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Courses/Lumen_Learning/Book%3A_English_Composition_I-3_(Lumen)/14%3A_Writing_Process%3A_Prewriting/13.6%3A_Developing_Your_Topic
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The following video demonstrates the process of selecting and refining a research process. (Notice that this video also makes use of a metaphor to describe this process!) A YouTube element has been excluded from this version of the text. You can view it online here: http://pb.libretexts.org/ec1/?p=158 CC licensed content, Original Video: Developing Your Topic. Provided by : Lumen Learning. License : CC BY: Attribution All rights reserved content Research Tutorial: Developing Your Topic. Authored by : UNC Library. Located at : https://youtu.be/x4XZxIqSuyY . License : All Rights Reserved . License Terms : Standard YouTube License
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Bookshelves/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/04%3A_Reactions_in_Aqueous_Solution/4.02%3A_Solution_Concentrations
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Learning Objectives To describe the concentrations of solutions quantitatively All of us have a qualitative idea of what is meant by concentration . Anyone who has made instant coffee or lemonade knows that too much powder gives a strongly flavored, highly concentrated drink, whereas too little results in a dilute solution that may be hard to distinguish from water. In chemistry, the concentration . The quantity of solute that is dissolved in a particular quantity of solvent or solution. of a solution describes the quantity of a solute that is contained in a particular quantity of solvent or solution. Knowing the concentration of solutes is important in controlling the stoichiometry of reactants for reactions that occur in solution. Chemists use many different ways to define concentrations, some of which are described in this section. Molarity The most common unit of concentration is molarity , which is also the most useful for calculations involving the stoichiometry of reactions in solution. The molarity (M) is a common unit of concentration and is the number of moles of solute present in exactly \(1 L\) of solution \((mol/L)\) of a solution is the number of moles of solute present in exactly \(1 L\) of solution. Molarity is also the number of millimoles of solute present in exactly 1 mL of solution: \( molarity = \dfrac{moles\: of\: solute}{liters\: of\: solution} = \dfrac{mmoles\: of\: solute} {milliliters\: of\: solution} \tag{ \(\PageIndex{1}\) }\) The units of molarity are therefore moles per liter of solution (mol/L), abbreviated as \(M\). An aqueous solution that contains 1 mol (342 g) of sucrose in enough water to give a final volume of 1.00 L has a sucrose concentration of 1.00 mol/L or 1.00 M. In chemical notation, square brackets around the name or formula of the solute represent the concentration of a solute. So \([\rm{sucrose}] = 1.00\: M\) is read as “the concentration of sucrose is 1.00 molar.” The relationships between volume, molarity, and moles may be expressed as either \( V_L M_{mol/L} = \cancel{L} \left( \dfrac{mol}{\cancel{L}} \right) = moles \tag{ \(\PageIndex{2}\) }\) or \( V_{mL} M_{mmol/mL} = \cancel{mL} \left( \dfrac{mmol} {\cancel{mL}} \right) = mmoles \tag{ \(\PageIndex{3}\) }\) Example \(\PageIndex{1}\) illustrates the use of Equation \(\PageIndex{2}\) and Equation \(\PageIndex{3}\). Example \(\PageIndex{1}\) Calculate the number of moles of sodium hydroxide (NaOH) in 2.50 L of 0.100 M NaOH. Given: identity of solute and volume and molarity of solution Asked for: amount of solute in moles Strategy: Use either Equation \(\PageIndex{2}\) or Equation \(\PageIndex{3}\), depending on the units given in the problem. Solution Because we are given the volume of the solution in liters and are asked for the number of moles of substance, Equation \(\PageIndex{2}\) is more useful: \( moles\: NaOH = V_L M_{mol/L} = (2 .50\: \cancel{L} ) \left( \dfrac{0.100\: mol } {\cancel{L}} \right) = 0 .250\: mol\: NaOH \) Exercise \(\PageIndex{1}\) Calculate the number of millimoles of alanine, a biologically important molecule, in 27.2 mL of 1.53 M alanine. Answer 41.6 mmol Concentrations are often reported on a mass-to-mass (m/m) basis or on a mass-to-volume (m/v) basis, particularly in clinical laboratories and engineering applications. A concentration expressed on an m/m basis is equal to the number of grams of solute per gram of solution; a concentration on an m/v basis is the number of grams of solute per milliliter of solution. Each measurement can be expressed as a percentage by multiplying the ratio by 100; the result is reported as percent m/m or percent m/v. The concentrations of very dilute solutions are often expressed in parts per million ( ppm ), which is grams of solute per 10 6 g of solution, or in parts per billion ( ppb ), which is grams of solute per 10 9 g of solution. For aqueous solutions at 20°C, 1 ppm corresponds to 1 μg per milliliter, and 1 ppb corresponds to 1 ng per milliliter. These concentrations and their units are summarized in Table \(\PageIndex{1}\). Table \(\PageIndex{1}\) Common Units of Concentration Concentration Units m/m g of solute/g of solution m/v g of solute/mL of solution ppm g of solute/106 g of solution ppm μg/mL ppb g of solute/109 g of solution ppb ng/mL The Preparation of Solutions To prepare a solution that contains a specified concentration of a substance, it is necessary to dissolve the desired number of moles of solute in enough solvent to give the desired final volume of solution. Figure \(\PageIndex{1}\) illustrates this procedure for a solution of cobalt(II) chloride dihydrate in ethanol. Note that the volume of the solvent is not specified. Because the solute occupies space in the solution, the volume of the solvent needed is almost always less than the desired volume of solution. For example, if the desired volume were 1.00 L, it would be incorrect to add 1.00 L of water to 342 g of sucrose because that would produce more than 1.00 L of solution. As shown in Figure \(\PageIndex{2}\), for some substances this effect can be significant, especially for concentrated solutions. Figure \(\PageIndex{1}\) Preparation of a Solution of Known Concentration Using a Solid Solute Figure \(\PageIndex{2}\) Preparation of 250 mL of a Solution of (NH 4 ) 2 Cr 2 O 7 in Water The solute occupies space in the solution, so less than 250 mL of water are needed to make 250 mL of solution. Example \(\PageIndex{2}\) The solution in Figure \(\PageIndex{1}\) contains 10.0 g of cobalt(II) chloride dihydrate, CoCl 2 ·2H 2 O, in enough ethanol to make exactly 500 mL of solution. What is the molar concentration of CoCl 2 ·2H 2 O? Given: mass of solute and volume of solution Asked for: concentration (M) Strategy: To find the number of moles of CoCl 2 ·2H 2 O, divide the mass of the compound by its molar mass. Calculate the molarity of the solution by dividing the number of moles of solute by the volume of the solution in liters. Solution The molar mass of CoCl 2 ·2H 2 O is 165.87 g/mol. Therefore, \( moles\: CoCl_2 \cdot 2H_2O = \left( \dfrac{10.0 \: \cancel{g}} {165 .87\: \cancel{g} /mol} \right) = 0 .0603\: mol \) The volume of the solution in liters is \( volume = 500\: \cancel{mL} \left( \dfrac{1\: L} {1000\: \cancel{mL}} \right) = 0 .500\: L \) Molarity is the number of moles of solute per liter of solution, so the molarity of the solution is \( molarity = \dfrac{0.0603\: mol} {0.500\: L} = 0.121\: M = CoCl_2 \cdot H_2O \) Exercise \(\PageIndex{2}\) The solution shown in Figure \(\PageIndex{2}\) contains 90.0 g of (NH 4 ) 2 Cr 2 O 7 in enough water to give a final volume of exactly 250 mL. What is the molar concentration of ammonium dichromate? Answer \((NH_4)_2Cr_2O_7 = 1.43\: M\) To prepare a particular volume of a solution that contains a specified concentration of a solute, we first need to calculate the number of moles of solute in the desired volume of solution using the relationship shown in Equation \(\PageIndex{2}\). We then convert the number of moles of solute to the corresponding mass of solute needed. This procedure is illustrated in Example 4. Example \(\PageIndex{3}\) The so-called D5W solution used for the intravenous replacement of body fluids contains 0.310 M glucose. (D5W is an approximately 5% solution of dextrose [the medical name for glucose] in water.) Calculate the mass of glucose necessary to prepare a 500 mL pouch of D5W. Glucose has a molar mass of 180.16 g/mol. Given: molarity, volume, and molar mass of solute Asked for: mass of solute Strategy: A Calculate the number of moles of glucose contained in the specified volume of solution by multiplying the volume of the solution by its molarity. B Obtain the mass of glucose needed by multiplying the number of moles of the compound by its molar mass. Solution We must first calculate the number of moles of glucose contained in 500 mL of a 0.310 M solution: \( V_L M_{mol/L} = moles \) A \( 500\: \cancel{mL} \left( \dfrac{1\: \cancel{L}} {1000\: \cancel{mL}} \right) \left( \dfrac{0 .310\: mol\: glucose} {1\: \cancel{L}} \right) = 0 .155\: mol\: glucose \) B We then convert the number of moles of glucose to the required mass of glucose: \( mass \: of \: glucose = 0.155 \: \cancel{mol\: glucose} \left( \dfrac{180.16 \: g\: glucose} {1\: \cancel{mol\: glucose}} \right) = 27.9 \: g \: glucose \) Exercise \(\PageIndex{3}\) Another solution commonly used for intravenous injections is normal saline, a 0.16 M solution of sodium chloride in water. Calculate the mass of sodium chloride needed to prepare 250 mL of normal saline solution. Answer 2.3 g NaCl A solution of a desired concentration can also be prepared by diluting a small volume of a more concentrated solution with additional solvent. A stock solution is a commercially prepared solution of known concentration and is a commercially prepared solution of known concentration, is often used for this purpose. Diluting a stock solution is preferred because the alternative method, weighing out tiny amounts of solute, is difficult to carry out with a high degree of accuracy. Dilution is also used to prepare solutions from substances that are sold as concentrated aqueous solutions, such as strong acids. The procedure for preparing a solution of known concentration from a stock solution is shown in Figure \(\PageIndex{3}\). It requires calculating the number of moles of solute desired in the final volume of the more dilute solution and then calculating the volume of the stock solution that contains this amount of solute. Remember that diluting a given quantity of stock solution with solvent does not change the number of moles of solute present. The relationship between the volume and concentration of the stock solution and the volume and concentration of the desired diluted solution is therefore \((V_s)(M_s) = moles\: of\: solute = (V_d)(M_d)\tag{ \(\PageIndex{4}\) }\) where the subscripts s and d indicate the stock and dilute solutions, respectively. Example 5 demonstrates the calculations involved in diluting a concentrated stock solution. Figure \(\PageIndex{3}\) Preparation of a Solution of Known Concentration by Diluting a Stock Solution (a) A volume ( V s ) containing the desired moles of solute (M s ) is measured from a stock solution of known concentration. (b) The measured volume of stock solution is transferred to a second volumetric flask. (c) The measured volume in the second flask is then diluted with solvent up to the volumetric mark [( V s )(M s ) = ( V d )(M d )]. Example \(\PageIndex{4}\) What volume of a 3.00 M glucose stock solution is necessary to prepare 2500 mL of the D5W solution in Example 4? Given: volume and molarity of dilute solution Asked for: volume of stock solution Strategy: A Calculate the number of moles of glucose contained in the indicated volume of dilute solution by multiplying the volume of the solution by its molarity. B To determine the volume of stock solution needed, divide the number of moles of glucose by the molarity of the stock solution. Solution: A The D5W solution in Example 4 was 0.310 M glucose. We begin by using Equation \(\PageIndex{4}\) to calculate the number of moles of glucose contained in 2500 mL of the solution: \( moles\: glucose = 2500\: \cancel{mL} \left( \dfrac{1\: \cancel{L}} {1000\: \cancel{mL}} \right) \left( \dfrac{0 .310\: mol\: glucose} {1\: \cancel{L}} \right) = 0 .775\: mol\: glucose \) B We must now determine the volume of the 3.00 M stock solution that contains this amount of glucose: \( volume\: of\: stock\: soln = 0 .775\: \cancel{mol\: glucose} \left( \dfrac{1\: L} {3 .00\: \cancel{mol\: glucose}} \right) = 0 .258\: L\: or\: 258\: mL \) In determining the volume of stock solution that was needed, we had to divide the desired number of moles of glucose by the concentration of the stock solution to obtain the appropriate units. Also, the number of moles of solute in 258 mL of the stock solution is the same as the number of moles in 2500 mL of the more dilute solution; only the amount of solvent has changed . The answer we obtained makes sense: diluting the stock solution about tenfold increases its volume by about a factor of 10 (258 mL → 2500 mL). Consequently, the concentration of the solute must decrease by about a factor of 10, as it does (3.00 M → 0.310 M). We could also have solved this problem in a single step by solving Equation \(\PageIndex{4}\) for V s and substituting the appropriate values: \( V_s = \dfrac{( V_d )(M_d )}{M_s} = \dfrac{(2 .500\: L)(0 .310\: \cancel{M} )} {3 .00\: \cancel{M}} = 0 .258\: L \) As we have noted, there is often more than one correct way to solve a problem. Exercise \(\PageIndex{4}\) What volume of a 5.0 M NaCl stock solution is necessary to prepare 500 mL of normal saline solution (0.16 M NaCl)? Answer 16 mL Ion Concentrations in Solution In Example 3, you calculated that the concentration of a solution containing 90.00 g of ammonium dichromate in a final volume of 250 mL is 1.43 M. Let’s consider in more detail exactly what that means. Ammonium dichromate is an ionic compound that contains two NH 4 + ions and one Cr 2 O 7 2− ion per formula unit. Like other ionic compounds, it is a strong electrolyte that dissociates in aqueous solution to give hydrated NH 4 + and Cr 2 O 7 2− ions: \( (NH_4 )_2 Cr_2 O_7 (s) \xrightarrow {H_2 O(l)} 2NH_4^+ (aq) + Cr_2 O_7^{2-} (aq)\tag{8.2.5} \) Thus 1 mol of ammonium dichromate formula units dissolves in water to produce 1 mol of Cr 2 O 7 2− anions and 2 mol of NH 4 + cations (see Figure \(\PageIndex{4}\) ). Figure \(\PageIndex{4}\) Dissolution of 1 mol of an Ionic Compound In this case, dissolving 1 mol of (NH 4 ) 2 Cr 2 O 7 produces a solution that contains 1 mol of Cr 2 O 7 2− ions and 2 mol of NH 4 + ions. (Water molecules are omitted from a molecular view of the solution for clarity.) When we carry out a chemical reaction using a solution of a salt such as ammonium dichromate, we need to know the concentration of each ion present in the solution. If a solution contains 1.43 M (NH 4 ) 2 Cr 2 O 7 , then the concentration of Cr 2 O 7 2− must also be 1.43 M because there is one Cr 2 O 7 2− ion per formula unit. However, there are two NH 4 + ions per formula unit, so the concentration of NH 4 + ions is 2 × 1.43 M = 2.86 M. Because each formula unit of (NH 4 ) 2 Cr 2 O 7 produces three ions when dissolved in water (2NH 4 + + 1Cr 2 O 7 2− ), the total concentration of ions in the solution is 3 × 1.43 M = 4.29 M. Example \(\PageIndex{5}\) What are the concentrations of all species derived from the solutes in these aqueous solutions? 0.21 M NaOH 3.7 M (CH 3 )CHOH 0.032 M In(NO 3 ) 3 Given: molarity Asked for: concentrations Strategy: A Classify each compound as either a strong electrolyte or a nonelectrolyte. B If the compound is a nonelectrolyte, its concentration is the same as the molarity of the solution. If the compound is a strong electrolyte, determine the number of each ion contained in one formula unit. Find the concentration of each species by multiplying the number of each ion by the molarity of the solution. Solution: Sodium hydroxide is an ionic compound that is a strong electrolyte (and a strong base) in aqueous solution: \( NaOH(s) \xrightarrow {H_2 O(l)} Na^+ (aq) + OH^- (aq) \) B Because each formula unit of NaOH produces one Na + ion and one OH − ion, the concentration of each ion is the same as the concentration of NaOH: [Na + ] = 0.21 M and [OH − ] = 0.21 M. A The formula (CH 3 ) 2 CHOH represents 2-propanol (isopropyl alcohol) and contains the –OH group, so it is an alcohol. Recall from Section 8.1 that alcohols are covalent compounds that dissolve in water to give solutions of neutral molecules. Thus alcohols are nonelectrolytes. B The only solute species in solution is therefore (CH 3 ) 2 CHOH molecules, so [(CH 3 ) 2 CHOH] = 3.7 M. A Indium nitrate is an ionic compound that contains In 3 + ions and NO 3 − ions, so we expect it to behave like a strong electrolyte in aqueous solution: \( In(NO _3 ) _3 (s) \xrightarrow {H_ 2 O(l)} In ^{3+} (aq) + 3NO _3^- (aq) \) B One formula unit of In(NO 3 ) 3 produces one In 3 + ion and three NO 3 − ions, so a 0.032 M In(NO 3 ) 3 solution contains 0.032 M In 3 + and 3 × 0.032 M = 0.096 M NO 3 – —that is, [In 3 + ] = 0.032 M and [NO 3 − ] = 0.096 M. Exercise \(\PageIndex{5}\) What are the concentrations of all species derived from the solutes in these aqueous solutions? Answer 0.0012 M Ba(OH) 2 0.17 M Na 2 SO 4 0.50 M (CH 3 ) 2 CO, commonly known as acetone Key Equations definition of molarity Equation \(\PageIndex{1}\) : \( molarity = \dfrac{moles\: of\: solute}{liters\: of\: solution} = \dfrac{mmoles\: of\: solute} {milliliters\: of\: solution} \) relationship among volume, molarity, and moles Equation \(\PageIndex{2}\) : \( V_L M_{mol/L} = \cancel{L} \left( \dfrac{mol}{\cancel{L}} \right) = moles \) relationship between volume and concentration of stock and dilute solutions Equation \(\PageIndex{4}\) : \((V_s)(M_s) = moles\: of\: solute = (V_d)(M_d)\) Summary The concentration of a substance is the quantity of solute present in a given quantity of solution. Concentrations are usually expressed as molarity , the number of moles of solute in 1 L of solution. Solutions of known concentration can be prepared either by dissolving a known mass of solute in a solvent and diluting to a desired final volume or by diluting the appropriate volume of a more concentrated solution (a stock solution ) to the desired final volume. Key Takeaway Solution concentrations are typically expressed as molarity and can be prepared by dissolving a known mass of solute in a solvent or diluting a stock solution. Conceptual Problems Which of the representations best corresponds to a 1 M aqueous solution of each compound? Justify your answers. NH 3 HF CH 3 CH 2 CH 2 OH Na 2 SO 4 Which of the representations shown in Problem 1 best corresponds to a 1 M aqueous solution of each compound? Justify your answers. CH 3 CO 2 H NaCl Na 2 S Na 3 PO 4 acetaldehyde Would you expect a 1.0 M solution of CaCl 2 to be a better conductor of electricity than a 1.0 M solution of NaCl? Why or why not? An alternative way to define the concentration of a solution is molality , abbreviated m . Molality is defined as the number of moles of solute in 1 kg of solvent . How is this different from molarity? Would you expect a 1 M solution of sucrose to be more or less concentrated than a 1 m solution of sucrose? Explain your answer. What are the advantages of using solutions for quantitative calculations? Answer If the amount of a substance required for a reaction is too small to be weighed accurately, the use of a solution of the substance, in which the solute is dispersed in a much larger mass of solvent, allows chemists to measure the quantity of the substance more accurately. Numerical Problems Calculate the number of grams of solute in 1.000 L of each solution. 0.2593 M NaBrO 3 1.592 M KNO 3 1.559 M acetic acid 0.943 M potassium iodate Calculate the number of grams of solute in 1.000 L of each solution. 0.1065 M BaI 2 1.135 M Na 2 SO 4 1.428 M NH 4 Br 0.889 M sodium acetate If all solutions contain the same solute, which solution contains the greater mass of solute? 1.40 L of a 0.334 M solution or 1.10 L of a 0.420 M solution 25.0 mL of a 0.134 M solution or 10.0 mL of a 0.295 M solution 250 mL of a 0.489 M solution or 150 mL of a 0.769 M solution Complete the following table for 500 mL of solution. Compound Mass (g) Moles Concentration (M) calcium sulfate 4.86 NaN NaN acetic acid NaN 3.62 NaN hydrogen iodide dihydrate NaN NaN 1.273 barium bromide 3.92 NaN NaN glucose NaN NaN 0.983 sodium acetate NaN 2.42 NaN What is the concentration of each species present in the following aqueous solutions? 0.489 mol of NiSO 4 in 600 mL of solution 1.045 mol of magnesium bromide in 500 mL of solution 0.146 mol of glucose in 800 mL of solution 0.479 mol of CeCl 3 in 700 mL of solution What is the concentration of each species present in the following aqueous solutions? 0.324 mol of K 2 MoO 4 in 250 mL of solution 0.528 mol of potassium formate in 300 mL of solution 0.477 mol of KClO 3 in 900 mL of solution 0.378 mol of potassium iodide in 750 mL of solution What is the molar concentration of each solution? 8.7 g of calcium bromide in 250 mL of solution 9.8 g of lithium sulfate in 300 mL of solution 12.4 g of sucrose (C 12 H 22 O 11 ) in 750 mL of solution 14.2 g of iron(III) nitrate hexahydrate in 300 mL of solution What is the molar concentration of each solution? 12.8 g of sodium hydrogen sulfate in 400 mL of solution 7.5 g of potassium hydrogen phosphate in 250 mL of solution 11.4 g of barium chloride in 350 mL of solution 4.3 g of tartaric acid (C 4 H 6 O 6 ) in 250 mL of solution Give the concentration of each reactant in the following equations, assuming 20.0 g of each and a solution volume of 250 mL for each reactant. BaCl 2 (aq) + Na 2 SO 4 (aq) → Ca(OH) 2 (aq) + H 3 PO 4 (aq) → Al(NO 3 ) 3 (aq) + H 2 SO 4 (aq) → Pb(NO 3 ) 2 (aq) + CuSO 4 (aq) → Al(CH 3 CO 2 ) 3 (aq) + NaOH(aq) → An experiment required 200.0 mL of a 0.330 M solution of Na 2 CrO 4 . A stock solution of Na 2 CrO 4 containing 20.0% solute by mass with a density of 1.19 g/cm 3 was used to prepare this solution. Describe how to prepare 200.0 mL of a 0.330 M solution of Na 2 CrO 4 using the stock solution. Calcium hypochlorite [Ca(OCl) 2 ] is an effective disinfectant for clothing and bedding. If a solution has a Ca(OCl) 2 concentration of 3.4 g per 100 mL of solution, what is the molarity of hypochlorite? Phenol (C 6 H 5 OH) is often used as an antiseptic in mouthwashes and throat lozenges. If a mouthwash has a phenol concentration of 1.5 g per 100 mL of solution, what is the molarity of phenol? If a tablet containing 100 mg of caffeine (C 8 H 10 N 4 O 2 ) is dissolved in water to give 10.0 oz of solution, what is the molar concentration of caffeine in the solution? A certain drug label carries instructions to add 10.0 mL of sterile water, stating that each milliliter of the resulting solution will contain 0.500 g of medication. If a patient has a prescribed dose of 900.0 mg, how many milliliters of the solution should be administered? Answers 0.48 M ClO − 1.74 × 10 −3 M caffeine Contributors Anonymous Modified by Joshua Halpern ( Howard University ) Thumbnail from Wikimedia
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Courses/El_Paso_Community_College/CHEM1306%3A_Health_Chemistry_I_(Rodriguez)/02%3A_Atoms/2.04%3A_Nuclei_of_Atoms
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Learning Objectives Define and differentiate between the atomic number and the mass number of an element. Explain how isotopes differ from one another. Now that we know how atoms are generally constructed, what do atoms of any particular element look like? How many protons, neutrons, and electrons are in a specific kind of atom? First, if an atom is electrically neutral overall, then the number of protons equals the number of electrons. Because these particles have the same but opposite charges, equal numbers cancel out, producing a neutral atom. Atomic Number In the 1910s, experiments with x-rays led to this useful conclusion: the magnitude of the positive charge in the nucleus of every atom of a particular element is the same. In other words, all atoms of the same element have the same number of protons. Furthermore, different elements have a different number of protons in their nuclei, so the number of protons in the nucleus of an atom is characteristic of a particular element. This discovery was so important to our understanding of atoms that the number of protons in the nucleus of an atom is called the atomic number (Z). For example, hydrogen has the atomic number 1; all hydrogen atoms have 1 proton in their nuclei. Helium has the atomic number 2; all helium atoms have 2 protons in their nuclei. There is no such thing as a hydrogen atom with 2 protons in its nucleus; a nucleus with 2 protons would be a helium atom. The atomic number defines an element. Table \(\PageIndex{1}\) lists some common elements and their atomic numbers. Based on its atomic number , you can determine the number of protons in the nucleus of an atom. The largest atoms have over 100 protons in their nuclei. Element Atomic Number Element.1 Atomic Nmbers aluminum (Al) 13.0 magnesium (Mg) 12.0 americium (Am) 95.0 manganese (Mn) 25.0 argon (Ar) 18.0 mercury (Hg) 80.0 barium (Ba) 56.0 neon (Ne) 10.0 beryllium (Be) 4.0 nickel (Ni) 28.0 bromine (Br) 35.0 nitrogen (N) 7.0 calcium (Ca) 20.0 oxygen (O) 8.0 carbon (C) 6.0 phosphorus (P) 15.0 chlorine (Cl) 17.0 platinum (Pt) 78.0 chromium (Cr) 24.0 potassium (K) 19.0 cesium (Cs) 55.0 radon (Rn) 86.0 fluorine (F) 9.0 silver (Ag) 47.0 gallium (Ga) 31.0 sodium (Na) 11.0 gold (Au) 79.0 strontium (Sr) 38.0 helium (He) 2.0 sulfur (S) 16.0 hydrogen (H) 1.0 titanium (Ti) 22.0 iron (Fe) 26.0 tungsten (W) 74.0 iodine (I) 53.0 uranium (U) 92.0 lead (Pb) 82.0 zinc (Zn) 30.0 lithium (Li) 3.0 zirconium (Zr) 40.0 NaN NaN NaN NaN Example \(\PageIndex{1}\) What is the number of protons in the nucleus of each element? aluminum iron carbon Answer a According to Table 2.4.1, aluminum has an atomic number of 13. Therefore, every aluminum atom has 13 protons in its nucleus. Answer b Iron has an atomic number of 26. Therefore, every iron atom has 26 protons in its nucleus. Answer c Carbon has an atomic number of 6. Therefore, every carbon atom has 6 protons in its nucleus. Exercise \(\PageIndex{1}\) What is the number of protons in the nucleus of each element? Use Table 2.4.1. sodium oxygen chlorine Answer a Sodium has 11 protons in its nucleus. Answer b Oxygen has 8 protons in its nucleus. Answer c Chlorine has 17 protons in its nucleus How many electrons are in an atom? Previously we said that for an electrically neutral atom, the number of electrons equals the number of protons , so the total opposite charges cancel. Thus, the atomic number of an element also gives the number of electrons in an atom of that element. (Later we will find that some elements may gain or lose electrons from their atoms, so those atoms will no longer be electrically neutral. Thus we will need a way to differentiate the number of electrons for those elements.) Example \(\PageIndex{2}\) How many electrons are present in the atoms of each element? sulfur tungsten argon Answer a The atomic number of sulfur is 16. Therefore, in a neutral atom of sulfur, there are 16 electrons. Answer b The atomic number of tungsten is 74. Therefore, in a neutral atom of tungsten, there are 74 electrons. Answer c The atomic number of argon is 18. Therefore, in a neutral atom of argon, there are 18 electrons. Exercise \(\PageIndex{2}\) How many electrons are present in the atoms of each element? magnesium potassium iodine Answer a Mg has 12 electrons. Answer b K has 19 electrons. Answer c I has 53 electrons. Isotopes How many neutrons are in atoms of a particular element? At first it was thought that the number of neutrons in a nucleus was also characteristic of an element. However, it was found that atoms of the same element can have different numbers of neutrons. Atoms of the same element (i.e., same atomic number, Z) that have different numbers of neutrons are called isotopes. For example, 99% of the carbon atoms on Earth have 6 neutrons and 6 protons in their nuclei; about 1% of the carbon atoms have 7 neutrons in their nuclei. Naturally occurring carbon on Earth, therefore, is actually a mixture of isotopes, albeit a mixture that is 99% carbon with 6 neutrons in each nucleus. An important series of isotopes is found with hydrogen atoms. Most hydrogen atoms have a nucleus with only a single proton. About 1 in 10,000 hydrogen nuclei, however, also has a neutron; this particular isotope is called deuterium . An extremely rare hydrogen isotope, tritium , has 1 proton and 2 neutrons in its nucleus. Figure \(\PageIndex{1}\) compares the three isotopes of hydrogen. The discovery of isotopes required a minor change in Dalton’s atomic theory. Dalton thought that all atoms of the same element were exactly the same. Most elements exist as mixtures of isotopes. In fact, there are currently over 3,500 isotopes known for all the elements. When scientists discuss individual isotopes, they need an efficient way to specify the number of neutrons in any particular nucleus. The mass number (A) of an atom is the sum of the numbers of protons and neutrons in the nucleus. Given the mass number for a nucleus (and knowing the atomic number of that particular atom), you can determine the number of neutrons by subtracting the atomic number from the mass number. A simple way of indicating the mass number of a particular isotope is to list it as a superscript on the left side of an element’s symbol. Atomic numbers are often listed as a subscript on the left side of an element’s symbol. Thus, we might see \[\mathrm{^{mass\: number\xrightarrow{\hspace{45px}} 56}_{atomic\: number \xrightarrow{\hspace{35px}} 26}Fe} \label{Eq1} \] which indicates a particular isotope of iron. The 26 is the atomic number (which is the same for all iron atoms), while the 56 is the mass number of the isotope. To determine the number of neutrons in this isotope, we subtract 26 from 56: 56 − 26 = 30, so there are 30 neutrons in this atom. Example \(\PageIndex{3}\) How many protons and neutrons are in each atom? \(\mathrm{^{35}_{17}Cl}\) \(\mathrm{^{127}_{53}I}\) Answer a In \(\mathrm{^{35}_{17}Cl}\) there are 17 protons, and 35 − 17 = 18 neutrons in each nucleus. Answer b In \(\mathrm{^{127}_{53}I}\) there are 53 protons, and 127 − 53 = 74 neutrons in each nucleus. Exercise \(\PageIndex{3}\) How many protons and neutrons are in each atom? \(\mathrm{^{197}_{79}Au}\) \(\mathrm{^{23}_{11}Na}\) Answer a In \(\mathrm{^{197}_{79}Au}\) there are 79 protons, and 197 − 79 = 118 neutrons in each nucleus. Answer b In \(\mathrm{^{23}_{11}Na}\) there are 11 protons, and 23 − 11 = 12 neutrons in each nucleus. It is not absolutely necessary to indicate the atomic number as a subscript because each element has its own unique atomic number. Many isotopes are indicated with a superscript only, such as 13 C or 235 U. You may also see isotopes represented in print as, for example, carbon-13 or uranium-235. Summary The atom consists of discrete particles that govern its chemical and physical behavior. Each atom of an element contains the same number of protons, which is the atomic number (Z) . Neutral atoms have the same number of electrons and protons. Atoms of an element that contain different numbers of neutrons are called isotopes . Each isotope of a given element has the same atomic number but a different mass number (A) , which is the sum of the numbers of protons and neutrons . Almost all of the mass of an atom is from the total protons and neutrons contained within a tiny (and therefore very dense) nucleus. The majority of the volume of an atom is the surrounding space in which the electrons reside. A representation of a carbon-12 atom is shown below in Figure \(\PageIndex{2}\). Concept Review Exercises Why is the atomic number so important to the identity of an atom? What is the relationship between the number of protons and the number of electrons in an atom? How do isotopes of an element differ from each other? What is the mass number of an element? Answers The atomic number defines the identity of an element. It describes the number of protons in the nucleus. In an electrically neutral atom, the number of protons equals the number of electrons. Isotopes of an element have the same number of protons but have different numbers of neutrons in their nuclei. The mass number is the sum of the numbers of protons and neutrons in the nucleus of an atom. Key Takeaways Each element is identified by its atomic number. The atomic number provides the element's location on the periodic table The isotopes of an element have different masses and are identified by their mass numbers.
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Bookshelves/Organic_Chemistry/Intermediate_Physical_Organic_(Morsch)/02%3A_Reaction_Kinetics/2.01%3A_Introduction_to_Reaction_Kinetics/2.1.05%3A_Second-Order_Reactions
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Many important biological reactions, such as the formation of double-stranded DNA from two complementary strands, can be described using second order kinetics. In a second-order reaction, the sum of the exponents in the rate law is equal to two. The two most common forms of second-order reactions will be discussed in detail in this section. To describe how the rate of a second-order reaction changes with concentration of reactants or products, the differential (derivative) rate equation is used as well as the integrated rate equation. The differential rate law can show us how the rate of the reaction changes in time, while the integrated rate equation shows how the concentration of species changes over time. The latter form, when graphed, yields a linear function and is, therefore, more convenient to look at. Nonetheless, both of these equations can be derived from the above expression for the reaction rate. Plotting these equations can also help us determine whether or not a certain reaction is second-order. Case 1: Identical Reactants Two of the same reactant (\(\ce{A}\)) combine in a single elementary step. \[\begin{align} \ce{A} + \ce{A} &\ce{->} \ce{P} \label{case1a} \\[4pt] \ce{2A} &\ce{->} \ce{P} \label{case1b} \end{align} \] The reaction rate for this step can be written as \[\text{Rate} = - \dfrac{1}{2} \dfrac{d[A]}{dt} = + \dfrac{d[P]}{dt} \nonumber \] and the rate of loss of reactant \(\ce{A}\) \[\dfrac{dA}{dt}= -k[A][A] = -k[A]^2 \label{2ndlaw} \] where \(k\) is a second order rate constant with units of \(\text{M}^{-1} \text{min}^{-1}\) or \(\text{M}^{-1} \text{s}^{-1}\). Therefore, doubling the concentration of reactant \(\ce{A}\) will quadruple the rate of the reaction. In this particular case, another reactant (\(B\)) could be present with \(A\); however, its concentration does not affect the rate of the reaction, i.e., the reaction order with respect to B is zero, and we can express the rate law as \(v = k[A]^2[B]^0\). Integration of Equation \ref{2ndlaw} yields \[ \dfrac{1}{[A]} = \dfrac{1}{[A]_0}+kt \nonumber \] which is easily rearranged into a form of the equation for a straight line and yields plots similar to the one shown below. The half-life is given by \[ t_{1/2}=\dfrac{1}{k[A_o]} \nonumber \] Notice that the half-life of a second-order reaction depends on the initial concentration, in contrast to first-order reactions. For this reason, the concept of half-life for a second-order reaction is far less useful. Reaction rates are discussed in more detail here. Reaction orders are defined here. Here are explanations of zero and first order reactions. For reactions that follow Equation \ref{case1a} or \ref{case1b}, the rate at which \(\ce{A}\) decreases can be expressed using the differential rate equation . \[-\dfrac{d[A]}{dt} = k[A]^2 \nonumber \] The equation can then be rearranged: \[\dfrac{d[A]}{[A]^2} = -k\,dt \nonumber \] Since we are interested in the change in concentration of A over a period of time, we integrate between \(t = 0\) and \(t\), the time of interest. \[ \int_{[A]_o}^{[A]_t} \dfrac{d[A]}{[A]^2} = -k \int_0^t dt \nonumber \] To solve this, we use the following rule of integration (power rule): \[\int \dfrac{dx}{x^2} = -\dfrac{1}{x} + constant \nonumber \] We then obtain the integrated rate equation . \[\dfrac{1}{[A]_t} - \dfrac{1}{[A]_o} = kt \nonumber \] Upon rearrangement of the integrated rate equation, we obtain an equation of the line: \[\dfrac{1}{[A]_t} = kt + \dfrac{1}{[A]_o} \nonumber \] The crucial part of this process is not understanding precisely how to derive the integrated rate law equation, rather it is important to understand how the equation directly relates to the graph which provides a linear relationship. In this case, and for all second order reactions, the linear plot of \(\dfrac{1}{[A]_t}\) versus time will yield the graph below. This graph is useful in a variety of ways. If we only know the concentrations at specific times for a reaction, we can attempt to create a graph similar to the one above. If the graph yields a straight line, then the reaction in question must be second order. In addition, with this graph we can find the slope of the line and this slope is \(k\), the reaction constant. The slope can be found be finding the "rise" and then dividing it by the "run" of the line. For an example of how to find the slope, please see the example section below. There are alternative graphs that could be drawn. The plot of \([A]_t\) versus time would result in a straight line if the reaction were zeroth order. It does, however, yield less information for a second order graph. This is because both the graphs of a first or second order reaction would look like exponential decays. The only obvious difference, as seen in the graph below, is that the concentration of reactants approaches zero more slowly in a second-order, compared to that in a first order reaction. Case 2: Second Order Reaction with Multiple Reactants Two different reactants (\(\ce{A}\) and \(\ce{B}\)) combine in a single elementary step: \[A + B \longrightarrow P \label{case2} \] The reaction rate for this step can be written as \[\text{Rate} = - \dfrac{d[A]}{dt}= - \dfrac{d[B]}{dt}= + \dfrac{d[P]}{dt} \nonumber \] and the rate of loss of reactant \(\ce{A}\) \[ \dfrac{d[A]}{dt}= - k[A][B] \nonumber \] where the reaction order with respect to each reactant is 1. This means that when the concentration of reactant A is doubled, the rate of the reaction will double, and quadrupling the concentration of reactant in a separate experiment will quadruple the rate. If we double the concentration of \(\ce{A}\) and quadruple the concentration of \(\ce{B}\) at the same time, then the reaction rate is increased by a factor of 8. This relationship holds true for any varying concentrations of \(\ce{A}\) or \(\ce{B}\). As before, the rate at which \(A\) decreases can be expressed using the differential rate equation: \[ \dfrac{d[A]}{dt} = -k[A][B] \nonumber \] Two situations can be identified. Situation 2a: \([A]_0 \neq [B]_0\) Situation 2a is the situation that the initial concentration of the two reactants are not equal. Let \(x\) be the concentration of each species reacted at time \(t\). Let \( [A]_0 =a\) and \([B]_0 =b\), then \([A]= a-x\) ;\( [B]= b-x\). The expression of rate law becomes: \[-\dfrac{dx}{dt} = -k([A]_o - x)([B]_o - x)\nonumber \] which can be rearranged to: \[\dfrac{dx}{([A]_o - x)([B]_o - x)} = kdt\nonumber \] We integrate between \(t = 0\) (when \(x = 0\)) and \(t\), the time of interest. \[ \int_0^x \dfrac{dx}{([A]_o - x)([B]_o - x)} = k \int_0^t dt \nonumber \] To solve this integral, we use the method of partial fractions . \[ \int_0^x \dfrac{1}{(a - x)(b -x)}dx = \dfrac{1}{b - a}\left(\ln\dfrac{1}{a - x} - \ln\dfrac{1}{b - x}\right)\nonumber \] Evaluating the integral gives us: \[ \int_0^x \dfrac{dx}{([A]_o - x)([B]_o - x)} = \dfrac{1}{[B]_o - [A]_o}\left(\ln\dfrac{[A]_o}{[A]_o - x} - \ln\dfrac{[B]_o}{[B]_o - x}\right) \nonumber \] Applying the rule of logarithm , the equation simplifies to: \[\int _0^x \dfrac{dx}{([A]_o - x)([B]_o - x)} = \dfrac{1}{[B]_o - [A]_o} \ln \dfrac{[B][A]_o}{[A][B]_o} \nonumber \] We then obtain the integrated rate equation (under the condition that [A] and [B] are not equal). \[ \dfrac{1}{[B]_o - [A]_o}\ln \dfrac{[B][A]_o}{[A][B]_o} = kt \nonumber \] Upon rearrangement of the integrated rate equation, we obtain: \[ \ln\dfrac{[B][A]_o}{[A][B]_o} = k([B]_o - [A]_o)t \nonumber \] Hence, from the last equation, we can see that a linear plot of \(\ln\dfrac{[A]_o[B]}{[A][B]_o}\) versus time is characteristic of second-order reactions. This graph can be used in the same manner as the graph in the section above or written in the other way: \[\ln\dfrac{[A]}{[B]} = k([A]_o - [B]_o)t+\ln\dfrac{[A]_o}{[B]_o}\nonumber \] in form \( y = ax + b\) with a slope of \(a= k([B]_0-[A]_0)\) and a y-intercept of \( b = \ln \dfrac{[A]_0}{[B]_0}\) Situation 2b: \([A]_0 =[B]_0\) Because \(A + B \rightarrow P\) Since \(A\) and \(B\) react with a 1 to 1 stoichiometry, \([A]= [A]_0 -x\) and \([B] = [B]_0 -x\) at any time \(t\), \([A] = [B]\) and the rate law will be, \[\text{rate} = k[A][B] = k[A][A] = k[A]^2.\nonumber \] Thus, it is assumed as the first case !!! Example \(\PageIndex{1}\) The following chemical equation reaction represents the thermal decomposition of gas \(E\) into \(K\) and \(G\) at 200° C \[\ce{ 5E(g) -> 4K(g) + G(g)} \nonumber \] This reaction follows a second order rate law with regards to \(\ce{E}\). For this reaction suppose that the rate constant at 200° C is equivalent to \(4.0 \times 10^{-2} M^{-1}s^{-1}\) and the initial concentration is \(0.050\; M\). W hat is the initial rate of decomposition of \(\ce{E}\) . Solution Start by defining the reaction rate in terms of the loss of reactants \[ \text{Rate (initial)} = - \dfrac{1}{5} \dfrac{d[E]}{dt}\nonumber \] and then use the rate law to define the rate of loss of \(\ce{E}\) \[ \dfrac{d[E]}{dt} = -k [A]_i^2 \nonumber \] We already know \(k\) and \([A]_i\) but we need to figure out \(x\). To do this look at the units of \(k\) and one sees it is M -1 s -1 which means the overall reaction is a second order reaction with \(x=2\). \[\begin{align*} \text{Initial rate} &= (4.0 \times 10^{-2} M^{-1}s^{-1})(0.050\,M)^2 \\[4pt] &= 1 \times 10^{-4} \, Ms^{-1}\end{align*} \nonumber \] Half-Life Another characteristic used to determine the order of a reaction from experimental data is the half-life (\(t_{1/2}\)). By definition, the half life of any reaction is the amount of time it takes to consume half of the starting material. For a second-order reaction, the half-life is inversely related to the initial concentration of the reactant (A). For a second-order reaction each half-life is twice as long as the life span of the one before. Consider the reaction \(2A \rightarrow P\): We can find an expression for the half-life of a second order reaction by using the previously derived integrated rate equation. \[\dfrac{1}{[A]_t} - \dfrac{1}{[A]_o} = kt \nonumber \] Since, \[[A]_{t_{1/2}} = \dfrac{1}{2}[A]_o \nonumber \] when \(t = t_{1/2} \). Our integrated rate equation becomes: \[\dfrac{1}{\dfrac{1}{2}[A]_o} - \dfrac{1}{[A]_o} = kt_{1/2} \nonumber \] After a series of algebraic steps, \[\begin{align*} \dfrac{2}{[A]_o} - \dfrac{1}{[A]_o} &= kt_{1/2} \\[4pt] \dfrac{1}{[A]_o} &= kt_{1/2} \end{align*} \nonumber \] We obtain the equation for the half-life of a second order reaction: \[t_{1/2} = \dfrac{1}{k[A]_o} \label{2nd halflife} \] This inverse relationship suggests that as the initial concentration of reactant is increased, there is a higher probability of the two reactant molecules interacting to form product. Consequently, the reactant will be consumed in a shorter amount of time, i.e. the reaction will have a shorter half-life. This equation also implies that since the half-life is longer when the concentrations are low, species decaying according to second-order kinetics may exist for a longer amount of time if their initial concentrations are small. Note that for the second scenario in which \(A + B \rightarrow P\), the half-life of the reaction cannot be determined. As stated earlier, \([A]_o\) cannot be equal to \([B]_o\). Hence, the time it takes to consume one-half of A is not the same as the time it takes to consume one-half of B. Because of this, we cannot define a general equation for the half-life of this type of second-order reaction. Example \(\PageIndex{2}\): Half-Life of a Second-Order Reaciton If the only reactant is the initial concentration of \(A\), and it is equivalent to \([A]_0=4.50 \times 10^{-5}\,M\) and the reaction is a second order with a rate constant \(k=0.89 M^{-1}s^{-1}\). What is the half-life of this reaction? Solution This is a direct application of Equation \ref{2nd halflife}. \[\begin{align*} \dfrac{1}{k[A]_0} &= \dfrac{1}{(4.50 \times 10^{-5} M)(0.89 M^{-1}{s^{-1})}} \\[4pt] &= 2.50 \times 10^4 \,s \end{align*} \nonumber \] Summary Unnamed: 0 \(2A \rightarrow P\) \(A + B \rightarrow P\) Differential Form \(-\dfrac{d[A]}{dt} = k[A]^2\) \(-\dfrac{d[A]}{dt} = k[A][B]\) Integral Form \(\dfrac{1}{[A]_t} = kt + \dfrac{1}{[A]_o}\) \(\dfrac{1}{[B]_o - [A]_o}\ln\dfrac{[B][A]_o}{[A][B]_o} = kt\) Half Life \(t_{1/2} = \dfrac{1}{k[A]_o}\) Cannot be easily defined; \(t_{1/2}\) for A and B are different. The graph below is the graph that tests if a reaction is second order. The reaction is second order if the graph has a straight line, as is in the example below. Exercise \(\PageIndex{3}\) Given the following information, determine the order of the reaction and the value of k, the reaction constant. Concentration (M) Time (s) 1.00 10 0.50 20 0.33 30 *Hint: Begin by graphing Answer Make graphs of concentration vs. time (zeroth order), natural log of concentration vs. time (first order), and one over concentration vs. time (second order). Determine which graph results in a straight line. This graph reflects the order of the reaction. For this problem, the straight line should be in the 3rd graph, meaning the reaction is second order. The numbers should have are: 1/Concentration(M-1) Time (s) 1 10 2 20 3 30 The slope can be found by taking the "rise" over the "run". This means taking two points, (10,1) and (20,2). The "rise" is the vertical distance between the points (2-1=1) and the "run" is the horizontal distance (20-10=10). Therefore the slope is 1/10=0.1. The value of k, therefore, is 0.1 M -2 s -1 . Exercise \(\PageIndex{4}\) Using the following information, determine the half life of this reaction, assuming there is a single reactant. Concentration (M) Time (s) 2.0000 0 1.3000 10 0.9633 20 Answer Determine the order of the reaction and the reaction constant, \(k\), for the reaction using the tactics described in the previous problem. The order of the reaction is second, and the value of k is 0.0269 M -2 s -1 . Since the reaction order is second, the formula for \(t_{1/2} = k^{-1}[A]_0^{-1}\). This means that the half life of the reaction is 0.0259 seconds. Exercise \(\PageIndex{5}\) Given the information from the previous problem, what is the concentration after 5 minutes? Answer Convert the time (5 minutes) to seconds. This means the time is 300 seconds. Use the integrated rate law to find the final concentration. The final concentration is 0.1167 M. References Atkins, P. W., & De Paula, J. (2006). Physical Chemistry for the Life Sciences. New York, NY: W. H. Freeman and Company. Petrucci, R. H., Harwood, W. S., & Herring, F. G. (2002). General Chemistry: Principles and Modern Applications. Upper Saddle River, NJ: Prentice-Hall, Inc.
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Courses/Widener_University/CHEM_176%3A_General_Chemistry_II_(Fischer-Drowos)/06%3A_Equilibria_of_Other_Reaction_Classes/6.03%3A_Lewis_Acids_and_Bases
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Learning Objectives By the end of this section, you will be able to: Explain the Lewis model of acid-base chemistry Write equations for the formation of adducts and complex ions In 1923, G. N. Lewis proposed a generalized definition of acid-base behavior in which acids and bases are identified by their ability to accept or to donate a pair of electrons and form a coordinate covalent bond. A coordinate covalent bond (or dative bond) occurs when one of the atoms in the bond provides both bonding electrons. For example, a coordinate covalent bond occurs when a water molecule combines with a hydrogen ion to form a hydronium ion. A coordinate covalent bond also results when an ammonia molecule combines with a hydrogen ion to form an ammonium ion. Both of these equations are shown here. Reactions involving the formation of coordinate covalent bonds are classified as Lewis acid-base chemistry . The species donating the electron pair that compose the bond is a Lewis base , the species accepting the electron pair is a Lewis acid , and the product of the reaction is a Lewis acid-base adduct . As the two examples above illustrate, Brønsted-Lowry acid-base reactions represent a subcategory of Lewis acid reactions, specifically, those in which the acid species is H + . A few examples involving other Lewis acids and bases are described below. The boron atom in boron trifluoride, BF 3 , has only six electrons in its valence shell. Being short of the preferred octet, BF 3 is a very good Lewis acid and reacts with many Lewis bases; a fluoride ion is the Lewis base in this reaction, donating one of its lone pairs: In the following reaction, each of two ammonia molecules, Lewis bases, donates a pair of electrons to a silver ion, the Lewis acid: Nonmetal oxides act as Lewis acids and react with oxide ions, Lewis bases, to form oxyanions: Many Lewis acid-base reactions are displacement reactions in which one Lewis base displaces another Lewis base from an acid-base adduct, or in which one Lewis acid displaces another Lewis acid: Another type of Lewis acid-base chemistry involves the formation of a complex ion (or a coordination complex) comprising a central atom, typically a transition metal cation, surrounded by ions or molecules called ligands . These ligands can be neutral molecules like H 2 O or NH 3 , or ions such as CN – or OH – . Often, the ligands act as Lewis bases, donating a pair of electrons to the central atom. These types of Lewis acid-base reactions are examples of a broad subdiscipline called coordination chemistry —the topic of another chapter in this text. The equilibrium constant for the reaction of a metal ion with one or more ligands to form a coordination complex is called a formation constant ( K f ) (sometimes called a stability constant). For example, the complex ion is produced by the reaction \[\ce{Cu^{+}(aq) + 2 CN^{-}(aq) <=> Cu(CN)2^{-}(aq)} \nonumber \] The formation constant for this reaction is \[K_{ f }=\frac{\left[ \ce{Cu(CN)2^{-}} \right]}{\left[ \ce{Cu^{+}} \right]\left[ \ce{CN^{-}}\right]^2} \nonumber \] Alternatively, the reverse reaction (decomposition of the complex ion) can be considered, in which case the equilibrium constant is a dissociation constant ( K d ) . Per the relation between equilibrium constants for reciprocal reactions described, the dissociation constant is the mathematical inverse of the formation constant, K d = K f –1 . A tabulation of formation constants is provided in Appendix K. As an example of dissolution by complex ion formation, let us consider what happens when we add aqueous ammonia to a mixture of silver chloride and water. Silver chloride dissolves slightly in water, giving a small concentration of Ag + (\([\ce{Ag^{+}}] = 1.3 \times 10^{–5} ~\text{M}\)): \[\ce{AgCl(s) <=> Ag^{+}(aq) + Cl^{-}(aq)} \nonumber \] However, if NH 3 is present in the water, the complex ion, \(\ce{Ag(NH3)2^{+}}\) can form according to the equation: \[\ce{Ag^{+}(aq) + 2 NH3(aq) <=> Ag(NH3)2^{+}(aq)} \nonumber \] with \[K_{ f }=\dfrac{[ \ce{Ag(NH3)2^{+}} ]}{[ \ce{Ag^{+}} ][\ce{NH3}]^2}=1.7 \times 10^7 \nonumber \] The large size of this formation constant indicates that most of the free silver ions produced by the dissolution of AgCl combine with NH 3 to form \(\ce{Ag(NH3)2^{+}}\). As a consequence, the concentration of silver ions, \(\ce{[Ag^{+}]}\), is reduced, and the reaction quotient for the dissolution of silver chloride, [Ag + ][Cl – ], falls below the solubility product of AgCl: \[Q=\left[ \ce{Ag^{+}} \right]\left[ \ce{Cl^{-}} \right]<K_{ sp } \nonumber \] More silver chloride then dissolves. If the concentration of ammonia is great enough, all of the silver chloride dissolves. Example \(\PageIndex{1}\): Dissociation of a Complex Ion Calculate the concentration of the silver ion in a solution that initially is 0.10 M with respect to \(\ce{Ag(NH3)2^{+}}\)). Solution Applying the standard ICE approach to this reaction yields the following: Substituting these equilibrium concentration terms into the K f expression gives \[\begin{align*} K_{ f } &=\dfrac{[ \ce{Ag(NH3)2^{+}} ]}{[ \ce{Ag^{+}} ][ \ce{NH3} ]^2} \\[4pt] 1.7 \times 10^7 &=\frac{0.10-x}{(x)(2 x)^2} \end{align*} \nonumber \] The very large equilibrium constant means the amount of the complex ion that will dissociate, x, will be very small. Assuming \(x \ll 0.1\) permits simplifying the above equation: \[\begin{align*} 1.7 \times 10^7 &=\frac{0.10}{(x)(2 x)^2} \\[4pt] x^3&=\frac{0.10}{4\left(1.7 \times 10^7\right)}=1.5 \times 10^{-9} \\[4pt] x&=\sqrt[3]{1.5 \times 10^{-9}}=1.1 \times 10^{-3} \end{align*} \nonumber \] Because only 1.1% of the \(\ce{Ag(NH3)2^{+}}\)) dissociates into \(\ce{Ag^{+}}\) and \(\ce{NH3}\), the assumption that x is small is justified. Using this value of x and the relations in the above ICE table allows calculation of all species’ equilibrium concentrations: \[\begin{align*} [ \ce{Ag^{+}} ] &=0+x=1.1 \times 10^{-3}~\text{M} \\[4pt] [ \ce{NH3}] &=0+2 x=2.2 \times 10^{-3} ~\text{M} \\[4pt] [ \ce{Ag(NH3)2^{+}}] &=0.10-x=0.10-0.0011 \\[4pt] &=0.099 ~\text{M} \end{align*} \nonumber \] The concentration of free silver ion in the solution is 0.0011 M . Exercise \(\PageIndex{1}\) Calculate the silver ion concentration, [Ag + ], of a solution prepared by dissolving 1.00 g of \(\ce{AgNO3}\) and 10.0 g of \(\ce{KCN}\) in sufficient water to make 1.00 L of solution. (Hint: Because K f is very large, assume the reaction goes to completion then calculate the [Ag + ] produced by dissociation of the complex.) Answer \(2.9 \times 10^{–22} ~\text{M}\)
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Courses/Lumen_Learning/Book%3A_US_History_II_(OS_Collection)_(Lumen)/11%3A_The_Jazz_Age%3A_Redefining_the_Nation%2C_1919-1929/11.6%3A_Video%3A_The_Roaring_20s
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This video teaches you about the United States in the 1920s. They were known as the roaring 20s, but not because there were lions running around everywhere. In the 1920s, America’s economy was booming, and all kinds of social changes were in progress. Hollywood, flappers, jazz, there was all kinds of stuff going on in the 20s. But as usual with Crash Course, things were about to take a turn for the worse. This video will teach you about the Charleston, the many Republican presidents of the 1920s, laissez-faire capitalism, jazz, consumer credit, the resurgent Klan, and all kinds of other stuff. A YouTube element has been excluded from this version of the text. You can view it online here: http://pb.libretexts.org/2os/?p=186 All rights reserved content The Roaring 20's: Crash Course US History #32. Provided by : Crash Course. Located at : https://youtu.be/VfOR1XCMf7A?t=1s . License : All Rights Reserved . License Terms : Standard YouTube License
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Bookshelves/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/11%3A_Coordination_Chemistry_III_-_Electronic_Spectra/11.03%3A_Electronic_Spectra_of_Coordination_Compounds
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Types of transitions related to the metal ion: d-d transitions : d-d transitions are electronic transitions that occur between the molecular orbitals (MOs) that are mostly metal in character: specifically, the orbitals that we think of as the d-orbitals of a transition metal complex. These transitions are useful in determining the energy of splitting and can be used to indicate coordination chemistry (geometry and ligand sets). In octahedral complexes, d-d transitions occur between the \(t_{2g}\) and \(e_g\) orbitals (across \(\Delta\)). These transitions can not occur in metal complexes where the d-orbital is completely empty (\(d^0\)) or completely full (\(d^{10}\)). In other words, a d-d transition is only possible in \(d^1 - d^9\) metal ions. In a UV-visible absorption spectrum, d-d transitions appear as relatively weak absorptions with extinction coefficients (\(\varepsilon\)) less than 1,000. Charge transfer (CT) transitions : Charge transfer transitions occur between MOs that are mostly metal in character and those that are mostly ligand in character. These transitions depend on the type of ligand: they occur only when the metal is bound to ligands that are \(\pi\)-donors or \(\pi\)-acceptors. And there are two types of CT transitions. If the metal is bound to a \(\pi\)-donor ligand, electrons from lower-energy MO's that are mostly ligand in character can become excited to MO's that are mostly metal in character. These are ligand to metal charge transfers (LMCT) transitions (Figure \(\PageIndex{1}\), left diagram). If the the metal is bound to ligands that are \(\pi\)-acceptors, electrons from the MO's that are mostly metal in character can become excited to higher-energy orbitals that are mostly ligand in character. These are metal to ligand charge transfer (MLCT) transitions (Figure \(\PageIndex{1}\), right diagram). In a UV-visible absorption spectrum, CT transitions appear as relatively intense absorptions with extinction coefficients (\(\varepsilon\)) much greater than 1,000. 11.3.1: Selection Rules 11.3.2: Correlation Diagrams 11.3.3: Tanabe-Sugano Diagrams 11.3.4: Symmetry labels for split terms 11.3.5: Applications of Tanabe-Sugano Diagrams 11.3.6: Tetrahedral Complexes 11.3.7: Charge-Transfer Spectra 11.3.8: Applications of Charge-Transfer
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Courses/Tennessee_State_University/CHEM_4210%3A_Inorganic_Chem_II_(Siddiquee)/05%3A_Reactions_of_Complexes/5.09%3A_Reactions_of_Coordinated_Ligands
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So far, our coverage of the reactions of metal complexes has focused only on the reactions that occur at the metal center. However, there are also reactions that can occur to the ligands bound at the metal center. When a ligand binds to an electropositive metal center, bonds within the ligand can be polarized in new ways, leading to new or enhanced reactivity. Some of these reactions are covered in the chapters on organometallic chemistry, and there are many examples of how biology employs metal ions at enzyme active sites to catalyze biological transformations. This section will briefly describe some of the most common types of reactions that can be catalyzed by metal ion binding.
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Courses/Duke_University/CHEM_310L%3A_Physical_Chemistry_I_Laboratory/CHEM310L_-_Physical_Chemistry_I_Lab_Manual/05%3A_Molecular_Spectroscopy_of_Iodine/5.05%3A_Experimental_Part_II_-_Emission_Spectrum_of_Iodine_Vapor
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In this part of the experiment you will use a HeNe laser at 543.5 nm to excite iodine to the B state, and a monochromator and photomultiplier to measure the spectrum of the emission back to the ground X state. It is worth noting that this technique of "laser-induced fluorescence" is an important spectroscopic tool, not only for physical chemists, but also for analytical chemists. The sensitivity is extremely high, and this has led to a wide variety of applications. These include the measurement of trace amounts of osmium and iridium in rock samples as part of an effort to test a theory for the sudden destruction of the dinosaurs by a catastrophic collision of the earth with an asteroid! Laser Safety Since the laser used in this experiment can be harmful, and the apparatus is relatively expensive, a number of safety precautions must be taken. Safety and Other Precautions Don't look into the laser ! This is a class IIIa/3R laser. If the beam is directed into your eye it will damage your eye before you can blink. However you will still want to avoid looking directly into the beam, or having the beam be directed at you. Any glass that the laser hits will reflect about 5% of the beam intensity. This specular reflection may accidentally be directed toward your eyes. Therefore, do not move items into or out of the laser path with the laser on. Wear laser goggles whenever the laser is on. Don't vent the cell. There is no danger to you if you vent it, but you won't be able to do the experiment until the cell is evacuated again. Don't kill the PMT. The Photomultiplier Tube (PMT) is an extremely sensitive light detector. If you can see the light, that's enough light to overload the PMT if there is voltage on it. Read and understand the experimental procedure before applying any voltage to the PMT. Do not turn on the voltage to the PMT if it is exposed to significant amounts of light. Also, turn up the voltage slowly as you look for signal. As long as the observed signal stays below 0.2 V the PMT will not be damaged. Again, no physical damage to you will result from overloading the PMT, only financial. Note The Helium-Neon laser frequency of 543.5 nm corresponds to one of the larger absorption peaks from the absorption spectrum obtained in Part I of the experiment. Notice that absorptions close to 543 nm in the absorption spectrum (Part I) are among the most intense absorptions. Preparation of Gas Sample Cell A cell containing solid iodine (\(\ce{I2}\)) will be put under high vacuum to vaporize the \(\ce{I2}\) for spectroscopic analysis. Vacuum-Line Start-Up Some of the following may be completed by the TA before the lab period. Check that all external valves on the working and main manifold are closed to the outside atmosphere. Attach an empty N 2 trap to the main manifold. Hold this trap in place until a vacuum has formed. Turn on the floor vacuum pump to start generating a vacuum. Turn on all valves in the manifold that are connected to gages (marked with purple tape). Turn on the diffusion pump. Ensure that the floor pump is on BEFORE the diffusion pump. Otherwise, the diffusion pump will generate heat and the oil within the pump will oxidize, causing long-term damage to the diffusion pump. Turn on Digivac electronic monitor. Use this to actively monitor the pressure at the end of the main manifold. Check the analog pressure gauge on the diffusion pump to ensure the pressure is stable. Throughout the experiment, the pressure should remain near 10 mtorr if the system is closed and stable. Fill a 4L dewar with liquid N 2 from the shared Chemistry department supply. Place a second dewar underneath the N 2 trap on the Main Manifold and fill with liquid N 2 such that the N 2 trap is submerged in liquid N 2 . Place the iodine trap on the main manifold and place a third (small) dewar under the trap such that the trap is submerged. Check the level of N 2 as the experiment progresses. Refill if necessary to ensure the trap remains cold. Check to ensure the pressure at the end of the main manifold is near 10 mtorr before proceeding. Use the high vacuum line, with help from your TA, to evacuate the iodine vapor cell that will be used in the emission experiment. Once the cell is well evacuated, place it on the sample holder on the laser table. The cell should be in the path of the laser and there should be nothing obstructing the light path. Vacuum Line Shutdown Turn off the diffusion pump. Wait 5 mins for the pump to cool. The pump should be cool before any oxygen is introduced to the system to prevent oil in the pump from oxidizing. Close all valves with purple tape. Turn off Digivac monitor. Turn off the floor pump. Pick an external valve and open it to release the vacuum and open both the working manifold and the main manifold to the air. Do this quickly after step 4. The system should not be left under vacuum without an active vacuum pump. Remove N 2 trap and transfer it to the fume hood to safely evaporate excess HCl/DCl gas. Do this quickly after step 5. Leaving the N 2 trap cold after the vacuum has been released may cause liquid O 2 to condense, which is highly volatile. Instrument Configuration The figure below shows a schematic diagram of the instrument set up. Laser : A Research Electro-Optics, Inc. HeNe laser (model 30968), which emits visible light (543.5 nm), is used to pump primarily transitions from \(\upsilon^{\prime\prime}=0\). Monochromator : A Triax 320 monochromator is used to disperse the emission signal from the sample. The entrance and exit slits of the monochromator control the signal intensity and resolution. It is best to use very narrow slits when looking for signal of unknown intensity so you don't overload the photomultiplier tube (PMT) detector with too much light. The PMT is mounted on the exit slit, which should be set to the same width as the entrance slit. There is a slide shutter on the exit port. The manual control unit (CD-2a) can be used for tuning the monochromator when you are setting up the experiment. The monochromator is computer controlled during data collection. PMT Power Supply : The PMT power supply allows you to introduce a voltage to the PMT. Typically you will use a voltage off 1100 volts for this specific PMT. Depending on the available time, your TA may show you the effect of PMT voltage on the observed signal. Lock-In Amplifier : In association with Stanford Research Systems (SRS) chopper an SRS510 Lock-in Amplifier is used to improve the signal to noise ratio of the emission spectrum. The interested student should read the Bentham Instruments introduction to the principles of a lock-in amplifier in the Course Documents section of your Sakai site, or at http://www.bentham.co.uk/pdf/F225.pdf . Spectra Acq2 : This is an analog to digital (A/D) converter which takes the analog signal (current from the PMT) and converts it to a digital signal that the SynerJY computer software can read and process. Operating the Laser Spectrometer Turn on the Chopper using the power switch on the back of the power supply. It should already be set to approximately 500 Hz. If necessary, adjust settings using the frequency adjust controls to bring it to approximately 500 Hz. Turn on the power for Amplifier and Photomultiplier Tube (PMT). Turn on the Pre-Amp power supply using the toggle switch on the box. Turn on the Lock-in Amplifier using the black switch on the front right of the box. Turn on the PMT Power Supply, but do NOT apply voltage yet. Turn off the lights in the room. With the lights off, turn on the laser (key on the laser power supply). Notice that while the laser beam itself is green (543.5 nm), the emitted light, seen as a beam along the length of the sample cell, is in the yellow orange, that is, at lower energy (higher wavelength) than the exiting laser. Of course what you see in the cell is the sum of all wavelengths of emitted light. To resolve that into a spectrum, you will send that light into the monochromator and scan an appropriate wavelength range. Click the SynerJY icon ( ) on the computer screen to load the SynerJY software. From the Collect menu choose Experiment Setup . Now, just wait until the system initializes. You will briefly see two windows come and go as the software connects with the monochromator. The first to appear is a Hardware Configuration window, and Triax320PMT should be highlighted in it. This window will disappear and be replaced by an Initialization Window. The software will initialize the monochromator and PMT and then that window will also disappear. The software will then load the Experiment Setup screen. Set up the optimization experiment: Type a name for your Experiment File (use YYYYMMDD_YourInitials_opt). Press SAVE . Choose a Start Wavelength of 540 nm and an End Wavelength of 550 nm . Set the Increment to 0.2 nm . Make sure Type is set to Wavelength, and Triax320 is selected under Scanning . Click the "Monos" button ( ) at the left of the screen and set the Front_Entrance and Front_Exit Slits each to 0.026 mm . Click on the Detectors button ( ) button at the left of the window and make sure there is a check in the PMT Active box and set the Integration Time to 3000 ms (i.e. 3 seconds). Leave the Gain at x1. Turn on the PMT Using the buttons on the PMT power supply, apply 900 volts to the PMT. Do this by typing "900" and then pressing the "Enter" button. Then switch the power to "on". On the software, click on the Real Time Control button (RTC, ). Make sure the Slits and Integration Time are set properly. Change the value in the Position box to 543.5 nm. Click on the "Run" button ( ). Wait a few seconds, then look at the digital readout on the lock-in. You should see less than a tenth of a millivolt signal at this point. Repeat the steps above to change the PMT voltage to 1000 V and watch the signal at the lock-in. Make sure it does not go off scale; decrease the sensitivity if needed. Bring the PMT voltage up to 1100 V, adjusting the lock-in sensitivity to obtain a good signal (several tenths of a mV) without going off scale. Click the "Cancel" button ( ) to return to the main Experiment screen. Run the optimization Experiment: Click on the "Run" button ( ) to start the short optimization experiment. You will see the spectrum appear in the Display window as the scan progresses. You should see a peak at the pump wavelength. When the scan is complete you may be asked about saving files. To be safe, save all files when asked. Your scan will then appear in the main window of the SynerJY software (Figure \(\PageIndex{10}\)). Double Click on the spectrum so that you can edit it. Then select Pick Peaks from the Analysis menu. To begin the peak analysis leave the default settings in the boxes and select Find Peaks at the bottom of the Peak Analyzer window. Then press Finish . If you do not like the results, alter the Height and Width values until you get a good result. In the example shown in Figure \(\PageIndex{10}\) the major peak is found at approx. 543.5 nm, demonstrating that our system is working properly. Do any final optimizing of the signal (check with your TA) and start the run. Create the Iodine Emission Experiment: Make sure the Increment is set to 0.2 nm. Make sure the Integration Time is still set at 3000 ms. Change the Start wavelength to 510 nm and End wavelength to 800 nm. Click the Real Time Control button (RTC, ) to set the monochromator to the start wavelength. Click the Run button ONCE only ( ) and wait a few seconds (this is equivalent to setting the monochromator to the start wavelength and allowing the detector signal to relax). Then click the Cancel Button ONCE only ( ) to return to Setup screen. Click the "Run" button ( ) from the setup screen to start the full scan. Watch the data collection for the first 5-10 minutes to be sure everything is running correctly. The total scan will take a at least 1.5 hours. If you leave the room, close the door behind you to be sure that stray light does not get into the experiment. At the end of the scan you may want to rescale the horizontal axis to see the spectrum more clearly. Repeat the Pick Peaks procedure you did earlier on the short scan. Likely you will need to adjust the parameters in the Pick Peaks dialog box to get most or all of the peaks. Below is a suggested set of options: Goal: Find Peaks Baseline Mode: Constant, Median Baseline Treatment: Auto subtract baseline Find Peaks: Enable Auto Find, Peak Filtering By Height, Threshhold at ~ 1-2% When you get a good listing of the peaks look at the bottom of the data window where you will see a list of the files associated with these data (Figure \(\PageIndex{15}\)). Click on the Peak_Centers1 worksheet and you will get a table of the peaks found in the spectrum. You can copy the appropriate data from there to an Excel spreedsheet. You should the save the spreadsheet and any screenshots that you need. Shutdown: Before you shut down, be sure you have collected all necessary data in your ELN and in excel. Then Send a copy of the ELN to you and your partners. Turn off the laser using the key. Turn off the power to the PMT and power off the power supply. Turn off the power to the Pre-amp Power Supply Turn off the power to the Lock-in Amplifier. Turn off the Chopper. Close the SynerJY software. Instructions for Data Analysis You may consult your instructors and classmates during data analysis, however all work should be your own. Assign at least 19 peaks in the emission spectrum based on the known pump transition. Remember, this is an emission spectrum. Longer wavelengths (lower energy) correspond to transitions from higher ground state vibrations. Convert the wavelengths of these peaks from nm to wavenumbers. Create a Birge-Sponer plot (see Equation 5.3.5 ) by plotting \(\Delta G_{\upsilon}^{\prime\prime}\) vs \(\left (\upsilon^{\prime\prime}+1 \right )\). From the intercept of the Birge-Sponer plot, calculate \(\omega_{e}^{\prime\prime}\) (using Equation 5.3.5 ). From the slope of the Birge-Sponer plot, calculate \(\chi_{e}^{\prime\prime}\) (using Equation 5.3.5 ). From the Morse Potential (using Equation 5.3.7 ), calculate \(D_{e}^{\prime\prime}\). Estimate the uncertainty in the peak positions using the standard deviations from the least squares analysis and calculate the uncertainties in \(\omega_{e}^{\prime\prime}\), \(\chi_{e}^{\prime\prime}\) and \(D_{e}^{\prime\prime}\). For comparison, calculate the value of \(D_{e}^{\prime\prime}\) using Equation 5.3.9 as follows: From the wavelengths of at least three absorption lines, calculate the corresponding transition energies \(\nu_{\left ( \upsilon^{\prime\prime},\upsilon^{\prime} \right )}\) (where \(\upsilon^{\prime\prime}=0\)) in units of cm –1 . For these same transitions, calculate (\(G_{\left ( \upsilon^{\prime} \right )}-G_{\left ( 0^{\prime\prime} \right )}\)) using Equation 5.3.4 and your values of \(\omega_{e}^{\prime\prime}\), \(\omega_{e}^{\prime}\), \(\chi_{e}^{\prime\prime}\) and \(\chi_{e}^{\prime}\). Now use Equation 5.3.8 to calculate the average value of \(\nu_{el}\) for the three transitions. Finally, use Equation 5.3.9 and your value of \(D_{e}^{\prime}\) from Part I to calculate \(D_{e}^{\prime\prime}\). The value of E(I*) is known to be 7603 cm –1 from atomic spectroscopy. 1 Which value of \(D_{e}^{\prime\prime}\) has the least uncertainty (neglect the error in E(I*))? Calculate the Morse parameter (a") from Equation 5.3.7. To plot the Morse potential wells, you will use the Matlab routine morse . Simply type morse and follow the screen prompts. Separate ground and excited state Morse potential curves will be plotted using your experimental data, and sent to the printer. For now, both \(R_{e}^{\prime\prime}\) and \(R_{e}^{\prime}\) are set at zero. Vibrational quantum levels will be shown for both ground and excited states. The maximum intensity in the absorption spectrum occurs as the excited state vibrational wave functions overlap vertically with the ground state wave functions. A Matlab script, overlap , is used to overlay the two plots and shift the ground and excited state potential wells such that the center of \(\upsilon^{\prime\prime}=0\) overlaps with the left edge of the \(\upsilon^{\prime}=25\). At the Matlab prompt type overlap . The script will animate the overlap of the ground and excited states and the vertical transition. It will also estimate the displacement (\(R_{e}^{\prime} - R_{e}^{\prime\prime}\)) (Ångstroms) between the minima of these two potential wells along the R-axis. Save a copy of this graph (or take a screenshot) for your ELN, and record the displacement. Note The ground and excited state potential wells are overlaid so that the displacement \(R_{e}^{\prime} - R_{e}^{\prime\prime}\) may be measured. However, this gives a misleading impression of the relative positions (energies) of the two potential wells. Clearly the excited state potential well is at much higher energy than the ground state potential well (see Figure 5.3.3 and note the discontinuity in the V(R) axis). A realistic picture is shown in reference [10]. In the Introduction to Matlab Excercises, you plotted the ground and excited state potential wells for iodine using literature values for D e , a, and R e . You will now repeat this assignment using your results. The functional form of the Morse potential may be represented by the equation \[ V\left ( R \right )=T_{e}+D_{e}\left [ 1-e^{-a\left ( R-R_{e} \right )} \right ]^2 \] where T e is the electronic term energy (the energy at the bottom of the potential well). \(T_{e}^{\prime\prime}\) and \(T_{e}^{\prime}\) represent the electronic term energies of the ground and excited states respectively. If we set \(T_{e}^{\prime\prime}=0\), then \(T_{e}^{\prime}=\nu_{el}\). Given that \(R_{e}^{\prime\prime}=2.66\) Å (obtained from IR spectroscopy), calculate \(R_{e}^{\prime}\) using your value of \(R_{e}^{\prime} - R_{e}^{\prime\prime}\). You will now use your values of \(T_{e}^{\prime\prime}\) (zero), \(T_{e}^{\prime}\) (\(=\nu_{el}\)), \(D_{e}^{\prime\prime}\), \(D_{e}^{\prime}\), a", a', \(R_{e}^{\prime\prime}\) and \(R_{e}^{\prime}\) to replot the Morse potential wells for the ground and excited states of iodine on the same graph. Return to your Matlab command window, and as instructed, press any key to continue. Enter your calculated \(T_{e}^{\prime}\). The two Morse potentials will be plotted on one graph and the vibrational quantum levels will be shown for both ground and excited states. Save a copy of this graph (or take a screenshot) for your ELN. Compare all of your results with literature values (See, for example, reference. 1, p. 430-432). Discussion Questions Include discussion of the following in your ELN. Why are the vibrational spacings different in the ground and excited states? Why is the shape of the Morse potential different in the ground and excited states? How is it possible to observe emission peaks at higher energies than the pump energy?
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Courses/Maryville_College/Essential_Chemistry_for_Poisons_Potions_and_Pharmaceuticals/09%3A_Poisons/9.06%3A_Carcinogens_and_Teratogens
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Learning Objectives Explain the process of cancer development. List the factors that may increase and lower the risk of cancer. Define teratogens and describe the birth defects that result from them. A carcinogen is any agent that directly increases the incidence of cancer. Most, but not all carcinogens are mutagens. Carcinogens that do not directly damage DNA include substances that accelerate cell division, thereby leaving less opportunity for cell to repair induced mutations, or errors in replication. Carcinogens that act as mutagens may be biological, physical, or chemical in nature, although the term is most often used in relation to chemical substances. Defining Cancer Cancer is actually a group of more than 100 diseases, all of which involve abnormal cell growth with the potential to invade or spread to other parts of the body. In general terms, cancer occurs when the cell cycle is no longer regulated due to DNA damage. The number of potential underlying causes of this DNA damage is great, so there are many different risk factors for cancer. Any cells that become cancerous divide more quickly than normal cells. They may form a mass of abnormal cells called a tumor. The rapidly dividing cells take up nutrients and space, damaging the normal cells around them. If the cancer cells spread to other parts of the body, they invade and damage other tissues and organs. They may eventually lead to death. By far, the most common of the 100-plus types of human cancer is basal cell carcinoma, the type of skin cancer Bernie Sanders had removed in 2016. Basal cell carcinoma makes up 40 percent of all new cancers each year in the United States. Other common types of cancer include lung, colorectal, prostate (in males), and breast (in females) cancers. These cancers are not as common as skin cancer, but they cause the majority of cancer deaths. How Cancer Develops Carcinogenesis, also called oncogenesis or tumorigenesis, is the formation of a cancer, whereby normal cells are transformed into cancer cells. The transformation of a normal cell into a cancer cell is a multi-step process that involves initiation, promotion, progression and finally malignancy (see Figure \(\PageIndex{2}\)). This process takes years and starts with a single cell in which the right genes are mutated so the cell does not appropriately die and begins to proliferate abnormally. Then, additional mutations occur that select for more rapidly growing cells within this population leading to a tumor with rapid growth and malignancy. By the time the cells are cancerous, proto-oncogenes have been activated and tumor suppressor genes inactivated. Even within the same tumor type, like colon cancer, the specific genes mutated can vary from person to person making cancer a unique disease for each individual. How Cancer Spreads Once a normal cell transforms into a cancer cell and starts dividing out of control, cancer cells can spread from the original site (called the primary tumor) to other tissues. This can occur in three different ways. One way is local spread, in which aggressively dividing cancer cells directly invade nearby tissues. Another way involves the lymphatic system. Cancer cells can spread to regional lymph nodes through lymph vessels that pass by the primary tumor. The third way cancer cells can spread is through the blood to distant sites. This is called metastasis, and the new cancers that form are called metastases. Although the blood can carry cancer cells to tissues everywhere in the body, cancer cells generally grow only in certain sites (Figure \(\PageIndex{3}\)). Different types of cancers tend to metastasize to particular organs. The most common places for metastases to occur are the brain, lungs, bones, and liver. Almost all cancers can metastasize, especially during the late stages of the disease. Cancer that has metastasized generally has the worst prognosis and is associated with most cancer deaths. Chemical Carcinogens Chemical carcinogens (Table \(\PageIndex{1}\)) can be either natural or synthetic compounds that, based on animal feeding trials or epidemiological (i.e. human population) studies, increase the incidence of cancer. The definition of a chemical as a carcinogen is problematic for several reasons. Some chemicals become carcinogenic only after they are metabolized into another compound in the body; not all species or individuals may metabolize chemicals in the same way. Also, the carcinogenic properties of a compound are usually dependent on its dose. It can be difficult to define a relevant dose for both lab animals and humans. Nevertheless, when a correlation between cancer incidence and chemical exposure is observed, it is usually possible to find ways to reduce exposure to that chemical. Table \(\PageIndex{1}\): Some Classes of Chemical Carcinogens. Class Examples and/or Sources PAHs (polycyclic aromatic hydrocarbons) benzo[a]pyrene and several other components of the smoke of cigarettes, wood, and fossil fuels Aromatic amines compounds formed in food when meat (including fish, poultry) are cooked at high temperature Nitrosamines and nitrosamides found in tobacco and in some smoked meat and fish Azo dyes various dyes and pigments used in textiles, leather, paints. Carbamates ethyl carbamate (urethane) found in some distilled beverages and fermented foods Halogenated compounds e.g. pentachlorophenol used in some wood preservatives and pesticides. Inorganic compounds asbestos; may induce chronic inflammation and reactive oxygen species Miscellaneous compounds e.g. alkylating agents, phenolics WEB LINK A detailed list of occupational carcinogens can be found on the link below https://bio.libretexts.org/Bookshelves/Human_Biology/Book:_Human_Biology_(Wakim_and_Grewal)/21:_Disease/21.7:_Cancer The carcinogens implicated as the main causative agents of the four most common cancers worldwide are given in the table below. These four cancers are lung, breast, colon, and stomach cancers. Together they account for about 41% of worldwide cancer incidence and 42% of cancer deaths. Table \(\PageIndex{2}\): Major Carcinogens Implicated In The Four Most Common Cancers Worldwide. Type of Cancer Carcinogen Lung Cancer Tobacco smoke Breast Cancer Estrogen Colon Cancer Tobacco smoke and bile acids: deoxycholic acid (DCA) or lithocholic acid (LCA) Stomach Cancer Heliobacter pylori WEB LINK Note: A more detailed description for each major carcinogen listed in Table \(\PageIndex{2}\) can be found on the link below https://en.Wikipedia.org/wiki/Carcin...cers_worldwide Anticarcinogens Dietary fiber and calorie restriction are two anti-carcinogen or anti-promoters that decrease the risk of tumor formation. Dietary fiber is both and is inversely associated with cancer, particularly colon cancer. So the more fiber you eat, the less risk you have of developing colon cancer. One mechanism by which fiber acts is hastening bile acid excretion. Fiber also increases the rate of passage of materials through the colon resulting in decreased production and exposure of the colon to cancer-causing agents, ie dilutes the concentration of carcinogens. Animal studies have shown that restricting caloric intake by 30% reduces tumor growth and increases life span. The mechanism is not known but may be due to less oxidation thus damage to DNA. Antioxidants can help block the action of initiators or promoters if their mode of action is to damage DNA by oxidation. Vitamin A, C, E, beta-carotene, and selenium are antioxidant nutrients. Some work locally, like vitamin E in the colon, while other work more globally like selenium and vitamin C. Vitamin A appears to work by keeping cells differentiated which slows the growth rate. Other compounds in food, particularly fruits and vegetables, have been shown to slow tumor formation. Cruciferous vegetables (eg broccoli, cauliflower, cabbage, and Brussel sprouts to name a few) are rich in nutrients, fiber, glucosinolates which are sulfur-containing chemicals, indoles, and isothiocyanates. Animal studies have found these substances inhibit the development of cancer in several organs in rats and mice (Hecht SS. Inhibition of carcinogenesis by isothiocyanates. Drug Metabolism Reviews 2000;32(3-4):395-411; Murillo G, Mehta RG. Cruciferous vegetables and cancer prevention. Nutrition and Cancer 2001;41(1-2):17-28). Indoles and isothiocyanates help protect cells from DNA damage; help inactivate carcinogens; have antiviral and antibacterial effects; have anti-inflammatory effects; induce cell death (apoptosis); and inhibit tumor blood vessel formation (angiogenesis) and tumor cell migration (needed for metastasis) (National Cancer Institute, Cruciferous Vegetables and Cancer Prevention, 2012, https://www.cancer.gov/about-cancer/...les-fact-sheet ). Studies in humans, however, have shown mixed results. Test for Carcinogens In genetics, a mutagen is a physical or chemical agent that permanently changes genetic material, usually DNA, in an organism and thus increases the frequency of mutations above the natural background level. As many mutations can cause cancer, such mutagens are therefore carcinogens, although not all necessarily are. Many different systems for detecting mutagen have been developed. Animal systems may more accurately reflect the metabolism of human, however, they are expensive and time-consuming (may take around three years to complete), they are therefore not used as a first screen for mutagenicity or carcinogenicity. The Ames test is a widely employed method that uses bacteria to test whether a given chemical can cause mutations in the DNA of the test organism. More formally, it is a biological assay to assess the mutagenic potential of chemical compounds. A positive test indicates that the chemical is mutagenic and therefore may act as a carcinogen, because cancer is often linked to mutation. The test serves as a quick and convenient assay to estimate the carcinogenic potential of a compound because standard carcinogen assays on mice and rats are time-consuming (taking two to three years to complete) and expensive. However, false-positives and false-negatives are known. Other tests, similar to the Ames tests have also been developed using yeast and other bacteria. Drosophila, plants, and cell cultures have been used in various test assays for mutagenecity of chemicals. In animal test systems, rodents are usually used in animal test. The chemicals under test are usually administered in the food and in the drinking water, but sometimes by dermal application, by gavage, or by inhalation, and carried out over the major part of the life span for rodents. Transgenic mouse assay using a mouse strain infected with a viral shuttle vector is another method for testing mutagens. Animals are first treated with suspected mutagen, the mouse DNA is then isolated and the phage segment recovered and used to infect E. coli . Did You Know? Video \(\PageIndex{1}\) Risk factors for cancer. WEB LINKS - American Cancer Society and NIH:MedlinePlus & National Cancer Institute Risk factors for Cancer https://www.cancer.gov/about-cancer/causes-prevention/patient-prevention-overview-pdq#_199 Common Cancer Myths and Misconceptions https://www.cancer.gov/about-cancer/causes-prevention/risk/myths List of Probable Carcinogens https://www.cancer.org/cancer/cancer-causes/general-info/known-and-probable-human-carcinogens.html?sitearea=who Cancer Prevention: Take Charge of Your Lifestyle https://medlineplus.gov/ency/patientinstructions/000825.htm Cruciferous Vegetables and Cancer Prevention https://www.cancer.gov/about-cancer/causes-prevention/risk/diet/cruciferous-vegetables-fact-sheet Birth Defects:Teratogens A teratogen is a compound that permanently deforms the function or structure of a developing embryo or fetus in utero. In general, the degree of teratogenicity depends on: The potency of the drug as a mutagen. The susceptibility of the fetus to teratogenesis. The dose of the teratogen. The duration of teratogen exposure. The time of exposure. The degree of transfer from maternal to fetal circulation. The global average of all live births complicated by malformation is 6% (Environmental Health Perspectives, (NIH), October 2009). The majority of these complications are due to unknown factors. The vast majority of recognized etiologies are genetic, with only 10% being attributed to environmental etiologies such as maternal health, infection, and toxicants. In general, the central nervous and skeletal systems are the most affected. Thalidomide (a sedative previously marketed in Europe to prevent morning sickness) see Figure \(\PageIndex{4}\)is a classic teratogen that caused limb defects in babies born to women who took this drug in the 1960s (see Figure \(\PageIndex{5}\)). Thalidomide was first marketed in 1957 in West Germany, where it was available over the counter. [5] [6] When first released, thalidomide was promoted for anxiety, trouble sleeping, "tension", and morning sickness. [6] [7] While initially deemed to be safe in pregnancy, concerns regarding birth defects were noted in 1961 and the medication was removed from the market in Europe that year. [6] [5] The total number of people affected by use during pregnancy is estimated at 10,000, of which about 40% died around the time of birth. [6] [3] Those who survived had limb, eye, urinary tract, and heart problems. [5] Its initial entry into the US market was prevented by Frances Kelsey at the FDA. [7] The birth defects of thalidomide led to the development of greater drug regulation and monitoring in many countries. in 2006 the U.S. Food and Drug Administration granted accelerated approval for thalidomide in combination with dexamethasone for the treatment of newly diagnosed multiple myeloma patients. Women may encounter a number of other teratogens. Smoking is most likely to cause growth retardation, but has also been implicated in the prelabor rupture of the membranes, preterm labor, abruption of the placenta, spontaneous abortion, perinatal morbidity and mortality, and sudden infant death syndrome. Smoking may exert its effects through competitive binding of carbon monoxide with hemoglobin and/or through the various other components found in cigarettes that cause adverse biological effects. Alcohol use in pregnancy may result in fetal alcohol syndrome (FAS) see Figure \(\PageIndex{6}\), which occurs in approximately 1% of all births. Children with FAS present with a flattened and thin upper lip, small palpebral fissures, epicanthal folds, flattened nasal bridge, and a short nose. They may also exhibit microcephaly, mental retardation, and have learning disabilities. It is not clear if there is any safe amount of alcohol consumption in pregnancy. Cocaine generally produces growth restriction, preterm delivery, microcephaly, spontaneous abortion, placental abruption, limb anomalies, and central nervous system abnormalities. Cocaine appear to exert a number of its effects through peripheral vasoconstriction that leads to fetal hypoxia. Women with indications for warfarin therapy should either abstain from pregnancy or switch to low molecular weight heparins. Warfarin typically produces mental retardation, growth restriction, nasal hypoplasia, and opthalmic abnormalities. Angiotensin converting enzyme (ACE) inhibitors will cause fetal renal failure and oligohydramnios that lead to pulmonary hypoplasia and limb contracture. Fetal cranial bone abnormalities are also common. Isotretinoin (Accutane) , used to treat acne, may cause cardiac, oral, otological, thymic, and central nervous system abnormalities. In one quarter of cases, it causes mental retardation. Other teratogenic substance classes and conditions include Various prescription drugs and nutrient deficiencies (e.g., insufficient folic acid). Chemical compounds such as methyl iodide (used in pesticides) and bisphenol A (used in plastics) are suspected teratogens. Summary A carcinogen is any agent that directly increases the incidence of cancer. Cancer is caused by changes to the DNA which could be inherited or acquired. Various tests have been developed to test for carcinogens. The American Cancer Society provides an extensive list of probable chemical carcinogens including those found in processed foods and beverages, various household products, pesticides etc. The effects of a teratogen on the fetus depend on several factors: the potency of the teratogen, the susceptibility of the fetus to the teratogen, the dose and duration of teratogen exposure, the degree of transfer from maternal to fetal circulation, and when during development the exposure occurs. Approximately 10% of congenital malformations are attributed to environmental factors, and 20% are due to genetic or hereditary factors. The rest have unknown causes or are due to a mix of different factors. Cigarette components, alcohol, cocaine, warfarin, ACE inhibitors, and Accutane are all teratogens that affect fetal development.
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Courses/Saint_Marys_College_Notre_Dame_IN/CHEM_431%3A_Inorganic_Chemistry_(Haas)/CHEM_431_Readings/04%3A_Molecular_Orbitals/4.03%3A_Heteronuclear_Diatomic_Molecules/4.3.01%3A_Polar_bonds
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Molecular orbital diagrams for heteronuclear diatomic molecules 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. Carbon monoxide MO diagram Carbon monoxide is an example of a heteronuclear diatomic molecule where both atoms are second-row elements. The valence molecular orbitals in both atoms are the \(2s\) and \(2p\) orbitals. The molecular orbital diagram for carbon monoxide (Figure \(\PageIndex{1}\)) is constructed similarly to how you would construct dicarbon or dioxygen, except that the oxygen orbitals have a lower potential energy than analogous carbon orbitals. The labeling of molecular orbitals in this diagram follows a convention by which orbitals are given serial labels according to type of orbital (\(\sigma\), \(\pi\), etc). The lowest energy orbitals of any type are assigned a value of 1 and higher energy orbitals of the same type are assigned by increasing intervals (..2, 3, 4...). The orbital labeling system described previously is inappropriate for heteronuclear diatomic molecules that cannot be assigned \(g\) and \(u\) subscripts. A consequence of unequal atomic orbital energy levels is that orbital mixing is significant. Notice the order of the molecular orbitals labeled \(1\pi\) and \(3\sigma\) in Figure \(\PageIndex{1}\). This is a similar order of \(\pi\) and \(\sigma\) orbitals as we saw in the case of the \(\sigma_g\) and \(\pi_u\) orbitals of \(N_2\) and lighter diatomics of the second period. Because the oxygen \(2p_z\) orbital is close in energy to both the carbon \(2p_z\) and carbon \(2s\), these three orbitals will have significant interaction (mixing). The result is an increase in the energy of the \(3\sigma\) orbital and a decrease in energy of the \(2\sigma^*\) orbital, resulting in the diagram shown in Figure \(\PageIndex{1}\). In the case of carbon monoxide (Figure \(\PageIndex{1}\)), atomic orbitals contribute unequally to each molecular orbital. For example, because the \(2s\) orbital of oxygen is very close in energy to the \(2\sigma\) moelcular orbital, it contributes to that molecular orbital more than the \(2s\) orbital from carbon. Notice the shape of this \(2\sigma\) orbital and how it is unevenly distributed over the two atoms; it is more heavily distributed on the oxygen because it is most like the oxygen \(2s\). This is in-line with the assumption that electron density is distributed more on oxygen because it is a more electronegative than carbon. Likewise, the \(1\pi\) orbitals are unevenly distributed, with more distribution close to the oxygen. Exercise \(\PageIndex{1}\) Examine the shape of the \(3\sigma\) orbital of carbon monoxide in Figure (\(\PageIndex{1}\)). Describe what ways this shape is different from the shape of the \(\sigma_g\) orbitals from second period homonuclear diatomic molecules ( see Fig. 5.2.1.1 ). Rationalize these differences. Both orbitals are re-created below for convenience. Answer The 3\(\sigma\) orbital is like the \(\sigma_g\) in that it has three lobes and two nodes distributed along the internuclear bond. They are different in their distribution. The two external lobes of \(\sigma_g\) are evenly distributed because the are an equal combination of two \(p_z\) orbitals (one from each atom). The \(3\sigma\) orbital is more heavily distributed toward the carbon atom, the less electronegative atom, than toward the oxygen. The unequal distribution of \(3\sigma\) is apparent in the unequal sizes of its exterior lobes, and the uneven shape of the interior lobe. The heavier distribution within the exterior lobe on carbon caused by the mixing of the carbon \(2s\) orbital with carbon and oxygen \(2p_z\) orbitals. The uneven shape of the interior lobe, where it leans toward oxygen, is best explained by the fact that the \(3\sigma\) orbital is closer in energy to the oxygen \(2p_z\) than the carbon \(2p_z\). Hydrogen fluoride MO diagram Hydrogen fluoride is an example of a heteronuclear diatomic molecule in which the two atoms are from different periods. In this case, the valence orbital of H is \(1s\) while those of F are \(2s\) and \(2p\). The molecular orbital diagram for HF is shown in Figure \(\PageIndex{2}\). Three of these orbitals have compatible symmetry for mixing; these are the hydrogen \(1s\), fluorine \(2s\), and fluorine \(2p\). However, the extent to which they will interact depends on their relative energies. Fluorine is more electronegative than H, and the fluorine atom has a higher first ionization energy than does hydrogen. From these trends, we can expect that the fluorine valence orbitals are lower in energy than that of hydrogen. From Table 5.3.1 , we find that the 1s orbital of H (-13.6 eV) is higher in energy than both fluorine orbitals (-18.7 and -40.2 eV, respectively for \(2p\) and \(2s\)). The energies of hydrogen \(1s\) and fluorine \(2p\) are a good match for combination, however, the \(2s\) orbital is much too different to create a productive interaction. Therefore, we expect that the fluorine \(2s\) will create a non-bonding molecular orbital, while the \(1s\) and \(2p_z\) orbital combine to make \(\sigma\) bonding and \(\sigma^*\) antibonding molecular orbitals. The remaining \(2p_x\) and \(2p_y\) orbitals do not have compatible symmetry for bonding with hydrogen, and they will form non-bonding \(\pi\) molecular orbitals. The non-bonding orbitals will have similar energy and character as their component atomic orbitals. Chemical reactions take place at the HOMO and LUMO orbitals Knowledge of molecular orbital diagrams, and the shapes of molecular orbitals, can be used to accurately explain and predict chemical reactivity. Chemical reactions take place using the highest occupied molecular orbitals (HOMO) of a nucleophile or Lewis base, and the lowest unoccupied molecular orbital (LUMO) of an electrophile or Lewis acid. Lewis bases react using electrons in the HOMO, while Lewis acids react using the empty LUMO. Example: Reactivity of CO with metal ions CO is an excellent ligand for many metal ions. In fact, the strong affinity between CO and the heme iron (Fe) ions in hemoglobin can explain the mechanism of carbon monoxide poisoning. When CO binds in place of \(O_2\) to hemoglobin, that hemoglobin can no longer carry \(O_2\) to tissue cells. CO binding to hemoglobin is strong and practically irreversible. When CO binds to metal ions, it does so through the carbon atom. This is contrary to expectations based on the Lewis structure and the known bond polarity, where electron density is polarized toward oxygen. The distribution of the electron density of the HOMO of CO can explain this observation! Exercise \(\PageIndex{2}\) Refer to the MO diagram for CO. Identify the HOMO and explain why CO bonds to metal ions through the carbon atom rather than through the oxygen atom. Answer Add texts here. Do not delete this text first. Curated or created by Kathryn Haas
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Courses/Millersville_University/CHEM_341-_Physical_Chemistry_I/08%3A_Phase_Equilibrium/8.07%3A_Liquid-Vapor_Systems_-_Raoults_Law
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Liquids tend to be volatile, and as such will enter the vapor phase when the temperature is increased to a high enough value (provided they do not decompose first!) A volatile liquid is one that has an appreciable vapor pressure at the specified temperature. An ideal mixture continuing at least one volatile liquid can be described using Raoult’s Law. Raoult’s Law Raoult’s law can be used to predict the total vapor pressure above a mixture of two volatile liquids. As it turns out, the composition of the vapor will be different than that of the two liquids, with the more volatile compound having a larger mole fraction in the vapor phase than in the liquid phase. This is summarized in the following theoretical diagram for an ideal mixture of two compounds, one having a pure vapor pressure of \(p_B^o = 450\, Torr\) and the other having a pure vapor pressure of \(p_B^o = 350\, Torr\). In Figure \(\PageIndex{1}\), the liquid phase is represented at the top of the graph where the pressure is higher. Oftentimes, it is desirable to depict the phase diagram at a single pressure so that temperature and composition are the variables included in the graphical representation. In such a diagram, the vapor, which exists at higher temperatures) is indicated at the top of the diagram, while the liquid is at the bottom. A typical temperature vs. composition diagram is depicted in Figure \(\PageIndex{2}\) for an ideal mixture of two volatile liquids. In this diagram, \(T_A^o\) and \(T_B^o\) represent the boiling points of pure compounds \(A\) and \(B\). If a system having the composition indicated by \(\chi_B^c\) has its temperature increased to that indicated by point c, The system will consist of two phases, a liquid phase, with a composition indicated by \(\chi_B^d\) and a vapor phase indicated with a composition indicated by \(\chi_B^b\). The relative amounts of material in each phase can be described by the lever rule, as described previously. Further, if the vapor with composition \(\chi_B^b\) is condensed (the temperature is lowered to that indicated by point b') and revaporized, the new vapor will have the composition consistent with \(\chi_B^{a}\). This demonstrates how the more volatile liquid (the one with the lower boiling temperature, which is A in the case of the above diagram) can be purified from the mixture by collecting and re-evaporating fractions of the vapor. If the liquid was the desired product, one would collect fractions of the residual liquid to achieve the desired result. This process is known as distillation.
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Courses/University_of_Alberta_Augustana_Campus/AUCHE_252_-_Organic_Chemistry_II/07%3A_Conjugated_Pi_Systems/7.07%3A_Aromatic_Ions
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Aromatic Ions As previously, noted a ring must be fully conjugated to have the potential to be aromatic. This means that every atom in the ring must have a p orbital which can overlap with adjacent p orbitals. Until now the atoms providing the p orbitals have been neutral sp 2 hybridized carbons, however it is also possible for sp 2 hybridized carbons to have a charge. There are several examples of cationic and anionic compounds with unexpected stabilities that suggest that they are aromatic . It is important to understand how charged carbons in these compounds will affect the determination of aromaticity. There are two main situations which need to be considered: The conjugation of a carbocation and the conjugation of a carbanion. A carbocation carbon is surrounded by three electron groups giving it sp 2 hybridization. The remaining unhybridized p orbital holds the carbocation's positive charge and is vacant of pi electrons. Although a carbocation is capable of extending conjugation it does not add to the compound's pi electron count. A carbanion carbon is surrounded by four electron groups and would normally be sp 3 hybridized. However, to obtain the stabilizing effects of conjugation, carbanion carbons can becomes sp 2 hybridized putting the set of lone pair electrons into the unhybridized p orbital. Because the carbanion's p orbital contains two electrons in the form of a set of lone pair electrons, it increases a compound's pi electron count by 2. Cyclopentadiene Ion One of the most well know examples of an aromatic ion is the 1,3-cyclopentadiene ion. 1,3-Cyclopentadiene is nonaromatic due to the presence of an intervening sp 3 hybridized -CH 2 - carbon atom which prevents pi electrons from delocalizing about the entire ring. Also, it only has 4 pi electrons which does not follow Hückel's 4n + 2 rule. However, if a proton is removed form the CH 2 group to form the cyclopentadienyl anion, the carbon atom becomes sp 2 hybridized and the two electrons of the resulting lone pair occupy the newly produced p orbital. This increases the number of pi electrons in the cyclopentadienyl anion to 6 which follows the 4n +2 rule. Moreover, this new p orbital overlaps with the p orbitals already present allowing for cyclic delocalization of pi electrons about the entire ring. The lone pair electrons and the negative charge are conjugated about the entire ring making each carbon in the cyclopentadienyl anion equivalent with 1/5 the negative charge. The resonance hybrid can be shown by drawing a series of five resonance form. Also, the electrostatic potential map of the cyclopentadienyl anion show the negative charge, seen in red/yellow is distributed over the entire ring. The fact that the red/yellow color is held in the center of the ring indicates that the extra electrons of the ion are involved in the aromatic p-electron system. The reason why the 4n +2 rule still works for a 5 p orbital ring system can be seen by looking the molecular orbital diagram of the cyclopentadienyl anion. As discussed in Section 15.3 , the molecular orbital diagram of a 5 p orbital system is made up of 3 bonding MO's and 2 antibonding MO's. The 6 pi electrons gained by forming an anion is enough to completely fill the bonding MO's in the diagram giving the cyclopentadienyl anion aromaticity. One of the effects of the aromaticity of the cyclopentadienyl anion is that the acidity of 1,3-cyclopentadiene is unusually strong. As previously discussed, stabilizing the conjugate base increases the acidity of the corresponding acid. In this case, the conjugate base of 1,3-cyclopentadiene, the cyclopentadienyl anion, is stabilized through aromaticity. This makes 1,3-cyclopentadiene one of the most acidic hydrocarbons known with a pK a of 16. This is almost 10 30 times more acidic than cyclopentane. Because of its acidity, cyclopentadiene can be deprotonated by moderately strong bases such as NaOH. Tropylium ion The molecule 1,3,5-cycloheptatriene has six pi electrons but is nonaromatic due the presence of an sp 3 hybridized -CH 2 - group which prevents cyclic delocalization. When 1,3,5-cycloheptatriene is reacted with a reagent that can remove a hydride ion (H: - ), the 1,3,5-cycloheptatrienyl cation, which is commonly known as the tropylium cation, is formed with unexpected ease. Despite the presence of an electron deficient carbocation, the tropylium cation, is unusually stable and can be isolated as a salt. Removal of a hydride from the -CH 2 - group in 1,3,5-cycloheptatriene creates an sp 2 hybridized carbocation with a vacant p orbital. The new p orbital allows for cyclic conjugation to occur among the seven p orbitals in the tropylium cation. The vacant p orbital does not change the pi electron count so the tropylium cation has 6 pi electrons which obeys the 4n + 2 rule for aromaticity. The molecular orbital diagram for the 7 p orbitals in the tropylium cation has three bonding MO's and 4 antibonding MO's. The tropylium cation's 6 pi electrons completely fill the bonding molecular orbitals which is consistent with the tropylium cation being aromatic and therefore unusually stable. As predicted with aromaticity, the positive charge is completely conjugated about the entire ring giving each carbon +1/7 charge. The true resonance hybrid can be depicted by drawing a series of seven resonance form. Also, the electrostatic potential map of the tropylium cation shows the positive charge is evenly distributed over the entire ring. The equivalency of the carbons in the seven membered ring is experimentally supported by an 1 H NMR spectrum of the tropylium cation which contains one peak showing that all seven protons are equivalent. Antiaromatic Ions In a similar fashion, cyclically conjugated ions with 4n pi electrons can be predicted to be antiaromatic and therefore highly unstable. An excellent example is the cyclopentadienyl cation. Above, the cyclopentadienyl anion was shown to be aromatic, however, the formation of a carbocation produces a different result. Although the vacant p orbital provided by the carbocation allows for cyclic conjugation to occur the compound only has 4 pi electrons. After placing these 4 pi electrons into the molecular orbital for a cyclic 5 p orbital species the bonding molecular orbitals remain unfilled. Two of the pi electrons are unpaired in degenerate molecular orbitals creating the highly unstable triplet state. As predicted by Hückel's Rule, the cyclopentadienyl cation has 4n pi electrons and should be anitaromatic and very unstable. Although there is some discussion as to whether the cyclopentadienyl cation is truly antiaromatic, there is experimental evidence which shows it is usually unstable. In particular, the compound 2,4-cyclopentadien-1-ol is usually resistant to S N 1 reactions with acid halides. The carbocation intermediate created during the mechanism of this reaction would be expected to easily form due to the resonance stabilization provided by the two double bonds. Because the carbocation intermediate is antiaromatic it does not form so the reaction does not occur.
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Courses/Providence_College/Organic_Chemistry_I/05%3A_Analytical_Methods_for_Structure_Elucidation/5.05%3A_Mass_Spectrometry
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A final technique in structure elucidation is mass spectrometry, which allows us to measure the mass of an ion via a mass-to-charge (m/ z ) ratio. The basic schematic of a mass spectrometer is shown in Figure. 5.5.1: There are four main components: Ionization : After injection of a sample and vaporization via the sample inlet, the sample is ionized by one of several mechanisms. Acceleration : The vaporized beam of ions is accelerated so that they have the same energy Deflection : In the mass analyzer (an electromagnet), the ions are sorted according to their size (m/ z ). Detection : Ions finally strike a plate and generate a current The type of mass spectrometer is normally determined by the mechanism of ionization: Electron Impact (EI) - high energy electrons (6700 kJ/mol) are bombarded at a neutral molecule. The resulting high energy radical cation can break apart into smaller fragments (this is called fragmentation - the molecular ion (M + ) breaks down into small daughter peaks ). Because electron impact is a high energy technique, one often does not see the molecular ion M + , only daughter peaks due to rapid fragmentation. Chemical Ionization (CI) - a cloud of high energy methane (CH 4 ) gas and a source of protons (H + ions) transfer a proton to a small molecule to generate a molecular ion [M+H] + (observed in the mass spectrum at a m/ z ratio of M+1). Electrospray ionization (ESI) - the most common technique droplets of a small molecule in solvent are created via an aerosol from the sample inlet. Normally, the solvent contains a proton source which protonates basic sites on the small molecule. As the solvent shell evaporates, a charged small molecule results. Matrix-Assisted Laser Desorption Ionization (MALDI) - a sample is mixed with a chemical matrix and deposited on a plate to dry and crystallize. In a vacuum chamber, the plate is pulsed with a laser that ionizes and desorbs ions, which are now in the gaseous state, usually [M+H] + or [M+Na] + . Isotopic patterns The natural abundance of certain isotopes can be useful in structure elucidation. For example, we can tell if a molecular ion has a chlorine or bromine atom in it just by loking at the isotopic pattern of the molecule ion or daughter peaks. To do this, we need a high resolution mass spectrum (HRMS). HRMS is necessary for accurate mass determination, and we use the following exact masses: 12 C = 12.000000 amu 1 H = 1.007825 amu 16 O = 15.99491 14 N = 14.00307 For calculating the mass-to-charge ratio of an ion, just add up all of the masses of the atoms in the ion using these values. Since these values are extremely accurate, only a handful (or maybe only one) of molecules can have that exact mass, down to an accuracy of less than 10 ppm. For example, we can distinguish each of the molecules below, which have a molecular weight of 82 g/mol, via HRMS (using EI). Getting back to isotopic patterns, we see that for the following molecule, there are two molecular ions because the natural abundance of chlorine is 35 Cl (75%) and 37 Cl (25%). Thus, is you have a sample of the molecule, 75% of it will have 35 Cl atoms and 25% of it will have 37 Cl atoms. You will see both bypes of species in a mass spectrum in a 3:1 ratio separated by two mass units. Other typical patterns in mass spectra are when bromine atoms appear. Here, one observes peaks that are two mass units apart and equal in intensity. This is because 79 Br and 81 Br are naturally abundant in a 1:1 ratio. Let's see what would happen if there were multiple halogen atoms, as in CH 2 Cl 2 (MW = 85 g/mol). Peak 1: m/ z 84 = CH 2 35 Cl 35 Cl = (0.75)*(0.75) = 56.25 = 56.25% Peak 2: m/ z 86 = 37.5% (second chlorine is 37 Cl) = CH 2 35 Cl 37 Cl = (0.75)*(0.25) = 18.75 = 18.75% (first chlorine is 37 Cl) = CH 2 37 Cl 35 Cl = (0.75)*(0.25) = 18.75 = 18.75% Peak 3: m/ z 88 = CH 2 37 Cl 37 Cl = (0.25)*(0.25) = 6.25 = 6.25% Biomedical Spotlight The United States Anti-Doping Agency (USADA) is the nation's premier non-profit organization charged with upholding the ideal of fair play by ensuring that athletic competitions are free from performance-enhancing drugs. In 2003, USADA was sent a syringe containing an unknown substance from an anonymous coach who claimed that there had been rampant use of a designer steroid among track and field athletes. In what then came widely known as the BALCO scandal, this same substance was implicated in Major League Baseball players, including Barry Bonds. A sample from the syringe was tested for all 70 steroids that are banned by USADA, but no matches were found. However, the mass spectrum had remarkable similarity to a known anabolic agent, gestrinone. In fact, the daughter peaks from electron impact ionization mass spectrometry (EI-MS) for a sample of the designer steroid completely matched that of gestrinone. However, the molecular ion for the designer steroid differed by 4 amu (EI-MS shown below). By using chemical intuition, anti-doping scientists were able to conclude that gestrinone had been partially and selectively hydrogenated to produce an ethyl group from the terminal alkyne. This new molecule had performance-enhancing capabilities, but had previously escaped detection. The new molecule, now known as tetrahydrogestrinone (THG), was quickly banned by the World Anti-Doping Agency.
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Courses/CSU_Chico/CSU_Chico%3A_CHEM_451_-_Biochemistry_I/CHEM_451_Test/03%3A_Lipid_Structure/3.1%3A_Lipid_Structure/Fatty_Acids
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Fatty acids can be saturated (contain no double bonds in the acyl chain), or unsaturated (with either one -monounsaturated - or multiple - polyunsaturated - double bond(s)) . The table below gives the names, in a variety of formats, of common fatty acids. Table: Names and structures of the most common fatty acids Symbol common name systematic name structure mp(C) 12:0 Lauric acid dodecanoic acid CH3(CH2)10COOH 44.2 14:0 Myristic acid tetradecanoic acid CH3(CH2)12COOH 52.0 16:0 Palmitic acid Hexadecanoic acid CH3(CH2)14COOH 63.1 18:0 Stearic acid Octadecanoic acid CH3(CH2)16COOH 69.6 20:0 Arachidic aicd Eicosanoic acid CH3(CH2)18COOH 75.4 Symbol common name systematic name structure mp(C) 16:1Δ9 Palmitoleic acid Hexadecenoic acid CH3(CH2)5CH=CH-(CH2)7COOH -0.5 18:1Δ9 Oleic acid 9-Octadecenoic acid CH3(CH2)7CH=CH-(CH2)7COOH 13.4 18:2Δ9,12 Linoleic acid 9,12 -Octadecadienoic acid CH3(CH2)4(CH=CHCH2)2(CH2)6COOH -9.0 18:3Δ9,12,15 α-Linolenic acid 9,12,15 -Octadecatrienoic acid CH3CH2(CH=CHCH2)3(CH2)6COOH -17.0 20:4Δ5,8,11,14 arachidonic acid 5,8,11,14- Eicosatetraenoic acid CH3(CH2)4(CH=CHCH2)4(CH2)2COOH -49.0 20:5Δ5,8,11,14,17 EPA 5,8,11,14,17-Eicosapentaenoic- acid CH3CH2(CH=CHCH2)5(CH2)2COOH -54.0 22:6 Δ4,7,10,13,16,19 DHA Docosohexaenoic acid 22:6w3 NaN FAT <16:0 16:1 18:0 18:1 18:2 18:3 20:0 22:1 22:2 . Coco-nut 87 . 3.0 7 2 . . . . . Canola 3 . NaN 11 13 10 . 7 50 2 Olive Oil 11 . 4.0 71 11 1 . . . . Butter-fat 50 4 12.0 26 4 1 2 . . . The figure below shows the relative conformations of saturated and unsaturated fatty acids, and in comparison, the conformations and potential energy graph for n-butane, which should provide insight into conformational changes in the nonpolar tail of fatty acids arising from rotation around C-C single bonds. We will explore this diagram a bit latter. Figure: Conformations of fatty acids and n-butane Jmol: conformations of ethane | conformations of propane | butane: the gauche conformation Fatty acids can be named in many ways. symbolic name: given as x:y Δ a,b,c where x is the number of C’s in the chain, y is the number of double bonds, and a, b, and c are the positions of the start of the double bonds counting from C1 - the carboxyl C. Saturated fatty acids contain no C-C double bonds. Monounsaturated fatty acids contain 1 C=C while polyunsaturated fatty acids contain more than 1 C=C. Double bonds are usual cis. systematic name using IUPAC nomenclature. The systematic name gives the number of Cs (e.g. hexadecanoic acid for 16:0). If the fatty acid is unsaturated, the base name reflects the number of double bonds (e.g. octadecenoic acid for 18:1 Δ 9 and octadecatrienoic acid for 18:3 Δ 9,12,15 ). common name: (e.g. oleic acid, which is found in high concentration in olive oil) You should know the common name, systematic name, and symbolic representations for these saturated fatty: lauric acid, dodecanoic acid, 12:0 palmitic acid, hexadecanoic acid, 16:0 stearic acid, octadecanic acid, 18:0. Learn the following unsaturated fatty acids - oleic acid, octadecenoic acid, 18:1 Δ 9 linoleic acid, octadecadienoic acid, 18:2 Δ 9,12 a-linolenic acid, octadecatrienoic acid, 18:3 Δ 9,12,15 (n-3) arachidonic acid, eicosatetraenoic acid, 20:4 Δ 5,8,11,14 (n-6) eicosapentenoic acid (EPA), 20:5 Δ 5,8,11,14,17 (n-3) Note: sometimes written as eicosapentaenoic docosahexenoic acid (DHA) 22:6 Δ4,7,10,13,16,19 (n-3) Note: sometimes written as docosahexaenoic There is an alternative to the symbolic representation of fatty acids, in which the Cs are numbered from the distal end (the n or w end) of the acyl chain (the opposite end from the carboxyl group). Hence 18:3 Δ 9,12,15 could be written as 18:3 (w -3) or 18:3 (n -3) where the terminal C is numbered one and the first double bond starts at C3. Arachidonic acid is an (w -6) fatty acid while docosahexaenoic acid is an (w -3) fatty acid. Note that all naturally occurring double bonds are cis (E), with a methylene spacer between double bonds - i.e. the double bonds are not conjugated. For saturated fatty acids, the melting point increases with C chain length, owing to increased likelihood of van der Waals (London or induced dipole) interactions between the overlapping and packed chains. Within chains of the same number of Cs, melting point decreases with increasing number of double bonds, owing to the kinking of the acyl chains, followed by decreased packing and reduced intermolecular forces (IMFs). Fatty acid composition differs in different organisms: animals have 5-7% of fatty acids with 20-22 carbons, while fish have 25-30% animals have <1% of their fatty acids with 5-6 double bonds, while plants have 5-6% and fish 15-30% Many studies support the claim the diets high in fish that contain abundant n-3 fatty acids, in particular EPA and DHA, reduce inflammation and cardiovascular disease. n-3 fatty acids are abundant in high oil fish (salmon, tuna, sardines), and lower in cod, flounder, snapper, shark, and tilapia. The most common polyunsaturated fats (PUFAs) in our diet are the n-3 and n-6 classes. Most abundant in the n-6 class in plant food is linoleic acid (18:2n-6, or 18:2 Δ 9 ,12 ), while linolenic acid (18:3n-3 or 18:3 Δ 9 ,12,15 ) is the most abundant in the n-3 class. These fatty acids are essential in that they are biological precursors for other PUFAs. Specifically, linoleic acid (18:2 n-6, or 18:2 Δ 9 ,12 ) is a biosynthetic precursor of arachidonic acid (20:4n-6 or 20:4 Δ 5 ,8,11,14) linolenic acid (18:3n-3, or 18:3 Δ 9 ,12,15 ) is a biosynthetic precursor of eicosapentaenoic acid (EPA, 20:5n-3 or 20:5 Δ 5 ,8,11,14,17 ) and to a much smaller extent, docosahexaenoic acid (DHA, 22:6n-3 or 22:6 Δ 4 ,7,10,13,16,19 ). Note These essential precursor fatty acids are substrates for intracellular enzymes such as elongases, desaturases, and beta-oxidation type enzymes in the endoplasmic reticulum and another organelle, the peroxisome (involved in oxidative metabolism of straight chain and branched fatty acids, peroxide metabolism, and cholesterol/bile salt synthesis). Animals fed diets high in plant 18:2(n-6) fats accumulate 20:4(n-6) fatty acids in their tissues while those fed diets high in plant 18:3(n-3) accumulate 22:6(n-3). Animals fed diets high in fish oils accumulate 20:5 (EPA) and 22:6 (DHA) at the expense of 20:4(n-6). Recent work has suggested that contrary to images of early hominids as hunters and scavengers of meat, human brain development might have required the consumption of fish which is highly enriched in arachidonic and docosahexaenoic acids. A large percent of the brain consists of lipids, which are highly enriched in these two fatty acids. These acids are necessary for the proper development of the human brain and in adults, deficiencies in these might contribute to cognitive disorders like ADHD, dementia, and dyslexia. These fatty acids are essential in the diet, and probably could not have been derived in high enough amounts from the eating of brains of other animals. The mechanism for the protective effects of n-3 fatty acids in health will be explored later in the course when we discuss prostaglandins synthesis and signal transduction. Saturated fatty acids chains can exist in many conformations resulting from free rotation around the C-C bonds of the acyl chains. A quick review of the conformations of n-butane shows that the energetically most favorable conformation is one in which the two CH3 groups attached to the 2 methylene C’s (C2 and C3) are trans to each other, which results in decreased steric strain. Looking at a Newman projection of n-butane shows the dihedral or torsional angle of this trans conformation to be 180 degrees. When the dihedral angle is 0 degrees, the two terminal CH 3 groups are syn to each other, which is the conformation of highest energy. When the angle is 60 (gauche+) or 300 (gauche-) degrees, a higher, local minimum is observed in the energy profile. At a given temperature and moment, a population of molecules of butane would consist of some in the g+ and g- state, with most in the t state. The same applies to fatty acids. To increase the number of chains with g+tg- conformations, for example, the temperature of the system can be increased.
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Bookshelves/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/17%3A_Lipids/17.04%3A_Steroids
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Learning Objectives To identify the functions of steroids produced in mammals. All the lipids discussed so far are saponifiable , reacting with aqueous alkali to yield simpler components, such as glycerol, fatty acids, amino alcohols, and sugars. Lipid samples extracted from cellular material, however, also contain a small but important fraction that does not react with alkali. The most important nonsaponifiable lipids are the steroids. These compounds include the bile salts , cholesterol and related compounds, and certain hormones (such as cortisone and the sex hormones). Steroids occur in plants, animals, yeasts, and molds but not in bacteria. They may exist in free form or combined with fatty acids or carbohydrates. All steroids have a characteristic structural component consisting of four fused rings. Chemists identify the rings by capital letters and number the carbon atoms as shown in Figure \(\PageIndex{1a}\). Slight variations in this structure or in the atoms or groups attached to it produce profound differences in biological activity. Cholesterol Cholesterol (Figure \(\PageIndex{1b}\)) does not occur in plants, but it is the most abundant steroid in the human body (240 g is a typical amount). Excess cholesterol is believed to be a primary factor in the development of atherosclerosis and heart disease, which are major health problems in the United States today. About half of the body’s cholesterol is interspersed in the lipid bilayer of cell membranes. Much of the rest is converted to cholic acid, which is used in the formation of bile salts. Cholesterol is also a precursor in the synthesis of sex hormones, adrenal hormones, and vitamin D. Excess cholesterol not metabolized by the body is released from the liver and transported by the blood to the gallbladder. Normally, it stays in solution there until being secreted into the intestine (as a component of bile) to be eliminated. Sometimes, however, cholesterol in the gallbladder precipitates in the form of gallstones (Figure \(\PageIndex{2}\)). Indeed, the name cholesterol is derived from the Greek chole , meaning “bile,” and stereos , meaning “solid.” To Your Health: Cholesterol and Heart Disease Heart disease is the leading cause of death in the United States for both men and women. The Centers for Disease Control and Prevention reported that heart disease claimed 631,636 lives in the United States (26% of all reported deaths) in 2006. Scientists agree that elevated cholesterol levels in the blood, as well as high blood pressure, obesity, diabetes, and cigarette smoking, are associated with an increased risk of heart disease. A long-term investigation by the National Institutes of Health showed that among men ages 30 to 49, the incidence of heart disease was five times greater for those whose cholesterol levels were above 260 mg/100 mL of serum than for those with cholesterol levels of 200 mg/100 mL or less. The cholesterol content of blood varies considerably with age, diet, and sex. Young adults average about 170 mg of cholesterol per 100 mL of blood, whereas males at age 55 may have cholesterol levels at 250 mg/100 mL or higher because the rate of cholesterol breakdown decreases with age. Females tend to have lower blood cholesterol levels than males. To understand the link between heart disease and cholesterol levels, it is important to understand how cholesterol and other lipids are transported in the body. Lipids, such as cholesterol, are not soluble in water and therefore cannot be transported in the blood (an aqueous medium) unless they are complexed with proteins that are soluble in water, forming assemblages called lipoproteins . Lipoproteins are classified according to their density, which is dependent on the relative amounts of protein and lipid they contain. Lipids are less dense than proteins, so lipoproteins containing a greater proportion of lipid are less dense than those containing a greater proportion of protein. Research on cholesterol and its role in heart disease has focused on serum levels of low-density lipoproteins (LDLs) and high-density lipoproteins (HDLs). One of the most fascinating discoveries is that high levels of HDLs reduce a person’s risk of developing heart disease, whereas high levels of LDLs increase that risk. Thus the serum LDL:HDL ratio is a better predictor of heart disease risk than the overall level of serum cholesterol. Persons who, because of hereditary or dietary factors, have high LDL:HDL ratios in their blood have a higher incidence of heart disease. How do HDLs reduce the risk of developing heart disease? No one knows for sure, but one role of HDLs appears to be the transport of excess cholesterol to the liver, where it can be metabolized. Therefore, HDLs aid in removing cholesterol from blood and from the smooth muscle cells of the arterial wall. Dietary modifications and increased physical activity can help lower total cholesterol and improve the LDL:HDL ratio. The average American consumes about 600 mg of cholesterol from animal products each day and also synthesizes approximately 1 g of cholesterol each day, mostly in the liver. The amount of cholesterol synthesized is controlled by the cholesterol level in the blood; when the blood cholesterol level exceeds 150 mg/100 mL, the rate of cholesterol biosynthesis is halved. Hence, if cholesterol is present in the diet, a feedback mechanism suppresses its synthesis in the liver. However, the ratio of suppression is not a 1:1 ratio; the reduction in biosynthesis does not equal the amount of cholesterol ingested. Thus, dietary substitutions of unsaturated fat for saturated fat, as well as a reduction in consumption of trans fatty acids, is recommended to help lower serum cholesterol and the risk of heart disease. Steroid Hormones Hormones are chemical messengers that are released in one tissue and transported through the circulatory system to one or more other tissues. One group of hormones is known as steroid hormones because these hormones are synthesized from cholesterol, which is also a steroid. There are two main groups of steroid hormones: adrenocortical hormones and sex hormones. The adrenocortical hormones, such as aldosterone and cortisol (Table \(\PageIndex{1}\)), are produced by the adrenal gland, which is located adjacent to each kidney. Aldosterone acts on most cells in the body, but it is particularly effective at enhancing the rate of reabsorption of sodium ions in the kidney tubules and increasing the secretion of potassium ions and/or hydrogen ions by the tubules. Because the concentration of sodium ions is the major factor influencing water retention in tissues, aldosterone promotes water retention and reduces urine output. Cortisol regulates several key metabolic reactions (for example, increasing glucose production and mobilizing fatty acids and amino acids). It also inhibits the inflammatory response of tissue to injury or stress. Cortisol and its analogs are therefore used pharmacologically as immunosuppressants after transplant operations and in the treatment of severe skin allergies and autoimmune diseases, such as rheumatoid arthritis. Hormone Effect NaN regulates salt metabolism; stimulates kidneys to retain sodium and excrete potassium NaN stimulates the conversion of proteins to carbohydrates NaN regulates the menstrual cycle; maintains pregnancy NaN stimulates female sex characteristics; regulates changes during the menstrual cycle NaN stimulates and maintains male sex characteristics The sex hormones are a class of steroid hormones secreted by the gonads (ovaries or testes), the placenta, and the adrenal glands. Testosterone and androstenedione are the primary male sex hormones, or androgens , controlling the primary sexual characteristics of males, or the development of the male genital organs and the continuous production of sperm. Androgens are also responsible for the development of secondary male characteristics, such as facial hair, deep voice, and muscle strength. Two kinds of sex hormones are of particular importance in females: progesterone, which prepares the uterus for pregnancy and prevents the further release of eggs from the ovaries during pregnancy, and the estrogens, which are mainly responsible for the development of female secondary sexual characteristics, such as breast development and increased deposition of fat tissue in the breasts, the buttocks, and the thighs. Both males and females produce androgens and estrogens, differing in the amounts of secreted hormones rather than in the presence or absence of one or the other. Sex hormones, both natural and synthetic, are sometimes used therapeutically. For example, a woman who has had her ovaries removed may be given female hormones to compensate. Some of the earliest chemical compounds employed in cancer chemotherapy were sex hormones. For example, estrogens are one treatment option for prostate cancer because they block the release and activity of testosterone. Testosterone enhances prostate cancer growth. Sex hormones are also administered in preparation for sex-change operations, to promote the development of the proper secondary sexual characteristics. Oral contraceptives are synthetic derivatives of the female sex hormones; they work by preventing ovulation. Bile Salts Bile is a yellowish green liquid (pH 7.8–8.6) produced in the liver. The most important constituents of bile are bile salts, which are sodium salts of amidelike combinations of bile acids, such as cholic acid (part (a) of Figure \(\PageIndex{3}\)) and an amine such as the amino acid glycine (part (b) of Figure \(\PageIndex{3}\)). They are synthesized from cholesterol in the liver, stored in the gallbladder, and then secreted in bile into the small intestine. In the gallbladder, the composition of bile gradually changes as water is absorbed and the other components become more concentrated. Because they contain both hydrophobic and hydrophilic groups, bile salts are highly effective detergents and emulsifying agents; they break down large fat globules into smaller ones and keep those smaller globules suspended in the aqueous digestive environment. Enzymes can then hydrolyze fat molecules more efficiently. Thus, the major function of bile salts is to aid in the digestion of dietary lipids. Surgical removal is often advised for a gallbladder that becomes infected, inflamed, or perforated. This surgery does not seriously affect digestion because bile is still produced by the liver, but the liver’s bile is more dilute and its secretion into the small intestine is not as closely tied to the arrival of food. Summary Steroids have a four-fused-ring structure and have a variety of functions. Cholesterol is a steroid found in mammals that is needed for the formation of cell membranes, bile acids, and several hormones. Bile salts are secreted into the small intestine to aid in the digestion of fats.
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Bookshelves/General_Chemistry/The_Video_Textbook_of_General_Chemistry_(Farmer)/10%3A_Electron_Configuration_of_Atoms/10.4_Writing_Quantum_Number_from_Electron_Configurations_(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 When writing quantum numbers for electrons in a specific subshell it is important to remember that the values of n and l are given in the orbital designation. Ml can be found be drawing the individual orbitals in a subshell and inserting electrons. Ms is still either +1/2 or -1/2 but remember their assignment follows Hund's rule and the Pauli exclusion principal. Link to Video Writing Quantum Number from Electron Configurations: https://youtu.be/SoUhs2_YnKU Attribution Prof. Steven Farmer ( Sonoma State University )
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Courses/Athabasca_University/Chemistry_360%3A_Organic_Chemistry_II/Chapter_19%3A_Aldehydes_and_Ketones%3A_Nucleophilic_Addition_Reactions/19.06_Nucleophilic_Addition_of_HCN%3A_Cyanohydrin_Formation
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Objectives After completing this section, you should be able to write an equation to describe the formation of a cyanohydrin from an aldehyde or ketone. identify the cyanohydrin formed from the reaction of a given aldehyde or ketone with hydrogen cyanide. identify the aldehyde or ketone, the reagents, or both, needed to prepare a given cyanohydrin. write the detailed mechanism for the addition of hydrogen cyanide to an aldehyde or ketone. write equations to represent the conversion of cyanohydrins to hydroxyamines and hydroxycarboxylic acids, and hence recognize the importance of cyanohydrin formation in organic synthesis. a. identify the product formed when a given cyanohydrin is either reduced with lithium aluminum hydride or hydrolyzed with mineral acid. b. identify the aldehyde or ketone, and other reagents needed to prepare a given α ‑hydroxyamine or α ‑hydroxycarboxylic acid. Key Terms Make certain that you can define, and use in context, the key term below. cyanohydrin Study Notes For successful cyanohydrin formation it is important to have free cyanide ions available to react with the ketone or aldehyde. This can be achieved by using a salt (e.g. KCN or NaCN) or a silylated (e.g. Me 3 SiCN) form of cyanide under acidic conditions or by using HCN with some base added to produce the needed CN − nucleophile. Cyanohydrins have the structural formula of R 2 C(OH)CN. The “R” on the formula represents an alkyl, aryl, or hydrogen. To form a cyanohydrin, a hydrogen cyanide adds reversibly to the carbonyl group of an organic compound thus forming a hydroxyalkanenitrile adducts (commonly known and called as cyanohydrins). Figure 19.6.1 : General structure of a cyanohydrin Hydrogen cyanide adds across the carbon-oxygen double bond in aldehydes and ketones to produce compounds known as hydroxynitriles. For example, with ethanal (an aldehyde) you get 2-hydroxypropanenitrile: With propanone (a ketone) you get 2-hydroxy-2-methylpropanenitrile: The reaction isn't normally done using hydrogen cyanide itself, because this is an extremely poisonous gas. Instead, the aldehyde or ketone is mixed with a solution of sodium or potassium cyanide in water to which a little sulphuric acid has been added. The pH of the solution is adjusted to about 4 - 5, because this gives the fastest reaction. The solution will contain hydrogen cyanide (from the reaction between the sodium or potassium cyanide and the sulphuric acid), but still contains some free cyanide ions. This is important for the mechanism. Mechanism of Cyanohydrin Formation In the first step of the mechanism, the cyanide ion acts as a nucleophile and forms a C-C bond with the electrophillic carbonyl carbon. The two electrons in the carbonyl pi bond are pushed on to the electronegative oxygen forming a tetrahedral alkoxide ion intermediate. In the second step, the alkoxide ion is protonated by HCN which regenerates the cyanide ion. Step 1: Nucleophilic attack Step 2: Protonation Acetone Cyanohydrins Acetone cyanohydrins (ACH) have the structural formula of (CH 3 ) 2 C(OH)CN. It is an organic compound serves in the production of methyl methacrylate (also known as acrylic). Figure 19.6.2: Acetone cyanohydrins It is classified as an extremely hazardous substance, since it rapidly decomposes when it's in contact with water. In ACH, sulfuric acid is treated to give the sulfate ester of the methacrylamid . Preparations of other cyanohydrins are also used from ACH: for HACN to Michael acceptors and for the formylation of arenas. The treatment with lithium hydride affords anhydrous lithium cyanide. Figure 19.6.3: Reduction of Acetone cyanohydrins Further Chemistry of Cyanohydrins Cyanohydrin functional groups often prove useful because of the further chemistry that can be carried out due to the presence of a hydroxyl and a nitrile functionality. In particular, dehydration can convert the hydroxyl group into an alkene ( Section 17.6 ). The nitrile can be converted into a carboxylic acid function group through reaction with a hot acidic aqueous solution ( Section 20.7 ). Also, the nitrile can be reduced by the addition of LiAlH 4 to form a primary amine. Other Cyanohydrins Other interesting cyanohydrins are: acetone cyanohydrin , and glycolonitrile . Acetone cyanohydrin has the structure, (CH 3 ) 2 C(OH)CN, and is used in the production of methyl methacrylate (also known as acrylic). Glycolonitrile is an organic compound with the structural formula of HOCH 2 CN, which is the simplest cyanohydrin that is derived by formaldehydes. Exercise Exercise 19.6.1 What product is formed in this reaction? Answer
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Courses/Monterey_Peninsula_College/CHEM_30A%3A_Introduction_to_Chemistry_for_Health_Sciences/02%3A_Measurements_Unit_Conversions_and_Density/2.04%3A_Converting_Units
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Learning Objective Convert from one unit to another unit of the same type. You may already know how to convert common units that you use in your everyday life. Continue to use that process when it works for you! However, in this course you may encounter unfamiliar problems that are more difficult to figure out. In that case, I recommend using the unit analysis method which will be demonstrated here. Consider a simple example: how many feet are there in 4 yards? Many people can answer that there are 12 feet in 4 yards. If there are 3 feet in 1 yard, and there are 4 yards, then there are 4 × 3 = 12 feet in 4 yards. To solve a problem like this one using unit analysis, we use equalities which are pairs of values that are equal to one another. In this case, 1 yard (yd) equals 3 feet (ft): \[1\, yd = 3\, ft\nonumber \] Conversion Factors In the unit analysis problem solving method we will use equalities as conversion factors by rewriting them as fractions equal to one. For example: \[1= \dfrac{3\,ft}{1\,yd}\nonumber \] or \[1= \dfrac{1\,yd}{3\,ft}\nonumber \] This is a strange way to write 1, but it makes sense: 3 ft equal 1 yd, so the quantities in the numerator and denominator are the same quantity, just expressed with different units. Unit Analysis To use the unit analysis method, start with the original quantity. This will generally be a measurement with a simple (not compound) unit that is not already part of a relationship or equality. In the problem above our original quantity was 4 yd. Next, multiply this quantity by a conversion factor that is equal to 1 (so that it does not change the value of the quantity). Remember that conversion factors are fractions made up of two values that are equal. Eventually we want the unit of yards to divide out leaving only a unit of feet for the answer, so we set up the conversion factor so that yards is on the bottom where it will cancel the original unit and feet is on the top. That will become the unit for the answer. \[4\,yd\times \dfrac{3\,ft}{1\,yd}\nonumber \] You'll follow this pattern when solving other unit analysis problems in the future. The unit for the original value will go on the bottom of the conversion factor and the desired unit will go on the top of the conversion factor. The 4 yd term can be thought of as 4yd/1; that is, it can be thought of as a fraction with 1 in the denominator. We are essentially multiplying fractions. If the same thing appears in the numerator and denominator of a fraction, they cancel. (You may have seen this with variables in algebra where an x that appears on both the top and bottom of a fraction divides out. It is the same with units). In this case, what cancels is the unit yard : \[4\,yd\times \dfrac{3\,ft}{1\,yd}\nonumber \] That is all that we can cancel. Now, multiply the fractions to get the final answer. To multiply fractions start with the number at the far left. Multiply by any values on the top of fractions and divide by any values on the bottom of fractions: \[\dfrac{4\times 3\, ft}{1}= \dfrac{12\,ft}{1}= 12\,ft\nonumber \] Again, we get an answer of 12 ft, just as we did originally. But in this case, we used a more formal procedure that is applicable to a variety of problems. How many millimeters are in 14.66 m? To answer this, we need to construct a conversion factor between millimeters and meters and apply it correctly to the original quantity. We start with the definition of a millimeter which is: \[1000\,mm=1\,m\nonumber \] Now we construct a conversion factor by dividing one quantity into both sides. Can you figureout which version of the conversion factor is needed for this problem? \[1=\dfrac{1\,m}{1000\,mm}\nonumber \] or \[\dfrac{1000\,mm}{1\,m}=1\nonumber \] The answer is based on what unit you want to get rid of in your initial quantity . The original unit of our quantity is meters, which we want to convert to millimeters. Because the original unit is assumed to be in the numerator, to get rid of it, we want the meter unit in the denominator ; then they will cancel. Therefore, we will use the second conversion factor. Canceling units and performing the mathematics, we get: \[14.66m\times \dfrac{1000\,mm}{1\,m}= 14660\,mm\nonumber \] Note how \(m\) cancels, leaving \(mm\), which is the unit of interest. The ability to construct and apply proper conversion factors is a very powerful mathematical technique in chemistry. You need to master this technique if you are going to be successful in this and future courses. Example \(\PageIndex{1}\) Convert 35.9 kL to liters. Convert 555 mm to meters. Solution We will use the fact that 1 kL = 1,000 L. Of the two conversion factors that can be defined, the one that will work is 1000L/ 1kL. Applying this conversion factor, we get: \[35.9\, kL\times \dfrac{1000\,L}{1\,kL}= 35,900\, L \nonumber \nonumber \] We will use the fact that 1,000 mm = 1 m. Of the two possible conversion factors, the appropriate one has the mm unit in the denominator: \[\dfrac{1\,m}{1000\,mm} \nonumber \nonumber \] Applying this conversion factor, we get: \[555\,mm\times \dfrac{1m}{1000 mm}= 0.555\,m \nonumber \nonumber \] Exercise \(\PageIndex{1}\) Convert 67.08 μL to liters. Convert 56.8 m to kilometers. Answer a 6.708 × 10 −5 L Answer b 5.68 × 10 −2 km Exercise \(\PageIndex{4}\) How many milliliters are in 607.8 kL? Answer 6.078 × 10 8 mL Chemistry is Everywhere: The Gimli Glider On July 23, 1983, an Air Canada Boeing 767 jet had to glide to an emergency landing at Gimli Industrial Park Airport in Gimli, Manitoba, because it unexpectedly ran out of fuel during flight. There was no loss of life in the course of the emergency landing, only some minor injuries associated in part with the evacuation of the craft after landing. For the remainder of its operational life (the plane was retired in 2008), the aircraft was nicknamed "the Gimli Glider." The Gimli Glider is the Boeing 767 that ran out of fuel and glided to safety at Gimli Airport. The aircraft ran out of fuel because of confusion over the units used to express the amount of fuel. Source: Photo courtesy of Will F., (CC BY-SA 2.5; Aero Icarus ). The 767 took off from Montreal on its way to Ottawa, ultimately heading for Edmonton, Canada. About halfway through the flight, all the engines on the plane began to shut down because of a lack of fuel. When the final engine cut off, all electricity (which was generated by the engines) was lost; the plane became, essentially, a powerless glider. Captain Robert Pearson was an experienced glider pilot, although he had never flown a glider the size of a 767. First Officer Maurice Quintal quickly determined that the aircraft would not be able make it to Winnipeg, the next large airport. He suggested his old Royal Air Force base at Gimli Station, one of whose runways was still being used as a community airport. Between the efforts of the pilots and the flight crew, they managed to get the airplane safely on the ground (although with buckled landing gear) and all passengers off safely. What happened? At the time, Canada was transitioning from the older English system to the metric system. The Boeing 767s were the first aircraft whose gauges were calibrated in the metric system of units (liters and kilograms) rather than the English system of units (gallons and pounds). Thus, when the fuel gauge read 22,300, the gauge meant kilograms, but the ground crew mistakenly fueled the plane with 22,300 pounds of fuel. This ended up being just less than half of the fuel needed to make the trip, causing the engines to quit about halfway to Ottawa. Quick thinking and extraordinary skill saved the lives of 61 passengers and 8 crew members—an incident that would not have occurred if people were watching their units. Key Takeaways Units can be converted to other units using the proper conversion factors. Conversion factors are constructed from equalities that relate two different units. Conversions can be a single step or multi-step. Unit conversion is a powerful mathematical technique in chemistry that must be mastered. Exact numbers do not affect the determination of significant figures.
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Courses/San_Diego_Miramar_College/Chem_103%3A_Fundamentals_of_GOB_Chemistry_(Garces)/11%3A_Acids_and_Bases/11.21%3A_Hydrolysis_of_Salts_-_Equations
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Baking seems easy with all the pre-mixed items available. ("just add water and stir"). However, there is a good amount of chemistry involved in baking with ingredients that are measured out. One important ingredient is baking powder. The fluffiness in the final product of a non-yeast recipe is usually due to the carbon dioxide formed by baking powder. One popular brand uses a mix of sodium bicarbonate and sodium aluminum sulfate to produce the \(\ce{CO_2}\). The reaction is: \[3 \ce{NaHCO_3} + \ce{NaAl(SO_4)_2} \rightarrow \ce{Al(OH)_3} + 2 \ce{Na_2SO_4} + 3 \ce{CO_2}\nonumber \] If all goes well, the biscuits rise, the pancakes are fluffy, and everybody is happy. Hydrolysis of Salts: Equations A salt is an ionic compound that is formed when an acid and a base neutralize each other. While it may seem that salt solutions are always neutral, they can frequently be either acidic or basic. Consider the salt formed when the weak acid hydrofluoric acid is neutralized by the strong base sodium hydroxide. The molecular and net ionic equations are shown below. \[\begin{align*} &\ce{HF} \left( aq \right) + \ce{NaOH} \left( aq \right) \rightarrow \ce{NaF} \left( aq \right) + \ce{H_2O} \left( l \right) \\ &\ce{HF} \left( aq \right) + \ce{OH^-} \left( aq \right) \rightarrow \ce{F^-} \left( aq \right) + \ce{H_2O} \left( l \right) \end{align*}\nonumber \] Since sodium fluoride is soluble, the sodium ion is a spectator ion in the neutralization reaction. The fluoride ion is capable of reacting, to a small extent, with water, accepting a proton. \[\ce{F^-} \left( aq \right) + \ce{H_2O} \left( l \right) \rightleftharpoons \ce{HF} \left( aq \right) + \ce{OH^-} \left( aq \right)\nonumber \] The fluoride ion is acting as a weak Brønsted-Lowry base. The hydroxide ion that is produced as a result of the above reaction makes the solution slightly basic. Salt hydrolysis is a reaction in which one of the ions from a salt reacts with water, forming either an acidic or basic solution. Salts That Form Basic Solutions When solid sodium fluoride is dissolved into water, it completely dissociates into sodium ions and fluoride ions. The sodium ions do not have any capability of hydrolyzing, but the fluoride ions hydrolyze to produce a small amount of hydrofluoric acid and hydroxide ion. \[\ce{F^-} \left( aq \right) + \ce{H_2O} \left( l \right) \rightleftharpoons \ce{HF} \left( aq \right) + \ce{OH^-} \left( aq \right)\nonumber \] Salts that are derived from the neutralization of a weak acid \(\left( \ce{HF} \right)\) by a strong base \(\left( \ce{NaOH} \right)\) will always produce salt solutions that are basic. Salts That Form Acidic Solutions Ammonium chloride \(\left( \ce{NH_4Cl} \right)\) is a salt that is formed when the strong acid \(\ce{HCl}\) is neutralized by the weak base \(\ce{NH_3}\). Ammonium chloride is soluble in water. The chloride ion produced is incapable of hydrolyzing because it is the conjugate base of the strong acid \(\ce{HCl}\). In other words, the \(\ce{Cl^-}\) ion cannot accept a proton from water to form \(\ce{HCl}\) and \(\ce{OH^-}\), as the fluoride ion did in the previous section. However, the ammonium ion is capable of reacting slightly with water, donating a proton and so acting as an acid. \[\ce{NH_4^+} \left( aq \right) + \ce{H_2O} \left( l \right) \rightleftharpoons \ce{H_3O^+} \left( aq \right) + \ce{NH_3} \left( aq \right)\nonumber \] Salts That Form Neutral Solutions A salt that is derived from the reaction of a strong acid with a strong base forms a solution that has a pH of 7. An example is sodium chloride, formed from the neutralization of \(\ce{HCl}\) by \(\ce{NaOH}\). A solution of \(\ce{NaCl}\) in water has no acidic or basic properties, since neither ion is capable of hydrolyzing. Other salts that form neutral solutions include potassium nitrate \(\left( \ce{KNO_3} \right)\) and lithium bromide \(\left( \ce{LiBr} \right)\). The table below summarizes how to determine the acidity or basicity of a salt solution. Salt formed from: Salt Solution Strong acid \(+\) Strong base Neutral Strong acid \(+\) Weak base Acidic Weak acid \(+\) Strong base Basic Salts formed from the reaction of a weak acid and a weak base are more difficult to analyze due to competing hydrolysis reactions between the cation and the anion. These salts are not considered in this chapter's concept. Summary Salt hydrolysis is a reaction in which one of the ions from a salt reacts with water, forming either an acidic or basic solution. Salts that are derived from the neutralization of a weak acid by a strong base will always produce salt solutions that are basic. Salts that are derived from the neutralization of a strong acid by a weak base will always produce salt solutions that are acidic. A salt that is derived from the reaction of a strong acid with a strong base forms a solution that has a pH of 7.
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Courses/Indiana_Tech/EWC%3A_CHEM_2300_-_Introductory_Organic_(Budhi)/4%3A_Organic_Acids_and_Bases_and_Some_of_Their_Derivatives/4.05%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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Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Pharmaceuticals/Drug_Receptor_Interactions
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The vast majority of drugs show a remarkably high correlation of structure and specificity to produce pharmacological effects. Experimental evidence indicates that drugs interact with receptor sites localized in macromolecules which have protein-like properties and specific three dimensional shapes. A minimum three point attachment of a drug to a receptor site is required. In most cases a rather specific chemical structure is required for the receptor site and a complementary drug structure. Slight changes in the molecular structure of the drug may drastically change specificity. Introduction Several chemical forces may result in a temporary binding of the drug to the receptor. Essentially any bond could be involved with the drug-receptor interaction. Covalent bonds would be very tight and practically irreversible. Since by definition the drug-receptor interaction is reversible, covalent bond formation is rather rare except in a rather toxic situation. Since many drugs contain acid or amine functional groups which are ionized at physiological pH, ionic bonds are formed by the attraction of opposite charges in the receptor site. Polar-polar interactions as in hydrogen bonding are a further extension of the attraction of opposite charges. The drug-receptor reaction is essentially an exchange of the hydrogen bond between a drug molecule, surrounding water, and the receptor site. Finally hydrophobic bonds are formed between non-polar hydrocarbon groups on the drug and those in the receptor site. These bonds are not very specific but the interactions do occur to exclude water molecules. Repulsive forces which decrease the stability of the drug-receptor interaction include repulsion of like charges and steric hindrance. Steric hindrance refers to certain 3-dimensional features where repulsion occurs between electron clouds, inflexible chemical bonds, or bulky alkyl groups. Drug Interaction with Receptor Site A neurotransmitter has a specific shape to fit into a receptor site and cause a pharmacological response such as a nerve impulse being sent. The neurotransmitter is similar to a substrate in an enzyme interaction. After attachment to a receptor site, a drug may either initiate a response or prevent a response from occurring. A drug must be a close "mimic" of the neurotransmitter. An agonist is a drug which produces a stimulation type response. The agonist is a very close mimic and "fits" with the receptor site and is thus able to initiate a response. An antagonist drug interacts with the receptor site and blocks or depresses the normal response for that receptor because it only partially fits the receptor site and can not produce an effect. However, it does block the site preventing any other agonist or the normal neurotransmitter from interacting with the receptor site. Contributors Charles Ophardt, Professor Emeritus, Elmhurst College; Virtual Chembook
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Bookshelves/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/12%3A_Coordination_Chemistry_IV_-_Reactions_and_Mechanisms
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12.1: Introduction to Reactions of Metal Complexes 12.2: Substitutions Reactions 12.2.1: Introduction to Substitution Reactions 12.2.2: Inert and Labile Complexes 12.2.3: Mechanistic Possibilities 12.3: Kinetics Hint at the Reaction Mechanism Since the different mechanisms have different rate-limiting steps, there are kinetic consequences for each type of mechanism. We can used kinetic data, therefore, to help distinguish the mechanism of a reaction. However, the information gleaned from rate laws alone can be ambiguous. For example, what does it mean if a reaction depends on on the concentration of incoming ligand? How can we distinguish between an associative mechanism and an associatively-activated dissociative mechanism? 12.3.1: Rate Law for Dissociative Mechanisms 12.3.2: Rate Laws for Interchange Mechanisms 12.3.3: Rate Law for Associative Mechanisms 12.3.4: Preassociation Complexes 12.3.5: Activation Parameters 12.3.6: Some Reasons for Differing Mechanisms 12.4: Experimental Evidence in Octahedral Substitutions This page reviews common experiments used to provide evidence supporting various mechanisms, specifically focusing on the reactions of octahedral complexes. 12.4.1: Dissociation 12.4.2: Linear Free Energy Relationships 12.4.3: Associative Mechanisms 12.4.4: The conjugate base mechanism 12.4.5: The Kinetic Chelate Effect 12.5: Stereochemistry of Octahedral Reactions 12.5.1: Substitution in trans-en octahedral complexes 12.5.2: Substitution in cis-en octahedral complexes 12.5.3: Isomerization of Chelate Rings 12.6: Substitutions in Square Planar Complexes 12.6.1: Kinetics and Stereochemistry of Square Planar Reactions 12.6.2: Evidence for Associative Reactions 12.7: The Trans Effect 12.8: Redox Mechanisms 12.8.1: Outer Sphere Electron Transfer 12.8.2: Inner Sphere Electron Transfer 12.9: Reactions of Coordinated Ligands 12.9.1: Metal-catalyzed Hydrolysis 12.9.2: Template Reactions 12.9.3: Electrophilic Substitutions
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Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Advanced_Statistical_Mechanics_(Tuckerman)/09%3A_Review_of_the_basic_postulates_of_quantum_mechanics
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9.1: Measurement The result of a measurement of the observable A must yield one of the eigenvalues of A^ . Thus, we see why A is required to be a hermitian operator: Hermitian operators have real eigenvalues. 9.2: Physical Observables 9.3: The Fundamental Postulates of Quantum Mechanics 9.4: The Heisenberg Picture In all of the above, notice that we have formulated the postulates of quantum mechanics such that the state vector evolves in time, but the operators corresponding to observables are taken to be stationary. This formulation of quantum mechanics is known as the Schrödinger picture. However, there is another, completely equivalent, picture in which the state vector remains stationary and the operators evolve in time. This picture is known as the Heisenberg picture. 9.5: The Heisenberg Uncertainty Principle Because the operators x and p are not compatible, [X,P]≠0, there is no measurement that can precisely determine both x and p simultaneously. Hence, there must be an uncertainty relation between them that specifies how uncertain we are about one quantity given a definite precision in the measurement of the other. Presumably, if one can be determined with infinite precision, then there will be an infinite uncertainty in the other. 9.6: The Physical State of a Quantum System The physical state of a quantum system is represented by a vector denoted |Ψ(t)⟩ which is a column vector, whose components are probability amplitudes for different states in which the system might be found if a measurement were made on it. 9.7: Time Evolution of the State Vector The time evolution of the state vector is prescribed by the Schrödinger equation. Template:HideTOC
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Courses/North_Central_State_College/CHEM_1010%3A_Introductory_Chemistry/12%3A_Acids_and_Bases
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Template:HideTOC Acids and bases are common substances found in many every day items, from fruit juices and soft drinks to soap. In this unit we'll examine what the properties are of acids and bases, and learn about the chemical nature of these important compounds. You'll learn what pH is and how to calculate the pH of a solution. 12.1: Prelude - Sour Patch Kids and International Spy Movies Sour Patch Kids are a soft candy with a coating of invert sugar and sour sugar (a combination of citric acid, tartaric acid and sugar). Its slogan, "Sour. Sweet. Gone.", refers to the sour-to-sweet taste of the candy. 12.2: Acids- Properties and Examples Acids are very common in some of the foods that we eat. Citrus fruits such as oranges and lemons contain citric acid and ascorbic acid, which is better known as vitamin C. Carbonated sodas contain phosphoric acid. Vinegar contains acetic acid. Your own stomach utilizes hydrochloric acid to digest food. Acids are a distinct class of compounds because of the properties of their aqueous solutions. 12.3: Bases- Properties and Examples A base is thought of as a substance which can accept protons or any chemical compound that yields hydroxide ions (OH-) in solution. It is also commonly referred to as any substance that can react with an acid to decrease or neutralize its acidic properties, change the color of indicators (e.g. turn red litmus paper blue), feel slippery to the touch when in solution, taste bitter, react with acids to form salts, and promote certain chemical reactions (e.g. base catalysis). 12.4: Molecular Definitions of Acids and Bases Although the properties of acids and bases had been recognized for a long time, it was Svante Arrhenius in the 1880's who determined that: the properties of acids were due to the presence of hydrogen ions, and the properties of bases were due to the presence of hydroxide ions. 12.5: Reactions of Acids and Bases When an acid and a base are combined, water and a salt are the products. Salts are ionic compounds containing a positive ion other than H+ and a negative ion other than the hydroxide ion, OH-. Double displacement reactions of this type are called neutralization reactions. Salt solutions do not always have a pH of 7, however. Through a process known as hydrolysis, the ions produced when an acid and base combine may react with the water to produce slightly acidic or basic solutions. 12.6: Strong and Weak Acids and Bases Acids are classified as either strong or weak, based on their ionization in water. A strong acid is an acid which is completely ionized in an aqueous solution. A weak acid is an acid that ionizes only slightly in an aqueous solution. Acetic acid (found in vinegar) is a very common weak acid. 12.7: Water - Acid and Base in One Water is an interesting compound in many respects. Here, we will consider its ability to behave as an acid or a base. In some circumstances, a water molecule will accept a proton and thus act as a Brønsted-Lowry base. 12.8: Buffers are Solutions that Resist pH Change A buffer is a solution that resists dramatic changes in pH. Buffers do so by being composed of certain pairs of solutes: either a weak acid plus a salt derived from that weak acid or a weak base plus a salt of that weak base. 12.9: The pH and pOH Scales - Ways to Express Acidity and Basicity pH and pOH are defined as the negative log of hydrogen ion concentration and hydroxide concentration, respectively. Knowledge of ether can be used to calculate either [H+] of [OH-]. pOH is related to pH and can be easily calculated from pH. 12.10: Acid–Base Titration Acid-base titrations are lab procedures used to determine the concentration of a solution. One of the standard laboratory exercises in General Chemistry is an acid-base titration. 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-]). Template:HideTOC
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Courses/Millersville_University/CHEM_341-_Physical_Chemistry_I/04%3A_Putting_the_First_Law_to_Work/4.03%3A_Compressibility_and_Expansivity
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Isothermal Compressibility (\(\kappa_T\)) A very important property of a substance is how compressible it is. Gases are very compressible, so when subjected to high pressures, their volumes decrease significantly (think Boyle’s Law!) Solids and liquids however are not as compressible. However, they are not entirely incompressible! High pressure will lead to a decrease in volume, even if it is only slight. And, of course, different substances are more compressible than others. To quantify just how compressible substances are, it is necessary to define the property. The isothermal compressibility is defined by the fractional differential change in volume due to a change in pressure. \[ \kappa_T \equiv - \dfrac{1}{V} \left( \dfrac{\partial V}{\partial p} \right)_T \label{compress} \] The negative sign is important in order to keep the value of \(\kappa_T\) positive, since an increase in pressure will lead to a decrease in volume. The \(1/V\) term is needed to make the property intensive so that it can be tabulated in a useful manner. Isobaric Thermal Expansivity (\(\alpha\)) Another very important property of a substance is how its volume will respond to changes in temperature. Again, gases respond profoundly to changes in temperature (think Charles’ Law!) whereas solids and liquid will have more modest (but not negligible) responses to changes in temperature. (For example, If mercury or alcohol didn’t expand with increasing temperature, we wouldn’t be able to use those substances in thermometers.) The definition of the isobaric thermal expansivity (or sometimes called the expansion coefficient) is \[ \alpha \equiv \dfrac{1}{V} \left( \dfrac{\partial V}{\partial T} \right)_p \label{expand} \] As was the case with the compressibility factor, the \(1/V\) term is needed to make the property intensive, and thus able to be tabulated in a useful fashion. In the case of expansion, volume tends to increase with increasing temperature, so the partial derivative is positive. Deriving an Expression for a Partial Derivative (Type I): The reciprocal rule Consider a system that is described by three variables, and for which one can write a mathematical constraint on the variables \[F(x, y, z) = 0 \nonumber \] Under these circumstances, one can specify the state of the system varying only two parameters independently because the third parameter will have a fixed value. As such one could define two functions: \(z(x, y)\) and \(y(x,z)\). This allows one to write the total differentials for \(dz\) and \(dy\) as follows \[dz = \left( \dfrac{\partial z}{\partial x} \right)_y dx + \left( \dfrac{\partial z}{\partial y} \right)_x dy \label{eq5} \] and \[dy= \left( \dfrac{\partial y}{\partial x} \right)_z dx + \left( \dfrac{\partial y}{\partial z} \right)_x dz \label{eq6} \] Substituting the Equation \ref{eq6} expression into Equation \ref{eq5}: \[ \begin{align} dz &= \left( \dfrac{\partial z}{\partial x} \right)_y dx + \left( \dfrac{\partial z}{\partial y} \right)_x \left[ \left( \dfrac{\partial y}{\partial x} \right)_z dx + \left( \dfrac{\partial y}{\partial z} \right)_x dz \right] \\[4pt] &= \left( \dfrac{\partial z}{\partial x} \right)_y dx + \left( \dfrac{\partial z}{\partial y} \right)_x \left( \dfrac{\partial y}{\partial x} \right)_z dx + \left( \dfrac{\partial z}{\partial y} \right)_x \left( \dfrac{\partial y}{\partial z} \right)_x dz \label{eq7} \end{align} \] If the system undergoes a change following a pathway where \(x\) is held constant (\(dx = 0\)), this expression simplifies to \[dz = \left( \dfrac{\partial z}{\partial y} \right)_x \left( \dfrac{\partial y}{\partial z} \right)_x dz \nonumber \] And so for changes for which \(dz \neq 0\), \[\left( \dfrac{\partial z}{\partial y} \right)_x = \dfrac{1}{\left( \dfrac{\partial y}{\partial z} \right)_x } \nonumber \] This reciprocal rule is very convenient in the manipulation of partial derivatives. But it can also be derived in a straight-forward, albeit less rigorous, manner. Begin by writing the total differential for \(z(x,y)\) (Equation \ref{eq5}): \[dz = \left( \dfrac{\partial z}{\partial x} \right)_y dx + \left( \dfrac{\partial z}{\partial y} \right)_x dy \nonumber \] Now, divide both sides by \(dz\) and constrain to constant \(x\). \[\left.\dfrac{dz}{dz} \right\rvert_{x}= \left( \dfrac{\partial z}{\partial x} \right)_y \left.\dfrac{dx}{dz} \right\rvert_{x} + \left( \dfrac{\partial z}{\partial y} \right)_x \left.\dfrac{dy}{dz} \right\rvert_{x} \label{eq10} \] Noting that \[\left.\dfrac{dz}{dz} \right\rvert_{x} =1 \nonumber \] \[ \left.\dfrac{dx}{dz} \right\rvert_{x} = 0 \nonumber \] and \[\left.\dfrac{dy}{dz} \right\rvert_{x} = \left( \dfrac{\partial y}{\partial z} \right)_{x} \nonumber \] Equation \ref{eq10} becomes \[ 1= \left( \dfrac{\partial z}{\partial y} \right)_z \left( \dfrac{\partial y}{\partial z} \right)_x \nonumber \] or \[ \left( \dfrac{\partial z}{\partial y} \right)_z = \dfrac{1}{\left( \dfrac{\partial y}{\partial z} \right)_x} \nonumber \] This “formal” method of partial derivative manipulation is convenient and useful, although it is not mathematically rigorous. However, it does work for the kind of partial derivatives encountered in thermodynamics because the variables are state variables and the differentials are exact . Deriving an Expression for a Partial Derivative (Type II): The Cyclic Permutation Rule This alternative derivation follow the initial steps in the derivation above to Equation \ref{eq7}: \[dz = \left( \dfrac{\partial z}{\partial x} \right)_y dx + \left( \dfrac{\partial z}{\partial y} \right)_x \left( \dfrac{\partial y}{\partial x} \right)_z dx + \left( \dfrac{\partial z}{\partial y} \right)_x \left( \dfrac{\partial y}{\partial z} \right)_x dz \nonumber \] If the system undergoes a change following a pathway where \(z\) is held constant (\(dz = 0\)), this expression simplifies to \[0 = \left( \dfrac{\partial z}{\partial x} \right)_y dy + \left( \dfrac{\partial z}{\partial y} \right)_x \left( \dfrac{\partial y}{\partial x} \right)_z dx \nonumber \] And so for and changes in which \(dx \neq 0\) \[\left( \dfrac{\partial z}{\partial x} \right)_y = - \left( \dfrac{\partial z}{\partial y} \right)_x \left( \dfrac{\partial y}{\partial x} \right)_z \nonumber \] This cyclic permutation rule is very convenient in the manipulation of partial derivatives. But it can also be derived in a straight-forward, albeit less rigorous, manner. As with the derivation above, we wegin by writing the total differential of \(z(x,y)\) \[dz = \left( \dfrac{\partial z}{\partial x} \right)_y dx + \left( \dfrac{\partial z}{\partial y} \right)_x dy \nonumber \] Now, divide both sides by \(dx\) and constrain to constant \(z\). \[\left.\dfrac{dz}{dx} \right\rvert_{z}= \left( \dfrac{\partial z}{\partial x} \right)_y \left.\dfrac{dx}{dx} \right\rvert_{z} + \left( \dfrac{\partial z}{\partial y} \right)_x \left.\dfrac{dy}{dx} \right\rvert_{z} \label{eq21} \] Note that \[\left.\dfrac{dz}{dx} \right\rvert_{z} =0 \nonumber \] \[ \left.\dfrac{dx}{dx} \right\rvert_{z} =1 \nonumber \] and \[\left.\dfrac{dy}{dx} \right\rvert_{z} = \left( \dfrac{\partial y}{\partial x} \right)_{z} \nonumber \] Equation \ref{eq21} becomes \[ 0 = \left( \dfrac{\partial z}{\partial x} \right)_y + \left( \dfrac{\partial z}{\partial y} \right)_x \left( \dfrac{\partial y}{\partial x} \right)_{z} \nonumber \] which is easily rearranged to \[ \left( \dfrac{\partial z}{\partial x} \right)_y = - \left( \dfrac{\partial z}{\partial y} \right)_x \left( \dfrac{\partial y}{\partial x} \right)_{z} \nonumber \] This type of transformation is very convenient, and will be used often in the manipulation of partial derivatives in thermodynamics. Example \(\PageIndex{1}\): Expanding Thermodynamic Functions Derive an expression for \[\dfrac{\alpha}{\kappa_T}. \label{e1} \] in terms of derivatives of thermodynamic functions using the definitions in Equations \ref{compress} and \ref{expand}. Solution Substituting Equations \ref{compress} and \ref{expand} into the Equation \ref{e1} \[\dfrac{\alpha}{\kappa_T}= \dfrac{\dfrac{1}{V} \left( \dfrac{\partial V}{\partial T} \right)_p}{- \dfrac{1}{V} \left( \dfrac{\partial V}{\partial p} \right)_T} \nonumber \] Simplifying (canceling the \(1/V\) terms and using transformation Type I to invert the partial derivative in the denominator) yields \[\dfrac{\alpha}{\kappa_T} = - \left( \dfrac{\partial V}{\partial T} \right)_p \left( \dfrac{\partial p}{\partial V} \right)_T \nonumber \] Applying Transformation Type II give the final result: \[ \dfrac{\alpha}{\kappa_T} = \left( \dfrac{\partial p}{\partial T} \right)_V \nonumber \]
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Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Nucleic_Acids/Nucleic_Acids/Nucleic_Acids
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The nucleic acids are informational molecules because their primary structure contains a code or set of directions by which they can duplicate themselves and guide the synthesis of proteins. The synthesis of proteins - most of which are enzymes - ultimately governs the metabolic activities of the cell. In 1953, Watson, an American biologist, and Crick, an English biologist, proposed the double helix structure for DNA. This development set the stage for a new and continuing era of chemical and biological investigation. The two main events in the life of a cell - dividing to make exact copies of themselves, and manufacturing proteins - both rely on blueprints coded in our genes. Introduction There are two types of nucleic acids which are polymers found in all living cells. Deoxyribonucleic Acid (DNA) is found mainly in the nucleus of the cell, while Ribonucleic Acid (RNA) is found mainly in the cytoplasm of the cell although it is usually synthesized in the nucleus. DNA contains the genetic codes to make RNA and the RNA in turn then contains the codes for the primary sequence of amino acids to make proteins. Nucleic Acid Parts List The best way to understand the structures of DNA and RNA is to identify and examine individual parts of the structures first. The complete hydrolysis of nucleic acids yields three major classes of compounds: pentose sugars, phosphates, and heterocyclic amines (or bases). Phosphate A major requirement of all living things is a suitable source of phosphorus. One of the major uses for phosphorus is as the phosphate ion which is incorporated into DNA and RNA. Pentose Sugars There are two types of pentose sugars found in nucleic acids. This difference is reflected in their names--deoxyribonucleic acid indicates the presence of deoxyribose; while ribonucleic acid indicates the presence of ribose. In the graphic below, the structures of both ribose and deoxyribose are shown. Note the red -OH on one and the red -H on the other are the only differences. The alpha and beta designations are interchangeable and are not a significant difference between the two. Heterocyclic Amines Heterocyclic amines are sometimes called nitrogen bases or simply bases . The heterocyclic amines are derived from two root structures: purines or pyrimidines. The purine root has both a six and a five member ring; the pyrimidine has a single six member ring. There are two major purines, adenine (A) and guanine (G), and three major pyrimidines, cytosine (C), uracil (U), and thymine (T). The structures are shown in the graphic on the left. As you can see, these structures are called "bases" because the amine groups as part of the ring or as a side chain have a basic property in water . A major difference between DNA and RNA is that DNA contains thymine, but not uracil, while RNA contains uracil but not thymine. The other three heterocyclic amines, adenine, guanine, and cytosine are found in both DNA and RNA. For convenience, you may remember, the list of heterocyclic amines in DNA by the words: The Amazing Gene Code (TAGC).
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Bookshelves/Introductory_Chemistry/Beginning_Chemistry_(Ball)/04%3A_Chemical_Reactions_and_Equations/4.03%3A_Types_of_Chemical_Reactions_-_Single_and_Double_Replacement_Reactions
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Learning Objectives Recognize chemical reactions as single-replacement reactions and double-replacement reactions. Use the periodic table, an activity series, or solubility rules to predict whether single-replacement reactions or double-replacement reactions will occur. Up until now, we have presented chemical reactions as a topic, but we have not discussed how the products of a chemical reaction can be predicted. Here we will begin our study of certain types of chemical reactions that allow us to predict what the products of the reaction will be. A single-replacement reaction is a chemical reaction in which one element is substituted for another element in a compound, generating a new element and a new compound as products. Presented below: \[\ce{2HCl(aq) + Zn(s) → ZnCl2(aq) + H2(g)}\nonumber \] is an example of a single-replacement reaction. The hydrogen atoms in \(\ce{HCl}\) are replaced by \(\ce{Zn}\) atoms, and in the process a new element—hydrogen—is formed. Another example of a single-replacement reaction is \[\ce{2NaCl(aq) + F2(g) → 2NaF(s) + Cl2(g)}\nonumber \] Here the negatively charged ion changes from chloride to fluoride. A typical characteristic of a single-replacement reaction is that there is one element as a reactant and another element as a product. Not all proposed single-replacement reactions will occur between two given reactants. This is most easily demonstrated with fluorine, chlorine, bromine, and iodine. Collectively, these elements are called the halogens and are in the next-to-last column on the periodic table (Figure \(\PageIndex{1}\)). The elements on top of the column will replace the elements below them on the periodic table, but not the other way around. Thus, the reaction represented by \[\ce{CaI2(s) + Cl2(g) → CaCl2(s) + I2(s)}\nonumber \] will occur; but the reaction \[\ce{CaF2(s) + Br2(ℓ) → CaBr2(s) + F2(g)}\nonumber \] will not, because bromine is below fluorine on the periodic table. This is just one of many ways the periodic table helps us to understand chemistry. Example \(\PageIndex{1}\) Will a single-replacement reaction occur? If so, identify the products. MgCl 2 + I 2 → ? CaBr 2 + F 2 → ? Solution Because iodine is below chlorine on the periodic table, a single-replacement reaction will not occur. Because fluorine is above bromine on the periodic table, a single-replacement reaction will occur, and the products of the reaction will be CaF 2 and Br 2 . Exercise \(\PageIndex{1}\) Will a single-replacement reaction occur? If so, identify the products. \[\ce{FeI2 + Cl2 → }\nonumber \] Answer Yes; FeCl 2 and I 2 Chemical reactivity trends are easy to predict when replacing anions in simple ionic compounds—simply use their relative positions on the periodic table. However, when replacing the cations, the trends are not as straightforward. This is partly because there are so many elements that can form cations; an element in one column on the periodic table may replace another element nearby, or it may not. A list called the activity series does the same thing the periodic table does for halogens: it lists the elements that will replace elements below them in single-replacement reactions. A simple activity series is shown below. Activity Series for Cation Replacement in Single-Replacement Reactions Li K Ba Sr Ca Na Mg Al Mn Zn Cr Fe Ni Sn Pb H 2 Cu Hg Ag Pd Pt Au Using the activity series is similar to using the positions of the halogens on the periodic table. An element on top will replace an element below it in compounds undergoing a single-replacement reaction. Elements will not replace elements above them in compounds. Example \(\PageIndex{2}\) Use the activity series to predict the products, if any, of each equation. FeCl 2 + Zn → ? HNO 3 + Au → ? Solution Because zinc is above iron in the activity series, it will replace iron in the compound. The products of this single-replacement reaction are ZnCl 2 and Fe. Gold is below hydrogen in the activity series. As such, it will not replace hydrogen in a compound with the nitrate ion. No reaction is predicted. Exercise \(\PageIndex{2}\) Use the activity series to predict the products, if any, of this equation. \[\ce{AlPO4 + Mg → }\nonumber \] Answer Mg 3 (PO 4 ) 2 and Al A double-replacement reaction occurs when parts of two ionic compounds are exchanged, making two new compounds. A characteristic of a double-replacement equation is that there are two compounds as reactants and two different compounds as products. An example is \[\ce{CuCl2(aq) + 2AgNO3(aq) → Cu(NO3)2(aq) + 2AgCl(s)}\nonumber \] There are two equivalent ways of considering a double-replacement equation: either the cations are swapped, or the anions are swapped. (You cannot swap both; you would end up with the same substances you started with.) Either perspective should allow you to predict the proper products, as long as you pair a cation with an anion, and not a cation with a cation or an anion with an anion. Example \(\PageIndex{3}\) Predict the products of this double-replacement equation: \[\ce{BaCl2 + Na2SO4 → }\nonumber \] Solution Thinking about the reaction as either switching the cations or switching the anions, we would expect the products to be BaSO 4 and NaCl. Exercise \(\PageIndex{3}\) Predict the products of this double-replacement equation: \[\ce{KBr + AgNO3 → }\nonumber \] Answer KNO 3 and AgBr Predicting whether a double-replacement reaction occurs is somewhat more difficult than predicting a single-replacement reaction. However, there is one type of double-replacement reaction that we can predict: the precipitation reaction. A precipitation reaction occurs when two ionic compounds are dissolved in water and form a new ionic compound that does not dissolve; this new compound falls out of solution as a solid precipitate. The formation of a solid precipitate is the driving force that makes the reaction proceed. 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 double-replacement 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/City_College_of_San_Francisco/Chemistry_101B/07%3A_Representative_Metals_Metalloids_and_Nonmetals/7.06%3A_Occurrence%2C_Preparation%2C_and_Properties_of_Carbonates
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Learning Objectives 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: \[ \begin{align} \ce{Na2O}(s)+\ce{CO2}(g) &⟶ \ce{Na2CO3}(s)\\[4pt] \ce{Ca(OH)2}(s)+\ce{CO2}(g) &⟶\ce{CaCO3}(s)+\ce{H2O}(l) \end{align}\] 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: \[ \begin{align} \ce{Ca^2+}(aq)+\ce{CO3^2-}(aq) &⟶\ce{CaCO3}(s) \\[4pt] \ce{Pb^2+}(aq)+\ce{CO3^2-}(aq) &⟶\ce{PbCO3}(s) \end{align}\] 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. 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: \[\ce{CaCO3}(s)+\ce{2HCl}(aq)⟶\ce{CaCl2}(aq)+\ce{CO2}(g)+\ce{H2O}(l)\] 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
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Bookshelves/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_II_(Wade)/26%3A_Lipids
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26.1: Introduction Although many lipids have complex structures, they are all insoluble in water. The major lipid classifications are: fatty acids, steroids, waxes, triglycerides (fats and oils), phospholipids, sphingolipids, glycolipids, terpenes, and prostaglandins (eicosanoids). 26.2: Waxes, Fats, and Oils Fats play an important role in human nutrition, and most people are aware of the desirability of limiting their dietary intake of saturated fats, as these compounds have been associated with heart disease. Unsaturated fats are generally considered to be much more desirable from the point of view of good health. Notice that all the fatty acids derived from naturally occurring fats have a Z (i.e., cis) configuration. 26.3: Saponification of Fats and Oils; Soaps and Detergents Soaps are the carboxylate salts of fatty acids, while detergents are sulfonate salts with long hydrocarbon tails. 26.4: Phospholipids Phospholipids are the main constituents of cell membranes. 26.5: Prostaglandins and other Eicosanoids Prostaglandins, are like hormones in that they act as chemical messengers, but do not move to other sites, but work right within the cells where they are synthesized. 26.6: Terpenes and Terpenoids The terpenoids are an estimated 60% of known natural products and a diverse group of lipids derived from five-carbon isoprene units assembled in thousands of combinations. Technically a terpenoid contains oxygen, while a terpene is a hydrocarbon, however, the two terms are commonly used to refer collectively to both groups. 26.7: Steroids Steroids may be recognized by their tetracyclic skeleton consisting of three fused six-membered and one five-membered ring. 26.8: Biosynthesis of Steroids A series of cation-like cyclizations and rearrangements, known as the Stork-Eschenmoser hypothesis, were identified in the biosynthesis of the triterpene lanosterol. Lanosterol is a precursor in the biosynthesis of steroids. Template:HideTOC
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Courses/Earlham_College/CHEM_361%3A_Inorganic_Chemistry_(Watson)/05%3A_Solid_State_Chemistry/5.06%3A_Prelude_to_Ionic_and_Covalent_Solids_-_Structures/5.6.03%3A_Tetrahedral_Structures
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In ccp and hcp lattices, there are two tetrahedral holes per packing atom. A stoichiometry of either M 2 X or MX 2 gives a structure that fills all tetrahedral sites, while an MX structure fills only half of the sites. An example of an MX 2 structure is fluorite , CaF 2 , whose structure is shown in the figure at the left. The packing atom in fluorite is Ca 2+ and the structure is composed of three interpenetrating fcc lattices. It should be noted that the Ca 2+ ion (gray spheres) as a packing atom defies our "rule" that anions are larger than cations and therefore must be the packing atoms. The fluorite structure is common for ionic MX 2 (MgF 2 , ZrO 2 , etc.) and M 2 X compounds (Li 2 O). In contrast, the hcp relative of the fluorite structure is quite rare because of unfavorable close contacts between like-charged ions. 0 The fluorite (CaF2) crystal structure showing the coordination environments of the Ca and F atoms In terms of geometry, Ca 2+ is in cubic coordination with eight F - neighbors, and the fluoride ions are tetrahedrally coordinated by four Ca 2+ ions. The 8:4 coordination geometry is consistent with the 1:2 Ca:F stoichiometry; in all crystal structures the ratio of the coordination numbers is the inverse of the stoichiometric ratio. The three interpenetrating fcc lattices have Ca at 0,0,0 , 1/2,1/2,0 , etc....F at 1/4,1/4,1/4 , 3/4,3/4,1/4 , etc... and F at 3/4,3/4,3/4 , 1/4,1/4/3/4 , etc. Looking more closely at the tetrahedral sites in fluorite, we see that they fall into two distinct groups: T + and T - . If a tetrahedron is oriented with a vertex pointing upwards along the stacking axis, the site is T + . Likewise, a tetrahedron with a vertex oriented downward is T - . The alternation of T + and T - sites allows for efficient packing of ions in the structure. The layer stacking sequence in this structure (including fluoride ions in the T + and T - sites) is: ------------ A - - -b- - - T + - - -a- - - T - ------------ B - - -c- - - T + - - -b- - - T - ------------ C - - -a- - - T + - - -c- - - T - ------------ A - - -b- - - T + - - -a- - - T - 0 Polyhedral view of the fluorite crystal structure, showing T+ and T- Ca4F tetrahedra. The Ca2+ ions are stacked ABCABC... along the body diagonal of the unit cell, which is the vertical direction in this image. Tetrahedrally bonded compounds with a 1:1 stoichiometry (MX compounds) have only half of the tetrahedral sites (either the T + or T - sites) filled. In this case, both the M and the X atoms are tetrahedrally coordinated. The zincblende and wurtzite structures of ZnS are 1:1 tetrahedral structures based on fcc and hcp lattices, respectively. Both structures are favored by p-block compounds that follow the octet rule, and these compounds are usually semiconductors or insulators. The zincblende structure, shown below, can be thought of as two interpenetrating fcc lattices, one of anions and one of cations, offset from each other by a translation of 1/4 along the body diagonal of the unit cell. Examples of compounds with the zincblende structure include CuCl, CuI, ZnSe, HgS, BeS, CdTe, AlP, GaP, SnSb, CSi, and diamond. Additionally, the compound CuInSe 2 is zincblende in an ordered, doubled unit cell (the chalcopyrite structure). The solid solution compounds CuIn 1-x Ga x Se 2 with this structure are among the most widely studied materials for use in efficient thin film photovoltaic cells. Using ZnS as a representative of zincblende, the coordination of both Zn and S atoms is tetrahedral. The layer sequence, which is AbBcCaAbBcC..., results in non-planar, six-membered ZnS rings which allows for a relatively long distance between opposite atoms in the ring. The sequence of close-packed layers in zincblende, filling only the T + sites and leaving the T - sites empty, is shown below: ------------ A - - -b- - - T + - - - - - - T - ------------ B - - -c- - - T + - - - - - - T - ------------ C - - -a- - - T + - - - - - - T - 0 The zincblende unit cell The wurtzite structure is a close relative of zinc blende, based on filling half the tetrahedral holes in the hcp lattice. Like zincblende, wurtzite contains planes of fused six-membered rings in the chair conformation. Unlike zincblende, however, the rings joining these planes contain non-planar, six-membered rings. The structure aligns the anions so that they are directly above the cations in the structure, a less favorable situation sterically but a more favorable one in terms of electrostatics. As a result, the wurtzite structure tends to favor more polar or ionic compounds (e.g., ZnO, NH 4 + F - ) than the zincblende structure. As with zincblende, both ions are in tetrahedral (4:4) coordination and there are typically eight valence electrons in the MX compound. Examples of compounds with this structure include: BeO, ZnO, MnS, CdSe, MgTe, AlN, and NH 4 F. The layered structure of wurtzite is AbBaAbB and the layer sequence with T + sites filled is illustrated below: ------------ A - - -b- - - T + - - - - - - T - ------------ B - - -a- - - T + - - - - - - T - ------------ A - - -b- - - T + - - - - - - T - ------------ B 0 The chair and boat conformations of six-membered ZnS rings in the wurtzite structure. An interesting consequence of the layer stacking in the wurtzite structure is that the crystals are polar. When cleaved along the c-axis (the stacking axis), crystals of ZnO, ZnS, and GaN have one negatively charged face and an opposite positively charged face. An applied electric field interacts with the crystal dipole, resulting in compression or elongation of the lattice along this direction. For this reason crystals of compounds in the wurtzite structure are typically piezoelectric (increasing the pressure on the material generates a voltage in the material). Some compounds are diamorphic and can have either the zincblende or wurtzite structure. Examples of these compounds that have intermediate polarities include CdS and ZnS. SiO 2 exists in polymorphs (crystobalite and tridymite) that resemble zincblende and wurtzite with O atoms midway between each of the Si atoms. The zincblende and wurtzite structures have efficient packing arrangements for tetrahedrally bonded networks and are commonly found in compounds that have tetrahedral bonding. Water, for example, has a tetrahedral hydrogen bonding network and is wurtzite-type. The undistorted wurtzite and zinc blende structures are typically found for AX compounds with eight valence electrons, which follow the octet rule. AX compounds with nine or ten electrons such as GaSe and GaAs crystallize in distorted variants of the wurtzite structure. In GaSe, the extra electrons form lone pairs and this creates layers in the structure, as can be seen in the figure below. To the right of GaSe, the structures of As, Sb, and SbAs show an ever further breakdown of the structure into layers as more valence electrons are added. Hexagonal ice is the most stable polymorph of ice, which is obtained upon freezing at 1 atmosphere pressure. This polymorph (ice-I) has a hcp wurtzite-type structure. Looking at the structure shown at the right, we see that there are irregular arrangements of the O-H---O bonds. In the structure, hydrogen bonding enforces the tetrahedral coordination of each water molecule, resulting in a relatively open structure that is less dense than liquid water. For this reason, ice floats in water.
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Bookshelves/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/10%3A_Alkynes/10.01%3A_Structure_and_Physical_Properties
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Learning Objective apply bonding theories to the structure of alkynes and distinguish between internal and terminal triple bonds Alkynes: Terminal vs Internal Alkynes are organic molecules with carbon-carbon triple bonds. They are unsaturated hydrocarbons with the empirical formula of CnH 2n-2 . The simplest alkyne is ethyne which has the common name acetylene. Acetylene is a common name to memorize. It is important to distinguish between terminal and internal alkynes because they can undergo different patterns of reactivity. Electronic Structure The sp hybridization of the carbon-carbon triple bond results in the perpendicular orientation of the sigma bond and two pi bonds. The close proximity of the electrons in this geometry orientation creates molecules with less stability. The structure of the carbon-carbon triple bond strongly influences the chemical reactivity of alkynes and the acidity of terminal alkynes . Because of its linear configuration (the bond angle of a sp-hybridized carbon is 180º), a ten-membered carbon ring is the smallest that can accommodate this function without excessive strain. Physical Properties Alkynes are nonpolar, unsaturated hydrocarbons with physical properties similar to alkanes and alkenes. Alkynes dissolve in organic solvents, have slight solubility in polar solvents, and are insoluble in water. Compared to alkanes and alkenes , alkynes have slightly higher boiling points. For example, ethane has a boiling point of -88.6 C, while e thene is -103.7 C and e thyne has a higher boiling point of -84.0 ?C. Exercise Arrange ethane, ethene, and acetylene in order of decreasing carbon-carbon length. How many pi bonds and sigma bonds are involved in the structure of ethyne ? What contribute to the weakness of the pi bonds in an alkyne? Arrange the following hydrocarbons in order of decreasing boiling point: 1-heptyne, 1-hexyne, 2-methyl-1-hexyne. Predict the solvent with greater 2-butyne solubility. a) water or 1-octanol? b) water or acetone? c) ethanol or hexane? Answer 1. relative carbon-carbon bond length: ethane < ethene < acetylene 2. There are three sigma bonds and two pi bonds. 3. The sigma bond and two pi bonds are all perpendicular to each other in the triple bond creating electron repulsion between the three pairs of bonding electrons in the triple bond. 4. 1-heptyne (99.7C) > 2-methyl-1-hexyne (91C) > 1-hexyne (71C) 5. a) 1-octanol b) acetone c) hexane Outside links www.ucc.ie/academic/chem/dolc...t/alkynes.html www.cliffsnotes.com/WileyCDA/...eId-22631.html References Bloch, D.R. Organic chemistry demystified, New York : McGraw-Hill, 2006. Vollhardt. Schore, Organic Chemistry Structure and Function Fifth Edition, New York: W.H. Freeman and Company, 2007.
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Bookshelves/General_Chemistry/Chem1_(Lower)/16%3A_Electrochemistry/16.07%3A_Timeline_of_Battery_Development
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Although the development practical batteries largely paralelled the expansion of electrical technology from about the mid-19th century on, it is now thought that a very primitive kind of battery was apparently in use more than 2000 years ago. The brief popularity of electrically powered automobiles in the 1920's encouraged storage battery development. The widespread use of portable "personal" electrical devices has kept the search for better batteries very much alive. 0 1 "Baghdad Battery" - 1000 BCE? Drawing of the three pieces. (CC-BY-SA 2.5; Ironie) Earthenware jars containing an iron rod surrounded by a copper cylinder were discovered near Baghdad in 1938. They are believed to have been used by the Parthian civilization that occupied the region about 2000 years ago as a source of electricity to plate gold onto silver. Allesandro Volta 1782 His "Voltaic pile", a stack of zinc and silver disks separated by a wet cloth containing a salt or a weak acid solution, was the first battery known to Western civilization. Sir Humphry Davy 1813 Davy builds a 2000-plate battery that occupies 889 square feet in the basement of Britain's Royal Society. His earlier batteries provided power for the first public demonstration of electric lighting (carbon arc). Michael Faraday, 1830's Faraday discovered the fundamentals of galvanic cells and electrolysis that put electrochemistry on a firm scientific basis. 1836 - Daniell cell (also known as a Crow's Foot or Gravity cell.) John Daniell (English chemist and meterologist) developed the first modern storage cell based on Faraday's principles. This consists of a large glass jar with a copper star-shaped electrode in the bottom and a zinc "crow's foot" shaped electrode suspended near the top. The bottom of the jar was filled with a concentrated copper sulfate solution. On top of this was poured dilute sulfuric acid, whose lower density kept it on top. This was the first practical battery to find wide use to power telegraphs and railway signaling systems and home doorbells. 1839 - William Grove (Welsh) Grove was best known in the 19th century for his "nitric acid battery" which came into wide use in early telegraphy. Now, however, he is most famous for his "gas voltaic battery" in which discovered "reverse electrolysis": the recombination of H2 and O2 following electrolysis of water at platinum electrodes. This was the first demonstration of what we now know as the hydrogen-oxygen fuel cell (see below.) 1859 - Gaston Planté (French) Invents the first lead-acid storage cell which consisted of two sheets of lead separated by a rubber sheet, rolled into a spiral and immersed in dilute sulfuric acid. 1866 - Georges Leclanché (French) By 1868 twenty thousand Leclanché cells were being used in telegraph systems. The original Leclanché cells were built in porous pots which were heavy and subject to breakage. Within twenty years other inventors had modified the design into what we now know as "dry cells" which became widely used in the first flashlights (1909) and in battery-powered radios of the 1920s. 1881 - Faure and others Development of the first practical lead-acid storage cell. The major improvement over Planté's design was the addition of a paste of PbSO4 to the positive plate. 1905 Nickel-iron cell (Thomas Edison) Edison, who was as much a chemist as an all-around inventor, thought that the lead in Planté-type cells made them too heavy, and that having acid in contact with any metal was an inherently bad idea. After much experimentation, he developed a successful alkaline battery. The Edison cell uses an iron anode, nickel oxide cathode, and KOH electrolyte. This cell is extremely rugged and is still used in certain industrial applications, but it was never able to displace the lead-acid cell as Edison had hoped. 1950s A similar cell, employing a nickel anode instead of iron, was the first rechargeable cell that was small enough to be used in portable consumer devices. Its main disadvantage is that it is ruined by complete discharge. 1949 - Alkaline dry cell - Lew Urry (Eveready Battery Co.) First commercial alkaline dry cell. These substitute KOH for the corrosive NH4Cl used in the older dry cells and last 5-8 times longer. 1947 - Mercury cell (Ruben and Mallory, 1950's) This was one of the first "button"-type cells which were widely used in cameras and hearing aids. The constancy of the 1.34 v output made them popular for use in sensitive instruments and cardiac pacemakers. The net cell reaction is Zn(s) + HgO(s) → ZnO(s) + Hg(l) Most countries have outlawed sales of these cells in order to reduce mercury contamination of the environment. Nickel-Cadmium (NiCad) cells The NiCad cell quickly become one of the most popular rechargeable batteries for small consumer devices. They can deliver high current and undergo hundreds of charge/discharge cycles. Because cadmium is an environmental toxin, their use is being discouraged. 1959 - Fuel cell - Francis Bacon (UK) The first practical fuel cell was developed by British engineer Francis Bacon (1904-1992). This hydrogen-oxygen cell used an alkaline electrolyte and inexpensive nickel electrodes. Late 1960's - Nickel-metal hydride cells The hydride ion H– would be an ideal cathode material except for the fact that its oxidation product H2 is a gas. The discovery that certain compounds such as LiNi5 and ZrNi2 can act as "hydrogen sponges" made it practical to employ metal hydrides as a cathode material. One peculiarity of Ni-MH cells is that recharging them is an exothermic process, so that proper dissipation of heat must be allowed for. These batteries are widely used in cell phones, computers, and portable power tools. The electrode reactions take place in a concentrated KOH electrolyte: Cathode (+): NiOOH + H2O + e–→ Ni(OH)2 + OH– Anode (-): (1/x) MHx + OH–→ (1/x) M + H2O + e– 1990s - Lithium cells (Sony Corp.) Lithium is an ideal anode material owing to its low density and high reduction potential, making Li-based cells the most compact ways of storing electrical energy. Lithium cells are used in wristwatches, cardiac pacemakers and digital cameras. Both primary (non-rechargeble) and rechargeable types have been available for some time. More recent applications are in portable power tools and— perhaps most importantly, in electric-powered or hybrid automobiles. Modern lithium cells operate by transporting Li+ ions between electrodes into which the ions can be inserted or intercalated. Cathodes are lithium transition-metal oxides such as LiCoO3, while anodes are lithium-containing carbon, LiC6. The species that undergoes oxidation-reduction is not lithium, but the transition metal, e.g. Co(III)-Co(IV). Lithium batteries as incendiary devices There have been numerous reports of fires and explosions associated with lithium batteries. In 2006, the Dell Corporation had to recall 4.1 million Sony batteries that had been shipped with Dell's laptop computers and were judged to be at risk owing to a manufacturing defect. This illustrates the difficulty of concentrating a large amount of chemical energy into a small package, which is of course the goal of all battery developers eager to meet commercial demands ranging from consumer personal electronics to electrically-powered cars. The fully-charged Li + -deficient lithium cobalt oxide cathodes are inherently unstable, held in check only by a thin insulating membrane which, if accidentally breached, can lead to thermal runaway involving gaseous oxygen, carbon, organic solvents, and (in some cases) lithium chlorate— all the components necessary for a fierce fire. Much research has gone into the development of fail-safe membranes. In one type, made by ExxonMobil and targeted at the automotive market, the pores are designed to close up and thus inhibit the passage of lithium ions when the temperature rises above a safe level. Biological Batteries Finally, we should mention the biological batteries that are found in a number of electric fish . The "electric organs" of these fish are modified muscle cells known as electrocytes which are arranged in long stacks. A neural signal from the brain causes all the electrocytes in a stack to become polarized simultaneously, in effect creating a battery made of series-connected cells. Most electric fish produce only a small voltage which they use for navigation, much in the way that bats use sound for echo-location of prey. The famous electric eel, however, is able to produce a 600-volt jolt that it employs to stun nearby prey.
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Courses/Williams_School/Chemistry_II/03%3A_Stereochemistry_at_Tetrahedral_Centers/3.07%3A_Racemic_Mixtures_and_the_Resolution_of_Enantiomers
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Objectives After completing this section, you should be able to describe a common process for separating a mixture of enantiomers. explain why racemic mixtures do not rotate plane-polarized light. Key Terms Make certain that you can define, and use in context, the key terms below. racemic mixture (or racemate) resolve Study Notes A racemic mixture is a 50:50 mixture of two enantiomers. Because they are mirror images, each enantiomer rotates plane-polarized light in an equal but opposite direction and is optically inactive. If the enantiomers are separated, the mixture is said to have been resolved . A common experiment in the laboratory component of introductory organic chemistry involves the resolution of a racemic mixture. The dramatic biochemical consequences of chirality are illustrated by the use, in the 1950s, of the drug Thalidomide, a sedative given to pregnant women to relieve morning sickness. It was later realized that while the (+)‑form of the molecule, was a safe and effective sedative, the (−)‑form was an active teratogen. The drug caused numerous birth abnormalities when taken in the early stages of pregnancy because it contained a mixture of the two forms. As noted earlier, chiral compounds synthesized from achiral starting materials and reagents are generally racemic (i.e. a 50:50 mixture of enantiomers). Separation of racemates into their component enantiomers is a process called resolution. Since enantiomers have identical physical properties, such as solubility and melting point, resolution is extremely difficult. Diastereomers, on the other hand, have different physical properties, and this fact is used to achieve resolution of racemates. Reaction of a racemate with an enantiomerically pure chiral reagent gives a mixture of diastereomers, which can be separated. For example, if a racemic mixture of a chiral alcohol is reacted with a enantiomerically pure carboxylic acid, the result is a mixture of diastereomers: in this case, because the pure (R) entantiomer of the acid was used, the product is a mixture of (R-R) and (R-S) diastereomeric esters, which can, in theory, be separated by their different physical properties. Subsequent hydrolysis of each separated ester will yield the 'resolved' (enantiomerically pure) alcohols. The used in this technique are known as 'Moscher's esters', after Harry Stone Moscher, a chemist who pioneered the method at Stanford University. As noted earlier, chiral compounds synthesized from achiral starting materials and reagents are generally racemic (i.e. a 50:50 mixture of enantiomers). Separation of racemates into their component enantiomers is a process called resolution . Since enantiomers have identical physical properties, such as solubility and melting point, resolution is extremely difficult. Diastereomers, on the other hand, have different physical properties, and this fact is used to achieve resolution of racemates. Reaction of a racemate with an enantiomerically pure chiral reagent gives a mixture of diastereomers, which can be separated. Reversing the first reaction then leads to the separated enantiomers plus the recovered reagent. Many kinds of chemical and physical reactions, including salt formation, may be used to achieve the diastereomeric intermediates needed for separation. Figure 5.8.1 illustrates this general principle by showing how a nut having a right-handed thread (R) could serve as a "reagent" to discriminate and separate a mixture of right- and left-handed bolts of identical size and weight. Only the two right-handed partners can interact to give a fully-threaded intermediate, so separation is fairly simple. The resolving moiety, i.e. the nut, is then removed, leaving the bolts separated into their right and left-handed forms. Chemical reactions of enantiomers are normally not so dramatically different, but a practical distinction is nevertheless possible. Because the physical properties of enantiomers are identical, they seldom can be separated by simple physical methods, such as fractional crystallization or distillation. It is only under the influence of another chiral substance that enantiomers behave differently, and almost all methods of resolution of enantiomers are based upon this fact. We include here a discussion of the primary methods of resolution. Chiral Amines as Resolving Agents and Resolution of Racemic Acids The most commonly used procedure for separating enantiomers is to convert them to a mixture of diastereomers that will have different physical properties: melting point, boiling point, solubility, and so on (Section 5-5). For example, if you have a racemic or ( R )/( S ) mixture of enantiomers of a carboxylic acid and convert this to a salt with a chiral amine base having the ( R ) configuration, the salt will be a mixture of two diastereomers, (( R)- acid・( R )-base) and (( S )-acid・( R )-base). These diastereomeric salts are not identical and they are not mirror images. Therefore they will differ to some degree in their physical properties, and a separation by physical methods, such as crystallization, may be possible. If the diastereomeric salts can be completely separated, the carboxylic acid regenerated from each salt will be either exclusively the ( R ) or the ( S ) enantiomer. Resolution of chiral acids through the formation of diastereomeric salts requires adequate supplies of suitable chiral bases. Brucine, strychnine, and quinine frequently are used for this purpose because they are readily available, naturally occurring chiral bases. Simpler amines of synthetic origin, such as 2-amino-1-butanol, amphetamine, and 1-phenylethanamine, also can be used, but first they must be resolved themselves. Worked Example \(\PageIndex{1}\) Show how ( S )-1-phenylethylamine can be used to resolve a racemic mixture of lactic acid. Please draw all the structures involved. Answer Resolution of Racemic Bases Chiral acids, such as (+)-tartaric acid, (-)-malic acid, (-)-mandelic acid, and (+)-camphor- 10-sulfonic acid, are used for the resolution of a racemic base. The principle is the same as for the resolution of a racemic acid with a chiral base, and the choice of acid will depend both on the ease of separation of the diastereomeric salts and, of course, on the availability of the acid for the scale of the resolution involved. Resolution methods of this kind can be tedious, because numerous recrystallizations in different solvents may be necessary to progressively enrich the crystals in the less-soluble diastereomer. To determine when the resolution is complete, the mixture of diastereomers is recrystallized until there is no further change in the measured optical rotation of the crystals. At this stage it is hoped that the crystalline salt is a pure diastereomer from which one pure enantiomer can be recovered. The optical rotation of this enantiomer will be a maximum value if it is "optically" pure because any amount of the other enantiomer could only reduce the magnitude of the measured rotation \(\alpha\). Resolution of Racemic Alcohols To resolve a racemic alcohol, a chiral acid can be used to convert the alcohol to a mixture of diastereomeric esters. This is not as generally useful as might be thought because esters tend to be liquids unless they are very high-molecularweight compounds. If the diastereomeric esters are not crystalline, they must be separated by some other method than fractional crystallization (for instance, by chromatography methods, Section 9-2). Two chiral acids that are useful resolving agents for alcohols are: The most common method of resolving an alcohol is to convert it to a half-ester of a dicarboxylic acid, such as butanedioic (succinic) or 1,2-benzenedicarboxylic (phthalic) acid, with the corresponding anhydride. The resulting half-ester has a free carboxyl function and may then be resolvable with a chiral base, usually brucine: Other Methods of Resolution One of the major goals in the field of organic chemistry is the development of reagents with the property of "chiral recognition" such that they can affect a clean separation of enantiomers in one operation without destroying either of the enantiomers. We have not achieved that ideal yet, but it may not be far in the future. Chromatographic methods, whereby the stationary phase is a chiral reagent that adsorbs one enantiomer more strongly than the other, have been used to resolve racemic compounds, but such resolutions seldom have led to both pure enantiomers on a preparative scale. Other methods, called kinetic resolutions, are excellent when applicable. The procedure takes advantage of differences in reaction rates of enantiomers with chiral reagents. One enantiomer may react more rapidly, thereby leaving an excess of the other enantiomer behind. For example, racemic tartaric acid can be resolved with the aid of certain penicillin molds that consume the dextrorotatory enantiomer faster than the levorotatory enantiomer. As a result, almost pure (-)-tartaric acid can be recovered from the mixture: (±)-tartaric acid + mold \(\rightarrow\) (-)-tartaric acid + more mold The crystallization procedure employed by Pasteur for his classical resolution of (±)-tartaric acid ( Section 5.4 ) has been successful only in a very few cases. This procedure depends on the formation of individual crystals of each enantiomer. Thus if the crystallization of sodium ammonium tartrate is carried out below 27", the usual racemate salt does not form; a mixture of crystals of the (+) and (-) salts forms instead. The two different kinds of crystals, which are related as an object to its mirror image, can be separated manually with the aid of a microscope and subsequently may be converted to the tartaric acid enantiomers by strong acid. A variation on this method of resolution is the seeding of a saturated solution of a racemic mixture with crystals of one pure enantiomer in the hope of causing crystallization of just that one enantiomer, thereby leaving the other in solution. Unfortunately, very few practical resolutions have been achieved in this way. Predicating the Chirality of the Product of a Reaction It important to understand the changes in chirality which occur during the formation of product during a reaction. A chiral reaction product, has the possibility of forming multiple stereoisomers which all need to be considered. Changes in chirality, if possible, will be discussed with each individual reaction as this textbook moves forward. Some possible situations which can occur are: A new chiral carbon is formed during a reaction. This commonly occurs when an sp 2 hybridized carbon in the reactant is converted to sp 3 hybridized chiral carbon in the product. When this occurs, a racemic mixture of the new chiral carbon is formed. A chiral carbon is lost during a reaction. This commonly occurs when an sp 3 hybridized chiral carbon in the reactant is converted to either a sp 2 or sp hybridized carbon in the product. An enantiomerically pure starting material is converted to a racemic mixture in the product. This commonly occurs when a sp 3 hybridized chiral carbon is temporarily converted to an sp 2 hybridized carbon during a reaction's mechanism. The chiral carbon is reformed as a racemic mixture. Chiral carbons remain unchanged during a reaction. If a chiral carbon is not directly involved in a reaction, it will move from a reactant to a product unchanged. Determining if a chiral carbon is involved in a given reaction is vital for determining which of these four situations is occurring. Worked Example \(\PageIndex{2}\) The following reaction involves the conversion of a carboxylic acid reacting with an alcohol to form an ester. If a pure sample of (R)-2-methylbutanoic acid is reacted with methanol to form an ester, what would be the stereochemistry of the product? Answer First it is important to identify the location of the chiral carbon and determine if it is directly involved in the reaction. In this case, the chiral carbon is not involved so the stereochemistry will be carried over into the product unchanged. Exercise \(\PageIndex{1}\) Indicate the reagents you could use to resolve the following compounds. Show the reactions involved and specify the physical method you believe would be the best to separate the diastereomers of 1 -phenyl-2-propanamine. Answer You could react the 1-phenyl-2-propanamine racemic mixture with a chiral acid such as (+)-tartaric acid ( R , R ). The reaction will produce a mixture of diastereomeric salts (i.e. R , R , R and S , R , R ). You can separate the diastereomers through crystallization and treat the salt with a strong base (e.g. KOH) to recover the pure enantiomeric amine. Exercise \(\PageIndex{2}\) Indicate the reagents you would use to resolve the following and discuss the reactions involved and specify the physical method you believe would be the best to separate the diastereomers of 2,3-pentadienedioic acid. Answer You could react the 2,3-pentadienedioic acid mixture with a chiral base such as ( R )‑1‑phenylethylamine. The reaction will produce a mixture of diastereomeric salts. Separate the diastereomers through crystallization and treat the resulting salt with strong acid (e.g. HCl) to recover the pure enantiomeric acid. Exercise \(\PageIndex{3}\) Indicate the reagents you would use to resolve the following and discuss the reactions involved and specify the physical method you believe would be the best to separate the diastereomers of 1 -phenylethanol. Answer You could react the 1-phenylethanol mixture with 1,2-benzenedicarboxylic anhydride. The reaction will produce a mixture of diastereomeric salts. You could then separate the diastereomers through crystallization and then alkaline hydrolysis treatment should recover the pure enantiomeric alcohol.
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Courses/Ursinus_College/CHEM322%3A_Inorganic_Chemistry/05%3A_Molecular_Orbitals/5.03%3A_Diatomic_MO_Diagrams/5.3.01%3A_Homonuclear_Diatomic_Molecules/5.3.1.03%3A_Diatomic_Molecules_of_the_First_and_Second_Periods
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First Period Homonuclear Diatomic Molecules In the first row of the periodic table, the valence atomic orbitals are \(1s\). There are two possible homonuclear diatomic molecules of the first period: Dihydrogen, H 2 \([\sigma_g^2(1s)]\): This is the simplest diatomic molecule. It has only two molecular orbitals (\(\sigma_g\) and \(\sigma_u^{*}\)), two electrons, a bond order of 1, and is diamagnetic. Its bond length is 74 pm . MO theory would lead us to expect bond order to decrease and bond length to increase if we either add or subtract one electron. The calculated bond length for the \(\ce{H_2^{+}}\) ion is approximately 105 pm . 1 Dihelium, He 2 \([\sigma_g^2\sigma_u^{*2}(1s)]\): This molecule has a bond order of zero due to equal number of electrons in bonding and antibonding orbitals. Like other nobel gases, He exists in the atomic form and does not form bonds at ordinary temperatures and pressures. Exercise \(\PageIndex{1}\) Draw the complete molecular orbital diagrams for \(\ce{H2}\) and for \(\ce{He2}\). Include sketches of the atomic and molecular orbitals. Answer A complete molecular orbital diagram includes all atomic orbitals and molecular orbitals, their symmetry labels, and electron filling. Second Period Homonuclear Diatomic Molecules The second period elements span from Li to Ne. The valence orbitals are 2s and 2p. In their molecular orbital diagrams, non-valence orbitals (1s in this case) are often disregarded in molecular orbital diagrams. Orbital mixing has significant consequences for the magnetic and spectroscopic properties of second period homonuclear diatomic molecules because it affects the order of filling of the \(\sigma_g(2p)\) and \(\pi_u(2p)\) orbitals. Early in period 2 (up to and including nitrogen), the \(\pi_u(2p)\) orbitals are lower in energy than the \(\sigma_g(2p)\) (see Figure \(\PageIndex{1})\). However, later in period 2, the \(\sigma_g(2p)\) orbitals are pulled to a lower energy. This lowering in energy of \(\sigma_g(2p)\) is not unique; all of the \(\sigma\) orbitals in the molecule are pulled to lower energy due to the increasing positive charge of the nucleus. The \(\pi\) orbitals in the molecule are also affected, but to a much lesser extent than \(\sigma\) orbitals. The reason has to do with the high penetration of \(s\) atomic orbitals compared to \(p\) atomic orbitals (recall our previous discussion on penetration and shielding , and its effect on periodic trends ). The \(\sigma\) molecular orbitals have more \(s\) character and thus their energy is more influenced by increasing nuclear charge. As nuclear charge increases, the energy of the \(\sigma_g(2p)\) orbital is lowered significantly more than the energy of the \(\pi_u(2p)\) orbitals (Figure \(\PageIndex{1})\). Dilithium, Li 2 \([\sigma_g^2(2s)]\): This molecule has a bond order of one and is observed experimentally in the gas phase to have one Li-Li bond. Diberylium, Be 2 \([\sigma_g^2\sigma_u^{*2}(2s)]\): This molecule has a bond order of zero due to the equal number of electrons in bonding and antibonding orbitals. Although Be 2 does not exist under ordinary conditions, it can be produced in a laboratory and its bond length measured (Figure \(\PageIndex{2}\)). Although the bond is very weak, its bond length is surprisingly ordinary for a covalent bond of the second period elements. 2 Diboron, B 2 \([\sigma_g^2\sigma_u^{*2}(2s)\pi_u^1\pi_u^1(2p)]\): The case of diboron is one that is much better described by molecular orbital theory than by Lewis structures or valence bond theory. This molecule has a bond order of one. The molecular orbital description of diboron also predicts, accurately, that diboron is paramagnetic. The paramagnetism is a consequence of orbital mixing, resulting in the \(\sigma_g\) orbital's being at a higher energy than the two degenerate \(\pi_u^*\) orbitals. Dicarbon, C 2 \([\sigma_g^2\sigma_u^{*2}(2s)\pi_u^2\pi_u^2(2p)]\): This molecule has a bond order of two. Molecular orbital theory predicts two bonds with \(\pi\) symmetry, and no \(\sigma\) bonding. C 2 is rare in nature because its allotrope, diamond, is much more stable. Dinitrogen, N 2 \([\sigma_g^2\sigma_u^{*2}(2s)\pi_u^2\pi_u^2\sigma_g^2(2p)]\): This molecule is predicted to have a triple bond. This prediction is consistent with its short bond length and bond dissociation energy. The energies of the \(\sigma_g(2p)\) and \(\pi_u(2p)\) orbitals are very close, and their relative energy levels have been a subject of some debate (see next section for discussion). Dioxygen, O 2 \([\sigma_g^2\sigma_u^{*2}(2s)\sigma_g^2\pi_u^2\pi_u^2\pi_g^{*1}\pi_g^{*1}(2p)]\): This is another case where valence bond theory fails to predict actual properties. Molecular orbital theory correctly predicts that dioxygen is paramagnetic, with a bond order of two. Here, the molecular orbital diagram returns to its "normal" order of orbitals where orbital mixing could be somewhat ignored, and where \(\sigma_g(2p)\) is lower in energy than \(\pi_u(2p)\). Difluorine, F 2 \([\sigma_g^2\sigma_u^{*2}(2s)\sigma_g^2\pi_u^2\pi_u^2\pi_g^{*2}\pi_g^{*2}(2p)]\): This molecule has a bond order of one and like oxygen, the \(\sigma_g(2p)\) is lower in energy than \(\pi_u(2p)\). Dineon, Ne 2 \([\sigma_g^2\sigma_u^{*2}(2s)\sigma_g^2\pi_u^2\pi_u^2\pi_g^{*2}\pi_g^{*2}\sigma_u^{*2}(2p)]\): Like other noble gases, Ne exists in the atomic form and does not form bonds at ordinary temperatures and pressures. Like Be 2 , Ne 2 is an unstable species that has been created in extreme laboratory conditions and its bond length has been measured (Figure \(\PageIndex{2}\)) Exercise \(\PageIndex{2}\) Draw the complete molecular orbital diagram for O 2 . Show calculation of its bond order and tell whether it is diamagnetic or paramagnetic. Answer O 2 is paramagnetic with a bond order of 2. Its \(\sigma_g(2p)\) molecular orbital is lower in energy that the set of \(\pi_{u}(2p)\) orbitals. Bond order \(=\frac{1}{2}\left[\left(\begin{array}{c}\text { 8 electrons in} \\ \text { valence bonding orbitals }\end{array}\right)-\left(\begin{array}{c}\text {4 electrons in} \\ \text { valence antibonding orbitals }\end{array}\right)\right]\) Exercise \(\PageIndex{3}\): The Peroxide Ion Use a qualitative molecular orbital energy-level diagram to predict the electron configuration, the bond order, and the number of unpaired electrons in the peroxide ion (O 2 2 − ). Answer This diagram looks similar to that of \(\ce{O2}\), except that there are two additional electrons. \( \left ( \sigma _{g}(2s) \right )^{2}\left ( \sigma_u ^{\star }(2s) \right )^{2}\left ( \sigma _g(2p) \right )^{2}\left ( \pi _{u}(2p) \right )^{4}\left ( \pi _g(2p) \right )^{4} \); bond order of 1; no unpaired electrons. Bond Lengths in Homonuclear Diatomic Molecules The trends in experimental bond lengths are predicted by molecular orbital theory, specifically by the calculated bond order. The values of bond order and experimental bond lengths for the second period diatomic molecules are given in Figure \(\PageIndex{1}\), and shown in graphical format on the plot in Figure \(\PageIndex{2}\). From the plot, we can see that bond length correlates well with bond order, with a minimum bond length occurring where the bond order is greatest (\(\ce{N2}\)). The shortest bond distance is at \(\ce{N2}\) due to its high bond order of 3. From \(\ce{N2}\) to \(\ce{F2}\) the bond distance increases despite the fact that atomic radius decreases. References NIST, Calculated Geometries available for H 2 + (Hydrogen cation) 2Σg+ D∞h, available at https://cccbdb.nist.gov/diatomicexpbondx.asp Merritt, J. M.; Bondybey, V. E.; Heaven, M. C., Beryllium Dimer-Caught in the Act of Bonding. Science 2009, 324 (5934), 1548-1551.
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Courses/Riverland_Community_College/CHEM_1000_-_Introduction_to_Chemistry_(Riverland)/04%3A_Atoms_Elements_and_the_Periodic_Table/4.07%3A_Elements-_Defined_by_Their_Number_of_Protons
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Learning Objectives Define atomic number. Define mass number. Determine the number of protons, neutrons, and electrons in an atom. It's important to be able to distinguish atoms of one element from atoms of another element. Elements are pure substances that make up all other matter, so each one is given a unique name. The names of elements are also represented by unique one- or two-letter symbols, such as \(\ce{H}\) for hydrogen, \(\ce{C}\) for carbon, or \(\ce{He}\) for helium. However, it would more powerful if these names could be used to identify the numbers of protons and neutrons in the atoms. That's where atomic number and mass number are useful. Atomic Number Scientists distinguish between different elements by counting the number of protons in the nucleus (Table \(\PageIndex{1}\)). If an atom has only one proton, we know that it's a hydrogen atom. An atom with two protons is always a helium atom. If scientists count four protons in an atom, they know it's a beryllium atom. An atom with three protons is a lithium atom, an atom with five protons is a boron atom, an atom with six protons is a carbon atom . . . the list goes on. Since an atom of one element can be distinguished from an atom of another element by the number of protons in its nucleus, scientists are always interested in this number, and how this number differs between different elements. The number of protons in an atom is called its atomic number (\(Z\)). This number is very important because it is unique for atoms of a given element. All atoms of an element have the same number of protons, and every element has a different number of protons in its atoms. For example, all helium atoms have two protons, and no other elements have atoms with two protons. Name Protons Neutrons Electrons Atomic Number (Z) Mass Number (A) Hydrogen 1 0 1 1 1 Helium 2 2 2 2 4 Lithium 3 4 3 3 7 Beryllium 4 5 4 4 9 Boron 5 6 5 5 11 Carbon 6 6 6 6 12 Of course, since neutral atoms have to have one electron for every proton, an element's atomic number also tells you how many electrons are in a neutral atom of that element. For example, hydrogen has an atomic number of 1. This means that an atom of hydrogen has one proton, and, if it's neutral, one electron as well. Gold, on the other hand, has an atomic number of 79, which means that an atom of gold has 79 protons, and, if it's neutral, 79 electrons as well. Neutral Atoms Atoms are neutral in electrical charge because they have the same number of negative electrons as positive protons (Table \(\PageIndex{1}\)). Therefore, the atomic number of an atom also tells you how many electrons the atom has. This, in turn, determines many of the atom's chemical properties. Mass Number The mass number (\(A\)) of an atom is the total number of protons and neutrons in its nucleus. The mass of the atom is a unit called the atomic mass unit \(\left( \text{amu} \right)\). One atomic mass unit is the mass of a proton, or about \(1.67 \times 10^{-27}\) kilograms, which is an extremely small mass. A neutron has just a tiny bit more mass than a proton, but its mass is often assumed to be one atomic mass unit as well. Because electrons have virtually no mass, just about all the mass of an atom is in its protons and neutrons. Therefore, the total number of protons and neutrons in an atom determines its mass in atomic mass units (Table \(\PageIndex{1}\)). Consider helium again. Most helium atoms have two neutrons in addition to two protons. Therefore the mass of most helium atoms is 4 atomic mass units (\(2 \: \text{amu}\) for the protons + \(2 \: \text{amu}\) for the neutrons). However, some helium atoms have more or less than two neutrons. Atoms with the same number of protons but different numbers of neutrons are called isotopes. Because the number of neutrons can vary for a given element, the mass numbers of different atoms of an element may also vary. For example, some helium atoms have three neutrons instead of two (these are called isotopes and are discussed in detail later on). Why do you think that the "mass number" includes protons and neutrons, but not electrons? You know that most of the mass of an atom is concentrated in its nucleus. The mass of an atom depends on the number of protons and neutrons. You have already learned that the mass of an electron is very, very small compared to the mass of either a proton or a neutron (like the mass of a penny compared to the mass of a bowling ball). Counting the number of protons and neutrons tells scientists about the total mass of an atom. \[\text{mass number} \: A = \left( \text{number of protons} \right) + \left( \text{number of neutrons} \right) \nonumber \] An atom's mass number is very easy to calculate, provided that you know the number of protons and neutrons in an atom. Example 4.5.1 What is the mass number of an atom of helium that contains 2 neutrons? Solution \(\left( \text{number of protons} \right) = 2\) (Remember that an atom of helium always has 2 protons.) \(\left( \text{number of neutrons} \right) = 2\) \(\text{mass number} = \left( \text{number of protons} \right) + \left( \text{number of neutrons} \right)\) \(\text{mass number} = 2 + 2 = 4\) A chemical symbol is a one- or two-letter designation of an element. Some examples of chemical symbols are \(\ce{O}\) for oxygen, \(\ce{Zn}\) for zinc, and \(\ce{Fe}\) for iron. The first letter of a symbol is always capitalized. If the symbol contains two letters, the second letter is lower case. The majority of elements have symbols that are based on their English names. However, some of the elements that have been known since ancient times have maintained symbols that are based on their Latin names, as shown in Table \(\PageIndex{2}\). Chemical Symbol Name Latin Name \(\ce{Na}\) Sodium Natrium \(\ce{K}\) Potassium Kalium \(\ce{Fe}\) Iron Ferrum \(\ce{Cu}\) Copper Cuprum \(\ce{Ag}\) Silver Argentum \(\ce{Sn}\) Tin Stannum \(\ce{Sb}\) Antimony Stibium \(\ce{Au}\) Gold Aurum \(\ce{Pb}\) Lead Plumbum Summary Elements are pure substances that make up all matter, so each one is given a unique name. The names of elements are also represented by unique one- or two-letter symbols. Each element has a unique number of protons. An element's atomic number is equal to the number of protons in the nuclei of any of its atoms. The mass number of an atom is the sum of the protons and neutrons in the atom. Isotopes are atoms of the same element (same number of protons) that have different numbers of neutrons in their atomic nuclei.
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Bookshelves/Organic_Chemistry/OCLUE%3A_Organic_Chemistry_Life_the_Universe_and_Everything_(Copper_and_Klymkowsky)/04%3A_Nucleophilic_Substitution_Part_II/4.01%3A_Kinetics_and_Mechanisms-
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One of the most powerful sources of evidence for how a particular reaction proceeds comes from the study of reaction rates or reaction kinetics. In such studies, how the rate of a reaction changes [1] is measured as a function of the concentration of each reactant. One of the most common ways of measuring this change is by using a spectroscopic technique. For example, if the compound absorbs in the UV-VIS region of the spectrum, the absorbance is proportional to the concentration. Therefore, if the concentration of the substance changes, it can be measured by changes in the absorbance. The reaction is carried out several times with all but one of the reactants set as constant; then a different concentration of the remaining reactant is added and the rate of the reaction measured. This is repeated for each reactant over several concentrations. The end result of such a study is to produce what is known as the rate equation; for a generic reaction \(\mathrm{A}+\mathrm{B} \rightarrow \mathrm{C}+\mathrm{D}\) the rate equation takes the form: \[\text { Rate }=k[\mathrm{A}]^{x}[\mathrm{B}]^{y}\] In this equation, \(k\) is the rate constant, and the exponents x and y tell us about how the concentration of each reactant influences the rate. The sum of the exponents (that is, \(x+y+\ldots\)) is the order of the reaction. For example, if \(x = 1\), then the rate is directly related to the concentration of \(\mathrm{A}\). If both \(x\) and \(y = 1\), then the rate is directly proportional to both \([\mathrm{A}]\) and \([\mathrm{B}]\), and the overall reaction order is = 2. If an exponent = 0 then the rate is not dependent on that reactant concentration, and that reactant can be removed from the rate law equation (since \([n]^{0}=1\) no matter what the value of concentration of n is). The most important idea to remember is that the rate equation only contains the reactants that are involved in the rate-determining step (that is, the slowest step) of the reaction. If the reaction proceeds by a number of steps, then the step with the highest activation energy will be rate-determining, and only those reactants that participate in this step will be present in the rate law. [2] Since the rate law is determined empirically, the rate law provides us with evidence about the mechanism of the reaction. Evidence for the \(\mathbf{S}_{\mathbf{N}} 2\) Mechanism: The reaction we discussed earlier in the course is known as an \(\mathrm{S}_{\mathrm{N}} 2\) reaction that is shorthand for S ubstitution, N ucleophilic, Second order. We proposed a mechanism for this reaction without providing any empirical evidence, but now let us use some of what you have learned to consider more carefully the evidence for this mechanism. The reaction is second order : the first piece of evidence comes from the kinetic rate law. The rate of reaction depends on both the concentration of the substrate and the nucleophile: \(\text { rate }=k[\mathrm{RX}][\mathrm{Nu}]\). This means that both must be present in the rate-determining step. The simplest explanation that is consistent with this finding is the one we have already proposed: the nucleophile attacks the electrophilic carbon at the same time as the leaving group leaves. That is, the reaction takes place in one continuous step. A reaction energy diagram in which we plot Energy v reaction progress looks like this (\(\rightarrow\)). In this reaction, there is only one energy barrier, only one maximum in the reaction pathway. The energy of this barrier is known as the activation energy \(\Delta \mathrm{E}_{+}^{+}\). Looking at the reaction diagram, we also note that the reaction is exothermic (or exergonic if we plot Gibbs energy), since the \(\Delta \mathrm{E}\) of the overall reaction is negative). The species at the peak of the activation energy barrier is known as the transition state , and its structure and associated energy determines the rate of the reaction. The structure of the substrate affects the rate : Perhaps you have noticed in our earlier discussions of nucleophilic substitution that the organic substrate was always either a methyl or primary carbon attached to a good leaving group. The reason was that in the reactions we have considered, both the rate and mechanism of reaction is highly dependent on the structure of the substrate. As the number of methyl groups attached to the primary carbon increases (from 0 for a methyl group itself to three [tertiary]) the reaction rate slows, as shown. The rate of an \(\mathrm{S}_{\mathrm{N}} 2\) reaction for a tertiary substrate is negligible. So two questions arise: first, why are these reaction rates different? and second, why is this change in reaction rates evidence for the \(\mathrm{S}_{\mathrm{N}} 2\) mechanism? Both can be answered by taking a closer look at the reaction from a molecular perspective. Remember, all of the reactants are dissolved in a solvent; thermal motion leads to their colliding with one another and with solvent molecules. For the nucleophile and the substrate to react with each other, they first have to collide with one another. For a reaction to occur, that collision has to transfer enough energy so that the complex (substrate + nucleophile) can form the transition state—moreover to form the transition-state molecule, the molecules must collide with one another in the correct orientation . Once formed, the transition state can decay to form the products of the reaction. Recall that our proposed structure for the transition state for this reaction has the central carbon connected to five groups: the incoming nucleophile, the leaving group, and the three other substituents that do not change during the reaction (they are not part of the reaction). As the bond forms between the nucleophile and the substrate carbon, and the bond breaks between the carbon and leaving group, the carbon changes its hybridization state. What does that mean? In the substrate molecule, the reacting carbon is attached to surrounding groups (\(\mathrm{H-}\) or \(\left.\mathrm{CH}_{3}-\right)\)) with bonds formed from \(\mathrm{sp}^{3}\) orbitals. In the transition state, this carbon is still attached to those groups that will remain in the product molecule, but now with bonds formed from \(\mathrm{sp}^{2}\) orbitals. Additionally, it is still attached to both the leaving group and the incoming nucleophile using a p orbital to form these partial bonds. You can think of this process as electron density being funneled from the nucleophile through the carbon and out the other side to the leaving group. However, for this to occur the nucleophile can only begin to bond when it approaches from the back of the bond to the leaving group \(\rightarrow\). At this point, you might well find yourself asking: what does all this have to do with the structure of the substrate? For a reaction to occur, the only productive collisions are those where the nucleophile begins to form a bond with the back part of the \(\mathrm{sp}^{3}\) hybrid orbital; but the structure of the substrate influences the probability of such an event. In the tertiary substrates (for example \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}\)) the approach to the substrate is hindered by the bulky alkyl groups such that the probability of the nucleophile interacting with the reactive center is low. This phenomenon is called steric hindrance and provides an explanation for the order of reaction of \(\mathrm{S}_{\mathrm{N}} 2\) reactions. \(\mathrm{S}_{\mathrm{N}} 2\) reactions at a chiral center: Another piece of evidence for the \(\mathrm{S}_{\mathrm{N}} 2\) mechanism is what happens when an \(\mathrm{S}_{\mathrm{N}} 2\) reaction takes place at a chiral center (within a molecule). It turns out that the configuration at that center is changed; the carbon inverts (like an umbrella blowing inside out in the wind) so that an S enantiomer is converted to an R enantiomer. In fact, it is possible to follow the progress of an S N 2 reaction involving a chiral center using a polarimeter (the instrument used to measure optical activity); as the reaction proceeds to completion, the optical rotation of the solution changes over time. For each particular substrate, the direction and magnitude of the rotation for the product will be different. This phenomenon is called the Walden inversion and provides another piece of evidence to support the proposed reaction mechanism. The role of solvent in an \(\mathrm{S}_{\mathrm{N}} 2\) reaction: \(\mathrm{S}_{\mathrm{N}} 2\) reactions are generally carried out in a solvent (why is that?). Empirical studies reveal that such reactions proceed more rapidly when carried out in what is known as a polar aprotic solvent. So what is a polar aprotic solvent? The term means that the solvent is polar but without acidic protons. Examples of polar aprotic solvents are acetone, dimethyl formamide (DMF), and dimethylsulfoxide (DMSO): each is polar, but lacks a potentially acidic proton such as the H that is bonded to the \(\mathrm{O}\) in ethanol \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) or in water \(\mathrm{H-O-H}\). Water (and methanol and ethanol) is a polar protic solvent. In a polar aprotic solvent, the negative end of the \(\mathrm{C=O}\) or \(\mathrm{S=O}\) dipole is localized to the \(\mathrm{O}\), while the positive end is diffuse and delocalized. For example, in acetone, the oxygen has a \(\delta-\) charge on the oxygen while the positive charge of the dipole is delocalized over both the \(\mathrm{C}\) and the methyl groups as shown in the electrostatic potential map (\(\rightarrow\)). In practice, polar aprotic solvents can solvate cations well through interactions with the localized negative end of the dipole, but they cannot solvate anions very well. Recall that solvation is an interaction that lowers the energy of the system, making it more stable (less reactive). Therefore, a solvent that leaves the nucleophile (the anion) unsolvated will make it more reactive. In contrast, a polar protic solvent (such as water or ethanol) can solvate the nucleophile, through interactions with the positive end of a dipole that is localized on an acidic \(\mathrm{H}\), stabilizing the nucleophile and making it less reactive. In summary, \(\mathrm{S}_{\mathrm{N}} 2\) reactions occur in one step with inversion at a chiral center. Such reactions are generally faster for unhindered substrates and are accelerated when carried out in polar aprotic solvents.
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Courses/Oregon_Institute_of_Technology/OIT%3A_CHE_202_-_General_Chemistry_II/Unit_7%3A_Intermolecular_and_Intramolecular_Forces_in_Action/7.2%3A_Vapor_Pressure
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Skills to Develop Define phase transitions and phase transition temperatures Explain the relation between phase transition temperatures and intermolecular attractive forces We witness and utilize changes of physical state, or phase transitions, in a great number of ways. As one example of global significance, consider the evaporation, condensation, freezing, and melting of water. These changes of state are essential aspects of our earth’s water cycle as well as many other natural phenomena and technological processes of central importance to our lives. In this module, the essential aspects of phase transitions are explored. Vaporization and Condensation When a liquid vaporizes in a closed container, gas molecules cannot escape. As these gas phase molecules move randomly about, they will occasionally collide with the surface of the condensed phase, and in some cases, these collisions will result in the molecules re-entering the condensed phase. The change from the gas phase to the liquid is called condensation . When the rate of condensation becomes equal to the rate of vaporization , neither the amount of the liquid nor the amount of the vapor in the container changes. The vapor in the container is then said to be in equilibrium with the liquid. Keep in mind that this is not a static situation, as molecules are continually exchanged between the condensed and gaseous phases. Such is an example of a dynamic equilibrium , the status of a system in which reciprocal processes (for example, vaporization and condensation) occur at equal rates. The pressure exerted by the vapor in equilibrium with a liquid in a closed container at a given temperature is called the liquid’s vapor pressure (or equilibrium vapor pressure). The area of the surface of the liquid in contact with a vapor and the size of the vessel have no effect on the vapor pressure, although they do affect the time required for the equilibrium to be reached. We can measure the vapor pressure of a liquid by placing a sample in a closed container, like that illustrated in Figure \(\PageIndex{1}\), and using a manometer to measure the increase in pressure that is due to the vapor in equilibrium with the condensed phase. Figure \(\PageIndex{1}\): In a closed container, dynamic equilibrium is reached when (a) the rate of molecules escaping from the liquid to become the gas (b) increases and eventually (c) equals the rate of gas molecules entering the liquid. When this equilibrium is reached, the vapor pressure of the gas is constant, although the vaporization and condensation processes continue. The chemical identities of the molecules in a liquid determine the types (and strengths) of intermolecular attractions possible; consequently, different substances will exhibit different equilibrium vapor pressures. Relatively strong intermolecular attractive forces will serve to impede vaporization as well as favoring “recapture” of gas-phase molecules when they collide with the liquid surface, resulting in a relatively low vapor pressure. Weak intermolecular attractions present less of a barrier to vaporization, and a reduced likelihood of gas recapture, yielding relatively high vapor pressures. The following example illustrates this dependence of vapor pressure on intermolecular attractive forces. Example \(\PageIndex{1}\): Explaining Vapor Pressure in Terms of IMFs Given the shown structural formulas for these four compounds, explain their relative vapor pressures in terms of types and extents of IMFs: Solution Diethyl ether has a very small dipole and most of its intermolecular attractions are London forces. Although this molecule is the largest of the four under consideration, its IMFs are the weakest and, as a result, its molecules most readily escape from the liquid. It also has the highest vapor pressure. Due to its smaller size, ethanol exhibits weaker dispersion forces than diethyl ether. However, ethanol is capable of hydrogen bonding and, therefore, exhibits stronger overall IMFs, which means that fewer molecules escape from the liquid at any given temperature, and so ethanol has a lower vapor pressure than diethyl ether. Water is much smaller than either of the previous substances and exhibits weaker dispersion forces, but its extensive hydrogen bonding provides stronger intermolecular attractions, fewer molecules escaping the liquid, and a lower vapor pressure than for either diethyl ether or ethanol. Ethylene glycol has two −OH groups, so, like water, it exhibits extensive hydrogen bonding. It is much larger than water and thus experiences larger London forces. Its overall IMFs are the largest of these four substances, which means its vaporization rate will be the slowest and, consequently, its vapor pressure the lowest. Exercise \(\PageIndex{1}\) At 20 °C, the vapor pressures of several alcohols are given in this table. Explain these vapor pressures in terms of types and extents of IMFs for these alcohols: Compound methanol CH3OH ethanol C2H5OH propanol C3H7OH butanol C4H9OH Vapor Pressure at 20 °C 11.9 kPa 5.95 kPa 2.67 kPa 0.56 kPa As temperature increases, the vapor pressure of a liquid also increases due to the increased average KE of its molecules. Recall that at any given temperature, the molecules of a substance experience a range of kinetic energies, with a certain fraction of molecules having a sufficient energy to overcome IMF and escape the liquid (vaporize). At a higher temperature, a greater fraction of molecules have enough energy to escape from the liquid, as shown in Figure \(\PageIndex{2}\). The escape of more molecules per unit of time and the greater average speed of the molecules that escape both contribute to the higher vapor pressure. Figure \(\PageIndex{2}\) : Temperature affects the distribution of kinetic energies for the molecules in a liquid. At the higher temperature, more molecules have the necessary kinetic energy, KE, to escape from the liquid into the gas phase. Boiling Points Video \(\PageIndex{1}\) : Mercury (Hg) boils at room pressure at 356 °C. When the vapor pressure increases enough to equal the external atmospheric pressure, the liquid reaches its boiling point. The boiling point of a liquid is the temperature at which its equilibrium vapor pressure is equal to the pressure exerted on the liquid by its gaseous surroundings. For liquids in open containers, this pressure is that due to the earth’s atmosphere. The normal boiling point of a liquid is defined as its boiling point when surrounding pressure is equal to 1 atm (101.3 kPa). Figure \(\PageIndex{3}\) shows the variation in vapor pressure with temperature for several different substances. Considering the definition of boiling point, these curves may be seen as depicting the dependence of a liquid’s boiling point on surrounding pressure. Figure \(\PageIndex{3}\) : The boiling points of liquids are the temperatures at which their equilibrium vapor pressures equal the pressure of the surrounding atmosphere. Normal boiling points are those corresponding to a pressure of 1 atm (101.3 kPa.) Example \(\PageIndex{2}\): A Boiling Point at Reduced Pressure A typical atmospheric pressure in Leadville, Colorado (elevation 10,200 feet) is 68 kPa. Use the graph in Figure \(\PageIndex{3}\) to determine the boiling point of water at this elevation. Solution The graph of the vapor pressure of water versus temperature in Figure \(\PageIndex{3}\) indicates that the vapor pressure of water is 68 kPa at about 90 °C. Thus, at about 90 °C, the vapor pressure of water will equal the atmospheric pressure in Leadville, and water will boil. Exercise \(\PageIndex{2}\) The boiling point of ethyl ether was measured to be 10 °C at a base camp on the slopes of Mount Everest. Use Figure \(\PageIndex{3}\) to determine the approximate atmospheric pressure at the camp. Answer Approximately 40 kPa (0.4 atm) The quantitative relation between a substance’s vapor pressure and its temperature is described by the Clausius-Clapeyron equation : \[P=Ae^{−ΔH_{vap}/RT} \label{10.4.1}\] where \(ΔH_{vap}\) is the enthalpy of vaporization for the liquid, \(R\) is the gas constant, and \(\ln A\) is a constant whose value depends on the chemical identity of the substance. Equation \(\ref{10.4.1}\) is often rearranged into logarithmic form to yield the linear equation: \[\ln P=−\dfrac{ΔH_\ce{vap}}{RT}+\ln A \label{10.4.2}\] This linear equation may be expressed in a two-point format that is convenient for use in various computations, as demonstrated in the examples and exercises that follow. If at temperature \(T_1\), the vapor pressure is \( P_1\), and at temperature \(T_2\), the vapor pressure is \(T_2\), the corresponding linear equations are: \[\ln P_1=−\dfrac{ΔH_\ce{vap}}{RT_1}+\ln A\] and \[\ln P_2=−\dfrac{ΔH_\ce{vap}}{RT_2}+\ln A \label{10.4.3}\] Since the constant, ln A, is the same, these two equations may be rearranged to isolate \(\ln A\) and then set them equal to one another: \(\ln P_1+\dfrac{ΔH_\ce{vap}}{RT_1}=\ln P_2+\dfrac{ΔH_\ce{vap}}{RT_2}\label{10.4.5}\) which can be combined into: \[\ln \left(\dfrac{P_2}{P_1}\right)=\dfrac{ΔH_\ce{vap}}{R} \left( \dfrac{1}{T_1}−\dfrac{1}{T_2}\right) \label{10.4.6}\] Example \(\PageIndex{3}\): Estimating Enthalpy of Vaporization Isooctane (2,2,4-trimethylpentane) has an octane rating of 100. It is used as one of the standards for the octane-rating system for gasoline. At 34.0 °C, the vapor pressure of isooctane is 10.0 kPa, and at 98.8 °C, its vapor pressure is 100.0 kPa. Use this information to estimate the enthalpy of vaporization for isooctane. Solution The enthalpy of vaporization, \(ΔH_\ce{vap}\), can be determined by using the Clausius-Clapeyron equation (Equation \(\ref{10.4.6}\)): \[\ln\left(\dfrac{P_2}{P_1}\right)=\dfrac{ΔH_\ce{vap}}{R}\left(\dfrac{1}{T_1}−\dfrac{1}{T_2}\right) \nonumber\] Since we have two vapor pressure-temperature values \(T_1 = 34.0^oC = 307.2\, K\) \(P_1 = 10.0\, kPa\) and \(T_2 = 98.8 ^oC = 372.0\, K\) \(P_2 = 100\, kPa\) we can substitute them into this equation and solve for \(ΔH_{vap}\). Rearranging the Clausius-Clapeyron equation and solving for \(ΔH_{vap}\) yields: \[ \begin{align*} ΔH_\ce{vap} &= \dfrac{R⋅\ln\left(\dfrac{P_2}{P_1}\right)}{\left(\dfrac{1}{T_1}−\dfrac{1}{T_2}\right)} \\[5pt] &= \dfrac{(8.3145\:J/mol⋅K)⋅\ln \left(\dfrac{100\: kPa}{10.0\: kPa}\right)}{\left(\dfrac{1}{307.2\:K}−\dfrac{1}{372.0\:K}\right)} \\[5pt] &=33,800\, J/mol =33.8\, kJ/mol \end{align*}\] Note that the pressure can be in any units, so long as they agree for both P values, but the temperature must be in kelvin for the Clausius-Clapeyron equation to be valid. Exercise \(\PageIndex{3}\) At 20.0 °C, the vapor pressure of ethanol is 5.95 kPa, and at 63.5 °C, its vapor pressure is 53.3 kPa. Use this information to estimate the enthalpy of vaporization for ethanol. Answer 47,782 J/mol = 47.8 kJ/mol Example \(\PageIndex{4}\): Estimating Temperature (or Vapor Pressure) For benzene (C 6 H 6 ), the normal boiling point is 80.1 °C and the enthalpy of vaporization is 30.8 kJ/mol. What is the boiling point of benzene in Denver, where atmospheric pressure = 83.4 kPa? Solution If the temperature and vapor pressure are known at one point, along with the enthalpy of vaporization, Δ H vap, then the temperature that corresponds to a different vapor pressure (or the vapor pressure that corresponds to a different temperature) can be determined by using the Clausius-Clapeyron equation (Equation \(\ref{10.4.1}\)) : \[\ln\left(\dfrac{P_2}{P_1}\right)=\dfrac{ΔH_\ce{vap}}{R}\left(\dfrac{1}{T_1}−\dfrac{1}{T_2}\right) \nonumber\] Since the normal boiling point is the temperature at which the vapor pressure equals atmospheric pressure at sea level, we know one vapor pressure-temperature value (\(T_1\) = 80.1 °C = 353.3 K, \(P_1\) = 101.3 kPa, \(ΔH_{vap}\) = 30.8 kJ/mol) and want to find the temperature (\(T_2\)) that corresponds to vapor pressure P 2 = 83.4 kPa. We can substitute these values into the Clausius-Clapeyron equation and then solve for \(T_2\). Rearranging the Clausius-Clapeyron equation and solving for \(T_2\) yields: \[\begin{align*} T_2 &=\left(\dfrac{−R⋅\ln\left(\dfrac{P_2}{P_1}\right)}{ΔH_\ce{vap}}+\dfrac{1}{T_1}\right)^{−1} \\[5pt] &=\mathrm{\left(\dfrac{−(8.3145\:J/mol⋅K)⋅\ln\left(\dfrac{83.4\:kPa}{101.3\:kPa}\right)}{30,800\: J/mol}+\dfrac{1}{353.3\:K}\right)^{−1}}\\[5pt] &=\mathrm{346.9\: K\: or\:73.8^\circ C} \end{align*}\] Exercise \(\PageIndex{4}\) For acetone \(\ce{(CH3)2CO}\), the normal boiling point is 56.5 °C and the enthalpy of vaporization is 31.3 kJ/mol. What is the vapor pressure of acetone at 25.0 °C? Answer 30.1 kPa Video \(\PageIndex{2}\): An overview of the role of vapor pressure on boiling point. Enthalpy of Vaporization Vaporization is an endothermic process. The cooling effect can be evident when you leave a swimming pool or a shower. When the water on your skin evaporates, it removes heat from your skin and causes you to feel cold. The energy change associated with the vaporization process is the enthalpy of vaporization, \(ΔH_{vap}\). For example, the vaporization of water at standard temperature is represented by: \[\ce{H2O}(l)⟶\ce{H2O}(g)\hspace{20px}ΔH_\ce{vap}=\mathrm{44.01\: kJ/mol}\] As described in the chapter on thermochemistry, the reverse of an endothermic process is exothermic. And so, the condensation of a gas releases heat: \[\ce{H2O}(g)⟶\ce{H2O}(l)\hspace{20px}ΔH_\ce{con}=−ΔH_\ce{vap}=\mathrm{−44.01\:kJ/mol}\] Example \(\PageIndex{5}\): Usi ng Enthalpy of Vaporization One way our body is cooled is by evaporation of the water in sweat (Figure \(\PageIndex{4}\)). In very hot climates, we can lose as much as 1.5 L of sweat per day. Although sweat is not pure water, we can get an approximate value of the amount of heat removed by evaporation by assuming that it is. How much heat is required to evaporate 1.5 L of water (1.5 kg) at T = 37 °C (normal body temperature); \(ΔH_{vap} = 43.46\, kJ/mol\) at 37 °C. Figure \(\PageIndex{4}\): Evaporation of sweat helps cool the body. (credit: “Kullez”/Flickr) Solution We start with the known volume of sweat (approximated as just water) and use the given information to convert to the amount of heat needed: \[\mathrm{1.5\cancel{L}×\dfrac{1000\cancel{g}}{1\cancel{L}}×\dfrac{1\cancel{mol}}{18\cancel{g}}×\dfrac{43.46\:kJ}{1\cancel{mol}}=3.6×10^3\:kJ} \nonumber\] Thus, 3600 kJ of heat are removed by the evaporation of 1.5 L of water. Exercise \(\PageIndex{5}\): Boiling Ammonia How much heat is required to evaporate 100.0 g of liquid ammonia, \(\ce{NH3}\), at its boiling point if its enthalpy of vaporization is 4.8 kJ/mol? Answer 28 kJ Melting and Freezing When we heat a crystalline solid, we increase the average energy of its atoms, molecules, or ions and the solid gets hotter. At some point, the added energy becomes large enough to partially overcome the forces holding the molecules or ions of the solid in their fixed positions, and the solid begins the process of transitioning to the liquid state, or melting . At this point, the temperature of the solid stops rising, despite the continual input of heat, and it remains constant until all of the solid is melted. Only after all of the solid has melted will continued heating increase the temperature of the liquid (Figure \(\PageIndex{5}\). Figure \(\PageIndex{5}\): (a) This beaker of ice has a temperature of −12.0 °C. (b) After 10 minutes the ice has absorbed enough heat from the air to warm to 0 °C. A small amount has melted. (c) Thirty minutes later, the ice has absorbed more heat, but its temperature is still 0 °C. The ice melts without changing its temperature. (d) Only after all the ice has melted does the heat absorbed cause the temperature to increase to 22.2 °C. (credit: modification of work by Mark Ott). If we stop heating during melting and place the mixture of solid and liquid in a perfectly insulated container so no heat can enter or escape, the solid and liquid phases remain in equilibrium. This is almost the situation with a mixture of ice and water in a very good thermos bottle; almost no heat gets in or out, and the mixture of solid ice and liquid water remains for hours. In a mixture of solid and liquid at equilibrium, the reciprocal processes of melting and freezing occur at equal rates, and the quantities of solid and liquid therefore remain constant. The temperature at which the solid and liquid phases of a given substance are in equilibrium is called the melting point of the solid or the freezing point of the liquid. Use of one term or the other is normally dictated by the direction of the phase transition being considered, for example, solid to liquid (melting) or liquid to solid (freezing). The enthalpy of fusion and the melting point of a crystalline solid depend on the strength of the attractive forces between the units present in the crystal. Molecules with weak attractive forces form crystals with low melting points. Crystals consisting of particles with stronger attractive forces melt at higher temperatures. The amount of heat required to change one mole of a substance from the solid state to the liquid state is the enthalpy of fusion, ΔHfus of the substance. The enthalpy of fusion of ice is 6.0 kJ/mol at 0 °C. Fusion (melting) is an endothermic process: \[\ce{H2O}_{(s)} \rightarrow \ce{H2O}_{(l)} \;\; ΔH_\ce{fus}=\mathrm{6.01\; kJ/mol} \label{10.4.9}\] The reciprocal process, freezing, is an exothermic process whose enthalpy change is −6.0 kJ/mol at 0 °C: \[\ce{H_2O}_{(l)} \rightarrow \ce{H_2O}_{(s)}\;\; ΔH_\ce{frz}=−ΔH_\ce{fus}=−6.01\;\mathrm{kJ/mol} \label{10.4.10}\] Sublimation and Deposition Some solids can transition directly into the gaseous state, bypassing the liquid state, via a process known as sublimation . At room temperature and standard pressure, a piece of dry ice (solid CO 2 ) sublimes, appearing to gradually disappear without ever forming any liquid. Snow and ice sublime at temperatures below the melting point of water, a slow process that may be accelerated by winds and the reduced atmospheric pressures at high altitudes. When solid iodine is warmed, the solid sublimes and a vivid purple vapor forms (Figure \(\PageIndex{6}\)). The reverse of sublimation is called deposition , a process in which gaseous substances condense directly into the solid state, bypassing the liquid state. The formation of frost is an example of deposition. Figure \(\PageIndex{6}\): Sublimation of solid iodine in the bottom of the tube produces a purple gas that subsequently deposits as solid iodine on the colder part of the tube above. (credit: modification of work by Mark Ott) Like vaporization, the process of sublimation requires an input of energy to overcome intermolecular attractions. The enthalpy of sublimation, ΔHsub, is the energy required to convert one mole of a substance from the solid to the gaseous state. For example, the sublimation of carbon dioxide is represented by: \[\ce{CO2}(s)⟶\ce{CO2}(g)\hspace{20px}ΔH_\ce{sub}=\mathrm{26.1\: kJ/mol}\] Likewise, the enthalpy change for the reverse process of deposition is equal in magnitude but opposite in sign to that for sublimation: \[\ce{CO2}(g)⟶\ce{CO2}(s)\hspace{20px}ΔH_\ce{dep}=−ΔH_\ce{sub}=\mathrm{−26.1\:kJ/mol}\] Consider the extent to which intermolecular attractions must be overcome to achieve a given phase transition. Converting a solid into a liquid requires that these attractions be only partially overcome; transition to the gaseous state requires that they be completely overcome. As a result, the enthalpy of fusion for a substance is less than its enthalpy of vaporization. This same logic can be used to derive an approximate relation between the enthalpies of all phase changes for a given substance. Though not an entirely accurate description, sublimation may be conveniently modeled as a sequential two-step process of melting followed by vaporization in order to apply Hess’s Law . \[\mathrm{solid⟶liquid}\hspace{20px}ΔH_\ce{fus}\\\underline{\mathrm{liquid⟶gas}\hspace{20px}ΔH_\ce{vap}}\\\mathrm{solid⟶gas}\hspace{20px}ΔH_\ce{sub}=ΔH_\ce{fus}+ΔH_\ce{vap}\] Viewed in this manner, the enthalpy of sublimation for a substance may be estimated as the sum of its enthalpies of fusion and vaporization, as illustrated in Figure \(\PageIndex{7}\). For example: Figure \(\PageIndex{7}\) : For a given substance, the sum of its enthalpy of fusion and enthalpy of vaporization is approximately equal to its enthalpy of sublimation. Summary Video \(\PageIndex{3}\) : An overview of phase changes with regards to kinetics. Phase transitions are processes that convert matter from one physical state into another. There are six phase transitions between the three phases of matter. Melting, vaporization, and sublimation are all endothermic processes, requiring an input of heat to overcome intermolecular attractions. The reciprocal transitions of freezing, condensation, and deposition are all exothermic processes, involving heat as intermolecular attractive forces are established or strengthened. The temperatures at which phase transitions occur are determined by the relative strengths of intermolecular attractions and are, therefore, dependent on the chemical identity of the substance. Key Equations \(P=Ae^{−ΔH_\ce{vap}/RT}\) \(\ln P=−\dfrac{ΔH_\ce{vap}}{RT}+\ln A\) \(\ln\left(\dfrac{P_2}{P_1}\right)=\dfrac{ΔH_\ce{vap}}{R}\left(\dfrac{1}{T_1}−\dfrac{1}{T_2}\right)\) Glossary boiling point temperature at which the vapor pressure of a liquid equals the pressure of the gas above it Clausius-Clapeyron equation mathematical relationship between the temperature, vapor pressure, and enthalpy of vaporization for a substance condensation change from a gaseous to a liquid state deposition change from a gaseous state directly to a solid state dynamic equilibrium state of a system in which reciprocal processes are occurring at equal rates freezing change from a liquid state to a solid state freezing point temperature at which the solid and liquid phases of a substance are in equilibrium; see also melting point melting change from a solid state to a liquid state melting point temperature at which the solid and liquid phases of a substance are in equilibrium; see also freezing point normal boiling point temperature at which a liquid’s vapor pressure equals 1 atm (760 torr) sublimation change from solid state directly to gaseous state vapor pressure (also, equilibrium vapor pressure) pressure exerted by a vapor in equilibrium with a solid or a liquid at a given temperature vaporization change from liquid state to gaseous state 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] ). Adelaide Clark, Oregon Institute of Technology Crash Course Chemistry: Crash Course is a division of Complexly and videos are free to stream for educational purposes. Guillotined Chemistry is a production of Mark Anticole and available for free on Youtube. Feedback Have feedback to give about this text? Click here . Found a typo and want extra credit? Click here .
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Courses/Chabot_College/Chem_12A%3A_Organic_Chemistry_Fall_2022/11%3A_Properties_and_Reactions_of_Alkynes/11.10%3A_An_Introduction_to_Multiple_Step_Synthesis
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Learning Objective use retrosynthetic analysis to design a multi-step synthesis with correct regiochemistry and stereochemistry using the reactions studied to date Introduction The study of organic chemistry introduces students to a wide range of interrelated reactions. Alkenes, for example, may be converted to structurally similar alkanes, alcohols, alkyl halides, epoxides, glycols and boranes; cleaved to smaller aldehydes, ketones and carboxylic acids; and enlarged by carbocation and radical additions as well as cycloadditions. Most of these reactions are shown in the Alkene Reaction Map below. All of these products may be subsequently transformed into a host of new compounds incorporating a wide variety of functional groups. Consequently, the logical conception of a multi-step synthesis for the construction of a designated compound from a specified starting material becomes one of the most challenging problems that may be posed. Functional group reaction maps like the one below for alkenes can be helpful in designing multi-step syntheses. It can be helpful to build and design your own reaction maps for each functional group studied. Alkene Reaction Map Please note: The reagents for each chemical transformation have been intentionally omitted so that this map can be used as a study tool. The answers are provided at the end of this section as part of the exercises. Simple Multi-Step Syntheses A one or two step sequence of simple reactions is not that difficult to deduce. For example, the synthesis of meso-3,4-hexanediol from 3-hexyne can occur by more than one multi-step pathway. One approach would be to reduce the alkyne to cis or trans-3-hexene before undertaking glycol formation. Permanaganate or osmium tetroxide hydroxylation of cis-3-hexene would form the desired meso isomer. From trans-3-hexene, it would be necessary to first epoxidize the alkene with a peracid followed by ring opening with acidic or basic hydrolysis. Longer multi-step syntheses require careful analysis and thought, since many options need to be considered. Like an expert chess player evaluating the long range pros and cons of potential moves, the chemist must appraise the potential success of various possible reaction paths, focusing on the scope and limitations constraining each of the individual reactions being employed. The skill is acquired by practice, experience, and often trial and error. Thinking it Through with 3 Examples The following three examples illustrate strategies for developing multi-step syntheses from the reactions studied in the first ten chapters of this text. It is helpful to systematically look for structural changes beginning with the carbon chain and brainstorm relevant functional group conversion reactions. Retro-synthesis is the approach of working backwards from the product to the starting material. In the first example, we are asked to synthesize 1-butanol from acetylene. The carbon chain doubles in size indicating an acetylide S N 2 reaction with an alkyl halide. Primary alcohol formation from an anti-Markovnikov alkene hydration reaction (hydroboration-oxidation) is more likely than a substitution reaction. Applying retro-synthesis, we work backwards from the alcohol to the alkene to the alkyne from an acetylide reaction that initially builds the carbon chain. Retro-Synthesis Working forwards, we specify the reagents needed for each transformation identified from the retro-synthesis. The ethylbromide must also be derived from acetylene so multiple reaction pathways are combined as shown below. In the second example, we are asked to synthesize 1,2-dibromobutane from acetylene. Once again there is an increase in the carbon chain length indicating an acetylide S N 2 reaction with an alkyl halide similar to the first example. The hydrohalogenation can be subtle to discern because the hydrogen atoms are not shown in bond-line structures. Comparing the chemical formulas of 1-butyne with 1,2-dibromobutane, there is a difference of two H atoms and two Br atoms indicating hydrohalogenation and not halogenation. The addition of both bromine atoms to the same carbon atom also supports the idea that hydrohalogenation occurs on an alkyne and not an alkene. The formation of the geminal dihalide also indicates hydrohalogenation instead of halogenatioin because halogenation produces vicinal dihalides. With this insight, the retro-synthesis indicates the following series of chemical transformations. Retro-Synthesis Working forwards, we specify the reagents needed for each transformation. In the third example, we are asked to produce 6-oxoheptanal from methylcyclohexane. Counting the carbons, the starting material and product both contain seven carbon atoms and there is a cleavage reaction of an alkene under reductive conditions. One important missing aspect of this reaction is a good leaving group (LG). Alkanes are chemically quite boring. We can burn them as fuel or perform free-radical halogenation to create alkyl halides with excellent leaving groups. With these observations, the following retro-synthesis is reasonable. Retro-Synthesis Working forwards, we specify the reagents needed for each reaction. For the initial free-radical halogenation of the alkane, we have the option of chlorine (Cl2) or bromine (Br2). Because methylcyclohexane has several different classifications of carbons, the selectivity of Br2 is more important than the faster reactivity of Cl2. An strong base with heat can be used for the second step to follow an E2 mechanism and form 1-methylcyclohexene. The aldhyde group on the final product indicates gentle oxidative cleavage by any of several reaction pathways. These reactions can be combined in to the following multi-step synthesis. Reaction Maps to Build Functional Group Conversion Mastery After working through the examples above, we can see how important it is to memorize all of the functional group reactions studied in the first ten chapters. We can apply the knowledge of these reactions to the wisdom of multi-step syntheses. Please note: The reagents for each chemical transformation have been intentionally omitted so that these maps can be used as a study tools. The answers are provided at the end of this section as part of the exercises. Alkane and Alkyl Halide Reaction Map Alkyne Reaction Map Exercise 1. Starting at 3-hexyne predict synthetic routes to achieve: a) trans -3-hexene b) 3,4-dibromohexane c) 3-hexanol. 2. Starting with acetylene and any alkyl halides propose a synthesis to make a) pentanal b) hexane. Answer 1. 2.
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Bookshelves/Analytical_Chemistry/Analytical_Chemistry_2.1_(Harvey)/06%3A_Equilibrium_Chemistry
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Regardless of the problem on which an analytical chemist is working, its solution requires a knowledge of chemistry and the ability to apply that knowledge to solve a problem. For example, an analytical chemist who is studying the effect of pollution on spruce trees needs to know, or know where to find, the chemical differences between p ‐hydroxybenzoic acid and p ‐hydroxyacetophenone, two common phenols found in the needles of spruce trees. The ability to “think as a chemist” is a product of your experience in the classroom and in the laboratory. For the most part, the material in this text assumes you are familiar with topics covered in earlier courses; however, because of its importance to analytical chemistry, this chapter provides a review of equilibrium chemistry. Much of the material in this chapter should be familiar to you, although some topics—ladder diagrams and activity, for example—likely afford you with new ways to look at equilibrium chemistry. 6.1: Reversible Reactions and Chemical Equilibria In 1798, chemist Claude Berthollet observed Na2CO3 deposits at Egypt's Natron Lakes, which contradicted existing chemical theory based on elective affinities that dictated reactions proceed in one direction only. Berthollet's insight into the reversibility of reactions, exemplified by the formation of Na2CO3 using CaCO3 and NaCl, contributed to the understanding of chemical equilibrium. 6.2: Thermodynamics and Equilibrium Chemistry The page discusses the principles of thermodynamics, focusing on chemical reactions and the factors influencing their equilibrium positions. It outlines the roles of Gibbs free energy, enthalpy, and entropy in determining whether reactions are thermodynamically favorable. The Gibbs free energy equation predicts reaction direction under specific conditions. The equilibrium constant (K) describes a reaction's equilibrium position using concentrations or partial pressures. 6.3: Manipulating Equilibrium Constants The page elaborates on two principles regarding equilibrium constants: reversing a reaction inverts its equilibrium constant, and combining reactions involves multiplying their equilibrium constants. An example and exercise demonstrate these principles. For the example, the equilibrium constant for a reaction is calculated by combining constants of related reactions, resulting in 0.10. In a similar exercise, the equilibrium constant for a different reaction is calculated to be approximately 31. 6.4: Equilibrium Constants for Chemical Reactions The document provides an in-depth overview of several essential chemical reactions relevant to analytical chemistry, such as precipitation, acid-base, complexation, and oxidation-reduction (redox) reactions. It explains the concepts of equilibrium constants like Ksp for precipitation reactions, Ka and Kb for acid-base reactions, and Kf for complexation reactions. The text discusses strong and weak acids and bases, amphiprotic species, the dissociation of water, and the pH scale. 6.5: Le Châtelier’s Principle The document explains the concept of chemical equilibria and Le Ch??telier's principle through examples involving acetic acid dissociation and silver chloride solubility. It discusses how adding reactants or products affects equilibrium, maintaining the equilibrium constant despite changes. It elaborates on how changing the concentration, such as adding sodium acetate or a ligand, affects reactions, and also how pressure and volume changes influence equilibrium through the ideal gas law. 6.6: Ladder Diagrams The page discusses the importance of considering chemical interactions, like pH and solubility, when developing or evaluating analytical methods. It critiques the inappropriate use of NH3 in precipitating AgCl due to its solubility-increasing effect. Key analytical errors often stem from overlooking chemical interferences. Ladder diagrams are introduced as tools for visualizing equilibrium chemistry, aiding in understanding reaction dynamics and evaluating changes in solution conditions. 6.7: Solving Equilibrium Problems This page discusses using ladder diagrams and algebraic solutions to evaluate and solve equilibrium problems related to chemical reactivity and solubility. It begins with a straightforward example of calculating the solubility of Pb(IO3)2 in deionized water and proceeds to more complex scenarios considering the common ion effect and the presence of ligands. 6.8: Buffer Solutions This page explains the different responses to adding HCl to pure water versus a solution with acetic acid and sodium acetate. It describes how buffers, like the acetic acid-sodium acetate mixture, resist changes in pH due to their equilibrium shifting. The Henderson-Hasselbalch equation is central to understanding buffer preparation and effectiveness. 6.9: Activity Effects The stability of the metal???ligand complex Fe(SCN)2+ decreases in the presence of inert ions, as seen when adding an inert salt like KNO3 to an equilibrium mixture of Fe3+ and SCN???. This leads to a shift in the equilibrium, reducing the concentration of Fe(SCN)2+ and lowering its formation constant, K1. The ionic strength of the solution, which is a measure of the concentration and charge of ions, affects the apparent formation constant. 6.10: Using Excel and R to Solve Equilibrium Problems The document discusses solving equilibrium problems in chemistry using tools like Excel and R, emphasizing the importance of simplifying assumptions to avoid complex equations. Excel's Solver function helps find polynomial roots and solve simultaneous equations, demonstrated with examples on solubility and pH calculations. R's uniroot command and custom functions can also solve these problems by iterating on solutions. 6.11: Some Final Thoughts on Equilibrium Calculations The chapter discusses tools for evaluating system composition at equilibrium, highlighting the importance of selecting the appropriate tool based on the precision required. It emphasizes the need to include all relevant equilibrium reactions to prevent errors. It introduces computational programs like Visual Minteq and CurTiPot for modeling equilibria and the R package CHNOSZ for thermodynamic calculations in aqueous geochemistry. 6.12: Problems This page contains a comprehensive set of chemistry problems related to equilibrium constants, redox reactions, solubility, acid-base equilibrium, buffer solutions, and complexation reactions. It starts with deriving equilibrium constant expressions for given chemical reactions, analyzing the favorability of reactions using ladder diagrams, and calculating potentials for redox systems. 6.13: Additional Resources The page provides a comprehensive list of references addressing various aspects of equilibrium chemistry. Topics covered include experimental determination of equilibrium constants, the impact of ionic strength, solubility products, and buffer capacity. Historical perspectives on the field are also offered. Additionally, the list encompasses instructional strategies, simulations for teaching, and critiques of conventional approaches to equilibrium concepts. 6.14: Chapter Summary and Key Terms The chapter discusses analytical chemistry as the application of chemistry to analyze samples, focusing on using chemical reactivity to dissolve samples, separate analytes, transform analytes, or provide a signal. Key reactions include precipitation, acid-base, metal-ligand complexation, and oxidation-reduction. It also covers equilibrium concepts, such as Le Ch??telier's principle, and solutions like buffers, using equilibrium constants, ladder diagrams, and activity coefficients.
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Courses/Oregon_Institute_of_Technology/OIT%3A_CHE_333_-_Organic_Chemistry_III_(Lund)/17%3A_The_Organic_Chemistry_of_Vitamins/17.E%3A_The_Oganic_Chemistry_of_Vitamins_(Exercises)
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P17.1 : Here is a chance to test your ability to recognize reactions catalyzed by enzymes using three coenzymes - thiamine diphosphate, pyridoxal phosphate, and folate - that we have studied in this chapter. For each generalized reaction, look carefully at the transformation that is taking place, and decide which of the three coenzymes is likely to be required. Then, draw the single mechanistic step by which the bond identified by an arrow is broken or formed. In the cases where a double bond is indicated, show the mechanistic step in which the s bond is formed. In each case, your drawing should include the structure of the reactive part of the coenzyme, and should clearly show the role it plays in catalyzing the mechanistic step you are drawing. P17.2 : The final step in the biosynthesis of the amino acid tryptophan is a PLP-dependent condensation between serine and indole, shown below (EC 4.2.1.20). The reaction mechanism involves steps that are familiar from this chapter, but also incorporates a reaction type we studied in chapter 14. Propose a mechanism. P17.3 : Draw a reasonable mechanism for the following reaction, identifying the species denoted by questions marks. (Biochemistry 2012, 51, 3059) P17.4 : Propose a mechanism for the reaction below, which is part of the anaerobic catabolism of alcohols in some species of bacteria. (ChemBioChem 2014, 15, 389) P17.5 : Identify cosubstrate A and propose a mechanism for the reaction shown below, which was reported to occur in the thermophilic bacterium Thermosporothrix hazakensis . (ChemBioChem 2014 15, 527). P17.6 : Propose a mechanism for each of the reactions below, being sure to show the role played by the coenzyme (you need to determine which coenzyme is needed in each case). P17.7 : Acetohydroxybutyrate is formed in a coenzyme-dependent reaction between pyruvate and a 4-carbon compound. What is a likely second substrate, coenzyme, and by-product (indicated below with a question mark)? http://www.sciencedirect.com/science...67593105001043 P17.8 : The 'benzoin condensation' reaction was discovered in the 19th century, and led eventually to a better understanding of \(ThDP\)-dependent reactions in the cell. In a traditional benzoin condensation reaction, cyanide ion (instead of ThDP) plays the role of electron acceptor. Enzyme-catalyzed benzoin condensation reactions are also known to occur in some bacteria: Pseudomonas fluorescens, for example, contains an enzyme that catalyzes the synthesis of (R)-benzoin. Draw a mechanism for the enzyme-catalyzed (\(ThDP\)-dependent) benzoin condensation reaction. Draw a mechanism for the cyanide-catalyzed benzoin condensation reaction (non-enzymatic, basic conditions). The following \(ThDP\)-reaction was recently reported to be part of the biosynthetic pathway for clavulanic acid, a compound that inhibits the action of b-lactamases (b-lactamases are bacterial enzymes that hydrolyze penicillin-based antibiotic drugs, rendering them ineffective). As is typical for \(ThDP\)-dependent reactions, the first step is addition of the ylide form of the coenzyme to the substrate carbonyl. The next steps are (in order): dehydration, tautomerization, elimination of phosphate, conjugate addition of arginine, and finally hydrolytic cleavage of the coenzyme-product bond. Draw out a complete mechanism that corresponds to this description. P17.9 : Practice with PLP-dependent reactions: Propose a mechanism for this reaction, which is part of the tryptophan degradation pathway (EC 3.7.1.3). Propose a mechanism for the final step of the threonine biosynthesis pathway (EC 4.2.3.1). Propose a mechanism for the reaction catalyzed by aspartate \(\beta \)-decarboxylase (EC 4.1.1.12), which converts aspartate to alanine in a PLP-dependent reaction. Sphingolipids are a type of membrane lipid found in the membranes of all eukaryotic cells, and are most abundant in the cells of central the central nervous system. Synthesis of sphingolipids involves the PLP-dependent reaction below, catalyzed by serine palmitoyl transferase (EC 2.3.1.50). Propose a mechanism. P17.10 : As we saw in this chapter, PLP-dependent enzymes usually catalyze reactions involving amino acid substrates. Here is an exception, a PLP-dependent \(\beta \)-elimination reaction in the folate biosynthetic pathway (EC 4.1.3.38). Propose a mechanism for this reaction. P17.11 : The final step in the degradation pathway for the amino acid glycine (also known as the 'glycine cleavage system') is shown below. Propose a likely mechanism, given that evidence suggests that \(CH_2NH_2^+\) is an intermediate. P17.12 : As we saw in chapter 15, the usual biochemical role of \(NAD^+\) is to act as a hydride acceptor in dehydrogenation reactions. An exception is the reaction catalyzed by the histidine degradation pathway enzyme urocanase (EC 4.2.1.49). In this reaction, \(NAD^+\) acts as a catalytic, electron-sink coenzyme - it temporarily accepts electrons from a pi bond in the substrate, resulting in a covalent substrate-\(NAD\) adduct. This allows a key isomerization step to occur on the substrate through a protonation-deprotonation mechanism, followed by addition of water, cleavage of the substrate-\(NAD\) adduct to regenerate \(NAD^+\), and finally tautomerization to the product. Propose a mechanism that fits this description, and involves the intermediate below. Pathway prediction problems P17.13 : Propose a multistep pathway for each of the following transformations. All involve at least one step requiring PLP, \(ThDP\), or folate. Below is portion of the biosynthesis of a modified membrane lipid in Salmonella and other pathogenic bacteria. The modified membrane confers antibiotic resistance to the bacterium. Biochemistry 2014 , 53 , 796 Below is the biosynthetic pathway for phenethanol in yeast. Phenethanol, which has a rose scent, is commonly used as a fragrance - this pathway has been proposed as a potential 'green' enzymatic synthesis to replace the traditional industrial synthesis, which uses toxic reagents. Below is an incomplete pathway diagram for the biosynthesis of the amino acid lysine, starting from aspartate. Fill in the missing steps and reactants/coenzymes to complete the diagram. The solid dot and dashed circle are provided to help you to trace two of the carbons from substrate to product. Below is the second half of the tryptophan degradation pathway. Fill in the missing steps and reactants/coenzymes to complete the diagram. Below is an incomplete pathway diagram for the biosynthesis of inosine monophosphate, a precursor to the nucleotides adenosine monophosphate (\(AMP\)) and guanosine monophosphate (\(GMP\)). Fill in missing steps and reactants/coenzymes to complete the diagram. Note that one enzymatic step is provided (this is a carboxylation reaction of a type that we have not studied). We begin our study of organic chemistry with a story about a hot pepper eating contest in Wisconsin (see the introduction the Chapter 1), and a compound called capsaicin which causes the 'hot' in hot peppers. As our last problem, let's try to predict some of the key steps in the biosynthesis of capsaicin. Phase 1: Phase 2: Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
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Courses/Los_Angeles_Trade_Technical_College/Foundations_of_Introductory_Chemistry-1/7%3A_Types_of_Chemical_Reactions/7.5%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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Courses/Athabasca_University/Chemistry_350%3A_Organic_Chemistry_I/05%3A_Stereochemistry_at_Tetrahedral_Centres/5.05%3A_Sequence_Rules_for_Specifying_Configuration
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Objectives After completing this section, you should be able to assign Cahn-Ingold-Prelog priorities to a given set of substituents. determine whether a given wedge-and-broken-line structure corresponds to an R or an S configuration, with or without the aid of molecular models. draw the wedge-and-broken-line structure for a compound, given its IUPAC name, complete with R or S designation. construct a stereochemically accurate model of a given enantiomer from either a wedge-and-broken-line structure or the IUPAC name of the compound, complete with R or S designation. Key Terms Make certain that you can define, and use in context, the key terms below. absolute configuration R configuration S configuration Study Notes When designating a structure as R or S , you must ensure that the atom or group with the lowest priority is pointing away from you, the observer. The easiest way to show this is to use the wedge-and-broken-line representation. You can then immediately determine whether you are observing an R configuration or an S configuration. 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, as such, 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 or ( 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. The "-H" (left) ranks lower than the "C-" (right) based on the relative molecular weights at the first point of difference. Example \(\PageIndex{1}\) Worked Exercise \(\PageIndex{1}\) For the following pairs of substituents, determine which would have the higher and lower priority based on the Cahn-Ingold-Prelog rules. Explain your answer. Answer 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: The "C-" (right) ranks higher than the "H-" (left) based on the first point of difference and their relative atomic numbers. However: In this case, even though the bold carbon on the right structure has two connections to a non-hydrogen atom (C), it is the lower priority. This is because one of the atoms attached to the bold carbon on the left molecule ranks higher than any of the atoms attached to the bold carbon on the right structure, since Br has a higher atomic number than C. 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. 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 5.5.2 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. Outside links http://www.youtube.com/watch?v=EphUiPiQiCo http://en.Wikipedia.org/wiki/Absolute_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. Exercises Exercise \(\PageIndex{1}\) Orient the following so that the least priority (4) atom is paced behind, then assign stereochemistry (( R ) or ( S )). Answer A = S and B = R Exercise \(\PageIndex{2}\) Draw ( R )-2-bromobutan-2-ol Answer Exercise \(\PageIndex{3}\) Assign R/S to the following molecule: Answer The stereo center is (R). Exercise \(\PageIndex{4}\) Identify which substituent in the following sets has a higher ranking. -H or -CH 3 -CH 2 CH 2 CH 3 or CH 2 CH 3 -CH 2 Cl or CH 2 OH Answer -CH 3 -CH 2 CH 2 CH 3 -CH 2 Cl Exercise \(\PageIndex{5}\) Identify which substituent in the following sets has a higher ranking. -NH 2 or -N=NH -CH 2 CH 2 OH or -CH 2 OH -CH=CH 2 or -CH 2 CH 3 Answer a) -N=NH b) -CH 2 OH c) -CH=CH 2 Exercise \(\PageIndex{6}\) Place the following sets of substituents in each group in order of lowest priority (1 st ) to highest priority (4 th ) -NH 2 , -F, -Br, -CH 3 -SH, -NH 2 , -F, -H Answer a) -CH 3 < -NH 2 < -F, < -Br b) -H < -NH 2 < -F, < -SH Exercise \(\PageIndex{7}\) Place the following sets of substituents in each group in order of lowest priority (1 st ) to highest priority (4 th ) -CH 2 CH 3 , -CN, -CH 2 CH 2 OH, -CH 2 CH 2 CH 2 OH -CH 2 NH 2 , -CH 2 SH, -C(CH 3 ) 3 , -CN Answer a) -CH 2 CH 3 < -CH 2 CH 2 CH 2 OH < -CH 2 CH 2 OH, < -CN b) -C(CH 3 ) 3 < -CH 2 NH 2 < -CN < -CH 2 SH Exercise \(\PageIndex{8}\) Assign the following chiral centers as ( R ) or ( S ). Answer a) ( S ) : I > Br > F > H. The lowest priority substituent is going backwards so following the highest priority, it goes left (counterclockwise). b) ( R ) : Br > Cl > CH 3 > H. Using a model kit, you need to rotate the H to the back position where the Br is. This causes the priority to go to the right (clockwise) when looking at it with the H in the back position. Alternatively, if you do not have a model kit, you can imagine the structure 3-dimensionally and since the lowest priority (H) is facing up (as drawn), if you look at it from below, starting with Br (1 st priority) and moving towards Cl (2 nd priority), you are moving right (clockwise) which represents ( R ) stereochemistry. c) Neither ( R ) or ( S ) : Since there are two identical substituents (H’s) the molecule is achiral and cannot be assigned ( R ) or ( S ).
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Courses/Sacramento_City_College/Chem_400%3A_General_Chemistry_I/13%3A_Solutions/13.8%3A_Colloids
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Learning Objectives To distinguish between true solutions and solutions with aggregate particles. Suspensions and colloids are two common types of mixtures whose properties are in many ways intermediate between those of true solutions and heterogeneous mixtures. A suspension is a heterogeneous mixture of particles with diameters of about 1 µm (1000 nm) that are distributed throughout a second phase. Common suspensions include paint, blood, and hot chocolate, which are solid particles in a liquid, and aerosol sprays, which are liquid particles in a gas. If the suspension is allowed to stand, the two phases will separate, which is why paints must be thoroughly stirred or shaken before use. A colloid is also a heterogeneous mixture, but the particles of a colloid are typically smaller than those of a suspension, generally in the range of 2 to about 500 nm in diameter. Colloids include fog and clouds (liquid particles in a gas), milk (solid particles in a liquid), and butter (solid particles in a solid). Other colloids are used industrially as catalysts. Unlike in a suspension, the particles in a colloid do not separate into two phases on standing. The only combination of substances that cannot produce a suspension or a colloid is a mixture of two gases because their particles are so small that they always form true solutions. The properties of suspensions, colloids, and solutions are summarized in Table \(\PageIndex{1}\). Type of Mixture Approximate Size of Particles (nm) Characteristic Properties Examples solution < 2 not filterable; does not separate on standing; does not scatter visible light air, white wine, gasoline, salt water colloid 2–500 scatters visible light; translucent or opaque; not filterable; does not separate on standing smoke, fog, ink, milk, butter, cheese suspension 500–1000 cloudy or opaque; filterable; separates on standing muddy water, hot cocoa, blood, paint Colloids and Suspensions Colloids were first characterized in about 1860 by Thomas Graham, who also gave us Graham’s law of diffusion and effusion. Although some substances, such as starch, gelatin, and glue, appear to dissolve in water to produce solutions, Graham found that they diffuse very slowly or not at all compared with solutions of substances such as salt and sugar. Graham coined the word colloid (from the Greek kólla, meaning “glue”) to describe these substances, as well as the words sol and gel to describe certain types of colloids in which all of the solvent has been absorbed by the solid particles, thus preventing the mixture from flowing readily, as we see in Jell-O. Two other important types of colloids are aerosols, which are dispersions of solid or liquid particles in a gas, and emulsions, which are dispersions of one liquid in another liquid with which it is immiscible. Colloids share many properties with solutions. For example, the particles in both are invisible without a powerful microscope, do not settle on standing, and pass through most filters. However, the particles in a colloid scatter a beam of visible light, a phenomenon known as the Tyndall effect,The effect is named after its discoverer, John Tyndall, an English physicist (1820–1893). whereas the particles of a solution do not. The Tyndall effect is responsible for the way the beams from automobile headlights are clearly visible from the side on a foggy night but cannot be seen from the side on a clear night. It is also responsible for the colored rays of light seen in many sunsets, where the sun’s light is scattered by water droplets and dust particles high in the atmosphere. An example of the Tyndall effect is shown in Figure \(\PageIndex{1}\). Although colloids and suspensions can have particles similar in size, the two differ in stability: the particles of a colloid remain dispersed indefinitely unless the temperature or chemical composition of the dispersing medium is changed. The chemical explanation for the stability of colloids depends on whether the colloidal particles are hydrophilic or hydrophobic. Most proteins, including those responsible for the properties of gelatin and glue, are hydrophilic because their exterior surface is largely covered with polar or charged groups. Starch, a long-branched polymer of glucose molecules, is also hydrophilic. A hydrophilic colloid particle interacts strongly with water, resulting in a shell of tightly bound water molecules that prevents the particles from aggregating when they collide. Heating such a colloid can cause aggregation because the particles collide with greater energy and disrupt the protective shell of solvent. Moreover, heat causes protein structures to unfold, exposing previously buried hydrophobic groups that can now interact with other hydrophobic groups and cause the particles to aggregate and precipitate from solution. When an egg is boiled, for example, the egg white, which is primarily a colloidal suspension of a protein called albumin, unfolds and exposes its hydrophobic groups, which aggregate and cause the albumin to precipitate as a white solid. In some cases, a stable colloid can be transformed to an aggregated suspension by a minor chemical modification. Consider, for example, the behavior of hemoglobin, a major component of red blood cells. Hemoglobin molecules normally form a colloidal suspension inside red blood cells, which typically have a “donut” shape and are easily deformed, allowing them to squeeze through the capillaries to deliver oxygen to tissues. In a common inherited disease called sickle-cell anemia, one of the amino acids in hemoglobin that has a hydrophilic carboxylic acid side chain (glutamate) is replaced by another amino acid that has a hydrophobic side chain (valine). Under some conditions, the abnormal hemoglobin molecules can aggregate to form long, rigid fibers that cause the red blood cells to deform, adopting a characteristic sickle shape that prevents them from passing through the capillaries (Figure \(\PageIndex{2}\)). The reduction in blood flow results in severe cramps, swollen joints, and liver damage. Until recently, many patients with sickle-cell anemia died before the age of 30 from infection, blood clots, or heart or kidney failure, although individuals with the sickle-cell genetic trait are more resistant to malaria than are those with “normal” hemoglobin. Aggregation and precipitation can also result when the outer, charged layer of a particle is neutralized by ions with the opposite charge. In inland waterways, clay particles, which have a charged surface, form a colloidal suspension. High salt concentrations in seawater neutralize the charge on the particles, causing them to precipitate and form land at the mouths of large rivers, as seen in the satellite view in Figure \(\PageIndex{3}\). Charge neutralization is also an important strategy for precipitating solid particles from gaseous colloids such as smoke, and it is widely used to reduce particulate emissions from power plants that burn fossil fuels. Emulsions Emulsions are colloids formed by the dispersion of a hydrophobic liquid in water, thereby bringing two mutually insoluble liquids, such as oil and water, in close contact. Various agents have been developed to stabilize emulsions, the most successful being molecules that combine a relatively long hydrophobic “tail” with a hydrophilic “head”: Examples of such emulsifying agents include soaps, which are salts of long-chain carboxylic acids, such as sodium stearate \(\ce{[CH_3(CH_2)_{16}CO_2−Na^{+}]}\), and detergents, such as sodium dodecyl sulfate \(\ce{[CH_3(CH_2)_{11}OSO_3−Na^{+}]}\), whose structures are as follows: When you wash your laundry, the hydrophobic tails of soaps and detergents interact with hydrophobic particles of dirt or grease through dispersion forces, dissolving in the interior of the hydrophobic particle. The hydrophilic group is then exposed at the surface of the particle, which enables it to interact with water through ion–dipole forces and hydrogen bonding. This causes the particles of dirt or grease to disperse in the wash water and allows them to be removed by rinsing. Similar agents are used in the food industry to stabilize emulsions such as mayonnaise. A related mechanism allows us to absorb and digest the fats in buttered popcorn and French fries. To solubilize the fats so that they can be absorbed, the gall bladder secretes a fluid called bile into the small intestine. Bile contains a variety of bile salts, detergent-like molecules that emulsify the fats. Micelles Detergents and soaps are surprisingly soluble in water in spite of their hydrophobic tails. The reason for their solubility is that they do not, in fact, form simple solutions. Instead, above a certain concentration they spontaneously form micelles, which are spherical or cylindrical aggregates that minimize contact between the hydrophobic tails and water. In a micelle, only the hydrophilic heads are in direct contact with water, and the hydrophobic tails are in the interior of the aggregate (Figure \(\PageIndex{4a}\)). A large class of biological molecules called phospholipids consists of detergent-like molecules with a hydrophilic head and two hydrophobic tails, as can be seen in the molecule of phosphatidylcholine. The additional tail results in a cylindrical shape that prevents phospholipids from forming a spherical micelle. Consequently, phospholipids form bilayers, extended sheets consisting of a double layer of molecules. As shown in Figure \(\PageIndex{4b}\), the hydrophobic tails are in the center of the bilayer, where they are not in contact with water, and the hydrophilic heads are on the two surfaces, in contact with the surrounding aqueous solution. A cell membrane is essentially a mixture of phospholipids that form a phospholipid bilayer. One definition of a cell is a collection of molecules surrounded by a phospholipid bilayer that is capable of reproducing itself. The simplest cells are bacteria, which consist of only a single compartment surrounded by a single membrane. Animal and plant cells are much more complex, however, and contain many different kinds of compartments, each surrounded by a membrane and able to carry out specialized tasks. Summary A suspension is a heterogeneous mixture of particles of one substance distributed throughout a second phase; the dispersed particles separate from the dispersing phase on standing. In contrast, the particles in a colloid are smaller and do not separate on standing. 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. Hydrophilic colloids contain an outer shell of groups that interact favorably with water, whereas hydrophobic colloids have an outer surface with little affinity for water. Emulsions are prepared by dispersing a hydrophobic liquid in water. In the absence of a dispersed hydrophobic liquid phase, solutions of detergents in water form organized spherical aggregates called micelles. Phospholipids are a class of detergent-like molecules that have two hydrophobic tails attached to a hydrophilic head. A bilayer is a two-dimensional sheet consisting of a double layer of phospholipid molecules arranged tail to tail with a hydrophobic interior and a hydrophilic exterior. Cells are collections of molecules that are surrounded by a phospholipid bilayer called a cell membrane and are able to reproduce themselves.
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Courses/Brevard_College/CHE_104%3A_Principles_of_Chemistry_II/08%3A_Radioactivity_and_Nuclear_Processes/8.07%3A_Nuclear_Fusion
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As we saw in the preceding section, when the nuclei of heavy atoms split, energy is released. For light atoms, the opposite is true; when these nuclei combine (fuse together), energy is released. This is the process of nuclear fusion. Fusion of light elements, mostly hydrogen, is the force that powers energy release in the sun and in sun-like stars. Imagine the sun as a huge sphere of hydrogen. Because a star is so massive, the gravitation pull on the hydrogen atoms is sufficient to overcome the repulsion between the two nuclei to force them together to form an unstable 2 2 He {\displaystyle {}_{2}^{2}{\text{He}}} nucleus. This immediately ejects a positron, leaving deuterium, 1 2 H {\displaystyle {}_{1}^{2}{\text{H}}} , and releasing a significant amount of energy. In the cascade of reactions deuterium fuses with another hydrogen to give 2 3 He {\displaystyle {}_{2}^{3}{\text{He}}} , and two of these combine to form helium, ejecting two high-energy protons in the process. In stars that are larger and heavier than our sun, the “triple alpha process” is the dominant nuclear reaction. In this, helium nuclei fuse to eventually form carbon, releasing significant energy in the process. One of the great challenges in physics and engineering today is to replicate fusion of this sort under controlled conditions, harvesting the energy released and converting it, indirectly, into electrical power. The extremely high temperatures and pressures that are required to initiate and sustain fusion reactions thwarted, thus far, attempts to build a fusion reactor that is “break even” in terms of the energy released relative to the energy required to produce the fusion events. Uncontrolled fusion is certainly possible, and fusion bombs exist, but these typically use an advanced fission bomb to create the temperatures and pressures necessary to promote the fusion of the lighter elements. Clearly, this approach does not work in the laboratory! Work on fusion reactors continues at a fast pace and includes novel approaches such as aneutronic fusion reactions that utilize proton-boron fusion to produce charged particles rather than a barrage of neutrons. The advantage here is that few neutrons are produced, reducing the need for shielding, and the charged particles formed can potentially be captured directly as electricity.
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Courses/Windward_Community_College/BIOC_141%3A_Fundamentals_of_Biochemistry_(Colmenares_and_Ashburn)/08%3A_Solids_Liquids_and_Gases/8.4%3A_Gas_Laws
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Learning Objectives To predict the properties of gases using the gas laws. Experience has shown that several properties of a gas can be related to each other under certain conditions. The properties are pressure ( P ), volume ( V ), temperature ( T , in kelvins), and amount of material expressed in moles ( n ). What we find is that a sample of gas cannot have any random values for these properties. Instead, only certain values, dictated by some simple mathematical relationships, will occur. Boyle’s Law The first simple relationship, referred to as a gas law, is between the pressure of a gas and its volume. If the amount of gas in a sample and its temperature are kept constant, then as the pressure of a gas is increased, the volume of the gas decreases proportionately. Mathematically, this is written as \[\mathrm{P \propto \dfrac{1}{V}} \nonumber \] where the “∝” symbol means “is proportional to.” This is one form of Boyle’s law, which relates the pressure of a gas to its volume. A more useful form of Boyle’s law involves a change in conditions of a gas. For a given amount of gas at a constant temperature, if we know the initial pressure and volume of a gas sample and the pressure or volume changes, we can calculate what the new volume or pressure will be. That form of Boyle’s law is written \[P_iV_i = P_fV_f \label{Eq1} \] where the subscript \(i\) refers to initial conditions and the subscript \(f\) refers to final conditions. To use \(\ref{Eq1}\), you need to know any three of the variables so that you can algebraically calculate the fourth variable. Also, the pressure quantities must have the same units, as must the two volume quantities. If the two similar variables don’t have the same variables, one value must be converted to the other value’s unit. Example \(\PageIndex{1}\): Increasing Pressure in a Gas What happens to the volume of a gas if its pressure is increased? Assume all other conditions remain the same. Solution If the pressure of a gas is increased, the volume decreases in response. Exercise \(\PageIndex{1}\): Increasing Volume in a Gas What happens to the pressure of a gas if its volume is increased? Assume all other conditions remain the same. Answer If the volume of a gas is increased, the pressure decreases. Example \(\PageIndex{2}\): Gas Compression If a sample of gas has an initial pressure of 1.56 atm and an initial volume of 7.02 L, what is the final volume if the pressure is reduced to 0.987 atm? Assume that the amount and the temperature of the gas remain constant. Solution The key in problems like this is to be able to identify which quantities represent which variables from the relevant equation. The way the question is worded, you should be able to tell that 1.56 atm is P i , 7.02 L is V i , and 0.987 atm is P f . What we are looking for is the final volume— V f . Therefore, substituting these values into P i V i = P f V f : (1.56 atm)(7.02 L) = (0.987 atm) × V f The expression has atmospheres on both sides of the equation, so they cancel algebraically: (1.56)(7.02 L) = (0.987) × V f Now we divide both sides of the expression by 0.987 to isolate V f , the quantity we are seeking: \(\mathrm{\dfrac{(1.56)(7.02\: L)}{0.987}=V_f}\) Performing the multiplication and division, we get the value of V f , which is 11.1 L. The volume increases. This should make sense because the pressure decreases, so pressure and volume are inversely related. Exercise \(\PageIndex{2}\) If a sample of gas has an initial pressure of 3.66 atm and an initial volume of 11.8 L, what is the final pressure if the volume is reduced to 5.09 L? Assume that the amount and the temperature of the gas remain constant. Answer 8.48 atm If the units of similar quantities are not the same, one of them must be converted to the other quantity’s units for the calculation to work out properly. It does not matter which quantity is converted to a different unit; the only thing that matters is that the conversion and subsequent algebra are performed properly. The following example illustrates this process. Example \(\PageIndex{3}\) If a sample of gas has an initial pressure of 1.56 atm and an initial volume of 7.02 L, what is the final volume if the pressure is changed to 1,775 torr? Does the answer make sense? Assume that the amount and the temperature of the gas remain constant. Solution This example is similar to Example \(\PageIndex{2}\), except now the final pressure is expressed in torr. For the math to work out properly, one of the pressure values must be converted to the other unit. Let us change the initial pressure to torr: \(\mathrm{1.56\: atm\times\dfrac{760\: torr}{1\: atm}=1,190\: torr}\) Now we can use Boyle’s law: (1,190 torr)(7.02 L) = (1,775 torr) × V f Torr cancels algebraically from both sides of the equation, leaving (1,190)(7.02 L) = (1,775) × V f Now we divide both sides of the equation by 1,775 to isolate V f on one side. Solving for the final volume, \(\mathrm{V_f=\dfrac{(1,190)(7.02\: L)}{1,775}=4.71\: L}\) Because the pressure increases, it makes sense that the volume decreases. The answer for the final volume is essentially the same if we converted the 1,775 torr to atmospheres: \(\mathrm{1,775\: torr\times\dfrac{1\: atm}{760\: torr}=2.336\: atm}\). Using Boyle’s law: (1.56 atm)(7.02 L) = (2.335 atm) × V f ; \(\mathrm{V_f=\dfrac{(1.56\: atm)(7.02\: L)}{2.336\: atm}=4.69\: L}\). Exercise \(\PageIndex{3}\) If a sample of gas has an initial pressure of 375 torr and an initial volume of 7.02 L, what is the final pressure if the volume is changed to 4,577 mL? Does the answer make sense? Assume that amount and the temperature of the gas remain constant. Answer 575 torr To Your Health: Breathing Breathing certainly is a major contribution to your health! Without breathing, we could not survive. Curiously, the act of breathing itself is little more than an application of Boyle’s law. The lungs are a series of ever-narrowing tubes that end in a myriad of tiny sacs called alveoli. It is in the alveoli that oxygen from the air transfers to the bloodstream and carbon dioxide from the bloodstream transfers to the lungs for exhalation. For air to move in and out of the lungs, the pressure inside the lungs must change, forcing the lungs to change volume—just as predicted by Boyle’s law. The pressure change is caused by the diaphragm, a muscle that covers the bottom of the lungs. When the diaphragm moves down, it expands the size of our lungs. When this happens, the air pressure inside our lungs decreases slightly. This causes new air to rush in, and we inhale. The pressure decrease is slight—only 3 torr, or about 0.4% of an atmosphere. We inhale only 0.5–1.0 L of air per normal breath. Exhaling air requires that we relax the diaphragm, which pushes against the lungs and slightly decreases the volume of the lungs. This slightly increases the pressure of the air in the lungs, and air is forced out; we exhale. Only 1–2 torr of extra pressure is needed to exhale. So with every breath, our own bodies are performing an experimental test of Boyle’s law. Charles’s Law Another simple gas law relates the volume of a gas to its temperature. Experiments indicate that as the temperature of a gas sample is increased, its volume increases as long as the pressure and the amount of gas remain constant. The way to write this mathematically is \[\mathrm V \propto T \nonumber \] At this point, the concept of temperature must be clarified. Although the Kelvin scale is the preferred temperature scale, the Celsius scale is also a common temperature scale used in science. The Celsius scale is based on the melting and boiling points of water and is actually the common temperature scale used by most countries around the world (except for the United States, which still uses the Fahrenheit scale). The value of a Celsius temperature is directly related to its Kelvin value by a simple expression: Kelvin temperature = Celsius temperature + 273 Thus, it is easy to convert from one temperature scale to another. The Kelvin scale is sometimes referred to as the absolute scale because the zero point on the Kelvin scale is at absolute zero, the coldest possible temperature. On the other temperature scales, absolute zero is −260°C or −459°F. The expression relating a gas volume to its temperature begs the following question: to which temperature scale is the volume of a gas related? The answer is that gas volumes are directly related to the Kelvin temperature . Therefore, the temperature of a gas sample should always be expressed in (or converted to) a Kelvin temperature. Example \(\PageIndex{4}\): Increasing Temperature What happens to the volume of a gas if its temperature is decreased? Assume that all other conditions remain constant. Solution If the temperature of a gas sample is decreased, the volume decreases as well. Exercise \(\PageIndex{4}\) What happens to the temperature of a gas if its volume is increased? Assume that all other conditions remain constant. Answer The temperature increases. As with Boyle’s law, the relationship between volume and temperature can be expressed in terms of initial and final values of volume and temperature, as follows: \[\mathrm{\dfrac{V_i}{T_i}=\dfrac{V_f}{T_f}} \nonumber \] where V i and T i are the initial volume and temperature, and V f and T f are the final volume and temperature. This is Charles’s law. The restriction on its use is that the pressure of the gas and the amount of gas must remain constant. (Charles’s law is sometimes referred to as Gay-Lussac’s law, after the scientist who promoted Charles’s work.) Example \(\PageIndex{5}\) A gas sample at 20°C has an initial volume of 20.0 L. What is its volume if the temperature is changed to 60°C? Does the answer make sense? Assume that the pressure and the amount of the gas remain constant. Solution Although the temperatures are given in degrees Celsius, we must convert them to the kelvins before we can use Charles’s law. Thus, 20°C + 273 = 293 K = T i 60°C + 273 = 333 K = T f Now we can substitute these values into Charles’s law, along with the initial volume of 20.0 L: \(\mathrm{\dfrac{20.0\: L}{293\: K}=\dfrac{V_f}{333\: K}}\) Multiplying the 333 K to the other side of the equation, we see that our temperature units will cancel: \(\mathrm{\dfrac{(333\: K)(20.0\: L)}{293\: K}=V_f}\) Solving for the final volume, V f = 22.7 L. So, as the temperature is increased, the volume increases. This makes sense because volume is directly proportional to the absolute temperature (as long as the pressure and the amount of the remain constant). Exercise \(\PageIndex{5}\) A gas sample at 35°C has an initial volume of 5.06 L. What is its volume if the temperature is changed to −35°C? Does the answer make sense? Assume that the pressure and the amount of the gas remain constant. Answer 3.91 L Combined Gas Law Other gas laws can be constructed, but we will focus on only two more. The combined gas law brings Boyle’s and Charles’s laws together to relate pressure, volume, and temperature changes of a gas sample: \[\mathrm{\dfrac{P_iV_i}{T_i}=\dfrac{P_fV_f}{T_f}} \nonumber \] To apply this gas law, the amount of gas should remain constant. As with the other gas laws, the temperature must be expressed in kelvins, and the units on the similar quantities should be the same. Because of the dependence on three quantities at the same time, it is difficult to tell in advance what will happen to one property of a gas sample as two other properties change. The best way to know is to work it out mathematically. Example \(\PageIndex{6}\) A sample of gas has P i = 1.50 atm, V i = 10.5 L, and T i = 300 K. What is the final volume if P f = 0.750 atm and T f = 350 K? Solution Using the combined gas law, substitute for five of the quantities: \(\mathrm{\dfrac{(1.50\: atm)(10.5\: L)}{300\: K}=\dfrac{(0.750\: atm)(V_f)}{350\: K}}\) We algebraically rearrange this expression to isolate V f on one side of the equation: \(\mathrm{V_f=\dfrac{(1.50\: atm)(10.5\: L)(350\: K)}{(300\: K)(0.750\: atm)}=24.5\: L}\) Note how all the units cancel except the unit for volume. A sample of gas has P i = 0.768 atm, V i = 10.5 L, and T i = 300 K. What is the final pressure if V f = 7.85 L and T f = 250 K? Answer 0.856 atm Example \(\PageIndex{7}\) A balloon containing a sample of gas has a temperature of 22°C and a pressure of 1.09 atm in an airport in Cleveland. The balloon has a volume of 1,070 mL. The balloon is transported by plane to Denver, where the temperature is 11°C and the pressure is 655 torr. What is the new volume of the balloon? Solution The first task is to convert all quantities to the proper and consistent units. The temperatures must be expressed in kelvins, and the pressure units are different so one of the quantities must be converted. Let us convert the atmospheres to torr: 22°C + 273 = 295 K = T i 11°C + 273 = 284 K = T f \(\mathrm{1.09\: atm\times\dfrac{760\: torr}{1\: atm}=828\: torr = P_i}\) Now we can substitute the quantities into the combined has law: \(\mathrm{\dfrac{(828\: torr)(1,070\: mL)}{295\: K}=\dfrac{(655\: torr)\times V_f}{284\: K}}\) To solve for V f , we multiply the 284 K in the denominator of the right side into the numerator on the left, and we divide 655 torr in the numerator of the right side into the denominator on the left: \(\mathrm{\dfrac{(828\: torr)(1,070\: mL)(284\: K)}{(295\: K)(655\: torr)}=V_f}\) Notice that torr and kelvins cancel, as they are found in both the numerator and denominator. The only unit that remains is milliliters, which is a unit of volume. So V f = 1,300 mL. The overall change is that the volume of the balloon has increased by 230 mL. A balloon used to lift weather instruments into the atmosphere contains gas having a volume of 1,150 L on the ground, where the pressure is 0.977 atm and the temperature is 18°C. Aloft, this gas has a pressure of 6.88 torr and a temperature of −15°C. What is the new volume of the gas? Answer 110,038 L The Ideal Gas Law So far, the gas laws we have used have focused on changing one or more properties of the gas, such as its volume, pressure, or temperature. There is one gas law that relates all the independent properties of a gas under any particular condition, rather than a change in conditions. This gas law is called the ideal gas law. The formula of this law is as follows: PV = nRT In this equation, P is pressure, V is volume, n is amount of moles, and T is temperature. R is called the ideal gas law constant and is a proportionality constant that relates the values of pressure, volume, amount, and temperature of a gas sample. The variables in this equation do not have the subscripts i and f to indicate an initial condition and a final condition. The ideal gas law relates the four independent properties of a gas under any conditions. The value of R depends on what units are used to express the other quantities. If volume is expressed in liters and pressure in atmospheres, then the proper value of R is as follows: \[\mathrm{R=0.08205 \: \dfrac{L\cdot atm}{mol\cdot K}} \nonumber \] This may seem like a strange unit, but that is what is required for the units to work out algebraically. Example \(\PageIndex{8}\) What is the volume in liters of 1.45 mol of N 2 gas at 298 K and 3.995 atm? Solution Using the ideal gas law where P = 3.995 atm, n = 1.45, and T = 298, \(\mathrm{(3.995\: atm)\times V=(1.45\: mol)\left(0.08205\: \dfrac{L\cdot atm}{mol\cdot K}\right)(298\: K)}\) On the right side, the moles and kelvins cancel. Also, because atmospheres appear in the numerator on both sides of the equation, they also cancel. The only remaining unit is liters, a unit of volume. So 3.995 × V = (1.45)(0.08205)(298) L Dividing both sides of the equation by 3.995 and evaluating, we get V = 8.87 L. Note that the conditions of the gas are not changing. Rather, the ideal gas law allows us to determine what the fourth property of a gas (here, volume) must be if three other properties (here, amount, pressure, and temperature) are known. What is the pressure of a sample of CO 2 gas if 0.557 mol is held in a 20.0 L container at 451 K? Answer 1.03 atm For convenience, scientists have selected 273 K (0°C) and 1.00 atm pressure as a set of standard conditions for gases. This combination of conditions is called standard temperature and pressure (STP). Under these conditions, 1 mol of any gas has about the same volume. We can use the ideal gas law to determine the volume of 1 mol of gas at STP: \[\mathrm{(1.00\: atm)\times V=(1.00\: mol)\left(0.08205\: \dfrac{L\cdot atm}{mol\cdot K}\right)(273\: K)} \nonumber \] This volume is 22.4 L. Because this volume is independent of the identity of a gas, the idea that 1 mol of gas has a volume of 22.4 L at STP makes a convenient conversion factor: 1 mol gas = 22.4 L (at STP) Example \(\PageIndex{9}\) Cyclopropane (C 3 H 6 ) is a gas that formerly was used as an anesthetic. How many moles of gas are there in a 100.0 L sample if the gas is at STP? Solution We can set up a simple, one-step conversion that relates moles and liters: \(\mathrm{100.0\: L\: C_3H_6\times \dfrac{1\: mol}{22.4\: L}=4.46\: mol\: C_3H_6}\) There are almost 4.5 mol of gas in 100.0 L. Note: Because of its flammability, cyclopropane is no longer used as an anesthetic gas. Freon is a trade name for a series of fluorine- and chlorine-containing gases that formerly were used in refrigeration systems. What volume does 8.75 mol of Freon have at STP? Note: Many gases known as Freon are no longer used because their presence in the atmosphere destroys the ozone layer, which protects us from ultraviolet light from the sun. Answer 196 L Airbags Airbags (Figure \(\PageIndex{3}\)) are a safety feature provided in most automobiles since the 1990s. The effective operation of an airbag requires that it be rapidly inflated with an appropriate amount (volume) of gas when the vehicle is involved in a collision. This requirement is satisfied in many automotive airbag systems through use of explosive chemical reactions, one common choice being the decomposition of sodium azide, NaN 3 . When sensors in the vehicle detect a collision, an electrical current is passed through a carefully measured amount of NaN 3 to initiate its decomposition: \[\ce{2NaN3}(s)\rightarrow \ce{3N2}(g)+\ce{2Na}(s) \nonumber \] This reaction is very rapid, generating gaseous nitrogen that can deploy and fully inflate a typical airbag in a fraction of a second (~0.03–0.1 s). Among many engineering considerations, the amount of sodium azide used must be appropriate for generating enough nitrogen gas to fully inflate the air bag and ensure its proper function. For example, a small mass (~100 g) of NaN 3 will generate approximately 50 L of N 2 . Dalton's Law of Partial Pressures The ideal gas equation of state applies to mixtures just as to pure gases. It was in fact with a gas mixture, ordinary air, that Boyle, Gay-Lussac and Charles did their early experiments. The only new concept we need in order to deal with gas mixtures is the partial pressure , a concept invented by the famous English chemist John Dalton (1766-1844). Dalton reasoned that the low density and high compressibility of gases indicates that they consist mostly of empty space; from this it follows that when two or more different gases occupy the same volume, they behave entirely independently. The contribution that each component of a gaseous mixture makes to the total pressure of the gas is known as the partial pressure of that gas. The definition of Dalton's Law of Partial Pressures that address this is: The total pressure of a gas is the sum of the partial pressures of its components which is expressed algebraically as \[P_{total}=P_1+P_2+P_3 ... = \sum_i P_i \nonumber \] or, equivalently \[ P_{total} = \dfrac{RT}{V} \sum_i n_i \nonumber \] There is also a similar relationship based on volume fractions , known as Amagat's law of partial volumes . It is exactly analogous to Dalton's law, in that it states that the total volume of a mixture is just the sum of the partial volumes of its components. But there are two important differences: Amagat's law holds only for ideal gases which must all be at the same temperature and pressure. Dalton's law has neither of these restrictions. Although Amagat's law seems intuitively obvious, it sometimes proves useful in chemical engineering applications. We will make no use of it in this course. Example \(\PageIndex{10}\) Three flasks having different volumes and containing different gases at various pressures are connected by stopcocks as shown. When the stopcocks are opened, What will be the pressure in the system? Which gas will be most abundant in the mixture? Assume that the temperature is uniform and that the volume of the connecting tubes is negligible. Solution The trick here is to note that the total number of moles n T and the temperature remain unchanged, so we can make use of Boyle's law PV = constant. We will work out the details for CO 2 only, denoted by subscripts a. For CO 2 , \[P_aV_a = (2.13\; atm)(1.50\; L) = 3.20\; L \cdot atm \nonumber \] Adding the PV products for each separate container, we obtain \[A\sum_i P_i V_i = 6.36\; L \cdot atm = n_T RT \nonumber \] We will call this sum P 1 V 1 . After the stopcocks have been opened and the gases mix, the new conditions are denoted by P 2 V 2 . From Boyle's law (\(\ref{Eq1}\), \[P_1V_1 = P_2V_2 = 6.36\; L \cdot atm \nonumber \] \[V_2 = \sum_i V_i = 4.50\; L \nonumber \] Solving for the final pressure P 2 we obtain (6.36 L-atm)/(4.50 L) = 1.41 atm . For part (b) , note that the number of moles of each gas is n = PV/RT . The mole fraction of any one gas is X i = n i / n T . For CO 2 , this works out to (3.19/ RT ) / (6.36/ RT ) = 0.501. Because this exceeds 0.5, we know that this is the most abundant gas in the final mixture. Dalton’s law states that in a gas mixture (\(P_{total}\)) each gas will exert a pressure independent of the other gases (\(P_n\)) and each gas will behave as if it alone occupies the total volume. By extension, the partial pressure of each gas can be calculated by multiplying the total pressure (\(P_{total}\)) by the gas percentage (%). \[P_{Total} = P_1 + P_2 + P_3 + P_4 + ... + P_n \nonumber \] or \[P_n = \dfrac{\text{% of individual gas}_n}{P_{Total}} \nonumber \] Gas Partial Pressure (mm Hg) Percentage (%) Nitrogen, (N_2\) \(P_{N_2}\) = 594 78.000 Oxygen, \(O_2\) \(P_{O_2}\)= 160 21.000 Carbon Dioxide, \(CO_2\) \(P_{CO_2}\) = 0.25 0.033 Water Vapor, \(H_2O\) \(P_{H_2O}\) = 5.7 0.750 Other trace gases \(P_{Other}\) = 0.05 0.220 Total air \(P_{Total}\) = 760 1.000 Application of Dalton's Law: Collecting Gases over Water A common laboratory method of collecting the gaseous product of a chemical reaction is to conduct it into an inverted tube or bottle filled with water, the opening of which is immersed in a larger container of water. This arrangement is called a pneumatic trough , and was widely used in the early days of chemistry. As the gas enters the bottle it displaces the water and becomes trapped in the upper part. The volume of the gas can be observed by means of a calibrated scale on the bottle, but what about its pressure? The total pressure confining the gas is just that of the atmosphere transmitting its force through the water. (An exact calculation would also have to take into account the height of the water column in the inverted tube.) But liquid water itself is always in equilibrium with its vapor, so the space in the top of the tube is a mixture of two gases: the gas being collected, and gaseous H 2 O. The partial pressure of H 2 O is known as the vapor pressure of water and it depends on the temperature. In order to determine the quantity of gas we have collected, we must use Dalton's Law to find the partial pressure of that gas. Example \(\PageIndex{11}\) Oxygen gas was collected over water as shown above. The atmospheric pressure was 754 torr, the temperature was 22°C, and the volume of the gas was 155 mL. The vapor pressure of water at 22°C is 19.8 torr. Use this information to estimate the number of moles of \(O_2\) produced. Solution From Dalton's law, \[P_{O_2} = P_{total} – P_{H_2O} = 754 – 19.8 = 734 \; torr = 0.966\; atm \nonumber \] Now use the Ideal Gas Law to convert to moles \[ n =\dfrac{PV}{RT} = \dfrac{(0.966\; atm)(0.155\;L)}{(0.082\; L atm mol^{-1} K^{-1})(295\; K)}= 0.00619 \; mol \nonumber \] Henry’s Law Henry's law is one of the gas laws formulated by William Henry in 1803. It states: "At a constant temperature, the amount of a given gas that dissolves in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid." An equivalent way of stating the law is that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. To explain this law, Henry derived the equation: \[ C =k P_{gas} \nonumber \] where Henry’s Law tells us that the greater the pressure of gas above the surface of a liquid, the higher the concentration of the gas in the liquid. Also, Henry’s law tells us that gases diffuse from areas of high gas concentration to areas of low gas concentration. Applicability of Henry's Law Henry's law only works if the molecules are at equilibrium. Henry's law does not work for gases at high pressures (e.g., \(N_{2\;(g)}\) at high pressure becomes very soluble and harmful when in the blood supply ). Henry's law does not work if there is a chemical reaction between the solute and solvent (e.g., \(HCl_{(g)}\) reacts with water by a dissociation reaction to generate \(H_3O^+\) and \(Cl^-\) ions). Application of Henry's Law: Scuba diving Our respiratory systems are designed to maintain the proper oxygen concentration in the blood when the partial pressure of O 2 is 0.21 atm, its normal sea-level value. Below the water surface, the pressure increases by 1 atm for each 10.3 m increase in depth; thus a scuba diver at 10.3 m experiences a total of 2 atm pressure pressing on the body. In order to prevent the lungs from collapsing, the air the diver breathes should also be at about the same pressure. But at a total pressure of 2 atm, the partial pressure of \(O_2\) in ordinary air would be 0.42 atm; at a depth of 100 ft (about 30 m), the \(O_2\) pressure of 0.8 atm would be far too high for health. For this reason, the air mixture in the pressurized tanks that scuba divers wear must contain a smaller fraction of \(O_2\). This can be achieved most simply by raising the nitrogen content, but high partial pressures of N 2 can also be dangerous, resulting in a condition known as nitrogen narcosis. The preferred diluting agent for sustained deep diving is helium, which has very little tendency to dissolve in the blood even at high pressures. Career Focus: Respiratory Therapist Certain diseases—such as emphysema, lung cancer, and severe asthma—primarily affect the lungs. Respiratory therapists help patients with breathing-related problems. They can evaluate, help diagnose, and treat breathing disorders and even help provide emergency assistance in acute illness where breathing is compromised. Most respiratory therapists must complete at least two years of college and earn an associate’s degree, although therapists can assume more responsibility if they have a college degree. Therapists must also pass state or national certification exams. Once certified, respiratory therapists can work in hospitals, doctor’s offices, nursing homes, or patient’s homes. Therapists work with equipment such as oxygen tanks and respirators, may sometimes dispense medication to aid in breathing, perform tests, and educate patients in breathing exercises and other therapy. Because respiratory therapists work directly with patients, the ability to work well with others is a must for this career. It is an important job because it deals with one of the most crucial functions of the body. Key Takeaway The physical properties of gases are predictable using mathematical formulas known as gas laws. \(C\) is the solubility of a gas at a fixed temperature in a particular solvent (in units of M or mL gas/L) \(k\) is Henry's law constant (often in units of M/atm) \(P_{gas}\) is the partial pressure of the gas (often in units of Atm)
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Courses/Nassau_Community_College/Organic_Chemistry_I_and_II/05%3A_An_Introduction_to_Organic_Reactions_using_Free_Radical_Halogenation_of_Alkanes
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Learning Objectives After reading the chapter and completing ALL the exercises and homework, a student can be able to: recognize and distinguish between the four major types of organic reactions (additions, eliminations, substitutions, and rearrangements) - refer to section 5.1 accurately and precisely use reaction mechanism notation and symbols including curved arrows to show the flow of electrons - refer to section 5.2 identify nucleophiles and electrophiles in polar reactions - refer to section 5.3 perform calculations using the equation \[ΔGº = –RT \ln K = –2.303 RT \log_{10} K \label{eq2}\] and explain the relationship between equilibrium and free energy - refer to section 5.4 calculate reaction enthalpies from bond dissociation energies - refer to section 5.5 draw Reaction Energy Diagrams from the thermodynamic and kinetic data/information - refer to se5.6 use a Reaction Energy Diagram to discuss transition states, Ea, intermediates & rate determining step - refer to section 5.6 draw the transition states & intermediates of a reaction - refer to section 5.6 describe the structure & relative stabilities of carbocations, free radicals and carbanions - refer to sections 5.7 - 5.9 respectively Explain the mechanism & energetics of the free-radical halogenation of alkanes - refer to section 5.10 Predict the products of chlorination & bromination reactions of alkanes based on relative reactivity and selectivity - refer to section 5.11 describe the similarities and differences between reactions performed in the lab with biochemical reactions - refer to section 5.12 5.1: Types of Organic Reactions The four main classes of organic reactions are additions, eliminations, substitutions, and rearrangements. 5.2: 5.2 Reaction Mechanism Notation and Symbols Arrows are used by chemists to communicate electron flow in mechanisms, reaction completion/equilibrium, and resonance relationships. It is important to use accuracy when selecting the type of arrow for reactions and precision in drawing the location of the arrow head and tail for the curved arrows of electron flow. 5.3: the Dance of the Nucleophile and Electrophile Sterics and electronics are the underlying driving forces for polar organic reactions. The electron rich nucleophile (Nu:) reacts with the electron poor electrophile through a variety of pathways that can be limited and/or influenced by steric hindrance. We explore and learn the polar reaction pathways in subsequent chapters. 5.4: Describing a Reaction - Equilibrium and Free Energy Changes The relationship between equilibrium and free energy is reviewed quantitatively and applied to organic reactions conceptually. 5.5: Homolytic Cleavage and Bond Dissociation Energies The products of homolytic cleavage are radicals and the energy that is required to break the bond homolytically is called the Bond Dissociation Energy (BDE) and is a measure of the strength of the bond. 5.6: Reaction Energy Diagrams and Transition States Reaction energy diagrams efficiently and effectively communicate the thermodynamics and kinetics of chemical reactions in a single diagram. They are a useful tool in learning organic chemistry. 5.7: 5.7 Reactive Intermediates - Carbocations A carbocation is a cation in which carbon has an empty p orbital and bears a positive charge creating a highly reactive intermediate. Comparing the relative stability of reaction intermediates helps elucidate reaction mechanisms and predict major and minor products. 5.8: 5.8 Reactive Intermediates - Radicals A radical (more precisely, a free radical) is an atom, molecule, or ion that has unpaired valence electron (half filled orbital) creating a highly reactive intermediate. 5.9: Reactive Intermediates: Carbanions and Carbon Acids A carbanion is an anion in which carbon has an unshared pair of electrons and bears a negative charge creating a highly reactive intermediate. 5.10: The Free-Radical Halogenation of Alkanes Free radical halogenation of alkanes is 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 opens a gateway to further chemical reactions. We will apply the reaction concepts discussed in this chapter to this reaction to show how empirical data supports these theories. 5.11: Reactivity and Selectivity In general, high reactivity correlates with low selectivity and vice versa. Depending on the structure of the substrate, reaction conditions can be optimized for high reactivity or high selectivity and occasionally for both. 5.12: A Comparison between Biological Reactions and Laboratory Reactions Biochemical reactions occur within our body fluids at a typical pH of 7.4 and temperature of 98.6C. Our biochemistry relies on enzymes to catalyze physiological reactions within this narrow range of environmental conditions. Synthetic organic chemists can create extreme conditions within reactions flasks to catalyze and promote chemical reactions. 5.13: Additional Exercises This section has additional exercises for the key learning objectives of this chapter. 5.14: Solutions to Additional Exercises This section has the solutions to the additional exercises from the previous section.
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Courses/Purdue/Purdue_Chem_26100%3A_Organic_Chemistry_I_(Wenthold)/Chapter_02._Structures_and_Properties_of_Organic_Molecules/2.4_Isomerism
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This page explains what structural isomerism is, and looks at some of the various ways that structural isomers can arise. What is structural isomerism? Isomers are molecules that have the same molecular formula, but have a different arrangement of the atoms in space. That excludes any different arrangements which are simply due to the molecule rotating as a whole, or rotating about particular bonds. For example, both of the following are the same molecule. They are not isomers. Both are butane. There are also endless other possible ways that this molecule could twist itself. There is completely free rotation around all the carbon-carbon single bonds. If you had a model of a molecule in front of you, you would have to take it to pieces and rebuild it if you wanted to make an isomer of that molecule. If you can make an apparently different molecule just by rotating single bonds, it's not different - it's still the same molecule. In structural isomerism, the atoms are arranged in a completely different order. This is easier to see with specific examples. What follows looks at some of the ways that structural isomers can arise. The names of the various forms of structural isomerism probably don't matter all that much, but you must be aware of the different possibilities when you come to draw isomers. Chain Isomerism These isomers arise because of the possibility of branching in carbon chains. For example, there are two isomers of butane, \(C_4H_{10}\). In one of them, the carbon atoms lie in a "straight chain" whereas in the other the chain is branched. Be careful not to draw "false" isomers which are just twisted versions of the original molecule. For example, this structure is just the straight chain version of butane rotated about the central carbon-carbon bond. You could easily see this with a model. This is the example we've already used at the top of this page. Example 1: Chain Isomers in Pentane Pentane, C 5 H 12 , has three chain isomers. If you think you can find any others, they are simply twisted versions of the ones below. If in doubt make some models. Position isomerism In position isomerism, the basic carbon skeleton remains unchanged, but important groups are moved around on that skeleton. Example 2: Positional Isomers in C 5 H 12 For example, there are two structural isomers with the molecular formula C3H7Br. In one of them the bromine atom is on the end of the chain, whereas in the other it's attached in the middle. If you made a model, there is no way that you could twist one molecule to turn it into the other one. You would have to break the bromine off the end and re-attach it in the middle. At the same time, you would have to move a hydrogen from the middle to the end. Another similar example occurs in alcohols such as \(C_4H_9OH\) These are the only two possibilities provided you keep to a four carbon chain, but there is no reason why you should do that. You can easily have a mixture of chain isomerism and position isomerism - you aren't restricted to one or the other. So two other isomers of butanol are: You can also get position isomers on benzene rings. Consider the molecular formula \(C_7H_7Cl\). There are four different isomers you could make depending on the position of the chlorine atom. In one case it is attached to the side-group carbon atom, and then there are three other possible positions it could have around the ring - next to the \(CH_3\) group, next-but-one to the \(CH_3\) group, or opposite the \(CH_3\) group. Functional group isomerism In this variety of structural isomerism, the isomers contain different functional groups - that is, they belong to different families of compounds (different homologous series). Example 3: Isomers in C 3 H 6 O A molecular formula \(C_3H_6O\) could be either propanal (an aldehyde) or propanone (a ketone). There are other possibilities as well for this same molecular formula - for example, you could have a carbon-carbon double bond (an alkene) and an -OH group (an alcohol) in the same molecule. Another common example is illustrated by the molecular formula \(C_3H_6O_2\). Amongst the several structural isomers of this are propanoic acid (a carboxylic acid) and methyl ethanoate (an ester). Contributors Jim Clark ( Chemguide.co.uk ) Further Reading Khan Academy Constitutional Isomers Cliffs Notes Structural Isomers and Stereoisomers Web Pages Information on Isomers Organizing Isomers Difference Between Isomers and Resonance Structures Lewis Formulas, Structural Isomerism, Resonance Structural Isomerism Videos Isomers Video Isomers Video Tutorial Drawing Isomers Drawing Isomers tutorial Practice Problems Isomer quiz Isomer quiz
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Courses/Fullerton_College/Beginning_Chemistry_(Chan)/17%3A_Organic_Chemistry/17.02%3A_Hydrocarbons
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Learning Objectives Identify alkanes, alkenes, alkynes, and aromatic compounds. List some properties of hydrocarbons. The simplest organic compounds are those composed of only two elements: carbon and hydrogen. These compounds are called hydrocarbons . Hydrocarbons themselves are separated into two types: aliphatic hydrocarbons and aromatic hydrocarbons. Aliphatic hydrocarbons are hydrocarbons based on chains of C atoms. There are three types of aliphatic hydrocarbons. Alkanes are aliphatic hydrocarbons with only single covalent bonds. Alkenes are hydrocarbons that contain at least one C–C double bond, and Alkynes are hydrocarbons that contain a C–C triple bond. Occasionally, we find an aliphatic hydrocarbon with a ring of C atoms; these hydrocarbons are called cycloalkanes (or cycloalkenes or cycloalkynes ). Aromatic hydrocarbons have a special six-carbon ring called a benzene ring. Electrons in the benzene ring have special energetic properties that give benzene physical and chemical properties that are markedly different from alkanes. Originally, the term aromatic was used to describe this class of compounds because they were particularly fragrant. However, in modern chemistry the term aromatic denotes the presence of a six-membered ring that imparts different and unique properties to a molecule. The simplest alkanes have their C atoms bonded in a straight chain; these are called normal alkanes. They are named according to the number of C atoms in the chain. The smallest alkane is methane: The next-largest alkane has two C atoms that are covalently bonded to each other. For each C atom to make four covalent bonds, each C atom must be bonded to three H atoms. The resulting molecule, whose formula is C 2 H 6 , is ethane: Propane has a backbone of three C atoms surrounded by H atoms. You should be able to verify that the molecular formula for propane is C 3 H 8 : The diagrams representing alkanes are called structural formulas because they show the structure of the molecule. As molecules get larger, structural formulas become more and more complex. One way around this is to use a condensed structural formula , which lists the formula of each C atom in the backbone of the molecule. For example, the condensed structural formula for ethane is CH 3 CH 3 , while for propane it is CH 3 CH 2 CH 3 . Table \(\PageIndex{1}\) - The First 10 Alkanes, gives the molecular formulas, the condensed structural formulas, and the names of the first 10 alkanes. Molecular Formula Condensed Structural Formula Name CH4 CH4 methane C2H6 CH3CH3 ethane C3H8 CH3CH2CH3 propane C4H10 CH3CH2CH2CH3 butane C5H12 CH3CH2CH2CH2CH3 pentane C6H14 CH3(CH2)4CH3 hexane C7H16 CH3(CH2)5CH3 heptane C8H18 CH3(CH2)6CH3 octane C9H20 CH3(CH2)7CH3 nonane C10H22 CH3(CH2)8CH3 decane Because alkanes have the maximum number of H atoms possible according to the rules of covalent bonds, alkanes are also referred to as saturated hydrocarbons . Alkenes have a C–C double bond. Because they have less than the maximum number of H atoms possible, they are unsaturated hydrocarbons . The smallest alkene—ethene—has two C atoms and is also known by its common name ethylene: The next largest alkene—propene—has three C atoms with a C–C double bond between two of the C atoms. It is also known as propylene: What do you notice about the names of alkanes and alkenes? The names of alkenes are the same as their corresponding alkanes except that the ending is - ene , rather than - ane . Using a stem to indicate the number of C atoms in a molecule and an ending to represent the type of organic compound is common in organic chemistry, as we shall see. With the introduction of the next alkene, butene, we begin to see a major issue with organic molecules: choices. With four C atoms, the C–C double bond can go between the first and second C atoms or between the second and third C atoms: 2 structural formulas for butene, with the first butene having the double bond on the first and second carbon from the left and the latter having its double bond on the second and third carbon from the left. (A double bond between the third and fourth C atoms is the same as having it between the first and second C atoms, only flipped over.) The rules of naming in organic chemistry require that these two substances have different names. The first molecule is named 1-butene , while the second molecule is named 2-butene . The number at the beginning of the name indicates where the double bond originates. The lowest possible number is used to number a feature in a molecule; hence, calling the second molecule 3-butene would be incorrect. Numbers are common parts of organic chemical names because they indicate which C atom in a chain contains a distinguishing feature. The compounds 1-butene and 2-butene have different physical and chemical properties, even though they have the same molecular formula—C 4 H 8 . Different molecules with the same molecular formula are called isomers . Isomers are common in organic chemistry and contribute to its complexity. Example \(\PageIndex{1}\) Based on the names for the butene molecules, propose a name for this molecule. Solution With five C atoms, we will use the pent - stem, and with a C–C double bond, this is an alkene, so this molecule is a pentene. In numbering the C atoms, we use the number 2 because it is the lower possible label. So this molecule is named 2-pentene. Exercise \(\PageIndex{1}\) Based on the names for the butene molecules, propose a name for this molecule. A structural formula of a six carbon molecule with a double bond on the third and fourth carbon from the left. There are twelve hydrogen atoms in total. Answer 3-hexene Alkynes, with a C–C triple bond, are named similarly to alkenes except their names end in - yne . The smallest alkyne is ethyne, which is also known as acetylene: Propyne has the structure Structural formula showing three carbon molecules with a triple bond present between the first and second carbon atom. The appropriate number of hydrogen atoms is attached to each carbon atom. With butyne, we need to start numbering the position of the triple bond, just as we did with alkenes: Two structural formula of butyne. One butyne has a triple bond between the first and second carbon atom, while two butyne has the triple bond between the second and third carbon atom. Aromatic compounds contain the benzene unit. Benzene itself is composed of six C atoms in a ring, with alternating single and double C–C bonds: The six carbons are arranged in a hexagon pattern with one hydrogen atom emerging outwards from each carbon atom. The presence of a double bond is alternated between every other carbon atom. The alternating single and double C–C bonds give the benzene ring a special stability, and it does not react like an alkene as might be suspected. Benzene has the molecular formula C 6 H 6 ; in larger aromatic compounds, a different atom replaces one or more of the H atoms. As fundamental as hydrocarbons are to organic chemistry, their properties and chemical reactions are rather mundane. Most hydrocarbons are nonpolar because of the close electronegativities of the C and H atoms. As such, they dissolve only sparingly in H 2 O and other polar solvents. Small hydrocarbons, such as methane and ethane, are gases at room temperature, while larger hydrocarbons, such as hexane and octane, are liquids. Even larger hydrocarbons are solids at room temperature and have a soft, waxy consistency. Hydrocarbons are rather unreactive, but they do participate in some classic chemical reactions. One common reaction is substitution with a halogen atom by combining a hydrocarbon with an elemental halogen. Light is sometimes used to promote the reaction, such as this one between methane and chlorine: \[CH_{4}+Cl_{2}\overset{light}{\rightarrow} CH_{3}Cl+HCl\nonumber \] Halogens can also react with alkenes and alkynes, but the reaction is different. In these cases, the halogen reacts with the C–C double or triple bond and inserts itself onto each C atom involved in the multiple bonds. This reaction is called an addition reaction . One example is The reaction conditions are usually mild; in many cases, the halogen reacts spontaneously with an alkene or an alkyne. Hydrogen can also be added across a multiple bond; this reaction is called a hydrogenation reaction . In this case, however, the reaction conditions may not be mild; high pressures of H 2 gas may be necessary. A platinum or palladium catalyst is usually employed to get the reaction to proceed at a reasonable pace: \[CH_{2}=CH_{2}+H_{2}\overset{metal\: catalyst}{\rightarrow} CH_{3}CH_{3}\nonumber \] By far the most common reaction of hydrocarbons is combustion , which is the combination of a hydrocarbon with O 2 to make CO 2 and H 2 O. The combustion of hydrocarbons is accompanied by a release of energy and is a primary source of energy production in our society (Figure \(\PageIndex{2}\) - Combustion). The combustion reaction for gasoline, for example, which can be represented by C 8 H 18 , is as follows: \[2C_{8}H_{18}+25O_{2}\rightarrow 16CO_{2}+18H_{2}O+\sim 5060kJ\nonumber \] Key Takeaways The simplest organic compounds are hydrocarbons and are composed of carbon and hydrogen. Hydrocarbons can be aliphatic or aromatic; aliphatic hydrocarbons are divided into alkanes, alkenes, and alkynes. The combustion of hydrocarbons is a primary source of energy for our society. Exercise \(\PageIndex{2}\) Define hydrocarbon . What are the two general types of hydrocarbons? What are the three different types of aliphatic hydrocarbons? How are they defined? Indicate whether each molecule is an aliphatic or an aromatic hydrocarbon; if aliphatic, identify the molecule as an alkane, an alkene, or an alkyne. Indicate whether each molecule is an aliphatic or an aromatic hydrocarbon; if aliphatic, identify the molecule as an alkane, an alkene, or an alkyne. Indicate whether each molecule is an aliphatic or an aromatic hydrocarbon; if aliphatic, identify the molecule as an alkane, an alkene, or an alkyne. Indicate whether each molecule is an aliphatic or an aromatic hydrocarbon; if aliphatic, identify the molecule as an alkane, an alkene, or an alkyne. Name and draw the structural formulas for the four smallest alkanes. Name and draw the structural formulas for the four smallest alkenes. What does the term aromatic imply about an organic molecule? What does the term normal imply when used for alkanes? Explain why the name 1-propene is incorrect. What is the proper name for this molecule? Explain why the name 3-butene is incorrect. What is the proper name for this molecule? Name and draw the structural formula of each isomer of pentene. Name and draw the structural formula of each isomer of hexyne. Write a chemical equation for the reaction between methane and bromine. Write a chemical equation for the reaction between ethane and chlorine. Draw the structure of the product of the reaction of bromine with propene. Draw the structure of the product of the reaction of chlorine with 2-butene. Draw the structure of the product of the reaction of hydrogen with 1-butene. Draw the structure of the product of the reaction of hydrogen with 1-butene. Write the balanced chemical equation for the combustion of heptane. Write the balanced chemical equation for the combustion of nonane. Nov 27, 2021, 2:38 PM Answers an organic compound composed of only carbon and hydrogen; aliphatic hydrocarbons and aromatic hydrocarbons aliphatic; alkane aromatic aliphatic; alkene aliphatic; alkane aliphatic; alkene aromatic Aromatic means that the molecule has a benzene ring. The 1 is not necessary. The name of the compound is simply propene . CH 4 + Br 2 → CH 3 Br + HBr C 7 H 16 + 11O 2 → 7CO 2 + 8H 2 O
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Bookshelves/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/10%3A_Alkynes/10.07%3A_Oxidation_of_Alkynes
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Learning Objective predict the products and specify the reagents for the oxidation of alkynes Alkynes, similar to alkenes, can be oxidized gently or strongly depending on the reaction environment. Since alkynes are less stable than alkenens, the reactions conditions can be gentler. For examples, alkynes form vicinal dicarbonyls in neutral permanganate solution. For the alkene reaction to vicinal dialcohols, the permanganate reaction requires a lightly basic environment for the reaction to occur. During strong oxidation with ozone or basic potassium permanganate, the alkyne is cleaved into two products. Because at least one of the reaction products is a carboxylic acid, it is important to consider the acid-base chemistry of the product in the reaction solution. Carboxylic acids are deprotonated in basic solutions to carboxylates. A second reaction step is required to protonate the carboxylate to the neutral form of the carboxylic acid. The generic reactions are summarized below for the different oxidative conditions - gentle or strong. Gentle Alkyne Oxidation Strong Alkyne Oxidation - Oxidative Cleavage Exercise 1. Draw the bond-line structures for the product(s) of the following reactions. Answer 1.
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Courses/Lubbock_Christian_University/LCU%3A_CHE_1305_-_Introductory_Chemistry/06%3A_Ionic_Bonds/6.1%3A_Writing_Formulas_for_Ionic_Compounds
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Learning Objectives Write the correct formula for an ionic compound. Recognize polyatomic ions in chemical formulas. Ionic compounds do not exist as molecules. In the solid state, ionic compounds are in crystal lattice containing many ions each of the cation and anion. An ionic formula, like \(\ce{NaCl}\), is an empirical formula. This formula merely indicates that sodium chloride is made of an equal number of sodium and chloride ions. Sodium sulfide, another ionic compound, has the formula \(\ce{Na_2S}\). This formula indicates that this compound is made up of twice as many sodium ions as sulfide ions. This section will teach you how to find the correct ratio of ions, so that you can write a correct formula. If you know the name of a binary ionic compound, you can write its chemical formula . Start by writing the metal ion with its charge, followed by the nonmetal ion with its charge. Because the overall compound must be electrically neutral, decide how many of each ion is needed in order for the positive and negative charges to cancel each other out. Example \(\PageIndex{1}\): Aluminum Nitride and Lithium Oxide Write the formulas for aluminum nitride and lithium oxide. Solution Unnamed: 0 Write the formula for aluminum nitride Write the formula for lithium oxide 1. Write the symbol and charge of the cation (metal) first and the anion (nonmetal) second. \(\ce{Al^{3+}} \: \: \: \: \: \ce{N^{3-}}\) \(\ce{Li^+} \: \: \: \: \: \ce{O^{2-}}\) 2. Use a multiplier to make the total charge of the cations and anions equal to each other. total charge of cations = total charge of anions 1(3+) = 1(3-) +3 = -3 total charge of cations = total charge of anions 2(1+) = 1(2-) +2 = -2 3. Use the multipliers as subscript for each ion. \(\ce{Al_1N_1}\) \(\ce{Li_2O_1}\) 4. Write the final formula. Leave out all charges and all subscripts that are 1. \(\ce{AlN}\) \(\ce{Li_2O}\) An alternative way to writing a correct formula for an ionic compound is to use the crisscross method . In this method, the numerical value of each of the ion charges is crossed over to become the subscript of the other ion. Signs of the charges are dropped. Example \(\PageIndex{2}\): The Crisscross Method for Lead (IV) Oxide Write the formula for lead (IV) oxide. Solution Crisscross Method Write the formula for lead (IV) oxide 1. Write the symbol and charge of the cation (metal) first and the anion (nonmetal) second. \(\ce{Pb^{4+}} \: \: \: \: \: \ce{O^{2-}}\) 2. Transpose only the number of the positive charge to become the subscript of the anion and the number only of the negative charge to become the subscript of the cation. NaN 3. Reduce to the lowest ratio. \(\ce{Pb_2O_4}\) 4. Write the final formula. Leave out all subscripts that are 1. \(\ce{PbO_2}\) Exercise \(\PageIndex{2}\) Write the chemical formula for an ionic compound composed of each pair of ions. the calcium ion and the oxygen ion the 2+ copper ion and the sulfur ion the 1+ copper ion and the sulfur ion Answer a: CaO Answer b: CuS Answer c: Cu 2 S Be aware that ionic compounds are empirical formulas and so must be written as the lowest ratio of the ions. Example \(\PageIndex{3}\): Sulfur Compound Write the formula for sodium combined with sulfur. Solution Crisscross Method Write the formula for sodium combined with sulfur 1. Write the symbol and charge of the cation (metal) first and the anion (nonmetal) second. \(\ce{Na^{+}} \: \: \: \: \: \ce{S^{2-}}\) 2. Transpose only the number of the positive charge to become the subscript of the anion and the number only of the negative charge to become the subscript of the cation. NaN 3. Reduce to the lowest ratio. This step is not necessary. 4. Write the final formula. Leave out all subscripts that are 1. \(\ce{Na_2S}\) Exercise \(\PageIndex{3}\) Write the formula for each ionic compound. sodium bromide lithium chloride magnesium oxide Answer a: NaBr Answer b: LiCl Answer c: MgO Polyatomic Ions Some ions consist of groups of atoms bonded together and have an overall electric charge. Because these ions contain more than one atom, they are called polyatomic ions. Polyatomic ions have characteristic formulas, names, and charges that should be memorized. For example, NO 3 − is the nitrate ion; it has one nitrogen atom and three oxygen atoms and an overall 1− charge. Table \(\PageIndex{1}\) lists the most common polyatomic ions. Name Formula ammonium ion NH4+ acetate ion C2H3O2− (also written CH3CO2−) carbonate ion CO32− chromate ion CrO42− dichromate ion Cr2O72− hydrogen carbonate ion (bicarbonate ion) HCO3− cyanide ion CN− hydroxide ion OH− nitrate ion NO3− nitrite ion NO2− permanganate ion MnO4− phosphate ion PO43− hydrogen phosphate ion HPO42− dihydrogen phosphate ion H2PO4− sulfate ion SO42− hydrogen sulfate ion (bisulfate ion) HSO4− sulfite ion SO32− The rule for constructing formulas for ionic compounds containing polyatomic ions is the same as for formulas containing monatomic (single-atom) ions: the positive and negative charges must balance. If more than one of a particular polyatomic ion is needed to balance the charge, the entire formula for the polyatomic ion must be enclosed in parentheses, and the numerical subscript is placed outside the parentheses. This is to show that the subscript applies to the entire polyatomic ion. An example is Ba(NO 3 ) 2 . Writing Formulas for Ionic Compounds Containing Polyatomic Ions Writing a formula for ionic compounds containing polyatomic ions also involves the same steps as for a binary ionic compound. Write the symbol and charge of the cation followed by the symbol and charge of the anion. Example \(\PageIndex{4}\): Calcium Nitrate Write the formula for calcium nitrate. Solution Crisscross Method Write the formula for calcium nitrate 1. Write the symbol and charge of the cation (metal) first and the anion (nonmetal) second. \(\ce{Ca^{2+}} \: \: \: \: \: \ce{NO_3^-}\) 2. Transpose only the number of the positive charge to become the subscript of the anion and the number only of the negative charge to become the subscript of the cation. The 2+ charge on Ca becomes the subscript of NO3 and the 1- charge on NO3 becomes the subscript of Ca. 3. Reduce to the lowest ratio. \(\ce{Ca_1(NO_3)_2}\) 4. Write the final formula. Leave out all subscripts that are 1. If there is only 1 of the polyatomic ion, leave off parentheses. \(\ce{Ca(NO_3)_2}\) Example \(\PageIndex{5}\) Write the chemical formula for an ionic compound composed of the potassium ion and the sulfate ion. Solution Explanation Answer Potassium ions have a charge of 1+, while sulfate ions have a charge of 2−. We will need two potassium ions to balance the charge on the sulfate ion, so the proper chemical formula is \(\ce{K2SO4}\). \(\ce{K_2SO_4}\) Exercise \(\PageIndex{5}\) Write the chemical formula for an ionic compound composed of each pair of ions. the magnesium ion and the carbonate ion the aluminum ion and the acetate ion Answer a: \(\ce{MgCO3}\) Answer b: \(\ce{Al(CH3COO)3}\) Recognizing Ionic Compounds There are two ways to recognize ionic compounds. Method 1 Compounds between metal and nonmetal elements are usually ionic. For example, \(\ce{CaBr2}\) contains a metallic element (calcium, a group 2 [or 2A] metal) and a nonmetallic element (bromine, a group 17 [or 7A] nonmetal). Therefore, it is most likely an ionic compound (in fact, it is ionic). In contrast, the compound \(\ce{NO2}\) contains two elements that are both nonmetals (nitrogen, from group 15 [or 5A] , and oxygen, from group 16 [or 6A] . It is not an ionic compound; it belongs to the category of covalent compounds discussed elsewhere. Also note that this combination of nitrogen and oxygen has no electric charge specified, so it is not the nitrite ion. Method 2 Second, if you recognize the formula of a polyatomic ion in a compound, the compound is ionic. For example, if you see the formula \(\ce{Ba(NO3)2}\), you may recognize the “\(\ce{NO3}\)” part as the nitrate ion, \(\ce{NO3^{-}}\). (Remember that the convention for writing formulas for ionic compounds is not to include the ionic charge.) This is a clue that the other part of the formula, \(\ce{Ba}\), is actually the \(\ce{Ba^{2+}}\) ion, with the 2+ charge balancing the overall 2− charge from the two nitrate ions. Thus, this compound is also ionic. Example \(\PageIndex{6}\) Identify each compound as ionic or not ionic. \(\ce{Na2O}\) \(\ce{PCl3}\) \(\ce{NH4Cl}\) \(\ce{OF2}\) Solution Explanation Answer a. Sodium is a metal, and oxygen is a nonmetal. Therefore, \(\ce{Na2O}\) is expected to be ionic via method 1. \(\ce{Na2O}\), ionic b. Both phosphorus and chlorine are nonmetals. Therefore, \(\ce{PCl3}\) is not ionic via method 1 \(\ce{PCl3}\), not ionic c. The \(\ce{NH4}\) in the formula represents the ammonium ion, \(\ce{NH4^{+}}\), which indicates that this compound is ionic via method 2 \(\ce{NH4Cl}\), ionic d. Both oxygen and fluorine are nonmetals. Therefore, \(\ce{OF2}\) is not ionic via method 1 \(\ce{OF2}\), not ionic Exercise \(\PageIndex{6}\) Identify each compound as ionic or not ionic. \(\ce{N2O}\) \(\ce{FeCl3}\) \(\ce{(NH4)3PO4}\) \(\ce{SOCl2}\) Answer a: not ionic Answer b: ionic Answer c: ionic Answer d: not ionic Summary Formulas for ionic compounds contain the symbols and number of each atom present in a compound in the lowest whole number ratio.
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Courses/Sacramento_City_College/SCC%3A_CHEM_300_-_Beginning_Chemistry/SCC%3A_CHEM_300_-_Beginning_Chemistry_(Alviar-Agnew)/04%3A_Atoms_and_Elements/4.E%3A_Homework_Chapter_4
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GENERAL QUESTIONS List the three statements that make up the modern atomic theory. 2. Define atomic number . What is the atomic number for a boron atom? 3. Which elements are alkali metals, alkaline earth metals, and halogens? a. sodium b. magnesium c. calcium d. chlorine e. potassium f. bromine 4. Two of the most well-known isotopes of carbon are carbon-12 and carbon-14. Answer the following: a. What is the atomic number of each isotope? b. What is the mass number of each isotope? ATOMIC AND NUCLEAR THEORY 5. What happened to most of the alpha particles in Rutherford’s experiment? What is the modern atomic theory? 7. Describe why Rutherford used the term planetary model to describe his model of atomic structure? PROTONS, NEUTRONS, AND ELECTRONS 8. What are the charges for each of the three subatomic particles? 9. Of the three subatomic particles, which is the least massive? Which of the subatomic particles are located inside the nucleus of the atom? 10. Circle one item in each parenthesis. Protons are among the (most, least) massive subatomic particles, and they are found (inside, outside) the nucleus. 11. Which subatomic particle has a positive charge? Which subatomic particle has a negative charge? 12. Which subatomic particle is electrically neutral? Does it exist inside or outside the nucleus? 13. Circle one item in each parenthesis. Electrons are among the (most, least) massive subatomic particles, and they are found (inside, outside) the nucleus. ELEMENTS, SYMBOLS, AND NAMES 14. Of the three subatomic particles that make up an element, which one can always be used to identify the element? 15. Identify the atomic symbols and names of the elements that correspond to the following atomic numbers: a. 8 b. 27 c. 33 d. 90 16. How many protons are in the nucleus of an atom of each element? a. fluorine b. magnesium c. iodine d. copper e. silicon 17. Which elements contain the following numbers of protons? a. 15 b. 42 c. 4 18. Which of the following substances are elements? a. sodium b. carbon c. gold d. water f. liquefied nitrogen 19. Write the chemical symbol for each element. a. silver b. sulfur c. nitrogen d. neon 20. List the name and the atomic number of each element. a. F b. Fe c. I d. Cr e. P 21. How many protons are in the nucleus of each element? a. radon b. tungsten c. chromium d. beryllium 22. Find the atomic number for each element. a. Ni b. Sr c. Se d. Cs THE PERIODIC TABLE 23. One of the main groups on the periodic tables is the Metals. How many different subgroups of metals are there on the periodic table? Name each of them. Where in the periodic table can each subgroup be located? Be specific (provide the column number). 24. Apart from Metals, there are Metalloids and Non-Metals as groups on the periodic table. Name at least two subgroups within the Non-Metals. Where on the periodic table are these subgroups located? Be specific (provide the column number). 25. How do the transition metals differ from the alkali and alkaline earth metals? 26. Of the following elements, which of the three categories, metal, non-metal and metalloid, do they each belong to? a. manganese b. bromine c. silicon d. uranium e. phosphorus 27. Which elements have chemical properties similar to those of magnesium? a. sodium b. fluorine c. calcium d. barium e. selenium 28. Which elements are alkali metals? a. sodium b. magnesium c. potassium d. aluminum e. beryllium 29. Which elements are alkaline earth metals? a. sodium b. magnesium c. potassium d. aluminum e. beryllium 30. Which elements are halogens? a. oxygen b. chlorine c. fluorine d. sulfur e. argon 31. Which elements are noble gases? a. helium b. hydrogen c. oxygen d. xenon e. neon 32. Which pairs of elements are located in the same period? a. H and Li b. H and He c. Na and S d. Na and Rb 33. Circle one item in each parenthesis. Scandium is a (metal, nonmetal, semimetal) and is a member of the (main group elements, transition metals). 34. Which of these sets of elements are all in the same group? a. sodium, rubidium, and barium b. nitrogen, phosphorus, and bismuth c. copper, silver, and gold d. magnesium, strontium, and samarium 35. Which elements are alkali metals? a. Rubidium b. Tin c. Gold d. Sodium 36. Provide the family or group name of each element. a. Li b. Ar c. Cl 37. To what group number does each element belong? a. sodium b. cesium c. argon d. tellurium 38. Which elements do you expect to be most like Strontium? a. beryllium b. radium c. vanadium d. potassium 39. Complete the table below 0 1 2 3 Chemical Symbol Group Number Group Name Metal or Non-metal Br NaN NaN NaN Sr 2A NaN NaN Ne NaN NaN Non-metal K NaN Alkali metal NaN 40. Which pair of elements do you expect to be most similar? a. V and W b. Mg and Ca c. Cr and Tc d. Ga and Si 41. Which elements are alkaline earth metals? a. radium b. scandium c. cesium d. beryllium 42. Which element is a main-group non-metal? a. S b. K c. Fe d. Ca 43. Which elements would you expect to lose electrons in chemical changes? a. copper b. barium c. fluorine d. sulfur IONS 44. What is the total number of electrons present in each ion? a. F − b. Rb + c. Ce 3+ d. Zr 4+ e. Zn 2+ 45. Predict how many electrons are in each ion. a. A copper ion with a +2 charge b. a molybdenum ion with a +4 charge c. an iodine ion with a −1 charge d. a gallium ion with a +3 charge 46. What is the total number of electrons present in each ion? a. Ca 2+ b. Se 2− c. In 3+ d. Sr 2+ e. As 3+ 47. Predict how many electrons are in each ion. a. an oxygen ion with a −2 charge b. a beryllium ion with a +2 charge c. a silver ion with a +1 charge d. a selenium ion with a +4 charge 48. Predict the charge on the most common monatomic ion formed by each element. a. chlorine b. phosphorus c. scandium d. magnesium 49. For each representation of a monatomic ion, identify the parent atom, write the formula of the ion using an appropriate superscript, and indicate the period and group on the periodic table in which the element is found. a. + b. - c. 3+ 50. Predict the charge on the most common monatomic ion formed by each element. a. sodium b. selenium c. barium d. aluminum 51. Predict how many electrons each element will most likely gain or lose. a. Ca b. N c. P d. Rb 52. Predict the ion formed by each element? a. Li b. Cl c. K d. O 53. Give the charge each atom takes when it forms an ion. If more than one charge is possible, list both. a. K b. O c. Co 54. Give the charge each atom takes when it forms an ion. If more than one charge is possible, list both. a. Ag b. Au c. Br What is the difference between SO 3 and SO 3 2− ? ISOTOPES 56. What is an isotope? How does one isotope differ from another within the same element? 57. What do all isotopes of nitrogen have in common? Is this true for isotopes of all elements? 58. Starting with polonium, atomic number 84, there are no stable, non-radioactive isotopes of any subsequent elements on the periodic table. Using the atomic symbol and the mass and atomic numbers, express the isotope of polonium-84. 59. Determine the atomic number and mass number for each isotope. a. the hydrogen isotope with 2 neutrons b. the calcium isotope with 22 neutrons c. the tantalum isotope with 109 neutrons d. the chromium isotope with 28 neutrons 60. State the number of protons, neutrons, and electrons in neutral atoms of each isotope. a. 131 I b. 40 K c. 201 Hg d. 19 F 61. Complete each sentence. a. 48 Ti has _____ neutrons. b. 40 Ar has _____ neutrons. c. 3 H has _____ neutrons. 62. What is the mass number of a gallium atom that has 38 neutrons in it? 63. Give the symbol for these isotopes. a. Fluorine-19 b. Helium-4 c. Terbium-159 d. Iodine-127 e. Gold-197 64. Identify each element, represented by X, that has the given symbols and determine the number of protons and neutrons in each isotope. a. b. c. d. 65. Determine the number of protons, neutrons, and electrons in a neutral atom of each isotope: a. 97 Tc b. 113 In c. 63 Ni d. 55 Fe 66. Both technetium-97 and americium-240 are produced in nuclear reactors. Determine the number of protons, neutrons, and electrons in the neutral atoms of each. 67. The following isotopes are important in archaeological research. How many protons, neutrons, and electrons does a neutral atom of each contain? a. 207 Pb b. 16 O c. 137 Cs d. 40 K 68. Give complete symbols of each atom, including the atomic number and the mass number. a. a magnesium atom with 12 protons and 12 neutrons b. a magnesium atom with 12 protons and 13 neutrons c. a xenon atom with 54 protons and 77 neutrons 69. How many protons and neutrons are in ? 70. Plutonium-239 is used in nuclear bombs. Determine the number of protons and neutrons in Plutonium-239 and write its symbol in the form . CUMULATIVE PROBLEMS 71. Using the given clues, determine the identity of the following elements: a. A noble gas with an atomic number less than 10 b. A metalloid from period 2 c. A group 5A element directly adjacent to a metalloid 72. Complete the following table. 0 1 2 3 4 Isotope Symbol Number of protons Number of neutrons Number of electrons Mass Number C-12 NaN NaN NaN NaN C-13 NaN NaN NaN NaN C-14 NaN NaN NaN NaN 73. Complete the following table. 0 1 2 3 Number of protons Number of neutrons Element name Isotope Symbol 80 120 NaN NaN NaN NaN NaN NaN NaN 2 Hydrogen NaN 74. Classify each element as a metal, a semimetal, or a nonmetal. If a metal, state whether it is an alkali metal, an alkaline earth metal, or a transition metal. a. iron b. tantalum c. sulfur d. silicon e. chlorine 75. Explain why it is improper to write CO as the chemical symbol for cobalt? 76. Write the appropriate symbol for each of the following ions (Example: 59 Co 3+ ): a. the ion with a 1+ charge, atomic number 55, and mass number 133 b. the ion with 54 electrons, 53 protons, and 74 neutrons c. the ion with atomic number 15, mass number 31, and a 3− charge d. the ion with 24 electrons, 30 neutrons, and a 3+ charge 77. The following are properties of isotopes of two elements that are essential in our diet. Determine the number of protons, neutrons and electrons in each and name each element. a. atomic number 26, mass number 58, charge of 2+ b. atomic number 53, mass number 127, charge of 1− 78. Fill in the blanks to complete the table. 0 1 2 3 Symbol Ion commonly formed Number of electrons in ion Number of protons in ion F NaN NaN 9 NaN Be2+ 2 NaN Br NaN 36 NaN Al NaN NaN 13 O NaN NaN NaN 79. An isotope of uranium has an atomic number of 92 and a mass number of 235. What are the number of protons and neutrons in the nucleus of this atom? 80. Under certain conditions, oxygen gas (O 2 ) has a density of 0.00134 g/mL. Find the volume occupied by 250.0 g of O 2 under the same conditions.
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Courses/Fullerton_College/Introductory_Biochemistry/14%3A_Carbohydrate_Metabolism/14.01%3A_Energy_Metabolism/14.1.07%3A_Gluconeogenesis
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The anabolic counterpart to glycolysis is gluconeogenesis, which occurs mostly in the cells of the liver and kidney. In seven of the eleven reactions of gluconeogenesis (starting from pyruvate), the same enzymes are used as in glycolysis, but the reaction directions are reversed. Notably, the \(\Delta\)G values of these reactions in the cell are typically near zero, meaning their direction can be readily controlled by changing substrate and product concentrations. Figure 6.4.1: Metabolic Redox The three regulated enzymes of glycolysis all catalyze reactions whose \(\Delta\)G values are not close to zero, making manipulation of reaction direction non-trivial. Consequently, cells employ “work-around" reactions catalyzed by four different enzymes to favor gluconeogenesis, when appropriate. Two of the enzymes (pyruvate carboxylase and PEP carboxykinase -PEPCK) catalyze reactions that bypass pyruvate kinase. F1,6BPase bypasses PFK and G6Pase bypasses hexokinase. Notably, pyruvate carboxylase and G6Pase are found in the mitochondria and endoplasmic reticulum, respectively, whereas the other two are found in the cytoplasm along with all of the enzymes of glycolysis. As a result, all of glycolysis and most of gluconeogenesis occurs in the cytoplasm. Controlling these pathways then becomes of critical importance because cells generally need to minimize the extent to which paired anabolic and catabolic pathways are occurring simultaneously, lest they waste energy and make no tangible product except heat. The mechanisms of controlling these pathways work, in some ways, in opposite fashions, called reciprocal regulation (see above). Figure 6.4.2: Gluconeogensis and Glycolysis Besides reciprocal regulation, other mechanisms help control gluconeogenesis. First, PEPCK is controlled largely at the level of synthesis. Overexpression of PEPCK (stimulated by glucagon, glucocorticoids, and cAMP and inhibited by insulin) causes symptoms of diabetes. Pyruvate carboxylase is sequestered in the mitochondrion and is sensitive to acetyl-CoA, which is an allosteric activator. Acetyl-CoA concentrations increase as the citric acid cycle activity decreases. Glucose-6-phosphatase is present in low concentrations in many tissues, but is found most abundantly and importantly in the major gluconeogenic organs – the liver and kidney cortex.
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Courses/Lumen_Learning/Book%3A_Microeconomics-2_(Lumen)/10%3A_Module-_Production/10.14%3A_Outcome-_Defining_Production
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What you’ll learn to do: define the term “production” and explain what a production function is In this section, we will learn more about the behavior of firms and how they make production decisions. The specific things you’ll learn to do in this section include: Define the term “production inputs,” and differentiate between labor, land, capital, entrepreneurship, technology LEARNING ACTIVITIES The learning activities for this section include: Reading: Cost and Industry Structure Reading: Factors of Production Self Check: Defining Production Take time to review and reflect on each of these activities in order to improve your performance on the assessment for this section. CC licensed content, Original Authored by : Steven Greenlaw and Lumen Learning. License : CC BY: Attribution
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Courses/Northern_Alberta_Institute_of_Technology/CHEM1130_Principles_in_Chemistry_I/2%3A_Quantum_Mechanical_Picture_of_the_Atom/2.03%3A_The_Photoelectric_Effect
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When light strikes materials, it can eject electrons from them. This is called the photoelectric effect , meaning that light ( photo ) produces electricity. One common use of the photoelectric effect is in light meters, such as those that adjust the automatic iris on various types of cameras. In a similar way, another use is in solar cells, as you probably have in your calculator or have seen on a roof top or a roadside sign. These make use of the photoelectric effect to convert light into electricity for running different devices. Figure \(\PageIndex{1}\): The photoelectric effect can be observed by allowing light to fall on the metal plate in this evacuated tube. Electrons ejected by the light are collected on the collector wire and measured as a current. A retarding voltage between the collector wire and plate can then be adjusted so as to determine the energy of the ejected electrons. For example, if it is sufficiently negative, no electrons will reach the wire. (credit: P.P. Urone) This effect has been known for more than a century and can be studied using a device such as that shown in Figure \(\PageIndex{1}\). This figure shows an evacuated tube with a metal plate and a collector wire that are connected by a variable voltage source, with the collector more negative than the plate. When light (or other EM radiation) strikes the plate in the evacuated tube, it may eject electrons. If the electrons have energy in electron volts (eV) greater than the potential difference between the plate and the wire in volts, some electrons will be collected on the wire. Since the electron energy in eV is \(qV\), where \(q\) is the electron charge and \(V\) is the potential difference, the electron energy can be measured by adjusting the retarding voltage between the wire and the plate. The voltage that stops the electrons from reaching the wire equals the energy in eV. For example, if \(-3.00 \, V\) barely stops the electrons, their energy is 3.00 eV. The number of electrons ejected can be determined by measuring the current between the wire and plate. The more light, the more electrons; a little circuitry allows this device to be used as a light meter. What is really important about the photoelectric effect is what Albert Einstein deduced from it. Einstein realized that there were several characteristics of the photoelectric effect that could be explained only if EM radiation is itself quantized : the apparently continuous stream of energy in an EM wave is actually composed of energy quanta called photons. In his explanation of the photoelectric effect, Einstein defined a quantized unit or quantum of EM energy, which we now call a photon , with an energy proportional to the frequency of EM radiation. In equation form, the photon energy is \[E = hf,\] where \(E\) is the energy of a photon of frequency \(f\) and \(h\) is Planck’s constant. This revolutionary idea looks similar to Planck’s quantization of energy states in blackbody oscillators, but it is quite different. It is the quantization of EM radiation itself. EM waves are composed of photons and are not continuous smooth waves as described in previous chapters on optics. Their energy is absorbed and emitted in lumps, not continuously. This is exactly consistent with Planck’s quantization of energy levels in blackbody oscillators, since these oscillators increase and decrease their energy in steps of \(hf\) by absorbing and emitting photons having \(E = hf\). We do not observe this with our eyes, because there are so many photons in common light sources that individual photons go unnoticed (Figure \(\PageIndex{2}\)). The next section of the text ( Photon Energies and the Electromagnetic Spectrum ) is devoted to a discussion of photons and some of their characteristics and implications. For now, we will use the photon concept to explain the photoelectric effect, much as Einstein did. Figure \(\PageIndex{2}\): An EM wave of frequency \(f\) is composed of photons, or individual quanta of EM radiation. The energy of each photon is \(E = hf\), where \(h\) is Planck’s constant and \(f\) is the frequency of the EM radiation. Higher intensity means more photons per unit area. The flashlight emits large numbers of photons of many different frequencies, hence others have energy \(E' = hf'\) and so on. The photoelectric effect has the properties discussed below. All these properties are consistent with the idea that individual photons of EM radiation are absorbed by individual electrons in a material, with the electron gaining the photon’s energy. Some of these properties are inconsistent with the idea that EM radiation is a simple wave. For simplicity, let us consider what happens with monochromatic EM radiation in which all photons have the same energy \(hf\). If we vary the frequency of the EM radiation falling on a material, we find the following: For a given material, there is a threshold frequency \(f_0\) for the EM radiation below which no electrons are ejected, regardless of intensity. Individual photons interact with individual electrons. Thus if the photon energy is too small to break an electron away, no electrons will be ejected. If EM radiation was a simple wave, sufficient energy could be obtained by increasing the intensity. Once EM radiation falls on a material, electrons are ejected without delay . As soon as an individual photon of a sufficiently high frequency is absorbed by an individual electron, the electron is ejected. If the EM radiation were a simple wave, several minutes would be required for sufficient energy to be deposited to the metal surface to eject an electron. The number of electrons ejected per unit time is proportional to the intensity of the EM radiation and to no other characteristic. High-intensity EM radiation consists of large numbers of photons per unit area, with all photons having the same characteristic energy \(hf\). If we vary the intensity of the EM radiation and measure the energy of ejected electrons, we find the following: The maximum kinetic energy of ejected electrons is independent of the intensity of the EM radiation . Since there are so many electrons in a material, it is extremely unlikely that two photons will interact with the same electron at the same time, thereby increasing the energy given it. Instead (as noted in 3 above), increased intensity results in more electrons of the same energy being ejected. If EM radiation were a simple wave, a higher intensity could give more energy, and higher-energy electrons would be ejected. The kinetic energy of an ejected electron equals the photon energy minus the binding energy of the electron in the specific material. An individual photon can give all of its energy to an electron. The photon’s energy is partly used to break the electron away from the material. The remainder goes into the ejected electron’s kinetic energy. In equation form, this is given by \[KE_e = hf - BE,\] where \(KE_e\) is the maximum kinetic energy of the ejected electron, \(hf\) is the photon’s energy, and BE is the binding energy of the electron to the particular material. (BE is sometimes called the work function of the material.) This equation, due to Einstein in 1905, explains the properties of the photoelectric effect quantitatively. An individual photon of EM radiation (it does not come any other way) interacts with an individual electron, supplying enough energy, BE, to break it away, with the remainder going to kinetic energy. The binding energy is \(BE = hf_0\), where \(f_0\) is the threshold frequency for the particular material. Figure \(\PageIndex{3}\) shows a graph of maximum \(KE_e\) versus the frequency of incident EM radiation falling on a particular material. Figure \(\PageIndex{3}\): Photoelectric effect. A graph of the kinetic energy of an ejected electron, \(KE_e\), versus the frequency of EM radiation impinging on a certain material. There is a threshold frequency below which no electrons are ejected, because the individual photon interacting with an individual electron has insufficient energy to break it away. Above the threshold energy, \(KE_e\), increases linearly with \(f\), consistent with \(KE_e = hf - BE\). The slope of this line is \(h\)—the data can be used to determine Planck’s constant experimentally. Einstein gave the first successful explanation of such data by proposing the idea of photons—quanta of EM radiation. Einstein’s idea that EM radiation is quantized was crucial to the beginnings of quantum mechanics. It is a far more general concept than its explanation of the photoelectric effect might imply. All EM radiation can also be modeled in the form of photons, and the characteristics of EM radiation are entirely consistent with this fact. (As we will see in the next section, many aspects of EM radiation, such as the hazards of ultraviolet (UV) radiation, can be explained only by photon properties.) More famous for modern relativity, Einstein planted an important seed for quantum mechanics in 1905, the same year he published his first paper on special relativity. His explanation of the photoelectric effect was the basis for the Nobel Prize awarded to him in 1921. Although his other contributions to theoretical physics were also noted in that award, special and general relativity were not fully recognized in spite of having been partially verified by experiment by 1921. Although hero-worshipped, this great man never received Nobel recognition for his most famous work—relativity. Example \(\PageIndex{1}\): Calculating Photon Energy and the Photoelectric Effect: A Violet Light What is the energy in joules and electron volts of a photon of 420-nm violet light? What is the maximum kinetic energy of electrons ejected from calcium by 420-nm violet light, given that the binding energy (or work function) of electrons for calcium metal is 2.71 eV? Strategy To solve part (a), note that the energy of a photon is given by \(E = hf\). For part (b), once the energy of the photon is calculated, it is a straightforward application of \(KE_e = hf - BE\) to find the ejected electron’s maximum kinetic energy, since BE is given. Solution for (a) Photon energy is given by \[E = hf\] Since we are given the wavelength rather than the frequency, we solve the familiar relationship \(c = f \lambda\) for the frequency, yielding \[f = \dfrac{c}{\lambda}.\] Now substituting known values yields \[E = \dfrac{(6.63 \times 10^{-34} \, J \cdot s)(3.00 \times 10^8 \, m/s)}{420 \times 10^{-9} \, m} = 4.74 \times 10^{-19} \, J. \nonumber\] Converting to eV, the energy of the photon is \[E = (4.74 \times 10^{-19} \, J) \dfrac{1 \, eV}{1.6 \times 10^{-19} \, J} = 2.96 \, eV. \nonumber\] Solution for (b) Finding the kinetic energy of the ejected electron is now a simple application of the equation \(KE_e = hf - BE\). Substituting the photon energy and binding energy yields \[KE_e = hf - BE = 2.96 \, eV - 2.71 \, eV = 0.246 \, eV. \nonumber\] Discussion The energy of this 420-nm photon of violet light is a tiny fraction of a joule, and so it is no wonder that a single photon would be difficult for us to sense directly—humans are more attuned to energies on the order of joules. But looking at the energy in electron volts, we can see that this photon has enough energy to affect atoms and molecules. A DNA molecule can be broken with about 1 eV of energy, for example, and typical atomic and molecular energies are on the order of eV, so that the UV photon in this example could have biological effects. The ejected electron (called a photoelectron ) has a rather low energy, and it would not travel far, except in a vacuum. The electron would be stopped by a retarding potential of but 0.26 eV. In fact, if the photon wavelength were longer and its energy less than 2.71 eV, then the formula would give a negative kinetic energy, an impossibility. This simply means that the 420-nm photons with their 2.96-eV energy are not much above the frequency threshold. You can show for yourself that the threshold wavelength is 459 nm (blue light). This means that if calcium metal is used in a light meter, the meter will be insensitive to wavelengths longer than those of blue light. Such a light meter would be completely insensitive to red light, for example. PHET EXPLORATIONS: PHOTOELECTRIC EFFECT See how light knocks electrons off a metal target, and recreate the experiment that spawned the field of quantum mechanics. Summary The photoelectric effect is the process in which EM radiation ejects electrons from a material. Einstein proposed photons to be quanta of EM radiation having energy \(E = hf\), where \(f\) is the frequency of the radiation. All EM radiation is composed of photons. As Einstein explained, all characteristics of the photoelectric effect are due to the interaction of individual photons with individual electrons. The maximum kinetic energy \(KE_e\) of ejected electrons (photoelectrons) is given by \(KE_e = hf - BE\), where \(hf\) is the photon energy and BE is the binding energy (or work function) of the electron to the particular material. Glossary photoelectric effect the phenomenon whereby some materials eject electrons when light is shined on them photon a quantum, or particle, of electromagnetic radiation photon energy the amount of energy a photon has; \(E = hf\) binding energy also called the work function ; the amount of energy necessary to eject an electron from a material Contributors Template:ContribOpenStaxCollege
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Courses/Earlham_College/CHEM_361%3A_Inorganic_Chemistry_(Watson)/05%3A_Solid_State_Chemistry/5.03%3A_Crystal_Structures_of_Metals/5.3.01%3A_Conduction_in_Metals
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In metals, the valence electrons are in molecular orbitals that extend over the entire crystal lattice. Metals are almost always crystalline and the individual crystal grains are typically micron size. This means that the spatial extent of the orbitals is very large compared to the size of the atoms or the unit cell. The diagram shows a generic plot of electron energy vs. density of states for a metal such as Na, Cu, or Ag. In these cases, there are N orbitals for N electrons, and each orbital can accommodate two electrons. Therefore the Fermi level, which corresponds to the energy of the highest occupied MO at zero temperature, is somewhere in the middle of the band of orbitals. The energy level spacing between orbitals is very small compared to the thermal energy kT, so we can think of the orbitals as forming a continuous band. Classically, if the electrons in this band were free to be thermally excited, we would expect them to have a specific heat of 3R per mole of electrons. However, experimentally we observe that C p is only about 0.02 R per mole. This suggests that only about 1% of the electrons in the metal can be thermally excited at room temperature. However, essentially all of the valence electrons are free to move in the crystal and contribute to electrical conduction. To understand this apparent paradox, we need to recall that the electrons exist in quantized energy levels. Because of quantization, electrons in metals have a Fermi-Dirac distribution of energies. In this distribution, most of the electrons are spin-paired, although the individual electrons in these pairs can be quite far apart since the orbitals extend over the entire crystal. A relatively small number of electrons at the top of the Fermi sea are unpaired by thermal excitation. This is the origin of the Pauli paramagnetism of metals. How fast are electrons traveling in a typical metal? Because of the bell shape of the E vs. DOS curve, most of the electrons have E ≈ E F . At the midpoint energy (E F ) of the band, the MO's have one node for every two atoms. We can calculate the de Broglie wavelength as twice the distance between nodes and thus: λ = 4a at the midpoint of the band. where a is the interatomic spacing. Since a typical value of a is about 2 Å, we obtain the de Broglie wavelength λ ≈ 8 Å . Using the de Broglie relation p = h/λ, we can write: \[\mathbf{p = \frac{h}{\lambda} = m_{e}v_{F}}\] where m e is the mass of the electron and v F is the velocity of electrons with energy E F . Solving for \(V_{F} = \frac{h}{m_{e}\lambda}\)we obtain \(\mathbf{v_{F}} = \frac{(6.62 \times 10^{-34} J s)}{(9.1 \times 10^{-31} kg)(8 \times 10^{-10}m)} = \mathbf{1.0 \times 10^{6} m/s}\) Experimental values of v F are 1.07 x 10 6 and 1.39 x 10 6 m/s for Na and Ag, respectively, so our approximations are pretty good. How fast are electrons moving in metals? Really, really fast!! 1,000,000 meters per second! This is about 1/300 the speed of light, and about 3000 times the speed of sound in air (3 x 10 2 m/s). However, the drift velocity of electrons in metals - the speed at which electrons move in applied electric field - is quite slow, on the order of 0.0001 m/s, or .01 cm/s. You can easily outrun an electron drifting in a metal, even if you have been drinking all night and have been personally reduced to a very slow crawl. In order to understand the great disparity between the Fermi velocity and the drift velocity of electrons in metals, we need to consider a picture for the scattering of electrons, and their acceleration in an electric field, as shown at the left. If we apply a voltage across a metal (e.g., a metal wire), the electrons are subjected to an electric field E , which is the voltage divided by the length of the wire. This electric field exerts a force on the electron, causing it to accelerate. However, the electron is frequently scattered, mostly by phonons (lattice vibrations). Each time the electron is scattered its acceleration starts all over again. The time between scattering events is τ and the distance the electrons travel between scattering events is the mean free path , λ . (Note that this is NOT the same λ as the de Broglie wavelength, they just unfortunately have the same symbol!) We can write the force on the electron as: \[\mathbf{F = eE = m_{e}a = \frac{m_{e}v_{drift}}{\tau}}\] In this equation, a is the acceleration in the electric field, m e is the mass, and v drift is the drift velocity of the electron. Experimentally, the mean free path is typically obtained by measuring the scattering time. For an electron in Cu metal at 300 K, the scattering time τ is about 2 x 10 -14 s. From this we can calculate the mean free path as: \[\mathbf{\lambda = v_{avg}\tau \approx v_{F}\tau} = (1 \times 10^{6} m/s)(2 \times 10^{-14}) = \mathbf{40 nm}\] The mean free path (40 nm = 400 Å) is quite long compared to the interatomic spacing (2 Å). To put it in perspective, if the interatomic spacing were scaled to the length of a football (0.3 m), the mean free path would be over half the length of the football field (60 m). Thus an electron travels a fairly long way between scattering events and scarcely notices the atomic structure of the metal in which it is traveling. To summarize, electrons are traveling in metals at the Fermi velocity v F , which is very, very fast (10 6 m/s), but the flux of electrons is the same in all directions. That is, they are going nowhere fast . In an electric field, a very small but directional drift velocity is superimposed on this fast random motion of valence electrons. From F = ma, we obtain the acceleration (a) as:We can calculate the drift velocity of electrons as the acceleration in the electric field times the scattering time: \[\mathbf{a = \frac{F}{m_{e}} = \frac{eE}{m_{e}}}\] And thus, \[\mathbf{v_{drift} = a\tau = \frac{eE\tau}{m_{e}}}\] If we divide both sides of this equation by the magnitude of the electric field ( E ), we obtain the mobility (μ) : \[\mathbf{\mu=\frac{v_{drift}}{E}= \frac{e\tau}{m_{e}}}\] μ has units of velocity/field = cm/s / V/cm = cm 2 /Vs An important consequence of the calculation of v drift is Ohm's Law , V = iR. From the equations above, we can see that the drift velocity increases linearly with the applied electric field. The drift velocity (cm/s) is proportional to the current (i, coul/s), and the electric field (E, V/cm) is proportional to the voltage (V): \[\mathbf{Current \: (i) = nev_{drift} \times area}\] \[\mathbf{Voltage \: (V) = E \times length}\] Here n is the density of valence electrons (#/cm 3 ) and e is the charge of the electron (coul). Combining these equations with our equation for v drift we obtain: \[V = i(\frac{m_{e}}{ne^{2}\tau})\frac({length}{area}) = iR\] Thus, V = iR, where R is the combination of the two terms in parentheses. The first of these is the resistivity , \(\rho\), and the second is a geometrical factor. The conductivity (σ) of a metal, which is the inverse of \(\rho\), is proportional to μ, which in turn is proportional to τ (and λ): \[\mathbf{\sigma = ne\mu = \frac{ne^{2}\tau}{m_{e}}}\] We can use this equation to work out the conductivity of a specific metal (Cu), for which n = 8.5 x 10 22 cm -3 and τ = 2 x 10 -14 s. Putting in the numbers for m e and e, we obtain σ = 7 x 10 5 Ω -1 cm -1 for Cu, in good agreement with the measured value (6 x 10 5 Ω -1 cm -1 ).
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Courses/Lumen_Learning/Book%3A_Western_Civilization_I_(Lumen)/00%3A_Front_Matter/03%3A_Table_of_Contents
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2: Start Here 2.1: Syllabus 2.2: Schedule of Work 3: Week 1: Pre-history, Neolithic Revolution, Agriculture, Writing 3.1: Introduction 3.2: Reading: The Study of History 3.3: Reading: Precursor to Civilization 3.4: Reading: The first Urban Civilizations 3.5: Assignments 4: Week 2: Ancient Near-east, Mesopotamia, Syria, Palestine, and Egypt 4.1: Introduction 4.2: Reading: Akkadian Empire 4.3: Reading: Babylonia 4.4: Reading: Ancient Societies on the Mediterranean 4.5: Assignments 5: Week 3: Ancient Egypt 5.1: Introduction 5.2: Reading: Introduction to Ancient Egypt 5.3: Reading: The Old Kingdom 5.4: Reading: The Middle Kingdom 5.5: Reading: The New Kingdom 5.6: Reading: Ancient Egyptian Society 5.7: Assignments 5.8: Primary Source Reading 1: Wenamen’s Journey 6: Week 4: Ancient Greece and the Indus River Valley 6.1: Introduction 6.2: Reading: The Indus River Valley Civilizations 6.3: Reading: Early Periods in Greek History 6.4: Assignments 7: Week 5: Persian Wars 7.1: Introduction 7.2: Reading: The Persian Empire 7.3: Reading: Sparta 7.4: Reading: The Persian Wars 7.5: Reading: Athens 7.6: Reading: Culture in Classical Greece 7.7: Reading: The Peloponnesian War 7.8: Assignments 7.9: Primary Source Reading 2: Hanno: Carthaginian Explorer 8: Week 6: Hellenistic Greece and Alexander the Great 8.1: Introduction 8.2: Reading: Macedonian Conquest 8.3: Assignments 9: Week 7: Etruscan Italy, and Early Republican Rome 9.1: Introduction 9.2: Reading: The Etruscans 9.3: Reading: Early Rome 9.4: Reading: The Roman Republic 9.5: Assignments 9.6: Primary Source Reading 3: Xenophon 10: Week 8: Rome 10.1: Introduction 10.2: Reading: The Roman Empire 10.3: Reading: The Flavian Dynasty 10.4: Reading: Nerva-Antonine Dynasty 10.5: Assignments 11: Week 9: Christianity, and the Late Roman Empire 11.1: Introduction 11.2: Reading: Christianity and the Late Roman Empire 11.3: Reading: Byzantium: The New Rome 11.4: Reading: The Heraclian and Isaurian Dynasties 11.5: Reading: The Germanic Tribes 11.6: Reading: The Catholic Church 11.7: Assignments 11.8: Primary Source Reading 4: Mt. Vesuvius Eruption 79 AD 12: Week 10: The Rise of Islam and the Byzantine Empire 12.1: Introduction 12.2: Reading: The Late Byzantine Empire 12.3: Reading: Pre-Islamic Arabia 12.4: Reading: Muhammad and the Rise of Islam 12.5: Assignments 13: Week 11: Islam, con’t, Charlemagne and Russia 13.1: Introduction 13.2: Reading: The Umayyad and Abbasid Empires 13.3: Reading: The Carolingian Dynasty 13.4: Reading: The Princes of Rus 13.5: Reading: The Grand Duchy of Moscow 13.6: Assignments 13.7: Primary Source Reading 5: Saga of Erik The Red 14: Week 12: The Holy Roman Empire and England 14.1: Introduction 14.2: Reading: The Holy Roman Empire 14.3: Reading: The Development of England 14.4: Reading: Medieval Life 14.5: Assignments 15: Week 13: The Crusades and The Late Middle Ages 15.1: Introduction 15.2: Reading: The Crusades 15.3: Assignments 15.4: Primary Source Reading 6: Travels of Sir John Mandeville (table of Contents) 15.5: Primary Source Reading 6: Sir John Mandeville (Excerpts) 16: Week 14: Italian City-states and Renaissance 16.1: Introduction 16.2: Reading: The Renaissance 16.3: Reading: Italy During the Renaissance 16.4: Reading: Humanist Thought 16.5: Reading: Art in the Renaissance 16.6: Reading: Literature in the Renaissance 16.7: Assignments 17: Week 15: Northern Renaissance and the Development of Absolutism 17.1: Introduction 17.2: Reading: The Northern Renaissance 17.3: Reading: Nation-States and Sovereignty 17.4: Reading: Spain and Catholicism 17.5: Reading: England and Parliamentary Monarchy 17.6: Assignments 17.7: Primary Source Reading 7: Margery Kempe, Explorer 18: Week 16: Presentations 18.1: Assignment Back Matter Index Glossary
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Courses/Modesto_Junior_College/Chemistry_143%3A_Introductory_College_Chemistry_(Brzezinski)/CHEM_143%3A_Text_(Brzezinski)/01%3A_Introduction/1.02%3A_Hypothesis%2C_Theories%2C_and_Laws/1.2.01%3A_The_Scientific_Method_-_How_Chemists_Think
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Learning Objectives Identify the components of the scientific method. Scientists search for answers to questions and solutions to problems by using a procedure called the scientific method. This procedure consists of making observations, formulating hypotheses, and designing experiments; which leads to additional observations, hypotheses, and experiments in repeated cycles (Figure \(\PageIndex{1}\)). Step 1: Make observations Observations can be qualitative or quantitative. Qualitative observations describe properties or occurrences in ways that do not rely on numbers. Examples of qualitative observations include the following: "the outside air temperature is cooler during the winter season," "table salt is a crystalline solid," "sulfur crystals are yellow," and "dissolving a penny in dilute nitric acid forms a blue solution and a brown gas." Quantitative observations are measurements, which by definition consist of both a number and a unit. Examples of quantitative observations include the following: "the melting point of crystalline sulfur is 115.21° Celsius," and "35.9 grams of table salt—the chemical name of which is sodium chloride—dissolve in 100 grams of water at 20° Celsius." For the question of the dinosaurs’ extinction, the initial observation was quantitative: iridium concentrations in sediments dating to 66 million years ago were 20–160 times higher than normal. Step 2: Formulate a hypothesis After deciding to learn more about an observation or a set of observations, scientists generally begin an investigation by forming a hypothesis, a tentative explanation for the observation(s). The hypothesis may not be correct, but it puts the scientist’s understanding of the system being studied into a form that can be tested. For example, the observation that we experience alternating periods of light and darkness corresponding to observed movements of the sun, moon, clouds, and shadows is consistent with either one of two hypotheses: Earth rotates on its axis every 24 hours, alternately exposing one side to the sun. The sun revolves around Earth every 24 hours. Suitable experiments can be designed to choose between these two alternatives. For the disappearance of the dinosaurs, the hypothesis was that the impact of a large extraterrestrial object caused their extinction. Unfortunately (or perhaps fortunately), this hypothesis does not lend itself to direct testing by any obvious experiment, but scientists can collect additional data that either support or refute it. Step 3: Design and perform experiments After a hypothesis has been formed, scientists conduct experiments to test its validity. Experiments are systematic observations or measurements, preferably made under controlled conditions—that is—under conditions in which a single variable changes. Step 4: Accept or modify the hypothesis A properly designed and executed experiment enables a scientist to determine whether or not the original hypothesis is valid. If the hypothesis is valid, the scientist can proceed to step 5. In other cases, experiments often demonstrate that the hypothesis is incorrect or that it must be modified and requires further experimentation. Step 5: Development into a law and/or theory More experimental data are then collected and analyzed, at which point a scientist may begin to think that the results are sufficiently reproducible (i.e., dependable) to merit being summarized in a law, a verbal or mathematical description of a phenomenon that allows for general predictions. A law simply states what happens; it does not address the question of why. One example of a law, the law of definite proportions , which was discovered by the French scientist Joseph Proust (1754–1826), states that a chemical substance always contains the same proportions of elements by mass. Thus, sodium chloride (table salt) always contains the same proportion by mass of sodium to chlorine, in this case 39.34% sodium and 60.66% chlorine by mass, and sucrose (table sugar) is always 42.11% carbon, 6.48% hydrogen, and 51.41% oxygen by mass. Whereas a law states only what happens, a theory attempts to explain why nature behaves as it does. Laws are unlikely to change greatly over time unless a major experimental error is discovered. In contrast, a theory, by definition, is incomplete and imperfect, evolving with time to explain new facts as they are discovered. Because scientists can enter the cycle shown in Figure \(\PageIndex{1}\) at any point, the actual application of the scientific method to different topics can take many different forms. For example, a scientist may start with a hypothesis formed by reading about work done by others in the field, rather than by making direct observations. Example \(\PageIndex{1}\) Classify each statement as a law, a theory, an experiment, a hypothesis, an observation. Ice always floats on liquid water. Birds evolved from dinosaurs. Hot air is less dense than cold air, probably because the components of hot air are moving more rapidly. When 10 g of ice were added to 100 mL of water at 25°C, the temperature of the water decreased to 15.5°C after the ice melted. The ingredients of Ivory soap were analyzed to see whether it really is 99.44% pure, as advertised. Solution This is a general statement of a relationship between the properties of liquid and solid water, so it is a law. This is a possible explanation for the origin of birds, so it is a hypothesis. This is a statement that tries to explain the relationship between the temperature and the density of air based on fundamental principles, so it is a theory. The temperature is measured before and after a change is made in a system, so these are observations. This is an analysis designed to test a hypothesis (in this case, the manufacturer’s claim of purity), so it is an experiment. Exercise \(\PageIndex{1}\) Classify each statement as a law, a theory, an experiment, a hypothesis, a qualitative observation, or a quantitative observation. Measured amounts of acid were added to a Rolaids tablet to see whether it really “consumes 47 times its weight in excess stomach acid.” Heat always flows from hot objects to cooler ones, not in the opposite direction. The universe was formed by a massive explosion that propelled matter into a vacuum. Michael Jordan is the greatest pure shooter to ever play professional basketball. Limestone is relatively insoluble in water, but dissolves readily in dilute acid with the evolution of a gas. Answer a experiment Answer b law Answer c theory Answer d hypothesis Answer e observation Summary The scientific method is a method of investigation involving experimentation and observation to acquire new knowledge, solve problems, and answer questions. The key steps in the scientific method include the following: Step 1: Make observations. Step 2: Formulate a hypothesis. Step 3: Test the hypothesis through experimentation. Step 4: Accept or modify the hypothesis. Step 5: Develop into a law and/or a theory. Contributions & Attributions Wikipedia
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Courses/Kenyon_College/Chemistry_231_and_232_-_Kenyon_College_(Getzler_Hofferberth_and_Hunsen)/17%3A_Aldehydes_and_Ketones_-_The_Carbonyl_Group/17.13%3A_Oxidation_by_Peroxycarboxylic_Acids%3A_The__Baeyer-_Villiger__Oxidation
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Baeyer-Villiger oxidation is the oxidation of a ketone to a carboxylic acid ester using a peroxyacid as the oxidizing agent. eg. 1: eg. 2: Mechanism When the two ligands on the carbonyl carbon in the ketone are different, Baeyer-Villiger oxidation is regioselective. Of the two alpha carbons in the ketone, the one that can stabilize a positive charge more effectively, which is the more highly substituted one, migrates from carbon to oxygen preferentially. eg. 1: eg. 2: References Gamini Gunawardena from the OChemPal site ( Utah Valley University )
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Courses/Cornell_College/CC_CHM_411%3A_Advanced_Analytical_Chemistry/03%3A_Mass_Spectrometry/3.10%3A_Mass_Spectrometry-_Isotope_Effects
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The example above is simple, but the same methods can be applied to determine isotope peaks in more complicated molecules as well. The molecule C 4 Br 1 O 2 H 5 has several isotope effects: 13 C, 2 H, 81 Br, 17 O, and 18 O all must be taken into account. First we will look at the (M+1) + peak in comparison with the M + peak. Only isotopes that will increase the value of M by 1 must be taken into consideration here – since 81 Br and 18 O would both increase M by 2, they can be ignored (the most abundant isotopes for Br and O are 79 Br and 16 O). Like the previous example, there are 1.08 13 C atoms for every 100 12 C atoms. However, there are 4 carbon atoms in our molecule, and any one of them being a 13 C atom would result in a molecule with mass (M+1). So it is necessary to multiply the probability of an atom being a 13 C atom by the number of C atoms in the molecule. Therefore, we have: 4C * 1.08 = 4.32 = molecules with a 13 C atom per 100 molecules We can repeat this analysis for 2 H and 17 O: 5H * 0.015 = 0.075 = molecules with a 2 H atom per 100 molecules 2O * 0.04 = 0.08 = molecules with a 17 O atom per 100 molecules Any of the three isotopes, 13 C, 2 H, or 17 O occurring in our molecule would result in an (M+1) + peak. To get the ratio of (M+1) + /M + , we need to add all three probabilities: 4.32 + 0.075 + 0.08 = 4.475 = (M+ 1) + molecules per 100 M + molecules We can say then that the (M+1) + peak is 4.475% as high as the M + peak. A similar analysis can be easily repeated for (M+2) + : 1Br * 98 = 98 = molecules with an 81 Br molecule per 100 molecules 2O * 0.2 = 0.4 = molecules with an 18 O molecule per 100 molecules 98 + 0.4 = 98.4 = (M+2) + molecules per 100 M + molecules The (M + 2) + peak is therefore 98.4% as tall as the M + peak. This method is useful because using isotopic differences, it is possible to differentiate two molecules of identical mass numbers. References Skoog, DA. Holler, FJ. Crouch, SR. Principles of Instrumental Analysis, 6 th Edition. Thomson Brooks/Cole (2007). Coursey, J. S., Schwab, D. J., Dragoset, R. A. (2001). Atomic Weights and Isotopic Compositions. National Institute of Standards and Technology, Gaithersburg, MD. Outside Links http://en.wikipedia.org/wiki/Mass_spectrometer This wikipedia page is about the Mass Spectrometer instrument. http://en.wikipedia.org/wiki/Mass_spectrum_analysis This wikipedia page is more directly related to isotope effects, as it focuses on reading mass spectra. http://www.chem.uoa.gr/applets/AppletMS/Appl_Ms2.html This applett is fun to play with. It generates isotope peaks in a specified mass fragment. Problems Predict the (M+1) + relative peak heights for meta-nitrobenzene. Why would this method of looking at isotope ratios relating to peak heights make distinguishing molecules with Chlorine and Bromine from other molecules very easy? Predict the (M+4) + relative peak heights for C 3 H 2 SCl 2 Predict the (M+1) + and (M+2) + relative peak heights for 1,1,1-tribromo-2-propene Contributors Morgan Kelley (UCD)
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Courses/CSU_Chico/CSU_Chico%3A_CHEM_451_-_Biochemistry_I/CHEM_451_Test/08%3A_Transport_and_Kinetics/8.5%3A_More_Complicated_Enzymes/Multi-Substrate_Ping-Pong_Mechanisms
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In this mechanism, one substrate binds first to the enzyme followed by product \(P\) release. Typically, product \(P\) is a fragment of the original substrate \(A\). The rest of the substrate is covalently attached to the enzyme \(E\), which is designated as \(E'\). Now the second reactant, \(B\), binds and reacts with the E' and forms a covalent bond to the fragment of \(A\) still attached to the enzyme, forming product \(Q\). This is now released and the enzyme is restored to its initial form, \(E\). This represents a ping-pong mechanism An abbreviated notation scheme is shown below for ping-pong mechanisms. For this mechanism, Lineweaver-Burk plots at varying \(A\) and different fixed values of \(B\) give a series of parallel lines. One example of a ping-pong enzyme is low molecular weight protein tyrosine phosphatase. It reacts with the small substrate p-initrophenylphosphate (A) which binds to the enzyme covalently with the expulsion of the product P, the p-nitrophenol leaving group. Water (B) then comes in and covalently attacks the enzyme, forming an adduct with the covalently bound phosphate releasing it as inorganic phosphate. In this particular example, however, you can't vary the water concentration and it would be impossible to generate the parallel Lineweaver-Burk plots characteristic of ping-pong kinetics. What are the meanings of the kinetic parameters, \(K_m\) and \(V_m\), for multisubstrate/multiproduct mechanisms? Consider a random sequential bi-bi reaction for a "simple" case in which the rapid equilibrium assumption defines the binding of substrates \(A\) and \(B\). By inspection, there would appear to be two types of "effective" dissociation constants for reactant \(A\). One describes the binding of \(A\) to E (\(K_{ia}\)) and the other the binding of \(A\) to \(EB\) (\(K_a\)). Using mass balance for \(E\) and relationship that \[v_0=k_{cat}[EAB]\] the following initial rate equation can be derived. Note that Kib does not appear in the final equation. How can that be? The answer lies in the fact that the final concentration of \(EAB\) can be derived from the path \(E\) to \(EA\) to \(EAB\) or from the path \(E\) to \(EB\) to \(EAB\). Assuming rapid equilibrium, \[K_{ia}K_b=K_{ib}K_a.\] The following equation can be derived from ping-pong bi-bi mechanism. \[ v =\dfrac{V_M AB}{K_bA + K_aB + AB}\] For simplicity, all of the enzyme kinetic equations have been derived assuming no products are present.
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Bookshelves/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/12%3A_Solids/12.01%3A_Crystalline_and_Amorphous_Solids
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Learning Objectives To understand the difference between a crystalline and an amorphous solid Crystalline solids have regular ordered arrays of components held together by uniform intermolecular forces, whereas the components of amorphous solids are not arranged in regular arrays. The learning objective of this module is to know the characteristic properties of crystalline and amorphous solids. Introduction With few exceptions, the particles that compose a solid material, whether ionic, molecular, covalent, or metallic, are held in place by strong attractive forces between them. When we discuss solids, therefore, we consider the positions of the atoms, molecules, or ions, which are essentially fixed in space, rather than their motions (which are more important in liquids and gases). The constituents of a solid can be arranged in two general ways: they can form a regular repeating three-dimensional structure called a crystal lattice, thus producing a crystalline solid, or they can aggregate with no particular order, in which case they form an amorphous solid (from the Greek ámorphos, meaning “shapeless”). (left) Crystalline faces . The faces of crystals can intersect at right angles, as in galena (PbS) and pyrite (FeS 2 ), or at other angles, as in quartz. (Right) Cleavage surfaces of an amorphous solid . Obsidian, a volcanic glass with the same chemical composition as granite (typically KAlSi 3 O 8 ), tends to have curved, irregular surfaces when cleaved. Crystalline solids, or crystals, have distinctive internal structures that in turn lead to distinctive flat surfaces, or faces. The faces intersect at angles that are characteristic of the substance. When exposed to x-rays, each structure also produces a distinctive pattern that can be used to identify the material. The characteristic angles do not depend on the size of the crystal; they reflect the regular repeating arrangement of the component atoms, molecules, or ions in space. When an ionic crystal is cleaved (Figure 12.1), for example, repulsive interactions cause it to break along fixed planes to produce new faces that intersect at the same angles as those in the original crystal. In a covalent solid such as a cut diamond, the angles at which the faces meet are also not arbitrary but are determined by the arrangement of the carbon atoms in the crystal. Crystals tend to have relatively sharp, well-defined melting points because all the component atoms, molecules, or ions are the same distance from the same number and type of neighbors; that is, the regularity of the crystalline lattice creates local environments that are the same. Thus the intermolecular forces holding the solid together are uniform, and the same amount of thermal energy is needed to break every interaction simultaneously. Amorphous solids have two characteristic properties. When cleaved or broken, they produce fragments with irregular, often curved surfaces; and they have poorly defined patterns when exposed to x-rays because their components are not arranged in a regular array. An amorphous, translucent solid is called a glass. Almost any substance can solidify in amorphous form if the liquid phase is cooled rapidly enough. Some solids, however, are intrinsically amorphous, because either their components cannot fit together well enough to form a stable crystalline lattice or they contain impurities that disrupt the lattice. For example, although the chemical composition and the basic structural units of a quartz crystal and quartz glass are the same—both are SiO 2 and both consist of linked SiO 4 tetrahedra—the arrangements of the atoms in space are not. Crystalline quartz contains a highly ordered arrangement of silicon and oxygen atoms, but in quartz glass the atoms are arranged almost randomly. When molten SiO 2 is cooled rapidly (4 K/min), it forms quartz glass, whereas the large, perfect quartz crystals sold in mineral shops have had cooling times of thousands of years. In contrast, aluminum crystallizes much more rapidly. Amorphous aluminum forms only when the liquid is cooled at the extraordinary rate of 4 × 10 13 K/s, which prevents the atoms from arranging themselves into a regular array. The lattice of crystalline quartz (SiO 2 ) . The atoms form a regular arrangement in a structure that consists of linked tetrahedra. In an amorphous solid, the local environment, including both the distances to neighboring units and the numbers of neighbors, varies throughout the material. Different amounts of thermal energy are needed to overcome these different interactions. Consequently, amorphous solids tend to soften slowly over a wide temperature range rather than having a well-defined melting point like a crystalline solid. If an amorphous solid is maintained at a temperature just below its melting point for long periods of time, the component molecules, atoms, or ions can gradually rearrange into a more highly ordered crystalline form. Note Crystals have sharp, well-defined melting points; amorphous solids do not. Summary Solids are characterized by an extended three-dimensional arrangement of atoms, ions, or molecules in which the components are generally locked into their positions. The components can be arranged in a regular repeating three-dimensional array (a crystal lattice), which results in a crystalline solid, or more or less randomly to produce an amorphous solid. Crystalline solids have well-defined edges and faces, diffract x-rays, and tend to have sharp melting points. In contrast, amorphous solids have irregular or curved surfaces, do not give well-resolved x-ray diffraction patterns, and melt over a wide range of temperatures. Conceptual Problems 1. Compare the solid and liquid states in terms of a. rigidity of structure. b. long-range order. c. short-range order. 2. How do amorphous solids differ from crystalline solids in each characteristic? Which of the two types of solid is most similar to a liquid? a. rigidity of structure b. long-range order c. short-range order 3. Why is the arrangement of the constituent atoms or molecules more important in determining the properties of a solid than a liquid or a gas? 4. Why are the structures of solids usually described in terms of the positions of the constituent atoms rather than their motion? 5. What physical characteristics distinguish a crystalline solid from an amorphous solid? Describe at least two ways to determine experimentally whether a material is crystalline or amorphous. 6. Explain why each characteristic would or would not favor the formation of an amorphous solid. a. slow cooling of pure molten material b. impurities in the liquid from which the solid is formed c. weak intermolecular attractive forces 7. A student obtained a solid product in a laboratory synthesis. To verify the identity of the solid, she measured its melting point and found that the material melted over a 12°C range. After it had cooled, she measured the melting point of the same sample again and found that this time the solid had a sharp melting point at the temperature that is characteristic of the desired product. Why were the two melting points different? What was responsible for the change in the melting point? Conceptual Answers 3. The arrangement of the atoms or molecules is more important in determining the properties of a solid because of the greater persistent long-range order of solids. Gases and liquids cannot readily be described by the spatial arrangement of their components because rapid molecular motion and rearrangement defines many of the properties of liquids and gases. 7. The initial solid contained the desired compound in an amorphous state, as indicated by the wide temperature range over which melting occurred. Slow cooling of the liquid caused it to crystallize, as evidenced by the sharp second melting point observed at the expected temperature.
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Courses/Chandler_Gilbert_Community_College/Fundamental_Organic_ala_Mech/05%3A_Unit_2/5.01%3A_Functional_Groups_and_Physical_Properties/5.1.02%3A_Polar_Covalent_Bonds_-_Dipole_Moments
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Just as individual bonds are often polar, molecules as a whole are often polar as well. Molecular polarity results from the vector summation of all individual bond polarities and lone-pair contributions in the molecule. As a practical matter, strongly polar substances are often soluble in polar solvents like water, whereas less polar substances are insoluble in water. Net polarity is measured by a quantity called the dipole moment and can be thought of in the following way: assume that there is a center of mass of all positive charges (nuclei) in a molecule and a center of mass of all negative charges (electrons). If these two centers don’t coincide, then the molecule has a net polarity. The dipole moment , μ (lowercase Greek letter mu), is defined as the magnitude of the charge Q at either end of the molecular dipole times the distance r between the charges, μ = Q × r . Dipole moments are expressed in debyes (D), where 1 D = 3.336 × 10 –30 coulomb meters (C · m) in SI units. For example, the unit charge on an electron is 1.60 × 10 –19 C. Thus, if one positive charge and one negative charge are separated by 100 pm (a bit less than the length of a typical covalent bond), the dipole moment is 1.60 × 10 –29 C · m, or 4.80 D. \[\begin{aligned} & \mu=Q \times r \\ & \mu=\left(1.60 \times 10^{-19} \mathrm{C}\right)\left(100 \times 10^{-12} \mathrm{~m}\right)\left(\dfrac{1 \mathrm{D}}{3.336 \times 10^{-35} \mathrm{C}-\mathrm{m}}\right)=4.80 \mathrm{D} \end{aligned}\] Dipole moments for some common substances are given in Table \(\PageIndex{1}\). Of the compounds shown in the table, sodium chloride has the largest dipole moment (9.00 D) because it is ionic. Even small molecules like water ( μ = 1.85 D), methanol (CH 3 OH; μ = 1.70 D), and ammonia ( μ = 1.47 D), have substantial dipole moments, however, both because they contain strongly electronegative atoms (oxygen and nitrogen) and because all three molecules have lone-pair electrons. The lone-pair electrons on oxygen and nitrogen stick out into space away from the positively charged nuclei, giving rise to a considerable charge separation and making a large contribution to the dipole moment. Compound Dipole moment (D) Compound.1 Dipole moment (D).1 NaCl 9.00 NH3 1.47 CH2O 2.33 CH3NH2 1.31 CH3Cl 1.87 CO2 0.00 H2O 1.85 CH4 0.00 CH3OH 1.70 CH3CH3 0.00 CH3CO2H 1.70 NaN 0.00 CH3SH 1.52 NaN NaN In contrast with water, methanol, and ammonia, molecules such as carbon dioxide, methane, ethane, and benzene have zero dipole moments. Because of the symmetrical structures of these molecules, the individual bond polarities and lone-pair contributions exactly cancel. Worked Example \(\PageIndex{1}\): Predicting the Direction of a Dipole Moment Make a three-dimensional drawing of methylamine, CH 3 NH 2 , and show the direction of its dipole moment ( μ = 1.31). Strategy Look for any lone-pair electrons, and identify any atom with an electronegativity substantially different from that of carbon. (Usually, this means O, N, F, Cl, or Br.) Electron density will be displaced in the general direction of the electronegative atoms and the lone pairs. Solution Methylamine contains an electronegative nitrogen atom with a lone pair of electrons. The dipole moment thus points generally from –CH 3 toward the lone pair. Exercise \(\PageIndex{1}\) Ethylene glycol, HOCH 2 CH 2 OH, may look nonpolar when drawn, but an internal hydrogen bond between the two –OH groups results in a dipole moment. Explain. Answer The two C–O dipoles cancel because of the symmetry of the molecule: Exercise \(\PageIndex{2}\) Make three-dimensional drawings of the following molecules, and predict whether each has a dipole moment. If you expect a dipole moment, show its direction. H 2 C=CH 2 CHCl 3 CH 2 Cl 2 H 2 C=CCl 2 Answer (a) (b) (c) (d)
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Bookshelves/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/24%3A_Nuclear_Chemistry/24.10%3A_Effects_of_Radiation
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Bacterial contamination in our food often makes the news. There are many bacteria present on raw food, especially raw meat. Campylobacter, salmonella, and other microorganisms can be present, even after cooking, if the meat has not been sufficiently exposed to the heat. Ionizing radiation can be used to disrupt the DNA-RNA-protein synthesis cycle that allows the bacteria to reproduce. Cobalt-60 is a common radiation source, as is cesium-137. But...just to be safe...order that burger well-done! Effects of Radiation In order to better understand how cellular radiation damage occurs, we need to take a quick review of how the cell functions. DNA in the nucleus is responsible for protein synthesis and for regulation of many cellular functions. In the process of protein synthesis, DNA partially unfolds to produce messenger RNA (mRNA). The mRNA leaves the nucleus and interacts with ribosomes, transfer RNA, amino acids, and other cellular constituents in the cytoplasm. Through a complex series of reactions, proteins are produced to carry out a number of specialized processes within the organism. Anything that disturbs this flow of reactions can produce damage to the cell. The major effect of ionizing radiation on the cell is the disruption of the DNA strand. With the DNA structure damaged, the cell cannot reproduce in its normal fashion. Protein synthesis is affected, as are a number of processes necessary for proper cell function. One common effect is the generation of cancer cells. These cells have an abnormal structure due to the damaged DNA. In addition, they usually grow rapidly since the normal control processes regulating cell growth have been changed by the altered composition of the DNA. Tissue damage is also common in people with severe exposure to radiation. Effects of Radiation on Humans We can see two general types of effects when humans are exposed to radiation. Low-level exposure can lead to development of cancer. The regulatory processes regulating cell growth are disrupted, leading to uncontrolled growth of abnormal cells. Acute exposure can produce nausea, weakness, skin burns, and internal tissue damage. Cancer patients receiving radiation therapy experience these symptoms; the radiation is targeted to a specific site in the body so that the damage is primarily to the cancer cells, and the patient is able to recover from the exposure. Summary Anything that affects DNA replication and protein synthesis can damage a cell. Effects of ionizing radiation on protein synthesis are listed. The impact of ionizing radiation on human health is discussed.
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Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Mechanics__in_Chemistry_(Simons_and_Nichols)/06%3A_Quantum_Mechanics_in_Reactions/6.01%3A_Reduction_in_Symmetry_Along_Reaction_Paths
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As fragments are brought together to form a larger molecule, the symmetry of the nuclear framework (recall the symmetry of the Coulombic potential experienced by electrons depends on the locations of the nuclei) changes. However, in some cases, certain symmetry elements persist throughout the path connecting the fragments and the product molecule. These preserved symmetry elements can be used to label the orbitals throughout the 'reaction'. The point-group, axial- and full-rotation group symmetries which arise in nonlinear molecules, linear molecules, and atoms, respectively, are seen to provide quantum numbers or symmetry labels which can be used to characterize the orbitals appropriate for each such species. In a physical event such as interaction with an external electric or magnetic field or a chemical process such as collision or reaction with another species, the atom or molecule can experience a change in environment which causes the electrostatic potential which its orbitals experience to be of lower symmetry than that of the isolated atom or molecule. For example, when an atom interacts with another atom to form a diatomic molecule or simply to exchange energy during a collision, each atom's environment changes from being spherically symmetric to being axially symmetric. When the formaldehyde molecule undergoes unimolecular decomposition to produce \(CO + H_2\) along a path that preserves \(C_{2v}\) symmetry, the orbitals of the CO moiety evolve from \(C_{2v}\) symmetry to axial symmetry. It is important, therefore to be able to label the orbitals of atoms, linear, and nonlinear molecules in terms of their full symmetries as well in terms of the groups appropriate to lower-symmetry situations. This can be done by knowing how the representations of a higher symmetry group decompose into representations of a lower group. For example, the \(Y_{l,m}\) functions appropriate for spherical symmetry, which belong to a 2l+1 fold degenerate set in this higher symmetry, decompose into doubly degenerate pairs of functions \(Y_{l,l} , Y_{l,- l} ; Y_{l,l-1} , Y_{l,-1+1}\); etc., plus a single non-degenerate function \(Y_{l,0}\), in axial symmetry. Moreover, because \(L^2\) no longer commutes with the Hamiltonian whereas \(L_z\) does, orbitals with different l-values but the same m-values can be coupled. As the \(N_2\) molecule is formed from two N atoms, the 2s and \(2p_z\) orbitals, both of which belong to the same \((\sigma)\) symmetry in the axial rotation group but which are of different symmetry in the isolated-atom spherical symmetry, can mix to form the sg bonding orbital, the su antibonding, as well as the \(\sigma_g\) and \(\sigma_u\) nonbonding lone-pair orbitals. The fact that 2s and 2p have different l-values no longer uncouples these orbitals as it did for the isolated atoms, because l is no longer a "good" quantum number. Another example of reduced symmetry is provided by the changes that occur as \(H_2O\) fragments into OH and H. The \(\sigma\) bonding orbitals \((a_1 \text{ and } b_2)\) and in-plane lone pair \((a_1)\) and the \(\sigma\)* antibonding \((a_1 \text{ and } b_2) \text{ of } H_2O\) become a' orbitals (see the Figure below); the out-of-plane \(b_1\) lone pair orbital becomes a'' (in Appendix IV of Electronic Spectra and Electronic Structure of Polyatomic Molecules , G. Herzberg, Van Nostrand Reinhold Co., New York, N.Y. (1966) tables are given which allow one to determine how particular symmetries of a higher group evolve into symmetries of a lower group). Figure 6.1.1: Insert caption here! To further illustrate these points dealing with orbital symmetry, consider the insertion of CO into \(H_2\) along a path which preserves \(C_{2v}\) symmetry. As the insertion occurs, the degenerate \(\pi\) bonding orbitals of CO become \(b1 \text{ and } b_2\) orbitals. The antibonding \(\pi\)* orbitals of CO also become \(b_1 \text{ and } b_2\). The \(\sigma_g\) bonding orbital of \(H_2 \text{ becomes } a_1\), and the \(\sigma_u \text{ antibonding } H_2 \text{ orbital becomes } b_2.\) The orbitals of the reactant \(H_2CO\) are energy-ordered and labeled according to \(C_{2v}\) symmetry in the Figure shown below as are the orbitals of the product \(H_2\) + CO. Figure 6.1.2: Insert caption here! When these orbitals are connected according to their symmetries as shown above, one reactant orbital to one product orbital starting with the low-energy orbitals and working to increasing energy, an orbital correlation diagram (OCD) is formed. These diagrams play essential roles in analyzing whether reactions will have symmetry-imposed energy barriers on their potential energy surfaces along the reaction path considered in the symmetry analysis. The essence of this analysis, which is covered in detail in Chapter 12, can be understood by noticing that the sixteen electrons of ground-state \(H_2CO\) do not occupy their orbitals with the same occupancy pattern, symmetry-by-symmetry, as do the sixteen electrons of ground-state \(H_2\) + CO. In particular, \(H_2CO\) places a pair of electrons in the second \(b_2 \text{ orbital while } H_2 + CO\) does not; on the other hand, \(H_2\) + CO places two electrons in the sixth \(a_1 \text{ orbital while } H_2CO\) does not. The mismatch of the orbitals near the \(5a_1, 6a_1, \text{ and } 2b_2\) orbitals is the source of the mismatch in the electronic configurations of the ground-states of \(H_2CO \text{ and } H_2\) + CO. These mismatches give rise, as shown in Chapter 12, to symmetry-caused energy barriers on the \(H_2CO \rightarrow H_2 + CO\) reaction potential energy surface.
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Courses/University_of_Illinois_UrbanaChampaign/Chem_2363A_Fundamental_Organic_Chemistry_I_(Chan)/11%3A_The_Chemistry_of_Ethers_Epoxides_Glycols_and_Sulfides/11.01%3A_Williamson_Ether_Synthesis
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Objectives After completing this section, you should be able to write an equation to illustrate the industrial preparation of simple symmetrical ethers. write an equation to illustrate the Williamson synthesis of ethers. identify the ether obtained from the reaction of a given alkyl halide with a given alkoxide ion. identify the reagents needed to prepare a given ether through a Williamson synthesis. identify the limitations of the Williamson synthesis, and make the appropriate choices when deciding how best to synthesize a given ether. write an equation to describe the formation of an alkoxide from an alcohol. identify silver(I) oxide as a reagent which can be used in a Williamson synthesis. write an equation to show how an ether can be prepared by the alkoxymercuration-demercuration of an alkene. identify the product formed from the alkoxymercuration-demercuration of a given alkene. identify the alkene, the reagents, or both, needed to prepare a given ether by the alkoxymercuration-demercuration process. write the detailed mechanism of the reaction between an alkene, an alcohol and mercury(II) trifluoroacetate. Key Terms Make certain that you can define, and use in context, the key terms below. alkoxymercuration oxymercuration Williamson ether synthesis Study Notes We studied oxymercuration as a method of converting an alkene to an alcohol in Section 8.5 . “Alkoxymercuration” is a very similar process, except that we are now converting an alkene into an ether. The two processes are compared below. Description oxymercuration alkoxymercuration we react an alkene alkene with water an alcohol in the presence of Hg(O2CCH3)2 Hg(O2CCF3)2 followed by treatment with NaBH4 NaBH4 to produce an alcohol ether Review the mechanism of the oxymercuration reaction in Section 8.5 , paying particular attention to the regiochemistry and the stereochemistry of the reaction. The mechanism is identical to alkoxymercuration. Ether Formation Though Dehydration Acid-catalyzed dehydration of small 1º-alcohols constitutes a specialized industrial method of preparing symmetrical ethers. This reaction cannot be employed to prepare unsymmetrical ethers because a mixture of products is likely to be obtained. Also, 2 o and 3 o alcohols cannot be used for this reaction because they dehydrate to form alkenes by an E1 mechanism ( Section 17-6 ). \[\ce{2 CH_3CH_2-OH + H_2SO_4 ->[130\;^oC] CH_3CH_2\bond{-}O\bond{-}CH_2CH_3 + H_2O} \tag{18.2.1} \] Mechanism In the first step of the reaction mechanism, one alcohol is protonated to become a good leaving group. In the second step, a second alcohol displaces water from the protonated alcohol during an S N 2 reaction yielding a protonated ether. In the final step, this intermediate is deprotonated to yield the symmetrical ether. Williamson Ether Synthesis One important procedure, known as the Williamson Ether Synthesis , proceeds by an S N 2 reaction of an alkoxide nucleophile with a primary alkyl halide or tosylate. As previously discussed in Section 17-2 , alkoxides are commonly created by deprotonating an alcohol with a strong base, such as sodium hydride (NaH). Simple alcohols can be used a solvent during a Williamson ether synthesis and with their alkoxide created through the addition of sodium metal (Na (s) ). Planning a Williamson Ether Synthesis The Williamson ether synthesis has the same limitations as other S N 2 reactions, as discussed in Section 11-3 . Since alkoxide anions are strong bases, utilizing 2 o or 3 o halogen leaving groups could possibly produce an E2 elimination product. When considering the synthesis of an unsymmetrical ether, there are two different combinations of reactants possible and each should be carefully considered. In general, the pathway which utilizes the least sterically hindered halogen will be preferred. The key bond cleavage in the target molecule involves a C-O bond. Because unsymmetrical ethers have two unique C-O bonds, each can be broken to provide a unique set of reactants. After cleavage, the fragment with the oxygen will become an alkoxide. The other fragment will become a halogen or tosylate. Worked Example \(\PageIndex{1}\) How would you prepare the following molecule using a Williamson Ether Synthesis? Answer Analysis: The ether is asymmetrical so each of the C-O bonds can be broken to create a different set of possible reactants. After cleavage of the C-O bond, pathway 1 shows a 3 o halogen as the starting material. This reaction will most likely not be effective due to alkoxides reacting with 3 o halogens to preferable form an alkenes by E2 elimination. Pathway 2 shows a 1 o halogen as a starting material which is favorable for S N 2 reactions. Pathway 1 Solution 1 Pathway 2 Solution 2 Ether Synthesis Using Silver Oxide A variation of the Williamson ether synthesis uses silver oxide (Ag 2 O) in the place of the strong base. The conditions of this variation are milder than the typical Willamson synthesis because a strong base and the formation of an alkoxide intermediate are not necessary. This reaction is particually useful when converting the -OH groups on a sugar into ethers. Mechanism During this reaction a partial positively charged silver in Ag 2 O gives draws electron density from the iodine in CH 3 I. This correspondingly removes electron density from the adjacent carbon increasing its partial positive charge which increases its electrophlicity. This allows the alcohol to act as a nucleophile in the subsequent S N 2 reaction. Ether Synthesis Using Alkoxymercuration Alkoxymercuration , is patterned after the oxymercuration reaction discussed in Section 8-4 . Reaction of an alkene with an alcohol in the presence of a trifluoroacetate mercury (II) salt [(CF 3 CO 2 ) 2 Hg] prodcues an alkoxymercuration product. Demercuration using sodium borohydride (NaBH 4 ) yields an ether product. Overall, this reaction allows for the Markovnikov addition of an alcohol to an alkene to create an ether. Note that the alcohol reactant is used as the solvent, and a trifluoroacetate mercury (II) salt is used in preference to the mercuric acetate (trifluoroacetate anion is a poorer nucleophile than acetate). Most 1 o , 2 o , 3 o alcohols can be successfully used for this reaction. Mechanism The mechanism of alkoxymercuration is similar to that of oxymercuration, with electrophillic addition of the mercuric species to the alkene. The alcohol nucleophile attacks the more substituted carbon of the three-membered ring via a S N 2 reaction. Finally, sodium borohydride (NaBH 4 ) provides a reductive demercuration to form the ether product. Planning the Synthesis of an Ether using Alkoxymercuration The key bond cleavage in the target molecule involves a C-O bond. Because unsymmetrical ethers have two unique C-O bonds, each can be broken to provide a unique set of reactants. After cleavage, the fragment with the oxygen will become an alcohol. The alkyl fragment will lose a hydrogen from a adjacent carbon to form an alkene. The main point to consider when choosing a possible synthesis pathways is the ability of the alkyl fragment to form an alkene. Worked Example \(\PageIndex{2}\) How would you prepare the following molecule using a alkoxymercuration? Answer Analysis: The ether is symmetrical so each C-O bond of the ether can be cleaved to produce a set of starting materials for consideration. Pathway one shows a set of starting material which should work well for this reaction. The alcohol, methanol, can easily be used as a solvent. Although the alkene does not have a defined more and less substituted side, its symmetry will prevent a mixture of product from forming. The fragmentation for pathway 2 shows starting material which are not viable for this reaction. The alkyl fragment only has one carbon which cannot be used to form an alkene starting material. This means pathway 2 is not a viable method for the synthesis of the target molecule. Pathway 1 Solution 1 Pathway 2 Exercise \(\PageIndex{1}\) When preparing ethers using the Williamson ether synthesis, what factors are important when considering the nucleophile and the electrophile? Answer The nucleophile ideally should be very basic, yet not sterically hindered. This will minimize any elimination reactions from occurring. The electrophile should have the characteristics of a good S N 2 electrophile, preferably primary to minimize any elimination reactions from occurring. Exercise \(\PageIndex{2}\) How would you synthesize the following ethers? Keep in mind there are multiple ways. The Williamson ether synthesis, alkoxymercuration of alkenes, and also the acid catalyzed substitution. (a) (b) (c) (d) (e) Answer The Williamson ether syntheses require added catalytic base. Also, most of the halides can be interchanged, say for example for a -Br or a -Cl. Although, typically -I is the best leaving group. (a) (b) (c) (d) Note, there is only one ether (also called a silyl ether, and often used as an alcohol protecting group.) The other group is an ester. (e) Exercise \(\PageIndex{3}\) Draw the electron arrow pushing mechanism for the formation of diethyl ether in the previous problem. Answer . Exercise \(\PageIndex{4}\) t-butoxycyclohexane can be prepared two different ways from an alkene and an alcohol, draw both possible reactions. Answer While both are possible, the top route is likely easier because both starting materials are a liquid. Exercise \(\PageIndex{5}\) Epoxides are often formed intramolecularly. Take for example this large ring, in a publication from 2016 [ J. Org. Chem. , 2016 , 81 (20) , pp 10029–10034]. If subjected to base, what epoxide would be formed? (Include stereochemistry) Answer Exercise \(\PageIndex{6}\) What reagents would you use to perform the following transformations? (a) (b) (c) Answer (a) (b) Note the cis addition (c) An oxidation to an alcohol through hydroboration, and subsequent substitution with 2-bromopropane could also work, but this route provides the least likelihood of an elimination reaction occurring. Exercise \(\PageIndex{7}\) Predict the product of the following. Answer The result is the production of dioxane, a common solvent.
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Courses/Lumen_Learning/Book%3A_US_History_II_(Lumen)/04%3A_Reconstruction_Reader/4.05%3A_Primary_Source-_General_Reynolds_Describes_Lawlessness_in_Texas_(1868)
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Most histories of the Civil War claim that the war ended in the summer of 1865 when Confederate armies surrendered. However, violent resistance and terrorism continued in the South for over a decade. In this report, General J.J. Reynolds describes the lawlessness of Texas during Reconstruction. General: I have the honor to forward herewith annual tabular statement of expeditions and souts, and reports of movements of the various regiments serving in this district, for the year ending September 30, 1868. Armed organizations, generally known as “Ku-Klux Klans,” exist, independently or in concert with other armed bands, in many parts of Texas, but are most numerous, bold, and aggressive east of Trinity River. The precise objects of the organizations cannot be readily explained, but seems, in this state, to be to disarm, rob, and in many cases murder Union men and negroes, and as occasion may offer, murder United States officers and soldiers; also to intimidate every one who knows anything of the organization but who will not join it. The civil law east of the Trinity River is almost a dead letter. In some counties the civil officers are all, or a portion of them, members of the Klan. In other counties where the civil officers will not join the Klan, or some other armed band, they have been compelled to leave their counties. Examples are Van Zandt, Smith, and Marion counties; (the county seat of the latter is Jefferson.) In many counties where the county officers have not been driven off their influence is scarcely felt. What political end, if any, is aimed at by these bands I cannot say, but they attend in large bodies the political meetings (barbecues) which have been and are still being held in various parts of this State under the auspices of the democratic clubs of the different counties. The speakers encourage their attendance, and in several counties men have been indicated by name from the speaker’s stand, as those selected for murder. The men thus pointed out have no course left them but to leave their homes or be murdered on the first convenient opportunity. The murder of negroes is so common as to render it impossible to keep an accurate account of them. Many of the members of these bands of outlaws are transient persons in the State; the absence of railroads and telegraphs and great length of time required to communicate between remote points facilitating their devilish purposes. These organizations are evidently countenanced, or at least not discouraged, by a majority of the white people in the counties where the bands are most numerous. They could not otherwise exist. I have given this matter close attention, and am satisfied that a remedy to be effective must be gradually applied and continued with the firm support of the army until these outlaws are punished or dispersed. They cannot be punished by the civil courts until some examples by military commissions show that men can be punished in Texas for murder and kindred crimes. Perpetrators of such crimes have not heretofore, except in very rare instances, been punished in this state at all. Free speech and a free press, as the terms are generally understood in other States, have never existed in Texas. In fact, the citizens of other states cannot appreciate the state of affairs in Texas without actually experiencing it. The official reports of lawlessness and crime, so far from being exaggerated, do not tell the whole truth. Jefferson is the center from which most of the trade, travel, and lawlessness of eastern Texas radiate, and at this point or its vicinity there should be stationed about a regiment of troops. The recent murder at Jefferson of Hon. G. W. Smith, a delegate to the constitutional convention, has made it necessary to order more troops to that point. This movement weakens the frontier posts to such an extent as to impair their efficiency for protection against Indians, but the bold, wholesale murdering in the interior of the state seems at present to present a more urgent demand for the troops than Indian depredations. The frontier posts should, however, be reinforced if possible, as it is not improbably that the Indians from the northwest, after having suffered defeat there, will make heavy incursions into Texas. To restore measurable peace and quiet to Texas will require, for a long time, that troops be stationed at many county seats, until, by their presence, and aid if necessary, the civil law can be placed in the hands of reliable officers, and executed. This will be the work of years, and will be fully accomplished only by an increase of population. “Report of Brevet Major General J. J. Reynolds, Commanding Fifth Military District” in Annual Report of the Secretary of War (Washington: 1868), 704-705. Available through Google Books CC licensed content, Original Revision and Adaptation. Authored by : Kimlisa Duchicela. License : CC BY: Attribution CC licensed content, Shared previously The American Yawp Reader. Located at : http://www.americanyawp.com/reader.html . License : CC BY-SA: Attribution-ShareAlike
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Courses/Lansing_Community_College/LCC%3A_Chem_151_-_General_Chemistry_I/Text/15%3A_AcidBase_Equilibria/15.09%3A_Acid-Base_Properties_of_Salt_Solutions
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Learning Objectives To recognize salts that will produce acidic, basic, or neutral solutions in water To understand the Lewis acidity of small, highly-charged metal ions in water A neutralization reaction can be defined as the reaction of an acid and a base to produce a salt and water. That is, another cation, such as \(Na^+\), replaces the proton on the acid. An example is the reaction of \(CH_3CO_2H\), a weak acid, with \(NaOH\), a strong base: \[\underset{acid}{CH_3CO_2H_{(l)}} +\underset{base}{NaOH_{(s)}} \overset{H_2O}{\longrightarrow} \underset{salt}{H_2OCH_3CO_2Na_{(aq)} }+\underset{water}{H_2O_{(l)}} \nonumber \] Depending on the acid–base properties of its component ions, however, a salt can dissolve in water to produce a neutral solution, a basic solution, or an acidic solution. When a salt such as \(NaCl\) dissolves in water, it produces \(Na^+_{(aq)}\) and \(Cl^−_{(aq)}\) ions. Using a Lewis approach, the \(Na^+\) ion can be viewed as an acid because it is an electron pair acceptor, although its low charge and relatively large radius make it a very weak acid. The \(Cl^−\) ion is the conjugate base of the strong acid \(HCl\), so it has essentially no basic character. Consequently, dissolving \(NaCl\) in water has no effect on the \(pH\) of a solution, and the solution remains neutral. Now let's compare this behavior to the behavior of aqueous solutions of potassium cyanide and sodium acetate. Again, the cations (\(K^+\) and \(Na^+\)) have essentially no acidic character, but the anions (\(CN^−\) and \(CH_3CO_2^−\)) are weak bases that can react with water because they are the conjugate bases of the weak acids \(HCN\) and acetic acid, respectively. \[ CN^-_{(aq)} + H_2O_{(l)} \ce{ <<=>} HCN_{(aq)} + OH^-_{(aq)} \nonumber \] \[ CH_3CO^2_{2(aq)} + H_2O_{(l)} \ce{<<=>} CH_3CO_2H_{(aq)} + OH^-_{(aq)} \nonumber \] Neither reaction proceeds very far to the right as written because the formation of the weaker acid–base pair is favored. Both \(HCN\) and acetic acid are stronger acids than water, and hydroxide is a stronger base than either acetate or cyanide, so in both cases, the equilibrium lies to the left. Nonetheless, each of these reactions generates enough hydroxide ions to produce a basic solution. For example, the \(pH\) of a 0.1 M solution of sodium acetate or potassium cyanide at 25°C is 8.8 or 11.1, respectively. From Table \(\PageIndex{1}\) and Figure \(\PageIndex{1}\), we can see that \(CN^−\) is a stronger base (\(pK_b = 4.79\)) than acetate (\(pK_b = 9.24\)), which is consistent with \(KCN\) producing a more basic solution than sodium acetate at the same concentration. In contrast, the conjugate acid of a weak base should be a weak acid (Equation \(\ref{16.2}\)). For example, ammonium chloride and pyridinium chloride are salts produced by reacting ammonia and pyridine, respectively, with \(HCl\). As you already know, the chloride ion is such a weak base that it does not react with water. In contrast, the cations of the two salts are weak acids that react with water as follows: \[ NH^+_{4(aq)} + H_2O_{(l)} \ce{ <<=>} HH_{3(aq)} + H_3O^+_{(aq)} \label{16.2} \] \[ C_5H_5NH^+_{(aq)} + H_2O_{(l)} \ce{<<=>} C_5H_5NH_{(aq)} + H_3O^+_{(aq)} \label{16.3} \] Equation \(\ref{16.2}\) indicates that \(H_3O^+\) is a stronger acid than either \(NH_4^+\) or \(C_5H_5NH^+\), and conversely, ammonia and pyridine are both stronger bases than water. The equilibrium will therefore lie far to the left in both cases, favoring the weaker acid–base pair. The \(H_3O^+\) concentration produced by the reactions is great enough, however, to decrease the \(pH\) of the solution significantly: the \(pH\) of a 0.10 M solution of ammonium chloride or pyridinium chloride at 25°C is 5.13 or 3.12, respectively. This is consistent with the information shown in Figure 16.2, indicating that the pyridinium ion is more acidic than the ammonium ion. What happens with aqueous solutions of a salt such as ammonium acetate, where both the cation and the anion can react separately with water to produce an acid and a base, respectively? According to Figure 16.10, the ammonium ion will lower the \(pH\), while according to Equation \(\ref{16.3}\), the acetate ion will raise the \(pH\). This particular case is unusual, in that the cation is as strong an acid as the anion is a base (pKa ≈ pKb). Consequently, the two effects cancel, and the solution remains neutral. With salts in which the cation is a stronger acid than the anion is a base, the final solution has a \(pH\) < 7.00. Conversely, if the cation is a weaker acid than the anion is a base, the final solution has a \(pH\) > 7.00. Solutions of simple salts of metal ions can also be acidic, even though a metal ion cannot donate a proton directly to water to produce \(H_3O^+\). Instead, a metal ion can act as a Lewis acid and interact with water, a Lewis base, by coordinating to a lone pair of electrons on the oxygen atom to form a hydrated metal ion (part (a) in Figure \(\PageIndex{1}\)). A water molecule coordinated to a metal ion is more acidic than a free water molecule for two reasons. First, repulsive electrostatic interactions between the positively charged metal ion and the partially positively charged hydrogen atoms of the coordinated water molecule make it easier for the coordinated water to lose a proton. The magnitude of this effect depends on the following two factors (Figure \(\PageIndex{2}\)): The charge on the metal ion. A divalent ion (\(M^{2+}\)) has approximately twice as strong an effect on the electron density in a coordinated water molecule as a monovalent ion (\(M^+\)) of the same radius. The radius of the metal ion. For metal ions with the same charge, the smaller the ion, the shorter the internuclear distance to the oxygen atom of the water molecule and the greater the effect of the metal on the electron density distribution in the water molecule. Thus aqueous solutions of small, highly charged metal ions, such as \(Al^{3+}\) and \(Fe^{3+}\), are acidic: \[[Al(H_2O)_6]^{3+}_{(aq)} \rightleftharpoons [Al(H_2O)_5(OH)]^{2+}_{(aq)}+H^+_{(aq)} \label{16.36} \] The \([Al(H_2O)_6]^{3+}\) ion has a \(pK_a\) of 5.0, making it almost as strong an acid as acetic acid. Because of the two factors described previously, the most important parameter for predicting the effect of a metal ion on the acidity of coordinated water molecules is the charge-to-radius ratio of the metal ion. A number of pairs of metal ions that lie on a diagonal line in the periodic table, such as \(Li^+\) and \(Mg^{2+}\) or \(Ca^{2+}\) and \(Y^{3+}\), have different sizes and charges, but similar charge-to-radius ratios. As a result, these pairs of metal ions have similar effects on the acidity of coordinated water molecules, and they often exhibit other significant similarities in chemistry as well. Solutions of small, highly charged metal ions in water are acidic. Reactions such as those discussed in this section, in which a salt reacts with water to give an acidic or basic solution, are often called hydrolysis reactions. Using a separate name for this type of reaction is unfortunate because it suggests that they are somehow different. In fact, hydrolysis reactions are just acid–base reactions in which the acid is a cation or the base is an anion; they obey the same principles and rules as all other acid–base reactions. A hydrolysis reaction is an acid–base reaction. Example \(\PageIndex{1}\) Predict whether aqueous solutions of these compounds are acidic, basic, or neutral. \(\ce{KNO_3}\) \(\ce{CrBr_3} cdot \ce{H_2O}\) \(\ce{Na_2SO_4}\) Given : compound Asked for : acidity or basicity of aqueous solution Strategy : Assess the acid–base properties of the cation and the anion. If the cation is a weak Lewis acid, it will not affect the \(pH\) of the solution. If the cation is the conjugate acid of a weak base or a relatively highly charged metal cation, however, it will react with water to produce an acidic solution. f the anion is the conjugate base of a strong acid, it will not affect the \(pH\) of the solution. If, however, the anion is the conjugate base of a weak acid, the solution will be basic. Solution : a The \(K^+\) cation has a small positive charge (+1) and a relatively large radius (because it is in the fourth row of the periodic table), so it is a very weak Lewis acid. The \(NO_3−\) anion is the conjugate base of a strong acid, so it has essentially no basic character (Table 16.1). Hence neither the cation nor the anion will react with water to produce \(H^+\) or \(OH^−\), and the solution will be neutral. b. The \(Cr^{3+}\) ion is a relatively highly charged metal cation that should behave similarly to the \(Al^{3+}\) ion and form the \([Cr(H2O)_6]^{3+}\) complex, which will behave as a weak acid: \[ Cr(H_2O)_6]^{3+}_{(aq)} \ce{ <=>>} Cr(H_2O)_5(OH)]^{2+}_{(aq)} + H^+_{(aq)}\nonumber \] The \(Br^−\) anion is a very weak base (it is the conjugate base of the strong acid \(HBr\)), so it does not affect the \(pH\) of the solution. Hence the solution will be acidic. c. The \(Na^+\) ion, like the \(K^+\), is a very weak acid, so it should not affect the acidity of the solution. In contrast, \(SO_4^{2−}\) is the conjugate base of \(HSO_4^−\), which is a weak acid. Hence the \(SO_4^{2−}\) ion will react with water as shown in Figure 16.6 to give a slightly basic solution. Exercise \(\PageIndex{1}\) Predict whether aqueous solutions of the following are acidic, basic, or neutral. \(KI\) \(Mg(ClO_4)_2\) \(NaHS\) Answer a neutral Answer b acidic Answer c basic (due to the reaction of \(\ce{HS^{-}}\) with water to form \(\ce{H_2S}\) and \(\ce{OH^{-}}\)) Summary A salt can dissolve in water to produce a neutral, a basic, or an acidic solution, depending on whether it contains the conjugate base of a weak acid as the anion (\(A^−\)), the conjugate acid of a weak base as the cation (\(BH^+\)), or both. Salts that contain small, highly charged metal ions produce acidic solutions in water. The reaction of a salt with water to produce an acidic or a basic solution is called a hydrolysis reaction. Key Takeaways Acid–base reactions always contain two conjugate acid–base pairs. Each acid and each base has an associated ionization constant that corresponds to its acid or base strength.
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Ancillary_Materials/Worksheets/Worksheets%3A_Inorganic_Chemistry/Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry/07%3A_Intermolecular_Attraction_and_Structure-Property_Relationships/7.14%3A_Application_problems
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Problem SP14.1. The cleaner, Goof Off (the MIRACLE remover)®, is advertised to remove grease, oil and wax from surfaces. Goof Off is composed of a mixture of xylene and ethyl benzene. The structures of these compounds are shown below: Grease, oil and wax are hydrocarbon compounds. Why does Goof Off® work better than water to remove these substances? To support your answer, use IMF tables to determine: a) the strongest IMF between water and water; b) Goof Off® and Goof Off; c) Goof Off® and grease; and d) strongest IMF between grease and water. Why does Goof-off work better to dissolve grease and oil than if you just use water? Problem SP14.2. Often, a pint of gas-line antifreeze (commonly called "HEET", after a popular brand) is added to the gas tanks of cars, especially in the coldest days of winter, to prevent gas-line freeze-up. (Small amounts of water in gasoline can freeze when it is very cold. This blocks the fuel line and stops the engine.) a) Assume that gasoline is pure octane (C 8 H 18 ). What is the strongest IMF force between octane molecules in gasoline? b) One type of gas-line antifreeze contains almost 100% isopropyl alcohol, shown below. What is the strongest IMF between two isopropyl alcohol molecules? c) Isopropyl alcohol is completely soluble in both water and gasoline. Why? Consider the IMF involved when isopropyl alcohol is: i) interacting with water ii) Interacting with octane d) When HEET is added to a tank of gas it tends to cause water to dissolve (or at least stay suspended) in gasoline. Show in a picture and explain how this happens, on the molecular level. e) All gasoline in Minnesota is required to contain 10% ethanol (CH 3 CH 2 OH). Given this formulation, would a Minnesotan need to use HEET? Explain. Problem SP14.3. The laboratory of Samuel Stupp at Northwestern designs molecules with potential biomedical applications. The compound below was made to promote bone regeneration at the cell surface. Bone is mostly composed of calcium phosphate, Ca 3 (PO 4 ) 2 (Adapted with permission from Liam C. Palmer, Christina J. Newcomb, Stuart R. Kaltz, Erik D. Spoerke and Samuel I. Stupp, Chem. Rev. 2008 , 108 , 4754-4783. Copyright 2018 American Chemical Society.). a) This molecule is a synthetic amphiphile; it has a polar end and a non-polar end. Put a rectangle around the polar part of this molecule. Circle the nonpolar part of the molecule. b) The amphiphile inserts into the cell membrane. A cartoon of a portion of a cell membrane is shown below. Add a cartoon of the synthetic amphiphile to the membrane. What is the IMF through which it interacts with the membrane? c) Based on its position in the periodic table, what is the charge on a calcium ion? d) Based on the overall neutral charge of calcium phosphate, Ca 3 (PO 4 ) 2 , what is the charge on phosphate? e) What is the advantage of bone being composed of calcium and phosphate rather than more common ions like sodium and chloride? f) Show how this amphiphile would promote bone growth. Problem SP14.4. Let's make some paper. Paper is made from cellulose fibers that stick together to form a sheet. Cellulose, in turn, is a polymer made from individual glucose molecules bonded together in a chain. a) Circle an individual glucose unit in the cellulose. b) Name and draw the IMF that holds two cellulose chains together. One of the problems in making paper is that the most important source of cellulose is generally wood pulp. Wood pulp differs from other plant materials in that it contains lignin. The lignin is pretty important in the plants -- it provides structural rigidity or stiffness that allows trees to grow a lot taller than other plants -- but it isn't useful in paper. It is left in brown paper bags (kraft paper), but those tear more easily than paper that has had lignin removed. c) Show why lignin would weaken the interaction between cellulose molecules. One of the problems with paper is that it can absorb water and swell; it can even dissolve if there is enough water. d) Show why paper absorbs water so easily. One way to combat moisture absorption is to add sizing to the paper. Sizing can be mixed in with the cellulose mash before it is rolled out into sheets ("internal sizing"), or it can be sprayed onto the paper sheet after it is rolled out ("surface sizing"). Rosin, also obtained from trees, is a common source of sizing; one of its major components is abietic acid. e) In order to modify the properties of the paper, the rosin will have to stick to the paper. Explain how it does that. f) How does rosin make the paper more water-resistant? A very old source of sizing was rabbitskin glue; gelatin is a related material that is still used as sizing today (in addition to being the basis for a jiggly dessert). These materials are mostly proteins, rich in glycine, proline, alanine, phenylalanine, leucine and valine, among other amino acids. g) How would a protein stick to the paper? h) What kind of amino acid residues would make the paper water-resistant? Raw paper is quite transparent, so opaquifiers are usually added. They make a more opaque material so that you can write on both sides. Titanium dioxide is a common opaquifier. i) In order to get to a noble gas configuration, what would be the charge on a Ti ion? An O ion? j) Therefore, what is the formula of titanium oxide (the ratio of the ions in the compound)? k) Titanium oxide does not dissolve very well in water. Why not? l) Titanium oxide does stick in the fabric of the paper. Name the IMF. m) In the picture, the O is dark and the Ti light (picture credit: Ben Mills; public domain). Calculate the number of each atom in the unit cell. n) Describe the type of unit cell formed by the Ti atoms alone. We will need to try out the paper by writing on it. In a ballpoint pen, the ink (such as eosin red, used to grade your compositions in English class) is suspended or dissolved in a light oil such as benzyl alcohol. o) How does eosin dissolve in benzyl alcohol? Show and name one kind of IMF on the drawing. p) Why will the eosin stick to the paper? Problem SP14.5. Let's make a Post-It Note©. Post-It Notes are basically sticky pieces of paper. If we are going to make one, we will need some paper. Once you have the paper, we will need an adhesive to make it sticky. Researchers in Teresa Reineke’s lab at University of Minnesota have developed new adhesives based on sugars, a renewable resource (Nasiri, M.; Saxon, D. L.; Reineke, T. M. Macromolecules 2018 , 51 , 2456-2465. Copyright 2018, American Chemical Society. Used with permission.). The adhesives are long polymer chains containing about 800 X groups in a row, with between 12 and 90 Y groups at each end. They tried two different kinds of Y groups: AAI and GATA. a) What kind of functional group is OAc? b) How many chiral centers are found in the AAI group? In the GATA group? A good glue has a balance between adhesion – sticking to something else – and cohesion – sticking together. If there are lots of these chains in the adhesive (like strands of spaghetti), they can stick to each other via intermolecular forces. c) Show and name the IMFs that would occur between (i) two alkyl chains on group X; (ii) the ester groups on GATA. d) What IMF leads to good adhesion of the polymer to paper? In a static shear test, a 1-pound weight is stuck on a vertical surface with the adhesive, and the time it takes to fall off is measured. The AAI version outperformed the GATA version (6000 vs. 800 minutes). However, the lab made a slight alteration to the GATA and the adhesion time doubled (to 1600 minutes). Thinking only of cohesion, why did the strength of the material improve? e) Show and name the IMF on a drawing of the modified GATA. Compare its strength to the IMF it replaces in the original GATA. The lab also performed a peel adhesion test. They measured the force required to peel the “post-it note” from a surface. Both GATA and AAI versions out-performed Scotch© Tape, duct tape, and genuine Post-It© Notes.
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Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/01%3A_Quantum_Fundamentals/1.32%3A_Simulating_the_Aharonov-Bohm_Effect
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The Aharonov–Bohm effect is a phenomenon by which an electron is affected by the vector potential, A , in regions in which both the magnetic field B , and electric field E are zero. The most commonly described case occurs when the wave function of an electron passing around a long solenoid experiences a phase shift as a result of the enclosed magnetic field, despite the magnetic field being zero in the region through which the particle passes. Schematic of double-slit experiment in which Aharonov–Bohm effect can be observed: electrons pass through two slits, interfering at an observation screen, with the interference pattern shifted when a magnetic field B is turned on in the cylindrical solenoid. (All of the above adapted from Wikipeida) The effect on the interference fringes is calculated and displayed below. Please consult other tutorials on the double-slit interference effect on my page for background information. 0 1 2 Slit positions: \(x_{L} : = 1 \quad x_{R} : = 2\) Slit width: \(\delta : = 0.2\) Relative phase shift: \(\phi : = \pi\) Momentum Distribution/Diffraction Pattern for B = 0: \[ \Psi(p) :=\frac{1}{\sqrt{2}}\left(\int_{x_{L}-\frac{\delta}{2}}^{x_{L}+\frac{\delta}{2}} \frac{1}{\sqrt{2 \cdot \pi}} \cdot \exp (-i \cdot p \cdot x) \cdot \frac{1}{\sqrt{\delta}} d x+\int_{x_{R}-\frac{\delta}{2}}^{x_{R}+\frac{\delta}{2}} \frac{1}{\sqrt{2 \cdot \pi}} \cdot \exp (-i \cdot p \cdot x) \cdot \frac{1}{\sqrt{\delta}} d x\right) \nonumber \] Relative phase shift, \(\phi\), introduced at right-hand slit for B not equal to zero: \[ \Phi(\mathrm{p}) :=\frac{1}{\sqrt{2}}\left(\int_{\mathrm{x}_{\mathrm{L}}-\frac{\delta}{2}}^{\mathrm{x}_{\mathrm{L}}+\frac{\delta}{2}} \frac{1}{\sqrt{2 \cdot \pi}} \cdot \exp (-\mathrm{i} \cdot \mathrm{p} \cdot \mathrm{x}) \cdot \frac{1}{\sqrt{\delta}} \mathrm{d} \mathrm{x}+\exp (\mathrm{i} \cdot \phi) \cdot \int_{\mathrm{x}_{\mathrm{R}} +\frac{\delta}{2}}^{\mathrm{x}_{\mathrm{R}}+\frac{\delta}{2}} \frac{1}{\sqrt{2 \cdot \pi}} \cdot \exp (-\mathrm{i} \cdot \mathrm{p} \cdot \mathrm{x}) \cdot \frac{1}{\sqrt{\delta}} \mathrm{d} \mathrm{x} \right) \nonumber \] Display both diffraction patterns:
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Bookshelves/Organic_Chemistry/Organic_Chemistry_(OpenStax)/25%3A_Biomolecules_-_Carbohydrates/25.06%3A_Reactions_of_Monosaccharides
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Because monosaccharides have only two kinds of functional groups, hydroxyls and carbonyls, most of the chemistry of monosaccharides is the familiar chemistry of these two groups. As we’ve seen, alcohols can be converted into esters and ethers and can be oxidized; carbonyl compounds can react with nucleophiles and can be reduced. Ester and Ether Formation Monosaccharides behave as simple alcohols in much of their chemistry. For example, carbohydrate –OH groups can be converted into esters and ethers, which are often easier to work with than the free sugars. Because of their many hydroxyl groups, monosaccharides are usually soluble in water but insoluble in organic solvents such as ether. They are also difficult to purify and have a tendency to form syrups rather than crystals when water is removed. Ester and ether derivatives, however, are soluble in organic solvents and are easily purified and crystallized. Esterification is normally carried out by treating a carbohydrate with an acid chloride or acid anhydride in the presence of a base (Section 21.4 and Section 21.5). All the –OH groups react, including the anomeric one. For example, β - D -glucopyranose is converted into its pentaacetate by treatment with acetic anhydride in pyridine solution. Carbohydrates are converted into ethers by treatment with an alkyl halide in the presence of base—the Williamson ether synthesis (Section 18.2). Standard Williamson conditions using a strong base tend to degrade sensitive sugar molecules, but silver oxide works well as a mild base and gives high yields of ethers. For example, α - D -glucopyranose is converted into its pentamethyl ether in 85% yield on reaction with iodomethane and Ag 2 O. Exercise \(\PageIndex{1}\) Draw the products you would obtain by the reaction of β-D-ribofuranose with: \(\ce{CH3I}\), \(\ce{Ag2O}\) \(\ce{(CH3CO)2O}\), pyridine Answer Glycoside Formation We saw in Section 19.10 that treatment of a hemiacetal with an alcohol and an acid catalyst yields an acetal. In the same way, treatment of a monosaccharide hemiacetal with an alcohol and an acid catalyst yields an acetal called a glycoside , in which the anomeric –OH has been replaced by an –OR group. For example, the reaction of β - D -glucopyranose with methanol gives a mixture of α and β methyl D -glucopyranosides. (Note that a gly coside is the functional group name for any sugar, whereas a glu coside is formed specifically from glucose.) Glycosides are named by first citing the alkyl group and then replacing the - ose ending of the sugar with - oside. Like all acetals, glycosides are stable in neutral water. They aren’t in equilibrium with an open-chain form, and they don’t show mutarotation. They can, however, be hydrolyzed to give back the free monosaccharide plus alcohol on treatment with aqueous acid (Section 19.10). Glycosides are abundant in nature, and many biologically important molecules contain glycosidic linkages. For example, digitoxin, the active component of the digitalis preparations used for treatment of heart disease, is a glycoside consisting of a steroid alcohol linked to a trisaccharide. Note also that the three sugars are linked to one another by glycoside bonds. The laboratory synthesis of glycosides can be difficult because of the numerous –OH groups on the sugar molecule. One method that is particularly suitable for preparing glucose β -glycosides involves treatment of glucose pentaacetate with HBr, followed by addition of the appropriate alcohol in the presence of silver oxide. Called the Koenigs–Knorr reaction , the sequence involves formation of a pyranosyl bromide, followed by nucleophilic substitution. For example, methylarbutin, a glycoside found in pears, has been prepared by reaction of tetraacetyl- α - D -glucopyranosyl bromide with p -methoxyphenol. Although the Koenigs–Knorr reaction appears to involve a simple backside S N 2 displacement of bromide ion by alkoxide ion, the situation is actually more complex. Both α and β anomers of tetraacetyl- D -glucopyranosyl bromide give the same β -glycoside product, implying that they react by a common pathway. This result can be understood by assuming that tetraacetyl- D -glucopyranosyl bromide (either α or β anomer) undergoes a spontaneous S N 1-like loss of Br – , followed by internal reaction with the ester group at C2 to form an oxonium ion. Since the acetate at C2 is on the bottom of the glucose ring, the C–O bond also forms from the bottom. Backside S N 2 displacement of the oxonium ion then occurs with the usual inversion of configuration, yielding a β -glycoside and regenerating the acetate at C2 (Figure \(\PageIndex{1}\)). The participation shown by the nearby acetate group in the Koenigs–Knorr reaction is referred to as a neighboring-group effect and is a common occurrence in organic chemistry. Neighboring-group effects are usually noticeable only because they affect the rate or stereochemistry of a reaction; the nearby group itself does not undergo any evident change during the reaction. Biological Ester Formation: Phosphorylation In living organisms, carbohydrates occur not only in the free form but also linked through their anomeric center to other molecules such as lipids (glycolipids) or proteins (glycoproteins). Collectively called glycoconjugates , these sugar-linked molecules are components of cell walls that are crucial to the mechanism by which different cell types recognize one another. Glycoconjugate formation occurs by reaction of the lipid or protein with a glycosyl nucleoside diphosphate. This diphosphate is itself formed by initial reaction of a monosaccharide with adenosine triphosphate (ATP) to give a glycosyl monophosphate, followed by reaction with uridine triphosphate (UTP), to give a glycosyl uridine diphosphate. (We’ll see the structures of nucleoside phosphates in Section 28.1 .) The purpose of the phosphorylation is to activate the anomeric –OH group of the sugar and make it a better leaving group in a nucleophilic substitution reaction by a protein or lipid (Figure \(\PageIndex{2}\)). Reduction of Monosaccharides Treatment of an aldose or ketose with NaBH 4 reduces it to a polyalcohol called an alditol . The reduction occurs by reaction of the open-chain form present in the aldehyde/ketone ⇄ ⇄ hemiacetal equilibrium. Although only a small amount of the open-chain form is present at any given time, that small amount is reduced, more is produced by opening of the pyranose form, that additional amount is reduced, and so on, until the entire sample has undergone reaction. D -Glucitol, the alditol produced by reduction of D -glucose, is itself a naturally occurring substance found in many fruits and berries. It is used under the name D -sorbitol as a sweetener and sugar substitute in many foods. Exercise \(\PageIndex{2}\) Reduction of D-glucose leads to an optically active alditol (D-glucitol), whereas reduction of D-galactose leads to an optically inactive alditol. Explain. Answer D-Galactitol has a plane of symmetry and is a meso compound, whereas D-glucitol is chiral. Exercise \(\PageIndex{3}\) Reduction of L-gulose with \(\ce{NaBH4}\) leads to the same alditol (D-glucitol) as reduction of D-glucose. Explain. Answer The \(\ce{−CHO}\) end of L-gulose corresponds to the \(\ce{−CH2OH}\) end of D-glucose after reduction. Oxidation of Monosaccharides Like other aldehydes, aldoses are easily oxidized to yield the corresponding carboxylic acids, called aldonic acids . A buffered solution of aqueous Br 2 is often used for this purpose. Historically, the oxidation of an aldose with either Ag + in aqueous ammonia (called Tollens’ reagent ) or Cu 2 + with aqueous sodium citrate ( Benedict’s reagent ) formed the basis of simple tests for what are called reducing sugars . ( Reducing because the aldose reduces the metal oxidizing agent.) Some simple diabetes self-test kits sold in drugstores still use Benedict’s reagent to detect glucose in urine, though more modern methods have largely replaced it. All aldoses are reducing sugars because they contain an aldehyde group, but some ketoses are reducing sugars as well. Fructose reduces Tollens’ reagent, for example, even though it contains no aldehyde group. Reduction occurs because fructose is readily isomerized to a mixture of aldoses (glucose and mannose) in basic solution by a series of keto–enol tautomeric shifts (Figure \(\PageIndex{3}\)). Glycosides, however, are nonreducing because the acetal group is not hydrolyzed to an aldehyde under basic conditions. If warm, dilute HNO 3 (nitric acid) is used as the oxidizing agent, an aldose is oxidized to a dicarboxylic acid called an aldaric acid . Both the aldehyde carbonyl and the terminal –CH 2 OH group are oxidized in this reaction. Finally, if only the –CH 2 OH end of the aldose is oxidized without affecting the –CHO group, the product is a monocarboxylic acid called a uronic acid . The reaction can only be done enzymatically; no chemical reagent is known that can accomplish this selective oxidation in the laboratory. Exercise \(\PageIndex{4}\) D-Glucose yields an optically active aldaric acid on treatment with \(\ce{HNO3}\), but D-allose yields an optically inactive aldaric acid. Explain. Answer D-Allaric acid has a symmetry plane and is a meso compound, but D-glucaric acid is chiral. Exercise \(\PageIndex{5}\) Which of the other six D aldohexoses yield optically active aldaric acids on oxidation, and which yield optically inactive (meso) aldaric acids? Answer D-Allose and D-galactose yield meso aldaric acids; the other six D-hexoses yield optically active aldaric acids. Chain Lengthening: The Kiliani–Fischer Synthesis Much early activity in carbohydrate chemistry was devoted to unraveling the stereochemical relationships among monosaccharides. One of the most important methods used was the Kiliani–Fischer synthesis , which results in the lengthening of an aldose chain by one carbon atom. The C1 aldehyde group of the starting sugar becomes C2 of the chain-lengthened sugar, and a new C1 carbon is added. For example, an aldopentose is converted by Kiliani–Fischer synthesis into two aldohexoses. Discovery of the chain-lengthening sequence was initiated by the observation of Heinrich Kiliani in 1886 that aldoses react with HCN to form cyanohydrins (Section 19.6). Emil Fischer immediately realized the importance of Kiliani’s discovery and devised a method for converting the cyanohydrin nitrile group into an aldehyde. Fischer’s original method for conversion of the nitrile into an aldehyde involved hydrolysis to a carboxylic acid, ring closure to a cyclic ester (lactone), and subsequent reduction. A modern improvement involves reducing the nitrile over a palladium catalyst, yielding an imine intermediate that is hydrolyzed to an aldehyde. Note that the cyanohydrin is formed as a mixture of stereoisomers at the new chirality center, so two new aldoses, differing only in their stereochemistry at C2, result from Kiliani–Fischer synthesis. Chain extension of D -arabinose, for example, yields a mixture of D -glucose and D -mannose. Exercise \(\PageIndex{6}\) What product(s) would you expect from Kiliani–Fischer reaction of D-ribose? Answer D-Allose + D-altrose Exercise \(\PageIndex{7}\) What aldopentose would give a mixture of L-gulose and L-idose on Kiliani–Fischer chain extension? Answer L-Xylose Chain Shortening: The Wohl Degradation Just as the Kiliani–Fischer synthesis lengthens an aldose chain by one carbon, the Wohl degradation shortens an aldose chain by one carbon. Wohl degradation is almost the exact opposite of the Kiliani–Fischer sequence. That is, the aldose aldehyde carbonyl group is first converted into a nitrile, and the resulting cyanohydrin loses HCN under basic conditions—the reverse of a nucleophilic addition reaction. Conversion of the aldehyde into a nitrile is accomplished by treatment of an aldose with hydroxylamine to give an imine called an oxime (Section 19.8), followed by dehydration of the oxime with acetic anhydride. The Wohl degradation does not give particularly high yields of chain-shortened aldoses, but the reaction is general for all aldopentoses and aldohexoses. For example, D -galactose is converted by Wohl degradation into D -lyxose. Exercise \(\PageIndex{8}\) Two of the four D aldopentoses yield D-threose on Wohl degradation. What are their structures? Answer D-Xylose and D-lyxose
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Courses/Prince_Georges_Community_College/CHEM_2000%3A_General_Chemistry_for_Engineers_-_F21/05%3A_Gases/5.02%3A_Pressure-_The_Result_of_Particle_Collisions
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Learning Objectives Define the property of pressure Define and convert among the units of pressure measurements Describe the operation of common tools for measuring gas pressure Calculate pressure from manometer data The earth’s atmosphere exerts a pressure, as does any other gas. Although we do not normally notice atmospheric pressure, we are sensitive to pressure changes—for example, when your ears “pop” during take-off and landing while flying, or when you dive underwater. Gas pressure is caused by the force exerted by gas molecules colliding with the surfaces of objects (Figure \(\PageIndex{1}\)). Although the force of each collision is very small, any surface of appreciable area experiences a large number of collisions in a short time, which can result in a high pressure. In fact, normal air pressure is strong enough to crush a metal container when not balanced by equal pressure from inside the container. Atmospheric pressure is caused by the weight of the column of air molecules in the atmosphere above an object, such as the tanker car. At sea level, this pressure is roughly the same as that exerted by a full-grown African elephant standing on a doormat, or a typical bowling ball resting on your thumbnail. These may seem like huge amounts, and they are, but life on earth has evolved under such atmospheric pressure. If you actually perch a bowling ball on your thumbnail, the pressure experienced is twice the usual pressure, and the sensation is unpleasant. A dramatic illustration of atmospheric pressure is provided in this brief video, which shows a railway tanker car imploding when its internal pressure is decreased. Pressure is defined as the force exerted on a given area: \[P=\dfrac{F}{A} \label{9.2.1} \] Since pressure is directly proportional to force and inversely proportional to area (Equation \ref{9.2.1}), pressure can be increased either by either increasing the amount of force or by decreasing the area over which it is applied. Correspondingly, pressure can be decreased by either decreasing the force or increasing the area. Let’s apply the definition of pressure (Equation \ref{9.2.1}) to determine which would be more likely to fall through thin ice in Figure \(\PageIndex{2}\).—the elephant or the figure skater? A large African elephant can weigh 7 tons, supported on four feet, each with a diameter of about 1.5 ft (footprint area of 250 in 2 ), so the pressure exerted by each foot is about 14 lb/in 2 : \[\mathrm{pressure\: per\: elephant\: foot=14,000\dfrac{lb}{elephant}×\dfrac{1\: elephant}{4\: feet}×\dfrac{1\: foot}{250\:in^2}=14\:lb/in^2} \label{9.2.2} \] The figure skater weighs about 120 lbs, supported on two skate blades, each with an area of about 2 in 2 , so the pressure exerted by each blade is about 30 lb/in 2 : \[\mathrm{pressure\: per\: skate\: blade=120\dfrac{lb}{skater}×\dfrac{1\: skater}{2\: blades}×\dfrac{1\: blade}{2\:in^2}=30\:lb/in^2} \label{9.2.3} \] Even though the elephant is more than one hundred times heavier than the skater, it exerts less than one-half of the pressure and would therefore be less likely to fall through thin ice. On the other hand, if the skater removes her skates and stands with bare feet (or regular footwear) on the ice, the larger area over which her weight is applied greatly reduces the pressure exerted: \[\mathrm{pressure\: per\: human\: foot=120\dfrac{lb}{skater}×\dfrac{1\: skater}{2\: feet}×\dfrac{1\: foot}{30\:in^2}=2\:lb/in^2} \label{9.2.4} \] The SI unit of pressure is the pascal (Pa) , with 1 Pa = 1 N/m 2 , where N is the newton, a unit of force defined as 1 kg m/s 2 . One pascal is a small pressure; in many cases, it is more convenient to use units of kilopascal (1 kPa = 1000 Pa) or bar (1 bar = 100,000 Pa). In the United States, pressure is often measured in pounds of force on an area of one square inch— pounds per square inch (psi) —for example, in car tires. Pressure can also be measured using the unit atmosphere (atm) , which originally represented the average sea level air pressure at the approximate latitude of Paris (45°). Table \(\PageIndex{1}\) provides some information on these and a few other common units for pressure measurements Unit Name and Abbreviation Definition or Relation to Other Unit Comment pascal (Pa) 1 Pa = 1 N/m2 recommended IUPAC unit kilopascal (kPa) 1 kPa = 1000 Pa NaN pounds per square inch (psi) air pressure at sea level is ~14.7 psi NaN atmosphere (atm) 1 atm = 101,325 Pa air pressure at sea level is ~1 atm bar (bar, or b) 1 bar = 100,000 Pa (exactly) commonly used in meteorology millibar (mbar, or mb) 1000 mbar = 1 bar NaN inches of mercury (in. Hg) 1 in. Hg = 3386 Pa used by aviation industry, also some weather reports torr \(\mathrm{1\: torr=\dfrac{1}{760}\:atm}\) named after Evangelista Torricelli, inventor of the barometer millimeters of mercury (mm Hg) 1 mm Hg ~1 torr NaN Example \(\PageIndex{1}\): Conversion of Pressure Units The United States National Weather Service reports pressure in both inches of Hg and millibars. Convert a pressure of 29.2 in. Hg into: torr atm kPa mbar Solution This is a unit conversion problem. The relationships between the various pressure units are given in Table 9.2.1. \(\mathrm{29.2\cancel{in\: Hg}×\dfrac{25.4\cancel{mm}}{1\cancel{in}} ×\dfrac{1\: torr}{1\cancel{mm\: Hg}} =742\: torr}\) \(\mathrm{742\cancel{torr}×\dfrac{1\: atm}{760\cancel{torr}}=0.976\: atm}\) \(\mathrm{742\cancel{torr}×\dfrac{101.325\: kPa}{760\cancel{torr}}=98.9\: kPa}\) \(\mathrm{98.9\cancel{kPa}×\dfrac{1000\cancel{Pa}}{1\cancel{kPa}} \times \dfrac{1\cancel{bar}}{100,000\cancel{Pa}} \times\dfrac{1000\: mbar}{1\cancel{bar}}=989\: mbar}\) Exercise \(\PageIndex{1}\) A typical barometric pressure in Kansas City is 740 torr. What is this pressure in atmospheres, in millimeters of mercury, in kilopascals, and in bar? Answer 0.974 atm; 740 mm Hg; 98.7 kPa; 0.987 bar We can measure atmospheric pressure, the force exerted by the atmosphere on the earth’s surface, with a barometer (Figure \(\PageIndex{3}\)). A barometer is a glass tube that is closed at one end, filled with a nonvolatile liquid such as mercury, and then inverted and immersed in a container of that liquid. The atmosphere exerts pressure on the liquid outside the tube, the column of liquid exerts pressure inside the tube, and the pressure at the liquid surface is the same inside and outside the tube. The height of the liquid in the tube is therefore proportional to the pressure exerted by the atmosphere. If the liquid is water, normal atmospheric pressure will support a column of water over 10 meters high, which is rather inconvenient for making (and reading) a barometer. Because mercury (Hg) is about 13.6-times denser than water, a mercury barometer only needs to be \(\dfrac{1}{13.6}\) as tall as a water barometer—a more suitable size. Standard atmospheric pressure of 1 atm at sea level (101,325 Pa) corresponds to a column of mercury that is about 760 mm (29.92 in.) high. The torr was originally intended to be a unit equal to one millimeter of mercury, but it no longer corresponds exactly. The pressure exerted by a fluid due to gravity is known as hydrostatic pressure , p : \[p=hρg \label{9.2.5} \] where \(h\) is the height of the fluid, \(ρ\) is the density of the fluid, and \(g\) is acceleration due to gravity. Example \(\PageIndex{2}\): Calculation of Barometric Pressure Show the calculation supporting the claim that atmospheric pressure near sea level corresponds to the pressure exerted by a column of mercury that is about 760 mm high. The density of mercury = \(13.6 \,g/cm^3\). Solution The hydrostatic pressure is given by Equation \ref{9.2.5}, with \(h = 760 \,mm\), \(ρ = 13.6\, g/cm^3\), and \(g = 9.81 \,m/s^2\). Plugging these values into the Equation \ref{9.2.5} and doing the necessary unit conversions will give us the value we seek. (Note: We are expecting to find a pressure of ~101,325 Pa:) \[\mathrm{101,325\:\mathit{N}/m^2=101,325\:\dfrac{kg·m/s^2}{m^2}=101,325\:\dfrac{kg}{m·s^2}} \nonumber \] \[\begin {align*} p&\mathrm{=\left(760\: mm×\dfrac{1\: m}{1000\: mm}\right)×\left(\dfrac{13.6\: g}{1\:cm^3}×\dfrac{1\: kg}{1000\: g}×\dfrac{( 100\: cm )^3}{( 1\: m )^3}\right)×\left(\dfrac{9.81\: m}{1\:s^2}\right)}\\[4pt] &\mathrm{=(0.760\: m)(13,600\:kg/m^3)(9.81\:m/s^2)=1.01 \times 10^5\:kg/ms^2=1.01×10^5\mathit{N}/m^2} \\[4pt] & \mathrm{=1.01×10^5\:Pa} \end {align*} \nonumber \] Exercise \(\PageIndex{2}\) Calculate the height of a column of water at 25 °C that corresponds to normal atmospheric pressure. The density of water at this temperature is 1.0 g/cm 3 . Answer 10.3 m A manometer is a device similar to a barometer that can be used to measure the pressure of a gas trapped in a container. A closed-end manometer is a U-shaped tube with one closed arm, one arm that connects to the gas to be measured, and a nonvolatile liquid (usually mercury) in between. As with a barometer, the distance between the liquid levels in the two arms of the tube ( h in the diagram) is proportional to the pressure of the gas in the container. An open-end manometer (Figure \(\PageIndex{3}\)) is the same as a closed-end manometer, but one of its arms is open to the atmosphere. In this case, the distance between the liquid levels corresponds to the difference in pressure between the gas in the container and the atmosphere. Example \(\PageIndex{3}\): Calculation of Pressure Using an Open-End Manometer The pressure of a sample of gas is measured at sea level with an open-end Hg (mercury) manometer, as shown below. Determine the pressure of the gas in: mm Hg atm kPa Solution The pressure of the gas equals the hydrostatic pressure due to a column of mercury of height 13.7 cm plus the pressure of the atmosphere at sea level. (The pressure at the bottom horizontal line is equal on both sides of the tube. The pressure on the left is due to the gas and the pressure on the right is due to 13.7 cm of Hg plus atmospheric pressure.) In mm Hg, this is: 137 mm Hg + 760 mm Hg = 897 mm Hg \(\mathrm{897\cancel{mm Hg}×\dfrac{1\: atm}{760\cancel{mm Hg}}=1.18\: atm}\) \(\mathrm{1.18\cancel{atm}×\dfrac{101.325\: kPa}{1\cancel{atm}}=1.20×10^2\:kPa}\) Exercise \(\PageIndex{3}\) The pressure of a sample of gas is measured at sea level with an open-end Hg manometer, as shown below Determine the pressure of the gas in: mm Hg atm kPa Answer a 642 mm Hg Answer b 0.845 atm Answer c 85.6 kPa Application: Measuring Blood Pressure Blood pressure is measured using a device called a sphygmomanometer (Greek sphygmos = “pulse”). It consists of an inflatable cuff to restrict blood flow, a manometer to measure the pressure, and a method of determining when blood flow begins and when it becomes impeded (Figure \(\PageIndex{5}\)). Since its invention in 1881, it has been an essential medical device. There are many types of sphygmomanometers: manual ones that require a stethoscope and are used by medical professionals; mercury ones, used when the most accuracy is required; less accurate mechanical ones; and digital ones that can be used with little training but that have limitations. When using a sphygmomanometer, the cuff is placed around the upper arm and inflated until blood flow is completely blocked, then slowly released. As the heart beats, blood forced through the arteries causes a rise in pressure. This rise in pressure at which blood flow begins is the systolic pressure— the peak pressure in the cardiac cycle. When the cuff’s pressure equals the arterial systolic pressure, blood flows past the cuff, creating audible sounds that can be heard using a stethoscope. This is followed by a decrease in pressure as the heart’s ventricles prepare for another beat. As cuff pressure continues to decrease, eventually sound is no longer heard; this is the diastolic pressure— the lowest pressure (resting phase) in the cardiac cycle. Blood pressure units from a sphygmomanometer are in terms of millimeters of mercury (mm Hg). Meteorology, Climatology, and Atmospheric Science Throughout the ages, people have observed clouds, winds, and precipitation, trying to discern patterns and make predictions: when it is best to plant and harvest; whether it is safe to set out on a sea voyage; and much more. We now face complex weather and atmosphere-related challenges that will have a major impact on our civilization and the ecosystem. Several different scientific disciplines use chemical principles to help us better understand weather, the atmosphere, and climate. These are meteorology, climatology, and atmospheric science. Meteorology is the study of the atmosphere, atmospheric phenomena, and atmospheric effects on earth’s weather. Meteorologists seek to understand and predict the weather in the short term, which can save lives and benefit the economy. Weather forecasts (Figure \(\PageIndex{5}\)) are the result of thousands of measurements of air pressure, temperature, and the like, which are compiled, modeled, and analyzed in weather centers worldwide. In terms of weather, low-pressure systems occur when the earth’s surface atmospheric pressure is lower than the surrounding environment: Moist air rises and condenses, producing clouds. Movement of moisture and air within various weather fronts instigates most weather events. The atmosphere is the gaseous layer that surrounds a planet. Earth’s atmosphere, which is roughly 100–125 km thick, consists of roughly 78.1% nitrogen and 21.0% oxygen, and can be subdivided further into the regions shown in Figure \(\PageIndex{7}\): the exosphere (furthest from earth, > 700 km above sea level), the thermosphere (80–700 km), the mesosphere (50–80 km), the stratosphere (second lowest level of our atmosphere, 12–50 km above sea level), and the troposphere (up to 12 km above sea level, roughly 80% of the earth’s atmosphere by mass and the layer where most weather events originate). As you go higher in the troposphere, air density and temperature both decrease. Climatology is the study of the climate, averaged weather conditions over long time periods, using atmospheric data. However, climatologists study patterns and effects that occur over decades, centuries, and millennia, rather than shorter time frames of hours, days, and weeks like meteorologists. Atmospheric science is an even broader field, combining meteorology, climatology, and other scientific disciplines that study the atmosphere. Summary Gases exert pressure, which is force per unit area. The pressure of a gas may be expressed in the SI unit of pascal or kilopascal, as well as in many other units including torr, atmosphere, and bar. Atmospheric pressure is measured using a barometer; other gas pressures can be measured using one of several types of manometers. Key Equations \(P=\dfrac{F}{A}\) p = hρg Glossary atmosphere (atm) unit of pressure; 1 atm = 101,325 Pa bar (bar or b) unit of pressure; 1 bar = 100,000 Pa barometer device used to measure atmospheric pressure hydrostatic pressure pressure exerted by a fluid due to gravity manometer device used to measure the pressure of a gas trapped in a container pascal (Pa) SI unit of pressure; 1 Pa = 1 N/m 2 pounds per square inch (psi) unit of pressure common in the US pressure force exerted per unit area torr unit of pressure; \(\mathrm{1\: torr=\dfrac{1}{760}\,atm}\)
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Courses/Woodland_Community_College/Chem_1A%3A_General_Chemistry_I/zz%3A_Back_Matter/20%3A_Glossary
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Words (or words that have the same definition) The definition is case sensitive (Optional) Image to display with the definition [Not displayed in Glossary, only in pop-up on pages] (Optional) Caption for Image (Optional) External or Internal Link (Optional) Source for Definition (Eg. "Genetic, Hereditary, DNA ...") (Eg. "Relating to genes or heredity") NaN The infamous double helix https://bio.libretexts.org/ CC-BY-SA; Delmar Larsen Word(s) Definition Image Caption Link Source Sample Word 1 Sample Definition 1 NaN NaN NaN NaN
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Courses/Ripon_College/CHM_321%3A_Inorganic_Chemistry/03%3A_Acid-Base_Chemistry_and_an_Introduction_to_Coordination_Compounds/3.07%3A_Coordination_Chemistry-_Bonding/3.7.05%3A_Metal-Metal_Bonds
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Molecular orbital considerations in Dinuclear Metal Complexes with Multiple M-M Bonds Let us begin a new chapter and think about dinuclear transition metal complexes with multiple metal-metal bonds. What is the maximum bond order that we could expect? The transition metals have five d-orbitals available, and in order to determine the maximum possible bond order we need to see how they overlap to form molecular orbitals (Fig. 11.1.1). If we define the M-M bond axis as the z-axis, then two d z 2 -orbitals can overlap in σ-fashion to form a bonding and an anti-bonding σ-molecular orbital. A d xz , and a d yz can overlap with another d xz and another d yz in π-fashion to form two degenerated bonding π and two degenerated anti-bonding π*-orbitals. π-overlap is smaller than σ-overlap, therefore the split in energy between the bonding and the anti-bonding π-orbitals is smaller than the split between the bonding and the anti-bonding σ-orbitals. The d xy orbital can overlap with another d xy orbital in \(\delta\)-fashion, and so can the d x 2 -y 2 with another d x 2 -y 2 . This gives two bonding \(\delta\)-orbitals and two anti-bonding \(\delta\)-orbitals. The \(\delta\)-overlap is even smaller than the π-overlap, therefore the energy split between the bonding and the anti-bonding \(\delta\)-orbitals is even smaller than those for the π and π*-orbitals. So what would be the maximum bond order that could be achieved? The maximum bond order would be achieved when all bonding MOs were full, and all anti-bonding MOs were empty. We have overall five bonding MOs that we could fill with ten electrons from two metal atoms with d 5 -electron configuration. The maximum bond order BO would therefore be BO=5. However, in practice only bond orders up to 4 are well known. This is because the d x 2 -y 2 orbital is usually too involved in the bonding with the ligands thereby becoming unavailable for metal-metal interactions. The d x 2 -y 2 orbital makes the strongest interactions with the ligands because most ligands approach on or near the x and the y-axes. Electron Configurations and Multiple M-M Bonds We can easily predict now which electron configuration leads to which metal-metal bond order. The bond order increases from 1 to 4 with the electron configuration changing from d 1 to d 4 . At d 4 all bonding MOs are full with eight electrons. An example of a complex with a bond order of 4 is the tetraacetatodiaquadichromium complex shown (Fig. 11.1.2). Cr is in the oxidation state +2, which makes the chromium a d 4 species. We can quickly show this by counting the valence electrons. A neutral Cr atom has 6 VE, and an electron configuration 3d 5 4s 1 . There are four acetate ligands having a 1- charge each which gives four negative charges overall. The complex is overall neutral which means that each Cr must have formally a 2+ charge, and the electron configuration is 3d 4 . With even more electrons in the metal d-orbitals the bond order begins to decrease again as anti-bonding MOs need to be filled (Fig. 11.1.3). The combination of two metals with d 5 electron configuration leads to a triple bond, two d 6 -metals give a double bond, and two d 7 metals give a single bond. A metal-metal bond should not exist for two d 8 -metals because then the bond order is zero. Evidence for M-M Multiple Bonds What experimental evidence can support the existence of a particular bond order (Fig. 11.1.4)? One argument is the bond length which can be obtained through the crystal structure determination of the complex. The shorter the bond, the higher the bond order. For instance, in the four tungsten complexes shown the bond lengths decrease from 272 pm, to 248 pm, to 230 pm to 221 pm corresponding to a single, double, triple, and quadruple bond, respectively. Another hint can be the conformation of a molecule (Fig. 11.1.5). For instance, the two square-planar units of the Re 2 Me 8 2 - complex anion show eclipsed conformation. Steric repulsion arguments would favor the staggered conformation, so there must be a reason why the two ReMe 4 units are eclipsed. The rhenium is in the oxidation state +3, thus it is a d 4 species, and we would argue that there may be a Re-Re quadruple bond. This quadruple bond can only form when the d xy orbitals are in eclipsed conformation, and this is only possible when the two ReMe 4 fragments are in eclipsed conformation. The very short bond length of 218 pm further supports the existence of the quadruple bond. Dr. Kai Landskron ( Lehigh University ). If you like this textbook, please consider to make a donation to support the author's research at Lehigh University: Click Here to Donate .
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Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Instrumentation_and_Analysis/Introduction_to_Lasers/03_Basic_Principles/02_Laser_Operation_and_Components
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Laser Operation and Components The process of light stimulated emission is fundamental to laser operation. Laser light is produced by an active medium , or gain medium inside the laser optical cavity . The active medium is a collection of atoms, or molecules that can undergo stimulated emission. The active medium can be in a gaseous, liquid or solid form. For lasing to take place, the active medium must be pumped into an excited state capable of undergoing stimulated emission. The energy required for excitation is often supplied by an electric current or an intense light source, such as a flashlamp. To induce stimulated emission, the laser cavity must provide a means to reflect, or feedback emitted light into the gain medium. A laser must have an output coupler to allow a portion of the laser light to leave the optical cavity. Laser Optical Cavity Sketch showing the main components of a laser optical (or resonator) cavity. The optical cavity is formed by a pair of mirrors that surround the gain medium and enable feedback of light into the medium. The output coupler is a partially reflective mirror that allows a portion of the laser radiation to leave the cavity. The gain medium is excited by an external source (not shown), such as a flash lamp, electric current or another laser. The light trapped between the mirrors forms standing wave structures called modes. Although beyond the scope of this discussion, the reader interested in cavity modes can consult References 7-10 and the “ Laser Radiation Properties ” section. Stimulated Emission 7-10, 12, 13 Stimulated emission occurs when a photon of light induces an atom or molecule to lose energy by producing a second photon. The second photon has the same phase, frequency, direction of travel and polarization state as the stimulating photon. Since from one photon a second identical photon is produced, stimulated emission leads to light amplification . Stimulated emission can be understood from an energy level diagram within the context of the competing optical processes of stimulated absorption and spontaneous emission . For stimulated emission to take place, a population inversion must be created in the laser gain medium. For more on stimulated emission, see energy level diagrams and subsequent sections. Energy Level Diagrams An energy level diagram displays states of an atom, molecule or material as levels ordered vertically according to energy. The states contain contributions from several sources, as appropriate for the matter considered. Sources include the orbital and spin angular momentum of electrons, vibrations of nuclei, molecular rotations, and spin contributions from nuclei. The lowest energy level is called the ground state . Absorption and emission of energy occurs when matter undergoes transitions between states . Energy level diagram showing states of a sodium atom. Each state is labeled by a term symbol and includes effects of electron orbital and spin angular momentum. Term Symbols Term symbols are a shorthand for describing the angular momentum and coupling interactions among electrons in atoms and molecules. As a starting point for understanding a term symbol, write the electron configuration for the state considered. For Na, the electron configuration of the ground state is: 1s 2 2s 2 2p 6 3s 1 The central letter describes the total orbital angular momentum. Only the valence electrons need to be considered. For Na, there is one valence electron, and it occupies an s-orbital. The angular momentum quantum number for an s-orbital is l = 0. The total orbital angular momentum for ground state Na is L = l = 0. Symbols are assigned to the values of L as follows: L = 0 (S), L = 1 (P), L = 2 (D), etc. The left superscript reflects the coupling of valence electron spin angular momentum and gives the degeneracy of spin states. For Na, s = 1/2 for the valence electron; therefore, the total spin, S = 1/2 and the degeneracy = (2 S + 1) = 2. The right subscript reflects the coupling between spin and orbital angular momentum. For ground state Na, J = L + S = 1/2. For a detailed discussion of term symbols, see Ref 11 . Absorption and Emission Processes and Transitions Between Energy States Stimulated absorption (a) occurs when light, or a photon of light (hν), excites matter to a higher energy (or excited ) state. Spontaneous emission (b) is a process whereby energy is spontaneously released from matter as light. Molecules typically transition to vibrationally excited levels within the excited electronic state. Following excitation, the vibrational energy is quickly released by non-radiative pathways (c) . In molecules, spontaneous emission known as fluorescence ( b ) occurs by transition from the lowest level in the excited electronic state, to upper vibrational levels of the lower electronic state. Energy level diagram for a typical dye molecule. The vibrational levels of each electronic state, labeled by S 0 and S 1 , are included. Stimulated Emission - Details 7-10, 12, 13 Laser radiation is produced when energy in atoms or molecules is released as stimulated emission ( c ). Stimulated emission requires a population inversion in the laser gain medium. A population inversion occurs when the number of atoms or molecules in an excited state exceeds the number in lower levels (usually the ground state). To create the population inversion, the gain medium must transition to a metastable state , which is long lived relative to spontaneous emission. The three-level diagram (below) shows excitation followed by non-radiative (nr) decay ( b ) to 2 E states. The 2 E states are long lived, because the transition to 4 A 2 requires a change in the electron spin state. A photon of the same energy as the 2 E → 4 A 2 transition can stimulate the emission of a second photon ( c ), leading to light amplification , or lasing. Three-level energy diagram. Simplified diagram showing transitions for Cr 3 + in a ruby laser. Three and Four Level Lasers 7-10, 12, 13 Three-level lasers require intense pumping to maintain the population inversion, because the lasing transition re-populates the ground state. Lasers based on transitions between four energy levels (see below), can be more efficiently pumped, because the lower level of the lasing transition is not the ground state. Only four-level lasers provide continuous output . HeNe and Nd:YAG are common four-level lasers. A population inversion is necessary for lasing, because without one, the photon inducing stimulated emission would instead have a greater probability of undergoing absorption in the gain medium. For more in depth information about laser transitions and population inversion, Refs 7-10, 12 (pg 96) and 13 can be consulted. Four-level energy diagram. Simplified diagram showing transitions for Nd 3 + in a Nd:YAG laser.
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Courses/CSU_San_Bernardino/CHEM_2200%3A_General_Chemistry_II_(Mink)/22%3A_Appendices/22.05%3A_Water_Properties
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Temperature Density (g/mL) 0 0.9998395 4 0.9999720 (density maximum) 10 0.9997026 15 0.9991026 20 0.9982071 22 0.9977735 25 0.9970479 30 0.9956502 40 0.9922 60 0.9832 80 0.9718 100 0.9584 Temperature Vapor Pressure (torr) Vapor Pressure (Pa) 0 4.6 613.2812 4 6.1 813.2642 10 9.2 1226.5620 15 12.8 1706.5220 20 17.5 2333.1350 22 19.8 2639.7760 25 23.8 3173.0640 30 31.8 4239.6400 35 42.2 5626.1880 40 55.3 7372.7070 45 71.9 9585.8520 50 92.5 12332.2900 55 118.0 15732.0000 60 149.4 19918.3100 65 187.5 24997.8800 70 233.7 31157.3500 75 289.1 38543.3900 80 355.1 47342.6400 85 433.6 57808.4200 90 525.8 70100.7100 95 633.9 84512.8200 100 760.0 101324.7000 Temperature Kw 10–14 pKw1 0 0.112 14.95 5 0.182 14.74 10 0.288 14.54 15 0.465 14.33 20 0.671 14.17 25 0.991 14.00 30 1.432 13.84 35 2.042 13.69 40 2.851 13.55 45 3.917 13.41 50 5.297 13.28 55 7.080 13.15 60 9.311 13.03 75 19.950 12.70 100 56.230 12.25 0 C°(H2O(l)) = 4.184 J∙g-1∙°C-1 C°(H2O(s)) = 1.864 J∙K−1∙g−1 C°(H2O(g)) = 2.093 J∙K−1∙g−1 Unnamed: 0 Temperature (K) ΔH (kJ/mol) melting 273.15 6.088 boiling 373.15 40.656 (44.016 at 298 K) 0 Kf = 1.86°C∙kg∙mol−1 (cryoscopic constant) Kb = 0.51°C∙kg∙mol−1 (ebullioscopic constant) Footnotes 1 pK w = –log 10 (K w )
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Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/10%3A_Electrochemistry/10.01%3A_Electricity
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Electricity has been known for some time. Ancient Egyptians, for example, referred to electric fish in the Nile River as early as 2750 BC (Moller & Kramer, 1991). In 1600, William Gilbert studied what would later be seen to be electrostatic attraction, by creating static charges rubbing amber (Stewart, 2001). And Benjamin Franklin’s famous experiment (although it is actually uncertain if he performed the experiment) of attaching a metal key to a kite string occurred in 1752, and showed that lightening is an electrical phenomenon (Uman, 1987). One of the biggest breakthroughs in the study of electricity as a chemical phenomenon was made by Alessandro Volta, who in 1799 showed that electricity could be generated by stacking copper and zinc disks submerged in sulfuric acid (Routledge, 1881). The reactions that Volta produced in his voltaic pile included both oxidation and reduction processes that could be considered as half-reactions . The half-reactions can be classified as oxidation (the loss of electrons) which happens at the anode and reduction (the gain of electrons) which occurs at the cathode . Those half reactions were \[\underbrace{Zn \rightarrow Zn^{2+} + 2 e^-}_{\text{aanode}} \nonumber \] \[\underbrace{2 H^+ + 2 e^- \rightarrow H_2}_{\text{cathode}} \nonumber \] The propensity of zinc to oxidize coupled with that of hydrogen to reduce creates a potential energy difference between the electrodes at which these processes occur. And like any potential energy difference, it can create a force which can be used to do work. In this case, the work is that of pushing electrons through a circuit. The work of such a process can be calculated by integrating \[ dw_e - -E \,dQ \nonumber \] where \(E\) is the potential energy difference, and \(dQ\) is an infinitesimal amount of charge carried through the circuit. The infinitesimal amount of charge carried through the circuit can be expressed as \[dQ = e\,dN \nonumber \] where \(e\) is the charge carried on one electron (\(1.6 \times 10^{-19} C\)) and \(dN\) is the infinitesimal change in the number of electrons. Thus, if the potential energy difference is constant \[w_e = -e\,E \int_o^{N} dN = -N\,e\,E \nonumber \] But since the number of electrons carried through a circuit is an enormous number, it would be far more convenient to express this in terms of the number of moles of electrons carried through the circuit. Noting that the number of moles (\(n\)) is given by \[n=\dfrac{N}{N_A} \nonumber \] and that the charge carried by one mole of electrons is given by \[ F = N_A e = 96484\,C \nonumber \] where \(F\) is Faraday's constant and has the magnitude of one Faraday (or the total charge carried by one mole of electrons.) The Faraday is named after Michael Faraday (1791-1867) (Doc, 2014), a British physicist who is credited with inventing the electric motor, among other accomplishments. Putting the pieces together, the total electrical work accomplished by pushing n moles of electrons through a circuit with a potential difference \(E\), is \[w_e = -nFE \nonumber \]
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Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Chirality/Meso_Compounds
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Meso compounds are achiral compounds that has multiple chiral centers. It is superimposed on its mirror image and is optically inactive despite its stereocenters. Introduction In general, a meso compound should contain two or more identical substituted stereocenters. Also, it has an internal symmetry plane that divides the compound in half. These two halves reflect each other by the internal mirror. The stereochemistry of stereocenters should "cancel out". What it means here is that when we have an internal plane that splits the compound into two symmetrical sides, the stereochemistry of both left and right side should be opposite to each other, and therefore, result in optically inactive . Cyclic compounds may also be meso. Identification If A is a meso compound, it should have two or more stereocenters, an internal plane, and the stereochemistry should be R and S . Look for an internal plane, or internal mirror, that lies in between the compound. The stereochemistry (e.g. R or S) is very crucial in determining whether it is a meso compound or not. As mentioned above, a meso compound is optically inactive, so their stereochemistry should cancel out. For instance, R cancels S out in a meso compound with two stereocenters. trans -1,2-dichloro-1,2-ethanediol (meso) -2,3-dibromobutane Tips: An interesting thing about single bonds or sp 3 -orbitals is that we can rotate the substituted groups that attached to a stereocenter around to recognize the internal plane. As the molecule is rotated, its stereochemistry does not change. For example: Another case is when we rotate the whole molecule by 180 degree. Both molecules below are still meso. Remember the internal plane here is depicted on two dimensions. However, in reality, it is three dimensions, so be aware of it when we identify the internal mirror. Example This molecule has a plane of symmetry (the horizontal plane going through the red broken line) and, therefore, is achiral; However, it has two chiral carbons and is consequentially a meso compound. Example 2 This molecules has a plane of symmetry (the vertical plane going through the red broken line perpendicular to the plane of the ring) and, therefore, is achiral, but has has two chiral centers. Thus, its is a meso compound. Other Examples of meso compounds Meso compounds can exist in many different forms such as pentane, butane, heptane, and even cyclobutane. They do not necessarily have to be two stereocenters, but can have more. Optical Activity Analysis When the optical activity of a meso compound is attempted to be determined with a polarimeter, the indicator will not show (+) or (-). It simply means there is no certain direction of rotation of the polarized light, neither levorotatory (-) and dexorotatory (+). Problems Beside meso, there are also other types of molecules: enantiomer, diastereomer , and identical. Determine if the following molecules are meso. Answer key: A C, D, E are meso compounds. References Vollhardt, K. P.C. & Shore, N. (2007). Organic Chemistry (5 th Ed.). New York: W. H. Freeman. (190-192) Shore, N. (2007). Study Guide and Solutions Manual for Organic Chemistry (5th Ed.). New York: W.H. Freeman. (70-80) Contributors Duy Dang Gamini Gunawardena from the OChemPal site ( Utah Valley University )
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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_1/08%3A_Mechanisms_of_Glycolysis/8.05%3A_Catalysis_in_Phase_One
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Reactions in biochemistry are usually catalysed by enzymes. In a catalysed reaction, an alternative pathway is available that makes it easier to get from reactants to products. That doesn't mean that there are fewer steps. In fact, normally there are more steps in a catalysed reaction than there are in an uncatalysed one. It does mean that the overall energy needed to traverse the catalysed barrier is lower than the energy needed to surpass the uncatalysed barrier. It's like taking the stairs up to the second floor rather than taking a running leap at the window: more steps, but overall it will save time. So far, the reactions we have seen in glycolysis are just the overall reactions. By looking at the overall reactions, we get a pretty good sense about what is happening at each stage of the pathway. We even get some sense of how those reactions might happen, because we can identify familar nucleophiles and electrophiles that appear to be involved. Here, we will take a more detailed look at the catalytic pathways taken during the first phase of glycolysis. Essentially every step of glycolysis involves catalysis, and so the reactions entail cofactors and detailed steps that we have glossed over until now. The first step, phosphorylation of glucose to afford glucose-6-phosphate, requires the consumption of ATP. During that step, the terminal hydroxy group of glucose takes up phosphate from ATP, leaving ADP. It seems like that first step should be pretty straightforward, because we think of ATP as this high-energy power source for the cell, so it must be really reactive. ATP is not quite as reactive as you might think, though. That's a good thing. If it reacted too readily, it couldn't travel around the cell at all; it would get hydrolysed the first time it encountered a water molecule, and there really are an awful lot of those in a typical cell. In order to react, the ATP needs to be activated. Part of the catalysis of the phosphorylation of glucose simply involves binding ATP to a magnesium ion. Once bound to the magnesium ion, the ATP becomes more electrophilic, because of that positive charge on the magnesium ion. Although a nucleophile, such as water, is unlikely to donate to ATP -- partly because of the negative charge on the ATP -- it is more likely to donate to ATP once coordination takes place, because the magnesium ion has leveled out that negative charge. Another aspect of catalysis in the phosphorylation of glucose involves the removal of a proton. A hydroxyl group is converted to a phosphate, and a proton is lost. Acid-base catalysis is quite common in biochemistry. There are only a handful of amino acids that commonly participate in deprotonation steps: aspartate, glutamate, lysine, and histidine. All of these residues have two structures in equilibrium: a protonated one and a non-protonated one. The non-protonated form is ready to remove a proton when needed. Similarly, acid-base catalysis is carried out by nearby amino acid residues in the active site of the enzyme that carries out the isomerisation of glucose-6-phosphate to fructose-6-phosphate. Phosphoglucoisomerase accomplishes this task by removing a proton from an alpha position, and also from an O-H group, as well as donation of protons to a different alpha position and a different oxygen. An enol intermediate is the halfway point between the two isomers. If the proton were to be removed from that end hydroxy group, and a proton put back in the alpha position again, the structure would return to the original G6P. If, instead, a proton is removed from the second hydroxyl group along the chain, a different structure results. Of course, the same sort of catalytic requirements arise again during the conversion of F6P to FBP. ATP must be activated by magnesium, and proton transfers must be carried out by acidic and basic amino acid residues. The magnesium will bind to the ATP in order to reduce the amount of negative charge. That way, the nucleophilic alcohol can donate electrons more easily. The proton from the alcohol group can be removed by an amino acid side chain, such as the negatively charged carboxylate of aspartate or glutamate, or a neutral histidine. A completely different kind of catalysis occurs during the scission step of phase one, when the six-carbon sugar is cleaved into a pair of three-carbon sugars. You may recall that this cleavage is accomplished via a retro-aldol reaction: an aldol reaction goes into reverse, spitting out an enolate or enol and a carbonyl. That retro-aldol step is accomplished via iminium ion catalysis. Very often in biochemical reactions, a lysine residue binds with a carbonyl to form either an iminium ion, containing an electrophilic C=N bond, or an enamine, with a nucleophilic N-C=C unit. The process begins with donation of the lysine, in its non-protonated form, to the carbonyl of FBP. Frequently, enzymatic reactions are presented in a more condensed form. Multiple steps are shown at once; but not just any steps. In the example below, the FBP reacts with both the lysine and the aspartate at the same time. We never draw three molecules coming together at once, because the probability of three molecules colliding at the same time is just about nil. In this case, that's not a problem; the aspartate and lysine are both part of the same molecule. Furthermore, by this point, the FBP has already bound to the enzyme, so the whole thing is one big assembly. Subsequently, the carbinolamine undergoes displacement of water to yield the imine, C=N. Imines and enamines are roughly equivalent to carbonyls and enolates, respectively. Enamines are very good nucleophiles, just like enolates. Enolates are a little better, because of the negative charge. However, even though they are neutral nucleophiles, enamines are almost as good, because the reaction is driven by a less electronegative nitrogen atom; which is more willing to donate its electrons. Furthermore, in the environment of the cell, enamines form much more easily than enolates. That's because there isn't a whole lot of LDA or even NaOH floating around inside your cells. As a result, enamines are often employed as nucleophile in cases where you might think of using an enolate. Exercise \(\PageIndex{1}\) Provide mechanisms for the following aldol-like reactions. The imine unit isn't an inherently better electrophile than a carbonyl; after all, it contains a less polar C=N bond instead of a C=O bond. However, the nitrogen in an imine is much more basic than the oxygen in a carbonyl. It can be protonated quite easily under biological conditions. The resulting iminium ion, containing the C=N-H + unit, is an activated electrophile. Of course it reacts much more quickly than a regular carbonyl. Because carbonyls can easily form either imines or enamines, they will often be converted into those compounds in order to do reactions. This process is called enamine and imine catalysis. Aldol reactions are actually reversible. A retro-aldol reaction is just the aldol reaction going in reverse. In that case, an enolate or an enol or an enamine might come out as a leaving group, rather than acting as the initial nucleophile. Exercise \(\PageIndex{2}\) Provide mechanisms for the following retro-aldol reactions. In the context of a retro-aldol reaction, we need to think about the catalysis backwards. Instead of an iminium ion acting as an activated intermediate to receive a nucleophile, it is accepting electrons to form a leaving group. Instead of having an enolate leaving group in the retro-aldol reaction, we have an enamine leaving group. The lysine group, which just came on board to catalyse the reaction, is liberated by addition of a water molecule. There is one last reaction in phase one, the fifth overall in glycolysis. It's the conversion of DHAP to G3P; but that's just another keto-enol tautomerism. The catalytic mechanisms will be very much like those seen in the conversion of G6P to F6P. Basic amino acid sites such as neutral histidine and anionic aspartate can readily remove a proton, whereas the corresponding conjugates can supply a proton. Tautomerism simply requires removal of a proton from an alpha position. When a proton is returned, it goes to the other end of the double bond in the enol.
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Bookshelves/Inorganic_Chemistry/Chemistry_of_the_Main_Group_Elements_(Barron)/02%3A_Hydrogen/2.05%3A_The_Proton
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The proton, H + , is the name given to hydrogen in the +1 oxidation state. Gas phase The proton can be formed from the photolysis of atomic hydrogen in the vapor phase at low pressure. \[H^._{(g)} + h\nu \rightarrow H^+_{(g)} + e^-_{(g)}\] The proton is more reactive than the hydrogen atom because of its high charge density. In addition, the proton's small ionic radius, 1.5 x 10 -15 cm, means that it can get close to other atoms and hence form strong bonds. \[H^+_{(g)} + NR_3 \rightarrow HNR^+_{3(g)}\] The strength of the bonding interaction is such that it is very hard to measure directly. Instead the relative bond strength between the proton and an appropriate base, B 1 , is measured in the presence of a competing base, B 2 . \[B_1H^+_{(g)} + B_{2(g)} \leftrightharpoons B_2H^+_{(g)} + B_{2(g)}\] In measuring the exchange reaction, the relative proton affinity of B 1 and B 2 is measured. This is also known as the gas phase acidity, and as such it is a measure of the inherent acidity of a species X-H because it obviates any solvent effects. Liquid and solution The high reactivity of the proton means that it does not exist free in solution. There are however many H + containing species. These are generally classified as acids . \[ B_1H^+_{(sol)} + B_{2(sol)} \leftrightharpoons B_2H^+_{(sol)} + B_{1(sol)} \\ \text{acid base acid base}\] The reaction between the acid and the base is a proton transfer reaction. While the proton travels from B 1 to B 2 , it is never free in solution. Instead a bridged transition state or intermediate is formed, B 1 ... H + ... B 2 . Acidity and pH The most common solvent for H+ is water. The acid form is usually defined as the hydronium ion or H 3 O + , (2.5.5). The terms oxonium, hydroxonium and oxidanium are also used for the H 3 O + . Although we commonly use H 3 O + it is known from spectroscopy that larger complexes are formed such as H 9 O 4 + (Figure \(\PageIndex{2}\).10). \[H_2O + H^+ \rightarrow H_3O^+ \] Acids and bases have been characterized in a number of different ways. In 1680 Robert Boyle (Figure \(\PageIndex{2}\).11) defined an acid as a compound that dissolved many other compounds, had a sour taste, and reacted with alkali (base). Boyle's simple observational description was rationalized by Danish physical chemist Johannes Brønsted (Figure \(\PageIndex{2}\).12). Brønsted proposed that acids are proton donors , and bases are proton acceptors . An acid-base reaction is one in which a proton is transferred from a proton donor (acid) to a proton acceptor (base). Based upon Brønsted's proposal simple acids contain an ionizable proton. Examples of simple acids include neutral molecules (HCl, H2SO4), anions (HSO 4 - , H 2 PO 4 - ), and cations (NH 4 + ). The most common Brønsted bases include metal hydroxides (MOH). Brønsted noted that when an acid donates a proton it forms a conjugate base. The following are examples of an acid and its conjugate base. \[ H_2O \space\space \rightarrow H^+ + OH^- \\ \text{acid conjugate base}\] \[ H_2SO_4 \space\space \rightarrow H^+ + HSO^-_4 \\ \text{acid conjugate base}\] \[NH^+_4 \space\space\space\space \rightarrow H^+ + NH_3 \\ \text{acid conjugate base}\] Exercise \(\PageIndex{1}\)\ What is the conjugate base of HCl? Answer Cl - Exercise \(\PageIndex{2}\)\ What is the conjugate base of HSO 4 - Answer SO 4 2- Exercise \(\PageIndex{3}\)\ What is the conjugate base of [Al(H 2 O) 6 ] 3+ Answer [Al(H 2 O) 5 (OH)] 2+ The same occurs when a base accepts a proton it forms a conjugate acid. The following are examples of a base and its conjugate acid. \[ H_2O \space\space + \space\space H^+ \rightarrow H_3O^+ \\ \text{base conjugate acid}\] \[HCO^-_3 \space\space + \space\space H^+ \rightarrow H_2CO_3 \\ \text{base conjugate acid}\] \[F^- \space\space + \space\space H^+ \rightarrow HF \\ \text{base conjugate acid}\] Exercise \(\PageIndex{4}\)\ What is the conjugate acid of NH 3 ? Answer NH 4 + Exercise \(\PageIndex{5}\)\ What is the conjugate acid of S 2- ? Answer HS - Exercise \(\PageIndex{6}\)\ What is the conjugate acid of CO 3 2- ? Answer HCO 3 - Thus, the reaction between an acid and a base results in the formation of the appropriate conjugate base and conjugate acid. \[acid_1 + base_1 \leftrightharpoons acid_2 + base_2 \] A specific example is as follows: \[HNO_3 + NH_3 \leftrightharpoons NH^+_4 + NO^-_3 \\ acid_1 \space\space\space\space base_1 \space\space\space\space\space\space\space acid_2 \space\space\space\space base_2\] What is the conjugate acid and base formed from the reaction of NH 4 + with S 2 + ? Answer \[NH_4^+ + S^{2-} \leftrightharpoons HS^- + NH_3 \\ acid_1 \space\space\space\space base_1 \space\space\space\space\space\space\space acid_2 \space\space\space\space base_2\] In the equilibrium reactions shown in (2.12) and (2.13) there is a competition between the two bases for the proton. As would be expected the strongest base wins. When a strong acid is added to (dissolved in) water it will react with the water as a base: \[HCl + H_2O \leftrightharpoons H_3O^+ + Cl^- \\ acid \space\space\space\space base \space\space\space\space\space\space\space acid \space\space\space\space base\] In contrast, when a strong base is added to (dissolved in) water it will react with the water as an acid: \[ H_2O + NH_3 \leftrightharpoons NH_4^+ + OH^- \\ acid \space\space\space\space base \space\space\space\space\space\space\space acid \space\space\space\space base\] pH a measure of acidity The acidity of a water (aqueous) solution depends on the concentration of the hydronium ion, i.e., [H 3 O + ]. The acidity of a solution is therefore the ability of the solution to donate a proton to a base. The acidity or pH of a solution is defined as: \[pH = -log [H_3O^+_a] \] It is important to note that the value is the activity of H 3 O + and not the concentration . Activity is a measure of the eective concentration of a species in a mixture. The dierence between activity and other measures of composition such as concentration arises because molecules in non-ideal gases or solutions interact with each other, either to attract or to repel each other. The activity of the H 3 O + ion can be measured by (a) A gas electrode (b) Acid-base indicators Proton Transfer Reactions The proton transfer reaction is one of the simplest reactions in chemistry. It involves no electrons and low mass transfer/change, giving it a low energy of activation. For proton transfer between O-H or N-H groups and their associated bases the reaction is very fast. The proton transfer occurs across a hydrogen-bonded pathway during which the proton is never free.
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Courses/University_of_Kentucky/UK%3A_General_Chemistry/11%3A_Solutions_and_Colloids/11.E%3A_Solutions_and_Colloids_(Exercises)
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11.2: The Dissolution Process Q11.2.1 How do solutions differ from compounds? From other mixtures? S11.2.1 A solution can vary in composition, while a compound cannot vary in composition. Solutions are homogeneous at the molecular level, while other mixtures are heterogeneous. Q11.2.2 Which of the principal characteristics of solutions can we see in the solutions of \(\ce{K2Cr2O7}\) shown in Figure: When potassium dichromate (\(\ce{K2Cr2O7}\)) is mixed with water, it forms a homogeneous orange solution. (credit: modification of work by Mark Ott) S11.2.2 The solutions are the same throughout (the color is constant throughout), and the composition of a solution of K 2 Cr 2 O 7 in water can vary. Q11.2.3 When KNO 3 is dissolved in water, the resulting solution is significantly colder than the water was originally. Is the dissolution of KNO 3 an endothermic or an exothermic process? What conclusions can you draw about the intermolecular attractions involved in the process? Is the resulting solution an ideal solution? S11.2.3 (a) The process is endothermic as the solution is consuming heat. (b) Attraction between the K + and \(\ce{NO3-}\) ions is stronger than between the ions and water molecules (the ion-ion interactions have a lower, more negative energy). Therefore, the dissolution process increases the energy of the molecular interactions, and it consumes the thermal energy of the solution to make up for the difference. (c) No, an ideal solution is formed with no appreciable heat release or consumption. Q11.2.4 Give an example of each of the following types of solutions: a gas in a liquid a gas in a gas a solid in a solid S11.2.4 (a) CO 2 in water; (b) O 2 in N 2 (air); (c) bronze (solution of tin or other metals in copper) Q11.2.5 Indicate the most important types of intermolecular attractions in each of the following solutions: The solution in Figure . NO( l ) in CO( l ) Cl 2 ( g ) in Br 2 ( l ) HCl( aq ) in benzene C 6 H 6 ( l ) Methanol CH 3 OH( l ) in H 2 O( l ) S11.2.5 (a) ion-dipole forces; (b) dipole-dipole forces; (c) dispersion forces; (d) dispersion forces; (e) hydrogen bonding Q11.2.5 Predict whether each of the following substances would be more soluble in water (polar solvent) or in a hydrocarbon such as heptane (C 7 H 16 , nonpolar solvent): vegetable oil (nonpolar) isopropyl alcohol (polar) potassium bromide (ionic) S11.2.5 (a) heptane; (b) water; (c) water Q11.2.6 Heat is released when some solutions form; heat is absorbed when other solutions form. Provide a molecular explanation for the difference between these two types of spontaneous processes. S11.2.6 Heat is released when the total intermolecular forces (IMFs) between the solute and solvent molecules are stronger than the total IMFs in the pure solute and in the pure solvent: Breaking weaker IMFs and forming stronger IMFs releases heat. Heat is absorbed when the total IMFs in the solution are weaker than the total of those in the pure solute and in the pure solvent: Breaking stronger IMFs and forming weaker IMFs absorbs heat. Q11.2.7 Solutions of hydrogen in palladium may be formed by exposing Pd metal to H 2 gas. The concentration of hydrogen in the palladium depends on the pressure of H 2 gas applied, but in a more complex fashion than can be described by Henry’s law. Under certain conditions, 0.94 g of hydrogen gas is dissolved in 215 g of palladium metal. Determine the molarity of this solution (solution density = 1.8 g/cm 3 ). Determine the molality of this solution (solution density = 1.8 g/cm 3 ). Determine the percent by mass of hydrogen atoms in this solution (solution density = 1.8 g/cm 3 ). S11.2.7 http://cnx.org/contents/ mH6aqegx @2/The-Dissolution-Process 11.3: Electrolytes Q11.3.1 Explain why the ions Na + and Cl − are strongly solvated in water but not in hexane, a solvent composed of nonpolar molecules. S11.3.2 Crystals of NaCl dissolve in water, a polar liquid with a very large dipole moment, and the individual ions become strongly solvated. Hexane is a nonpolar liquid with a dipole moment of zero and, therefore, does not significantly interact with the ions of the NaCl crystals. Q11.3.2 Explain why solutions of HBr in benzene (a nonpolar solvent) are nonconductive, while solutions in water (a polar solvent) are conductive. S11.3.2 HBr is an acid and so its molecules react with water molecules to form H 3 O + and Br − ions that provide conductivity to the solution. Though HBr is soluble in benzene, it does not react chemically but remains dissolved as neutral HBr molecules. With no ions present in the benzene solution, it is electrically nonconductive. Q11.3.3 Consider the solutions presented: (a) Which of the following sketches best represents the ions in a solution of Fe(NO 3 ) 3 ( aq )? (b) Write a balanced chemical equation showing the products of the dissolution of Fe(NO 3 ) 3 . S11.3.3 (a) Fe(NO 3 ) 3 is a strong electrolyte, thus it should completely dissociate into Fe 3+ and \(\ce{(NO3- )}\) ions. Therefore, (z) best represents the solution. (b) \(\ce{Fe(NO3)3}(s)⟶\ce{Fe^3+}(aq)+\ce{3NO3- }(aq)\) Q11.3.4 Compare the processes that occur when methanol (CH 3 OH), hydrogen chloride (HCl), and sodium hydroxide (NaOH) dissolve in water. Write equations and prepare sketches showing the form in which each of these compounds is present in its respective solution. S11.3.3 Methanol, \(CH_3OH\), dissolves in water in all proportions, interacting via hydrogen bonding. Methanol: \[CH_3OH_{(l)}+H_2O_{(l)}⟶CH_3OH_{(aq)}\] Hydrogen chloride, HCl, dissolves in and reacts with water to yield hydronium cations and chloride anions that are solvated by strong ion-dipole interactions. Hydrogen chloride: \[HCl{(g)}+H_2O_{(l)} \rightarrow H_3O^+_{(aq)}+Cl^−_{(aq)}\] Sodium hydroxide, NaOH, dissolves in water and dissociates to yield sodium cations and hydroxide anions that are strongly solvated by ion-dipole interactions and hydrogen bonding, respectively. Sodium hydroxide: \[NaOH_{(s)} \rightarrow Na^+_{(aq)} + OH^−_{(aq)}\] Q11.3.5 What is the expected electrical conductivity of the following solutions? NaOH( aq ) HCl( aq ) C 6 H 12 O 6 ( aq ) (glucose) NH 3 ( l ) S11.3.5 (a) high conductivity (solute is an ionic compound that will dissociate when dissolved); (b) high conductivity (solute is a strong acid and will ionize completely when dissolved); (c) nonconductive (solute is a covalent compound, neither acid nor base, unreactive towards water); (d) low conductivity (solute is a weak base and will partially ionize when dissolved) Q11.3.6 Why are most solid ionic compounds electrically nonconductive, whereas aqueous solutions of ionic compounds are good conductors? Would you expect a liquid (molten) ionic compound to be electrically conductive or nonconductive? Explain. S11.3.6 A medium must contain freely mobile, charged entities to be electrically conductive. The ions present in a typical ionic solid are immobilized in a crystalline lattice and so the solid is not able to support an electrical current. When the ions are mobilized, either by melting the solid or dissolving it in water to dissociate the ions, current may flow and these forms of the ionic compound are conductive. Q11.3.7 Indicate the most important type of intermolecular attraction responsible for solvation in each of the following solutions: the solutions in Figure methanol, CH 3 OH, dissolved in ethanol, C 2 H 5 OH methane, CH 4 , dissolved in benzene, C 6 H 6 the polar halocarbon CF 2 Cl 2 dissolved in the polar halocarbon CF 2 ClCFCl 2 O 2 ( l ) in N 2 ( l ) S11.3.7 (a) ion-dipole; (b) hydrogen bonds; (c) dispersion forces; (d) dipole-dipole attractions; (e) dispersion forces 11.4: Solubility Q11.4.1 Suppose you are presented with a clear solution of sodium thiosulfate, Na 2 S 2 O 3 . How could you determine whether the solution is unsaturated, saturated, or supersaturated? S11.4.1 Add a small crystal of \(Na_2S_2O_3\). It will dissolve in an unsaturated solution, remain apparently unchanged in a saturated solution, or initiate precipitation in a supersaturated solution. Q11.4.2 Supersaturated solutions of most solids in water are prepared by cooling saturated solutions. Supersaturated solutions of most gases in water are prepared by heating saturated solutions. Explain the reasons for the difference in the two procedures. S11.4.2 The solubility of solids usually decreases upon cooling a solution, while the solubility of gases usually decreases upon heating. Q11.4.3 Suggest an explanation for the observations that ethanol, C 2 H 5 OH, is completely miscible with water and that ethanethiol, C 2 H 5 SH, is soluble only to the extent of 1.5 g per 100 mL of water. S11.4.3 The hydrogen bonds between water and C 2 H 5 OH are much stronger than the intermolecular attractions between water and C 2 H 5 SH. Q11.4.4 Calculate the percent by mass of KBr in a saturated solution of KBr in water at 10 °C using the following figure for useful data, and report the computed percentage to one significant digit. This graph shows how the solubility of several solids changes with temperature. S11.4.4 At 10 °C, the solubility of KBr in water is approximately 60 g per 100 g of water. \[\%\; KBr =\dfrac{60\; g\; KBr}{(60+100)\;g\; solution} = 40\%\] Q11.4.5 Which of the following gases is expected to be most soluble in water? Explain your reasoning. CH 4 CCl 4 CHCl 3 S11.4.5 (c) CHCl 3 is expected to be most soluble in water. Of the three gases, only this one is polar and thus capable of experiencing relatively strong dipole-dipole attraction to water molecules. Q11.4.6 At 0 °C and 1.00 atm, as much as 0.70 g of O 2 can dissolve in 1 L of water. At 0 °C and 4.00 atm, how many grams of O 2 dissolve in 1 L of water? S11.4.6 This problem requires the application of Henry’s law. The governing equation is \(C_g = kP_g\). \[k=\dfrac{C_g}{P_g}=\dfrac{0.70\;g}{1.00\; atm} =0.70\;g\; atm^{−1}\] Under the new conditions, \(C_g=0.70\;g\;atm^{−1} \times 4.00\; atm = 2.80\; g\). Q11.4.7 Refer to following figure for the following three questions: How did the concentration of dissolved CO 2 in the beverage change when the bottle was opened? What caused this change? Is the beverage unsaturated, saturated, or supersaturated with CO 2 ? S11.4.7 (a) It decreased as some of the CO 2 gas left the solution (evidenced by effervescence). (b) Opening the bottle released the high-pressure CO 2 gas above the beverage. The reduced CO 2 gas pressure, per Henry’s law, lowers the solubility for CO 2 . (c) The dissolved CO 2 concentration will continue to slowly decrease until equilibrium is reestablished between the beverage and the very low CO 2 gas pressure in the opened bottle. Immediately after opening, the beverage, therefore, contains dissolved CO 2 at a concentration greater than its solubility, a nonequilibrium condition, and is said to be supersaturated. Q11.4.8 The Henry’s law constant for CO 2 is \(3.4 \times 10^{−2}\; M/atm\) at 25 °C. What pressure of carbon dioxide is needed to maintain a CO 2 concentration of 0.10 M in a can of lemon-lime soda? S11.4.8 \[P_g=\dfrac{C_g}{k}=\dfrac{0.10\; M}{3.4 \times 10^{−2}\;M/atm} =2.9\; atm\] Q11.4.9 The Henry’s law constant for O 2 is \(1.3\times 10^{−3}\; M/atm\) at 25 °C. What mass of oxygen would be dissolved in a 40-L aquarium at 25 °C, assuming an atmospheric pressure of 1.00 atm, and that the partial pressure of O 2 is 0.21 atm? S11.4.9 Start with Henry's law \[C_g=kP_g\] and apply it to \(O_2\) \[C(O_2)=(1.3 \times 10^{−3}\; M/atm) (0.21\;atm)=2.7 \times 10^{−4}\;mol/L\] The total amount is \((2.7 \times 10^{−4}\; mol/L)(40\;L=1.08 \times 10^{−2} \;mol\] The mass of oxygen is \((1.08 \times 10^{−2}\; mol)(32.0\; g/mol)=0.346\;g\) or, using two significant figures, \(0.35\; g\). Q11.4.10 How many liters of HCl gas, measured at 30.0 °C and 745 torr, are required to prepare 1.25 L of a 3.20- M solution of hydrochloric acid? S11.4.10 First, calculate the moles of HCl needed. Then use the ideal gas law to find the volume required. M = mol L−1 3.20M=xmol1.25L x = 4.00 mol HCl Before using the ideal gas law, change pressure to atmospheres and convert temperature from °C to kelvin . \[1\;atmx = 760torr745torr x = 0.9803 atm V= nRTP =(4.000molHCl)(0.08206LatmK−1mol−1)(303.15K)0.9803atm=102 L HCl 102 L HCl more http://cnx.org/contents/ 2488fW6W @2/Solubility 11.5: Colligative Properties Q11.5.1 Which is/are part of the macroscopic domain of solutions and which is/are part of the microscopic domain: boiling point elevation, Henry’s law, hydrogen bond, ion-dipole attraction, molarity, nonelectrolyte, nonstoichiometric compound, osmosis, solvated ion? Q11.5.2 What is the microscopic explanation for the macroscopic behavior illustrated in [link] ? S11.5.2 The strength of the bonds between like molecules is stronger than the strength between unlike molecules. Therefore, some regions will exist in which the water molecules will exclude oil molecules and other regions will exist in which oil molecules will exclude water molecules, forming a heterogeneous region. Q11.5.3 Sketch a qualitative graph of the pressure versus time for water vapor above a sample of pure water and a sugar solution, as the liquids evaporate to half their original volume. Q11.5.4 A solution of potassium nitrate, an electrolyte, and a solution of glycerin (C 3 H 5 (OH) 3 ), a nonelectrolyte, both boil at 100.3 °C. What other physical properties of the two solutions are identical? S11.5.4 Both form homogeneous solutions; their boiling point elevations are the same, as are their lowering of vapor pressures. Osmotic pressure and the lowering of the freezing point are also the same for both solutions. Q11.5.5 What are the mole fractions of H 3 PO 4 and water in a solution of 14.5 g of H 3 PO 4 in 125 g of water? Q11.5.6 What are the mole fractions of HNO 3 and water in a concentrated solution of nitric acid (68.0% HNO 3 by mass)? S11.5.6 Find number of moles of HNO 3 and H 2 O in 100 g of the solution. Find the mole fractions for the components. The mole fraction of HNO 3 is 0.378. The mole fraction of H 2 O is 0.622. Q11.5.7 Calculate the mole fraction of each solute and solvent: 583 g of H 2 SO 4 in 1.50 kg of water—the acid solution used in an automobile battery 0.86 g of NaCl in 1.00 × 10 2 g of water—a solution of sodium chloride for intravenous injection 46.85 g of codeine, C 18 H 21 NO 3 , in 125.5 g of ethanol, C 2 H 5 OH 25 g of I 2 in 125 g of ethanol, C 2 H 5 OH S11.5.7 a. \(583\:g\:\ce{H2SO4}\times\dfrac{1\:mole\:\ce{H2SO4}}{98.08\:g\:\ce{H2SO4}}=5.94\:mole\:\ce{H2SO4}\) \(\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\) \(1.50\:kg\:\ce{H2O}\times\dfrac{1000\:g}{1\:kg}\times\dfrac{1\:mole\:\ce{H2O}}{18.02\:g\:\ce{H2O}}=83.2\:moles\:\ce{H2O}\) Q11.5.8 Calculate the mole fraction of each solute and solvent: 0.710 kg of sodium carbonate (washing soda), Na 2 CO 3 , in 10.0 kg of water—a saturated solution at 0 °C 125 g of NH 4 NO 3 in 275 g of water—a mixture used to make an instant ice pack 25 g of Cl 2 in 125 g of dichloromethane, CH 2 Cl 2 0.372 g of histamine, C 5 H 9 N, in 125 g of chloroform, CHCl 3 S11.5.8 \(X_\mathrm{Na_2CO_3}=0.0119\); \(X_\mathrm{H_2O}=0.988\); \(X_\mathrm{NH_4NO_3}=0.9927\); \(X_\mathrm{H_2O}=0.907\); \(X_\mathrm{Cl_2}=0.192\); \(X_\mathrm{CH_2CI_2}=0.808\); \(X_\mathrm{C_5H_9N}=0.00426\); \(X_\mathrm{CHCl_3}=0.997\) Q11.5.9 Calculate the mole fractions of methanol, CH 3 OH; ethanol, C 2 H 5 OH; and water in a solution that is 40% methanol, 40% ethanol, and 20% water by mass. (Assume the data are good to two significant figures.) Q11.5.10 What is the difference between a 1 M solution and a 1 m solution? S11.5.10 In a 1 M solution, the mole is contained in exactly 1 L of solution. In a 1 m solution, the mole is contained in exactly 1 kg of solvent. Q11.5.11 What is the molality of phosphoric acid, H 3 PO 4 , in a solution of 14.5 g of H 3 PO 4 in 125 g of water? Q11.5.12 What is the molality of nitric acid in a concentrated solution of nitric acid (68.0% HNO 3 by mass)? S11.5.12 (a) Determine the molar mass of HNO 3 . Determine the number of moles of acid in the solution. From the number of moles and the mass of solvent, determine the molality. (b) 33.7 m Q11.5.13 Calculate the molality of each of the following solutions: 583 g of H 2 SO 4 in 1.50 kg of water—the acid solution used in an automobile battery 0.86 g of NaCl in 1.00 × 10 2 g of water—a solution of sodium chloride for intravenous injection 46.85 g of codeine, C 18 H 21 NO 3 , in 125.5 g of ethanol, C 2 H 5 OH 25 g of I 2 in 125 g of ethanol, C 2 H 5 OH Q11.5.14 Calculate the molality of each of the following solutions: 0.710 kg of sodium carbonate (washing soda), Na 2 CO 3 , in 10.0 kg of water—a saturated solution at 0°C 125 g of NH 4 NO 3 in 275 g of water—a mixture used to make an instant ice pack 25 g of Cl 2 in 125 g of dichloromethane, CH 2 Cl 2 0.372 g of histamine, C 5 H 9 N, in 125 g of chloroform, CHCl 3 S11.5.14 (a) 6.70 × 10 −1 m ; (b) 5.67 m ; (c) 2.8 m ; (d) 0.0358 m Q11.5.15 The concentration of glucose, C 6 H 12 O 6 , in normal spinal fluid is \(\mathrm{\dfrac{75\:mg}{100\:g}}\). What is the molality of the solution? Q11.5.16 A 13.0% solution of K 2 CO 3 by mass has a density of 1.09 g/cm 3 . Calculate the molality of the solution. S11.5.16 1.08 m Q11.5.17 Why does 1 mol of sodium chloride depress the freezing point of 1 kg of water almost twice as much as 1 mol of glycerin? What is the boiling point of a solution of 115.0 g of sucrose, C 12 H 22 O 11 , in 350.0 g of water? S11.5.17 Determine the molar mass of sucrose; determine the number of moles of sucrose in the solution; convert the mass of solvent to units of kilograms; from the number of moles and the mass of solvent, determine the molality; determine the difference between the boiling point of water and the boiling point of the solution; determine the new boiling point. 100.5 °C Q11.5.18 What is the boiling point of a solution of 9.04 g of I 2 in 75.5 g of benzene, assuming the I 2 is nonvolatile? Q11.5.19 What is the freezing temperature of a solution of 115.0 g of sucrose, C 12 H 22 O 11 , in 350.0 g of water, which freezes at 0.0 °C when pure? S11.5.19 (a) Determine the molar mass of sucrose; determine the number of moles of sucrose in the solution; convert the mass of solvent to units of kilograms; from the number of moles and the mass of solvent, determine the molality; determine the difference between the freezing temperature of water and the freezing temperature of the solution; determine the new freezing temperature. (b) −1.8 °C Q11.5.20 What is the freezing point of a solution of 9.04 g of I 2 in 75.5 g of benzene? Q11.5.21 What is the osmotic pressure of an aqueous solution of 1.64 g of Ca(NO 3 ) 2 in water at 25 °C? The volume of the solution is 275 mL. S11.5.21 (a) Determine the molar mass of Ca(NO 3 ) 2 ; determine the number of moles of Ca(NO 3 ) 2 in the solution; determine the number of moles of ions in the solution; determine the molarity of ions, then the osmotic pressure. (b) 2.67 atm Q11.5.22 What is osmotic pressure of a solution of bovine insulin (molar mass, 5700 g mol −1 ) at 18 °C if 100.0 mL of the solution contains 0.103 g of the insulin? Q11.5.23 What is the molar mass of a solution of 5.00 g of a compound in 25.00 g of carbon tetrachloride (bp 76.8 °C; K b = 5.02 °C/ m ) that boils at 81.5 °C at 1 atm? S11.5.24 (a) Determine the molal concentration from the change in boiling point and K b ; determine the moles of solute in the solution from the molal concentration and mass of solvent; determine the molar mass from the number of moles and the mass of solute. (b) 2.1 × 10 2 g mol −1 Q11.5.25 A sample of an organic compound (a nonelectrolyte) weighing 1.35 g lowered the freezing point of 10.0 g of benzene by 3.66 °C. Calculate the molar mass of the compound. Q11.5.26 A 1.0 m solution of HCl in benzene has a freezing point of 0.4 °C. Is HCl an electrolyte in benzene? Explain. S11.5.26 No. Pure benzene freezes at 5.5 °C, and so the observed freezing point of this solution is depressed by Δ T f = 5.5 − 0.4 = 5.1 °C. The value computed, assuming no ionization of HCl, is Δ T f = (1.0 m)(5.14 °C/ m ) = 5.1 °C. Agreement of these values supports the assumption that HCl is not ionized. Q11.5.27 A solution contains 5.00 g of urea, CO(NH 2 ) 2 , a nonvolatile compound, dissolved in 0.100 kg of water. If the vapor pressure of pure water at 25 °C is 23.7 torr, what is the vapor pressure of the solution? Q11.5.28 A 12.0-g sample of a nonelectrolyte is dissolved in 80.0 g of water. The solution freezes at −1.94 °C. Calculate the molar mass of the substance. S11.5.28 144 g mol −1 Q11.5.29 Arrange the following solutions in order by their decreasing freezing points: 0.1 m Na 3 PO 4 , 0.1 m C 2 H 5 OH, 0.01 m CO 2 , 0.15 m NaCl, and 0.2 m CaCl 2 . Q11.5.30 Calculate the boiling point elevation of 0.100 kg of water containing 0.010 mol of NaCl, 0.020 mol of Na 2 SO 4 , and 0.030 mol of MgCl 2 , assuming complete dissociation of these electrolytes. S11.5.30 0.870 °C Q11.5.31 How could you prepare a 3.08 m aqueous solution of glycerin, C 3 H 8 O 3 ? What is the freezing point of this solution? Q11.5.32 A sample of sulfur weighing 0.210 g was dissolved in 17.8 g of carbon disulfide, CS 2 ( K b = 2.43 °C/ m ). If the boiling point elevation was 0.107 °C, what is the formula of a sulfur molecule in carbon disulfide? S11.5.32 S 8 Q11.5.33 In a significant experiment performed many years ago, 5.6977 g of cadmium iodide in 44.69 g of water raised the boiling point 0.181 °C. What does this suggest about the nature of a solution of CdI 2 ? Q11.5.34 Lysozyme is an enzyme that cleaves cell walls. A 0.100-L sample of a solution of lysozyme that contains 0.0750 g of the enzyme exhibits an osmotic pressure of 1.32 × 10 −3 atm at 25 °C. What is the molar mass of lysozyme? S11.5.34 1.39 × 10 4 g mol −1 Q11.5.35 The osmotic pressure of a solution containing 7.0 g of insulin per liter is 23 torr at 25 °C. What is the molar mass of insulin? Q11.5.36 The osmotic pressure of human blood is 7.6 atm at 37 °C. What mass of glucose, C 6 H 12 O 6 , is required to make 1.00 L of aqueous solution for intravenous feeding if the solution must have the same osmotic pressure as blood at body temperature, 37 °C? S11.5.36 54 g Q11.5.37 What is the freezing point of a solution of dibromobenzene, C 6 H 4 Br 2 , in 0.250 kg of benzene, if the solution boils at 83.5 °C? Q11.5.38 What is the boiling point of a solution of NaCl in water if the solution freezes at −0.93 °C? S11.5.38 100.26 °C Q11.5.39 The sugar fructose contains 40.0% C, 6.7% H, and 53.3% O by mass. A solution of 11.7 g of fructose in 325 g of ethanol has a boiling point of 78.59 °C. The boiling point of ethanol is 78.35 °C, and K b for ethanol is 1.20 °C/ m . What is the molecular formula of fructose? Q11.5.40 The vapor pressure of methanol, CH 3 OH, is 94 torr at 20 °C. The vapor pressure of ethanol, C 2 H 5 OH, is 44 torr at the same temperature. Calculate the mole fraction of methanol and of ethanol in a solution of 50.0 g of methanol and 50.0 g of ethanol. Ethanol and methanol form a solution that behaves like an ideal solution. Calculate the vapor pressure of methanol and of ethanol above the solution at 20 °C. Calculate the mole fraction of methanol and of ethanol in the vapor above the solution. S11.5.40 (a) \(X_\mathrm{CH_3OH}=0.590\); \(X_\mathrm{C_2H_5OH}=0.410\); (b) Vapor pressures are: CH 3 OH: 55 torr; C 2 H 5 OH: 18 torr; (c) CH 3 OH: 0.75; C 2 H 5 OH: 0.25 Q11.5.41 The triple point of air-free water is defined as 273.15 K. Why is it important that the water be free of air? Q11.5.42 Meat can be classified as fresh (not frozen) even though it is stored at −1 °C. Why wouldn’t meat freeze at this temperature? S11.5.42 The ions and compounds present in the water in the beef lower the freezing point of the beef below −1 °C. Q11.5.43 An organic compound has a composition of 93.46% C and 6.54% H by mass. A solution of 0.090 g of this compound in 1.10 g of camphor melts at 158.4 °C. The melting point of pure camphor is 178.4 °C. K f for camphor is 37.7 °C/ m . What is the molecular formula of the solute? Show your calculations. Q11.5.44 A sample of HgCl 2 weighing 9.41 g is dissolved in 32.75 g of ethanol, C 2 H 5 OH ( K b = 1.20 °C/ m ). The boiling point elevation of the solution is 1.27 °C. Is HgCl 2 an electrolyte in ethanol? Show your calculations. S11.5.44 \(\mathrm{Δbp}=K_\ce{b}m=(1.20\:°\ce C/m)\mathrm{\left(\dfrac{9.41\:g×\dfrac{1\:mol\: HgCl_2}{271.496\:g}}{0.03275\:kg}\right)=1.27\:°\ce C}\) The observed change equals the theoretical change; therefore, no dissociation occurs. Q11.5.45 A salt is known to be an alkali metal fluoride. A quick approximate determination of freezing point indicates that 4 g of the salt dissolved in 100 g of water produces a solution that freezes at about −1.4 °C. What is the formula of the salt? Show your calculations. 11.6: Colloids Q11.6.1 Identify the dispersed phase and the dispersion medium in each of the following colloidal systems: starch dispersion, smoke, fog, pearl, whipped cream, floating soap, jelly, milk, and ruby. S11.6.1 Colloidal System Dispersed Phase Dispersion Medium starch dispersion starch water smoke solid particles air fog water air pearl water calcium carbonate (CaCO3) whipped cream air cream floating soap air soap jelly fruit juice pectin gel milk butterfat water ruby chromium(III) oxide (Cr2O3) aluminum oxide (Al2O3) Q11.6.2 Distinguish between dispersion methods and condensation methods for preparing colloidal systems. S11.6.2 Dispersion methods use a grinding device or some other means to bring about the subdivision of larger particles. Condensation methods bring smaller units together to form a larger unit. For example, water molecules in the vapor state come together to form very small droplets that we see as clouds. Q11.6.3 How do colloids differ from solutions with regard to dispersed particle size and homogeneity? S11.6.3 Colloidal dispersions consist of particles that are much bigger than the solutes of typical solutions. Colloidal particles are either very large molecules or aggregates of smaller species that usually are big enough to scatter light. Colloids are homogeneous on a macroscopic (visual) scale, while solutions are homogeneous on a microscopic (molecular) scale. Q11.6.4 Explain the cleansing action of soap. S11.6.4 Soap molecules have both a hydrophobic and a hydrophilic end. The charged (hydrophilic) end, which is usually associated with an alkali metal ion, ensures water solubility The hydrophobic end permits attraction to oil, grease, and other similar nonpolar substances that normally do not dissolve in water but are pulled into the solution by the soap molecules. Q11.6.5 How can it be demonstrated that colloidal particles are electrically charged? S11.6.5 If they are placed in an electrolytic cell, dispersed particles will move toward the electrode that carries a charge opposite to their own charge. At this electrode, the charged particles will be neutralized and will coagulate as a precipitate.
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Courses/SUNY_Oneonta/Organic_Chemistry_with_a_Biological_Emphasis_(SUNY_Oneonta)/01%3A_Introduction_to_Organic_Structure_and_Bonding_I/1.04%3A_Structures_of_some_important_biomolecules/1.4.04%3A_Introduction_to_nucleic_acid_(DNA_and_RNA)_structure
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Deoxyribonucleic acid (DNA) and ribonucleic acid (RNA) are polymers composed of monomers called nucleotides . An RNA nucleotide consists of a five-carbon sugar phosphate linked to one of four nucleic acid bases : guanine (G), cytosine (C), adenine (A) and uracil (U). In a DNA nucleototide, the sugar is missing the hydroxyl group at the 2' position, and the thymine base (T) is used instead of uracil. The conventional numbering system used for DNA and RNA is shown here for reference - the prime (') symbol is used to distinguish the sugar carbon numbers from the base carbon numbers. The two 'hooks' on the RNA or DNA monomer are the 5' phosphate and the 3' hydroxyl on the sugar, which in DNA polymer synthesis are linked by a 'phosphate diester' group. By convention, DNA and RNA sequences are written in 5' to 3' direction. Back to chapter 1 main page ⇒
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Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/16%3A_Acids_and_Bases/16.7%3A_Ions_as_Acids_and_Bases
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Learning Objectives Extend previously introduced equilibrium concepts to acids and bases that may donate or accept more than one proton We can also use the relative strengths of conjugate acid–base pairs to understand the acid–base properties of solutions of salts. A neutralization reaction can be defined as the reaction of an acid and a base to produce a salt and water. That is, another cation, such as \(Na^+\), replaces the proton on the acid. An example is the reaction of \(\ce{CH3CO2H}\), a weak acid, with \(\ce{NaOH}\), a strong base: \[\underset{acid}{\ce{CH3CO2H(l)}} +\underset{base}{\ce{NaOH(s)}} \overset{\ce{H2O}}{\longrightarrow} \underset{salt}{\ce{H2OCH3CO2Na(aq)} }+\underset{water}{\ce{H2O(l)}} \label{16.35}\] Depending on the acid–base properties of its component ions, however, a salt can dissolve in water to produce a neutral solution, a basic solution, or an acidic solution. When a salt such as \(NaCl\) dissolves in water, it produces \(Na^+_{(aq)}\) and \(Cl^−_{(aq)}\) ions. Using a Lewis approach, the \(Na^+\) ion can be viewed as an acid because it is an electron pair acceptor, although its low charge and relatively large radius make it a very weak acid. The \(Cl^−\) ion is the conjugate base of the strong acid \(HCl\), so it has essentially no basic character. Consequently, dissolving \(NaCl\) in water has no effect on the \(pH\) of a solution, and the solution remains neutral. Now let's compare this behavior to the behavior of aqueous solutions of potassium cyanide and sodium acetate. Again, the cations (\(K^+\) and \(Na^+\)) have essentially no acidic character, but the anions (\(CN^−\) and \(CH_3CO_2^−\)) are weak bases that can react with water because they are the conjugate bases of the weak acids \(HCN\) and acetic acid, respectively. \[ \ce{CN^{-}(aq) + H2O(l) <<=> HCN(aq) + OH^{-}(aq)}\] \[ \ce{CH3CO^{-}2(aq) + H2O(l) <<=> CH3CO2H(aq) + OH^{-}(aq)}\] Neither reaction proceeds very far to the right as written because the formation of the weaker acid–base pair is favored. Both \(HCN\) and acetic acid are stronger acids than water, and hydroxide is a stronger base than either acetate or cyanide, so in both cases, the equilibrium lies to the left. Nonetheless, each of these reactions generates enough hydroxide ions to produce a basic solution. For example, the \(pH\) of a 0.1 M solution of sodium acetate or potassium cyanide at 25°C is 8.8 or 11.1, respectively. From Table E1 and E2, we can see that \(CN^−\) is a stronger base (\(pK_b = 4.79\)) than acetate (\(pK_b = 9.24\)), which is consistent with \(KCN\) producing a more basic solution than sodium acetate at the same concentration. In contrast, the conjugate acid of a weak base should be a weak acid. For example, ammonium chloride and pyridinium chloride are salts produced by reacting ammonia and pyridine, respectively, with \(HCl\). As you already know, the chloride ion is such a weak base that it does not react with water. In contrast, the cations of the two salts are weak acids that react with water as follows: \[ \ce{NH^{+}4(aq) + H2O(l) <<=> NH3(aq) + H3O^{+}(aq)}\] \[ \ce{C5H5NH^{+}(aq) + H2O(l) <<=> C5H5NH(aq) + H3O^{+}(aq)}\] Figure \(\PageIndex{2}\) shows that \(H_3O^+\) is a stronger acid than either \(NH_4^+\) or \(C_5H_5NH^+\), and conversely, ammonia and pyridine are both stronger bases than water. The equilibrium will therefore lie far to the left in both cases, favoring the weaker acid–base pair. The \(H_3O^+\) concentration produced by the reactions is great enough, however, to decrease the \(pH\) of the solution significantly: the \(pH\) of a 0.10 M solution of ammonium chloride or pyridinium chloride at 25°C is 5.13 or 3.12, respectively. What happens with aqueous solutions of a salt such as ammonium acetate, where both the cation and the anion can react separately with water to produce an acid and a base, respectively? According to Figure 16.10, the ammonium ion will lower the \(pH\), while according to Figure 16.9, the acetate ion will raise the \(pH\). This particular case is unusual, in that the cation is as strong an acid as the anion is a base (pKa ≈ pKb). Consequently, the two effects cancel, and the solution remains neutral. With salts in which the cation is a stronger acid than the anion is a base, the final solution has a \(pH\) < 7.00. Conversely, if the cation is a weaker acid than the anion is a base, the final solution has a \(pH\) > 7.00. Ions as Acids and Bases: https://youtu.be/XYAGNonPSow Acidic Metal Ions Solutions of simple salts of metal ions can also be acidic, even though a metal ion cannot donate a proton directly to water to produce \(H_3O^+\). Instead, a metal ion can act as a Lewis acid and interact with water, a Lewis base, by coordinating to a lone pair of electrons on the oxygen atom to form a hydrated metal ion (Figure \(\PageIndex{1a}\)). A water molecule coordinated to a metal ion is more acidic than a free water molecule for two reasons. First, repulsive electrostatic interactions between the positively charged metal ion and the partially positively charged hydrogen atoms of the coordinated water molecule make it easier for the coordinated water to lose a proton. Figure \(\PageIndex{1}\) Second, the positive charge on the \(Al^{3+}\) ion attracts electron density from the oxygen atoms of the water molecules, which decreases the electron density in the \(\ce{O–H}\) bonds, as shown in Figure \(\PageIndex{1b}\). With less electron density between the \(O\) atoms and the H atoms, the \(\ce{O–H}\) bonds are weaker than in a free \(H_2O\) molecule, making it easier to lose a \(H^+\) ion. The magnitude of this effect depends on the following two factors (Figure \(\PageIndex{3}\)): The charge on the metal ion. A divalent ion (\(M^{2+}\)) has approximately twice as strong an effect on the electron density in a coordinated water molecule as a monovalent ion (\(M^+\)) of the same radius. The radius of the metal ion. For metal ions with the same charge, the smaller the ion, the shorter the internuclear distance to the oxygen atom of the water molecule and the greater the effect of the metal on the electron density distribution in the water molecule. Thus aqueous solutions of small, highly charged metal ions, such as \(Al^{3+}\) and \(Fe^{3+}\), are acidic: \[[Al(H_2O)_6]^{3+}_{(aq)} \rightleftharpoons [Al(H_2O)_5(OH)]^{2+}_{(aq)}+H^+_{(aq)} \label{16.36}\] The \([Al(H_2O)_6]^{3+}\) ion has a \(pK_a\) of 5.0, making it almost as strong an acid as acetic acid. Because of the two factors described previously, the most important parameter for predicting the effect of a metal ion on the acidity of coordinated water molecules is the charge-to-radius ratio of the metal ion. A number of pairs of metal ions that lie on a diagonal line in the periodic table, such as \(Li^+\) and \(Mg^{2+}\) or \(Ca^{2+}\) and \(Y^{3+}\), have different sizes and charges, but similar charge-to-radius ratios. As a result, these pairs of metal ions have similar effects on the acidity of coordinated water molecules, and they often exhibit other significant similarities in chemistry as well. Acidic Metal Ions Solutions of small, highly charged metal ions in water are acidic. Reactions such as those discussed in this section, in which a salt reacts with water to give an acidic or basic solution, are often called hydrolysis reactions. Using a separate name for this type of reaction is unfortunate because it suggests that they are somehow different. In fact, hydrolysis reactions are just acid–base reactions in which the acid is a cation or the base is an anion; they obey the same principles and rules as all other acid–base reactions. Hydrolysis A hydrolysis reaction is an acid–base reaction. Example \(\PageIndex{1}\) Predict whether aqueous solutions of these compounds are acidic, basic, or neutral. \(KNO_3\) \(CrBr_3 \cdot H_2O\) \(Na_2SO_4\) Given : compound Asked for : acidity or basicity of aqueous solution Strategy : Assess the acid–base properties of the cation and the anion. If the cation is a weak Lewis acid, it will not affect the \(pH\) of the solution. If the cation is the conjugate acid of a weak base or a relatively highly charged metal cation, however, it will react with water to produce an acidic solution. f the anion is the conjugate base of a strong acid, it will not affect the \(pH\) of the solution. If, however, the anion is the conjugate base of a weak acid, the solution will be basic. Solution : a The \(K^+\) cation has a small positive charge (+1) and a relatively large radius (because it is in the fourth row of the periodic table), so it is a very weak Lewis acid. The \(NO_3−\) anion is the conjugate base of a strong acid, so it has essentially no basic character (Table 16.1). Hence neither the cation nor the anion will react with water to produce \(H^+\) or \(OH^−\), and the solution will be neutral. b. The \(Cr^{3+}\) ion is a relatively highly charged metal cation that should behave similarly to the \(Al^{3+}\) ion and form the [Cr(H2O)6]3+ complex, which will behave as a weak acid: \[ Cr(H_2O)_6]^{3+}_{(aq)} \ce{ <=>>} Cr(H_2O)_5(OH)]^{2+}_{(aq)} + H^+_({aq)}\] The \(Br^−\) anion is a very weak base (it is the conjugate base of the strong acid \(HBr\)), so it does not affect the \(pH\) of the solution. Hence the solution will be acidic. c. The \(Na^+\) ion, like the \(K^+\), is a very weak acid, so it should not affect the acidity of the solution. In contrast, \(SO_4^{2−}\) is the conjugate base of \(HSO_4^−\), which is a weak acid. Hence the \(SO_4^{2−}\) ion will react with water as shown in Figure 16.6 to give a slightly basic solution. Exercise \(\PageIndex{1}\) Predict whether aqueous solutions of the following are acidic, basic, or neutral. \(KI\) \(Mg(ClO_4)_2\) \(NaHS\) Answer a neutral Answer b acidic Answer c basic (due to the reaction of \(HS^−\) with water to form \(H_2S\) and \(OH^−\)) Summary A salt can dissolve in water to produce a neutral, a basic, or an acidic solution, depending on whether it contains the conjugate base of a weak acid as the anion (\(A^−\)), the conjugate acid of a weak base as the cation (\(BH^+\)), or both. Salts that contain small, highly charged metal ions produce acidic solutions in water. The reaction of a salt with water to produce an acidic or a basic solution is called a hydrolysis reaction.
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Courses/BethuneCookman_University/B-CU%3A_CH-345_Quantitative_Analysis/Problem_Sets%3A_Quantitative_Analysis/13.E%3A_Kinetic_Methods_(Exercises)
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1. Equation 13.18 shows how [A] 0 is determined using a two-point fixed-time integral method in which the concentration of A for the pseudo-first-order reaction \[\mathrm{A + R → P}\] is measured at times t 1 and t 2 . Derive a similar equation for the case where the product is monitored under pseudo-first order conditions. 2. The concentration of phenylacetate can be determined from the kinetics of its pseudo-first order hydrolysis reaction in an ethylamine buffer. When a standard solution of 0.55 mM phenylacetate is analyzed, the concentration of phenylacetate after 60 s is 0.17 mM. When an sample is analyzed the concentration of phenylacetate remaining after 60 s is 0.23 mM. What is the initial concentration of phenylacetate in the sample? 3. In the presence of acid, solutions of iodide are oxidized by hydrogen peroxide \[\ce{2I-}(aq) + \ce{H2O2}(aq) + \ce{2H3O+}(aq) → \ce{4H2O}(l) + \ce{I2}(aq)\] When I – and H 3 O + are present in excess we can use the kinetics of the reaction—which is pseudo-first order in H 2 O 2 —to determine the concentration of H 2 O 2 by following the production of I 2 with time. In one analysis the absorbance of the solution was measured after 240 s at 348 nm. Analysis of a set of standard solutions of H 2 O 2 gives the results shown in the following table. 0 1 [H2O2] (μM) absorbance 100.0 0.236 200.0 0.471 400.0 0.933 800.0 1.872 What is the concentration of H 2 O 2 in a sample if its absorbance is 0.669 after 240 s? 4. The concentration of chromic acid can be determined by reducing it with an alcohol under conditions that are pseudo-first order in analyte. One approach is to monitor the reaction mixture’s absorbance at a wavelength of 355 nm. A standard of 5.1 × 10 –4 M chromic acid yields absorbances of 0.855 and 0.709 at 100 s and 300 s after the reaction’s initiation. When a sample with an unknown amount of chromic acid is analyzed under identical conditions, the absorbances are 0.883 and 0.706. What is the concentration of chromic acid in the sample? 5. Malmstadt and Pardue developed a variable time method for the determination of glucose based on its oxidation by the enzyme glucose oxidase. 22 To monitor the reaction’s progress, iodide is added to the samples and standards. The H 2 O 2 produced by the oxidation of glucose reacts with I – , forming I 2 as a product. The time required to produce a fixed amount of I 2 is determined spectrophotometrically. The following data was reported for a set of calibration standards 0 1 2 3 [glucose] (ppm) time (s) time (s) time (s) 5.0 146.5 150.0 149.6 10.0 69.2 67.1 66.0 20.0 34.8 35.0 34.0 30.0 22.3 22.7 22.6 40.0 16.7 16.5 17.0 50.0 13.3 13.3 13.8 To verify the method a 1.00-mL aliquot of a standard solution of 40.0 ppm glucose was added to 1.00 mL of the combined reagents, requiring 34.6 s to produce the same extent of reaction. Determine the concentration of glucose in the standard and the percent error for the analysis. 6. Deming and Pardue studied the kinetics for the hydrolysis of p -nitrophenyl phosphate by the enzyme alkaline phosphatase. 23 The reaction’s progress was monitored by measuring the absorbance of p -nitrophenol, which is one of the reaction’s products. A plot of the reaction’s rate (with units of μmol mL –1 sec –1 ) versus the volume, V , in milliliters of a serum calibration standard containing the enzyme, yielded a straight line with the following equation. \[\mathrm{rate} = 2.7×10^{-7} + 3.485 × 10^{-5}\: V\] A 10.00-mL sample of serum is analyzed, yielding a rate of 6.84 × 10 –5 μmol mL –1 sec –1 . How much more dilute is the enzyme in the serum sample than in the serum calibration standard? 7. The following data were collected for a reaction known to be pseudo-first order in analyte, A, during the time in which the reaction is monitored. 0 1 time (s) [A]t (mM) 2 1.36 4 1.24 6 1.12 8 1.02 10 0.924 12 0.838 14 0.760 16 0.690 18 0.626 20 0.568 What are the rate constant and the initial concentration of analyte in the sample? 8. The enzyme acetylcholinesterase catalyzes the decomposition of acetylcholine to choline and acetic acid. Under a given set of conditions the enzyme has a K m of 9 × 10 –5 M and a k 2 of 1.4 × 10 4 s –1 . What is the concentration of acetylcholine in a sample if the rate of reaction in the presence of 6.61 × 10 –7 M enzyme is 12.33 μM s –1 ? You may assume that the concentration of acetylcholine is significantly smaller than K m . 9. The enzyme fumarase catalyzes the stereospecific addition of water to fumarate to form l-malate. A standard 0.150 μM solution of fumarase has a rate of reaction of 2.00 μM min –1 under conditions in which the substrate’s concentration is significantly greater than K m . The rate of reaction for a sample under identical condition is 1.15 μM min –1 . What is the concentration of fumarase in the sample? 10. The enzyme urease catalyzes the hydrolysis of urea. The rate of this reaction was determined for a series of solutions in which the concentration of urea was changed while maintaining a fixed urease concentration of 5.0 μM. The following data were obtained. 0 1 [urea] (mM) rate (mM s–1) 0.100 6.25 0.200 12.5 0.300 18.8 0.400 25.0 0.500 31.2 0.600 37.5 0.700 43.7 0.800 50.0 0.900 56.2 1.00 62.5 Determine the values of V max , k 2 , and K m for urease. 11. To study the effect of an enzyme inhibitor V max and K m are measured for several concentrations of inhibitor. As the concentration of the inhibitor increases V max remains essentially constant, but the value of K m increases. Which mechanism for enzyme inhibition is in effect? 12. In the case of competitive inhibition, the equilibrium between the enzyme, E, the inhibitor, I, and the enzyme–inhibitor complex, EI, is described by the equilibrium constant K I . Show that for competitive inhibition the equation for the rate of reaction is \[\dfrac{d[\ce P]}{dt} =\dfrac{V_\ce{max}[\ce S]}{K_\ce{m}\left\{1 + ([\ce I] / K_\ce{I})\right\} + [\ce S]}\] where K I is the formation constant for the EI complex \[\mathrm{E + I ⇋ EI}\] You may assume that k 2 << k –1 . 13. Analytes A and B react with a common reagent R with first-order kinetics. If 99.9% of A must react before 0.1% of B has reacted, what is the minimum acceptable ratio for their respective rate constants? 14. A mixture of two analytes, A and B, is analyzed simultaneously by monitoring their combined concentration, C = [A] + [B], as a function of time when they react with a common reagent. Both A and B are known to follow first-order kinetics with the reagent, and A is known to react faster than B. Given the data in the following table, determine the initial concentrations of A and B, and the first-order rate constants, k A and k B . 0 1 time (min) [C] (mM) 1 0.313 6 0.200 11 0.136 16 0.098 21 0.074 26 0.058 31 0.047 36 0.038 41 0.032 46 0.027 51 0.023 56 0.019 61 0.016 66 0.014 71 0.012 15. Table 13.1 provides a list of several isotopes commonly used as tracers. The half-lives for these isotopes also are listed. What is the rate constant for the radioactive decay of each isotope? 16. 60 Co is a long-lived isotope ( t 1 /2 = 5.3 yr) frequently used as a radiotracer. The activity in a 5.00-mL sample of a solution of 60 Co is 2.1 × 10 7 disintegrations/sec. What is the molar concentration of 60 Co in the sample? 17. The concentration of Ni a new alloy is determined by a neutron activation analysis. A 0.500-g sample of the alloy and a 1.000-g sample of a standard alloy that is 5.93% w/w Ni are irradiated with neutrons in a nuclear reactor. When irradiation is complete, the sample and standard are allowed to cool and the gamma ray activities measured. Given that the activity is 1020 cpm for the sample and 3540 cpm for the standard, determine the %w/w Ni in the alloy. 18. The vitamin B 12 content of a multivitamin tablet is determined by the following procedure. A sample of 10 tablets is dissolved in water and diluted to volume in a 100-mL volumetric flask. A 50.00-mL portion is removed and treated with 0.500 mg of radioactive vitamin B 12 having an activity of 572 cpm. After homogenization, the vitamin B 12 in the sample is isolated and purified, producing 18.6 mg with an activity of 361 cpm. Calculate the milligrams of vitamin B 12 in a multivitamin tablet. 19. The oldest sample that can be dated by 14 C is approximately 30 000 years. What percentage of the 14 C still remains after this time span? 20. Potassium–argon dating is based on the nuclear decay of 40 K to 40 Ar ( t 1 /2 = 1.3 × yr). If no 40 Ar is originally present in the rock, and if 40 Ar can not escape to the atmosphere, then the relative amounts of 40K and 40Ar can be used to determine the age of the rock. When a 100.0-mg rock sample was analyzed it contains 4.63 × 10 –6 mol of 40 K and 2.09 × 10 –6 mol 40 Ar. How old is the rock sample? 21. The steady state activity for 14 C is 15 cpm of carbon. If counting is limited to 1 hr, what mass of carbon is needed to give a percent relative standard deviation of 1% for the sample’s activity? How long must we monitor the radioactive decay from a 0.50-gram sample of carbon to give a percent relative standard deviation of 1% for the activity? 22. To improve the sensitivity of a FIA analysis you might do any of the following: inject a larger volume of sample, increase the flow rate, decrease the length and the diameter of the manifold’s tubing, or merge separate channels before injecting the sample. For each action, explain why it leads to an improvement in sensitivity. 23. Figure 13.31 shows a fiagram for a solution of 50.0-ppm PO 4 3– using the method in Representative Method 13.2 . Determine values for h , t a , T , t ′, ∆ t , and T ′. What is the sensitivity of this FIA method, assuming a linear relationship between absorbance and concentration? How many samples can be analyzed per hour? 23. A sensitive method for the flow injection analysis of Cu 2 + is based on its ability to catalyze the oxidation of di-2-pyridyl ketone hydrazone (DPKH). 24 The product of the reaction is fluorescent and can be used to generate a signal when using a fluorimeter as a detector. The yield of the reaction is at a maximum when the solution is made basic with NaOH. The fluorescence, however, is greatest in the presence of HCl. Sketch an appropriate FIA manifold for this analysis. 24. The concentration of chloride in seawater can be determined by a flow injection analysis. The analysis of a set of calibration standards gives the following results. 0 1 [Cl−] (ppm) absorbance 5.00 0.057 10.00 0.099 20.00 0.230 30.00 0.354 40.00 0.478 50.00 0.594 75.00 0.840 A 1.00-mL sample of seawater is placed in a 500-mL volumetric flask and diluted to volume with distilled water. When injected into the flow injection analyzer an absorbance of 0.317 is measured. What is the concentration of Cl – in the sample? 25. Ramsing and co-workers developed an FIA method for acid–base titrations using a carrier stream consisting of 2.0 × 10 –3 M NaOH and the acid–base indicator bromothymol blue. 25 Standard solutions of HCl were injected, and the following values of ∆ t were measured from the resulting fiagrams. 0 1 2 3 [HCl] (M) ∆t (s) [HCl] (M) ∆t (s) 0.008 3.13 0.080 7.71 0.010 3.59 0.100 8.13 0.020 5.11 0.200 9.27 0.040 6.39 0.400 10.45 0.060 7.06 0.600 11.40 A sample with an unknown concentration of HCl is analyzed five times, giving values of 7.43, 7.28, 7.41, 7.37, and 7.33 s for ∆ t . Determine the concentration of HCl in the sample and its 95% confidence interval. 26. Milardovíc and colleagues used a flow injection analysis method with an amperometric biosensor to determine the concentration of glucose in blood. 26 Given that a blood sample that is 6.93 mM in glucose has a signal of 7.13 nA, what is the concentration of glucose in a sample of blood if its signal is 11.50 nA? 27. Fernández-Abedul and Costa-García developed an FIA method for detecting cocaine in samples using an amperometric detector. 27 The following signals (arbitrary units) were collected for 12 replicate injections of a 6.2 × 10 –6 M sample of cocaine. 0 1 2 24.5 24.1 24.1 23.8 23.9 25.1 23.9 24.8 23.7 23.3 23.2 23.2 (a) What is the relative standard deviation for this sample? (b) If it took 10.5 min to obtain these replicate analyses, what is the expected sample throughput in samples per hour? (c) The following calibration data are available 0 1 [cocaine] (μM) signal (arb. units) 0.18 0.8 0.36 2.1 0.60 2.4 0.81 3.2 1.0 4.5 2.0 8.1 4.0 14.4 6.0 21.6 8.0 27.1 10.0 32.9 In a typical analysis a 10.0-mg sample is dissolved in water and diluted to volume in a 25-mL volumetric flask. A 125-μL aliquot is transferred to a 25-mL volumetric flask and diluted to volume with a pH 9 buffer. When injected into the flow injection apparatus a signal of 21.4 (arb. units) is obtained. What is the %w/w cocaine in the sample? 28. Holman, Christian, and Ruzicka described an FIA method for determining the concentration of H 2 SO 4 in nonaqueous solvents. 28 Agarose beads (22–45 μm diameter) with a bonded acid–base indicator are soaked in NaOH and immobilized in the detector’s flow cell. Samples of H 2 SO 4 in n -butanol are injected into the carrier stream. As a sample passes through the flow cell, an acid–base reaction takes place between H 2 SO 4 and NaOH. The endpoint of the neutralization reaction is signaled by a change in the bound indicator’s color and is detected spectrophotometrically. The elution volume needed to reach the titration’s endpoint is inversely proportional to the concentration of H 2 SO 4 ; thus, a plot of endpoint volume versus [H 2 SO 4 ] –1 is linear. The following data is typical of that obtained using a set of external standards. 0 1 [H2SO4] (mM) end point volume (mL) 0.358 0.266 0.436 0.227 0.560 0.176 0.752 0.136 1.38 0.075 2.98 0.037 5.62 0.017 What is the concentration of H 2 SO 4 in a sample if its endpoint volume is 0.157 mL? 13.5.3 Solutions to Practice Exercises Practice Exercise 13.1 Figure 13.32 shows the calibration curve and calibration equation for the external standards. Substituting 2.21 × 10 –3 M for [CH 3 NO 2 ] t =2s gives [CH 3 NO 2 ] 0 as 5.21 × 10 –2 M. Click here to return to the chapter. Practice Exercise 13.2 Figure 13.33 shows the calibration curve and calibration equation for the external standards. Substituting 3.52 × 10 –2 M for [Fe(SCN) 2+ ] t =10s gives [SCN − ] 0 as 6.87 × 10 –2 M. Click here to return to the chapter. Practice Exercise 13.3 Figure 13.34 shows the Lineweaver–Burk plot and equation for the data in Practice Exercise 13.3. The y -intercept of 9.974 min/∆AU is equivalent to 1/ V max ; thus, V max is 0.10 ∆AU/min. The slope of 11.89 min/∆AU•mM is equivalent to K m / V max ; thus, K m is 1.2 mM. Click here to return to the chapter. Practice Exercise 13.4 Figure 13.35 shows the Lineweaver–Burk plots for the two sets of data in Practice Exercise 13.4. The nearly identical x -intercepts suggests that phenylthiourea is a noncompetitive inhibitor. Click here to return to the chapter. Contributors David Harvey (DePauw University)
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Bookshelves/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/10%3A_Gases/10.05%3A_Gas_Mixtures
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Learning Objectives To determine the contribution of each component gas to the total pressure of a mixture of gases. In our use of the ideal gas law thus far, we have focused entirely on the properties of pure gases with only a single chemical species. But what happens when two or more gases are mixed? In this section, we describe how to determine the contribution of each gas present to the total pressure of the mixture. Partial Pressures The ideal gas law assumes that all gases behave identically and that their behavior is independent of attractive and repulsive forces. If volume and temperature are held constant, the ideal gas equation can be rearranged to show that the pressure of a sample of gas is directly proportional to the number of moles of gas present: \[P=n \bigg(\dfrac{RT}{V}\bigg) = n \times \rm const. \tag{10.5.1}\] Nothing in the equation depends on the nature of the gas—only the amount. With this assumption, let’s suppose we have a mixture of two ideal gases that are present in equal amounts. What is the total pressure of the mixture? Because the pressure depends on only the total number of particles of gas present, the total pressure of the mixture will simply be twice the pressure of either component. More generally, the total pressure exerted by a mixture of gases at a given temperature and volume is the sum of the pressures exerted by each gas alone. Furthermore, if we know the volume, the temperature, and the number of moles of each gas in a mixture, then we can calculate the pressure exerted by each gas individually, which is its partial pressure The pressure a gas in a mixture would exert if it were the only one present (at the same temperature and volume). , the pressure the gas would exert if it were the only one present (at the same temperature and volume). To summarize, the total pressure exerted by a mixture of gases is the sum of the partial pressures of component gases . This law was first discovered by John Dalton, the father of the atomic theory of matter. It is now known as Dalton’s law of partial pressures A law that states that the total pressure exerted by a mixture of gases is the sum of the partial pressures of component gases. . We can write it mathematically as \[P_{tot}= P_1+P_2+P_3+P_4 \; ... = \sum_{i=1}^n{P_i} \tag{10.5.2}\] where P t is the total pressure and the other terms are the partial pressures of the individual gases ( Figure 10.5.1 ). Figure 10.5.1 Dalton’s Law. The total pressure of a mixture of gases is the sum of the partial pressures of the individual gases. For a mixture of two ideal gases, A and B, we can write an expression for the total pressure: \[P_{tot}=P_A+P_B=n_A\bigg(\dfrac{RT}{V}\bigg) + n_B\bigg(\dfrac{RT}{V}\bigg)=(n_A+n_B)\bigg(\dfrac{RT}{V}\bigg) \tag{10.5.3}\] More generally, for a mixture of i components, the total pressure is given by \[P_{tot}=(P_1+P_2+P_3+ \; \cdots +P_n)\bigg(\dfrac{RT}{V}\bigg)\tag{10.5.4a}\] \[P_{tot}=\sum_{i=1}^n{n_i}\bigg(\dfrac{RT}{V}\bigg)\tag{10.5.4b}\] Equation 10.5.4 . restates Equation 10.5.3 in a more general form and makes it explicitly clear that, at constant temperature and volume, the pressure exerted by a gas depends on only the total number of moles of gas present, whether the gas is a single chemical species or a mixture of dozens or even hundreds of gaseous species. For Equation 10.27 to be valid, the identity of the particles present cannot have an effect. Thus an ideal gas must be one whose properties are not affected by either the size of the particles or their intermolecular interactions because both will vary from one gas to another. The calculation of total and partial pressures for mixtures of gases is illustrated in Example 11. Example 10.5.1 For reasons that we will examine in Chapter 15 , deep-sea divers must use special gas mixtures in their tanks, rather than compressed air, to avoid serious problems, most notably a condition called “the bends.” At depths of about 350 ft, divers are subject to a pressure of approximately 10 atm. A typical gas cylinder used for such depths contains 51.2 g of O 2 and 326.4 g of He and has a volume of 10.0 L. What is the partial pressure of each gas at 20.00°C, and what is the total pressure in the cylinder at this temperature? Given: masses of components, total volume, and temperature Asked for: partial pressures and total pressure Strategy: A Calculate the number of moles of He and O 2 present. B Use the ideal gas law to calculate the partial pressure of each gas. Then add together the partial pressures to obtain the total pressure of the gaseous mixture. Solution: A The number of moles of He is \[n_{\rm He}=\rm\dfrac{326.4\;g}{4.003\;g/mol}=81.54\;mol \notag \] The number of moles of O 2 is \[n_{\rm O_2}=\rm \dfrac{51.2\;g}{32.00\;g/mol}=1.60\;mol \notag \] B We can now use the ideal gas law to calculate the partial pressure of each: \[P_{\rm He}=\dfrac{n_{\rm He}RT}{V}=\rm\dfrac{81.54\;mol\times0.08206\;\dfrac{atm\cdot L}{mol\cdot K}\times293.15\;K}{10.0\;L}=196.2\;atm \notag \] \[P_{\rm O_2}=\dfrac{n_{\rm O_2}RT}{V}=\rm\dfrac{1.60\;mol\times0.08206\;\dfrac{atm\cdot L}{mol\cdot K}\times293.15\;K}{10.0\;L}=3.85\;atm \notag \] The total pressure is the sum of the two partial pressures: \[P_{\rm tot}=P_{\rm He}+P_{\rm O_2}=\rm(196.2+3.85)\;atm=200.1\;atm \notag \] Exercise A cylinder of compressed natural gas has a volume of 20.0 L and contains 1813 g of methane and 336 g of ethane. Calculate the partial pressure of each gas at 22.0°C and the total pressure in the cylinder. Answer: P(CH 4 ) = 137 atm; P(C 2 H 6 ) = 13.4 atm; P t = 151 atm. Mole Fractions of Gas Mixtures The composition of a gas mixture can be described by the mole fractions of the gases present. The mole fraction ( X ) The ratio of the number of moles of any component of a mixture to the total number of moles of all species present in the mixture. of any component of a mixture is the ratio of the number of moles of that component to the total number of moles of all the species present in the mixture ( n t ): \[x_A=\dfrac{\text{moles of A}}{\text{total moles}}= \dfrac{n_A}{n_{tot}} =\dfrac{n_A}{n_A+n_B+\cdots}\tag{10.5.6}\] The mole fraction is a dimensionless quantity between 0 and 1. If X A = 1.0, then the sample is pure A, not a mixture. If X A = 0, then no A is present in the mixture. The sum of the mole fractions of all the components present must equal 1. To see how mole fractions can help us understand the properties of gas mixtures, let’s evaluate the ratio of the pressure of a gas A to the total pressure of a gas mixture that contains A. We can use the ideal gas law to describe the pressures of both gas A and the mixture: P A = n A RT / V and P t = n t RT / V . The ratio of the two is thus \[\dfrac{P_A}{P_{tot}}=\dfrac{n_ART/V}{n_{tot}RT/V} = \dfrac{n_A}{n_{tot}}=x_A \tag{10.5.7}\] Rearranging this equation gives \[P_A = x_AP_{tot} \tag{10.5.8}\] That is, the partial pressure of any gas in a mixture is the total pressure multiplied by the mole fraction of that gas. This conclusion is a direct result of the ideal gas law, which assumes that all gas particles behave ideally. Consequently, the pressure of a gas in a mixture depends on only the percentage of particles in the mixture that are of that type, not their specific physical or chemical properties. Recall from Chapter 7 ( Table 7.6.1 ) that by volume, Earth’s atmosphere is about 78% N 2 , 21% O 2 , and 0.9% Ar, with trace amounts of gases such as CO 2 , H 2 O, and others. This means that 78% of the particles present in the atmosphere are N 2 ; hence the mole fraction of N 2 is 78%/100% = 0.78. Similarly, the mole fractions of O 2 and Ar are 0.21 and 0.009, respectively. Using Equation 10.5 .4, we therefore know that the partial pressure of N 2 is 0.78 atm (assuming an atmospheric pressure of exactly 760 mmHg) and, similarly, the partial pressures of O 2 and Ar are 0.21 and 0.009 atm, respectively. Example 10.5.2 We have just calculated the partial pressures of the major gases in the air we inhale. Experiments that measure the composition of the air we exhale yield different results, however. The following table gives the measured pressures of the major gases in both inhaled and exhaled air. Calculate the mole fractions of the gases in exhaled air. Unnamed: 0 Inhaled Air (mmHg) Exhaled Air (mmHg) \(P_{\rm N_2}\) 597.0 568 \(P_{\rm O_2}\) 158.0 116 \(P_{\rm H_2O}\) 0.3 28 \(P_{\rm CO_2}\) 5.0 48 P Ar 8.0 8 P t 767.0 767 Given: pressures of gases in inhaled and exhaled air Asked for: mole fractions of gases in exhaled air Strategy: Calculate the mole fraction of each gas using Equation 10.5.8 . Solution: The mole fraction of any gas A is given by \[x_A=\dfrac{P_A}{P_{tot}} \notag \] where P A is the partial pressure of A and P t is the total pressure. In this case, \[x_{\rm CO_2}=\rm\dfrac{48\;mmHg}{767\;mmHg}=0.063 \notag \] The following table gives the values of P A and X A for exhaled air. P A P A.1 X A \({\rm N_2}\) \( \left ( 568 \; \cancel{mmHg} \right )\dfrac{1 \; atm}{760 \; \cancel{mmHg}}= 0.747 \; atm \) \( \dfrac{0.747 \; \cancel{atm}}{1.01 \; \cancel{atm}} = 0.740 \) \({\rm O_2}\) \( \left (116 \; \cancel{mmHg} \right )\dfrac{1 \; atm}{760 \; \cancel{mmHg}}= 0.153 \; atm \) \( \dfrac{0.153 \; \cancel{atm}}{1.01 \; \cancel{atm}} = 0.151 \) \({\rm H_2O}\) \( \left (28 \; \cancel{mmHg} \right )\dfrac{1 \; atm}{760 \; \cancel{mmHg}}= 0.037 \; atm \) \( \dfrac{0.031 \; \cancel{atm}}{1.01 \; \cancel{atm}} = 0.031 \) \({\rm CO_2}\) \( \left (48 \; \cancel{mmHg} \right )\dfrac{1 \; atm}{760 \; \cancel{mmHg}}= 0.063 \; atm \) \( \dfrac{0.063 \; \cancel{atm}}{1.01 \; \cancel{atm}} = 0.061 \) \({\rm Ar}\) \( \left (8 \; \cancel{mmHg} \right )\dfrac{1 \; atm}{760 \; \cancel{mmHg}}= 0.011 \; atm \) \( \dfrac{0.011 \; \cancel{atm}}{1.01 \; \cancel{atm}} = 0.011 \) Exercise We saw in Example 10 that Venus is an inhospitable place, with a surface temperature of 560°C and a surface pressure of 90 atm. The atmosphere consists of about 96% CO 2 and 3% N 2 , with trace amounts of other gases, including water, sulfur dioxide, and sulfuric acid. Calculate the partial pressures of CO 2 and N 2 . Answer: \(P_{\rm CO_2}=\rm86\; atm\), \(P_{\rm N_2}=\rm2.7\;atm\) Summary The pressure exerted by each gas in a gas mixture (its partial pressure ) is independent of the pressure exerted by all other gases present. Consequently, the total pressure exerted by a mixture of gases is the sum of the partial pressures of the components ( Dalton’s law of partial pressures ). The amount of gas present in a mixture may be described by its partial pressure or its mole fraction. The mole fraction of any component of a mixture is the ratio of the number of moles of that substance to the total number of moles of all substances present. In a mixture of gases, the partial pressure of each gas is the product of the total pressure and the mole fraction of that gas. Key Takeaway The partial pressure of each gas in a mixture is proportional to its mole fraction. Key Equations Mole fraction Equation 10.5.7 : \( X_{A}= \dfrac{moles A}{total moles} = \dfrac{n_{A}}{n_{t}} \) Relationship between partial pressure and mole fraction Equation 10.5 .8: P A = X A P t Conceptual Problems Dalton’s law of partial pressures makes one key assumption about the nature of the intermolecular interactions in a mixture of gases. What is it? What is the relationship between the partial pressure of a gas and its mole fraction in a mixture? Numerical Problems What is the partial pressure of each gas if the following amounts of substances are placed in a 25.0 L container at 25°C? What is the total pressure of each mixture? 1.570 mol of CH 4 and 0.870 mol of CO 2 2.63 g of CO and 1.24 g of NO 2 1.78 kg of CH 3 Cl and 0.92 kg of SO 2 What is the partial pressure of each gas in the following 3.0 L mixtures at 37°C as well as the total pressure? 0.128 mol of SO 2 and 0.098 mol of methane (CH 4 ) 3.40 g of acetylene (C 2 H 2 ) and 1.54 g of He 0.267 g of NO, 4.3 g of Ar, and 0.872 g of SO 2 In a mixture of helium, oxygen, and methane in a 2.00 L container, the partial pressures of He and O 2 are 13.6 kPa and 29.2 kPa, respectively, and the total pressure inside the container is 95.4 kPa. What is the partial pressure of methane? If the methane is ignited to initiate its combustion with oxygen and the system is then cooled to the original temperature of 30°C, what is the final pressure inside the container (in kilopascals)? A 2.00 L flask originally contains 1.00 g of ethane (C 2 H 6 ) and 32.0 g of oxygen at 21°C. During ignition, the ethane reacts completely with oxygen to produce CO 2 and water vapor, and the temperature of the flask increases to 200°C. Determine the total pressure and the partial pressure of each gas before and after the reaction. If a 20.0 L cylinder at 19°C is charged with 5.0 g each of sulfur dioxide and oxygen, what is the partial pressure of each gas? The sulfur dioxide is ignited in the oxygen to produce sulfur trioxide gas, and the mixture is allowed to cool to 19°C at constant pressure. What is the final volume of the cylinder? What is the partial pressure of each gas in the piston? The highest point on the continent of Europe is Mt. Elbrus in Russia, with an elevation of 18,476 ft. The highest point on the continent of South America is Mt. Aconcagua in Argentina, with an elevation of 22,841 ft. The following table shows the variation of atmospheric pressure with elevation. Use the data in the table to construct a plot of pressure versus elevation. Elevation (km) Pressure in Summer (mmHg) Pressure in Winter (mmHg) 0.0 760.0 760.0 1.0 674.8 670.6 1.5 635.4 629.6 2.0 598.0 590.8 3.0 528.9 519.7 5.0 410.6 398.7 7.0 314.9 301.6 9.0 237.8 224.1 Use your graph to estimate the pressures in millimeters of mercury during the summer and the winter at the top of both mountains in both atmospheres and kilopascals. Given that air is 20.95% O 2 by volume, what is the partial pressure of oxygen in atmospheres during the summer at each location? Answers P(CH 4 ) = 1.54 atm, P(CO 2 ) = 0.851 atm, P T = 2.39 atm P(CO) = 0.0918 atm, P(NO 2 ) = 0.0264 atm, P T = 0.1182 atm P( CH3Cl) = 34.5 atm, P(SO 2 ) = 14 atm, P T = 49 atm 52.6 kPa, 66.2 kPa Contributors Anonymous Modified by Joshua Halpern Thumbnail from Wikimedia
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Courses/Fordham_University/Chem1102%3A_Drug_Discovery_-_From_the_Laboratory_to_the_Clinic/11%3A_Molecular_Genetics/11.05%3A_Protein_Synthesis_and_the_Genetic_Code
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Learning Objectives describe the characteristics of the genetic code. describe how a protein is synthesized from mRNA. One of the definitions of a gene is as follows: a segment of deoxyribonucleic acid (DNA) carrying the code for a specific polypeptide. Each molecule of messenger RNA (mRNA) is a transcribed copy of a gene that is used by a cell for synthesizing a polypeptide chain. If a protein contains two or more different polypeptide chains, each chain is coded by a different gene. We turn now to the question of how the sequence of nucleotides in a molecule of ribonucleic acid (RNA) is translated into an amino acid sequence. How can a molecule containing just 4 different nucleotides specify the sequence of the 20 amino acids that occur in proteins? If each nucleotide coded for 1 amino acid, then obviously the nucleic acids could code for only 4 amino acids. What if amino acids were coded for by groups of 2 nucleotides? There are 4 2 , or 16, different combinations of 2 nucleotides (AA, AU , AC , AG , UU , and so on). Such a code is more extensive but still not adequate to code for 20 amino acids. However, if the nucleotides are arranged in groups of 3, the number of different possible combinations is 4 3 , or 64. Here we have a code that is extensive enough to direct the synthesis of the primary structure of a protein molecule. Video : NDSU Virtual Cell Animations project animation "Translation". For more information, see VCell, NDSU(opens in new window) [vcell.ndsu.nodak.edu] The genetic code can therefore be described as the identification of each group of three nucleotides and its particular amino acid. The sequence of these triplet groups in the mRNA dictates the sequence of the amino acids in the protein. Each individual three-nucleotide coding unit, as we have seen, is called a codon . Protein synthesis is accomplished by orderly interactions between mRNA and the other ribonucleic acids (transfer RNA [tRNA] and ribosomal RNA [rRNA]), the ribosome, and more than 100 enzymes. The mRNA formed in the nucleus during transcription is transported across the nuclear membrane into the cytoplasm to the ribosomes—carrying with it the genetic instructions. The process in which the information encoded in the mRNA is used to direct the sequencing of amino acids and thus ultimately to synthesize a protein is referred to as translation. Before an amino acid can be incorporated into a polypeptide chain, it must be attached to its unique tRNA. This crucial process requires an enzyme known as aminoacyl-tRNA synthetase (Figure \(\PageIndex{1}\)). There is a specific aminoacyl-tRNA synthetase for each amino acid. This high degree of specificity is vital to the incorporation of the correct amino acid into a protein. After the amino acid molecule has been bound to its tRNA carrier, protein synthesis can take place. Figure \(\PageIndex{2}\) depicts a schematic stepwise representation of this all-important process. Early experimenters were faced with the task of determining which of the 64 possible codons stood for each of the 20 amino acids. The cracking of the genetic code was the joint accomplishment of several well-known geneticists—notably Har Khorana, Marshall Nirenberg, Philip Leder, and Severo Ochoa—from 1961 to 1964. The genetic dictionary they compiled, summarized in Figure \(\PageIndex{3}\), shows that 61 codons code for amino acids, and 3 codons serve as signals for the termination of polypeptide synthesis (much like the period at the end of a sentence). Notice that only methionine (AUG) and tryptophan (UGG) have single codons. All other amino acids have two or more codons. Example \(\PageIndex{1}\): Using the Genetic Code A portion of an mRNA molecule has the sequence 5′‑AUGCCACGAGUUGAC‑3′. What amino acid sequence does this code for? Solution Use Figure \(\PageIndex{3}\) to determine what amino acid each set of three nucleotides (codon) codes for. Remember that the sequence is read starting from the 5′ end and that a protein is synthesized starting with the N-terminal amino acid. The sequence 5′‑AUGCCACGAGUUGAC‑3′ codes for met-pro-arg-val-asp. Exercise \(\PageIndex{4}\) A portion of an RNA molecule has the sequence 5′‑AUGCUGAAUUGCGUAGGA‑3′. What amino acid sequence does this code for? Further experimentation threw much light on the nature of the genetic code, as follows: The code is virtually universal; animal, plant, and bacterial cells use the same codons to specify each amino acid (with a few exceptions). The code is “degenerate”; in all but two cases (methionine and tryptophan), more than one triplet codes for a given amino acid. The first two bases of each codon are most significant; the third base often varies. This suggests that a change in the third base by a mutation may still permit the correct incorporation of a given amino acid into a protein. The third base is sometimes called the “wobble” base. The code is continuous and nonoverlapping; there are no nucleotides between codons, and adjacent codons do not overlap. The three termination codons are read by special proteins called release factors, which signal the end of the translation process. The codon AUG codes for methionine and is also the initiation codon. Thus methionine is the first amino acid in each newly synthesized polypeptide. This first amino acid is usually removed enzymatically before the polypeptide chain is completed; the vast majority of polypeptides do not begin with methionine. Summary In translation, the information in mRNA directs the order of amino acids in protein synthesis. A set of three nucleotides (codon) codes for a specific amino acid.
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Courses/CSU_San_Bernardino/CHEM_2200%3A_General_Chemistry_II_(Mink)/14%3A_Acid-Base_Equilibria/14.03%3A_pH_and_pOH
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Learning Objectives By the end of this section, you will be able to: Explain the characterization of aqueous solutions as acidic, basic, or neutral Express hydronium and hydroxide ion concentrations on the pH and pOH scales Perform calculations relating pH and pOH As discussed earlier, hydronium and hydroxide ions are present both in pure water and in all aqueous solutions, and their concentrations are inversely proportional as determined by the ion product of water (\(K_w\)). The concentrations of these ions in a solution are often critical determinants of the solution’s properties and the chemical behaviors of its other solutes, and specific vocabulary has been developed to describe these concentrations in relative terms. A solution is neutral if it contains equal concentrations of hydronium and hydroxide ions; acidic if it contains a greater concentration of hydronium ions than hydroxide ions; and basic if it contains a lesser concentration of hydronium ions than hydroxide ions. A common means of expressing quantities that may span many orders of magnitude is to use a logarithmic scale. One such scale that is very popular for chemical concentrations and equilibrium constants is based on the p-function, defined as shown where “\(X\)” is the quantity of interest and “log” is the base-10 logarithm: \[pX =-\log X \nonumber \] The pH of a solution is therefore defined as shown here, where [H 3 O + ] is the molar concentration of hydronium ion in the solution: \[pH =-\log \left[ H_3 O^{+}\right] \nonumber \] Rearranging this equation to isolate the hydronium ion molarity yields the equivalent expression: \[\left[ H_3 O^{+}\right]=10^{- pH } \nonumber \] Likewise, the hydroxide ion molarity may be expressed as a p-function, or pOH : \[pOH =-\log \left[ OH^{-}\right] \nonumber \] or \[\left[ OH^{-}\right]=10^{- pOH } \nonumber \] Finally, the relation between these two ion concentration expressed as p-functions is easily derived from the \(K_w\) expression: \[\begin{align} K_{ w } &=\left[ \ce{H3O^{+}} \right]\left[\ce{OH^{-}}\right] \nonumber \\[4pt] -\log K_{ w } &=-\log \left(\left[ \ce{H_3O^{+}}\right]\left[\ce{OH^{-}}\right]\right) \nonumber \\[4pt] &=-\log \left[ \ce{H3O^{+}}\right] + -\log \left[\ce{OH^{-}}\right] \nonumber \\[4pt] pK_{ w } &= pH + pOH \label{pHpOH} \end{align} \] At 25 °C, the value of \(K_w\) is \(1.0 \times 10^{−14}\), and so: \[14.00= pH + pOH \nonumber \] As was demonstrated previously, the hydronium ion molarity in pure water (or any neutral solution) is \(1.0 \times 10^{−7}~\text{M}\) at 25 °C. The pH and pOH of a neutral solution at this temperature are therefore: \[pH =-\log \left[ H_3 O^{+}\right]=-\log \left(1.0 \times 10^{-7}\right)=7.00 \nonumber \] \[pOH =-\log \left[ OH^{-}\right]=-\log \left(1.0 \times 10^{-7}\right)=7.00 \nonumber \] And so, at this temperature , acidic solutions are those with hydronium ion molarities greater than \(1.0 \times 10^{−7} ~\text{M}\) and hydroxide ion molarities less than \(1.0 \times 10^{−7} ~\text{M}\) (corresponding to pH values less than 7.00 and pOH values greater than 7.00). Basic solutions are those with hydronium ion molarities less than \(1.0 \times 10^{−7} ~\text{M}\) and hydroxide ion molarities greater than \(1.0 \times 10^{−7} ~\text{M}\) (corresponding to pH values greater than 7.00 and pOH values less than 7.00). Since the autoionization constant K w is temperature dependent, these correlations between pH values and the acidic/neutral/basic adjectives will be different at temperatures other than 25 °C. For example, Exercise \(\PageIndex{1}\) showed the hydronium molarity of pure water at 80 °C is \(4.9 \times 10^{−7} ~\text{M}\), which corresponds to pH and pOH values of: \[pH =-\log \left[ H_3 O^{+}\right]=-\log \left(4.9 \times 10^{-7}\right)=6.31 \nonumber \] \[pOH =-\log \left[ OH^{-}\right]=-\log \left(4.9 \times 10^{-7}\right)=6.31 \nonumber \] At this temperature, then, neutral solutions exhibit pH = pOH = 6.31, acidic solutions exhibit pH less than 6.31 and pOH greater than 6.31, whereas basic solutions exhibit pH greater than 6.31 and pOH less than 6.31. This distinction can be important when studying certain processes that occur at other temperatures, such as enzyme reactions in warm-blooded organisms at a temperature around 36–40 °C. Unless otherwise noted, references to pH values are presumed to be those at 25 °C (Table \(\PageIndex{1}\)). Classification Relative Ion Concentrations pH at 25 °C acidic [H3O+] > [OH−] pH < 7 neutral [H3O+] = [OH−] pH = 7 basic [H3O+] < [OH−] pH > 7 Figure \(\PageIndex{1}\): shows the relationships between [H 3 O + ], [OH − ], pH, and pOH for solutions classified as acidic, basic, and neutral. Example \(\PageIndex{1}\): Calculation of pH from [ H 3 O + ] What is the pH of stomach acid, a solution of \(\ce{HCl}\) with a hydronium ion concentration of \(1.2 \times 10^{−3}\, M\)? Solution \[\begin{align*} pH &=-\log \left[\ce{H3O^{+}}\right] \\[4pt] &=-\log \left(1.2 \times 10^{-3}\right) \\[4pt] &=-(-2.92)=2.92 \end{align*} \nonumber \] (The use of logarithms is explained in Appendix B. When taking the log of a value, keep as many decimal places in the result as there are significant figures in the value.) Exercise \(\PageIndex{1}\) Water exposed to air contains carbonic acid, \(\ce{H2CO3}\), due to the reaction between carbon dioxide and water: \[\ce{CO2(aq) + H2O(l) <=> H2CO3(aq)} \nonumber \] Air-saturated water has a hydronium ion concentration caused by the dissolved \(\ce{CO2}\) of \(2.0 \times 10^{−6} ~\text{M}\), about 20-times larger than that of pure water. Calculate the pH of the solution at 25 °C. Answer 5.70 Example \(\PageIndex{2}\) : Calculation of Hydronium Ion Concentration from pH Calculate the hydronium ion concentration of blood, the pH of which is 7.3. Solution \[\begin{align*} pH =-\log \left[ H_3 O^{+}\right] &=7.3 \\[4pt] \log \left[ H_3O^{+}\right] &=-7.3 \\[4pt] \left[ H_3 O^{+}\right] &=10^{-7.3} \end{align*} \nonumber \] or \[\begin{align*} [ \ce{H3O^{+}} ] &= \text {antilog of } -7.3 \\[4pt] &=5 \times 10^{-8} M \end{align*} \nonumber \] (On a calculator take the antilog, or the “inverse” log, of −7.3, or calculate \(10^{−7.3}\).) Exercise \(\PageIndex{2}\) Calculate the hydronium ion concentration of a solution with a pH of −1.07. Answer 12 M How Sciences Interconnect: Environmental Science Normal rainwater has a pH between 5 and 6 due to the presence of dissolved \(\ce{CO2}\) which forms carbonic acid: \[\begin{align*} \ce{H2O(l) + CO2(g) & -> H2CO3(aq)} \\[4pt] \ce{H2CO3(aq) &<=> H^{+}(aq) + HCO3^{-}(aq)}\end{align*} \nonumber \] Acid rain is rainwater that has a pH of less than 5, due to a variety of nonmetal oxides, including \(\ce{CO2}\), \(\ce{SO2}\), \(\ce{SO3}\), \(\ce{NO}\), and \(\ce{NO2}\) being dissolved in the water and reacting with it to form not only carbonic acid, but sulfuric acid and nitric acid. The formation and subsequent ionization of sulfuric acid are shown here: \[\begin{align*} \ce{H2O(l) + SO3(g) &-> H2SO4(aq)} \\[4pt] \ce{H2SO4(aq) &-> H^{+}(aq) + HSO_4^{-}(aq)} \end{align*} \nonumber \] Carbon dioxide is naturally present in the atmosphere because most organisms produce it as a waste product of metabolism. Carbon dioxide is also formed when fires release carbon stored in vegetation or fossil fuels. Sulfur trioxide in the atmosphere is naturally produced by volcanic activity, but it also originates from burning fossil fuels, which have traces of sulfur, and from the process of “roasting” ores of metal sulfides in metal-refining processes. Oxides of nitrogen are formed in internal combustion engines where the high temperatures make it possible for the nitrogen and oxygen in air to chemically combine. Acid rain is a particular problem in industrial areas where the products of combustion and smelting are released into the air without being stripped of sulfur and nitrogen oxides. In North America and Europe until the 1980s, it was responsible for the destruction of forests and freshwater lakes, when the acidity of the rain actually killed trees, damaged soil, and made lakes uninhabitable for all but the most acid-tolerant species. Acid rain also corrodes statuary and building facades that are made of marble and limestone (Figure \(\PageIndex{2}\)). Regulations limiting the amount of sulfur and nitrogen oxides that can be released into the atmosphere by industry and automobiles have reduced the severity of acid damage to both natural and manmade environments in North America and Europe. It is now a growing problem in industrial areas of China and India. For further information on acid rain, visit this website hosted by the US Environmental Protection Agency. Example \(\PageIndex{3}\) : Calculation of pOH What are the \(\text{pOH}\) and the \(\text{pH} of a 0.0125- M solution of potassium hydroxide, \(\ce{KOH}\)? Solution Potassium hydroxide is a highly soluble ionic compound and completely dissociates when dissolved in dilute solution, yielding [OH − ] = 0.0125 M : \[\begin{align*} \text{pOH} & =-\log \left[ OH^{-}\right]=-\log 0.0125 \\[4pt] & =-(-1.903)=1.903 \end{align*} \nonumber \] The pH can be found from the pOH: \[\begin{align*} \text{pH} + \text{pOH} &=14.00 \\[4pt] pH &=14.00- pOH \\[4pt] &=14.00-1.903=12.10 \end{align*} \nonumber \] Exercise \(\PageIndex{3}\) The hydronium ion concentration of vinegar is approximately \(4 \times 10^{−3} ~\text{M}\). What are the corresponding values of \(\text{pOH}\) and \(\text{pH}\)? Answer pOH = 11.6, pH = 2.4 The acidity of a solution is typically assessed experimentally by measurement of its pH. The pOH of a solution is not usually measured, as it is easily calculated from an experimentally determined pH value. The pH of a solution can be directly measured using a pH meter (Figure \(\PageIndex{3}\)). The pH of a solution may also be visually estimated using colored indicators (Figure \(\PageIndex{4}\)). The acid-base equilibria that enable use of these indicator dyes for pH measurements are described in a later section of this chapter.
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Courses/Howard_University/General_Chemistry%3A_An_Atoms_First_Approach/Unit_1%3A__Atomic_Structure/Chapter_1%3A_Introduction
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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 Chapter 1.1: Chemistry in the Modern World Chapter 1.3: A Description of Matter Chapter 1.4: A Brief History of Chemistry This page outlines a general chemistry course focusing on atomic theory, molecular structure, stoichiometry, thermochemistry, states of matter, kinetics, equilibria, and materials. It highlights the development of atomic theory from ancient Greek thought to modern science, discussing key figures like Democritus, Robert Boyle, Joseph Priestley, and Antoine Lavoisier. It explains fundamental chemical laws, Dalton's atomic theory, Avogadro's hypothesis, and the law of multiple proportions. Chapter 1.5: The Atom Chapter 1.6: Isotopes and Atomic Masses Chapter 1.7: The Mole and Molar Mass This page is a detailed exploration of fundamental chemistry concepts, focusing on atomic theory, molecular mass, the mole, and molar mass. It outlines methods to calculate molecular mass, using examples like ethanol and trichlorofluoromethane. The mole, a core chemical unit, is discussed in depth, including its definition, significance, and Avogadro's number. Chapter 1.8: Essential Skills I
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Bookshelves/Organic_Chemistry/Book%3A_Organic_Chemistry_with_a_Biological_Emphasis_v2.0_(Soderberg)/01%3A_Introduction_to_Organic_Structure_and_Bonding_I/1.05%3A_Solutions_to_Chapter_1_Exercises
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E1.1: a) The atomic number of P (phosphorus) is 15, meaning there are 15 protons. The mass number for the 31 P isotope is 31, so: 15 protons + 16 neutrons = mass number 31 (recall that mass number is number of protons and neutrons). (for parts b-d, use the same reasoning as above) b) 15 protons + 17 neutrons = mass number 32 c) 17 protons + 20 neutrons = mass number 37 d) 1 proton + 2 neutrons = mass number 3 e) 6 protons + 8 neutrons = mass number 14 E1.2: a) 1 s 2 2 s 2 2 p 3 b) 1 s 2 2 s 2 2 p 4 c) 1 s 2 2s 2 2 p 5 d) 1 s 2 2s 2 2 p 6 3 s 2 e) 1 s 2 2s 2 2 p 6 (same as Neon atom) f) 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 4 s 1 g) 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 (same as Argon atom) h) 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 (same as Argon atom) i) 1 s 2 2 s 2 2 p 6 3 s 2 3 p 4 j) 1 s 2 (same as Helium atom) k) 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 (same as Argon atom) E1.3: E1.4: E1.6: Below are full structural drawings, showing all carbons and hydrogens: E1.8: E1.10: There is only one constitutional isomer of ethanol: dimethyl ether CH 3 OCH 3 E1.11: E1.12: a) carboxylate, sulfide, aromatic, two amide groups (one of which is cyclic) b) tertiary alcohol, thioester c) carboxylate, ketone d) ether, primary amine, alkene E1.14: acetic acid: ethanoic acid chloroform: trichloromethane acetone: propanone (not 2-propanone, because the '2' in this case would be redundant: if the carbonyl carbon were not in the #2 position, the compound would be an aldehyde not a ketone) E1.17: The linking group is a phosphate diester
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Bookshelves/Organic_Chemistry/Introduction_to_Organic_Spectroscopy/zz%3A_Back_Matter/20%3A_Glossary
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Words (or words that have the same definition) The definition is case sensitive (Optional) Image to display with the definition [Not displayed in Glossary, only in pop-up on pages] (Optional) Caption for Image (Optional) External or Internal Link (Optional) Source for Definition (Eg. "Genetic, Hereditary, DNA ...") (Eg. "Relating to genes or heredity") NaN The infamous double helix https://bio.libretexts.org/ CC-BY-SA; Delmar Larsen Word(s) Definition Image Caption Link Source Sample Word 1 Sample Definition 1 NaN NaN NaN NaN
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Courses/Pacific_Union_College/Kinetics/09%3A_Enzyme_Kinetics/9.02%3A_The_Equations_of_Enzyme_Kinetics
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In biological systems, enzymes act as catalysts and play a critical role in accelerating reactions, anywhere from \(10^3\) to \(10^{17}\) times faster than the reaction would normally proceed. Enzymes are high-molecular weight proteins that act on a substrate, or reactant molecule, to form one or more products. Michaelis-Menten Enzyme Kinetics Enzymes are highly specific catalysts for biochemical reactions, with each enzyme showing a selectivity for a single reactant, or substrate . For example, the enzyme acetylcholinesterase catalyzes the decomposition of the neurotransmitter acetylcholine to choline and acetic acid. Many enzyme–substrate reactions follow a simple mechanism that consists of the initial formation of an enzyme–substrate complex, \(ES\), which subsequently decomposes to form product, releasing the enzyme to react again. This is described within the following multi-step mechanism \[ E + S \underset{k_{-1}}{\overset{k_1}{\rightleftharpoons}} ES \underset{k_{-2}}{\overset{k_2}{\rightleftharpoons}} E + P \label{13.20}\] where \(k_1\), \(k_{–1}\), \(k_2\), and \(k_{–2}\) are rate constants. The reaction’s rate law for generating the product \([P]\) is \[ rate = \dfrac{d[P]}{dt} = k_2[ES] - k_{-2}[E][P] \label{13.21A}\] However, if we make measurement early in the reaction, the concentration of products is negligible, i.e., \[[P] \approx 0\] and we can ignore the back reaction (second term in right side of Equation \(\ref{13.21A}\)). Then under these conditions, the reaction’s rate is \[ rate = \dfrac{d[P]}{dt} = k_2[ES] \label{13.21}\] To be analytically useful we need to write Equation \(\ref{13.21}\) in terms of the reactants (e.g., the concentrations of enzyme and substrate). To do this we use the steady-state approximation , in which we assume that the concentration of \(ES\) remains essentially constant. Following an initial period, during which the enzyme–substrate complex first forms, the rate at which \(ES\) forms \[ \dfrac{d[ES]}{dt} =k_1[E] [S] = k_1([E]_0 − [ES])[S] \label{13.22}\] is equal to the rate at which it disappears \[ − \dfrac{d[ES]}{dt} = k_{−1}[ES] + k_2[ES] \label{13.23}\] where \([E]_0\) is the enzyme’s original concentration. Combining Equations \(\ref{13.22}\) and \(\ref{13.23}\) gives \[k_1([E]_0 − [ES]) [S] = k_{−1}[ES] + k_2[ES]\] which we solve for the concentration of the enzyme–substrate complex \[ [ES] = \dfrac{ [E]_0[S] }{ \dfrac{k_{−1} + k_2}{k_1} + [S] } = \dfrac{[E]_0[S] }{K_m + [S]} \label{Eq13.24}\] where \(K_m\) is the Michaelis constant . Substituting Equation \(\ref{Eq13.24}\) into Equation \(\ref{13.21}\) leaves us with our final rate equation. \[ \dfrac{d[P]}{dt} = \dfrac{k_2[E]_0[S]}{K_m + [S]} \label{Eq13.25} \] A plot of Equation \(\ref{Eq13.25}\), as shown in Figure \(\PageIndex{1}\), is instructive for defining conditions where we can use the rate of an enzymatic reaction for the quantitative analysis of an enzyme or substrate. For high substrate concentrations, where \([S] \gg K_m\), Equation \(\ref{Eq13.25}\) simplifies to \[ \dfrac{d[P]}{dt} = \dfrac{k_2[E]_0[S]}{ K_m + [S]} \approx \dfrac{k_2[E]_0[S]}{[S]} = k_2[E]_0 = V_{max} \label{Eq13.26}\] where \(V_{max}\) is the maximum rate for the catalyzed reaction. Under these conditions the reaction is zero-order in substrate and we can use \(V_{max}\) to calculate the enzyme’s concentration, typically using a variable-time method. At lower substrate concentrations, where \([S] \ll K_m\), Equation \(\ref{Eq13.25}\) becomes \[ \dfrac{d[P]}{dt} = \dfrac{k_2[E]_0[S]}{K_m + [S]} \approx \dfrac{k_2[E]_0[S]}{ K_m} =\dfrac{V_{max}[S]}{K_m} \label{13.27}\] The reaction is now first-order in substrate, and we can use the rate of the reaction to determine the substrate’s concentration by a fixed-time method. The Michaelis constant \(K_m\) is the substrate concentration at which the reaction rate is at half-maximum, and is an inverse measure of the substrate's affinity for the enzyme—as a small \(K_m\) indicates high affinity, meaning that the rate will approach \(V_{max}\) more quickly. The value of \(K_m\) is dependent on both the enzyme and the substrate, as well as conditions such as temperature and pH. The Michaelis constant \(K_m\) is the substrate concentration at which the reaction rate is at half-maximum. From the last two terms in Equation \(\ref{13.27}\), we can express \(V_{max}\) in terms of a turnover number (\(k_{cat}\)): \[ V_{max} = k_{cat}[E]_o\] where \([E]_0\) is the enzyme concentration and \(k_{cat}\) is the turnover number, defined as the maximum number of substrate molecules converted to product per enzyme molecule per second. Hence, the turnover number is defined as the maximum number of chemical conversions of substrate molecules per second that a single catalytic site will execute for a given enzyme concentration \([E]_o\). Example \(\PageIndex{1}\): Turnover number of acetylcholinesterase Acetylcholinesterase (AChE) may be one of the fastest enzymes. It hydrolyzes acetylcholine to choline and an acetate group. One of the earliest values of the turnover number was \(3 \times 10^7\) (molecules of acetylcholine) per minute per molecule of enzyme. A more recent value at 25°C, pH = 7.0, acetylcholine concentration of \(2.5 \times 10^{−3}\; M\), was found to be \(7.4 \times 10^5\; min^{−1}\) ( J Biol Chem. 236 (8): 2292–5. ). There may be some 30 active centers per molecule. AChE is a serine hydrolase that reacts with acetylcholine at close to the diffusion-controlled rate . A diffusion-controlled reaction occurs so quickly that the reaction rate is the rate of transport of the reactants through the solution; a s quickly as the reactants encounter each other, they react. The Significance of \(K_M\) and \(V_{max}\) The Michaelis-Menten model is used in a variety of biochemical situations other than enzyme-substrate interaction, including antigen-antibody binding, DNA-DNA hybridization, and protein-protein interaction. It can be used to characterize a generic biochemical reaction, in the same way that the Langmuir equation can be used to model generic adsorption of biomolecular species. When an empirical equation of this form is applied to microbial growth. The experimentally determined parameters values vary wildly between enzymes (Table \(\PageIndex{1}\)): Enzyme \(K_m\) (M) \(k_{cat}\) (1/s) \(k_{cat}/K_m\) (1/M.s) Chymotrypsin 1.5 × 10−2 0.14 9.3 Pepsin 3.0 × 10−4 0.50 1.7 × 103 Tyrosyl-tRNA synthetase 9.0 × 10−4 7.6 8.4 × 103 Ribonuclease 7.9 × 10−3 7.9 × 102 1.0 × 105 Carbonic anhydrase 2.6 × 10−2 4.0 × 105 1.5 × 107 Fumarase 5.0 × 10−6 8.0 × 102 1.6 × 108 While \(K_m\) is equal to the substrate concentration at which the enzyme converts substrates into products at half its maximal rate and hence is related to the affinity of the substrate for the enzyme. The catalytic rate \(k_{cat}\) is the rate of product formation when the enzyme is saturated with substrate and therefore reflects the enzyme's maximum rate. The rate of product formation is dependent on both how well the enzyme binds substrate and how fast the enzyme converts substrate into product once substrate is bound. For a kinetically perfect enzyme, every encounter between enzyme and substrate leads to product and hence the reaction velocity is only limited by the rate the enzyme encounters substrate in solution. From Equation \(\ref{Eq13.24}\), the catalytic efficiency of a protein can be evaluated. \[ \dfrac{k_{cat}}{K_m} = \dfrac{k_2}{K_m} = \dfrac{k_1k_2}{k_{-1} + k_2}\] This \(k_{cat}/K_m\) ratio is called the specificity constant measure of how efficiently an enzyme converts a substrate into product. It has a theoretical upper limit of 10 8 – 10 10 /M.s ; enzymes working close to this, such as fumarase, are termed superefficient (Table \(\PageIndex{1}\)). Determining \(V_m\) and \(K_m\) from experimental data can be difficult and the most common way is to determine initial rates, \(v_0\), from experimental values of \([P]\) or \([S]\) as a function of time. Hyperbolic graphs of \(v_0\) vs. \([S]\) can be fit or transformed as we explored with the different mathematical transformations of the hyperbolic binding equation to determine \(K_d\). These included: nonlinear hyperbolic fit (e.g., Figure \(\PageIndex{1}\)) double reciprocal plot (e.g., Lineweaver–Burk plot discussed below Eadie-Hofstee plot Lineweaver–Burk plot Another commonly-used plot in examining enzyme kinetics is the Lineweaver-Burk plot , in with the inverse of the reaction rate, \(1/r\), is plotted against the inverse of the substrate concentration \(1/\left[ \text{S} \right]\). Rearranging Equation \(\ref{Eq13.26}\), \[\dfrac{1}{r} = \dfrac{K_M + \left[ \text{S} \right]}{k_2 \left[ \text{E} \right]_0 \left[ \text{S} \right]} = \dfrac{K_M}{k_2 \left[ \text{E} \right]_0} \dfrac{1}{\left[ \text{S} \right]} + \dfrac{1}{k_2 \left[ \text{E} \right]_0} \label{Eq28}\] Tthe Lineweaver–Burk plot (or double reciprocal plot) is a graphical representation of the Lineweaver–Burk equation of enzyme kinetics, described by Hans Lineweaver and Dean Burk in 1934 (Figure \(\PageIndex{2}\)). The Lineweaver-Burk plot results in a straight line with the slope equal to \(K_M/k_2 \left[ \text{E} \right]_0\) and \(y\)-intercept equal to \(1/k_2 \left[ \text{E} \right]_0\) which is \(1/V_{max}\) via Equation \(\ref{Eq13.26}\). The plot provides a useful graphical method for analysis of the Michaelis–Menten equation: \[V = \dfrac{V_{\max} [S]}{K_m + [S]}\] Taking the reciprocal gives \[\dfrac{1}{V} = \dfrac {K_m + [S]} {V_{max}[S]} = \dfrac{K_m}{V_{max}} \dfrac{1}{[S]} + \dfrac{1}{V_{max}}\] where \(V\) is the reaction velocity (the reaction rate), \(K_m\) is the Michaelis–Menten constant, \(V_{max}\) is the maximum reaction velocity, and \([S]\) is the substrate concentration. The Lineweaver–Burk plot was widely used to determine important terms in enzyme kinetics, such as \(K_m\) and \(V_{max}\), before the wide availability of powerful computers and non-linear regression software. The y-intercept of such a graph is equivalent to the inverse of \(V_{max}\); the x-intercept of the graph represents \(−1/K_m\). It also gives a quick, visual impression of the different forms of enzyme inhibition. Example \(\PageIndex{2}\) The reaction between nicotineamide mononucleotide and ATP to form nicotineamide–adenine dinucleotide and pyrophosphate is catalyzed by the enzyme nicotinamide mononucleotide adenylyltransferase. The following table provides typical data obtained at a pH of 4.95. The substrate, S, is nicotinamide mononucleotide and the initial rate, v , is the μmol of nicotinamide–adenine dinucleotide formed in a 3-min reaction period. [S] (mM) v (μmol) [S] (mM).1 v (μmol).1 0.138 0.148 0.560 0.324 0.220 0.171 0.766 0.390 0.291 0.234 1.460 0.493 Determine values for V max and K m . Solution Figure 13.12 shows the Lineweaver–Burk plot for this data and the resulting regression equation. Using the y -intercept, we calculate V max as V max = 1 / y −intercept = 1 / 1.708 mol = 0.585 mol and using the slope we find that K m is K m = slope × V max = 0.7528 molimM × 0.585 mol = 0.440 mM Exercise \(\PageIndex{2}\): o-diphenyl oxidase The following data are for the oxidation of catechol (the substrate) to o -quinone by the enzyme o -diphenyl oxidase. The reaction is followed by monitoring the change in absorbance at 540 nm. The data in this exercise are adapted from jkimball. 0 1 2 3 4 [catechol] (mM) 0.30 0.600 1.200 4.800 rate (∆AU/min) 0.02 0.035 0.048 0.081 Determine values for V max and K m . Click here to review your answer to this exercise. The double reciprocal plot distorts the error structure of the data, and it is therefore unreliable for the determination of enzyme kinetic parameters. Although it is still used for representation of kinetic data, non-linear regression or alternative linear forms of the Michaelis–Menten equation such as the Hanes-Woolf plot or Eadie–Hofstee plot are generally used for the calculation of parameters. Problems with the Method The Lineweaver–Burk plot is classically used in older texts, but is prone to error, as the y -axis takes the reciprocal of the rate of reaction – in turn increasing any small errors in measurement. Also, most points on the plot are found far to the right of the y -axis (due to limiting solubility not allowing for large values of \([S]\) and hence no small values for \(1/[S]\)), calling for a large extrapolation back to obtain x - and y -intercepts . When used for determining the type of enzyme inhibition, the Lineweaver–Burk plot can distinguish competitive, non-competitive and uncompetitive inhibitors. Competitive inhibitors have the same y-intercept as uninhibited enzyme (since \(V_{max}\) is unaffected by competitive inhibitors the inverse of \(V_{max}\) also doesn't change) but there are different slopes and x-intercepts between the two data sets. Non-competitive inhibition produces plots with the same x-intercept as uninhibited enzyme (\(K_m\) is unaffected) but different slopes and y-intercepts. Uncompetitive inhibition causes different intercepts on both the y- and x-axes but the same slope. Eadie–Hofstee Plot The Eadie–Hofstee plot is a graphical representation of enzyme kinetics in which reaction rate is plotted as a function of the ratio between rate and substrate concentration and can be derived from the Michaelis–Menten equation (\(\ref{Eq13.25}\)) by inverting and multiplying with \(V_{max}\): \[ \dfrac{V_{max}}{v} = \dfrac{V_{max}(K_m+[S])}{V_{max}[S]} = \dfrac{K_m+[S]}{[S]}\] Rearrange: \[V_{max} = \dfrac{vK_m}{[S]} + \dfrac{v[S]}{[S]} = \dfrac{v K_m}{[S]} + v\] Isolate v: \[v = -K_m \dfrac{v}{[S]} + V_{max}\] A plot of v against \(v/[S]\) will hence yield \(V_{max}\) as the y-intercept, \(V_{max}/K_m\) as the x-intercept, and \(K_m\) as the negative slope (Figure \(\PageIndex{3}\)). Like other techniques that linearize the Michaelis–Menten equation, the Eadie-Hofstee plot was used historically for rapid identification of important kinetic terms like \(K_m\) and \(V_{max}\), but has been superseded by nonlinear regression methods that are significantly more accurate and no longer computationally inaccessible. It is also more robust against error-prone data than the Lineweaver–Burk plot, particularly because it gives equal weight to data points in any range of substrate concentration or reaction rate (the Lineweaver–Burk plot unevenly weights such points). Both Eadie-Hofstee and Lineweaver–Burk plots remain useful as a means to present data graphically. Problems with the Method One drawback from the Eadie–Hofstee approach is that neither ordinate nor abscissa represent independent variables: both are dependent on reaction rate. Thus any experimental error will be present in both axes. Also, experimental error or uncertainty will propagate unevenly and become larger over the abscissa thereby giving more weight to smaller values of v/[S]. Therefore, the typical measure of goodness of fit for linear regression, the correlation coefficient R, is not applicable. Contributors David Harvey (DePauw University) World Public Library Wikipedia Dr. S.K. Khare ( IIT Delhi ) via NPTEL
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Bookshelves/General_Chemistry/Concept_Development_Studies_in_Chemistry_(Hutchinson)/16%3A_Reaction_Rates
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Foundation We will assume an understanding of the postulates of the Kinetic Molecular Theory and of the energetics of chemical reactions. We will also assume an understanding of phase equilibrium and reaction equilibrium, including the temperature dependence of equilibrium constants. Goals We have carefully examined the observation that chemical reactions come to equilibrium. Depending on the reaction, the equilibrium conditions can be such that there is a mixture of reactants and products, or virtually all products, or virtually all reactants. We have not considered the time scale for the reaction to achieve these conditions, however. In many cases, the speed of the reaction might be of more interest than the final equilibrium conditions of the reaction. Some reactions proceed so slowly towards equilibrium as to appear not to occur at all. For example, metallic iron will eventually oxidize in the presence of aqueous salt solutions, but the time is sufficiently long for this process that we can reasonably expect to build a boat out of iron. On the other hand, some reactions may be so rapid as to pose a hazard. For example, hydrogen gas will react with oxygen gas so rapidly as to cause an explosion. In addition, the time scale for a reaction can depend very strongly on the amounts of reactants and their temperature. In this concept development study, we seek an understanding of the rates of chemical reactions. We will define and measure reaction rates and develop a quantitative analysis of the dependence of the reaction rates on the conditions of the reaction, including concentration of reactants and temperature. This quantitative analysis will provide us insight into the process of a chemical reaction and thus lead us to develop a model to provide an understanding of the significance of reactant concentration and temperature. We will find that many reactions proceed quite simply, with reactant molecules colliding and exchanging atoms. In other cases, we will find that the process of reaction can be quite complicated, involving many molecular collisions and rearrangements leading from reactant molecules to product molecules. The rate of the chemical reaction is determined by these steps. Observation 1: Reaction Rates We begin by considering a fairly simple reaction on a rather elegant molecule. One oxidized form of buckminsterfullerene, \(\ce{C_{60}}\), is \(\ce{C_{60}O_3}\), with a three oxygen bridge as shown in Figure 16.1. Figure 16.1: Oxidized buckminsterfullerene \(\ce{C_{60}O_3}\) is prepared from \(\ce{C_{60}}\) dissolved in toluene solution at temperatures of \(0^\text{o} \text{C}\) or below. When the solution is warmed, \(\ce{C_{60}O_3}\) decomposes, releasing \(\ce{O_2}\) and creating \(\ce{C_{60}O}\) in a reaction which goes essentially to completion. We can actually watch this process happen in time by measuring the amount of light of a specific frequency absorbed by the \(\ce{C_{60}O_3}\) molecules, called the absorbance . The absorbance is proportional to the concentration of the \(\ce{C_{60}O_3}\) in the toluene solution, so observing the absorbance as a function of time is essentially the same as observing the concentration as a function of time. One such set of data is given in Table 16.1, and is shown in the graph in Figure 16.2. Time (minutes) \(\ce{C_{60}O_3}\) absorbance 3 0.04241 9 0.03634 15 0.03121 21 0.02680 27 0.02311 33 0.01992 39 0.01721 45 0.01484 51 0.01286 57 0.01106 63 0.00955 69 0.00827 75 0.00710 81 0.00616 87 0.00534 93 0.00461 99 0.00395 Figure 16.2: Oxidized Buckminsterfullerene Absorbance The rate at which the decomposition reaction is occurring is clearly related to the rate of change of the concentration \(\left[ \ce{C_{60}O_3} \right]\), which is proportional to the slope of the graph in Figure 16.2. Therefore, we define the rate of this reaction as \[\text{Rate} = -\frac{d \left[ \ce{C_{60}O_3} \right]}{dt} \cong -\frac{\Delta \left[ \ce{C_{60}O_3} \right]}{\Delta t}\] We want the rate of reaction to be positive, since the reaction is proceeding forward. However, because we are measuring the rate of disappearance of the reactant in this case, that rate is negative. We include a negative sign in this definition of rate so that the rate in the equation is a positive number. Note also that the slope of the graph in Figure 16.2 should be taken as the derivative of the graph, since the graph is not a straight line. We will approximate that derivative by estimating the slope at each time in the data, taking the change in the absorbance of the \(\ce{C_{60}O_3}\) divided by the change in time at each time step. The rate, calculated in this way, is plotted as a function of time in Figure 16.3. Figure 16.3: Rate of Decomposition of \(\ce{C_{60}O_3}\) It is clear that the slope of the graph in Figure 16.2 changes over the course of time. Correspondingly, Figure 16.3 shows that the rate of the reaction decreases as the reaction proceeds. The reaction is at first very fast but then slows considerably as the reactant \(\ce{C_{60}O_3}\) is depleted. The shape of the graph for rate versus time (Figure 16.3) is very similar to the shape of the graph for concentration versus time (Figure 16.2). This suggests that the rate of the reaction is related to the concentration of \(\ce{C_{60}O_3}\) at each time. Therefore, in Figure 16.4, we plot the rate of the reaction, defined in the equation above and shown in Figure 16.3, versus the absorbance of the \(\ce{C_{60}O_3}\). Figure 16.4: Rate versus Concentration for \(\ce{C_{60}O_3}\) Decomposition We find that there is a very simple proportional relationship between the rate of the reaction and the concentration of the reactant. Therefore we can write \[\begin{align} \text{Rate} &= -\frac{d \left[ \ce{C_{60}O_3} \right]}{dt} \\ &= k \left[ \ce{C_{60}O_3} \right] \end{align}\] where \(k\) is a proportionality constant. This equation shows that, early in the reaction when \(\left[ \ce{C_{60}O_3} \right]\) is large, the reaction proceeds rapidly, and that as \(\ce{C_{60}O_3}\) is consumed, the reaction slows down. The rate equation above is an example of a rate law , expressing the relationship between the rate of a reaction and the concentrations of the reactant or reactants. Rate laws are expressions of the relationship between experimentally observed rates and concentrations. As a second example of a reaction rate, we consider the dimerization reaction of butadiene gas, \(\ce{CH_2=CH-CH=CH_2}\). Two butadiene molecules can combine to form vinylcyclohexene, shown in Figure 16.5. Figure 16.5: Dimerization of Butadiene to Vinylcyclohexene Table 16.2 provides experimental data on the gas phase concentration of butadiene \(\left[ \ce{C_4H_6} \right]\) as a function of time at \(T = 250^\text{o} \text{C}\). Time \(\left( \text{s} \right)\) \(\left[ \ce{C_4H_6} \right] \: \left( \text{M} \right)\) Rate \(\left( \text{M/s} \right)\) \(\frac{\text{Rate}}{\left[ \ce{C_4H_6} \right]}\) \(\frac{\text{Rate}}{\left[ \ce{C_4H_6} \right]^2}\) 0 0.0917 \(9.48 \times 10^{-6}\) \(1.03 \times 10^{-4}\) \(1.13 \times 10^{-3}\) 500 0.0870 \(8.55 \times 10^{-6}\) \(9.84 \times 10^{-5}\) \(1.13 \times 10^{-3}\) 1000 0.0827 \(7.75 \times 10^{-6}\) \(9.37 \times 10^{-5}\) \(1.13 \times 10^{-3}\) 1500 0.0788 \(7.05 \times 10^{-6}\) \(8.95 \times 10^{-5}\) \(1.14 \times 10^{-3}\) 2000 0.0753 \(6.45 \times 10^{-6}\) \(8.57 \times 10^{-5}\) \(1.14 \times 10^{-3}\) 2500 0.0720 \(5.92 \times 10^{-6}\) \(8.22 \times 10^{-5}\) \(1.14 \times 10^{-3}\) 3000 0.0691 \(5.45 \times 10^{-6}\) \(7.90 \times 10^{-5}\) \(1.14 \times 10^{-3}\) 3500 0.0664 \(5.04 \times 10^{-6}\) \(7.60 \times 10^{-5}\) \(1.14 \times 10^{-3}\) 4000 0.0638 \(4.67 \times 10^{-6}\) \(7.32 \times 10^{-5}\) \(1.15 \times 10^{-3}\) We can estimate the rate of reaction at each time step as in the rate equation shown earlier, and these data are presented in Table 16.2 as well. Again we see that the rate of reaction decreases as the concentration of butadiene decreases. This suggests that the rate is given by an expression like the rate law. To test this, we calculate \(\frac{\text{Rate}}{\left[ \ce{C_4H_6} \right]}\) in Table 16.2 for each time step. We note that this is not a constant, so the rate law above does not describe the relationship between the rate of reaction and the concentration of butadiene. Instead we calculate \(\frac{\text{Rate}}{\left[ \ce{C_4H_6} \right]^2}\) in Table 16.2. We discover that this ratio is a constant throughout the reaction. Therefore, the relationship between the rate of the reaction and the concentration of the reactant in this case is given by \[\begin{align} \text{Rate} &= -\frac{d \left[ \ce{C_4H_6} \right]}{dt} \\ &= k \left[ \ce{C_4H_6} \right]^2 \end{align}\] which is the rate law for the reaction in Figure 16.5. This is a very interesting result when compared to the rate law given above. In both cases, the results demonstrate that the rate of reaction depends on the concentration of the reactant. However, we now also know that the way in which the rate varies with the concentration depends on what the reaction is. Each reaction has its own rate law, observed experimentally. Observation 2: Rate Laws and the Order of Reaction We would like to understand what determines the specific dependence of the reaction rate on concentration in each reaction. In the first case considered above, the rate depends on the concentration of the reactant to the first power. We refer to this as a first order reaction . In the second case above, the rate depends on the concentration of the reactant to the second power, so this is called a second order reaction. There are also third order reactions , and even zeroth order reactions whose rates do not depend on the amount of the reactant. We need more observations of rate laws for different reactions. The approach used in the previous section to determine a reaction's rate law is fairly clumsy and at this point difficult to apply. We consider here a more systematic approach. First, consider the decomposition of \(\ce{N_2O_5} \left( g \right)\). \[2 \ce{N_2O_5} \left( g \right) \rightarrow 4 \ce{NO_2} \left( g \right) + \ce{O_2} \left( g \right)\] We can create an initial concentration of \(\ce{N_2O_5}\) in a flask and measure the rate at which the \(\ce{N_2O_5}\) first decomposes. We can then create a different initial concentration of \(\ce{N_2O_5}\) and measure the new rate at which the \(\ce{N_2O_5}\) decomposes. By comparing these rates, we can find the order of the decomposition reaction. The rate law for decomposition of \(\ce{N_2O_5} \left( g \right)\) is of the general form: \[\text{Rate} = k \left[ \ce{N_2O_5} \right]^m\] so we need to determine the exponent \(m\). For example, at \(25^\text{o} \text{C}\) we observe that the rate of decomposition is \(1.4 \times 10^{-3} \: \frac{\text{M}}{\text{s}}\) when the concentration of \(\ce{N_2O_5}\) is \(0.020 \: \text{M}\). If instead we begin with \(\left[ \ce{N_2O_5} \right] = 0.010 \: \text{M}\), we observe that the rate of decomposition is \(7.0 \times 10^{-4} \: \frac{\text{M}}{\text{s}}\). We can compare the rate from the first measurement, Rate 1, to the rate from the second measurement, Rate 2. From the equation above, we can write that \[\begin{align} \frac{\text{Rate 1}}{\text{Rate 2}} &= \frac{k \left[ \ce{N_2O_5} \right]^m_1}{k \left[ \ce{N_2O_5} \right]^m_2} \\ &= \frac{1.4 \times 10^{-3} \: \frac{\text{M}}{\text{s}}}{7.0 \times 10^{-4} \: \frac{\text{M}}{\text{s}}} \\ &= \frac{k \left( 0.020 \: \text{M} \right)^m}{k \left( 0.010 \: \text{M} \right)^m} \end{align}\] This can be simplified on both sides of the equation to give \[2.0 = 2.0^m\] Clearly, then, \(m = 1\) and the decomposition is a first order reaction. We can also then find the first order rate constant \(k\) for this reaction by simply plugging in one of the initial rate measurements to the rate law equation. We find that \(k = 0.070 \: \text{s}^{-1}\). This approach to finding reaction order is called the method of initial rates, since it relies on fixing the concentration at specific initial values and measuring the initial rate associated with each concentration. So far we have considered only reactions which have a single reactant. Consider a second example of the method of initial rates involving the reaction of hydrogen gas and iodine gas: \[\ce{H_2} \left( g \right) + \ce{I_2} \left( g \right) \rightarrow 2 \ce{HI} \left( g \right)\] In this case, we expect to find that the rate of the reaction depends on the concentrations for both reactants. As such, we need more initial rate observations to determine the rate law. In Table 16.3, observations are reported for the initial rate for three sets of initial concentrations of \(\ce{H_2}\) and \(\ce{I_2}\). Experiment \(\left[ \ce{H_2} \right]_0 \: \left( \text{M} \right)\) \(\left[ \ce{I_2} \right]_0 \: \left( \text{M} \right)\) Rate \(\left( \text{M/s} \right)\) 1 0.1 0.1 \(3.00 \times 10^{-4}\) 2 0.2 0.1 \(6.00 \times 10^{-4}\) 3 0.2 0.2 \(1.19 \times 10^{-3}\) Following the same process we used in the \(\ce{N_2O_5}\) example, we write the general rate law for the reaction as \[\text{Rate} = k \left[ \ce{H_2} \right]^n \left[ \ce{I_2} \right]^m\] By comparing Experiment 1 to Experiment 2, we can write \[\begin{align} \frac{\text{Rate 1}}{\text{Rate 2}} &= \frac{k \left[ \ce{H_2} \right]^n_1 \left[ \ce{I_2} \right]^m_1}{k \left[ \ce{H_2} \right]^n_2 \left[ \ce{I_2} \right]^m_2} \\ &= \frac{3.00 \times 10^{-4} \: \frac{\text{M}}{\text{s}}}{6.00 \times 10^{-4} \: \frac{\text{M}}{\text{s}}} \\ &= \frac{k \left( 0.10 \: \text{M} \right)^n \left( 0.10 \: \text{M} \right)^m}{k \left( 0.20 \: \text{M} \right)^n \left( 0.10 \: \text{M} \right)^m} \end{align}\] This simplifies to \[0.50 = 0.50^m 1.00^n\] from which it is clear that \(m = 1\). Similarly, we can find than \(n = 1\). The reaction is therefore first order in each reactant and is second order overall. \[\text{Rate} = k \left[ \ce{H_2} \right] \left[ \ce{I_2} \right]\] Once we know the rate law, we can use any of the data from Table 16.3 to determine the rate constant, simply by plugging in concentrations and rate into the rate law equation. We find that \(k = 3.00 \times 10^{-2} \: \frac{1}{\text{Ms}}\). This procedure can be applied to any number of reactions. The challenge is preparing the initial conditions and measuring the initial change in concentration precisely versus time. Table 16.4 provides an overview of the rate laws for several reactions. A variety of reaction orders are observed, and they cannot be easily correlated with the stoichiometry of the reaction. Reaction Rate Law \(2 \ce{NO} \left( g \right) + \ce{O_2} \left( g \right) \rightarrow 2 \ce{NO_2} \left( g \right)\) \(\text{Rate} = k \left[ \ce{NO} \right]^2 \left[ \ce{O_2} \right]\) \(2 \ce{NO} \left( g \right) + 2 \ce{H_2} \left( g \right) \rightarrow 2 \ce{N_2} \left( g \right) + 2 \ce{H_2O} \left( g \right)\) \(\text{Rate} = k \left[ \ce{NO} \right]^2 \left[ \ce{H_2} \right]\) \(2 \ce{ICl} \left( g \right) + \ce{H_2} \left( g \right) \rightarrow 2 \ce{HCl} \left( g \right) + \ce{I_2} \left( g \right)\) \(\text{Rate} = k \left[ \ce{ICl} \right] \left[ \ce{H_2} \right]\) \(2 \ce{N_2O_5} \left( g \right) \rightarrow 4 \ce{NO_2} \left( g \right) + \ce{O_2} \left( g \right)\) \(\text{Rate} = k \left[ \ce{N_2O_5} \right]\) \(2 \ce{NO_2} \left( g \right) + \ce{F_2} \left( g \right) \rightarrow 2 \ce{NO_2F} \left( g \right)\) \(\text{Rate} = k \left[ \ce{NO_2} \right] \left[ \ce{F_2} \right]\) \(2 \ce{H_2O_2} \left( aq \right) \rightarrow 2 \ce{H_2O} \left( l \right) + \ce{O_2} \left( g \right)\) \(\text{Rate} = k \left[ \ce{H_2O_2} \right]\) \(\ce{H_2} \left( g \right) + \ce{Br_2} \left( g \right) \rightarrow 2 \ce{HBr} \left( g \right)\) \(\text{Rate} = k \left[ \ce{H_2} \right] \left[ \ce{Br_2} \right]^\frac{1}{2}\) \(\ce{O_3} \left( g \right) + \ce{Cl} \left( g \right) \rightarrow \ce{O_2} \left( g \right) + \ce{ClO} \left( g \right)\) \(\text{Rate} = k \left[ \ce{O_3} \right] \left[ \ce{Cl} \right]\) Concentrations as a Function of Time and the Reaction Half-life Once we know the rate law for a reaction, we should be able to predict how fast a reaction will proceed. From this, we should also be able to predict how much reactant remains or how much product has been produced at any given time in the reaction. We will focus on the reactions with a single reactant to illustrate these ideas. Consider a first order reaction like \(\ce{A} \rightarrow \text{products}\), for which the rate law must be \[\begin{align} \text{Rate} &= -\frac{d \ce{A}}{dt} \\ &= k \left[ \ce{A} \right] \end{align}\] From calculus, it is possible to use the above equation to find the function \(\left[ \ce{A} \right] \left( t \right)\) which tells us the concentration \(\left[ \ce{A} \right]\) as a function of time. The result is \[\left[ \ce{A} \right] = \left[ \ce{A} \right]_0 e^{-kt}\] or equivalently \[\text{ln} \left( \left[ \ce{A} \right] \right) = \text{ln} \left( \left[ \ce{A} \right]_0 \right) - kt\] The above equation reveals that, if a reaction is first order, we can plot \(\text{ln} \left( \left[ \ce{A} \right]_0 \right)\) versus time and get a straight line with slope equal to \(-k\). Moreover, if we know the rate constant and the initial concentration, we can predict the concentration at any time during the reaction. An interesting point in the reaction is the time at which exactly half of the original concentration of \(\ce{A}\) has been consumed. We call this time the half life of the reaction and denote it as \(t_\frac{1}{2}\). At that time, \(\left[ \ce{A} \right] = \frac{1}{2} \left[ \ce{A} \right]_0\). From the above equation and using the properties of logarithms, we find that, for a first order reaction \[t_\frac{1}{2} = \frac{\text{ln} \left( 2 \right)}{k}\] This equation tells us that the half-life of a first order reaction does not depend on how much material we start with. It takes exactly the same amount of time for the reaction to proceed from all of the starting material to half of the starting material as it does to proceed from half of the starting material to one-fourth of the starting material. In each case, we halve the remaining material in time equal to the constant half-life shown in the equation above. These conclusions are only valid for first order reactions. Consider then a second order reaction, such as the butadiene dimerization discussed above. The general second order reaction \(\ce{A} \rightarrow \text{products}\) has the rate law \[\begin{align} \text{Rate} &= -\frac{d \left[ \ce{A} \right]}{dt} \\ &= k \left[ \ce{A} \right]^2 \end{align}\] Again, we can use calculus to find the function \(\left[ \ce{A} \right] \left( t \right)\) from the above equation. The result is most easily written as \[\frac{1}{\left[ \ce{A} \right]} = \frac{1}{\left[ \ce{A} \right]_0} + k \left( t \right)\] Note that, as \(t\) increases, \(\frac{1}{\left[ \ce{A} \right]}\) increases, so \(\left[ \ce{A} \right]\) decreases. The equation reveals that, for a reaction which is second order in the reactant \(\ce{A}\), we can plot \(\frac{1}{\left[ \ce{A} \right]}\) as a function of time to get a straight line with slope equal to \(k\). Again, if we know the rate constant and the initial concentration, we can find the concentration \(\left[ \ce{A} \right]\) at any time of interest during the reaction. The half-life of a second order reaction differs from the half-life of a first order reaction. From the above equation, if we take \(\left[ \ce{A} \right] = \frac{1}{2} \left[ \ce{A} \right]_0\), we get \[t_\frac{1}{2} = \frac{1}{k \left[ \ce{A} \right]_0}\] This shows that, unlike a first order reaction, the half-life for a second order reaction depends on how much material we start with. From this equation, the the more concentrated the reactant is, the shorter the half-life. Observation 3: Temperature Dependence of Reaction Rates It is a common observation that reactions tend to proceed more rapidly with increasing temperature. Similarly, cooling reactants can have the effect of slowing a reaction to a near halt. How is this change in rate reflected in the rate law equation? One possibility is that there is a slight dependence on temperature of the concentrations, since volumes do vary with temperature. However, this is insufficient to account for the dramatic changes in rate typically observed. Therefore, the temperature dependence of reaction rate is primarily found in the rate constant, \(k\). Consider for example the reaction of hydrogen gas with iodine gas at high temperatures. The rate constant of this reaction at each temperature can be found using the method of initial rates, as discussed above, and we find in Table 16.5 that the rate constant increases dramatically as the temperature increases. \(T \: \left( \text{K} \right)\) \(k \: \left( \frac{1}{\text{Ms}} \right)\) 667 \(6.80 \times 10^{-3}\) 675 \(9.87 \times 10^{-3}\) 700 \(3.00 \times 10^{-2}\) 725 \(8.43 \times 10^{-2}\) 750 \(2.21 \times 10^{-1}\) 775 \(5.46 \times 10^{-1}\) 800 1.27 As shown in Figure 16.6, the rate constant appears to increase exponentially with temperature. After a little experimentation with the data, we find in Figure 16.7 that there is a simple linear relationship between \(\text{ln} \left( k \right)\) and \(\frac{1}{T}\). Figure 16.6: Rate constant vs. Temperature for reaction of \(\ce{H_2}\) and \(\ce{I_2}\) Figure 16.7: Rate constant vs. Inverse Temperature for reaction of \(\ce{H_2}\) and \(\ce{I_2}\) From Figure 16.7, we can see that the data in Table 16.5 fit the equation \[\text{ln} \left( k \right) = a \frac{1}{T} + b\] where \(a\) and \(b\) are constant for this reaction. It turns out that, for our purposes, all reactions have rate constants which fit this equation, but with different constants \(a\) and \(b\) for each reaction. Figure 16.7 is referred to as an Arrhenius plot , after Svante Arrhenius. It is very important to note that the form of the above equation and the appearance of Figure 16.7 are both the same as the equations and graphs for the temperature dependence of the equilibrium constant for an endothermic reaction. This suggests a model to account for the temperature dependence of the rate constant, based on the energetics of the reaction. In particular, it appears that the reaction rate is related to the amount of energy required for the reaction to occur. We will develop this further in the next section. Collision Model for Reaction Rates At this point, we have only observed the dependence of reaction rates on concentration of reactants and on temperature, and we have fit these data to equations called rate laws. Although this is very convenient, it does not provide us insight into why a particular reaction has a specific rate law or why the temperature dependence should obey the equation shown above. Nor does it provide any physical insights into the order of the reaction or the meaning of the constants \(a\) and \(b\) in the equation. We begin by asking why the reaction rate should depend on the concentration of the reactants. To answer this, we consider a simple reaction between two molecules in which atoms are transferred between the molecules during the reaction. For example, a reaction important in the decomposition of ozone \(\ce{O_3}\) by aerosis is \[\ce{O_3} \left( g \right) + \ce{Cl} \left( g \right) \rightarrow \ce{O_2} \left( g \right) + \ce{ClO} \left( g \right)\] What must happen for a reaction to occur between an \(\ce{O_3}\) molecule and a \(\ce{Cl}\) atom? Obviously, for these two particles to react, they must come into close proximity to one another so that an \(\ce{O}\) atom can be transferred from one to the other. In general, two molecules cannot trade atoms to produce new produce molecules unless they are close together for the atoms of the two molecules to interact. This requires a collision between molecules. The rate of collisions depends on the concentrations of the reactants, since the more molecules there are in a confined space, the more likely they are to run into each other. To write this relationship in an equation, we can think in terms of probability, and we consider the reaction above. The probability for an \(\ce{O_3}\) molecule to be near a specific point increases with the number of \(\ce{O_3}\) molecules, and therefore increases with the concentration of \(\ce{O_3}\) molecules. The probability for a \(\ce{Cl}\) atom to be near that specific point is also proportional to the concentration of \(\ce{Cl}\) atoms. Therefore, the probability for an \(\ce{O_3}\) molecule and a \(\ce{Cl}\) atom to be in close proximity to the same specific point at the same time is proportional to the \(\left[ \ce{O_3} \right]\) times \(\left[ \ce{Cl} \right]\). It is important to remember that not all collisions between \(\ce{O_3}\) molecules and \(\ce{Cl}\) atoms will result in a reaction. There are other factors to consider including how the molecules approach one another. The atoms may not be positioned properly to exchange between molecules, in which case the molecules will simply bounce off of one another without reacting. For example, if the \(\ce{Cl}\) atom approaches the center \(\ce{O}\) atom of the \(\ce{O_3}\) molecule, that \(\ce{O}\) atom will not transfer. Another factor is energy associated with the reaction. Clearly, though, a collision must occur for the reaction to occur, and therefore the rate of the reaction can be no faster than the rate of collisions between the reactant molecules. Therefore, we can say that, in a bimolecular reaction , where two molecules collide and react, the rate of the reaction will be proportional to the product of the concentrations of the reactants. For the reaction of \(\ce{O_3}\) with \(\ce{Cl}\), the rate must therefore be proportional to \(\left[ \ce{O_3} \right] \left[ \ce{Cl} \right]\), and we observe this in the experimental rate law in Table 16.4. Thus, it appears that we can understand the rate law by understanding the collisions which must occur for the reaction to take place. We also need our model to account for the temperature dependence of the rate constant. As noted at the end of the last section, the temperature dependence of the rate constant is the same as the temperature dependence of the equilibrium constant for an endothermic reaction. This suggests that the temperature dependence is due to an energetic factor required for the reaction to occur. However, we find experimentally that the rate constant equation describes the rate constant temperature dependence regardless of whether the reaction is endothermic or exothermic. Therefore, whatever the energetic factor is that is required for the reaction to occur, it is not just the endothermicity of the reaction. It must be that all reactions, regardless of the overall change in energy, require energy to occur. A model to account for this is the concept of activation energy . For a reaction to occur, at least some bonds in the reactant molecule must be broken, so that atoms can rearrange and new bonds can be created. At the time of collision, bonds are stretched and broken as new bonds are made. Breaking these bonds and rearranging the atoms during the collision requires the input of energy. The minimum amount of energy required for the reaction to occur is called the activation energy, \(E_a\). This is illustrated in Figure 16.8, showing conceptually how the energy of the reactants varies as the reaction proceeds. In Figure 16.8a, the energy is low early in the reaction, when the molecules are still arranged as reactants. As the molecules approach and begin to rearrange, the energy rises sharply, rising to a maximum in the middle of the reaction. This sharp rise in energy is the activation energy, as illustrated. After the middle of the reaction has passed and the molecules are arranged more as products than reactants, the energy begins to fall again. However, the energy does not fall to its original value, so this is an endothermic reaction. Figure 16.8b shows the analogous situation for an exothermic reaction. Again, as the reactants approach one another, the energy rises as the atoms begin to rearrange. At the middle of the collision, the energy maximizes and then falls as the product molecules form. In an exothermic reaction, the product energy is lower than the reactant energy. Figure 16.8 thus shows that an energy barrier must be surmounted for the reaction to occur, regardless of whether the energy of the products is greater than (Figure 16.8a) or less than (Figure 16.8b) the energy of the reactants. This barrier accounts for the temperature dependence of the reaction rate. We know from the kinetic molecular theory that as temperature increases the average energy of the molecules in a sample increases. Therefore, as temperature increases, the fraction of molecules with sufficient energy to surmount the reaction activation barrier increases. a. b. Figure 16.8: Reaction energy for (a) an endothermic reaction and (b) an exothermic reaction. Although we will not show it here, kinetic molecular theory shows that the fraction of molecules with energy greater than \(E_a\) at temperature \(T\) is proportional to \(e^{-\frac{E_a}{RT}}\). This means that the reaction rate and therefore also the rate constant must be proportional to \(e^{-\frac{E_a}{RT}}\). Therefore we can write \[k \left( T \right) = Ae^{-\frac{E_a}{RT}}\] where \(A\) is a proportionality constant. If we take the logarithm of both sides of the equation, we find that \[\text{ln} \left( k \left( T \right) \right) = -\frac{E_a}{RT} + \text{ln} \left( A \right)\] This equation matches the experimentally observed equation. We recall that a graph of \(\text{ln} \left( k \right)\) versus \(\frac{1}{T}\) is observed to be linear. Now we can see that the slope of that graph is equal to \(-\frac{E_a}{R}\). As a final note on the above equation, the constant \(A\) must have some physical significance. We have accounted for the probability of collision between two molecules and we have accounted for the energetic requirement for a successful reactive collision. We have not accounted for the probability that a collision will have the appropriate orientation of reactant molecules during the collision. Moreover, not every collision which occurs with proper orientation and sufficient energy will actually result in a reaction. There are other random factors relating to the internal structure of each molecule at the instant of collision. The factor \(A\) takes account for all of these factors, and is essentially the probability that a collision with sufficient energy for reaction will indeed lead to reaction. \(A\) is commonly called the frequency factor . Observation 4: Rate Laws for More Complicated Reaction Processes Our collision model in the previous section accounts for the concentration and temperature dependence of the reaction rate, as expressed by the rate law. The concentration dependence arises from calculating the probability of the reactant molecules being in the same vicinity at the same instant. Therefore, we should be able to predict the rate law for any reaction by simply multiplying together the concentrations of all reactant molecules in the balanced stoichiometric equation. The order of the reaction should therefore be simply related to the stoichiometric coefficients in the reaction. However, Table 16.4 shows that this is incorrect for many reactions. Consider for example the apparently simple reaction \[2 \ce{ICl} \left( g \right) + \ce{H_2} \left( g \right) \rightarrow 2 \ce{HCl} \left( g \right) + \ce{I_2} \left( g \right)\] Based on the collision model, we would assume that the reaction occurs by \(2 \ce{ICl}\) molecules colliding with a single \(\ce{H_2}\) molecule. The probability for such a collision should be proportional to \(\left[ \ce{ICl} \right]^2 \left[ \ce{H_2} \right]\). However, experimentally we observe (see Table 16.4) that the rate law for this reaction is \[\text{Rate} = k \left[ \ce{ICl} \right] \left[ \ce{H_2} \right]\] As a second example, consider the reaction \[\ce{NO_2} \left( g \right) + \ce{CO} \left( g \right) \rightarrow \ce{NO} \left( g \right) + \ce{CO_2} \left( g \right)\] It would seem reasonable to assume that this reaction occurs as a single collision in which an oxygen atom is exchanged between the two molecules. However, the experimentally observed rate law for this reaction is \[\text{Rate} = k \left[ \ce{NO_2} \right]^2\] In this case, the \(\left[ \ce{CO} \right]\) concentration does not affect the rate of the reaction at all, and the \(\left[ \ce{NO_2} \right]\) concentration is squared. These examples demonstrate that the rate law for a reaction cannot be predicted from the stoichiometric coefficients and therefore that the collision model does not account for the rate of the reaction. There must be something seriously incomplete with the collision model. The key assumption of the collision model is that the reaction occurs by a single collision. Since this assumption leads to incorrect predictions of rate laws in some cases, the assumption must be invalid in at least those cases. It may well be that reactions require more than a single collision to occur, even in reactions involving just two types of molecules. Moreover, if more than two molecules are involved, the chance of a single collision involving all of the reactive molecules becomes very small. We conclude that many reactions must occur as a result of several collisions occurring in sequence, rather than a single collision. The rate of the chemical reaction must be determined by the rates of the individual steps in the reaction. Each step in a complex reaction is a single collision, often referred to as an elementary process . In a single collision process step, our collision model should correctly predict the rate of that step. The sequence of such elementary processes leading to the overall reaction is referred to as the reaction mechanism . Determining the mechanism for a reaction can require gaining substantially more information than simply the rate data we have considered here. However, we can gain some progress just from the rate law. Consider for example the reaction of nitrogen dioxide and carbon monoxide. Since the rate law involved \(\left[ \ce{NO_2} \right]^2\), one step in the reaction mechanism must involve the collision of two \(\ce{NO_2}\) molecules. Furthermore, this step must determine the rate of the overall reaction. Why would that be? In any multi-step process, if one step is considerably slower than all of the other steps, the rate of the multi-step process is determined entirely by that slowest step, because the overall process cannot go any faster than the slowest step. It does not matter how rapidly the rapid steps occur. Therefore, the slowest step in a multi-step process is thus called the rate determining or rate limiting step. This argument suggests that the reaction proceeds via a slow step in which two \(\ce{NO_2}\) molecules collide, followed by at least one other rapid step leading to the products. A possible mechanism is therefore Step 1 \[\ce{NO_2} + \ce{NO_2} \rightarrow \ce{NO_3} + \ce{NO}\] Step 2 \[\ce{NO_3} + \ce{CO} \rightarrow \ce{NO_2} + \ce{CO_2}\] If Step 1 is much slower than Step 2, the rate of the reaction is entirely determined by the rate of Step 1. From our collision model, the rate law for Step 1 must be \(k \left[ \ce{NO_2} \right]^2\), which is consistent with the experimentally observed rate law for the overall reaction. This suggests that the mechanism is the correct description of the reaction process, with the first step as the rate determining step. There are a few important notes about the mechanism. First, one product of the reactions is produced in the first step, and the other is produced in the second step. Therefore, the mechanism does lead to the overall reaction, consuming the correct amount of reactants and producing the correct amount of products. Second, the first reaction produces a new molecule, \(\ce{NO_3}\), which is neither a reactant nor a product. The second step then consumes that molecule, and \(\ce{NO_3}\) therefore does not appear in the overall reaction. As such, \(\ce{NO_3}\) is called a reaction intermediate . Intermediates play important roles in the rates of many reactions. If the first step in a mechanism is rate determining as in this case, it is easy to find the rate law for the overall expression from the mechanism. If the second step or later steps are rate determining, determining the rate law is slightly more involved. Review and Discussion Questions When \(\ce{C_{60}O_3}\) in toluene solution decomposes, \(\ce{O_2}\) is released leaving \(\ce{C_{60}O}\) in solution. Based on the data in Figure 16.2 and Figure 16.3, plot the concentration of \(\ce{C_{60}O}\) as a function of time. How would you define the rate of the reaction in terms of the slope of the graph from Figure 16.3? How is the rate of appearance of \(\ce{C_{60}O}\) related to the rate of disappearance of \(\ce{C_{60}O_3}\)? Based on this, plot the rate of appearance of \(\ce{C_{60}O}\) as a function of time. The reaction \(2 \ce{N_2O_5} \left( g \right) \rightarrow 4 \ce{NO_2} \left( g \right) + \ce{O_2} \left( g \right)\) was found in this study to have rate law given by \(\text{Rate} = k \left[ \ce{N_2O_5} \right]\) with \(k = 0.070 \: \text{s}^{-1}\). How is the rate of appearance of \(\ce{NO_2}\) related to the rate of disappearance of \(\ce{N_2O_5}\)? Which rate is larger? Based on the rate law and rate constant, sketch a plot of \(\left[ \ce{N_2O_5} \right]\), \(\left[ \ce{NO_2} \right]\), and \(\left[ \ce{O_2} \right]\) versus time all on the same graph. For which of the reactions listed in Table 16.4 can you be certain that the reaction does not occur as a single step collision? Explain your reasoning. Consider two decomposition reactions for two hypothetical materials, \(\ce{A}\) and \(\ce{B}\). The decomposition of \(\ce{A}\) is found to be first order, and the decomposition of \(\ce{B}\) is found to be second order. Assuming that the two reactions have the same rate constant at the same temperature, sketch \(\left[ \ce{A} \right]\) and \(\left[ \ce{B} \right]\) versus time on the same graph for the same initial conditions, i.e. \(\left[ \ce{A} \right]_0 = \left[ \ce{B} \right]_0\). Compare the half-lives of the two reactions. Under what conditions will the half-life of \(\ce{B}\) be less than the half-life of \(\ce{A}\)? Under what conditions will the half-life of \(\ce{B}\) be greater than the half-life of \(\ce{A}\)? A graph of the logarithm of the equilibrium constant for a reaction versus \(\frac{1}{T}\) is linear but can have either a negative slope or a positive slope, depending on the reaction. However, the graph of the logarithm of the rate constant for a reaction versus \(\frac{1}{T}\) has a negative slope for essentially every reaction. Using equilibrium arguments, explain why the graph for the rate constant must have a negative slope. Using the rate constant equation involving activation energy and the data in Table 16.5, determine the activation energy for the reaction \(\ce{H_2} \left( g \right) + \ce{I_2} \left( g \right) \rightarrow 2 \ce{HI} \left( g \right)\). We found that the rate law for the reaction \(\ce{H_2} \left( g \right) + \ce{I_2} \left( g \right) \rightarrow 2 \ce{HI} \left( g \right)\) is \(\text{Rate} = k \left[ \ce{H_2} \right] \left[ \ce{I_2} \right]\). Therefore, the reaction is second order overall but first order in \(\ce{H_2}\). Imagine that we start with \(\left[ \ce{H_2} \right]_0 = \left[ \ce{I_2} \right]_0\) and we measure \(\left[ \ce{H_2} \right]\) versus time. Will a graph of \(\text{ln} \left( \left[ \ce{H_2} \right] \right)\) versus time be linear or will a graph of \(\frac{1}{\left[ \ce{H_2} \right]}\) versus time be linear? Explain your reasoning. As a rough estimate, chemists often assume a rule of thumb that the rate of any reaction will double when the temperature is increased by \(10^\text{o} \text{C}\). What does this suggest about the activation energies of reactions?Using the rate constant equation involving activation energy, calculate the activation energy of a reaction whose rate doubles when the temperature is raised from \(25^\text{o} \text{C}\) to \(35^\text{o} \text{C}\). Does this rule of thumb estimate depend on the temperature range? To find out, calculate the factor by which the rate constant increases when the temperature is raised from \(100^\text{o} \text{C}\) to \(110^\text{o} \text{C}\), assuming the same activation energy you found above. Does the rate double in this case? Consider a very simple hypothetical reaction \(\ce{A} + \ce{B} \leftrightarrow 2 \ce{C}\) which comes to equilibrium. At equilibrium, what must be the relationship between the rate of the forward reaction, \(\ce{A} + \ce{B} \rightarrow 2 \ce{C}\) and the reverse reaction \(2 \ce{C} \rightarrow \ce{A} + \ce{B}\)? Assume that both the forward and reverse reactions are elementary processes occurring by a single collision. What is the rate law for the forward reaction? What is the rate law for the reverse reaction? Using these results, show that the equilibrium constant for this reaction can be calculated from \(K_c = \frac{k_f}{k_r}\), where \(k_f\) is the rate constant for the forward reaction and \(k_r\) is the rate constant for the reverse reaction. Consider a very simple hypothetical reaction \(\ce{A} + \ce{B} \leftrightarrow \ce{C} + \ce{D}\). By examining Figure 16.8, provide and explain the relationship between the activation energy in the forward direction, \(E_{a,f}\), and in the reverse direction, \(E_{a,r}\). Does this relationship depend on whether the reaction is endothermic (Figure 16.8a) or exothermic (Figure 16.8b)? Explain. For the reaction \(\ce{H_2} \left( g \right) + \ce{I_2} \left( g \right) \rightarrow 2 \ce{HI} \left( g \right)\), the rate law is \(\text{Rate} = k \left[ \ce{H_2} \right] \left[ \ce{I_2} \right]\). Although this suggests that the reaction is a one-step elementary process, there is evidence that the reaction occurs in two steps, and the second step is the rate determining step: Step 1 \[\ce{I_2} <=>2 \ce{I}\] Step 2 \[\ce{H_2} + 2 \ce{I} \rightarrow 2 \ce{HI}\] Where Step 1 is fast and Step 2 is slow. If both the forward and reverse reactions in Step 1 are much faster than Step 2, explain why Step 1 can be considered to be at equilibrium. What is the rate law for the rate determining step? Since this rate law depends on the concentration of an intermediate \(\ce{I}\), we need to find that intermediate. Calculate \(\left[ \ce{I} \right]\) from Step 1, assuming that Step 1 is at equilibrium. Substitute \(\left[ \ce{I} \right]\) into the rate law found previously to find the overall rate law for the reaction. Is this result consistent with the experimental observation?
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Courses/Lumen_Learning/Book%3A_Western_Civilization_(Lumen)/Ch._12_The_Rise_of_Nation-States/13.3%3A_The_Reconquista
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Learning Objective Explain how the Reconquista led to Spain’s increasing commitment to Catholicism Key Points The Reconquista is a period in the history of the Iberian Peninsula, spanning approximately 770 years, between the initial Umayyad conquest of Hispania in the 710s and the fall of the Emirate of Granada, the last Islamic state on the peninsula, to expanding Christian kingdoms in 1492. By 718 the Muslims were in control of nearly the whole Iberian Peninsula. The advance into Western Europe was only stopped in what is now north-central France by the West Germanic Franks at the Battle of Tours in 732. The Kingdom of Asturias became the main base for Christian resistance to Islamic rule in the Iberian Peninsula for several centuries. Medieval Spain was the scene of almost constant warfare between Muslims and Christians. By 1250, nearly all of Iberia was back under Christian rule, with the exception of the Muslim kingdom of Granada—the only independent Muslim realm in Spain that would last until 1492. Around 1480, Catholic Monarchs Ferdinand II of Aragon and Isabella I of Castile (known as the Catholic Monarchs) established what would be known as the Spanish Inquisition. It was intended to maintain Catholic orthodoxy in their kingdoms. In the aftermath of the Reconquista and the Inquisition, Catholicism dominated the politics, social relations, and culture of Spain, shaping Spain as a state and the Spanish as a nation. Terms Alhambra Decree An edict issued on March 31, 1492, by the joint Catholic Monarchs of Spain (Isabella I of Castile and Ferdinand II of Aragon) ordering the expulsion of practicing Jews from the Kingdoms of Castile and Aragon and its territories and possessions by July 31 of that year. Battle of Covadonga The first victory by a Christian military force in Iberia following the Islamic conquest of Visigothic Hispania in 711–718. It was fought most likely in 722. The battle was followed by the creation of an independent Christian principality in the mountains of Asturias that became a bastion of Christian resistance to the expansion of Muslim rule. It was from there that the return of Christian rule to the entire Iberian peninsula began. Visigothic Kingdom A kingdom that occupied what is now southwestern France and the Iberian Peninsula from the 5th century to the 8th century. One of the Germanic successor states to the Western Roman Empire, the kingdom maintained independence from the Eastern Roman, or Byzantine, Empire. During its existence, Catholicism coalesced in Spain. Kingdom of Asturias A kingdom in the Iberian Peninsula founded in 718 by the nobleman Pelagius of Asturias. In 718 or 722, Pelagius defeated an Umayyad patrol at the Battle of Covadonga, in what is usually regarded as the beginning of the Reconquista. It transitioned to the Kingdom of León in 924 and became the main base for Christian resistance to Islamic rule in the Iberian Peninsula for several centuries. the Catholic Monarchs The joint title used in history for Queen Isabella I of Castile and King Ferdinand II of Aragon. They were both from the House of Trastámara and were second cousins, both descended from John I of Castile; on marriage they were given a papal dispensation to deal with consanguinity by Sixtus IV. They established the Spanish Inquisition around 1480. Arianism A Christian belief that asserts that Jesus Christ is the Son of God who was created by God the Father at a point in time, is distinct from the Father, and is therefore subordinate to the Father. Arian teachings were first attributed to Arius (c. 250–336 CE), a Christian presbyter in Alexandria, Egypt. They gained popularity in the Iberian Peninsula before Catholicism became the predominant religion of the region. Background The Reconquista (“reconquest”) is a period in the history of the Iberian Peninsula, spanning approximately 770 years, between the initial Umayyad conquest of Hispania in the 710s and the fall of the Emirate of Granada, the last Islamic state on the peninsula, to expanding Christian kingdoms in 1492. Historians traditionally mark the beginning of the Reconquista with the Battle of Covadonga (most likely in 722), and its end is associated with Portuguese and Spanish colonization of the Americas. The Arab Islamic conquest had dominated most of North Africa by 710 CE. In 711 an Islamic Berber raiding party, led by Tariq ibn Ziyad, was sent to Iberia to intervene in a civil war in the Visigothic Kingdom. Tariq’s army crossed the Strait of Gibraltar and won a decisive victory in the summer of 711 when the Visigothic King Roderic was defeated and killed at the Battle of Guadalete. Tariq’s commander, Musa, quickly crossed with Arab reinforcements, and by 718 the Muslims were in control of nearly the whole Iberian Peninsula. The advance into Western Europe was only stopped in what is now north-central France by the West Germanic Franks at the Battle of Tours in 732. A decisive victory for the Christians took place at Covadonga, in the north of the Iberian Peninsula, in the summer of 722. In a minor battle known as the Battle of Covadonga, a Muslim force sent to put down the Christian rebels in the northern mountains was defeated by Pelagius of Asturias, who established the monarchy of the Christian Kingdom of Asturias. In 739, a rebellion in Galicia, assisted by the Asturians, drove out Muslim forces, and it joined the Asturian kingdom. The Kingdom of Asturias became the main base for Christian resistance to Islamic rule in the Iberian Peninsula for several centuries. Warfare between Muslims and Christians Muslim interest in the peninsula returned in force around when Al-Mansur sacked Barcelona in 985. Under his son, other Christian cities were subjected to numerous raids. After his son’s death, the caliphate plunged into a civil war and splintered into the so-called “Taifa Kingdoms.” Medieval Spain was the scene of almost constant warfare between Muslims and Christians. The Almohads, who had taken control of the Almoravids’ Maghribi and al-Andalus territories by 1147, surpassed the Almoravides in fundamentalist Islamic outlook, and they treated the non-believer dhimmis harshly. Faced with the choice of death, conversion, or emigration, many Jews and Christians left. The Taifa kingdoms lost ground to the Christian realms in the north. After the loss of Toledo in 1085, the Muslim rulers reluctantly invited the Almoravides, who invaded Al-Andalus from North Africa and established an empire. In the 12th century the Almoravid empire broke up again, only to be taken over by the invasion of the Almohads, who were defeated by an alliance of the Christian kingdoms in the decisive battle of Las Navas de Tolosa in 1212. By 1250, nearly all of Iberia was back under Christian rule, with the exception of the Muslim kingdom of Granada—the only independent Muslim realm in Spain that would last until 1492. Francisco Pradilla Ortiz, The Capitulation of Granada (1882). The Capitulation of Granada shows Muhammad XII confronting Ferdinand and Isabella. Despite the decline in Muslim-controlled kingdoms, it is important to note the lasting effects exerted on the peninsula by Muslims in technology, culture, and society. Spanish Inquisition Around 1480, Ferdinand II of Aragon and Isabella I of Castile, known as the Catholic Monarchs, established what would be known as the Spanish Inquisition. It was intended to maintain Catholic orthodoxy in their kingdoms and to replace the Medieval Inquisition, which was under Papal control. It covered Spain and all the Spanish colonies and territories, which included the Canary Islands, the Spanish Netherlands, the Kingdom of Naples, and all Spanish possessions in North, Central, and South America. People who converted to Catholicism were not subject to expulsion, but between 1480 and 1492 hundreds of those who had converted ( conversos and moriscos ) were accused of secretly practicing their original religion (crypto-Judaism or crypto-Islam) and arrested, imprisoned, interrogated under torture, and in some cases burned to death, in both Castile and Aragon. In 1492 Ferdinand and Isabella ordered segregation of communities to create closed quarters that became what were later called “ghettos.” They also furthered economic pressures upon Jews and other non-Christians by increasing taxes and social restrictions. In 1492 the monarchs issued a decree of expulsion of Jews, known formally as the Alhambra Decree, which gave Jews in Spain four months to either convert to Catholicism or leave Spain. Tens of thousands of Jews emigrated to other lands such as Portugal, North Africa, the Low Countries, Italy, and the Ottoman Empire. Later in 1492, Ferdinand issued a letter addressed to the Jews who had left Castile and Aragon, inviting them back to Spain if they had become Christians. The Inquisition was not definitively abolished until 1834, during the reign of Isabella II, after a period of declining influence in the preceding century. Most of the descendants of the Muslims who submitted to conversion to Christianity rather than exile during the early periods of the Spanish and Portuguese Inquisition, the Moriscos, were later expelled from Spain after serious social upheaval, when the Inquisition was at its height. The expulsions were carried out more severely in eastern Spain (Valencia and Aragon) due to local animosity towards Muslims and Moriscos perceived as economic rivals; local workers saw them as cheap labor undermining their bargaining position with the landlords. Those that the Spanish Inquisition found to be secretly practicing Islam or Judaism were executed, imprisoned, or expelled. Nevertheless, all those deemed to be “New Christians” were perpetually suspected of various crimes against the Spanish state, including continued practice of Islam or Judaism. The Inquisition Tribunal as illustrated by Francisco de Goya (1808/1812). Catholicism Although the period of rule by the Visigothic Kingdom (c. 5th–8th centuries) saw the brief spread of Arianism, Catholic religion coalesced in Spain at the time. The Councils of Toledo debated creed and liturgy in orthodox Catholicism, and the Council of Lerida in 546 constrained the clergy and extended the power of law over them under the blessings of Rome. In 587, the Visigothic king at Toledo, Reccared, converted to Catholicism and launched a movement in Spain to unify the various religious doctrines that existed in the land. This put an end to dissension on the question of Arianism. The period of Reconquista and the Spanish Inquisition that followed turned Catholicism into the dominant religion of Spain, which has shaped the development of the Spanish state and national identity. Sources CC licensed content, Shared previously Boundless World History. Authored by : Boundless. Located at : https://www.boundless.com/world-history/textbooks/boundless-world-history-textbook/ . License : CC BY-SA: Attribution-ShareAlike
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Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/01%3A_Quantum_Fundamentals/1.99%3A_Visualizing_the_Difference_Between_a_Superposition_and_a_Mixture
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The superposition principle, as Feynman said, is at the heart of quantum mechanics. While its mathematical expression is simple, its true meaning is difficult to grasp. For example, given a linear superposition (not normalized) of two states, \[ |\Psi\rangle=|\phi_{1}\rangle+\left|\phi_{2}\right\rangle \nonumber \] one might assume that it represents a mixture of \(\phi_{1}\) and \(\phi_{2}\). In other words, half of the quons [1] are in state \(\phi_{1}\) and half in \(\phi_{2}\). However, the correct quantum mechanical interpretation of this equation is that the system represented by \(\Psi\) is simultaneously in the states \(\phi_{1}\) and \(\phi_{2}\), properly weighted. A mixture, half \(\phi_{1}\) and half \(\phi_{2}\), or any other ratio, cannot be represented by a wavefunction. It requires a density operator, which is a more general quantum mechanical construct that can be used to represent both pure states (superpositions) and mixtures, as shown below. \[ \hat{\rho}_{\max }=|\Psi\rangle\langle\Psi|\qquad \hat{\rho}_{\min d}=\sum p_{i}| \Psi_{i}\rangle\langle\Psi_{i}| \nonumber \] In the equation on the right, p i is the fraction of the mixture in the state \(\Psi_{i}\). To illustrate how these equations distinguish between a mixture and a superposition, we will consider a superposition and a mixture of equally weighted gaussian functions representing one-dimensional wave packets. The normalization constants are omitted in the interest of mathematical clarity. The gaussians are centered at x = \(\pm\) 4. \[ \phi_{1}(x) :=\exp \left[-(x+4)^{2}\right] \qquad \phi_{2}(x) :=\exp \left[-(x-4)^{2}\right] \nonumber \] To visualize how the density operator discriminates between a superposition and a mixture, we calculate its matrix elements in coordinate space for the 50-50 superposition and mixture of \(\phi_{1}\) and \(\phi_{2}\). The superposition is considered first. \[ \Psi(x) :=\phi_{1}(x)+\phi_{2}(x) \nonumber \] The matrix elements of this pure state are calculated as follows. \[ \rho_{\text {pure}}=\left\langle x|\hat{\rho}_{\text {pure}}| x^{\prime}\right\rangle=\langle x | \Psi\rangle\left\langle\Psi | x^{\prime}\right\rangle \nonumber \] Looking at the right side we see that the matrix elements are the product of the probability amplitudes of a quon in state \(\Psi\) being at x and xʹ. Next we display the density matrix graphically. \[ \operatorname{DensityMatrixPure}\left(x, x^{\prime}\right) :=\Psi(x) \cdot \Psi\left(x^{\prime}\right) \nonumber \] \[ x_{0}=8 \qquad N :=80 \qquad i :=0 \ldots N \\ \mathrm{x}_{\mathrm{i}} :=-\mathrm{x}_{0}+\frac{2 \cdot \mathrm{x}_{0} \cdot \mathrm{i}}{\mathrm{N}} \qquad \mathrm{j} :=0 \ldots \mathrm{N} \qquad \mathrm{x}_{\mathrm{j}}^{\prime} :=-\mathrm{x}_{0}+\frac{2 \cdot \mathrm{x}_{0} \cdot \mathrm{j}}{\mathrm{N}} \nonumber \] \[ \operatorname{DensityMatrixPure}_{\mathrm{i},\mathrm{j}} : = \operatorname{DensityMatrixPure}\left(x, x^{\prime}\right) \nonumber \] The presence of off-diagonal elements in this density matrix is the signature of a quantum mechanical superposition. For example, from the quantum mechanical perspective bi-location is possible. Now we turn our attention to the density matrix of a mixture of gaussian states. \[ \rho_{\operatorname{mix}}=\left\langle x\left|\hat{\rho}_{\operatorname{mix}}\right| x^{\prime}\right\rangle=\sum_{i} p_{i}\left\langle x | \phi_{i}\right\rangle\left\langle\phi_{i} | x^{\prime}\right\rangle=\frac{1}{2}\left\langle x | \phi_{1}\right\rangle\left\langle\phi_{1} | x^{\prime}\right\rangle+\frac{1}{2}\left\langle x | \phi_{2}\right\rangle\left\langle\phi_{2} | x^{\prime}\right\rangle \nonumber \] \[ \operatorname{DensityMatrixMix}(\mathrm{x}, \mathrm{x'}) :=\frac{\phi_{1}(\mathrm{x}) \cdot \phi_{1}(\mathrm{x'})+\phi_{2}(\mathrm{x}) \cdot \phi_{2}\left(\mathrm{x'}^{\prime}\right)}{2} \nonumber \] \[ \operatorname{DensityMatrixMix}_{\mathrm{i},\mathrm{j}} : = \operatorname{DensityMatrixMix}\left(x, x^{\prime}\right) \nonumber \] The obvious difference between a superposition and a mixture is the absence of off-diagonal elements, \(\phi_{1}(\mathrm{x}) \cdot \phi_{2}\left(\mathrm{x}^{\prime}\right)+\phi_{2}(\mathrm{x}) \cdot \phi_{1}\left(\mathrm{x}^{\prime}\right)\), in the mixed state. This indicates the mixture is in a definite but unknown state; it is an example of classical ignorance. An equivalent way to describe the difference between a superposition and a mixture, is to say that to calculate the probability of measurement outcomes for a superpostion you add the probability amplitudes and square the sum. For a mixture you square the individual probability amplitudes and sum the squares. 1. Nick Herbert ( Quantum Reality , page 64) suggested ʺquonʺ be used to stand for a generic quantum object. ʺA quon is any entity, no matter how immense, that exhibits both wave and particle aspects in the peculiar quantum manner.
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Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/05%3A_Stereoisomerism_of_Organic_Molecules/5.02%3A_Configurational_Isomers
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Geometric Isomerism We have defined isomers in a very general way as nonidentical molecules that possess the same number and kind of atoms. However, there are several ways in which isomers can be nonidentical. Among the alkenes, 1- and 2-butene are position isomers, because in these compounds the double bond has a different position in the carbon chain: Most, but not all alkenes, have stereoisomers that are not identical because of different spatial arrangements of the component atoms. Thus there are two stereoisomers of 2-butene that differ in the geometric arrangement of the groups attached to the double bond. In one isomer, both methyl groups are on the same side of the double bond ( cis -2-butene) and in the other, the methyl groups are on opposite sides of the double bond ( trans -2-butene): It should be clear to you that there will be no cis-trans isomers of alkenes in which one end of the double bond carries identical groups. Thus we don not expect there to be cis-trans isomers of 1-butene or 2-methylpropene, and indeed none are known: You may wish to verify this by making ball-and-stick models of these substances. Ring formation also confers rigidity on molecular structure such that rotation about the ring bonds is prevented. As a result, stereoisomerism of the cis-trans type is possible. For example, 1,2-dimethylcyclopropane exists in two forms that differ in the arrangement of the two methyl groups with respect to the ring. In the cis isomer, the methyl groups both are situated above (or below) the plane of the ring and in the trans isomer they are situated one above and one below, as shown in Figure 5-2. Interconversion of these isomers does not occur without breaking one or more chemical bonds. Stereoisomers that do not interconvert rapidly under normal conditions, and therefore are stable enough to be separated, specifically are called configurational isomers . Thus cis - and trans -2-butene are configurational isomers, as are cis - and trans -1,2-dimethylcyclopropane. The terms cis-trans isomerism or geometric isomerism commonly are used to describe configurational isomerism in compounds with double bonds and rings. When referring to the configuration of a particular isomer, we mean to specify its geometry. For instance, the isomer of 1,2-dichloroethene shown below has the trans configuration; the isomer of 1,3-dichlorocyclobutane has the cis configuration: Cis-trans isomerism is encountered very frequently. By one convention, the configuration of a complex alkene is taken to correspond to the configuration of the longest continuous chain as it passes through the double bond . Thus the following compound is trans -4-ethyl-3-methyl-3-heptene, despite the fact that two identical groups are cis with respect to each other, because the longest continuous chain is trans as it passes through the double bond: Notice that cis-trans isomerism is not possible at a carbon-carbon triple bond, as for 2-butyne, because the bonding arrangement at the triply bonded carbons is linear: Many compounds have more than one double bond and each may have the potential for the cis or trans arrangement. For example, 2,4-hexadiene has three different configurations, which are designated as trans-trans, cis-cis, and trans-cis. Because the two ends of this molecule are identically substituted, the trans-cis becomes identical with cis-trans: Chirality The most important type of stereoisomerism is that which arises when molecules possess two structures that are not identical and also are mirror images of one another. This is not a difficult or unfamiliar concept. Many things around us, such as our hands, and pairs of shoes, are not identical and also are mirror images of one another. In the same way, nonidentical molecules exist in which the only distinction between them is that one is the mirror image of the other. A common statement is that such isomers are mirror images of one another, but these images are not "superimposable." A simple example of this type of stereoisomerism is 2-chlorobutane, , which can exist in two spatial configurations, \(1\) and \(2\), that correspond to reflections of each other. These isomers are specifically called enantiomers . Compounds that lack reflection symmetry - meaning that they are not identical with their mirror images - are said to be chiral (pronounced "ki-rall", rhymes with spiral). This term is derived from the Greek word \(\chi \epsilon \iota \rho =\) hand; and "handedness" or chirality is a property of dissymmetric molecules such that two configurational isomers are possible that are nonidentical mirror images. Compounds that possess reflection symmetry - meaning that they are identical with their mirror images - are said to be achiral . Enantiomers are not possible for achiral compounds. An enantiomeric pair is a pair of substances whose molecules are nonidentical mirror images. The pressing question at this point is how can we tell whether a substances will be chiral or achiral. The most common origin of chirality in molecules, and the one originally recognized by van't Hoff and Le Bel, is the presence of one or more atoms, usually carbon atoms, each of which forms coplanar bonds to four different atoms or groups . This is the case for 2-chlorobutane, because the second tetrahedral carbon along the chain is bonded to four different groups: hydrogen, chlorine, methyl, and ethyl. Therefore there is a pair of enantiomers, \(1\) and \(2\). Another example is 2-bromo-2-chloro-1,1,1-trifluoroethane, which is a widely used inhalation anaesthetic. The four different groups in this case are hydrogen, chlorine, bromine, and trifluoromethyl; the pair of enantiomers is shown in Structures \(3\) and \(4\): The atom that carries the four different substituents in \(1\) and \(2\), or \(3\) and \(4\), is called a chiral atoms or chiral center . The latter is the more general term because, as we shall see later (Section 13-5A), dissymmetry in molecules need not be centered at an atom.\(^1\) In evaluating a chemical structure for chirality, you should look for carbons carrying four different attached groups. There may be more than one chiral carbon, and you should be alert to the fact that structural differences in the attached groups do not necessarily show up at the first, or even the second, position along a chain. As an example, consider the chirality of 1,1,3-trimethylcyclohexane, Carbons \(C2\), \(C4\), \(C5\), and \(C6\) are clearly achiral because each is connected to two identical groups, which for these atoms are hydrogens. The same is true for \(C1\) because it is connected to two \(CH_3\) groups. You might conclude that \(C3\) also is an achiral position because it is connected to two \(CH_2\) groups. But this would be wrong. If you look farther, you will see that the groups attached to \(C3\) actually are different and are \(H\), \(CH_3\), \(-CH_2CH_2CH_2-\), and \(-CH_2C \left( CH_3 \right)_2\). Therefore 1,1,3-trimethylcyclohexane has a chiral center at \(C3\). In contrast, the 1,1,4-isomer has no chiral centers because the groups attached to the ring at \(C4\) are identical: Several other terms that we shall use frequently in addition to chirality are racemic mixture , resolution , and racemization . A mixture of equal amounts of both enantiomers is a racemic mixture ; separation of a racemic mixture into its component enantiomers is a resolution , and the conversion of either enantiomer into equal parts of both is called racemization . Optical Activity Until recently, the phenomenon of chirality has been better known as optical isomerism , and configurational isomers that are enantiomers were referred to as optical antipodes . The reasons for this are mainly historical. It was discovered early in the nineteenth century that many compounds, whether solid, liquid, or gas, have the property of rotating the plane of polarization of polarized light and can be said to be " optically active ." A satisfactory explanation of the origin of optical activity came much later and developed in its modern form from the classic researches of Louis Pasteur, and from the concept of the three-dimensional carbon atoms expressed independently by J. H. van't Hoff and J. A. Le Bel.\(^2\) Pasteur's contribution to stereochemistry came as a result of his studies of the shapes of crystals of tartaric acid, \(\ce{HO_2C-CHOH-CHOH-CO_2H}\), and its salts. Tartaric acid, a by-product of wine production, was known to be optically active, and Pasteur showed that it, and nineteen different salts of it, all formed crystals that were not identical with their mirror images. A different substance known as "racemic acid," for which we can write the same condensed formula, \(\ce{HO_2C-CHOH-CHOH-CO_2H}\), was known to be optically inactive , and Pasteur expected that when he crystallized this acid or its salts he would obtain crystals that would be identical with their mirror images. However, crystallization of the sodium ammonium salt of racemic acid from water at temperatures below \(28^\text{o}\) gave crystals of two different shapes and these shapes were mirror images of one another. Pasteur carefully picked apart the two kinds of crystals and showed that one of them was identical with the corresponding salt of tartaric acid, except that it rotated the plane of polarization of polarized light in the opposite direction. This separation of racemic acid into two optically active forms now is called a "resolution of racemic acid." On the basis of his discoveries, Pasteur postulated that "optical isomerism" had to be related to the molecular dissymmetry of substances such that nonidentical mirror-image forms could exist. However, it remained for van' t Hoff and Le Bel to provide, almost simultaneously, a satisfactory explanation at the molecular level. In his first published work on tetrahedral carbon van't Hoff said "...it appears more and more that the present constitutional formulae are incapable of explaining certain cases of isomerism; the reason for this is perhaps the fact that we need a more definite statement about the actual positions of the atoms."\(^3\) He goes on to discuss the consequences of the tetrahedral arrangements of atoms about carbon, explicitly in connection with optical isomerism and geometric, or cis-trans, isomerism. It is not easy for the chemist of today to appreciate fully the contributions of these early chemists because we have long accepted the tetrahedral carbon as an experimentally established fact. At the time the concept was enunciated, however, even the existence of atoms and molecules was questioned openly by many scientists, and to ascribe "shapes" to what in the first place seemed like metaphysical conceptions was too much for many to accept. Properties of Enantiomers Optical activity is an experimentally useful property and usually is measured as the angle of rotation (\(\alpha\)) of the plane of polarization of polarized light passing through solutions of the substances under investigation (Figure 5-4). Where measurable optical activity is present, it is found that one enantiomer rotates the plane of polarization in one direction, whereas the other causes the plane to rotate equally but in the opposite direction. With reference to the plane of incident light, the enantiomer that rotates the plane to the right is called dextrorotatory and is symbolized by either d or (\(+\)); the enantiomer that rotates the plane to the left is levorotatory , symbolized by l or (\(-\)). A racemic mixture then can be designated as dl or (\(\pm\)), and will have no net optical rotation. It is very important to know that d , l , (\(+\)), or (\(-\)) do not designate configurations. Thus, although (\(+\))-2-butanol actually has configuration \(5\) and (\(-\))-2-butanol has configuration \(6\), there is no simple way to predict that a particular sign of rotation will be associated with a particular configuration. Methods used in assigning the true configurations to enantiomers will be discussed later. A very important point to keep in mind about any pair of enantiomers is that they will have identical chemical and physical properties, except for the signs of their optical rotations, with one important proviso: All of the properties to be compared must be determined using achiral reagents in a solvent made up of achiral molecules or, in short, in an achiral environment . Thus the melting and boiling points (but not the optical rotations) of \(5\) and \(6\) will be identical in an achiral environment. How a chiral environment or chiral reagents influence the properties of substances such as \(5\) and \(6\) will be considered in Chapter 19 . \(^1\)In the older literature, chiral centers often are called asymmetric centers and you may be confused by the difference between asymmetric and dissymmetric . Both asymmetric and dissymmetric molecules (or objects) are chiral. An asymmetric object has no symmetry at all and looks different from all angles of view. Formulas \(3\) and \(4\) represent asymmetric molecules. A dissymmetric molecule is chiral, but looks the same from more than one angle of view. A helical spring is dissymmetric - it looks the same from each end. We will encounter dissymmetric molecules later. \(^2\)The tetrahedral carbon was first proposed by E. Paterno in 1869 (see Section1-1E), but he apparently did not recognize its implications for chirality. These implications were recognized first by van't Hoff and Le Bel, with van't Hoff proceeding on the basis of bonds to carbon being directed to the corners of a regular tetrahedron. Le Bel was opposed to such a rigid formulation of the bonds to carbon. \(^3\)An interesting account and references to van't Hoff's early work can be found in "The Reception of J. H. van't Hoff's Theory of the Asymmetric Carbon" by H. A. M. Snelders, J. Chem. Educ. 51 , 2 (1974). A century has passed since van't Hoff first published his theory, which he did before he obtained his doctoral degree from the University of Utrecht. van't Hoff was the first recipient of the Nobel Prize in chemistry (1901) for his later work in thermodynamics and chemical kinetics.
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Courses/City_College_of_San_Francisco/CCSF_Chemistry_Resources/03%3A_CHE_202_-_General_Chemistry_II/3.02%3A_Periodic_Patterns/3.2.02%3A_Trends_in_Ionization_Energy
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Learning Objectives Describe and explain the observed trends in ionization energy of the elements Video \(\PageIndex{1}\) : A brief overview of ionization energy. Variation in Ionization Energies The amount of energy required to remove the most loosely bound electron from a gaseous atom in its ground state is called its first ionization energy (IE 1 ). The first ionization energy for an element, X, is the energy required to form a cation with +1 charge: \[\ce{X}(g)⟶\ce{X+}(g)+\ce{e-}\hspace{20px}\ce{IE_1}\] The energy required to remove the second most loosely bound electron is called the second ionization energy (IE 2 ). \[\ce{X+}(g)⟶\ce{X^2+}(g)+\ce{e-}\hspace{20px}\ce{IE_2}\] The energy required to remove the third electron is the third ionization energy, and so on. Energy is always required to remove electrons from atoms or ions, so ionization processes are endothermic and IE values are always positive. For larger atoms, the most loosely bound electron is located farther from the nucleus and so is easier to remove. Thus, as size (atomic radius) increases, the ionization energy should decrease. Relating this logic to what we have just learned about radii, we would expect first ionization energies to decrease down a group and to increase across a period. Figure \(\PageIndex{1}\) : The first ionization energy of the elements in the first five periods are plotted against their atomic number. Figure \(\PageIndex{1}\) graphs the relationship between the first ionization energy and the atomic number of several elements. Within a period, the values of first ionization energy for the elements (IE 1 ) generally increases with increasing Z . Down a group, the IE 1 value generally decreases with increasing Z . There are some systematic deviations from this trend, however. Note that the ionization energy of boron (atomic number 5) is less than that of beryllium (atomic number 4) even though the nuclear charge of boron is greater by one proton. This can be explained because the energy of the subshells increases as l increases, due to penetration and shielding (as discussed previously in this chapter). Within any one shell, the s electrons are lower in energy than the p electrons. This means that an s electron is harder to remove from an atom than a p electron in the same shell. The electron removed during the ionization of beryllium ([He]2 s 2 ) is an s electron, whereas the electron removed during the ionization of boron ([He]2 s 2 2 p 1 ) is a p electron; this results in a lower first ionization energy for boron, even though its nuclear charge is greater by one proton. Thus, we see a small deviation from the predicted trend occurring each time a new subshell begins. Figure \(\PageIndex{2}\) : This version of the periodic table shows the first ionization energy of (IE: 1 ), in kJ/mol, of selected elements. Another deviation occurs as orbitals become more than one-half filled. The first ionization energy for oxygen is slightly less than that for nitrogen, despite the trend in increasing IE 1 values across a period. Looking at the orbital diagram of oxygen, we can see that removing one electron will eliminate the electron–electron repulsion caused by pairing the electrons in the 2 p orbital and will result in a half-filled orbital (which is energetically favorable). Analogous changes occur in succeeding periods (note the dip for sulfur after phosphorus in Figure \(\PageIndex{2}\). Removing an electron from a cation is more difficult than removing an electron from a neutral atom because of the greater electrostatic attraction to the cation. Likewise, removing an electron from a cation with a higher positive charge is more difficult than removing an electron from an ion with a lower charge. Thus, successive ionization energies for one element always increase. As seen in Table \(\PageIndex{1}\), there is a large increase in the ionization energies (color change) for each element. This jump corresponds to removal of the core electrons, which are harder to remove than the valence electrons. For example, Sc and Ga both have three valence electrons, so the rapid increase in ionization energy occurs after the third ionization. Element IE1 IE2 IE3 IE4 IE5 IE6 IE7 K 418.8 3051.8 4419.6 5876.9 7975.5 9590.6 11343 Ca 589.8 1145.4 4912.4 6490.6 8153.0 10495.7 12272.9 Sc 633.1 1235.0 2388.7 7090.6 8842.9 10679.0 13315.0 Ga 578.8 1979.4 2964.6 6180.0 8298.7 10873.9 13594.8 Ge 762.2 1537.5 3302.1 4410.6 9021.4 Not available Not available As 944.5 1793.6 2735.5 4836.8 6042.9 12311.5 Not available Example \(\PageIndex{1}\): Ranking Ionization Energies Predict the order of increasing energy for the following processes: IE 1 for Al, IE 1 for Tl, IE 2 for Na, IE 3 for Al. Solution Removing the 6 p 1 electron from Tl is easier than removing the 3 p 1 electron from Al because the higher n orbital is farther from the nucleus, so IE 1 (Tl) < IE 1 (Al). Ionizing the third electron from \[\ce{Al}\hspace{20px}\ce{(Al^2+⟶Al^3+ + e- )} \nonumber\] requires more energy because the cation Al 2+ exerts a stronger pull on the electron than the neutral Al atom, so IE 1 (Al) < IE 3 (Al). The second ionization energy for sodium removes a core electron, which is a much higher energy process than removing valence electrons. Putting this all together, we obtain: IE 1 (Tl) < IE 1 (Al) < IE 3 (Al) < IE 2 (Na). Exercise \(\PageIndex{1}\) Which has the lowest value for IE 1 : O, Po, Pb, or Ba? Answer Ba Summary Video \(\PageIndex{2}\) : A brief review of ionization energy. Ionization energy (the energy associated with forming a cation) decreases down a group and mostly increases across a period because it is easier to remove an electron from a larger, higher energy orbital. Glossary ionization energy energy required to remove an electron from a gaseous atom or ion. The associated number (e.g., second ionization energy) corresponds to the charge of the ion produced (X 2+ )
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Courses/Los_Angeles_Trade_Technical_College/LATTC_Hybrid_Chem_51/10%3A_Solutions
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A solution is a homogeneous mixture —a mixture of two or more substances that are so intimately mixed that the mixture behaves in many ways like a single substance. Many chemical reactions occur when the reactants are dissolved in solution. In this chapter, we will introduce concepts that are applicable to solutions and the chemical reactions that occur in them. 10.1: Definitions Solutions are composed of a solvent (major component) and a solute (minor component). Concentration is the expression of the amount of solute in a given amount of solvent and can be described by several qualitative terms. Solubility is a specific amount of solute that can dissolve in a given amount of solvent. “Like dissolves like” is a useful rule for deciding if a solute will be soluble in a solvent. 10.2: Quantitative Units of Concentration Quantitative units of concentration include molarity, molality, mass percentage, parts per thousand, parts per million, and parts per billion. 10.3: Dilutions and Concentrations Calculate the new concentration or volume for a dilution or concentration of a solution. 10.4: Concentrations as Conversion Factors Know how to apply concentration units as conversion factors.
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Courses/Athabasca_University/Chemistry_360%3A_Organic_Chemistry_II/Chapter_22%3A_Carbonyl_Alpha-Substitution_Reactions/22.03_Alpha_Halogenation_of_Aldehydes_and_Ketones
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Objectives After completing this section, you should be able to write an equation to illustrate the alpha halogenation of aldehydes and ketones. identify the product formed from the alpha halogenation of a given aldehyde or ketone. identify the carbonyl compound, the reagents, or both, needed to prepare a given α ‑halogenated aldehyde or ketone. illustrate the importance of the alpha halogenation of carbonyl compounds as an intermediate step in the synthesis of α , β ‑unsaturated aldehydes and ketones. write a detailed mechanism for the acid‑catalyzed halogenation of a ketone. describe the evidence provided by kinetic experiments supporting the suggestion that the acid‑catalyzed, alpha halogenation of ketones proceeds via the rate‑determining formation of an enol. Study Notes Note: α ‑bromo ketones are a good starting material to generate α , β ‑unsaturated ketones by dehydrobromination. A carbonyl containing compound with α hydrogens can undergo a substitution reaction with halogens. This reaction comes about because of the tendency of carbonyl compounds to form enolates in basic condition and enols in acidic condition. In these cases even weak bases, such as the hydroxide anion, is sufficient enough to cause the reaction to occur because it is not necessary for a complete conversion to the enolate. For this reaction Cl 2 , Br 2 or I 2 can be used as the halogens. General reaction Example 22.3.1 Acid Catalyzed Mechanism Under acidic conditions the reaction occurs thought the formation of an enol which then reacts with the halogen. 1) Protonation of the carbonyl 2) Enol formation 3) S N 2 attack 4) Deprotonation Kinetic studies provide some evidence for the mechanism shown above. The rate law for the alpha-halogenation of a ketone can be given by: rate = [ketone][H + ] The implication is that the rate determining step is dependent on the concentrations of the ketone and acid catalyst and therefore associated with the enol formation part of the mechanism. The halogen does not even appear in the rate law. Indeed, the overall rate is completely independent of the concentration of the halogen and suggests the halogenation step occurs rapidly. Base Catalyzed Mechanism Under basic conditions the enolate forms and then reacts with the halogen. Note! This is base promoted and not base catalyzed because an entire equivalent of base is required. 1) Enolate formation 2) S N 2 attack Overreaction during base promoted α halogenation The fact that an electronegative halogen is placed on an α carbon means that the product of a base promoted α halogenation is actually more reactive than the starting material. The electron withdrawing effect of the halogen makes the α carbon even more acidic and therefor promotes further reaction. Because of this multiple halogenations can occur. This effect is exploited in the haloform reaction discussed later. If a monohalo product is required then acidic conditions are usually used. Deuterium Exchange Due to the acidic nature of α hydrogens they can be exchanged with deuterium by reaction with D 2 O (heavy water). The process is accelerated by the addition of an acid or base; an excess of D 2 O is required. The end result is the complete exchange of all α hydrogens with deuteriums. Example 22.3.2 Mechanism in basic conditions 1) Enolate Formation 2) Deuteration Exercises Exercise 22.3.1 Please draw the products of the following reaction. Answer Exercise 22.3.2 Draw out the mechanism for the following reaction. Answer Exercise 22.3.3 How might you form 2-hepten-4-one from 4-heptanone? Answer 1) Br 2 , H 3 O + 2) Pyridine, Heat Exercise 22.3.4 Show the products of the following reactions: Answer Exercise 22.3.5 The following compound was reacted with D 3 O + . The only signals that could be found in the 1H NMR spectrum of the product were at 3.9 ppm (3H) and 6.6-6.9 ppm (4H). Please explain the results of the NMR. Answer A deuterium exchange reaction occurred. All of the alpha-hydrogens in the molecule have been exchanged with deuterium. Because detueriums do not appear in a typical 1 H NMR, only the remaining hydrogens appear.
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Bookshelves/General_Chemistry/Chemistry_2e_(OpenStax)/17%3A_Electrochemistry/17.06%3A_Corrosion
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Learning Objectives By the end of this section, you will be able to: Define corrosion List some of the methods used to prevent or slow corrosion Corrosion is usually defined as the degradation of metals by a naturally occurring electrochemical process. The formation of rust on iron, tarnish on silver, and the blue-green patina that develops on copper are all examples of corrosion. The total cost of corrosion remediation in the United States is significant, with estimates in excess of half a trillion dollars a year. Chemistry in Everyday Life: Statue of Liberty: Changing Colors The Statue of Liberty is a landmark every American recognizes. The Statue of Liberty is easily identified by its height, stance, and unique blue-green color (Figure \(\PageIndex{1}\)). When this statue was first delivered from France, its appearance was not green. It was brown, the color of its copper “skin.” So how did the Statue of Liberty change colors? The change in appearance was a direct result of corrosion. The copper that is the primary component of the statue slowly underwent oxidation from the air. The oxidation-reduction reactions of copper metal in the environment occur in several steps. Copper metal is oxidized to copper(I) oxide (Cu 2 O), which is red, and then to copper(II) oxide, which is black \[\begin{align} \ce{2 Cu(s) + 1/2 O2(g) -> Cu_2 O (s)} \tag{red} \\[4pt] \ce{2 Cu2O(s) + 1/2 O2(g) -> 2CuO (s)} \tag{black} \end{align} \] Coal, which was often high in sulfur, was burned extensively in the early part of the last century. As a result, atmospheric sulfur trioxide, carbon dioxide, and water all reacted with the \(\ce{CuO}\): \[\begin{align*} \ce{2 CuO(s) + CO2(g) + H2O (l) &-> Cu2CO3(OH)2(s)} \tag{green}\\[4pt] \ce{3 CuO(s) +2 CO2(g) + H2O (l) &-> Cu2(CO3)2(OH)2(s)} \tag{blue} \\[4pt] \ce{4 CuO(s) + SO3(g) + 3 H2O (l) &-> Cu4SO4(OH)6(s)} \tag{green} \end{align*} \nonumber \] These three compounds are responsible for the characteristic blue-green patina seen on the Statue of Liberty (and other outdoor copper structures). Fortunately, formation of patina creates a protective layer on the copper surface, preventing further corrosion of the underlying copper. The formation of the protective layer is called passivation , a phenomenon discussed further in another chapter of this text. Perhaps the most familiar example of corrosion is the formation of rust on iron. Iron will rust when it is exposed to oxygen and water. Rust formation involves the creation of a galvanic cell at an iron surface, as illustrated in Figure \(\PageIndex{1}\). The relevant redox reactions are described by the following equations: \[\begin{align*} &\text { anode: } & \ce{Fe(s) \longrightarrow Fe^{2+}(aq) + 2 e^{-}} && E_{\ce{Fe^{2+} / Fe} }^{\circ}=-0.44\,\text{V} \\[4pt] &\text { cathode: } & \ce{O2(g) + 4 H^{+}(aq) + 4 e^{-} -> 2 H2O (l)} && E_{\ce{O2 /O^{2-}}}^{\circ}=+1.23\,\text{V} \\[4pt] \hline &\text { overall: } & \ce{2 Fe(s) + O2(g) + 4 H^{+}(aq) -> 2 Fe^{2+}(aq) + 2 H2O (l)} && E_{\text {cell }}^{\circ}=+1.67\,\text{V} \end{align*} \nonumber \] Further reaction of the iron(II) product in humid air results in the production of an iron(III) oxide hydrate known as rust: \[\ce{4 Fe^{2+}(aq) + O2(g) + (4+2 x) ~H2O(l) -> 2 Fe2O3 \cdot x~ H2O(s) + 8 H^{+}(aq)} \nonumber \] The stoichiometry of the hydrate varies, as indicated by the use of x in the compound formula. Unlike the patina on copper, the formation of rust does not create a protective layer and so corrosion of the iron continues as the rust flakes off and exposes fresh iron to the atmosphere. One way to keep iron from corroding is to keep it painted. The layer of paint prevents the water and oxygen necessary for rust formation from coming into contact with the iron. As long as the paint remains intact, the iron is protected from corrosion. Other strategies include alloying the iron with other metals. For example, stainless steel is an alloy of iron containing a small amount of chromium. The chromium tends to collect near the surface, where it corrodes and forms a passivating an oxide layer that protects the iron. Iron and other metals may also be protected from corrosion by galvanization , a process in which the metal to be protected is coated with a layer of a more readily oxidized metal, usually zinc. When the zinc layer is intact, it prevents air from contacting the underlying iron and thus prevents corrosion. If the zinc layer is breached by either corrosion or mechanical abrasion, the iron may still be protected from corrosion by a cathodic protection process, which is described in the next paragraph. Another important way to protect metal is to make it the cathode in a galvanic cell. This is cathodic protection and can be used for metals other than just iron. For example, the rusting of underground iron storage tanks and pipes can be prevented or greatly reduced by connecting them to a more active metal such as zinc or magnesium (Figure \(\PageIndex{3}\)). This is also used to protect the metal parts in water heaters. The more active metals (lower reduction potential) are called sacrificial anodes because as they get used up as they corrode (oxidize) at the anode. The metal being protected serves as the cathode for the reduction of oxygen in air, and so it simply serves to conduct (not react with) the electrons being transferred. When the anodes are properly monitored and periodically replaced, the useful lifetime of the iron storage tank can be greatly extended.
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Courses/Lumen_Learning/Book%3A_English_Composition_I-3_(Lumen)/21%3A_Research_Process%3A_Source_Analysis/20.4%3A_Finding_Sources
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Beginning your research with Google or another search engine is an easy way to quickly get an overview on your topic. Even more effective than Google Search is Google Advanced Search , and even better than that for academic resources is Google Scholar . Let’s consider Marvin’s experience. Finding Sources Marvin: So can I just use Google or Bing to find sources? O-Prof: Internet search engines can help you find sources, but they aren’t always the best route to getting to a good source. Try entering the search term “bottled water quality” into Google, without quotation marks around the term. How many hits do you get? Marvin types it in. Marvin: 1,180,000. That’s pretty much what I get whenever I do an Internet search. Too many results. O-Prof: Which is one of the drawbacks of using only Internet search engines. The Internet may have cut down on the physical walking needed to find good sources, but it’s made up for the time savings by pointing you to more places than you could possibly go! But there are some ways you can narrow your search to get fewer, more focused results. Marvin: Yeah, I know. Sometimes I add extra words in and it helps weed down the hits. O-Prof: By combining search terms with certain words or symbols, you can control what the search engine looks for. If you put more than one term into a Google search box, the search engine will only give you sites that include both terms, since it uses the Boolean operator AND as the default for its searches. If you put OR between two search terms, you’ll end up getting even more results, because Google will look for all websites containing either of the terms. Using a minus sign in front of a term eliminates things you’re not interested in. It’s the Google equivalent of the Boolean operator NOT. Try entering bottled water quality health -teeth. Marvin types in the words, remembering suddenly that he has to make an appointment with the dentist. Marvin: 784,000 hits. O-Prof: Still a lot. You can also put quotation marks around groups of words and the search engine will look only for sites that contain all of those words in the exact order you’ve given. And you can combine this strategy with the other ways of limiting your search. Try “bottled water quality” (in quotation marks) health -teeth. Marvin: 225,000. That’s a little better. O-Prof: Now try adding what type of website you are looking for, maybe a .gov or an .edu. Try typing “bottled water quality” heath -teeth site:.edu Marvin: Wow, under 6,000 results now. O-Prof: Yes, a definite improvement. Sometimes you want to be careful though not to narrow it so far that you miss useful sources. You have to play around with your search terms to get to what you need. A bigger problem with Internet search engines, though, is that they won’t necessarily lead you to the sources considered most valuable for college writing. Marvin: My professor said something about using peer-reviewed articles in scholarly journals. O-Prof: Professors will often want you to use such sources. Articles in scholarly journals are written by experts; and if a journal’s peer-reviewed, its articles have been screened by other experts (the authors’ peers) before being published. Marvin: So that would make peer-reviewed articles pretty reliable. Where do I find them? O-Prof: Google’s got a specialized search engine, Google Scholar, that will search for scholarly articles that might be useful. But often the best place is the college library’s bibliographic databases. To be continued. . . Google Scholar Google Scholar is Google’s academic search engine that searches across scholarly literature. It has extensive coverage, retrieving information from academic publishers, professional organizations, university repositories, professional websites, and government websites. The benefits of searching within Google Scholar are numerous, but a search solely using Google Scholar will be insufficient for your research. Consider the following benefits of Google Scholar and library databases. Google Scholar benefits Library Databases benefits Find content not available in library databasesFind more government resources than available in library databasesFind case law instead of or in addition to other contentLimit searches to papers written in a specific language (13 options) and display Google tips in a larger variety of languagesShow library access links for up to five libraries Find content not available in Google ScholarLimit results to full text contentSpecify more information fields (subject headings,abstract, author, etc.) where you want your search terms foundLimit search results by subject headings suggested on the search results pagesLimit results to peer-reviewed, scholarly, or academic journal articles Like other Google search products, Google Scholar starts with a basic search blank. Because researchers are more likely to need the results of more specific searches, the Advanced Search link is accessed via a down-arrow in the search blank. Google Scholar Advanced Search. Keep in mind that Google is not transparent about the journals or time ranges it indexes, and publishers occasionally request that Google Scholar not index their publications. Non-scholarly and/or non-peer reviewed material may also appear in Google Scholar, so it is best used in conjunction with other search tools. One of the greatest features of Google Scholar is the “Cited by” link found below each search result. If you find one article you like, you can click on the link to find other articles that reference that same work. While the following video is specific to the University of West Florida, the same tips and principles still apply to most institutions. Contact your library to ensure you can set up your library account through Google Scholar so you have greater access to articles housed behind paywalls. An interactive or media element has been excluded from this version of the text. You can view it online here: http://pb.libretexts.org/ec1/?p=262 18 – Google Scholar from Joshua Vossler on Vimeo . Google Scholar Search Results Click on the links below to see how search results vary when using different search parameters. caffeine health : A simple Google Scholar search for caffeine and health. caffeine health coffee OR “green tea” OR “black tea” : Results of the search listed above, where the search specifies articles covering caffeine and health, and noting coffee, black tea, or green tea. caffeine health author:”RR McCusker” : A search for articles on a topic by a specific author, Rachel R. McCusker. caffeine health [2012-2016] : These results have been limited to those published from 2012-2016. CC licensed content, Shared previously Walk, Talk, Cook, Eat: A Guide to Using Sources. Authored by : Cynthia R. Haller. Located at : http://www.saylor.org/site/wp-content/uploads/2013/01/writing-spaces-readings-on-writing-vol-2.pdf . Project : Writing Spaces: Readings on Writing Vol. 2. License : CC BY-NC-SA: Attribution-NonCommercial-ShareAlike Google Tips and Tricks: Google Scholar. Provided by : The University of Rhode Island University Libraries. Located at : http://uri.libguides.com/c.php?g=42527&p=269108 . License : CC BY-SA: Attribution-ShareAlike Making the Most of your Search: Google Scholar. Provided by : University of Texas at Arlington Libraries. Located at : http://libguides.uta.edu/c.php?g=472514&p=3232594 . License : CC BY-NC: Attribution-NonCommercial
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