rule. Hydrogen only needs to form one bond. This is the reason why H is always a terminal atom and never a central atom. Figure \(\PageIndex{1}\) shows the number of covalent bonds various atoms typically form. The transition elements and inner transition elements also do not follow the octet rule since they have d and f electrons involved in their valence shells. Example \(\PageIndex{2}\) Examine the Lewis structure of OF 2 below. Count the number of bonds formed by each element. Based on the element's location in the periodic table, does it correspond to the expected number of bonds shown in Table 4.1? Does the Lewis structure below follow the octet rule? Solution Yes. F (group 7A) forms one bond and O (group 6A) forms 2 bonds. Each atom is surrounded by 8 electrons. This structure satisfies the octet rule. Examine the Lewis structure of NCl 3 below. Count the number of bonds formed by each element. Based on the element's location in the periodic table, does it correspond to the expected number of bonds shown in Table 4.1? Does the Lewis structure below follow the octet rule? Answer Both Cl and N form the expected number of bonds. Cl (group 7A) has one bond and 3 lone pairs. The central atom N (group 5A) has 3 bonds and one lone pair. Yes, the Lewis structure of NCl 3 follows the octet rule. Key Takeaways A covalent bond is formed between two atoms by sharing electrons. The number of bonds an element forms in a covalent compound is determined by the number of electrons it needs to reach octet. Hydrogen is an exception to the octet rule. H forms only one bond because it needs only two electrons.
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Courses/Sophia_College_for_Women/Thermal_Methods_of_Analysis_(Shetty)/03%3A_Differential_Thermal_Analysis
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Learning Objectives After completing this chapter, you should be able to: Explain the principle and working of a differential thermal analyser. Draw and interpret DTA thermogram. Compare and contrast TG and DTA results. Explain the applications of DTA and simultaneous TG-DTA analysis. When a molecule undergoes a physical or chemical transition, heat is either absorbed or liberated. Two thermal methods, DTA and DSC are particularly useful for investigating these physical and chemical changes. 3.1 Principle and Instrumentation: In DTA, difference in temperature between the sample and an inert reference (∆T) is measured as the sample and the reference are heated or cooled in a controlled manner. \[∆T = T_S – T_R\] with \(T_S\) as the temperature of sample and \(T_R\) is temperature of reference which is thermally stable. Please go through the Chemlibre link to understand the principle and working of DTA. https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Instrumental_Analysis_(LibreTexts)/31%3A_Thermal_Methods/31.02%3A_Differential_Thermal_Analysis Instrumentation and working: A typical DTA set up is shown in the figure given below. In differential thermal analysis (DTA), the difference in temperature between the sample and a thermally inert reference material is measured as a function of temperature (usually the sample temperature). Any transition that the sample undergoes results in the liberation or absorption of energy by the sample with a corresponding deviation of its temperature from that of the reference. A plot of the differential temperature, ΔT, versus the programmed temperature, T, indicates the transition temperature(s) and whether the transition is exothermic or endothermic. DTA and thermogravimetric analyses (measurement of the change in weight as a function of temperature) are often run simultaneously on a single sample. 3.2 DTA curve and its interpretation \[∆T = T_S – T_R\] T S = temperature of sample T R = temperature of thermally stable reference A typical DTA curve is represented below Brain Teaser: Can you suggest why we adopt ∆T > 0 for exothermic and ∆T < 0 for endothermic transition? Activity \(\PageIndex{1}\) The following figure represents thermal investigation. Fill in the blanks by choosing the correct option. (James & Tonge, 2008) The record shown is that of a_________________ experiment since the_________ plot ∆T which is a ______________temperature. The ___________direction of the peak indicates that a __________________reaction has occurred. This in turn implies that the corresponding _________________ change ∆H must have been _________ ie the value of enthalpy ________________the thermal effect was ____________ than its value _____________.This means that the sample ____________ heat during the reaction. Furthermore, there is evidence of a change _______________in the temperature is increased beyond the thermal transition. This is shown by the ___________of the just beyond the end. Answer Select from the following list [upward/downward, free energy/heat capacity, greater/less, DTG/DTA, base-line/background, derivative/differential, took in/gave out, negative/positive, enthalpy/entropy, before/after/during, exothermic/endothermic/isothermal, abscissa/ordinate, distortion/displacement.] Activity 3.2: (James & Tonge, 2008) During the Falklands campaign in 1982 many naval personnel suffered serious burns. In 1985 the Admiralty decreed that all men on active service in Navy should be issued cotton uniforms instead of polyester ones which had been issued in 1982. Does the thermal data suggest a reason for the decision? Justify your answer. [Hint: Cotton decomposes at 345 °C whereas polyester melts at 255 °C and decomposes at 420 °C] 3.3 Comparison with TGA Unnamed: 0 TGA DTA 1 It measures change (loss or gain) in weight as the sample is subjected to controlled heating program It is a technique in which the difference in temperature between the sample and an inert reference material, is measured as a function of temperature 2 It will detect only those physical and chemical transitions which are accompanied by change in weight. It can detect all physical and chemical transitions. 3 This does not require an inert reference. This requires an inert reference material. 4 It is a quantitative method It is a semi-qualitative method. 5 This technique is generally used to study decomposition reactions. This technique is used to study phase transitions Query \(
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Bookshelves/Analytical_Chemistry
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Analytical chemistry spans nearly all areas of chemistry but involves the development of tools and methods to measure physical properties of substances and apply those techniques to the identification of their presence (qualitative analysis) and quantify the amount present (quantitative analysis) of species in a wide variety of settings. Supplemental Modules (Analytical Chemistry) Analytical Chemistry 2.1 (Harvey) Chemometrics Using R (Harvey) Instrumental Analysis (LibreTexts) Physical Methods in Chemistry and Nano Science (Barron) Non-Isothermal Kinetic Methods (Arhangel'skii et al.) Molecular and Atomic Spectroscopy (Wenzel) Qualitative Analysis of Common Cations in Water (Malik) An Introduction to Mass Spectrometry (Van Bramer) Crystallography in a Nutshell (Ripoll and Cano) Analytical Chemistry Volume I (Harvey) Analytical Chemistry Volume II (Harvey)
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Courses/Nassau_Community_College/Organic_Chemistry_I_and_II/01%3A_Introduction_and_Review
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Learning Objectives After reading this chapter and completing ALL the exercises, a student can b e able to discuss the origins of organic chemistry - refer to section 1.1 use and apply the language of Atomic Structure (atomic number, mass number, isotopes) - refer to section 1.2 draw, interpret, and convert between Lewis (Kekule), Condensed, and Bond-line Structures - refer to sections 1.3, 1.4, 1.5, and 1.6 apply bonding patterns and polarity to organic compounds - refer to section 1.7 and 1.8 identify polar bonds and compounds - refer to section 1.9 draw resonance forms and predict the relative contribution of each resonance form to the overall structure of the compound or ion - refer to section 1.10 recognize acids and bases - refer to sections 1.11 and 1.12 use the definition of Lewis Acids and Bases to recognize electron movement in reactions - refer to section 1.13 predict reaction products of acid-base reactions - refer to sections 1.11, 1.12 ,and 1.13 determine relative strengths of acids and bases from their pK a values - refer to section 1.14 determine the form of an acid or base at a specified pH (given the pK a ) - refer to section 1.14 predict relative strengths of acids and bases from their structure, bonding and resonance - refer to section 1.15 determine the e mpirical and molecular formulas from combustion data - refer to section 1.16 1.1: The Origins of Organic Chemistry The term "organic chemistry" was first used in the 1800's to distinguish reactions performed by chemists from those occurring within living organisms. 1.2: Principles of Atomic Structure (Review) Atomic structure is applied to several topics of organic chemistry. We will use atomic numbers and isotopes to determine priorities in nomenclature and stereoisomerism and mass numbers and isotopic distribution will be applicable in mass spectrometry. 1.3: Electronic Structure (Review) A basic knowledge of the electronic structure of atoms requires an understanding of the properties of waves and electromagnetic radiation. 1.4: Electron Configurations The electron configuration of an atom indicates the number of valence electrons. Valence electrons determine the unique chemistry of each element. 1.5: Octet Rule - Ionic and Covalent Bonding (Review) By now we can all recite the Octet Rule in our sleep: atoms gain, lose, or share electrons to fill their s and p subshells to create stable compounds and ions. 1.6: Lewis Structures Lewis structures show us how atoms come together to create compounds and ions according to the octet rule. Recognizing the formal charge of some common bonding patterns will be helpful in learning reaction mechanisms. 1.7: Common Bonding Patterns for Organic Chemistry For organic chemistry, the common bonding patterns of carbon, oxygen, and nitrogen have useful applications to chemical structure and reactivity. 1.8: Structural Formulas - Lewis, Kekule, Bond-line, Condensed, Here you will learn how to understand, write, draw, and talk-the-talk of organic molecules. Organic molecules can get complicated and large, so o-chemists have developed short hand notations to communicate structure. 1.9: Electronegativity and Bond Polarity (Review) Recognizing and distinguishing between polar and nonpolar compounds is an essential skill for organic chemistry. Polarity is the underlying factor of intermolecular forces and polar bonds are often a source of chemical reactivity. 1.10: Resonance Resonance delocalizes shared electrons over three or more atoms to lower the electron density and stabilize the compound or ion. 1.11: Arrhenius Acids and Bases (Review) Arrhenius acids form hydrogen ions in aqueous solution with Arrhenius bases forming hydroxide ions. 1.12: Brønsted-Lowry Acids and Bases (Review) A Brønsted-Lowry acid is a proton donor, while a Brønsted-Lowry base is a proton acceptor. 1.13: Lewis Acids and Bases Lewis acids are electron acceptors and Lewis bases are electron donors. The electron flow in Lewis acid-base chemistry prepares us for the electron flow of many organic reactions. 1.14: Distinguishing between pH and pKa The pKa of a compound is empirically determined similar to the melting point or boiling point. The pH communicates the environment of the sample as a measure of the hydrogen ion concentration and can be changed. 1.15: Predicting Relative Acidity To determine the relative acidity of compounds, we compare the relative stabilities of their conjugate bases. This skill can be applied to organic reactions when comparing leaving group stability to evaluate reaction pathways. 1.16: Molecular Formulas and Empirical Formulas (Review) Identification of unknown organic compounds often begins with the determination of the empirical and molecular formulas. 1.17: Additional Exercises This section has additional exercises for the key learning objectives of this chapter. 1.18: Solutions to Additional Exercises This section has the solutions to the additional exercises from the previous section.
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Courses/Mt._San_Antonio_College/Chem_10_-_Chemistry_for_Allied_Health_Majors_(1st_semester)/01%3A_Numbers_and_Measurements/1.04%3A_Measurement_and_Significant_Figures
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Learning Objectives Identify the number of significant figures in a reported value. Apply the concept of significant figures to report a measurement with the proper number of digits. Scientists have established certain conventions for communicating the degree of precision of a measurement, which is dependent on the measuring device used (See Figure \(\PageIndex{1}\)). Imagine, for example, that you are using a meterstick to measure the width of a table. The centimeters (cm) marked on the meterstick, tell you how many centimeters wide the table is. Many metersticks also have markings for millimeters (mm), so we can measure the table to the nearest millimeter . The measurement made using millimeters is more precise, it is closer to the actual length of the table. Most metersticks do not have any smaller (or more precise) markings indicated, so you cannot report the measured width of the table any more precise than to the nearest millimeter. The concept of significant figures takes the limitation of measuring devices into account. The significant figures of a measured quantity are defined as all the digits known with certainty (those indicated by the markings on the measuring device) and the first uncertain, or estimated, digit (one digit past the smallest marking on the measuring device). It makes no sense to estimate and report any digits after the first uncertain one, so it is the last significant digit reported in a measurement. Zeros are used when needed to place the significant figures in their correct positions. Thus, zeros are sometimes counted as significant figures but are sometimes only used as placeholders (see the rules for significant figures below for more details). “Sig figs” is a common abbreviation for significant figures. Rules for Determination of Significant Figures Consider the earlier example of measuring the width of a table with a meterstick. If the table is measured and reported as being 1,357 mm wide, the number 1,357 has four significant figures. The 1 (thousands place), the 3 (hundreds place), and the 5 (tens place) are certain; the 7 (ones place) is assumed to have been estimated. It would make no sense to report such a measurement as 1,357.0 (five Sig Figs) or 1,357.00 (six Sig Figs) because that would suggest the measuring device was able to determine the width to the nearest tenth or hundredth of a millimeter, when in fact it shows only tens of millimeters and therefore the ones place was estimated. On the other hand, if a measurement is reported as 150 mm, the 1 (hundreds) and the 5 (tens) are known to be significant, but how do we know whether the zero is or is not significant? The measuring device could have had marks indicating every 100 mm or marks indicating every 10 mm. How can you determine if the zero is significant (the estimated digit), or if the 5 is significant and the zero a value placeholder? The rules for deciding which digits in a measurement are significant are as follows: Rule 1: All nonzero digits in a measurement are significant. 237 has three significant figures. 1.897 has four significant figures. Rule 2: Zeros that appear between other nonzero digits (i.e., " middle zeros ") are always significant. 39,004 has five significant figures. 5.02 has three significant figures. Rule 3: Zeros that appear in front of all of the nonzero digits are called leading zeros . Leading zeros are never significant. 0.008 has one significant figure. 0.000416 has three significant figures. Rule 4: Zeros that appear after all nonzero digits are called trailing zeros . A number with trailing zeros that lacks a decimal point may or may not be significant. 1400 is ambiguous. \(1.4 \times 10^3\) has two significant figures. \(1.40 \times 10^3\) three significant figures. \(1.400 \times 10^3\) has four significant figures. Rule 5: Trailing zeros in a number with a decimal point are significant. This is true whether the zeros occur before or after the decimal point. 620.0 has four significant figures. 19.000 has five significant figures. It needs to be emphasized that just because a certain digit is not significant does not mean that it is not important or that it can be left out. Though the zero in a measurement of 140 may not be significant, the value cannot simply be reported as 14. An insignificant zero functions as a placeholder for the decimal point. When numbers are written in scientific notation, this becomes more apparent. The measurement 140 can be written as \(1.4 \times 10^2\), with two significant figures in the coefficient or as \(1.40 \times 10^3\), with three significant figures. A number less than one, such as 0.000416, can be written in scientific notation as \(4.16 \times 10^{-4}\), which has 3 significant figures. In some cases, scientific notation is the only way to correctly indicate the correct number of significant figures. In order to report a value of 15,000,00 with four significant figures, it would need to be written as \(1.500 \times 10^7\). Exact Quantities When numbers are known exactly, the significant figure rules do not apply. This occurs when objects are counted rather than measured. For example, a carton of eggs has 12 eggs. The actual value cannot be 11.8 eggs, since we count eggs in whole number quantities. So the 12 is an exact quantity. Exact quantities are considered to have an infinite number of significant figures; the importance of this concept will be seen later when we begin looking at how significant figures are dealt with during calculations. Numbers in many conversion factors, especially for simple unit conversions, are also exact quantities and have infinite significant figures. There are exactly 100 centimeters in 1 meter and exactly 60 seconds in 1 minute. Those values are definitions and are not the result of a measurement. Example \(\PageIndex{2}\) Give the number of significant figures in each. Identify the rule for each. 5.87 0.031 52.90 00.2001 500 6 atoms Solution Explanation Answer a. All three numbers are significant (rule 1). 5.87 , three significant figures b. The leading zeros are not significant (rule 3). The 3 and the 1 are significant (rule 1) 0.031, two significant figures c. The 5, the 2 and the 9 are significant (rule 1). The trailing zero is also significant (rule 5). 52.90, four significant figures d. The leading zeros are not significant (rule 3). The 2 and the 1 are significant (rule 1) and the middle zeros are also significant (rule 2). 00.2001, four significant figures e. The number is ambiguous. It could have one, two or three significant figures. 500, ambiguous f. The 6 is a counting number. A counting number is an exact number. 6, infinite Exercise \(\PageIndex{2}\) Give the number of significant figures in each. 36.7 m 0.006606 s 2,002 kg 306,490,000 people 3,800 g Answer a: three significant figures. Answer b: four significant figures. Answer c: four significant figures. Answer d: Infinite (Exact number) Answer e: Ambiguous, could be two, three or four significant figure. Summary Uncertainty exists in all measurements. The degree of uncertainty is affected in part by the quality of the measuring tool. Significant figures give an indication of the certainty of a measurement. Rules allow decisions to be made about how many digits to use in any given situation.
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Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Topics_in_Thermodynamics_of_Solutions_and_Liquid_Mixtures/01%3A_Modules/1.24%3A_Misc/1.14.63%3A_Solubilities_of_Solids_in_Liquids
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This very large subject can be divided into two groups. The first group concerns the solubility of a given solid substance \(j\) in a given solvent, liquid \(\ell_{1}\). The second group involves comparison of the solubilities of a given solid in two liquids, \(\ell_{1}\) and \(\ell_{2}\). A closed system (at defined \(\mathrm{T}\) and \(\mathrm{p}\), the latter being close to the standard pressure) contains solid substance \(j\) in equilibrium with an aqueous solution containing solute \(j\). The system is characterized by the (equilibrium) solubility, \(\mathrm{m}_{\mathrm{j}}^{\mathrm{eq}}(\mathrm{aq})\). At equilibrium, \[\mu_{\mathrm{j}}^{*}(\mathrm{~s})=\mu_{\mathrm{j}}^{0}(\mathrm{aq})+\mathrm{R} \, \mathrm{T} \, \ln \left[\mathrm{m}_{\mathrm{j}}^{\mathrm{eq}}(\mathrm{aq}) \, \gamma_{\mathrm{j}}^{\mathrm{eq}}(\mathrm{aq}) / \mathrm{m}^{0}\right] \nonumber \] Then \[\Delta \mu_{\mathrm{j}}^{0}=\mu_{\mathrm{j}}^{0}(\mathrm{aq})-\mu_{\mathrm{j}}^{*}(\mathrm{~s})=-\mathrm{R} \, \mathrm{T} \, \ln \left[\mathrm{m}_{\mathrm{j}}^{\mathrm{eq}}(\mathrm{aq}) \, \gamma_{\mathrm{j}}^{\mathrm{eq}}(\mathrm{aq}) / \mathrm{m}^{0}\right] \nonumber \] If the aqueous solution is dilute and the solubility is low, it can often be assumed that the properties of the solution are ideal. Hence, \[\Delta \mu_{\mathrm{j}}^{0}=\mu_{\mathrm{j}}^{0}(\mathrm{aq})-\mu_{\mathrm{j}}^{\mathrm{*}}(\mathrm{s})=-\mathrm{R} \, \mathrm{T} \, \ln \left[\mathrm{m}_{\mathrm{j}}^{\mathrm{eq}}(\mathrm{aq}) / \mathrm{m}^{0}\right] \nonumber \] It should be noted that the sign of \(\Delta \mu_{j}^{0}\) depends on whether or not \(m_{j}^{e q}(a q)\) is larger or less than unity. We illustrate the second approach by considering a combination of the experiment described above and an experiment where the solvent is a binary aqueous mixture, mole fraction composition \(\mathrm{x}_{2}\). At equilibrium, \[\mu_{j}^{*}(s)=\mu_{j}^{0}\left(s \ln ; x_{2}\right)+R \, T \, \ln \left[m_{j}^{\mathrm{eq}}\left(s \ln ; x_{2}\right) \, \gamma_{j}^{\mathrm{eq}}\left(s \ln ; x_{2}\right) / m^{0}\right] \nonumber \] \[\begin{aligned} \Delta\left(\mathrm{aq} \rightarrow \mathrm{x}_{2}\right) \mu_{\mathrm{j}}^{0}=\mu_{\mathrm{j}}^{0}\left(\mathrm{~s} \ln ; \mathrm{x}_{2}\right)-\mu_{\mathrm{j}}^{0}(\mathrm{aq}) \\ =-\mathrm{R} \, \mathrm{T} \, \ln \left[\mathrm{m}_{\mathrm{j}}^{\mathrm{eq}}\left(\mathrm{s} \ln ; \mathrm{x}_{2}\right) \, \gamma_{\mathrm{j}}^{\mathrm{eq}}\left(\mathrm{s} \ln ; \mathrm{x}_{2}\right) / \mathrm{m}_{\mathrm{j}}^{\mathrm{eq}}(\mathrm{aq}) \, \gamma_{\mathrm{j}}^{\mathrm{eq}}(\mathrm{aq})\right] \end{aligned} \nonumber \] If both solutions are dilute in substance \(j\), the ratio, \(\gamma_{j}^{\mathrm{eq}}\left(\mathrm{s} \ln ; \mathrm{x}_{2}\right) / \gamma_{\mathrm{j}}^{\mathrm{eq}}(\mathrm{aq})\) can be assumed to be close to unity. In fact this is a better approximation than assuming both activity coefficients are unity. Then \[\Delta\left(\mathrm{aq} \rightarrow \mathrm{x}_{2}\right) \mu_{\mathrm{j}}^{0}=-\mathrm{R} \, \mathrm{T} \, \ln \left[\mathrm{m}_{\mathrm{j}}^{\mathrm{eq}}\left(\mathrm{s} \ln ; \mathrm{x}_{2}\right) / \mathrm{m}_{\mathrm{j}}^{\mathrm{eq}}(\mathrm{aq})\right] \nonumber \] In other words if the solubility of substance \(j\) increases on adding solvent component 2 then \(\Delta\left(\mathrm{aq} \rightarrow \mathrm{x}_{2}\right) \mu_{\mathrm{j}}^{\mathrm{c}}\) is negative. This stabilization is a consequence of a difference in solute-solvent interactions.
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Courses/University_of_Illinois_UrbanaChampaign/Chem_2363A_Fundamental_Organic_Chemistry_I_(Chan)/08%3A_Introduction_to_Alkyl_Halides_Alcohols_Ethers_Thiols_and_Sulfides/8.02%3A_Haloalkanes/8.2.01%3A_Alkyl_Halide_Occurrence
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Halogen containing organic compounds are relatively rare in terrestrial plants and animals. The thyroid hormones T 3 and T 4 are exceptions; as is fluoroacetate, the toxic agent in the South African shrub Dichapetalum cymosum , known as "gifblaar". However, the halogen rich environment of the ocean has produced many interesting natural products incorporating large amounts of halogen. Some examples are shown below. The ocean is the largest known source for atmospheric methyl bromide and methyl iodide. Furthermore, the ocean is also estimated to supply 10-20% of atmospheric methyl chloride, with other significant contributions coming from biomass burning, salt marshes and wood-rotting fungi. Many subsequent chemical and biological processes produce poly-halogenated methanes. Synthetic organic halogen compounds are readily available by direct halogenation of hydrocarbons and by addition reactions to alkenes and alkynes. Many of these have proven useful as intermediates in traditional synthetic processes. Some halogen compounds, shown in the box. have been used as pesticides, but their persistence in the environment, once applied, has led to restrictions, including banning, of their use in developed countries. Because DDT is a cheap and effective mosquito control agent, underdeveloped countries in Africa and Latin America have experienced a dramatic increase in malaria deaths following its removal, and arguments are made for returning it to limited use. 2,4,5-T and 2,4-D are common herbicides that are sold by most garden stores. Other organic halogen compounds that have been implicated in environmental damage include the polychloro- and polybromo-biphenyls (PCBs and PBBs), used as heat transfer fluids and fire retardants; and freons (e.g. CCl 2 F 2 and other chlorofluorocarbons) used as refrigeration gases and fire extinguishing agents. Contributors William Reusch, Professor Emeritus ( Michigan State U. ), Virtual Textbook of Organic Chemistry
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Courses/University_of_Kansas/KU%3A_CHEM_110_GOB_Chemistry_(Sharpe_Elles)/01%3A_Chemistry_Matter_and_Measurement/1.07%3A_Converting_Units
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Skills to Develop To convert a value reported in one unit to a corresponding value in a different unit. The ability to convert from one unit to another is an important skill. For example, a nurse with 50 mg aspirin tablets who must administer 0.2 g of aspirin to a patient needs to know that 0.2 g equals 200 mg, so 4 tablets are needed. Fortunately, there is a simple way to convert from one unit to another. Conversion Factors If you learned the SI units and prefixes described, then you know that 1 cm is 1/100th of a meter. \[ 1\; \rm{cm} = \dfrac{1}{100} \; \rm{m}\] or \[100\; \rm{cm} = 1\; \rm{m}\] Suppose we divide both sides of the equation by 1 m (both the number and the unit): \[\mathrm{\dfrac{100\:cm}{1\:m}=\dfrac{1\:m}{1\:m}}\] As long as we perform the same operation on both sides of the equals sign, the expression remains an equality. Look at the right side of the equation; it now has the same quantity in the numerator (the top) as it has in the denominator (the bottom). Any fraction that has the same quantity in the numerator and the denominator has a value of 1: We know that 100 cm is 1 m, so we have the same quantity on the top and the bottom of our fraction, although it is expressed in different units. A fraction that has equivalent quantities in the numerator and the denominator but expressed in different units is called a conversion factor. Here is a simple example. How many centimeters are there in 3.55 m? Perhaps you can determine the answer in your head. If there are 100 cm in every meter, then 3.55 m equals 355 cm. To solve the problem more formally with a conversion factor, we first write the quantity we are given, 3.55 m. Then we multiply this quantity by a conversion factor, which is the same as multiplying it by 1. We can write 1 as \(\mathrm{\frac{100\:cm}{1\:m}}\) and multiply: \[ 3.55 \; \rm{m} \times \dfrac{100 \; \rm{cm}}{1\; \rm{m}}\] The 3.55 m can be thought of as a fraction with a 1 in the denominator. Because m, the abbreviation for meters, occurs in both the numerator and the denominator of our expression, they cancel out: \[\dfrac{3.55 \; \cancel{\rm{m}}}{ 1} \times \dfrac{100 \; \rm{cm}}{1 \; \cancel{\rm{m}}}\] The final step is to perform the calculation that remains once the units have been canceled: \[ \dfrac{3.55}{1} \times \dfrac{100 \; \rm{cm}}{1} = 355 \; \rm{cm} \label{Ex1}\] In the final answer, we omit the 1 in the denominator. Thus, by a more formal procedure, we find that 3.55 m equals 355 cm. A generalized description of this process is as follows: \[\text{quantity (in old units)} \times \text{conversion factor} = \text{quantity (in new units)} \nonumber\] You may be wondering why we use a seemingly complicated procedure for a straightforward conversion. In later studies, the conversion problems you will encounter will not always be so simple . If you can master the technique of applying conversion factors, you will be able to solve a large variety of problems. In the previous example (Equation \ref{Ex1}), we used the fraction \(\frac{100 \; \rm{cm}}{1 \; \rm{m}}\) as a conversion factor. Does the conversion factor \(\frac{1 \; \rm m}{100 \; \rm{cm}}\) also equal 1? Yes, it does; it has the same quantity in the numerator as in the denominator (except that they are expressed in different units). Why did we not use that conversion factor? If we had used the second conversion factor, the original unit would not have canceled, and the result would have been meaningless. Here is what we would have gotten: \[ 3.55 \; \rm{m} \times \dfrac{1\; \rm{m}}{100 \; \rm{cm}} = 0.0355 \dfrac{\rm{m}^2}{\rm{cm}}\] For the answer to be meaningful, we have to construct the conversion factor in a form that causes the original unit to cancel out . Figure \(\PageIndex{1}\) shows a concept map for constructing a proper conversion. Figure \(\PageIndex{1}\): A Concept Map for Conversions. This is how you construct a conversion factor to convert from one unit to another. Significant Figures in Conversions How do conversion factors affect the determination of significant figures? Numbers in conversion factors based on prefix changes, such as kilograms to grams, are not considered in the determination of significant figures in a calculation because the numbers in such conversion factors are exact. Exact numbers are defined or counted numbers, not measured numbers, and can be considered as having an infinite number of significant figures. (In other words, 1 kg is exactly 1,000 g, by the definition of kilo-.) Counted numbers are also exact. If there are 16 students in a classroom, the number 16 is exact. In contrast, conversion factors that come from measurements (such as density, as we will see shortly) or are approximations have a limited number of significant figures and should be considered in determining the significant figures of the final answer. Example \(\PageIndex{1}\) The average volume of blood in an adult male is 4.7 L. What is this volume in milliliters? A hummingbird can flap its wings once in 18 ms. How many seconds are in 18 ms? SOLUTION We start with what we are given, 4.7 L. We want to change the unit from liters to milliliters. There are 1,000 mL in 1 L. From this relationship, we can construct two conversion factors: \[ \dfrac{1\; \rm{L}}{1,000\; \rm{mL}} \; \text{ or } \; \dfrac{1,000 \; \rm{mL}}{1\; \rm{L}} \nonumber\] We use the conversion factor that will cancel out the original unit, liters, and introduce the unit we are converting to, which is milliliters. The conversion factor that does this is the one on the right. \[ 4.7 \cancel{\rm{L}} \times \dfrac{1,000 \; \rm{mL}}{1\; \cancel{\rm{L}}} = 4,700\; \rm{mL} \nonumber\] Because the numbers in the conversion factor are exact, we do not consider them when determining the number of significant figures in the final answer. Thus, we report two significant figures in the final answer. We can construct two conversion factors from the relationships between milliseconds and seconds: \[ \dfrac{1,000 \; \rm{ms}}{1\; \rm{s}} \; \text{ or } \; \dfrac{1\; \rm{s}}{1,000 \; \rm{ms}} \nonumber\] To convert 18 ms to seconds, we choose the conversion factor that will cancel out milliseconds and introduce seconds. The conversion factor on the right is the appropriate one. We set up the conversion as follows: \[ 18 \; \cancel{\rm{ms}} \times \dfrac{1\; \rm{s}}{1,000 \; \cancel{\rm{ms}}} = 0.018\; \rm{s} \nonumber\] The conversion factor’s numerical values do not affect our determination of the number of significant figures in the final answer. Exercise \(\PageIndex{1}\) Perform each conversion. 101,000 ns to seconds 32.08 kg to grams Conversion factors can also be constructed for converting between different kinds of units. For example, density can be used to convert between the mass and the volume of a substance. Consider mercury, which is a liquid at room temperature and has a density of 13.6 g/mL. The density tells us that 13.6 g of mercury have a volume of 1 mL. We can write that relationship as follows: 13.6 g mercury = 1 mL mercury This relationship can be used to construct two conversion factors: \[\mathrm{\dfrac{13.6\:g}{1\:mL}\:and\:\dfrac{1\:mL}{13.6\:g}}\] Which one do we use? It depends, as usual, on the units we need to cancel and introduce. For example, suppose we want to know the mass of 16 mL of mercury. We would use the conversion factor that has milliliters on the bottom (so that the milliliter unit cancels) and grams on top so that our final answer has a unit of mass: \[ \begin{align*} \mathrm{16\:\cancel{mL}\times\dfrac{13.6\:g}{1\:\cancel{mL}}} &= \mathrm{217.6\:g} \\[5pt] &\approx \mathrm{220\:g} \end{align*} \] In the last step, we limit our final answer to two significant figures because the volume quantity has only two significant figures; the 1 in the volume unit is considered an exact number, so it does not affect the number of significant figures. The other conversion factor would be useful if we were given a mass and asked to find volume, as the following example illustrates. Density can be used as a conversion factor between mass and volume. Example \(\PageIndex{2}\): Mercury Thermometer A mercury thermometer for measuring a patient’s temperature contains 0.750 g of mercury. What is the volume of this mass of mercury? SOLUTION Because we are starting with grams, we want to use the conversion factor that has grams in the denominator. The gram unit will cancel algebraically, and milliliters will be introduced in the numerator. \[ \begin{align*} 0.750 \; \cancel{\rm{g}} \times \dfrac{1\; \rm{mL}}{13.6 \; \cancel{\rm{g}}} &= 0.055147 \ldots \; \rm{mL} \\[5pt] &\approx 0.0551\; \rm{mL} \end{align*} \] We have limited the final answer to three significant figures. Exercise \(\PageIndex{2}\) What is the volume of 100.0 g of air if its density is 1.3 g/L? Looking Closer: Density and the Body The densities of many components and products of the body have a bearing on our health. Bones. Bone density is important because bone tissue of lower-than-normal density is mechanically weaker and susceptible to breaking. The density of bone is, in part, related to the amount of calcium in one’s diet; people who have a diet deficient in calcium, which is an important component of bones, tend to have weaker bones. Dietary supplements or adding dairy products to the diet seems to help strengthen bones. As a group, women experience a decrease in bone density as they age. It has been estimated that fully half of women over age 50 suffer from excessive bone loss, a condition known as osteoporosis. Exact bone densities vary within the body, but for a healthy 30-year-old female, it is about 0.95–1.05 g/cm 3 . Osteoporosis is diagnosed if the bone density is below 0.6–0.7 g/cm 3 . Urine. The density of urine can be affected by a variety of medical conditions. Sufferers of diabetes produce an abnormally large volume of urine with a relatively low density. In another form of diabetes, called diabetes mellitus, there is excess glucose dissolved in the urine, so that the density of urine is abnormally high. The density of urine may also be abnormally high because of excess protein in the urine, which can be caused by congestive heart failure or certain renal (kidney) problems. Thus, a urine density test can provide clues to various kinds of health problems. The density of urine is commonly expressed as a specific gravity, which is a unitless quantity defined as \[ \dfrac{\text{density of some material}}{\text{density of water}} \nonumber\] Normal values for the specific gravity of urine range from 1.002 to 1.028. Body Fat. The overall density of the body is one indicator of a person’s total body fat. Fat is less dense than muscle and other tissues, so as it accumulates, the overall density of the body decreases. Measurements of a person’s weight and volume provide the overall body density, which can then be correlated to the percentage of body fat. (The body’s volume can be measured by immersion in a large tank of water. The amount of water displaced is equal to the volume of the body.) Multiple Conversions Sometimes you will have to perform more than one conversion to obtain the desired unit. For example, suppose you want to convert 54.7 km into millimeters. You can either memorize the relationship between kilometers and millimeters, or you can do the conversion in steps. Most people prefer to convert in steps. To do a stepwise conversion, we first convert the given amount to the base unit. In this example, the base unit is meters. We know that there are 1,000 m in 1 km: \[ 54.7\; \cancel{\rm{km}} \times \dfrac{1,000 \; \rm{m}}{1\; \cancel{\rm{km}}} = 54,700\; \rm{m} \nonumber\] Then we take the result (54,700 m) and convert it to millimeters, remembering that there are \(1,000\; \rm{mm}\) for every \(1\; \rm{m}\): \[ \begin{align*} 54,700 \; \cancel{\rm{m}} \times \dfrac{1,000 \; \rm{mm}}{1\; \cancel{\rm{m}}} &= 54,700,000 \; \rm{mm} \\[5pt] &= 5.47 \times 10^7\; \rm{mm} \end{align*} \] We have expressed the final answer in scientific notation. As a shortcut, both steps in the conversion can be combined into a single, multistep expression: \[ \begin{align*} 54.7\; \cancel{\rm{km}} \times \dfrac{1,000 \; \cancel{\rm{m}}}{1\; \cancel{\rm{km}}} \times \dfrac{1,000 \; \rm{mm}}{1\; \cancel{\rm{m}}} &= 54,700,000 \; \rm{mm} \\[5pt] &= 5.47 \times 10^7\; \rm{mm} \end{align*} \] Either method—one step at a time or all the steps together—is acceptable. If you do all the steps together, the restriction for the proper number of significant figures should be done after the last step. As long as the math is performed correctly, you should get the same answer no matter which method you use. Example \(\PageIndex{3}\) Convert 58.2 ms to megaseconds in one multistep calculation. SOLUTION First, convert the given unit to the base unit—in this case, seconds—and then convert seconds to the final unit, megaseconds: \[ \begin{align*} 58.2 \; \cancel{\rm{ms}} \times \dfrac{\cancel{1 \rm{s}}}{1,000\; \cancel{\rm{ms}}} \times \dfrac{1\; \rm{Ms}}{1,000,000\; \cancel{ \rm{s}}} &=0.0000000582\; \rm{Ms} \\[5pt] &= 5.82 \times 10^{-8}\; \rm{mS} \end{align*} \] Neither conversion factor affects the number of significant figures in the final answer. Exercise \(\PageIndex{3}\) Convert 43.007 ng to kilograms in one multistep calculation. Career Focus: Pharmacist A pharmacist dispenses drugs that have been prescribed by a doctor. Although that may sound straightforward, pharmacists in the United States must hold a doctorate in pharmacy and be licensed by the state in which they work. Most pharmacy programs require four years of education in a specialty pharmacy school. Pharmacists must know a lot of chemistry and biology so they can understand the effects that drugs (which are chemicals, after all) have on the body. Pharmacists can advise physicians on the selection, dosage, interactions, and side effects of drugs. They can also advise patients on the proper use of their medications, including when and how to take specific drugs properly. Pharmacists can be found in drugstores, hospitals, and other medical facilities. Curiously, an outdated name for pharmacist is chemist , which was used when pharmacists formerly did a lot of drug preparation, or compounding . In modern times, pharmacists rarely compound their own drugs, but their knowledge of the sciences, including chemistry, helps them provide valuable services in support of everyone’s health. Key Takeaway A unit can be converted to another unit of the same type with a conversion factor. Concept Review Exercises How do you determine which quantity in a conversion factor goes in the denominator of the fraction? State the guidelines for determining significant figures when using a conversion factor. Write a concept map (a plan) for how you would convert \(1.0 \times 10^{12}\) nano liters (nL) to kilo liters (kL). Answers The unit you want to cancel from the numerator goes in the denominator of the conversion factor. Exact numbers that appear in many conversion factors do not affect the number of significant figures; otherwise, the normal rules of multiplication and division for significant figures apply.
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Courses/Erie_Community_College/ECC%3A_Introduction_to_General_Organic_and_Biochemistry_(Sorrentino)/Text/15%3A_Organic_Acids_and_Bases_and_Some_of_Their_Derivatives/15.06%3A_Esters_-_Structures_and_Names
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Learning Objectives Identify the general structure for an ester. Use common names to name esters. Name esters according to the IUPAC system. Esters have the general formula RCOOR ′, where R may be a hydrogen atom, an alkyl group, or an aryl group, and R′ may be an alkyl group or an aryl group but not a hydrogen atom. (If it were hydrogen atom, the compound would be a carboxylic acid.) Figure \(\PageIndex{1}\) shows models for two common esters. Esters occur widely in nature. Unlike carboxylic acids, esters generally have pleasant odors and are often responsible for the characteristic fragrances of fruits and flowers. Once a flower or fruit has been chemically analyzed, flavor chemists can attempt to duplicate the natural odor or taste. Both natural and synthetic esters are used in perfumes and as flavoring agents. Fats and vegetable oils are esters of long-chain fatty acids and glycerol. Esters of phosphoric acid are of the utmost importance to life. Names of Esters Although esters are covalent compounds and salts are ionic, esters are named in a manner similar to that used for naming salts. The group name of the alkyl or aryl portion is given first and is followed by the name of the acid portion. In both common and International Union of Pure and Applied Chemistry (IUPAC) nomenclature, the - ic ending of the parent acid is replaced by the suffix - ate (Table \(\PageIndex{1}\)). Condensed Structural Formula Common Name IUPAC Name HCOOCH3 methyl formate methyl methanoate CH3COOCH3 methyl acetate methyl ethanoate CH3COOCH2CH3 ethyl acetate ethyl ethanoate CH3CH2COOCH2CH3 ethyl propionate ethyl propanoate CH3CH2CH2COOCH(CH3)2 isopropyl butyrate isopropyl butanoate NaN ethyl benzoate ethyl benzoate Example \(\PageIndex{1}\) Give the common and IUPAC names for each compound. Solution The alkyl group attached directly to the oxygen atom is a butyl group (in green). The part of the molecule derived from the carboxylic acid (in red) has three carbon atoms. It is called propionate (common) or propanoate (IUPAC). The ester is therefore butyl propionate or butyl propanoate. An alkyl group (in green) is attached directly to the oxygen atom by its middle carbon atom; it is an isopropyl group. The part derived from the acid (that is, the benzene ring and the carbonyl group, in red) is benzoate. The ester is therefore isopropyl benzoate (both the common name and the IUPAC name). Exercise \(\PageIndex{1}\) Give the common and IUPAC names for each compound. Example \(\PageIndex{2}\) Draw the structure for ethyl pentanoate. Solution Start with the portion from the acid. Draw the pentanoate (five carbon atoms) group first; keeping in mind that the last carbon atom is a part of the carboxyl group. Then attach the ethyl group to the bond that ordinarily holds the hydrogen atom in the carboxyl group. Exercise \(\PageIndex{2}\) Draw the structure for phenyl pentanoate. Key Takeaway An ester has an OR group attached to the carbon atom of a carbonyl group.
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Courses/Los_Angeles_Trade_Technical_College/LATTC_Hybrid_Chem_51/11%3A_Acids_and_Bases/11.6%3A_Autoionization_of_Water
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Learning Objectives Describe the autoionization of water. Calculate the concentrations of H + and OH − in solutions, knowing the other concentration. We have already seen that H 2 O can act as an acid or a base: NH 3 + H 2 O → NH 4 + + OH − (H 2 O acts as an acid) HCl + H 2 O → H 3 O + + Cl − (H 2 O acts as a base) It may not be surprising to learn, then, that within any given sample of water, some H 2 O molecules are acting as acids, and other H 2 O molecules are acting as bases. The chemical equation is as follows: H 2 O + H 2 O → H 3 O + + OH − This occurs only to a very small degree: only about 6 in 10 8 H 2 O molecules are participating in this process, which is called the autoionization of water . At this level, the concentration of both H + (aq) and OH − (aq) in a sample of pure H 2 O is about 1.0 × 10 −7 M. If we use square brackets—[ ]—around a dissolved species to imply the molar concentration of that species, we have [H + ] = [OH − ] = 1.0 × 10 −7 M for any sample of pure water because H 2 O can act as both an acid and a base. The product of these two concentrations is 1.0 × 10 −14 : [H + ] × [OH − ] = (1.0 × 10 −7 )(1.0 × 10 −7 ) = 1.0 × 10 −14 In acids, the concentration of H + (aq)—[H + ]—is greater than 1.0 × 10 −7 M, while for bases the concentration of OH − (aq)—[OH − ]—is greater than 1.0 × 10 −7 M. However, the product of the two concentrations—[H + ][OH − ]—is always equal to 1.0 × 10 −14 , no matter whether the aqueous solution is an acid, a base, or neutral: [H + ][OH − ] = 1.0 × 10 −14 This value of the product of concentrations is so important for aqueous solutions that it is called the autoionization constant of water and is denoted K w : K w = [H + ][OH − ] = 1.0 × 10 −14 This means that if you know [H + ] for a solution, you can calculate what [OH − ] has to be for the product to equal 1.0 × 10 −14 , or if you know [OH − ], you can calculate [H + ]. This also implies that as one concentration goes up, the other must go down to compensate so that their product always equals the value of K w . Example \(\PageIndex{1}\) What is [OH − ] of an aqueous solution if [H + ] is 1.0 × 10 −4 M? Solution Using the expression and known value for K w , K w = [H + ][OH − ] = 1.0 × 10 −14 = (1.0 × 10 −4 )[OH − ] We solve by dividing both sides of the equation by 1.0 × 10 −4 : \[\left [ OH^{-} \right ]=\frac{1.0\times 10^{-14}}{1.0\times 10^{-4}}=1.0\times 10^{-10}M\nonumber \] It is assumed that the concentration unit is molarity, so [OH − ] is 1.0 × 10 −10 M. Exercise \(\PageIndex{1}\) What is [H + ] of an aqueous solution if [OH − ] is 1.0 × 10 −9 M? Answer 1.0 × 10 −5 M When you have a solution of a particular acid or base, you need to look at the formula of the acid or base to determine the number of H + or OH − ions in the formula unit because [H + ] or [OH − ] may not be the same as the concentration of the acid or base itself. Example \(\PageIndex{2}\) What is [H + ] in a 0.0044 M solution of Ca(OH) 2 ? Solution We begin by determining [OH − ]. The concentration of the solute is 0.0044 M, but because Ca(OH) 2 is a strong base, there are two OH − ions in solution for every formula unit dissolved, so the actual [OH − ] is two times this, or 2 × 0.0044 M = 0.0088 M. Now we can use the K w expression: [H + ][OH − ] = 1.0 × 10 −14 = [H + ](0.0088 M) Divide both sides by 0.0088: \[\left [ H^{+} \right ]=\frac{1.0\times 10^{-14}}{(0.0088)}=1.1\times 10^{-12}M\nonumber \] [H + ] has decreased significantly in this basic solution. Exercise \(\PageIndex{2}\) What is [OH − ] in a 0.00032 M solution of H 2 SO 4 ? (Hint: assume both H + ions ionize.) Answer 1.6 × 10 −11 M For strong acids and bases, [H + ] and [OH − ] can be determined directly from the concentration of the acid or base itself because these ions are 100% ionized by definition. However, for weak acids and bases, this is not so. The degree, or percentage, of ionization would need to be known before we can determine [H + ] and [OH − ]. Example \(\PageIndex{3}\) A 0.0788 M solution of HC 2 H 3 O 2 is 3.0% ionized into H + ions and C 2 H 3 O 2 − ions. What are [H + ] and [OH − ] for this solution? Solution Because the acid is only 3.0% ionized, we can determine [H + ] from the concentration of the acid. Recall that 3.0% is 0.030 in decimal form: [H + ] = 0.030 × 0.0788 = 0.00236 M With this [H + ], then [OH − ] can be calculated as follows: \[\left [ OH^{-} \right ]=\frac{1.0\times 10^{-14}}{0.00236}=4.2\times 10^{-12}M\nonumber \] This is about 30 times higher than would be expected for a strong acid of the same concentration. Exercise \(\PageIndex{3}\) A 0.0222 M solution of pyridine (C 5 H 5 N) is 0.44% ionized into pyridinium ions (C 5 H 5 NH + ) and OH − ions. What are [OH − ] and [H + ] for this solution? Answer [OH − ] = 9.77 × 10 −5 M; [H + ] = 1.02 × 10 −10 M Summary In any aqueous solution, the product of [H+] and [OH−] equals \(1.0 \times 10^{−14}\) (at room temperature).
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Courses/Saint_Francis_University/CHEM_113%3A_Human_Chemistry_I_(Zovinka)/03%3A_Ionic_Compounds/3.01%3A_Ions
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Learning Objectives Describe how an ion is formed. Distinguish the difference between the two types of ions. Ions As introduced previously, atoms contain a nucleus with neutrons and positively charged protons, surrounded by negatively charged electrons. In an atom, the total number of electrons, negative charge, equals the total number of protons, positive charge, and therefore, atoms are electrically neutral or uncharged. If an atom loses or gains electrons, it will become a positively or negatively charged particle, called an ion . The loss of one or more electrons results in more protons than electrons and an overall positively charged ion , called a cation . For example, a sodium atom with one less electron is a cation, Na + , with a +1 charge (Figure \(\PageIndex{1}\)). When an atom gains one or more electrons, it becomes a negatively charged a nion , because there are more electrons than protons. When chlorine gains one electron it forms a chloride ion, Cl – , with a –1 charge (Figures \(\PageIndex{2}\)) The names for positive and negative ions are pronounced CAT-eye-ons (cations) and ANN-eye-ons (anions), respectively. Note Naming Ions Cations are named using the element name plus " ion " to indicate it is charged. Anions are named by changing the element name ending to " ide ". For example, a magnesium ion is formed when neutral magnesium loses electrons and a fluor ide ion is formed when neutral fluorine gains electrons. Example \(\PageIndex{1}\) A calcium (Ca) atom loses two electrons and a sulfur (S) atom gains two electrons. Determine if the resulting ions are cations or anions? Write the ion symbols for each. Solution When calcium ( Z = 20) gains two electrons the resulting ion will have 18 electrons and 20 protons and therefore a charge of +2 (there are two more positive protons than negative electrons), it is a cation. The symbol for a calcium ion is Ca 2 + . When sulfur ( Z = 16) gains two electrons the resulting ion will have 18 electrons and 16 protons and therefore a charge of –2 (there are two more negative electrons than positive protons), it is an anion. The symbol for a sulf ide ion is S 2 – . Key Takeaways Ions are formed when atoms gain or lose electrons. Ions can be positively charged (cations) or negatively charged (anions). Contributors Lisa Sharpe Elles, University of Kansas
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Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Thermodynamics_and_Chemical_Equilibrium_(Ellgen)/20%3A_Boltzmann_Statistics/20.15%3A_Problems
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1. Three non-degenerate energy levels are available to a set of five distinguishable molecules, \(\{A,\ B,\ C,\ D,\ E\}\). The energies of these levels are \(1\), \(2\), and \(3\), in arbitrary units. Find all of the population sets that are possible in this system. For each population set, find the system energy, \(E\), and the number of microstates, \(W\). For each system energy, \(E\), list the associated population sets and the total number of microstates. How many population sets are there? What is \(W_{max}\)? If this system is isolated with \(E=10\), how many population sets are possible? What is \({\mathit{\Omega}}_E\) for \(E=10\)? 2. For the particle in a box, the allowed energies are proportional to the squares of the successive integers. What population sets are possible for the distinguishable molecules, \(\{A,\ B,\ C,\ D,\ E\}\), if they can occupy three quantum states whose energies are \(1\), \(4\), and \(9\)? For each population set, find the system energy, \(E\), and the number of microstates. For each system energy, \(E\), list the associated population sets and the total number of microstates. How many population sets are there? What is \(W_{max}\)? If this system is isolated with \(E=24\), how many population sets are possible? What is \({\mathit{\Omega}}_E\) for \(E=24\)? 3. Consider the results you obtained in problem 2. In general, when the allowed energies are proportional to the squares of successive integers, how many population sets do you think will be associated with each system energy? 4. (a) Compare \(W\) for the population set \(\{3,3,3\}\) to \(W\) for the population set \(\{2,5,2\}\). The energy levels are non-degenerate. (b) Consider an \(N\)-molecule system that has a finite number, \(M\), of quantum states. Show that \(W\) is (at least locally) a maximum when \(N_1=N_2=\dots =N_M={N}/{M}\). (Hint: Let \(U={N}/{M}\), and assume that \(N\) can be chosen so that \(U\) is an integer. Let \[W_U={N!}/{\left[U!U!\prod^{i=M-2}_{i=1}{U!}\right]} \nonumber \] and let \[W_O={N!}/{\left[\left(U+1\right)!\left(U-1\right)!\prod^{i=M-2}_{i=1}{U!}\right]} \nonumber \] Show that \({W_O}/{W_U}<1\).) 5. The energy levels available to isomer \(A\) are \({\epsilon }_0=1\), \({\epsilon }_2=2\), and \({\epsilon }_4=3\), in arbitrary units. The energy levels available to isomer B are \({\epsilon }_1=2\), \({\epsilon }_3=3\), and \({\epsilon }_5=4\). The energy levels are non-degenerate. (a) A system contains five molecules. The energy of the system is \(10\). List the population sets that are consistent with \(N=5\) and \(E=10\). Find \(W\) for each of these population sets. What are \(W^{max}_{A,B}\), \(W^{max}_A\), and \(W^{max}_B\)? What is the total number of microstates, \(=\mathit{\Omega}_{A,B}\), available to the system in all of the cases in which \(A\) and \(B\) molecules are present? What is the ratio \(\mathit{\Omega}_{A,B}/W^{max}_{A,B}\)? (b) Repeat this analysis for a system that contains six molecules and whose energy is \(12\). (c) Would the ratio \(\mathit{\Omega}_{A,B}/W^{max}_{A,B}\) be larger or smaller for a system with \(N=50\) and \(E=100\)? (d) What would happen to this ratio if the number of molecules became very large, while the average energy per molecule remained the same? 6. In Section 20.11 , we assume that all of the energy levels available to an isomeric pair of molecules have the same degeneracy. We then argue that the thermodynamic probabilities of a mixture of the isomers must be greater than the thermodynamic probability of either pure isomer: \(W^{max}_{A,B}>W^{max}_A\) and \(W^{max}_{A,B}>W^{max}_B\). Implicitly, we assume that many energy levels are multiply occupied: \(N_i>1\) for many energy levels \({\epsilon }_i\). Now consider the case that \(g_i>1\) for most \({\epsilon }_i\), but that nearly all energy levels are either unoccupied or contain only one molecule: \(N_i=0\) or \(N_i=1\). Show that under this assumption also, we must have \(W^{max}_{A,B}>W^{max}_A\) and \(W^{max}_{A,B}>W^{max}_B\). Notes \({}^{1}\)The statistical-mechanical procedures that have been developed for finding the energy levels available to a molecule express molecular energies as the difference between the molecule energy and the energy that its constituent atoms have when they are motionless. This is usually effected in two steps. The molecular energy levels are first expressed relative to the energy of the molecule’s own lowest energy state. The energy released when the molecules is formed in its lowest energy state from the isolated constituent atoms is then added. The energy of each level is then equal to the work done on the component atoms when they are brought together from infinite separation to form the molecule in that energy level. (Since energy is released in the formation of a stable molecule, the work done on the atoms and the energy of the resulting molecule are less than zero.) In our present discussion, we suppose that we can solve the Schrödinger equation to find the energies of the allowed quantum states. This corresponds to choosing the isolated constituent electrons and nuclei as the zero of energy for both isomers.
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Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/04%3A_Alkanes/4.05%3A_Halogenation_of_Alkanes._Energies_and_Rates_of_Reactions
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The economies of the highly industrialized nations of the world are based in large part on energy and chemicals produced from petroleum. Although the most important and versatile intermediates for conversion of petroleum to chemicals are compounds with double or triple bonds, it also is possible to prepare many valuable substances by substitution reactions of alkanes . In such substitutions, a hydrogen is removed from a carbon chain and another atom or group of atoms becomes attached in its place. A simple example of a substitution reaction is the formation of chloromethane and chlorine: \[ \ce{CH_4 + Cl_2 \rightarrow CH_3Cl + HCl}\] The equation for the reaction is simple, the ingredients are cheap, and the product is useful. However, if we want to decide in advance whether such a reaction is actually feasible, we have to know more. Particularly, we have to know whether the reaction proceeds in the direction it is written and, if so, whether conditions can be found under which it proceeds at a convenient rate. Obviously, if one were to mix methane and chlorine and find that, at most, only \(1 \%\) conversion to the desired product occurred and that the \(1 \%\) conversion could be achieved only after a day or so of strong heating, this reaction would be both too unfavorable and too slow for an industrial process. One way of visualizing the problems involved is with energy diagrams, which show the energy in terms of some arbitrary reaction coordinate that is a measure of progress between the initial and final states (Figure 4-4). Diagrams such as Figure 4-4 may not be familiar to you, and a mechanical analogy may be helpful to provide better understanding of the very important ideas involved. Consider a two-level box containing a number of tennis balls. An analog to an energetically favorable reaction would be to have all of the balls on the upper level where any disturbance would cause them to roll down to the lower level under the influence of gravity, thereby losing energy. If the upper level is modified and a low fence added to hold the balls in place, it will be just as energetically favorable as when the fence is not there for the balls to be at the lower level. The difference is that the process will not occur spontaneously without some major disturbance. We can say there is an energy barrier to occurrence of the favorable process. Now, if we shake the box hard enough, the balls on the upper level can acquire enough energy to bounce over the barrier and drop to the lower level. The balls then can be said to acquire enough activation energy to surmount the barrier. At the molecular level, the activation energy must be acquired either by collisions between molecules as the result of their thermal motions, or from some external agency, to permit the reactants to get over the barrier and be transformed into products. We shortly will discuss this more, but first we wish to illustrate another important concept with our mechanical analogy, that of equilibrium and equilibration . With gentle shaking of our two-level box, all of the balls on the upper level are expected to wind up on the lower level. There will not be enough activation to have them go from the lower to the upper level. In this circumstance, we can say that the balls are not equilibrated between the lower and upper levels. However, if we shake the box vigorously and continuously , no matter whether we start with all of the balls on the lower or upper level, an equilibrium will be set up with, on the average, most of the balls in the energetically more favorable lower level, but some in the upper level as well. To maintain a constant average fraction of the balls at each level with vigorous and continued shaking, the rate at which balls go from the upper to the lower level must be equal to the rate that they go in the opposite direction. The balls now will be equilibrated between the two levels. At equilibrium, the fraction of the balls on each of the two levels is wholly independent of the height of the barrier, just as long as the activation (shaking) is sufficient to permit the balls to go both ways. The diagrams of Figure 4-4 are to be interpreted in the same general way. If thermal agitation of the molecules is sufficient, then equilibrium can be expected to be established between the reactants and the products, whether the overall reaction is energetically favorable (left side of Figure 4-4) or energetically unfavorable (right side of Figure 4-4). But as with our analogy, when equilibrium is established we expect the major portion of the molecules to be in the more favorable energy state. What happens when methane is mixed with chlorine? No measurable reaction occurs when the gases are mixed and kept in the dark at room temperature. Clearly, either the reaction is energetically unfavorable or the energy barrier is high. The answer as to which becomes clear when the mixture is heated to temperatures in excess of \(300^\text{o}\) or when exposed to strong violet or ultraviolet light, whereby a rapid or even explosive reaction takes place. Therefore the reaction is energetically favorable, but the activation energy is greater than can be attained by thermal agitation alone at room temperature. Heat or light therefore must initiate a pathway for the reactants to be converted to products that has a low barrier or activation energy. Could we have predicted the results of this experiment ahead of time? First, we must recognize that there really are several questions here. Could we have decided whether the reaction was energetically favorable? That the dark reaction would be slow at room temperature? That light would cause the reaction to be fast? We consider these and some related questions in detail because they are important questions and the answers to them are relevant in one way or another to the study of all reactions in organic chemistry. The Question of the Equilibrium Constant Presumably, methane could react with chlorine to give chloromethane and hydrogen chloride, or chloromethane could react with hydrogen chloride to give methane and chlorine. If conditions were found for which both reactions proceeded at a finite rate, equilibrium finally would be established when the rates of the reactions in each direction became equal: \[ \ce{CH_4 + Cl_2 \rightleftharpoons CH_3Cl + HCl}\] At equilibrium, the relationship among the amounts of reactants and products is given by the equilibrium constant expression \[K_{eq} = \dfrac{[CH_3Cl][HCl]}{[CH_4][Cl_2]} \label{4-1}\] in which \(K_\text{eq}\) is the equilibrium constant. The quantities within the brackets of Equation \(\ref{4-1}\) denote either concentrations for liquid reactants or partial pressures for gaseous substances. If the equilibrium constant \(K_\text{eq}\) is greater than \(1\), then on mixing equal volumes of each of the participant substances (all are gases above \(-24^\text{o}\)), reaction to the right will be initially faster than reaction to the left, until equilibrium is established; at this point there will be more chloromethane and hydrogen chloride present than methane and chlorine. However, if the equilibrium constant were less than \(1\), the reaction initially would proceed faster to the left and, at equilibrium, there would be more methane and chlorine present than chloromethane and hydrogen chloride.\(^4\) For methane chlorination, we know from experiment that the reaction goes to the right and that \(K_\text{eq}\) is much greater than unity. Naturally, it would be helpful in planning other organic preparations to be able to estimate \(K_\text{eq}\) in advance. It is a common experience to associate chemical reactions with equilibrium constants greater than one with the evolution of heat, in other words, with negative \(\Delta H^\text{0}\) values. There are, in fact, many striking examples. Formation of chloromethane and hydrogen chloride from methane and chlorine has a \(K_\text{eq}\) of \(10^{18}\) and \(\Delta H^\text{0}\) of \(-24 \: \text{kcal}\) per mole of \(CH_3Cl\) formed at \(25^\text{o}\). Combustion of hydrogen with oxygen to give water has a \(K_\text{eq}\) of \(10^{40}\) and \(\Delta H^\text{0} = -57 \: \text{kcal}\) per mole of water formed at \(25^\text{o}\). However, this correlation between \(K_\text{eq}\) and \(\Delta H^\text{0}\) is neither universal nor rigorous. Reactions are known that absorb heat (are endothermic) and yet have \(K_\text{eq} > 1\). Other reactions have large \(\Delta H^\text{0}\) values and equilibrium constants much less than \(1\). The problem is that the energy change that correlates with \(K_\text{eq}\) is not \(\Delta H^\text{0}\) but \(\Delta G^\text{0}\) (the so-called change of " standard Gibbs energy ")\(^5\), and if we know \(\Delta G^\text{0}\), we can calculate \(K_\text{eq}\) by the equation \[ \Delta G^o =-2.303 RT \log_{10} K_{eq} \label{4-2}\] in which \(R\) is the gas constant and \(T\) is the absolute temperature in degrees Kelvin. For our calculations, we shall use \(R\) as \(1.987 \: \text{cal} \: \text{deg}^{-1} \: \text{mol}^{-1}\) and you should not forget to convert \(\Delta G^\text{0}\) to \(\text{cal}\). Tables of \(\Delta G^\text{0}\) values for formation of particular compounds (at various temperatures and states) from the elements are available in handbooks and the literature. With these, we can calculate equilibrium constants quite accurately. For example, handbooks give the following data, which are useful for methane chlorination: Combining these with proper regard for sign gives and \(\text{log} \: K_\text{eq} = -\left( -24.7 \times 1000 \right)/ \left(2.303 \times 1.987 \times 298.2 \right)\), so \(K_\text{eq} = 1.3 \times 10^{18}\). Unfortunately, insufficient \(\Delta G^\text{0}\) values for formation reactions are available to make this a widely applicable method for calculating \(K_\text{eq}\) values. The situation is not wholly hopeless, because there is a relationship between \(\Delta G^\text{0}\) and \(\Delta H^\text{0}\) that also involves \(T\) and another quantity, \(\Delta S^\text{0}\), the standard entropy change of the process: \[ \Delta G^o = \Delta H^o -T \Delta S^o \label{4-3}\] This equation shows that \(\Delta G^\text{0}\) and \(\Delta H^\text{0}\) are equal when \(\Delta S^\text{0}\) is zero. Therefore the sign and magnitude of \(T \Delta S^\text{0}\) determine how well \(K_\text{eq}\) correlates with \(\Delta H^\text{0}\). Now, we have to give attention to whether we can estimate \(T \Delta S^\text{0}\) values well enough to decide whether the \(\Delta H^\text{0}\) of a given reaction (calculated from bond energies or other information) will give a good or poor measure of \(\Delta G^\text{0}\). Entropy and Molecular Disorder To decide whether we need to worry about \(\Delta S^\text{0}\) with regard to any particular reaction, we have to have some idea what physical meaning entropy has. To be very detailed about this subject is beyond the scope of this book, but you should try to understand the physical basis of entropy, because if you do, then you will be able to predict at least qualitatively whether \(\Delta H^\text{0}\) will be about the same or very different from \(\Delta G^\text{0}\). Essentially, the entropy of a chemical system is a measure of its molecular disorder or randomness . Other things being the same, the more random the system is, the more favorable the system is. Different kinds of molecules have different degrees of translational, vibrational, and rotational freedom and, hence, different average degrees of molecular disorder or randomness. Now, if for a chemical reaction the degree of molecular disorder is different for the products than for the reactants, there will be a change in entropy and \(\Delta S^\text{0} \neq 0\). A spectacular example of the effect of molecular disorder in contributing to the difference between \(\Delta H^\text{0}\) and \(\Delta G^\text{0}\) is afforded by the formation of liquid nonane, \(C_9H_{20}\), from solid carbon and hydrogen gas at \(25^\text{o}\): \[\ce{9C(s) + 10H_2(g) \rightarrow C_910_{20}(l)}\] with \(\Delta H^o = -54.7 \, kcal\) and \(\Delta S^o = 5.0 \, kcal\). Equations \(\ref{4-2}\) and \(\ref{4-3}\) can be rearranged to calculate \(\Delta S^\text{0}\) and \(K_\text{eq}\) from \(\Delta H^\text{0}\) and \(\Delta G^\text{0}\): and \[K_{eq} = 10^{-\Delta G^o/2.303 \,RT} = 10^{-5.900/(2.303 \times 1.987 \times 298.2)} = 4.7 \times 10^{-5}\] These \(\Delta H^\text{0}\), \(\Delta S^\text{0}\), and \(K_\text{eq}\) values can be compared to those for \(H_2 + \frac{1}{2} O_2 \longrightarrow H_2O\), for which \(\Delta H^\text{0}\) is \(-57 \: \text{kcal}\), \(\Delta S^\text{0}\) is \(8.6 \: \text{e.u.}\), and \(K_\text{eq}\) is \(10^{40}\). Obviously, there is something unfavorable about the entropy change from carbon and hydrogen to nonane. The important thing is that there is a great difference in the constraints on the atoms on each side of the equation. In particular, hydrogen molecules in the gaseous state have great translational freedom and a high degree of disorder, the greater part of which is lost when the hydrogen atoms become attached to a chain of carbons. This makes for a large negative \(\Delta S^\text{0}\), which corresponds to a decrease in \(K_\text{eq}\). The differences in constraints of the carbons are less important. Solid carbon has an ordered, rigid structure with little freedom of motion of the individual carbon atoms. These carbons are less constrained in nonane, and this would tend to make \(\Delta S^\text{0}\) more positive and \(\Delta G^\text{0}\) more negative, corresponding to an increase in \(K_\text{eq}\) (Equations \(\ref{4-2}\) and \(\ref{4-3}\)). However, this is a small effect on \(\Delta S^\text{0}\) compared to the enormous difference in the degree of disorder of hydrogen between hydrogen gas and hydrogen bound to carbon in nonane. Negative entropy effects usually are observed in ring-closure reactions such as the formation of cyclohexane from 1-hexene, which occur with substantial loss of rotational freedom (disorder) about the \(C-C\) bonds: There is an even greater loss in entropy on forming cyclohexane from ethene because substantially more freedom is lost in orienting three ethene molecules to form a ring: For simple reactions, with the same number of molecules on each side of the equation, with no ring formation or other unusual changes in the constraints between the products and reactants, \(\Delta S^\text{0}\) usually is relatively small. In general, for such processes, we know from experience that \(K_\text{eq}\) usually is greater than 1 if \(\Delta H^\text{0}\) is more negative than \(-15 \: \text{kcal}\) and usually is less than 1 for \(\Delta H^\text{0}\) more positive than \(+15 \: \text{kcal}\). We can use this as a "rule of thumb" to predict whether \(K_\text{eq}\) should be greater or less than unity for vapor-phase reactions involving simple molecules. Some idea of the degree of success to be expected from this rule may be inferred from the examples in Table 4-5, which also contains a further comparison of some experimental \(\Delta H^\text{0}\) values with those calculated from bond energies. Suppose \(\Delta G^\text{0}\) is positive, what hope do we have of obtaining a useful conversion to a desired product? There is no simple straightforward and general answer to this question. When the reaction is reversible the classic procedure of removing one or more of the products to prevent equilibrium from being established has many applications in organic chemistry, as will be seen later. When this approach is inapplicable, a change in reagents is necessary. Thus, iodine does not give a useful conversion with 2,2-dimethylpropane, \(1\), to give 1-iodo-2,2-dimethylpropane, \(2\), because the position of equilibrium is too far to the left (\(K_\text{eq} \cong 10^{-5}\)): Alternative routes with favorable \(\Delta G^\text{0}\) values are required. Development of ways to make indirectly, by efficient processes, what cannot be made directly is one of the most interesting and challenging activities of organic chemists. Why Do Methane and Chlorine Fail to React in the Dark at \(25^\text{o}\)? To reach an understanding of why methane and chlorine do not react in the dark, we must consider the details of how the reaction occurs - that is, the reaction mechanism . The simplest mechanism would be for a chlorine molecule to collide with a methane molecule in such a way as to have chloromethane and hydrogen chloride formed directly as a result of a concerted breaking of the \(Cl-Cl\) and \(C-H\) bonds and making of the \(C-Cl\) and \(H-Cl\) bonds (see Figure 4-5). The failure to react indicates that there must be an energy barrier too high for this mechanism to operate. Why should this be so? First, this mechanism involves a very precisely oriented "four-center" collision between chlorine and methane that would have a low probability of occurrence (i.e., a large decrease in entropy because a precise orientation means high molecular ordering). Second, it requires pushing a chlorine molecule sufficiently deeply into a methane molecule so one of the chlorine atoms comes close enough to the carbon to form a bond and yield chloromethane. Generally, to bring nonbonded atoms to near-bonding distances (\(1.2 \: \text{A}\) to \(1.8 \: \text{A}\)) requires a large expenditure of energy, as can be seen in Figure 4-6. Interatomic repulsive forces increase rapidly at short distances, and pushing a chlorine molecule into a methane molecule to attain distances similar to the \(1.77\)-\(\text{A}\) carbon-chlorine bond distance in chloromethane would require a considerable amount of compression (see Figure 4-7). Valuable information about interatomic repulsions can be obtained with space-filling models of the CPK type ( Section 2-2 ), which have radii scaled to correspond to actual atomic interference radii, that is, the interatomic distance at the point where curves of the type of Figure 4-6 start to rise steeply. With such models, the degree of atomic compression required to bring the nonbonded atoms to within near-bonding distance is more evident than with the ball-and-stick models. It may be noted that four-center reactions of the type postulated in Figure 4-5 are encountered only rarely. If the concerted four-center mechanism for formation of chloromethane and hydrogen chloride from chlorine and methane is discarded, all the remaining possibilities are stepwise reaction mechanisms . A slow stepwise reaction is dynamically analogous to the flow of sand through a succession of funnels with different stem diameters. The funnel with the smallest stem will be the most important bottleneck and, if its stem diameter is much smaller than the others, it alone will determine the flow rate. Generally, a multistep chemical reaction will have a slow rate-determining step (analogous to the funnel with the small stem) and other relatively fast steps , which may occur either before or after the slow step. A possible set of steps for the chlorination of methane follows: Reactions (1) and (2) involve dissociation of chlorine into chlorine atoms and the breaking of a \(C-H\) bond of methane to give a methyl radical and a hydrogen atom. The methyl radical, like chlorine and hydrogen atoms, has one election not involved in bonding. Atoms and radicals usually are highly reactive, so formation of chloromethane and hydrogen chloride should proceed readily by Reactions (3) and (4). The crux then will be whether Steps (1) and (2) are reasonable under the reaction conditions. In the absence of some external stimulus , only collisions due to the usual thermal motions of the molecules can provide the energy needed to break the bonds. At temperatures below \(100^\text{o}\), it is very rare indeed that the thermal agitation alone can supply sufficient energy to break any significant number of bonds stronger than \(30\) to \(35 \: \text{kcal mol}^{-1}\). The \(Cl-Cl\) bond energy from Table 4-3 is \(58.1 \: \text{kcal}\), which is much too great to allow bond breaking from thermal agitation at \(25^\text{o}\) in accord with Reaction (1). For Reaction (2) it is not advisable to use the \(98.7 \: \text{kcal} \: C-H\) bond energy from Table 4-3 because this is one fourth of the energy required to break all four \(C-H\) bonds ( Section 4-3 ). More specific bond-dissociation energies are given in Table 4-5, and it will be seen that to break one \(C-H\) bond of methane requires \(104 \: \text{kcal}\) at \(25^\text{o}\), which again is too much to be gained by thermal agitation. Therefore we can conclude that Reactions (1)-(4) can not be an important mechanism for chlorination of methane at room temperature. One might ask whether dissociation into ions would provide viable mechanisms for methane chlorination. Part of the answer certainly is: Not in the vapor phase, as the following thermochemical data show: Ionic dissociation simply does not occur at ordinarily accessible temperatures by collisions between molecules in the vapor state. What is needed for formation of ions is either a highly energetic external stimulus, such as bombardment with fast-moving electrons, or an ionizing solvent that will assist ionization. Both of these processes will be discussed later. The point here is that ionic dissociation is not a viable step for the vapor-phase chlorination of methane. Why Does Light Induce the Chlorination of Methane? First, we should make clear that the light does more than provide energy merely to lift the molecules of methane and chlorine over the barrier of Figure 4-4. This is evident from the fact that very little light is needed, far less than one light photon per molecule of chloromethane produced. The light could activate either methane or chlorine, or both. However, methane is colorless and chlorine is yellow-green. This indicates that chlorine, not methane, interacts with visible light. A photon of near-ultraviolet light, such as is absorbed by chlorine gas, provides more than enough energy to split the molecule into two chlorine atoms: Once produced, a chlorine atom can remove a hydrogen atom from a methane molecule and form a methyl radical and a hydrogen chloride molecule. The bond-dissociation energies of \(CH_4\) (\(104 \: \text{kcal}\)) and \(HCl\) (\(103.1 \: \text{kcal}\)) suggest that this reaction is endothermic by about \(1 \: \text{kcal}\): Use of bond-dissociation energies gives a calculated \(\Delta H^\text{0}\) of \(-26 \: \text{kcal}\) for this reaction, which is certainly large enough, by our rule of thumb, to predict that \(K_\text{eq}\) will be greater than 1. Attack of a methyl radical on molecular chlorine is expected to require somewhat more oriented collision than for a chlorine atom reacting with methane (the chlorine molecule probably should be endwise, not sidewise, to the radical) but the interatomic repulsion probably should not be much different. The net result of \(CH_4 + Cl \cdot \longrightarrow CH_3 \cdot + HCl\) and \(CH_3 \cdot + Cl_2 \longrightarrow CH_3Cl + Cl \cdot\) is formation of chloromethane and hydrogen chloride from methane and chlorine. Notice that the chlorine atom consumed in the first step is replaced by another one in the second step. This kind of sequence of reactions is called a chain reaction because, in principle, one atom can induce the reaction of an infinite number of molecules through operation of a "chain" or cycle of reactions. In our example, chlorine atoms formed by the action of light on \(Cl_2\) can induce the chlorination of methane by the chain-propagating steps : In practice, chain reactions are limited by so-called termination processes. In our example, chlorine atoms or methyl radicals are destroyed by reacting with one another, as shown in the following equations: Chain reactions may be considered to involve three phases: First, chain initiation must occur, which for methane chlorination is activation and conversion of chlorine molecules to chlorine atoms by light. Second, chain-propagation steps convert reactants to products with no net consumption of atoms or radicals. The propagation reactions occur in competition with chain-terminating steps, which result in destruction of atoms or radicals. Putting everything together, we can write: The chain-termination reactions are expected to be exceedingly fast because atoms and radicals have electrons in unfilled shells that normally are bonding. As a result, bond formation can begin as soon as the atoms or radicals approach one another closely, without need for other bonds to begin to break. The evidence is strong that bond-forming reactions between atoms and radicals usually are diffusion-controlled , that there is almost no barrier or activation energy required, and the rates of combination are simply the rates at which encounters between radicals or atoms occur. If the rates of combination of radicals or atoms are so fast, you might well wonder how chain propagation ever could compete. Of course, competition will be possible if the propagation reactions themselves are fast, but another important consideration is the fact that the atom or radical concentrations are very low . Suppose that the concentration of \(Cl \cdot\) is \(10^{-11} \: \text{M}\) and the \(CH_4\) concentration \(1 \: \text{M}\). The probability of encounters between two \(Cl \cdot\) atoms will be proportional to \(10^{-11} \times 10^{-11}\), and between \(CH_4\) and \(Cl \cdot\) atoms it will be \(10^{-11} \times 1\). Thus, other things being the same, \(CH_4 + Cl \cdot \longrightarrow CH_3 \cdot + HCl\) (propagation) would be favored over \(2Cl \cdot \longrightarrow Cl_2\) (termination) by a factor of \(10^{11}\). Under favorable conditions, the methane-chlorination chain may go through 100 to 10,000 cycles before termination occurs by radical or atom combination. Consequently the efficiency (or quantum yield ) of the reaction is very high in terms of the amount of chlorination that occurs relative to the amount of the light absorbed. The overall rates of chain reactions usually are slowed very much by substances that can combine with atoms or radicals and convert them into species incapable of participating in the chain-propagation steps. Such substances are called radical traps , or inhibitors . Oxygen acts as an inhibitor in the chlorination of methane by rapidly combining with a methyl radical to form the comparatively stable (less reactive) peroxymethyl radical, \(CH_3OO \cdot\). This effectively terminates the chain: Can We Predict Whether Reactions Will Be Fast or Slow? To a considerable degree, we can predict relative reactivities, provided we use common sense to limit our efforts to reasonable situations. In the preceding section, we argued that reactions in which atoms or radicals combine can well be expected to be extremely fast because each entity has a potentially bonding electron in an outer unfilled shell, and bringing these together to form a bond does not require that other bonds be broken: The difference between the average energy of the reactants and the energy of the transition state is called the activation energy (Figure 4-4). We expect this energy to be smaller (lower barrier) if a weak bond is being broken and a strong bond is being made. The perceptive reader will notice that we are suggesting a parallel between reaction rate and \(\Delta H^\text{0}\) because \(\Delta H^\text{0}\) depends on the difference in strengths of the bonds being broken and formed. Yet previously ( Section 4-4A ), we pointed out that the energy barrier for a reaction need bear no relationship to how energetically feasible the reaction is, and this is indeed true for complex reactions involving many steps. But our intuitive parallel between rate and \(\Delta H^\text{0}\) usually works quite well for the rates of individual steps. This is borne out by experimental data on rates of removal of a hydrogen atom from methane by atoms or radicals (\(X \cdot\)), such as \(F \cdot\), \(Cl \cdot\), \(Br \cdot\), \(HO \cdot\), \(H_2N \cdot\), which generally parallel the strength of the new bond formed: Similarly, if we look at the \(H-C\) bond-dissociation energies of the hydrocarbons shown in Table 4-6, we would infer that \(Cl \cdot\) would remove a hydrogen most rapidly from the carbon forming the weakest \(C-H\) bond and, again, this is very much in accord with experience. For example, the chlorination of methylbenzene (toluene) in sunlight leads to the substitution of a methyl hydrogen rather than a ring hydrogen for the reason that the methyl \(C-H\) bonds are weaker and are attacked more rapidly than the ring \(C-H\) bonds. This can be seen explicitly in the \(\Delta H^\text{0}\) values for the chain-propagation steps calculated from the bond-dissociation energies of Table 4-6. Methyl substitution (observed): Ring substitution (not observed): The \(\Delta H^\text{0}\) of ring-hydrogen abstraction is unfavorable by \(+7 \: \text{kcal}\) because of the high \(C-H\) bond energy (\(110 \: \text{kcal}\)). Thus this step is not observed. It is too slow in comparison with the more favorable reaction at the methyl group even though the second propagation step is energetically favorable by \(-37 \: \text{kcal}\) and presumably would occur very rapidly. Use of bond-dissociation energies to predict relative reaction rates becomes much less valid when we try to compare different kinds of reactions. To illustrate, ethane might react with \(F \cdot\) to give fluoromethane or hydrogen fluoride: It is not a good idea to try to predict the relative rates of these two reactions on the basis of their overall \(\Delta H^\text{0}\) values because the nature of the bonds made and broken is too different. How Should We Go about Formulating a Reaction Mechanism? Faced with proposing a mechanism for a reaction that involves overall making or breaking of more than two bonds, the beginner almost invariably tries to concoct a process wherein, with a single step, all of the right bonds break and all of the right bonds form. Such mechanisms, called concerted mechanisms , have three disadvantages. First, they are almost impossible to prove correct. Second, prediction of the relative rates of reactions involving concerted mechanisms is especially difficult. Third, concerted mechanisms have a certain sterility in that one has no control over what happens while they are taking place, except an overall control of rate by regulating concentrations, temperature, pressure, choice of solvents, and so on. To illustrate, suppose that methane chlorination appeared to proceed by way of a one-step concerted mechanism: At the instant of reaction, the reactant molecules in effect would disappear into a dark closet and later emerge as product molecules. There is no way to prove experimentally that all of the bonds were made and formed simultaneously. All one could do would be to use the most searching possible tests to probe for the existence of discrete steps. If these tests fail, the reaction still would not be proved concerted because other, still more searching tests might be developed later that would give a different answer. The fact is, once you accept that a particular reaction is concerted, you, in effect, accept the proposition that further work on its mechanism is futile, no matter how important you might feel that other studies would be regarding the factors affecting the reaction rate. The experienced practitioner in reaction mechanisms accepts a concerted mechanism for a reaction involving the breaking and making of more than two bonds as a last resort. He first will try to analyze the overall transformation in terms of discrete steps that are individually simple enough surely to be concerted and that also involves energetically reasonable intermediates. Such an analysis of a reaction in terms of discrete mechanistic steps offers many possibilities for experimental studies, especially in development of procedures for detecting the existence, even if highly transitory, of the proposed intermediates. We shall give many examples of the fruitfulness of this kind of approach in subsequent discussions. \(^4\)If calculations based on chemical equilibrium constants are unfamiliar to you, we suggest you study one of the general chemistry texts listed for supplemental reading at the end of Chapter 1. \(^5\)Many books and references use \(\Delta F^\text{0}\) instead of \(\Delta G^\text{0}\). The difference between standard Gibbs energy \(\Delta G^\text{0}\) and the Gibbs energy \(\Delta G\) is that \(\Delta G^\text{0}\) is defined as the value of the free energy when all of the participants are in standard states. The free energy for \(\Delta G\) for a reaction \(\text{A} + \text{B} + \cdots \longrightarrow \text{X} + \text{Y} + \cdots\) is equal to \(\Delta G^\text{0} - 2.303 RT \: \text{log} \: \frac{\left[ \text{X} \right] \left[ \text{Y} \right] \cdots}{\left[ \text{A} \right] \left[ \text{B} \right] \cdots}\) where the products, \(\left[ \text{X} \right], \left[ \text{Y} \right] \cdots\), and the reactants, \(\left[ \text{A} \right], \left[ \text{B} \right] \cdots\), do not have to be in standard states. We shall use only \(\Delta G^\text{0}\) in this book. \(^6\)The entropy unit \(\text{e.u.}\) has the dimensions calorie per degree or \(\text{cal deg}^{-1}\).
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Courses/Saint_Marys_College_Notre_Dame_IN/CHEM_118_(Under_Construction)/CHEM_118_Textbook/10%3A_Proteins/10.5%3A_Enzyme_Activity
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Learning Objectives To describe how pH, temperature, and the concentration of an enzyme and its substrate influence enzyme activity. The single most important property of enzymes is the ability to increase the rates of reactions occurring in living organisms, a property known as catalytic activity . Because most enzymes are proteins, their activity is affected by factors that disrupt protein structure, as well as by factors that affect catalysts in general. Factors that disrupt protein structure include temperature and pH; factors that affect catalysts in general include reactant or substrate concentration and catalyst or enzyme concentration. The activity of an enzyme can be measured by monitoring either the rate at which a substrate disappears or the rate at which a product forms. Concentration of Substrate In the presence of a given amount of enzyme, the rate of an enzymatic reaction increases as the substrate concentration increases until a limiting rate is reached, after which further increase in the substrate concentration produces no significant change in the reaction rate (part (a) of Figure \(\PageIndex{1}\)). At this point, so much substrate is present that essentially all of the enzyme active sites have substrate bound to them. In other words, the enzyme molecules are saturated with substrate. The excess substrate molecules cannot react until the substrate already bound to the enzymes has reacted and been released (or been released without reacting). Let’s consider an analogy. Ten taxis (enzyme molecules) are waiting at a taxi stand to take people (substrate) on a 10-minute trip to a concert hall, one passenger at a time. If only 5 people are present at the stand, the rate of their arrival at the concert hall is 5 people in 10 minutes. If the number of people at the stand is increased to 10, the rate increases to 10 arrivals in 10 minutes. With 20 people at the stand, the rate would still be 10 arrivals in 10 minutes. The taxis have been “saturated.” If the taxis could carry 2 or 3 passengers each, the same principle would apply. The rate would simply be higher (20 or 30 people in 10 minutes) before it leveled off. Concentration of Enzyme When the concentration of the enzyme is significantly lower than the concentration of the substrate (as when the number of taxis is far lower than the number of waiting passengers), the rate of an enzyme-catalyzed reaction is directly dependent on the enzyme concentration (part (b) of Figure \(\PageIndex{1}\)). This is true for any catalyst; the reaction rate increases as the concentration of the catalyst is increased. Temperature A general rule of thumb for most chemical reactions is that a temperature rise of 10°C approximately doubles the reaction rate. To some extent, this rule holds for all enzymatic reactions. After a certain point, however, an increase in temperature causes a decrease in the reaction rate, due to denaturation of the protein structure and disruption of the active site (part (a) of Figure \(\PageIndex{2}\)). For many proteins, denaturation occurs between 45°C and 55°C. Furthermore, even though an enzyme may appear to have a maximum reaction rate between 40°C and 50°C, most biochemical reactions are carried out at lower temperatures because enzymes are not stable at these higher temperatures and will denature after a few minutes. At 0°C and 100°C, the rate of enzyme-catalyzed reactions is nearly zero. This fact has several practical applications. We sterilize objects by placing them in boiling water, which denatures the enzymes of any bacteria that may be in or on them. We preserve our food by refrigerating or freezing it, which slows enzyme activity. When animals go into hibernation in winter, their body temperature drops, decreasing the rates of their metabolic processes to levels that can be maintained by the amount of energy stored in the fat reserves in the animals’ tissues. Hydrogen Ion Concentration (pH) Because most enzymes are proteins, they are sensitive to changes in the hydrogen ion concentration or pH. Enzymes may be denatured by extreme levels of hydrogen ions (whether high or low); any change in pH, even a small one, alters the degree of ionization of an enzyme’s acidic and basic side groups and the substrate components as well. Ionizable side groups located in the active site must have a certain charge for the enzyme to bind its substrate. Neutralization of even one of these charges alters an enzyme’s catalytic activity. An enzyme exhibits maximum activity over the narrow pH range in which a molecule exists in its properly charged form. The median value of this pH range is called the optimum pH of the enzyme (part (b) of Figure \(\PageIndex{2}\)). With the notable exception of gastric juice (the fluids secreted in the stomach), most body fluids have pH values between 6 and 8. Not surprisingly, most enzymes exhibit optimal activity in this pH range. However, a few enzymes have optimum pH values outside this range. For example, the optimum pH for pepsin, an enzyme that is active in the stomach, is 2.0. Summary Initially, an increase in substrate concentration leads to an increase in the rate of an enzyme-catalyzed reaction. As the enzyme molecules become saturated with substrate, this increase in reaction rate levels off. The rate of an enzyme-catalyzed reaction increases with an increase in the concentration of an enzyme. At low temperatures, an increase in temperature increases the rate of an enzyme-catalyzed reaction. At higher temperatures, the protein is denatured, and the rate of the reaction dramatically decreases. An enzyme has an optimum pH range in which it exhibits maximum activity.
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Bookshelves/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/13%3A_Solutions/13.05%3A_Colligative_Properties_of_Solutions
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Learning Objectives To describe the relationship between solute concentration and the physical properties of a solution. To understand that the total number of nonvolatile solute particles determines the decrease in vapor pressure, increase in boiling point, and decrease in freezing point of a solution versus the pure solvent. Many of the physical properties of solutions differ significantly from those of the pure substances discussed in earlier chapters, and these differences have important consequences. For example, the limited temperature range of liquid water (0°C–100°C) severely limits its use. Aqueous solutions have both a lower freezing point and a higher boiling point than pure water. Probably one of the most familiar applications of this phenomenon is the addition of ethylene glycol (“antifreeze”) to the water in an automobile radiator. This solute lowers the freezing point of the water, preventing the engine from cracking in very cold weather from the expansion of pure water on freezing. Antifreeze also enables the cooling system to operate at temperatures greater than 100°C without generating enough pressure to explode. Changes in the freezing point and boiling point of a solution depend primarily on the number of solute particles present rather than the kind of particles. Such properties of solutions are called colligative properties (from the Latin colligatus, meaning “bound together” as in a quantity). As we will see, the vapor pressure and osmotic pressure of solutions are also colligative properties. Counting concentrations When we determine the number of particles in a solution, it is important to remember that not all solutions with the same molarity contain the same concentration of solute particles. Consider, for example, 0.01 M aqueous solutions of sucrose, \(\ce{NaCl}\), and \(\ce{CaCl_2}\). Because sucrose dissolves to give a solution of neutral molecules, the concentration of solute particles in a 0.01 M sucrose solution is 0.01 M. In contrast, both \(\ce{NaCl}\) and \(\ce{CaCl_2}\) are ionic compounds that dissociate in water to yield solvated ions. As a result, a 0.01 M aqueous solution of \(\ce{NaCl}\) contains 0.01 M \(\ce{Na^{+}}\) ions and 0.01 M \(\ce{Cl^{−}}\) ions, for a total particle concentration of 0.02 M. Similarly, the \(\ce{CaCl_2}\) solution contains 0.01 M \(\ce{Ca^{2+}}\) ions and 0.02 M \(\ce{Cl^{−}}\) ions, for a total particle concentration of 0.03 M. These values are correct for dilute solutions, where the dissociation of the compounds to form separately solvated ions is complete. At higher concentrations (typically >1 M), especially with salts of small, highly charged ions (such as \(\ce{Mg^{2+}}\) or \(\ce{Al^{3+}}\)), or in solutions with less polar solvents, dissociation to give separate ions is often incomplete. The sum of the concentrations of the dissolved solute particles dictates the physical properties of a solution. In the following discussion, we must therefore keep the chemical nature of the solute firmly in mind. Vapor Pressure of Solutions and Raoult’s Law Adding a nonvolatile solute, one whose vapor pressure is too low to measure readily, to a volatile solvent decreases the vapor pressure of the solvent. We can understand this phenomenon qualitatively by examining Figure \(\PageIndex{1}\), which is a schematic diagram of the surface of a solution of glucose in water. In an aqueous solution of glucose, a portion of the surface area is occupied by nonvolatile glucose molecules rather than by volatile water molecules. As a result, fewer water molecules can enter the vapor phase per unit time, even though the surface water molecules have the same kinetic energy distribution as they would in pure water. At the same time, the rate at which water molecules in the vapor phase collide with the surface and reenter the solution is unaffected. The net effect is to shift the dynamic equilibrium between water in the vapor and the liquid phases, decreasing the vapor pressure of the solution compared with the vapor pressure of the pure solvent. Figure \(\PageIndex{2}\) shows two beakers, one containing pure water and one containing an aqueous glucose solution, in a sealed chamber. We can view the system as having two competing equilibria: water vapor will condense in both beakers at the same rate, but water molecules will evaporate more slowly from the glucose solution because fewer water molecules are at the surface. Eventually all of the water will evaporate from the beaker containing the liquid with the higher vapor pressure (pure water) and condense in the beaker containing the liquid with the lower vapor pressure (the glucose solution). If the system consisted of only a beaker of water inside a sealed container, equilibrium between the liquid and vapor would be achieved rather rapidly, and the amount of liquid water in the beaker would remain constant. If the particles of a solute are essentially the same size as those of the solvent and both solute and solvent have roughly equal probabilities of being at the surface of the solution, then the effect of a solute on the vapor pressure of the solvent is proportional to the number of sites occupied by solute particles at the surface of the solution. Doubling the concentration of a given solute causes twice as many surface sites to be occupied by solute molecules, resulting in twice the decrease in vapor pressure. The relationship between solution composition and vapor pressure is therefore \[P_A=\chi_AP^0_A \label{13.5.1} \] where \(P_A\) is the vapor pressure of component A of the solution (in this case the solvent), \(\chi_A\) is the mole fraction of \(A\) in solution, and \(P^0_A\) is the vapor pressure of pure \(A\). Equation \(\ref{13.5.1}\) is known as Raoult’s law , after the French chemist who developed it. If the solution contains only a single nonvolatile solute (B), then \(\chi_A + \chi_B = 1\), and we can substitute \(\chi_A = 1 − \chi_B\) to obtain \[\begin{align} P_A &=(1−\chi_B)P^0_A \\[4pt] &=P^0_A−\chi_BP^0_A \label{13.5.2} \end{align} \] Rearranging and defining \(ΔP_A=P^0_A−P_A\), we obtain a relationship between the decrease in vapor pressure and the mole fraction of nonvolatile solute: \[ \begin{align} P^0_A−P_A &=ΔP_A \\[4pt] &=\chi_BP^0_A \label{13.5.3} \end{align} \] We can solve vapor pressure problems in either of two ways: by using Equation \(\ref{13.5.1}\) to calculate the actual vapor pressure above a solution of a nonvolatile solute, or by using Equation \(\ref{13.5.3}\) to calculate the decrease in vapor pressure caused by a specified amount of a nonvolatile solute. Example \(\PageIndex{1}\): Anti-Freeze Ethylene glycol (\(\ce{HOCH_2CH_2OH}\)), the major ingredient in commercial automotive antifreeze, increases the boiling point of radiator fluid by lowering its vapor pressure. At 100°C, the vapor pressure of pure water is 760 mmHg. Calculate the vapor pressure of an aqueous solution containing 30.2% ethylene glycol by mass, a concentration commonly used in climates that do not get extremely cold in winter. Given : identity of solute, percentage by mass, and vapor pressure of pure solvent Asked for : vapor pressure of solution Strategy : Calculate the number of moles of ethylene glycol in an arbitrary quantity of water, and then calculate the mole fraction of water. Use Raoult’s law to calculate the vapor pressure of the solution. Solution : A A 30.2% solution of ethylene glycol contains 302 g of ethylene glycol per kilogram of solution; the remainder (698 g) is water. To use Raoult’s law to calculate the vapor pressure of the solution, we must know the mole fraction of water. Thus we must first calculate the number of moles of both ethylene glycol (EG) and water present: \[moles \;EG=(302 \;\cancel{g}) \left( \dfrac{1\; mol}{62.07\; \cancel{g}} \right)=4.87\; mol\; EG \nonumber \] \[moles \; H_2O=(698 \;\cancel{g}) \left( \dfrac{1\; mol}{18.02\; \cancel{g}} \right)=38.7\; mol\; H_2O \nonumber \] The mole fraction of water is thus \[\chi_{H_2O}=\dfrac{38.7\; \cancel{mol} \; H_2O}{38.7\; \cancel{mol}\; H_2O +4.87 \cancel{mol}\; EG} =0.888 \nonumber \] B From Raoult’s law (Equation \(\ref{13.5.1}\)), the vapor pressure of the solution is \[ \begin{align*} P_{H_2O} &=(\chi_{H2_O})(P^0_{H_2O}) \\[4pt] &=(0.888)(760\; mmHg) =675 \;mmHg \end{align*} \nonumber \] Alternatively, we could solve this problem by calculating the mole fraction of ethylene glycol and then using Equation \(\ref{13.5.3}\) to calculate the resulting decrease in vapor pressure: \[\chi_{EG}=\dfrac{4.87\; mol\; EG}{4.87\; mol\; EG+38.7\; mol\; H_2O}=0.112 \nonumber \] \[ΔP_{H2_O}=(\chi_{EG})(P^0_{H_2O})=(0.112)(760\; mmHg)=85.1\; mmHg \nonumber \] \[P_{H_2O}=P^0_{H_2O}−ΔP_{H_2O}=760\; mmHg−85.1\; mmHg=675\; mmHg \nonumber \] The same result is obtained using either method. Exercise \(\PageIndex{1}\) Seawater is an approximately 3.0% aqueous solution of \(\ce{NaCl}\) by mass with about 0.5% of other salts by mass. Calculate the decrease in the vapor pressure of water at 25°C caused by this concentration of \(\ce{NaCl}\), remembering that 1 mol of \(\ce{NaCl}\) produces 2 mol of solute particles. The vapor pressure of pure water at 25°C is 23.8 mmHg. Answer 0.45 mmHg. This may seem like a small amount, but it constitutes about a 2% decrease in the vapor pressure of water and accounts in part for the higher humidity in the north-central United States near the Great Lakes, which are freshwater lakes. The decrease therefore has important implications for climate modeling. Even when a solute is volatile, meaning that it has a measurable vapor pressure, we can still use Raoult’s law. In this case, we calculate the vapor pressure of each component separately. The total vapor pressure of the solution (\(P_{tot}\)) is the sum of the vapor pressures of the components: \[P_{tot}=P_A+P_B=\chi_AP^0_A+\chi_BP^0_B \label{13.5.4} \] Because \(\chi_B = 1 − \chi_A\) for a two-component system, \[P_{tot}=\chi_AP^0_A+(1−\chi_A)P^0_B \label{13.5.5} \] Thus we need to specify the mole fraction of only one of the components in a two-component system. Consider, for example, the vapor pressure of solutions of benzene and toluene of various compositions. At 20°C, the vapor pressures of pure benzene and toluene are 74.7 and 22.3 mmHg, respectively. The vapor pressure of benzene in a benzene–toluene solution is \[P_{C_6H_6}=\chi_{C_6H_6}P^0_{C_6H_6} \label{13.5.6} \] and the vapor pressure of toluene in the solution is \[P{C_6H_5CH_3}=\chi_{C_6H_5CH_3}P^0_{C_6H_5CH3} \label{13.5.7} \] Equations \(\ref{13.5.6}\) and \(\ref{13.5.7}\) are both in the form of the equation for a straight line: \(y = mx + b\), where \(b = 0\). Plots of the vapor pressures of both components versus the mole fractions are therefore straight lines that pass through the origin, as shown in Figure \(\PageIndex{3}\). Furthermore, a plot of the total vapor pressure of the solution versus the mole fraction is a straight line that represents the sum of the vapor pressures of the pure components. Thus the vapor pressure of the solution is always greater than the vapor pressure of either component. A solution of two volatile components that behaves like the solution in Figure \(\PageIndex{3}\), which is defined as a solution that obeys Raoult’s law. Like an ideal gas, an ideal solution is a hypothetical system whose properties can be described in terms of a simple model. Mixtures of benzene and toluene approximate an ideal solution because the intermolecular forces in the two pure liquids are almost identical in both kind and magnitude. Consequently, the change in enthalpy on solution formation is essentially zero (\(ΔH_{soln} ≈ 0\)), which is one of the defining characteristics of an ideal solution. Ideal solutions and ideal gases are both simple models that ignore intermolecular interactions. Most real solutions, however, do not obey Raoult’s law precisely, just as most real gases do not obey the ideal gas law exactly. Real solutions generally deviate from Raoult’s law because the intermolecular interactions between the two components A and B differ. We can distinguish between two general kinds of behavior, depending on whether the intermolecular interactions between molecules A and B are stronger or weaker than the A–A and B–B interactions in the pure components. If the A–B interactions are stronger than the A–A and B–B interactions, each component of the solution exhibits a lower vapor pressure than expected for an ideal solution, as does the solution as a whole. The favorable A–B interactions effectively stabilize the solution compared with the vapor. This kind of behavior is called a negative deviation from Raoult’s law. Systems stabilized by hydrogen bonding between two molecules, such as acetone and ethanol, exhibit negative deviations from Raoult’s law. Conversely, if the A–B interactions are weaker than the A–A and B–B interactions yet the entropy increase is enough to allow the solution to form, both A and B have an increased tendency to escape from the solution into the vapor phase. The result is a higher vapor pressure than expected for an ideal solution, producing a positive deviation from Raoult’s law. In a solution of \(\ce{CCl_4}\) and methanol, for example, the nonpolar \(\ce{CCl_4}\) molecules interrupt the extensive hydrogen bonding network in methanol, and the lighter methanol molecules have weaker London dispersion forces than the heavier \(\ce{CCl_4}\) molecules. Consequently, solutions of \(\ce{CCl_4}\) and methanol exhibit positive deviations from Raoult’s law. Example \(\PageIndex{2}\) For each system, compare the intermolecular interactions in the pure liquids and in the solution to decide whether the vapor pressure will be greater than that predicted by Raoult’s law (positive deviation), approximately equal to that predicted by Raoult’s law (an ideal solution), or less than the pressure predicted by Raoult’s law (negative deviation). cyclohexane and ethanol methanol and acetone n-hexane and isooctane (2,2,4-trimethylpentane) Given : identity of pure liquids Asked for : predicted deviation from Raoult’s law (Equation \ref{13.5.1}) Strategy : Identify whether each liquid is polar or nonpolar, and then predict the type of intermolecular interactions that occur in solution. Solution : Liquid ethanol contains an extensive hydrogen bonding network, and cyclohexane is nonpolar. Because the cyclohexane molecules cannot interact favorably with the polar ethanol molecules, they will disrupt the hydrogen bonding. Hence the A–B interactions will be weaker than the A–A and B–B interactions, leading to a higher vapor pressure than predicted by Raoult’s law (a positive deviation). Methanol contains an extensive hydrogen bonding network, but in this case the polar acetone molecules create A–B interactions that are stronger than the A–A or B–B interactions, leading to a negative enthalpy of solution and a lower vapor pressure than predicted by Raoult’s law (a negative deviation). Hexane and isooctane are both nonpolar molecules (isooctane actually has a very small dipole moment, but it is so small that it can be ignored). Hence the predominant intermolecular forces in both liquids are London dispersion forces. We expect the A–B interactions to be comparable in strength to the A–A and B–B interactions, leading to a vapor pressure in good agreement with that predicted by Raoult’s law (an ideal solution). Exercise \(\PageIndex{2}\) For each system, compare the intermolecular interactions in the pure liquids with those in the solution to decide whether the vapor pressure will be greater than that predicted by Raoult’s law (positive deviation), approximately equal to that predicted by Raoult’s law (an ideal solution), or less than the pressure predicted by Raoult’s law (negative deviation): benzene and n-hexane ethylene glycol and \(\ce{CCl_4}\) acetic acid and n-propanol Answer a approximately equal Answer b positive deviation (vapor pressure greater than predicted) Answer c negative deviation (vapor pressure less than predicted) A Video Discussing Roult's Law. Video Link: Introduction to the Vapor Pressure of a Solution (Raoult's Law), YouTube(opens in new window) [youtu.be] A Video Discussing How to find the Vapor Pressure of a Solution. Video Link: Finding the Vapor Pressure of a Solution (Nonionic-Nonvolatile Solute), YouTube(opens in new window) [youtu.be] Boiling Point Elevation Recall that the normal boiling point of a substance is the temperature at which the vapor pressure equals 1 atm. If a nonvolatile solute lowers the vapor pressure of a solvent, it must also affect the boiling point. Because the vapor pressure of the solution at a given temperature is less than the vapor pressure of the pure solvent, achieving a vapor pressure of 1 atm for the solution requires a higher temperature than the normal boiling point of the solvent. Thus the boiling point of a solution is always greater than that of the pure solvent. We can see why this must be true by comparing the phase diagram for an aqueous solution with the phase diagram for pure water (Figure \(\PageIndex{4}\)). The vapor pressure of the solution is less than that of pure water at all temperatures. Consequently, the liquid–vapor curve for the solution crosses the horizontal line corresponding to P = 1 atm at a higher temperature than does the curve for pure water. Figure \(\PageIndex{4}\): Phase Diagrams of Pure Water and an Aqueous Solution of a Nonvolatile Solute. The vaporization curve for the solution lies below the curve for pure water at all temperatures, which results in an increase in the boiling point and a decrease in the freezing point of the solution. The boiling point of a solution with a nonvolatile solute is always greater than the boiling point of the pure solvent. The magnitude of the increase in the boiling point is related to the magnitude of the decrease in the vapor pressure. As we have just discussed, the decrease in the vapor pressure is proportional to the concentration of the solute in the solution. Hence the magnitude of the increase in the boiling point must also be proportional to the concentration of the solute (Figure \(\PageIndex{5}\)). We can define the boiling point elevation (\(ΔT_b\)) as the difference between the boiling points of the solution and the pure solvent: \[ΔT_b=T_b−T^0_b \label{13.5.8} \] where \(T_b\) is the boiling point of the solution and \(T^0_b\) is the boiling point of the pure solvent. We can express the relationship between \(ΔT_b\) and concentration as follows \[ΔT_b = mK_b \label{13.5.9} \] where m is the concentration of the solute expressed in molality, and \(K_b\) is the molal boiling point elevation constant of the solvent, which has units of °C/m. Table \(\PageIndex{1}\) lists characteristic \(K_b\) values for several commonly used solvents. For relatively dilute solutions, the magnitude of both properties is proportional to the solute concentration. Solvent Boiling Point (°C) Kb (°C/m) Freezing Point (°C) Kf (°C/m) acetic acid 117.90 3.22 16.64 3.63 benzene 80.09 2.64 5.49 5.07 d-(+)-camphor 207.40 4.91 178.8 37.80 carbon disulfide 46.20 2.42 −112.1 3.74 carbon tetrachloride 76.80 5.26 −22.62 31.40 chloroform 61.17 3.80 −63.41 4.60 nitrobenzene 210.80 5.24 5.70 6.87 water 100.00 0.51 0.00 1.86 The concentration of the solute is typically expressed as molality rather than mole fraction or molarity for two reasons. First, because the density of a solution changes with temperature, the value of molarity also varies with temperature. If the boiling point depends on the solute concentration, then by definition the system is not maintained at a constant temperature. Second, molality and mole fraction are proportional for relatively dilute solutions, but molality has a larger numerical value (a mole fraction can be only between zero and one). Using molality allows us to eliminate nonsignificant zeros. According to Table \(\PageIndex{1}\), the molal boiling point elevation constant for water is 0.51°C/m. Thus a 1.00 m aqueous solution of a nonvolatile molecular solute such as glucose or sucrose will have an increase in boiling point of 0.51°C, to give a boiling point of 100.51°C at 1.00 atm. The increase in the boiling point of a 1.00 m aqueous \(\ce{NaCl}\) solution will be approximately twice as large as that of the glucose or sucrose solution because 1 mol of \(\ce{NaCl}\) produces 2 mol of dissolved ions. Hence a 1.00 m \(\ce{NaCl}\) solution will have a boiling point of about 101.02°C. Example \(\PageIndex{3}\) In Example \(\PageIndex{1}\), we calculated that the vapor pressure of a 30.2% aqueous solution of ethylene glycol at 100°C is 85.1 mmHg less than the vapor pressure of pure water. We stated (without offering proof) that this should result in a higher boiling point for the solution compared with pure water. Now that we have seen why this assertion is correct, calculate the boiling point of the aqueous ethylene glycol solution. Given : composition of solution Asked for : boiling point Strategy : Calculate the molality of ethylene glycol in the 30.2% solution. Then use Equation \(\ref{13.5.9}\) to calculate the increase in boiling point. Solution : From Example \(\PageIndex{1}\), we know that a 30.2% solution of ethylene glycol in water contains 302 g of ethylene glycol (4.87 mol) per 698 g of water. The molality of the solution is thus \[\text{molality of ethylene glycol}= \left(\dfrac{4.87 \;mol}{698 \; \cancel{g} \;H_2O} \right) \left(\dfrac{1000\; \cancel{g}}{1 \;kg} \right)=6.98\, m \nonumber \] From Equation \(\ref{13.5.9}\), the increase in boiling point is therefore \[ΔT_b=m K_b=(6.98 \cancel{m})(0.51°C/\cancel{m})=3.6°C \nonumber \] The boiling point of the solution is thus predicted to be 104°C. With a solute concentration of almost 7 m, however, the assumption of a dilute solution used to obtain Equation \(\ref{13.5.9}\) may not be valid. Exercise \(\PageIndex{3}\) Assume that a tablespoon (5.00 g) of \(\ce{NaCl}\) is added to 2.00 L of water at 20.0°C, which is then brought to a boil to cook spaghetti. At what temperature will the water boil? Answer 100.04°C, or 100°C to three significant figures. (Recall that 1 mol of \(\ce{NaCl}\) produces 2 mol of dissolved particles. The small increase in temperature means that adding salt to the water used to cook pasta has essentially no effect on the cooking time.) A Video Discussing Boiling Point Elevation and Freezing Point Depression. Video Link: Boiling Point Elevation and Freezing Point Depression, YouTube(opens in new window) [youtu.be] (opens in new window) Freezing Point Depression The phase diagram in Figure \(\PageIndex{4}\) shows that dissolving a nonvolatile solute in water not only raises the boiling point of the water but also lowers its freezing point. The solid–liquid curve for the solution crosses the line corresponding to \(P = 1\,atm\) at a lower temperature than the curve for pure water. We can understand this result by imagining that we have a sample of water at the normal freezing point temperature, where there is a dynamic equilibrium between solid and liquid. Water molecules are continuously colliding with the ice surface and entering the solid phase at the same rate that water molecules are leaving the surface of the ice and entering the liquid phase. If we dissolve a nonvolatile solute such as glucose in the liquid, the dissolved glucose molecules will reduce the number of collisions per unit time between water molecules and the ice surface because some of the molecules colliding with the ice will be glucose. Glucose, though, has a very different structure than water, and it cannot fit into the ice lattice. Consequently, the presence of glucose molecules in the solution can only decrease the rate at which water molecules in the liquid collide with the ice surface and solidify. Meanwhile, the rate at which the water molecules leave the surface of the ice and enter the liquid phase is unchanged. The net effect is to cause the ice to melt. The only way to reestablish a dynamic equilibrium between solid and liquid water is to lower the temperature of the system, which decreases the rate at which water molecules leave the surface of the ice crystals until it equals the rate at which water molecules in the solution collide with the ice. By analogy to our treatment of boiling point elevation,the freezing point depression (\(ΔT_f\)) is defined as the difference between the freezing point of the pure solvent and the freezing point of the solution: \[ ΔT_f=T^0_f−T_f \label{13.5.10} \] where \(T^0_f\) is the freezing point of the pure solvent and \(T_f\) is the freezing point of the solution. The order of the terms is reversed compared with Equation \(\ref{13.5.8}\) to express the freezing point depression as a positive number. The relationship between \(ΔT_f\) and the solute concentration is given by an equation analogous to Equation \(\ref{13.5.9}\): \[ΔT_f = mK_f \label{13.5.11} \] where \(m\) is the molality of the solution and \(K_f\) is the molal freezing point depression constant for the solvent (in units of °C/m). Like \(K_b\), each solvent has a characteristic value of \(K_f\) (see Table \(\PageIndex{1}\)). Freezing point depression depends on the total number of dissolved nonvolatile solute particles, just as with boiling point elevation. Thus an aqueous \(\ce{NaCl}\) solution has twice as large a freezing point depression as a glucose solution of the same molality. People who live in cold climates use freezing point depression to their advantage in many ways. For example, salt is used to melt ice and snow on roads and sidewalks, ethylene glycol is added to engine coolant water to prevent an automobile engine from being destroyed, and methanol is added to windshield washer fluid to prevent the fluid from freezing. The decrease in vapor pressure, increase in boiling point, and decrease in freezing point of a solution versus a pure liquid all depend on the total number of dissolved nonvolatile solute particles. Example \(\PageIndex{4}\): Salting the Roads In colder regions of the United States, \(\ce{NaCl}\) or \(\ce{CaCl_2}\) is often sprinkled on icy roads in winter to melt the ice and make driving safer. Use the data in the Figure below to estimate the concentrations of two saturated solutions at 0°C, one of \(\ce{NaCl}\) and one of \(\ce{CaCl_2}\), and calculate the freezing points of both solutions to see which salt is likely to be more effective at melting ice. Given : solubilities of two compounds Asked for : concentrations and freezing points Strategy : Estimate the solubility of each salt in 100 g of water from the figure. Determine the number of moles of each in 100 g and calculate the molalities. Determine the concentrations of the dissolved salts in the solutions. Substitute these values into Equation \(\ref{13.5.11}\) to calculate the freezing point depressions of the solutions. Solution : A From Figure above, we can estimate the solubilities of \(\ce{NaCl}\) and \(\ce{CaCl_2}\) to be about 36 g and 60 g, respectively, per 100 g of water at 0°C. The corresponding concentrations in molality are \[m_{NaCl}=\left(\dfrac{36 \; \cancel{g \;NaCl}}{100 \;\cancel{g} \;H_2O}\right)\left(\dfrac{1\; mol\; NaCl}{58.44\; \cancel{ g\; NaCl}}\right)\left(\dfrac{1000\; \cancel{g}}{1\; kg}\right)=6.2\; m \nonumber \] \[m_{CaCl_2}=\left(\dfrac{60\; \cancel{g\; CaCl_2}}{100\;\cancel{g}\; H_2O}\right)\left(\dfrac{1\; mol\; CaCl_2}{110.98\; \cancel{g\; CaCl_2}}\right)\left(\dfrac{1000 \;\cancel{g}}{1 kg}\right)=5.4\; m \nonumber \] The lower formula mass of \(\ce{NaCl}\) more than compensates for its lower solubility, resulting in a saturated solution that has a slightly higher concentration than \(CaCl_2\). B Because these salts are ionic compounds that dissociate in water to yield two and three ions per formula unit of \(\ce{NaCl}\) and \(CaCl_2\), respectively, the actual concentrations of the dissolved species in the two saturated solutions are 2 × 6.2 m = 12 m for \(\ce{NaCl}\) and 3 × 5.4 m = 16 m for \(CaCl_2\). The resulting freezing point depressions can be calculated using Equation \(\ref{13.5.11}\): \[\ce{NaCl}: ΔT_f=mK_f=(12\; \cancel{m})(1.86°C/\cancel{m})=22°C \nonumber \] \[\ce{CaCl2}: ΔT_f=mK_f=(16\;\cancel{m})(1.86°C/\cancel{m})=30°C \nonumber \] Because the freezing point of pure water is 0°C, the actual freezing points of the solutions are −22°C and −30°C, respectively. Note that \(CaCl_2\) is substantially more effective at lowering the freezing point of water because its solutions contain three ions per formula unit. In fact, \(CaCl_2\) is the salt usually sold for home use, and it is also often used on highways. Because the solubilities of both salts decrease with decreasing temperature, the freezing point can be depressed by only a certain amount, regardless of how much salt is spread on an icy road. If the temperature is significantly below the minimum temperature at which one of these salts will cause ice to melt (say −35°C), there is no point in using salt until it gets warmer Exercise \(\PageIndex{4}\) Calculate the freezing point of the 30.2% solution of ethylene glycol in water whose vapor pressure and boiling point we calculated in Examples \(\PageIndex{5}\) and \(\PageIndex{5}\). Answer −13.0°C Example \(\PageIndex{5}\) Arrange these aqueous solutions in order of decreasing freezing points: 0.1 m \(KCl\), 0.1 m glucose, 0.1 m SrCl2, 0.1 m ethylene glycol, 0.1 m benzoic acid, and 0.1 m HCl. Given : molalities of six solutions Asked for: relative freezing points Strategy : Identify each solute as a strong, weak, or nonelectrolyte, and use this information to determine the number of solute particles produced. Multiply this number by the concentration of the solution to obtain the effective concentration of solute particles. The solution with the highest effective concentration of solute particles has the largest freezing point depression. Solution : A Because the molal concentrations of all six solutions are the same, we must focus on which of the substances are strong electrolytes, which are weak electrolytes, and which are nonelectrolytes to determine the actual numbers of particles in solution. \(\ce{KCl}\), \(\ce{SrCl_2}\), and \(\ce{HCl}\) are strong electrolytes , producing two, three, and two ions per formula unit, respectively. Benzoic acid is a weak electrolyte (approximately one particle per molecule), and glucose and ethylene glycol are both nonelectrolytes (one particle per molecule). B The molalities of the solutions in terms of the total particles of solute are: \(\ce{KCl}\) and \(\ce{HCl}\), 0.2 m; \(SrCl_2\), 0.3 m; glucose and ethylene glycol, 0.1 m; and benzoic acid, 0.1–0.2 m. Because the magnitude of the decrease in freezing point is proportional to the concentration of dissolved particles, the order of freezing points of the solutions is: glucose and ethylene glycol (highest freezing point, smallest freezing point depression) > benzoic acid > \(\ce{HCl}\) = \(\ce{KCl}\) > \(\ce{SrCl_2}\). Exercise \(\PageIndex{5}\) Arrange these aqueous solutions in order of increasing freezing points: 0.2 m \(\ce{NaCl}\), 0.3 m acetic acid, 0.1 m \(\ce{CaCl_2}\), and 0.2 m sucrose. Answer 0.2 m \(\ce{NaCl}\) (lowest freezing point) < 0.3 m acetic acid ≈ 0.1 m \(\ce{CaCl_2}\) < 0.2 m sucrose (highest freezing point) Colligative properties can also be used to determine the molar mass of an unknown compound. One method that can be carried out in the laboratory with minimal equipment is to measure the freezing point of a solution with a known mass of solute. This method is accurate for dilute solutions (≤1% by mass) because changes in the freezing point are usually large enough to measure accurately and precisely. By comparing \(K_b\) and \(K_f\) values in Table \(\PageIndex{1}\), we see that changes in the boiling point are smaller than changes in the freezing point for a given solvent. Boiling point elevations are thus more difficult to measure precisely. For this reason, freezing point depression is more commonly used to determine molar mass than is boiling point elevation. Because of its very large value of \(K_f\) (37.8°C/m), d-(+)-camphor (Table \(\PageIndex{1}\)) is often used to determine the molar mass of organic compounds by this method. Example \(\PageIndex{6}\): Sulfur A 7.08 g sample of elemental sulfur is dissolved in 75.0 g of \(CS_2\) to create a solution whose freezing point is −113.5°C. Use these data to calculate the molar mass of elemental sulfur and thus the formula of the dissolved \(\ce{S_n}\) molecules (i.e., what is the value of \(n\)?). Given : masses of solute and solvent and freezing point Asked for : molar mass and number of \(\ce{S}\) atoms per molecule Strategy : Use Equation \(\ref{13.5.10}\), the measured freezing point of the solution, and the freezing point of \(CS_2\) from Table \(\PageIndex{1}\) to calculate the freezing point depression. Then use Equation \(\ref{13.5.11}\) and the value of \(K_f\) from Table \(\PageIndex{1}\) to calculate the molality of the solution. From the calculated molality, determine the number of moles of solute present. Use the mass and number of moles of the solute to calculate the molar mass of sulfur in solution. Divide the result by the molar mass of atomic sulfur to obtain \(n\), the number of sulfur atoms per mole of dissolved sulfur. Solution : A The first step is to calculate the freezing point depression using Equation \(\ref{13.5.10}\): \[ΔT_f=T^0_f−T_f=−112.1°C−(−113.5°C)=1.4°C \nonumber \] Then Equation \(\ref{13.5.11}\) gives \[m=\dfrac{ΔT_f}{K_f}=\dfrac{1.4° \cancel{C}}{3.74° \cancel{C}/m}=0.37\;m \nonumber \] B The total number of moles of solute present in the solution is \[\text{moles solute}=\left(\dfrac{0.37 mol}{\cancel{kg}}\right) (75.0\; g) \left(\dfrac{1 kg}{1000\; g}\right)=0.028 \;mol \nonumber \] C We now know that 0.708 g of elemental sulfur corresponds to 0.028 mol of solute. The molar mass of dissolved sulfur is thus \[\text{molar mass}=\dfrac{7.08\; g}{0.028\; mol}=260\; g/mol \nonumber \] The molar mass of atomic sulfur is 32 g/mol, so there must be 260/32 = 8.1 sulfur atoms per mole, corresponding to a formula of \(\ce{S_8}\). Exercise \(\PageIndex{6}\) One of the byproducts formed during the synthesis of \(C_{60}\) is a deep red solid containing only carbon. A solution of 205 mg of this compound in 10.0 g of \(CCl_4\) has a freezing point of −23.38°C. What are the molar mass and most probable formula of the substance? Answer 847 g/mol; \(\ce{C_{70}}\) A Video Discussing how to find the Molecular Weight of an Unknown using Colligative Properties. Video Link: Finding the Molecular Weight of an Unknown using Colligative Properties, YouTube(opens in new window) [youtu.be] Osmotic Pressure Osmotic pressure is a colligative property of solutions that is observed using a semipermeable membrane, a barrier with pores small enough to allow solvent molecules to pass through but not solute molecules or ions. The net flow of solvent through a semipermeable membrane is called osmosis (from the Greek osmós, meaning “push”). The direction of net solvent flow is always from the side with the lower concentration of solute to the side with the higher concentration. Osmosis can be demonstrated using a U-tube like the one shown in Figure \(\PageIndex{6}\), which contains pure water in the left arm and a dilute aqueous solution of glucose in the right arm. A net flow of water through the membrane occurs until the levels in the arms eventually stop changing, which indicates that equilibrium has been reached. The osmotic pressure (\(\Pi\)) of the glucose solution is the difference in the pressure between the two sides, in this case the heights of the two columns. Although the semipermeable membrane allows water molecules to flow through in either direction, the rate of flow is not the same in both directions because the concentration of water is not the same in the two arms. The net flow of water through the membrane can be prevented by applying a pressure to the right arm that is equal to the osmotic pressure of the glucose solution. Just as with any other colligative property, the osmotic pressure of a solution depends on the concentration of dissolved solute particles. Osmotic pressure obeys a law that resembles the ideal gas equation: \[\Pi=\dfrac{nRT}{V}=MRT \label{13.5.12} \] where \(M\) is the number of moles of solute per unit volume of solution (i.e., the molarity of the solution), \(R\) is the ideal gas constant, and \(T\) is the absolute temperature. As shown in Example \(\PageIndex{7}\), osmotic pressures tend to be quite high, even for rather dilute solutions. Example \(\PageIndex{7}\) When placed in a concentrated salt solution, certain yeasts are able to produce high internal concentrations of glycerol to counteract the osmotic pressure of the surrounding medium. Suppose that the yeast cells are placed in an aqueous solution containing 4.0% \(\ce{NaCl}\) by mass; the solution density is 1.02 g/mL at 25°C. Calculate the osmotic pressure of a 4.0% aqueous \(\ce{NaCl}\) solution at 25°C. If the normal osmotic pressure inside a yeast cell is 7.3 atm, corresponding to a total concentration of dissolved particles of 0.30 M, what concentration of glycerol must the cells synthesize to exactly balance the external osmotic pressure at 25°C? Given : concentration, density, and temperature of \(\ce{NaCl}\) solution; internal osmotic pressure of cell Asked for : osmotic pressure of \(\ce{NaCl}\) solution and concentration of glycerol needed Strategy : Calculate the molarity of the \(\ce{NaCl}\) solution using the formula mass of the solute and the density of the solution. Then calculate the total concentration of dissolved particles. Use Equation \ref{13.5.12} to calculate the osmotic pressure of the solution. Subtract the normal osmotic pressure of the cells from the osmotic pressure of the salt solution to obtain the additional pressure needed to balance the two. Use Equation \ref{13.5.12} to calculate the molarity of glycerol needed to create this osmotic pressure. Solution : A The solution contains 4.0 g of \(\ce{NaCl}\) per 100 g of solution. Using the formula mass of \(\ce{NaCl}\) (58.44 g/mol) and the density of the solution (1.02 g/mL), we can calculate the molarity: \[ \begin{align*} M_{NaCl} &=\dfrac{moles\; NaCl}{\text{liter solution}} \\[4pt] &=\left(\dfrac{4.0 \; \cancel{g} \;NaCl}{58.44\; \cancel{g}/mol\; NaCl}\right)\left(\dfrac{1}{100\; \cancel{g \;solution}}\right)\left(\dfrac{1.02\; \cancel{g\; solution}}{1.00\; \cancel{mL}\; solution}\right)\left(\dfrac{1000\; \cancel{mL}}{1\; L}\right) \\[4pt] &= 0.70\; M\; \ce{NaCl} \end{align*} \nonumber \] Because 1 mol of \(\ce{NaCl}\) produces 2 mol of particles in solution, the total concentration of dissolved particles in the solution is (2)(0.70 M) = 1.4 M. B Now we can use Equation \ref{13.5.12} to calculate the osmotic pressure of the solution: \[ \begin{align*} \Pi &=MRT \\[4pt] &=(1.4 \;mol/L)\left[ 0.0821\; (L⋅atm)/(K⋅mol) \right ] (298\; K)\\[4pt] &=34 \;atm\end{align*} \nonumber \] C If the yeast cells are to exactly balance the external osmotic pressure, they must produce enough glycerol to give an additional internal pressure of (34 atm − 7.3 atm) = 27 atm. Glycerol is a nonelectrolyte, so we can solve Equation \ref{13.5.12} for the molarity corresponding to this osmotic pressure: \[ \begin{align*} M&=\dfrac{\Pi}{RT}\\[4pt] &=\dfrac{27\; \cancel{atm}}{[0.0821(L⋅\cancel{atm})/(\cancel{K}⋅mol)] (298 \;\cancel{K})}\\[4pt] &=1.1 \;M \;\text{glycerol} \end{align*} \nonumber \] In solving this problem, we could also have recognized that the only way the osmotic pressures can be the same inside the cells and in the solution is if the concentrations of dissolved particles are the same. We are given that the normal concentration of dissolved particles in the cells is 0.3 M, and we have calculated that the \(\ce{NaCl}\) solution is effectively 1.4 M in dissolved particles. The yeast cells must therefore synthesize enough glycerol to increase the internal concentration of dissolved particles from 0.3 M to 1.4 M—that is, an additional 1.1 M concentration of glycerol. Exercise \(\PageIndex{7}\) Assume that the fluids inside a sausage are approximately 0.80 M in dissolved particles due to the salt and sodium nitrite used to prepare them. Calculate the osmotic pressure inside the sausage at 100°C to learn why experienced cooks pierce the semipermeable skin of sausages before boiling them. Answer 24 atm Because of the large magnitude of osmotic pressures, osmosis is extraordinarily important in biochemistry, biology, and medicine. Virtually every barrier that separates an organism or cell from its environment acts like a semipermeable membrane, permitting the flow of water but not solutes. The same is true of the compartments inside an organism or cell. Some specialized barriers, such as those in your kidneys, are slightly more permeable and use a related process called dialysis, which permits both water and small molecules to pass through but not large molecules such as proteins. The same principle has long been used to preserve fruits and their essential vitamins over the long winter. High concentrations of sugar are used in jams and jellies not for sweetness alone but because they greatly increase the osmotic pressure. Thus any bacteria not killed in the cooking process are dehydrated, which keeps them from multiplying in an otherwise rich medium for bacterial growth. A similar process using salt prevents bacteria from growing in ham, bacon, salt pork, salt cod, and other preserved meats. The effect of osmotic pressure is dramatically illustrated in Figure \(\PageIndex{7}\), which shows what happens when red blood cells are placed in a solution whose osmotic pressure is much lower or much higher than the internal pressure of the cells. In addition to capillary action, trees use osmotic pressure to transport water and other nutrients from the roots to the upper branches. Evaporation of water from the leaves results in a local increase in the salt concentration, which generates an osmotic pressure that pulls water up the trunk of the tree to the leaves. Finally, a process called reverse osmosis can be used to produce pure water from seawater. As shown schematically in Figure \(\PageIndex{8}\), applying high pressure to seawater forces water molecules to flow through a semipermeable membrane that separates pure water from the solution, leaving the dissolved salt behind. Large-scale desalinization plants that can produce hundreds of thousands of gallons of freshwater per day are common in the desert lands of the Middle East, where they supply a large proportion of the freshwater needed by the population. Similar facilities are now being used to supply freshwater in southern California. Small, hand-operated reverse osmosis units can produce approximately 5 L of freshwater per hour, enough to keep 25 people alive, and are now standard equipment on US Navy lifeboats. A Video Discussing Osmotic Pressure. Video Link: Osmotic Pressure, YouTube(opens in new window) [youtu.be] (opens in new window) Colligative Properties of Electrolyte Solutions Thus far we have assumed that we could simply multiply the molar concentration of a solute by the number of ions per formula unit to obtain the actual concentration of dissolved particles in an electrolyte solution. We have used this simple model to predict such properties as freezing points, melting points, vapor pressure, and osmotic pressure. If this model were perfectly correct, we would expect the freezing point depression of a 0.10 m solution of sodium chloride, with 2 mol of ions per mole of \(\ce{NaCl}\) in solution, to be exactly twice that of a 0.10 m solution of glucose, with only 1 mol of molecules per mole of glucose in solution. In reality, this is not always the case. Instead, the observed change in freezing points for 0.10 m aqueous solutions of \(\ce{NaCl}\) and \(\ce{KCl}\) are significantly less than expected (−0.348°C and −0.344°C, respectively, rather than −0.372°C), which suggests that fewer particles than we expected are present in solution. The relationship between the actual number of moles of solute added to form a solution and the apparent number as determined by colligative properties is called the van’t Hoff factor (\(i\)) and is defined as follows: \[i=\dfrac{\text{apparent number of particles in solution}}{\text{ number of moles of solute dissolved}} \label{13.5.13} \] Named for Jacobus Hendricus van’t Hoff (1852–1911), a Dutch chemistry professor at the University of Amsterdam who won the first Nobel Prize in Chemistry (1901) for his work on thermodynamics and solutions. As the solute concentration increases, the van’t Hoff factor decreases. The van’t Hoff factor is therefore a measure of a deviation from ideal behavior . The lower the van’t Hoff factor, the greater the deviation. As the data in Table \(\PageIndex{2}\) show, the van’t Hoff factors for ionic compounds are somewhat lower than expected; that is, their solutions apparently contain fewer particles than predicted by the number of ions per formula unit. As the concentration of the solute increases, the van’t Hoff factor decreases because ionic compounds generally do not totally dissociate in aqueous solution. Compound i (measured) i (ideal) glucose 1.0 1.0 sucrose 1.0 1.0 \(NaCl\) 1.9 2.0 \(HCl\) 1.9 2.0 \(MgCl_2\) 2.7 3.0 \(FeCl_3\) 3.4 4.0 \(Ca(NO_3)_2\) 2.5 3.0 \(AlCl_3\) 3.2 4.0 \(MgSO_4\) 1.4 2.0 Instead, some of the ions exist as ion pairs, a cation and an anion that for a brief time are associated with each other without an intervening shell of water molecules (Figure \(\PageIndex{9}\)). Each of these temporary units behaves like a single dissolved particle until it dissociates. Highly charged ions such as \(Mg^{2+}\), \(Al^{3+}\), \(SO_4^{2−}\), and \(PO_4^{3−}\) have a greater tendency to form ion pairs because of their strong electrostatic interactions. The actual number of solvated ions present in a solution can be determined by measuring a colligative property at several solute concentrations. Example \(\PageIndex{8}\): Iron Chloride in Water A 0.0500 M aqueous solution of \(FeCl_3\) has an osmotic pressure of 4.15 atm at 25°C. Calculate the van’t Hoff factor \(i\) for the solution. Given : solute concentration, osmotic pressure, and temperature Asked for : van’t Hoff factor Strategy : Use Equation \(\ref{13.5.12}\) to calculate the expected osmotic pressure of the solution based on the effective concentration of dissolved particles in the solvent. Calculate the ratio of the observed osmotic pressure to the expected value. Multiply this number by the number of ions of solute per formula unit, and then use Equation \(\ref{13.5.13}\) to calculate the van’t Hoff factor. Solution : A If \(\ce{FeCl_3}\) dissociated completely in aqueous solution, it would produce four ions per formula unit [Fe3+(aq) plus 3Cl−(aq)] for an effective concentration of dissolved particles of 4 × 0.0500 M = 0.200 M. The osmotic pressure would be \[\begin{align*} \Pi &=MRT \\[4pt] &=(0.200 \;mol/L) \left[0.0821\;(L⋅atm)/(K⋅mol) \right] (298\; K)=4.89\; atm \end{align*} \nonumber \] B The observed osmotic pressure is only 4.15 atm, presumably due to ion pair formation. The ratio of the observed osmotic pressure to the calculated value is 4.15 atm/4.89 atm = 0.849, which indicates that the solution contains (0.849)(4) = 3.40 particles per mole of \(\ce{FeCl_3}\) dissolved. Alternatively, we can calculate the observed particle concentration from the osmotic pressure of 4.15 atm: \[4.15\; atm=M \left( \dfrac{0.0821 \;(L⋅atm)}{(K⋅mol)}\right) (298 \;K) \nonumber \] or after rearranging \[M = 0.170 mol \nonumber \] The ratio of this value to the expected value of 0.200 M is 0.170 M/0.200 M = 0.850, which again gives us (0.850)(4) = 3.40 particles per mole of \(\ce{FeCl_3}\) dissolved. From Equation \(\ref{13.5.13}\), the van’t Hoff factor for the solution is \[i=\dfrac{\text{3.40 particles observed}}{\text{1 formula unit}\; \ce{FeCl_3}}=3.40 \nonumber \] Exercise \(\PageIndex{8}\): Magnesium Chloride in Water Calculate the van’t Hoff factor for a 0.050 m aqueous solution of \(MgCl_2\) that has a measured freezing point of −0.25°C. Answer 2.7 (versus an ideal value of 3). A Video Discussing the Colligative Properties in Solutions. Video Link: Colligative Properties in Solutions, YouTube(opens in new window) [youtu.be] Summary The colligative properties of a solution depend on only the total number of dissolved particles in solution, not on their chemical identity. Colligative properties include vapor pressure, boiling point, freezing point, and osmotic pressure. The addition of a nonvolatile solute (one without a measurable vapor pressure) decreases the vapor pressure of the solvent. The vapor pressure of the solution is proportional to the mole fraction of solvent in the solution, a relationship known as Raoult’s law . Solutions that obey Raoult’s law are called ideal solutions. Most real solutions exhibit positive or negative deviations from Raoult’s law. The boiling point elevation (\(ΔT_b\)) and freezing point depression (\(ΔT_f\)) of a solution are defined as the differences between the boiling and freezing points, respectively, of the solution and the pure solvent. Both are proportional to the molality of the solute. When a solution and a pure solvent are separated by a semipermeable membrane, a barrier that allows solvent molecules but not solute molecules to pass through, the flow of solvent in opposing directions is unequal and produces an osmotic pressure, which is the difference in pressure between the two sides of the membrane. Osmosis is the net flow of solvent through such a membrane due to different solute concentrations. Dialysis uses a semipermeable membrane with pores that allow only small solute molecules and solvent molecules to pass through. In more concentrated solutions, or in solutions of salts with highly charged ions, the cations and anions can associate to form ion pairs, which decreases their effect on the colligative properties of the solution. The extent of ion pair formation is given by the van’t Hoff factor (i), the ratio of the apparent number of particles in solution to the number predicted by the stoichiometry of the salt. Henry’s law: \[C = kP \nonumber \] Raoult’s law : \[P_A=\chi_AP^0_A \nonumber \] vapor pressure lowering : \[P^0_A−P_A=ΔP_A=\chi_BP^0_A \nonumber \] vapor pressure of a system containing two volatile components : \[P_{tot}=\chi_AP^0_A+(1−\chi_A)P^0_B \nonumber \] boiling point elevation : \[ΔT_b = mK_b \nonumber \] freezing point depression: \[ΔT_f = mK_f \nonumber \] osmotic pressure : \[\Pi=nRTV=MRT \nonumber \] van ’t Hoff factor : \[i=\dfrac{\text{apparent number of particles in solution}}{\text{ number of moles of solute dissolved}} \nonumber \]
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Courses/Ursinus_College/CHEM322%3A_Inorganic_Chemistry/04%3A_Group_Theory/4.03%3A_Application_to_Vibrational_Spectroscopy/4.3.01%3A_Molecular_Vibrations
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Symmetry and group theory can be applied to understand molecular vibrations. This is particularly useful in the context of predicting the number of peaks expected in the infrared (IR) and Raman spectra of a given compound. We will use water as a case study to illustrate how group theory is used to predict the number of peaks in IR and Raman spectra. How many IR and Raman peaks would we expect for \(H_2O\)? To answer this question with group theory, a pre-requisite is that you assign the molecule's point group and assign an axis system to the entire molecule. By convention, the \(z\) axis is collinear with the principle axis, the \(x\) axis is in-plane with the molecule or the most number of atoms. It is a good idea to stick with this convention (see Figure \(\PageIndex{1}\)). What is the point group for \(H_2O\)? (click to see answer) \(H_2O\) has the following operations: \(E\), \(C_2\), \(\sigma_v\), \(\sigma_v'\). The point group is \(C_{2v}\). Now that we know the molecule's point group, we can use group theory to determine the symmetry of all motions in the molecule, or the symmetry of each of its degrees of freedom. Then we will subtract rotational and translational degrees of freedom to find the vibrational degrees of freedom. The number of degrees of freedom depends on the number of atoms (\(N\)) in a molecule. Each atom in the molecule can move in three dimensions (\(x,y,z\)), and so the number of degrees of freedom is three dimensions times \(N\) number of atoms, or \(3N\). The total degrees of freedom include a number of vibrations, three translations (in \(x\), \(y\), and \(z\)), and either two or three rotations. Linear molecules have two rotational degrees of freedom, while non-linear molecules have three. The vibrational modes are represented by the following expressions: \[\begin{array}{ccc} \text{Linear Molecule Degrees of Freedom} & = & 3N - 5 \\ \text{Non-Linear Molecule Degrees of Freedom} & = & 3N-6 \end{array} \nonumber \] Our goal is to find the symmetry of all degrees of freedom, and then determine which are vibrations that are IR- and Raman-active. STEP 1: Find the reducible representation for all normal modes \(\Gamma_{modes}\). The first major step is to find a reducible representation (\(\Gamma\)) for the movement of all atoms in the molecule (including rotational, translational, and vibrational degrees of freedom). We'll refer to this as \(\Gamma_{modes}\). To find normal modes using group theory, assign an axis system to each individual atom to represent the three dimensions in which each atom can move. Each axis on each atom should be consistent with the conventional axis system you previously assigned to the entire molecule (see Figure \(\PageIndex{1}\)). \(\Gamma_{modes}\) is the sum of the characters (trace) of the transformation matrix for the entire molecule (in the case of water, there are 9 degrees of freedom and this is now a 9x9 matrix). Let's walk through this step-by-step. The transformation matrix of \(E\) and \(C_2\) are shown below: \[E=\begin{pmatrix} \color{red}1&0&0&0&0&0&0&0&0 \\ 0&\color{red}1&0&0&0&0&0&0&0 \\ 0&0&\color{red}1&0&0&0&0&0&0 \\0&0&0&\color{red}1&0&0&0&0&0 \\ 0&0&0&0&\color{red}1&0&0&0&0 \\ 0&0&0&0&0&\color{red}1&0&0&0 \\ 0&0&0&0&0&0&\color{red}1&0&0 \\ 0&0&0&0&0&0&0&\color{red}1&0 \\ 0&0&0&0&0&0&0&0&\color{red}1 \\ \end{pmatrix} \begin{pmatrix} x_{oxygen} \\ y_{oxygen} \\ z_{oxygen} \\ x_{hydrogen-a} \\ y_{hydrogen-a} \\ z_{hydrogen-a} \\ x_{hydrogen-b} \\ y_{hydrogen-b} \\ z_{hydrogen-b} \end{pmatrix} = \begin{pmatrix} x'_{oxygen} \\ y'_{oxygen} \\ z'_{oxygen} \\ x'_{hydrogen-a} \\ y'_{hydrogen-a} \\ z'_{hydrogen-a} \\ x'_{hydrogen-b} \\ y'_{hydrogen-b} \\ z'_{hydrogen-b} \end{pmatrix}, \chi=9 \nonumber \] \[C_2=\begin{pmatrix} \color{red}-1&0&0&0&0&0&0&0&0 \\ 0&\color{red}-1&0&0&0&0&0&0&0 \\ 0&0&\color{red}1&0&0&0&0&0&0 \\0&0&0&\color{red}0&0&0&-1&0&0 \\ 0&0&0&0&\color{red}0&0&0&-1&0 \\ 0&0&0&0&0&\color{red}0&0&0&1 \\ 0&0&0&-1&0&0&\color{red}0&0&0 \\ 0&0&0&0&-1&0&0&\color{red}0&0 \\ 0&0&0&0&0&1&0&0&\color{red}0 \\ \end{pmatrix} \begin{pmatrix} x_{oxygen} \\ y_{oxygen} \\ z_{oxygen} \\ x_{hydrogen-a} \\ y_{hydrogen-a} \\ z_{hydrogen-a} \\ x_{hydrogen-b} \\ y_{hydrogen-b} \\ z_{hydrogen-b} \end{pmatrix} = \begin{pmatrix} x'_{oxygen} \\ y'_{oxygen} \\ z'_{oxygen} \\ x'_{hydrogen-a} \\ y'_{hydrogen-a} \\ z'_{hydrogen-a} \\ x'_{hydrogen-b} \\ y'_{hydrogen-b} \\ z'_{hydrogen-b} \end{pmatrix}, \chi=1 \nonumber \] It is unnecessary to find the transformation matrix for each operation since it is only the TRACE that gives us the character, and any off-diagonal entries do not contribute to \(\Gamma_{modes}\). The values that contribute to the trace can be found simply by performing each operation in the point group and assigning a value to each individual atom to represent how it is changed by that operation. If the atom moves away from itself, that atom gets a character of zero (this is because any non-zero characters of the transformation matrix are off of the diagonal). If the atom remains in place, each of its three dimensions is assigned a value of \(\cos \theta\) . For the example of \(H_2O\) under the \(C_{2v}\) point group, the axes that remain unchanged (\(\theta = 0^{\circ}\)) are assigned a value of \(\cos(0^{\circ})=1\), while those that are moved into the negative of themselves (rotated or reflected to \( \theta = 180^{\circ}\)) are assigned \(\cos(180^{\circ}) = -1\). The character for \(\Gamma\) is the sum of the values for each transformation. Let's walk through the steps to assign characters of \(\Gamma_{modes}\) for \(H_2O\) to illustrate how this works: For the operation \(E\) , performed on \(H_2O\), all three atoms remain in place. The three axes \(x,y,z\) on each atom remain unchanged. Thus, each of the three axes on each of three atom (nine axes) is assigned the value \(\cos(0^{\circ})=1\), resulting in a sum of \(\chi=9\) for the \(\Gamma_{modes}\). For the operation \(C_2\) , the two hydrogen atoms are moved away from their original position, and so the hydrogens are assigned a value of zero. The oxygen remains in place; the \(z\)-axis on oxygen is unchanged (\(\cos(0^{\circ})=1\)), while the \(x\) and \(y\) axes are inverted (\(\cos(180^{\circ})\)). The sum of these characters gives \(\chi=-1\) in the \(\Gamma_{modes}\). Now you try! Find the characters of \(\sigma_{v(xz)}\) and \(\sigma_{v(yz)}\) under the \(C_{2v}\) point group. Compare what you find to the \(\Gamma_{modes}\) for all normal modes given below. \[\begin{array}{l|llll} C_{2v} & E & C_2 & \sigma_v & \sigma_v' \\ \hline \Gamma_{modes} & 9 & -1 & 3 & 1 \end{array} \label{gammamodes} \] STEP 2: Break \(\Gamma_{modes}\) into its component irreducible representations. Now that we've found the \(\Gamma_{modes}\) (\(\ref{gammamodes}\)), we need to break it down into the individual irreducible representations (\(i,j,k...\)) for the point group. We can do this systematically using the following formula: \[\text{# of } i = \frac{1}{h}\sum(\text{# of operations in class)}\times(\chi_{\Gamma}) \times (\chi_i) \label{irs} \] In other words, the number of irreducible representations of type \(i\) is equal to the sum of the number of operations in the class \(\times\) the character of the \(\Gamma_{modes}\) \(\times\) the character of \(i\), and that sum is divided by the order of the group (\(h\)). Using equation \(\ref{irs}\), we find that for all normal modes of \(H_2O\): \[\Gamma_{modes}=3A_1+1A_2+3B_1+2B_2 \label{water} \]. Notice there are 9 irreducible representations in Equation \ref{water}. These irreducible representations represent the symmetries of all 9 motions of the molecule: vibrations, rotations, and translations. Exercise \(\PageIndex{1}\): Derive the irreducible representation in equation \(\ref{water}\). Derive the nine irreducible representations of \(\Gamma_{modes}\) for \(H_2O\), expression \(\ref{water}\). Hint To find the number of each kind of irreducible representation that combine to form the \(\Gamma_{modes}\), we need the characters of \(\Gamma_{modes}\) that we found above (\(\ref{gammamodes}\)), the \(C_{2v}\) character table (below), and equation \(\ref{irs}\). \(\begin{array}{l|llll|l|l} C_{2v} & {\color{red}1}E & {\color{red}1}C_2 & {\color{red}1}\sigma_v & {\color{red}1}\sigma_v' & \color{orange}h=4\\ \hline \color{green}A_1 & \color{green}1 & \color{green}1 & \color{green}1 & \color{green}1 & \color{green}z & \color{green}x^2,y^2,z^2\\ \color{green}A_2 & \color{green}1 & \color{green}1 & \color{green}-1 & \color{green}-1 & \color{green}R_z & \color{green}xy \\ \color{green}B_1 & \color{green}1 & \color{green}-1&\color{green}1&\color{green}-1 & \color{green}x,R_y & \color{green}xz \\ \color{green}B_2 & \color{green}1 & \color{green}-1 & \color{green}-1 & \color{green}1 & {\color{green}y} ,\color{green}R_x & \color{green}yz \end{array} \) In the \(C_{2v}\) point group, each class has only one operation, so the number of operations in each class (from equation \(\ref{irs}\)) is \({\color{red}1}\) for each class. This has been explicitly added to the character table above for emphasis. Answer The number of \(A_1\) = \(\frac{1}{\color{orange}4} \left[ ({\color{green}1} \times 9 \times {\color{red}1}) + ({\color{green}1} \times (-1) \times {\color{red}1}) + ({\color{green}1} \times 3 \times {\color{red}1}) + ({\color{green}1} \times 1 \times {\color{red}1})\right] = 3A_1 \) The number of \(A_2\) = \(\frac{1}{\color{orange}4} \left[ ({\color{green}1} \times 9 \times {\color{red}1}) + ({\color{green}1} \times (-1) \times {\color{red}1}) + ({\color{green}(-1)} \times 3 \times {\color{red}1}) + ({\color{green}(-1)} \times 1 \times {\color{red}1})\right] = 1A_2 \) The number of \(B_1\) = \(\frac{1}{\color{orange}4} \left[ ({\color{green}1} \times 9 \times {\color{red}1}) + ({\color{green}(-1)} \times (-1) \times {\color{red}1}) + ({\color{green}1} \times 3 \times {\color{red}1}) + ({\color{green}(-1)} \times 1 \times {\color{red}1})\right] = 3B_1 \) The number of \(B_2\) = \(\frac{1}{\color{orange}4} \left[ ({\color{green}1} \times 9 \times {\color{red}1}) + ({\color{green}(-1)} \times (-1) \times {\color{red}1}) + ({\color{green}(-1)} \times 3 \times {\color{red}1}) + ({\color{green}1} \times 1 \times {\color{red}1})\right] = 2B_2 \) STEP 3: Subtract rotations and translations to find vibrational modes. Because we are interested in molecular vibrations, we need to subtract the rotations and translations from the total degrees of freedom. \[\text{Vibrations } = \Gamma_{modes}-\text{ Rotations } - \text{ Translations } \nonumber \] In the example of \(H_2O\), the total degrees of freedom are given above in equation \(\ref{water}\), and therefore the vibrational degrees of freedom can be found by: \[H_2O\text{ vibrations} = \left(3A_1 + 1A_2 + 3B_1 + 2B_2\right) - \text{ Rotations } - \text{ Translations } \label{watervib} \] But which of the irreducible representations are ones that represent rotations and translations? The symmetry of rotational and translational degree modes can be found by inspecting the right-hand columns of any character table. Rotational modes correspond to irreducible representations that include \(R_x\), \(R_y\), and \(R_z\) in the table, while each of the three translational modes has the same symmetry as the \(x\), \(y\) and \(z\) axes. For a non-linear molecule, subtract three rotational irreducible representations and three translational irreducible representations from the total \(\Gamma_{modes}\). In the specific case of water, we refer to the \(C_{2v}\) character table: \[\begin{array}{l|llll|l|l} C_{2v} & E & C_2 & \sigma_v & \sigma_v' & h=4\\ \hline A_1 &1 & 1 & 1 & 1 & \color{red}z & x^2,y^2,z^2\\ A_2 & 1 & 1 & -1 & -1 & \color{red}R_z & xy \\ B_1 &1 & -1&1&-1 & \color{red}x,R_y &xz \\ B_2 & 1 & -1 &-1 & 1 & \color{red}y ,R_x & yz \end{array} \nonumber \] In \(C_{2v}\), translations correspond to \(B_1\), \(B_2\), and \(A_1\) (respectively for \(x,yz\)), and rotations correspond to \(B_2\), \(B_1\), and \(A_1\) (respectively for \(R_x,R_y,R_z\)). Subtracting these six irreducible representations from \(\Gamma_{modes}\) will leave us with the irreducible representations for vibrations. \[\begin{array}{lll} H_2O\text{ vibrations} &=& \Gamma_{modes} - \text{ Rotations } - \text{ Translations }\\ &=& \left(3A_1 + 1A_2 + 3B_1 + 2B_2\right) - (A_1 + B_1 + B_2) -(A_2 + B_1 + B_2) \\ &=& 2A_1 + 1B_1 \end{array} \nonumber \] The three vibrational modes for \(H_2O\) are \(2A_1 + 1B_1\). Note that we have the correct number of vibrational modes based on the expectation of \(3N-6\) vibrations for a non-linear molecule. STEP 4: Determine which of the vibrational modes are IR-active and Raman-active. The next step is to determine which of the vibrational modes is IR-active and Raman-active. To do this, we apply the IR and Raman Selection Rules below: IR and Raman Selection Rules Infrared selection rules: If a vibration results in the change in the molecular dipole moment, it is IR-active. In the character table, we can recognize the vibrational modes that are IR-active by those with symmetry of the \(x,y\), and \(z\) axes. In \(C_{2v}\), any vibrations with \(A_1\), \(B_1\) or \(B_2\) symmetry would be IR-active. Raman selection rules: If a vibration results in a change in the molecular polarizability, it will be Raman-active. In the character table, we can recognize the vibrational modes that are Raman-active by those with symmetry of any of the binary products (\(xy\), \(xz\), \(yz\), \(x^2\), \(y^2\), and \(z^2\)) or a linear combination of binary products (e.g. \(x^2-y^2\)). In \(C_{2v}\), any vibrations with \(A_1\), \(A_2\), \(B_1\) or \(B_2\) symmetry would be Raman-active. In our \(H_2O\) example, we found that of the three vibrational modes, two have \(A_1\) and one has \(B_1\) symmetry. Both \(A_1\) and \(B_1\) are IR-active, and both are also Raman-active. There are two possible IR peaks and three possible Raman peaks expected for water .* *It is important to note that this prediction tells only what is possible, but not what we might actually see in the IR and Raman spectra. For example, if the two IR peaks overlap, we might actually notice only one peak in the spectrum. Or, if one or more peaks is off-scale, we wouldn't see it in actual data. Group theory tells us what is possible and allows us to make predictions or interpretations of spectra. Summary of Analysis for Water Each molecular motion for water, or any molecule, can be assigned a symmetry under the molecule's point group. For water, we found that there are a total of 9 molecular motions; \(3A_1 + A_2 +3B_1 + 2B_2\). Six of these motions are not the translations and rotations. The remaining motions are vibrations; two with \(A_1\) symmetry and one with \(B_1\) symmetry. We can tell what these vibrations would look like based on their symmetries. The two \(A_1\) vibrations must be completely symmetric, while the \(B_1\) vibration is antisymmetric with respect to the principal \(C_2\) axis. All Motions (step 2 above) Translations (x,y,z) Rotations (\(R_x,R_y,R_z\)) Remaining Vibrations Description of Vibration \(3A_1\) \(1A_1\) NaN \(2A_1\) One is a symmetric stretch. The other is a symmetric bend. Both are IR-active and Raman-active \(A_2\) NaN \(1A_2\) NaN NaN \(3B_1\) \(1B_1\) \(1B_1\) \(1B_1\) Antisymmetric stretch that is IR-active and Raman-active. \(2B_2\) \(1B_2\) \(1B_2\) NaN NaN Exercise \(\PageIndex{2}\) Find the symmetries of all motions of the square planar complex, tetrachloroplatinate (II). Determine which are rotations, translations, and vibrations. Determine which vibrations are IR and Raman active. Answer The point group of \(\ce{[PtCl4]^2-}\) is \(D_{4h}\) ( refer to its character table ). There are five atoms and 15 vectors (\(x,y,z\) for each atom \(\times\) 5 atoms). STEP 1: The first major step is to find a reducible representation (\(\Gamma\)) for the movement of all atoms in the molecule. \[\begin{array}{l|rrrrrrrrrr} C_{2v} & E & 2C_4 & C_2 & 2C_2' & 2C_2" & i & 2S_4 & \sigma_h & 2\sigma_v & 2\sigma_d \\ \hline \Gamma_{modes} & 15 & 1 & -1 & -3 & -1 & -3 & -1 & 5 & 3 & 1 \end{array} \label{gammamodes2} \]. STEP 2: Break \(\Gamma_{modes}\) into its component irreducible representations. Following the process described earlier, we come to \(A_{1g} + A_{2g} + B_{1g} + B_{2g} + E_g + 2A_{2u} + B_{2u} + 3E_u\). This accounts for all modes of movement, including rotations and translations. STEP 3 : Subtract rotations and translations to find vibrational modes. The translations are \(A_{2u}+E_u\) and the rotations are \(A_{2g}+E_g\). The remaining normal modes are: \( A_{1g} + B_{1g} + B_{2g} + A_{2u} + B_{2u} + 2 E_u \) STEP 4: Determine which of the vibrational modes are IR-active and Raman-active: \(A_{2u} + E_u\) are IR-active. Since \(A_{2u}\) is singly degenerate and \(E_u\) is doubly degenerate, we expect three possible IR bands. \(A_{1g} + B_{1g} + B_{2g}\) are Raman-Active. Each of these is singly degenerate, so we expect three possible Raman bands. Selected Vibrational Modes The interpretation of CO stretching vibrations in an IR spectrum is particularly useful. Symmetry and group theory can be applied to predict the number of CO stretching bands that appear in a vibrational spectrum for a given metal coordination complex. A classic example of this application is in distinguishing isomers of metal-carbonyl complexes. For example, the cis- and trans- isomers of square planar metal dicarbonyl complexes (ML 2 (CO) 2 ) have a different number of IR stretches that can be predicted and interpreted using symmetry and group theory. Another example is the case of mer- and fac- isomers of octahedral metal tricarbonyl complexes (ML 3 (CO) 3 ). Structures of the two types of metal carbonyl configurations and their isomers are shown in Figure \(\PageIndex{1}\). The isomers in each case can be distinguished using vibrational spectroscopy. EXAMPLE 1: Distinguishing cis- and trans- isomers of square planar metal dicarbonyl complexes General structures of the cis- and trans- isomers of square planar metal dicarbonyl complexes (ML 2 (CO) 2 ) are shown in the left box in Figure \(\PageIndex{1}\). We can use symmetry and group theory to predict how many carbonyl stretches we should expect for each isomer following the steps below. Step 1: Assign the point group and Cartesian coordinates for each isomer. The cis- isomer has \(C_{2v}\) symmetry and the trans- isomer has \(D_{2h}\) symmetry. We assign the Cartesian coordinates so that \(z\) is colinear with the principle axis in each case. For the \(D_2{h}\) isomer, there are several orientations of the \(z\) axis possible. The axes shown in Figure \(\PageIndex{2}\) will be used here. Step 2: Produce a reducible representation (\(\Gamma\)) for CO stretches in each isomer First, assign a vector along each C—O bond in the molecule to represent the direction of C—O stretching motions, as shown in Figure \(\PageIndex{2}\) (red arrows → ). These vectors are used to produce a reducible representation (\(\Gamma\)) for the C—O stretching motions in each molecule. Using the symmetry operations under the appropriate character table, assign a value of 1 to each vector that remains in place during the operation, and a value of 0 if the vector moves out of place. There will be no occasion where a vector remains in place but is inverted, so a value of -1 will not occur. cis - ML 2 (CO) 2 : For cis - ML 2 (CO) 2 , the point group is \(C_{2v}\) and so we use the operations under the \(C_{2v}\) character table to create the \(\Gamma_{cis-CO}\). \[\begin{array}{|c|cccc|} \hline \bf{C_{2v}} & E & C_2 &\sigma_v (xz) & \sigma_v' (yz) \\ \hline \bf{\Gamma_{cis-CO}} & 2 & 0 & 2 & 0 \\ \hline \end{array} \nonumber \] trans - ML 2 (CO) 2 : For trans - ML 2 (CO) 2 , the point group is \(D_{2h}\) and so we use the operations under the \(D_{2h}\) character table to create the \(\Gamma_{trans-CO}\). \[\begin{array}{|c|cccccccc|} \hline \bf{D_{2h}} & E & C_2(z) & C_2(y) &C_2(x) & i &\sigma(xy) & \sigma(xz) & \sigma(yz) \\ \hline \bf{\Gamma_{trans-CO}} & 2 & 0 & 0 & 2 & 0 & 2 & 2 & 0\\ \hline \end{array} \nonumber \] Step 3: Break each \(\Gamma\) into its component irreducible representations Each \(\Gamma\) can be reduced using inspection or by the systematic method described previously . In the case of the cis - ML 2 (CO) 2 , the CO stretching vibrations are represented by \(A_1\) and \(B_1\) irreducible representations: \[\begin{array}{|c|cccc|cc|} \hline \bf{C_{2v}} & E & C_2 &\sigma_v (xz) & \sigma_v' (yz) \\ \hline \bf{\Gamma_{cis-CO}} & 2 & 0 & 2 & 0 & & \\ \hline A_1 & 1 & 1 & 1 & 1 & z & x^2, y^2, z^2 \\ B_1 & 1 & -1 & 1 & -1 & x, R_y & xz \\ \hline \end{array} \label{c2v} \] In the case of trans - ML 2 (CO) 2 , the CO stretching vibrations are represented by \(A_1\) and \(B_{3u}\) irreducible representations: \[\begin{array}{|c|cccccccc|cc|} \hline \bf{D_{2h}} & E & C_2(z) & C_2(y) &C_2(x) & i &\sigma(xy) & \sigma(xz) & \sigma(yz) \\ \hline \bf{\Gamma_{trans-CO}} & 2 & 0 & 0 & 2 & 0 & 2 & 2 & 0 & & \\ \hline A_{g} & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & & x^2, \; y^2, \; z^2\\ B_{3u} & 1 & -1 & -1 & 1 & -1 & 1 & 1 & -1 & x & \\ \hline \end{array} \nonumber \] These irreducible representations correspond to the symmetries of only the selected C—O vibrations. Since these motions are isolated to the C—O group, they do not include any rotations or translations of the entire molecule, and so we do not need to find and subtract rotationals or translations (unlike the previous cases where all motions were considered). Step 4: Determine which vibrational modes are IR-active and/or Raman-active Apply the infrared selection rules described previously to determine which of the CO vibrational motions are IR-active and Raman-active. The two isomers of ML 2 (CO) 2 are described below. In the case of the cis - ML 2 (CO) 2 , the CO stretching vibrations are represented by \(A_1\) and \(B_1\) irreducible representations. The characters of both representations and their functions are shown above, in \ref{c2v} (and can be found in the \(C_{2v}\) character table). Under \(C_{2v}\), both the \(A_1\) and \(B_1\) CO vibrational modes are IR-active and Raman-active. Therefore, two bands in the IR spectrum and two bands in the Raman spectrum are possible. In the case of the trans - ML 2 (CO) 2 , the CO stretching vibrations are represented by \(A_g\) and \(B_{3u}\) irreducible representations. The characters of both representations and their functions are shown above, in \ref{c2v} (and can be found in the \(D_{2h}\) character table). Under \(D_{2h}\), the \(A_g\) vibrational mode is is Raman-active only, while the \(B_{3u}\) vibrational mode is IR-active only. Therefore, only one IR band and one Raman band are possible for this isomer. Summary It is possible to distinguish between the two isomers of square planar ML 2 (CO) 2 using either IR or Raman vibrational spectroscopy. The cis - ML 2 (CO) 2 can produce two CO stretches in an IR or Raman spectrum, while the trans - ML 2 (CO) 2 isomer can produce only one band in either type of vibrational spectrum. If a sample of ML 2 (CO) 2 produced two CO stretching bands, we could rule out the possibility of a pure sample of trans- ML 2 (CO) 2 . Exercise \(\PageIndex{3}\) Repeat the steps outlined above to determine how many CO vibrations are possible for mer- ML 3 (CO) 3 and fac- ML 3 (CO) 3 isomers (see Figure \(\PageIndex{1}\)) in both IR and Raman spectra. Could either of these vibrational spectroscopies be used to distinguish the two isomers? Answer Step 1: Assign the point group and Cartesian coordinates for each isomer. The fac -isomer is \(C_{3v}\). The mer -isomer is \(C_{2v}\). Step 2: Produce a reducible representation for CO stretches in each isomer. Step 3: Break each into its component irreducible representations. Step 4: Determine which vibrational modes are IR-active and/or Raman-active. For fac- ML 3 (CO) 3 , the point group is \(C_{3v}\) and so we use the operations under the \(C_{3v}\) character table to create the \(\Gamma_{fac-CO}\). Then break it into its irreducible representations and determine which are IR and Raman active: \[\begin{array}{|c|ccc|} \hline \bf{C_{3v}} & E & 2C_2 &3\sigma_v \\ \hline \bf{\Gamma_{fac-CO}} & 3 & 0 & 1 \\ \hline \end{array} \nonumber \] This reduces to \(A_1 + E\). Both of these are IR active, and since one is singly degenerate while the other is doubly degenerate, we expect three possible IR bands from this isomer. Both vibrational modes are also Raman active, and again we would expect three possible bands in the Raman spectrum. For mer - ML 3 (CO) 3 , the point group is \(C_{2v}\) and so we use the operations under the \(C_{2v}\) character table to create the \(\Gamma_{mer-CO}\). Then break it into its irreducible representations and determine which are IR and Raman active: \[\begin{array}{|c|cccc|} \hline \bf{C_{2v}} & E & C_2 &\sigma_v (xz) & \sigma_v' (yz) \\ \hline \bf{\Gamma_{mer-CO}} & 3 & 1 & 3 & 1 \\ \hline \end{array} \nonumber \] This reduces to \(2A_1+B_1\); both \(A_1\) and \(B_1\) are IR and Raman active. So this isomer would have three possible IR bands and three possible Raman Bands. These two isomers have the same number of possible bands in both IR and Raman spectroscopy. It would not be straightforward to distinguish them from each other based on the number of possible bands in the vibrational spectrum.
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Courses/CSU_San_Bernardino/CHEM_2200%3A_General_Chemistry_II_(Mink)/22%3A_Appendices/22.14%3A_Answer_Key/22.14.18%3A_Chapter_18
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1. The alkali metals all have a single s electron in their outermost shell. In contrast, the alkaline earth metals have a completed s subshell in their outermost shell. In general, the alkali metals react faster and are more reactive than the corresponding alkaline earth metals in the same period. 3. 5. The possible ways of distinguishing between the two include infrared spectroscopy by comparison of known compounds, a flame test that gives the characteristic yellow color for sodium (strontium has a red flame), or comparison of their solubilities in water. At 20 °C, NaCl dissolves to the extent of compared with for SrCl 2 . Heating to 100 °C provides an easy test, since the solubility of NaCl is but that of SrCl 2 is Density determination on a solid is sometimes difficult, but there is enough difference (2.165 g/mL NaCl and 3.052 g/mL SrCl 2 ) that this method would be viable and perhaps the easiest and least expensive test to perform. 7. (a) (b) (c) (d) (e) 9. 11 lb 11. Yes, tin reacts with hydrochloric acid to produce hydrogen gas. 13. In PbCl 2 , the bonding is ionic, as indicated by its melting point of 501 °C. In PbCl 4 , the bonding is covalent, as evidenced by it being an unstable liquid at room temperature. 15. 17. Cathode (reduction): Anode (oxidation): Overall reaction: 19. 0.5035 g H 2 21. Despite its reactivity, magnesium can be used in construction even when the magnesium is going to come in contact with a flame because a protective oxide coating is formed, preventing gross oxidation. Only if the metal is finely subdivided or present in a thin sheet will a high-intensity flame cause its rapid burning. 23. Extract from ore: Recover: Sinter: Dissolve in Na 3 AlF 6 ( l ) and electrolyze: 25. 25.83% 27. 39 kg 29. (a) H 3 BPH 3 : (b) (c) BBr 3 : (d) B(CH 3 ) 3 : (e) B(OH) 3 : 31. 1 s 2 2 s 2 2 p 6 3 s 2 3 p 2 3 d 0 . 33. (a) (CH 3 ) 3 SiH: sp 3 bonding about Si; the structure is tetrahedral; (b) sp 3 bonding about Si; the structure is tetrahedral; (c) Si 2 H 6 : sp 3 bonding about each Si; the structure is linear along the Si-Si bond; (d) Si(OH) 4 : sp 3 bonding about Si; the structure is tetrahedral; (e) sp 3 d 2 bonding about Si; the structure is octahedral 35. (a) nonpolar; (b) nonpolar; (c) polar; (d) nonpolar; (e) polar 37. (a) tellurium dioxide or tellurium(IV) oxide; (b) antimony(III) sulfide; (c) germanium(IV) fluoride; (d) silane or silicon(IV) hydride; (e) germanium(IV) hydride 39. Boron has only s and p orbitals available, which can accommodate a maximum of four electron pairs. Unlike silicon, no d orbitals are available in boron. 41. (a) Δ H ° = 87 kJ; Δ G ° = 44 kJ; (b) Δ H ° = −109.9 kJ; Δ G ° = −154.7 kJ; (c) Δ H ° = −510 kJ; Δ G ° = −601.5 kJ 43. A mild solution of hydrofluoric acid would dissolve the silicate and would not harm the diamond. 45. In the N 2 molecule, the nitrogen atoms have an σ bond and two π bonds holding the two atoms together. The presence of three strong bonds makes N 2 a very stable molecule. Phosphorus is a third-period element, and as such, does not form π bonds efficiently; therefore, it must fulfill its bonding requirement by forming three σ bonds. 47. (a) H = 1+, C = 2+, and N = 3−; (b) O = 2+ and F = 1−; (c) As = 3+ and Cl = 1− 49. S < Cl < O < F 51. The electronegativity of the nonmetals is greater than that of hydrogen. Thus, the negative charge is better represented on the nonmetal, which has the greater tendency to attract electrons in the bond to itself. 53. Hydrogen has only one orbital with which to bond to other atoms. Consequently, only one two-electron bond can form. 55. 0.43 g H 2 57. (a) (b) (c) 59. (a) NH 2− : (b) N 2 F 4 : (c) (d) NF 3 : (e) 61. Ammonia acts as a Brønsted base because it readily accepts protons and as a Lewis base in that it has an electron pair to donate. Brønsted base: Lewis base: 63. (a) NO 2 : Nitrogen is sp 2 hybridized. The molecule has a bent geometry with an ONO bond angle of approximately 120°. (b) Nitrogen is sp 2 hybridized. The molecule has a bent geometry with an ONO bond angle slightly less than 120°. (c) Nitrogen is sp hybridized. The molecule has a linear geometry with an ONO bond angle of 180°. 65. Nitrogen cannot form a NF 5 molecule because it does not have d orbitals to bond with the additional two fluorine atoms. 67. (a) (b) (c) (d) (e) 69. (a) (b) (c) (d) or (e) or (f) 71. 291 mL 73. 28 tons 75. (a) (b) (c) (d) 77. (a) P = 3+; (b) P = 5+; (c) P = 3+; (d) P = 5+; (e) P = 3−; (f) P = 5+ 79. FrO 2 81. (a) (b) (c) (d) 83. 85. (a) (b) (c) (d) (e) 87. HClO 4 is the stronger acid because, in a series of oxyacids with similar formulas, the higher the electronegativity of the central atom, the stronger is the attraction of the central atom for the electrons of the oxygen(s). The stronger attraction of the oxygen electron results in a stronger attraction of oxygen for the electrons in the O-H bond, making the hydrogen more easily released. The weaker this bond, the stronger the acid. 89. As H 2 SO 4 and H 2 SeO 4 are both oxyacids and their central atoms both have the same oxidation number, the acid strength depends on the relative electronegativity of the central atom. As sulfur is more electronegative than selenium, H 2 SO 4 is the stronger acid. 91. SO 2 , sp 2 4+; SO 3 , sp 2 , 6+; H 2 SO 4 , sp 3 , 6+ 93. SF 6 : S = 6+; SO 2 F 2 : S = 6+; KHS: S = 2− 95. Sulfur is able to form double bonds only at high temperatures (substantially endothermic conditions), which is not the case for oxygen. 97. There are many possible answers including: and 99. 5.1 10 4 g 101. SnCl 4 is not a salt because it is covalently bonded. A salt must have ionic bonds. 103. In oxyacids with similar formulas, the acid strength increases as the electronegativity of the central atom increases. HClO 3 is stronger than HBrO 3 ; Cl is more electronegative than Br. 105. (a) (b) (c) (d) (e) 107. (a) bromine trifluoride; (b) sodium bromate; (c) phosphorus pentabromide; (d) sodium perchlorate; (e) potassium hypochlorite 109. (a) I: 7+; (b) I: 7+; (c) Cl: 4+; (d) I: 3+; Cl: 1−; (e) F: 0 111. (a) sp 3 d hybridized; (b) sp 3 d 2 hybridized; (c) sp 3 hybridized; (d) sp 3 hybridized; (e) sp 3 d 2 hybridized; 113. (a) nonpolar; (b) nonpolar; (c) polar; (d) nonpolar; (e) polar 115. The empirical formula is XeF 6 , and the balanced reactions are:
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Bookshelves/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/16%3A_The_Organic_Chemistry_of_Amino_Acids_Peptides_and_Proteins/16.08%3A_An_Introduction_to_Protein_Structure
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Secondary structure refers to the shape of a folding protein due exclusively to hydrogen bonding between its backbone amide and carbonyl groups. Secondary structure does not include bonding between the R-groups of amino acids, hydrophobic interactions, or other interactions associated with tertiary structure. The two most commonly encountered secondary structures of a polypeptide chain are α-helices and beta-pleated sheets. These structures are the first major steps in the folding of a polypeptide chain, and they establish important topological motifs that dictate subsequent tertiary structure and the ultimate function of the protein. Protein Folding Secondary Structure: α-Helices An α-helix is a right-handed coil of amino-acid residues on a polypeptide chain, typically ranging between 4 and 40 residues. This coil is held together by hydrogen bonds between the oxygen of C=O on top coil and the hydrogen of N-H on the bottom coil. Secondary Structure: β-Pleated Sheet This structure occurs when two (or more, e.g. ψ-loop) segments of a polypeptide chain overlap one another and form a row of hydrogen bonds with each other. This can happen in a parallel arrangement or in anti-parallel arrangement. Parallel and anti-parallel arrangement is the direct consequence of the directionality of the polypeptide chain. Secondary Structure: α-Pleated Sheet A similar structure to the beta-pleated sheet is the α-pleated sheet. This structure is energetically less favorable than the beta-pleated sheet, and is fairly uncommon in proteins. An α-pleated sheet is characterized by the alignment of its carbonyl and amino groups; the carbonyl groups are all aligned in one direction, while all the N-H groups are aligned in the opposite direction. The Structure of Proteins This page explains how amino acids combine to make proteins and what is meant by the primary, secondary and tertiary structures of proteins. Quaternary structure isn't covered. It only applies to proteins consisting of more than one polypeptide chain. Thumbnail: Structure of human hemoglobin. The proteins α and β subunits are in red and blue, and the iron-containing heme groups in green. (CC BY-SA 3.0; Zephyris ).
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