The ChemPile
Collection
8 items
•
Updated
Dataset Viewer
url
stringlengths 0
550
| text
stringlengths 501
228k
|
|---|---|
Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Chemical_Thermodynamics_(Supplement_to_Shepherd_et_al.)/04%3A_Fundamental_2_-_Counting_Configurations/4.04%3A_Applying_the_Laws_of_Probability | The laws of probability apply to events that are independent. If the result of one trial depends on the result of another trial, we may still be able to use the laws of probability. However, to do so, we must know the nature of the interdependence. If the activity associated with event C precedes the activity associated with event D , the probability of D may depend on whether C occurs. Suppose that the first activity is tossing a coin and that the second activity is drawing a card from a deck; however, the deck we use depends on whether the coin comes up heads or tails. If the coin is heads, we draw a card from an ordinary deck; if the coin is tails, we draw a coin from a deck with the face cards removed. Now we ask about the probability of drawing an ace. If the coin is heads, the probability of drawing an ace is \({4}/{52}={1}/{13}\). If the coin is tails, the probability of drawing an ace is \({4}/{40}={1}/{10}\). The combination coin is heads and card is ace has probability: \(\left({1}/{2}\right)\left({1}/{13}\right)={1}/{26}\). The combination coin is tails and card is ace has probability \(\left({1}/{2}\right)\left({1}/{10}\right)={1}/{20}\). In this case, the probability of drawing an ace depends on the modification we make to the deck based on the outcome of the coin toss. Applying the laws of probability is straightforward. An example that illustrates the application of these laws in a transparent way is provided by villages First, Second, Third, and Fourth, which are separated by rivers. (See Figure 1.) Bridges \(1\), \(2\), and \(3\) span the river between First and Second. Bridges \(a\) and \(b\) span the river between Second and Third. Bridges \(A\), \(B\), \(C\), and \(D\) span the river between Third and Fourth. A traveler from First to Fourth who is free to take any route he pleases has a choice from among \(3\times 2\times 4=24\) possible combinations. Let us consider the probabilities associated with various events: There are 24 possible routes. If a traveler chooses his route at random, the probability that he will take any particular route is \({1}/{24}\). This illustrates our assumption that each event in a set of \(N\) exhaustive and mutually exclusive events occurs with probability \({1}/{N}\). If he chooses a route at random, the probability that he goes from First to Second by either bridge \(1\) or bridge \(2\) is \(P\left(1\right)+P\left(2\right)=\ {1}/{3}+{1}/{3}={2}/{3}\). This illustrates the calculation of the probability of alternative events. The probability of the particular route \(2\to a\to C\) is \(P\left(2\right)\times P\left(a\right)\times P\left(C\right)=\left({1}/{3}\right)\left({1}/{2}\right)\left({1}/{4}\right)={1}/{24}\), and we calculate the same probability for any other route from First to Fourth. This illustrates the calculation of the probability of a compound event. If he crosses bridge \(1\), the probability that his route will be \(2\to a\to C\) is zero, of course. The probability of an event that has already occurred is 1, and the probability of any alternative is zero. If he crosses bridge \(1,\) \(P\left(1\right)=1\), and \(P\left(2\right)=P\left(3\right)=0\). Given that a traveler has used bridge \(1\), the probability of the route \(1\to a\to C\) becomes the probability of path \(a\to C\), which is \(P\left(a\right)\times P\left(C\right)=\left({1}/{2}\right)\left({1}/{4}\right)={1}/{8}\). Since \(P\left(1\right)=1\), the probability of the compound event \(1\to a\to C\) is the probability of the compound event \(a\to C\). The outcomes of rolling dice, rolling provide more illustrations. If we roll two dice, we can classify the possible outcomes according to the sums of the outcomes for the individual dice. There are thirty-six possible outcomes. They are displayed in Table 1. Table 1: Outcomes from tossing two dice 0 1 2 3 4 5 6 7 NaN Outcome for first die Outcome for first die Outcome for first die Outcome for first die Outcome for first die Outcome for first die Outcome for first die Outcome for second die NaN 1 2 3 4 5 6 Outcome for second die 1 2 3 4 5 6 7 Outcome for second die 2 3 4 5 6 7 8 Outcome for second die 3 4 5 6 7 8 9 Outcome for second die 4 5 6 7 8 9 10 Outcome for second die 5 6 7 8 9 10 11 Outcome for second die 6 7 8 9 10 11 12 Let us consider the probabilities associated with various dice-throwing events: The probability of any given outcome, say the first die shows \(2\) and the second die shows \(3\), is \({1}/{36}\). Since the probability that the first die shows \(3\) while the second die shows \(2\) is also \({1}/{36}\), the probability that one die shows \(2\) and the other shows \(3\) is \[P\left(3\right)\times P\left(2\right)+P\left(2\right)\times P\left(3\right) =\left({1}/{36}\right)+\left({1}/{36}\right) ={1}/{18}. \nonumber \] Four different outcomes correspond to the event that the score is \(5\). Therefore, the probability of rolling \(5\) is \[P\left(1\right)\times P\left(4\right)+P\left(2\right)\times P\left(3\right) +P\left(3\right)\times P\left(2\right)+P\left(4\right)\times P\left(1\right) ={1}/{9} \nonumber \] The probability of rolling a score of three or less is the probability of rolling \(2\), plus the probability of rolling \(3\) which is \(\left({1}/{36}\right)+\left({2}/{36}\right)={3}/{36}={1}/{12}\) Suppose we roll the dice one at a time and that the first die shows \(2\). The probability of rolling \(7\) when the second die is thrown is now \({1}/{6}\), because only rolling a \(5\) can make the score 7, and there is a probability of \({1}/{6}\) that a \(5\) will come up when the second die is thrown. Suppose the first die is red and the second die is green. The probability that the red die comes up \(2\) and the green die comes up \(3\) is \(\left({1}/{6}\right)\left({1}/{6}\right)={1}/{36}\). Above we looked at the number of outcomes associated with a score of \(3\) to find that the probability of this event is \({1}/{18}\). We can use another argument to get this result. The probability that two dice roll a score of three is equal to the probability that the first die shows \(1\) or \(2\) times the probability that the second die shows whatever score is necessary to make the total equal to three. This is: \[\begin{align*} P\left(first\ die\ shows\ 1\ or\ 2\right)\times \left({1}/{6}\right) &= \left[\left({1}/{6}\right)+\left({1}/{6}\right)\right]\times {1}/{6} \\[4pt] &={2}/{36} \\[4pt]& ={1}/{18} \end{align*} \nonumber \] Application of the laws of probability is frequently made easier by recognizing a simple restatement of the requirement that events be mutually exclusive. In a given trial, either an event occurs or it does not. Let the probability that an event A occurs be \(P\left(A\right)\). Let the probability that event A does not occur be \(P\left(\sim A\right)\). Since in any given trial, the outcome must belong either to event A or to event \(\sim A\), we have \[P\left(A\right)+P\left(\sim A\right)=1 \nonumber \] For example, if the probability of success in a single trial is \({2}/{3}\), the probability of failure is \({1}/{3}\). If we consider the outcomes of two successive trials, we can group them into four events. Event SS: First trial is a success; second trial is a success. Event SF: First trial is a success; second trial is a failure. Event FS: First trial is a failure; second trial is a success. Event FF: First trial is a failure; second trial is a failure. Using the laws of probability, we have \[ \begin{align*} 1 &=P\left(Event\ SS\right)+P\left(Event\ SF\right)+P\left(Event\ FS\right)+\ P(Event\ FF) \\[4pt] &=P_1\left(S\right)\times P_2\left(S\right)+P_1\left(S\right)\times P_2\left(F\right) +P_1(F)\times P_2(S)+P_1(F)\times P_2(F) \end{align*} \nonumber \] where \(P_1\left(X\right)\) and \(P_2\left(X\right)\) are the probability of event \(X\) in the first and second trials, respectively. This situation can be mapped onto a simple diagram. We represent the possible outcomes of the first trial by line segments on one side of a unit square \(P_1\left(S\right)+P_1\left(F\right)=1\). We represent the outcomes of the second trial by line segments along an adjoining side of the unit square. The four possible events are now represented by the areas of four mutually exclusive and exhaustive portions of the unit square as shown in Figure 2. |
Courses/University_of_Georgia/CHEM_3212%3A_Physical_Chemistry_II/07%3A_Entropy_Part_II/7.03%3A_The_Third_Law_of_Thermodynamics | One important consequence of Botlzmann’s proposal is that a perfectly ordered crystal (i.e. one that has only one energetic arrangement in its lowest energy state) will have an entropy of 0. This makes entropy qualitatively different than other thermodynamic functions. For example, in the case of enthalpy, it is impossible have a zero to the scale without setting an arbitrary reference (which is that the enthalpy of formation of elements in their standard states is zero.) But entropy has a natural zero! It is the state at which a system has perfect order. This also has another important consequence, in that it suggests that there must also be a zero to the temperature scale. These consequences are summed up in the Third Law of Thermodynamics . The entropy of a perfectly ordered crystal at 0 K is zero. This also suggests that absolute molar entropies can be calculated by \[S = \int_o^{T} \dfrac{C}{T} dT \nonumber \] where \(C\) is the heat capacity. An entropy value determined in this manner is called a Third Law Entropy . Naturally, the heat capacity will have some temperature dependence. It will also change abruptly if the substance undergoes a phase change. Unfortunately, it is exceedingly difficult to measure heat capacities very near zero K. Fortunately, many substances follow the Debye Extrapolation in that at very low temperatures, their heat capacities are proportional to T 3 . Using this assumption, we have a temperature dependence model that allows us to extrapolate absolute zero based on the heat capacity measured at as low a temperature as can be found. Example \(\PageIndex{1}\) SiO 2 is found to have a molar heat capacity of 0.777 J mol -1 K -1 at 15 K (Yamashita, et al., 2001). Calculate the molar entropy of SiO 2 at 15 K. Solution Using the Debye model, the heat capacity is given by The value of a can be determined by The entropy is then calculated by Calculating a third Law Entropy Start at 0 K, and go from there! |
Bookshelves/General_Chemistry/Chemistry_2e_(OpenStax)/05%3A_Thermochemistry/5.07%3A_Exercises | 1. A burning match and a bonfire may have the same temperature, yet you would not sit around a burning match on a fall evening to stay warm. Why not? 2. Prepare a table identifying several energy transitions that take place during the typical operation of an automobile. 3. Explain the difference between heat capacity and specific heat of a substance. 4. Calculate the heat capacity, in joules and in calories per degree, of the following: 28.4 g of water 1.00 oz of lead 5. Calculate the heat capacity, in joules and in calories per degree, of the following: 45.8 g of nitrogen gas 1.00 pound of aluminum metal 6. How much heat, in joules and in calories, must be added to a 75.0–g iron block with a specific heat of 0.449 J/g °C to increase its temperature from 25 °C to its melting temperature of 1535 °C? 7. How much heat, in joules and in calories, is required to heat a 28.4-g (1-oz) ice cube from −23.0 °C to −1.0 °C? 8. How much would the temperature of 275 g of water increase if 36.5 kJ of heat were added? 9. If 14.5 kJ of heat were added to 485 g of liquid water, how much would its temperature increase? 10. A piece of unknown substance weighs 44.7 g and requires 2110 J to increase its temperature from 23.2 °C to 89.6 °C. What is the specific heat of the substance? If it is one of the substances found in Table 5.1, what is its likely identity? 11. A piece of unknown solid substance weighs 437.2 g, and requires 8460 J to increase its temperature from 19.3 °C to 68.9 °C. What is the specific heat of the substance? If it is one of the substances found in Table 5.1, what is its likely identity? 12. An aluminum kettle weighs 1.05 kg. What is the heat capacity of the kettle? How much heat is required to increase the temperature of this kettle from 23.0 °C to 99.0 °C? How much heat is required to heat this kettle from 23.0 °C to 99.0 °C if it contains 1.25 L of water (density of 0.997 g/mL and a specific heat of 4.184 J/g °C)? 13. Most people find waterbeds uncomfortable unless the water temperature is maintained at about 85 °F. Unless it is heated, a waterbed that contains 892 L of water cools from 85 °F to 72 °F in 24 hours. Estimate the amount of electrical energy required over 24 hours, in kWh, to keep the bed from cooling. Note that 1 kilowatt-hour (kWh) = 3.6 10 6 J, and assume that the density of water is 1.0 g/mL (independent of temperature). What other assumptions did you make? How did they affect your calculated result (i.e., were they likely to yield “positive” or “negative” errors)? 14. A 500-mL bottle of water at room temperature and a 2-L bottle of water at the same temperature were placed in a refrigerator. After 30 minutes, the 500-mL bottle of water had cooled to the temperature of the refrigerator. An hour later, the 2-L of water had cooled to the same temperature. When asked which sample of water lost the most heat, one student replied that both bottles lost the same amount of heat because they started at the same temperature and finished at the same temperature. A second student thought that the 2-L bottle of water lost more heat because there was more water. A third student believed that the 500-mL bottle of water lost more heat because it cooled more quickly. A fourth student thought that it was not possible to tell because we do not know the initial temperature and the final temperature of the water. Indicate which of these answers is correct and describe the error in each of the other answers. 15. Would the amount of heat measured for the reaction in Example 5.5 be greater, lesser, or remain the same if we used a calorimeter that was a poorer insulator than a coffee cup calorimeter? Explain your answer. 16. Would the amount of heat absorbed by the dissolution in Example 5.6 appear greater, lesser, or remain the same if the experimenter used a calorimeter that was a poorer insulator than a coffee cup calorimeter? Explain your answer. 17. Would the amount of heat absorbed by the dissolution in Example 5.6 appear greater, lesser, or remain the same if the heat capacity of the calorimeter were taken into account? Explain your answer. 18. How many milliliters of water at 23 °C with a density of 1.00 g/mL must be mixed with 180 mL (about 6 oz) of coffee at 95 °C so that the resulting combination will have a temperature of 60 °C? Assume that coffee and water have the same density and the same specific heat. 19. How much will the temperature of a cup (180 g) of coffee at 95 °C be reduced when a 45 g silver spoon (specific heat 0.24 J/g °C) at 25 °C is placed in the coffee and the two are allowed to reach the same temperature? Assume that the coffee has the same density and specific heat as water. 20. A 45-g aluminum spoon (specific heat 0.88 J/g °C) at 24 °C is placed in 180 mL (180 g) of coffee at 85 °C and the temperature of the two become equal. What is the final temperature when the two become equal? Assume that coffee has the same specific heat as water. The first time a student solved this problem she got an answer of 88 °C. Explain why this is clearly an incorrect answer. 21. The temperature of the cooling water as it leaves the hot engine of an automobile is 240 °F. After it passes through the radiator it has a temperature of 175 °F. Calculate the amount of heat transferred from the engine to the surroundings by one gallon of water with a specific heat of 4.184 J/g °C. 22. A 70.0-g piece of metal at 80.0 °C is placed in 100 g of water at 22.0 °C contained in a calorimeter like that shown in Figure 5.12. The metal and water come to the same temperature at 24.6 °C. How much heat did the metal give up to the water? What is the specific heat of the metal? 23. If a reaction produces 1.506 kJ of heat, which is trapped in 30.0 g of water initially at 26.5 °C in a calorimeter like that in Figure 5.12, what is the resulting temperature of the water? 24. A 0.500-g sample of KCl is added to 50.0 g of water in a calorimeter (Figure 5.12). If the temperature decreases by 1.05 °C, what is the approximate amount of heat involved in the dissolution of the KCl, assuming the specific heat of the resulting solution is 4.18 J/g °C? Is the reaction exothermic or endothermic? 25. Dissolving 3.0 g of CaCl 2 ( s ) in 150.0 g of water in a calorimeter (Figure 5.12) at 22.4 °C causes the temperature to rise to 25.8 °C. What is the approximate amount of heat involved in the dissolution, assuming the specific heat of the resulting solution is 4.18 J/g °C? Is the reaction exothermic or endothermic? 26. When 50.0 g of 0.200 M NaCl( aq ) at 24.1 °C is added to 100.0 g of 0.100 M AgNO 3 ( aq ) at 24.1 °C in a calorimeter, the temperature increases to 25.2 °C as AgCl( s ) forms. Assuming the specific heat of the solution and products is 4.20 J/g °C, calculate the approximate amount of heat in joules produced. 27. The addition of 3.15 g of Ba(OH) 2 ·8H 2 O to a solution of 1.52 g of NH 4 SCN in 100 g of water in a calorimeter caused the temperature to fall by 3.1 °C. Assuming the specific heat of the solution and products is 4.20 J/g °C, calculate the approximate amount of heat absorbed by the reaction, which can be represented by the following equation: Ba(OH) 2 ·8H 2 O( s ) + 2NH 4 SCN( aq ) ⟶ Ba(SCN) 2 ( aq ) + 2NH 3 ( aq ) + 10H 2 O( l ) 28. The reaction of 50 mL of acid and 50 mL of base described in Example 5.5 increased the temperature of the solution by 6.9 ºC. How much would the temperature have increased if 100 mL of acid and 100 mL of base had been used in the same calorimeter starting at the same temperature of 22.0 ºC? Explain your answer. 29. If the 3.21 g of NH 4 NO 3 in Example 5.6 were dissolved in 100.0 g of water under the same conditions, how much would the temperature change? Explain your answer. 30. When 1.0 g of fructose, C 6 H 12 O 6 ( s ), a sugar commonly found in fruits, is burned in oxygen in a bomb calorimeter, the temperature of the calorimeter increases by 1.58 °C. If the heat capacity of the calorimeter and its contents is 9.90 kJ/°C, what is q for this combustion? 31. When a 0.740-g sample of trinitrotoluene (TNT), C 7 H 5 N 2 O 6 , is burned in a bomb calorimeter, the temperature increases from 23.4 °C to 26.9 °C. The heat capacity of the calorimeter is 534 J/°C, and it contains 675 mL of water. How much heat was produced by the combustion of the TNT sample? 32. One method of generating electricity is by burning coal to heat water, which produces steam that drives an electric generator. To determine the rate at which coal is to be fed into the burner in this type of plant, the heat of combustion per ton of coal must be determined using a bomb calorimeter. When 1.00 g of coal is burned in a bomb calorimeter (Figure 5.17), the temperature increases by 1.48 °C. If the heat capacity of the calorimeter is 21.6 kJ/°C, determine the heat produced by combustion of a ton of coal (2.000 10 3 pounds). 33. The amount of fat recommended for someone with a daily diet of 2000 Calories is 65 g. What percent of the calories in this diet would be supplied by this amount of fat if the average number of Calories for fat is 9.1 Calories/g? 34. A teaspoon of the carbohydrate sucrose (common sugar) contains 16 Calories (16 kcal). What is the mass of one teaspoon of sucrose if the average number of Calories for carbohydrates is 4.1 Calories/g? 35. What is the maximum mass of carbohydrate in a 6-oz serving of diet soda that contains less than 1 Calorie per can if the average number of Calories for carbohydrates is 4.1 Calories/g? 36. A pint of premium ice cream can contain 1100 Calories. What mass of fat, in grams and pounds, must be produced in the body to store an extra 1.1 10 3 Calories if the average number of Calories for fat is 9.1 Calories/g? 37. A serving of a breakfast cereal contains 3 g of protein, 18 g of carbohydrates, and 6 g of fat. What is the Calorie content of a serving of this cereal if the average number of Calories for fat is 9.1 Calories/g, for carbohydrates is 4.1 Calories/g, and for protein is 4.1 Calories/g? 38. Which is the least expensive source of energy in kilojoules per dollar: a box of breakfast cereal that weighs 32 ounces and costs $4.23, or a liter of isooctane (density, 0.6919 g/mL) that costs $0.45? Compare the nutritional value of the cereal with the heat produced by combustion of the isooctane under standard conditions. A 1.0-ounce serving of the cereal provides 130 Calories. 39. Explain how the heat measured in Example 5.5 differs from the enthalpy change for the exothermic reaction described by the following equation: 40. Using the data in the check your learning section of Example 5.5, calculate Δ H in kJ/mol of AgNO 3 ( aq ) for the reaction: 41. Calculate the enthalpy of solution (Δ H for the dissolution) per mole of NH 4 NO 3 under the conditions described in Example 5.6. 42. Calculate Δ H for the reaction described by the equation. ( Hint : Use the value for the approximate amount of heat absorbed by the reaction that you calculated in a previous exercise.) 43. Calculate the enthalpy of solution (Δ H for the dissolution) per mole of CaCl 2 (refer to Exercise 5.25). 44. Although the gas used in an oxyacetylene torch (Figure 5.7) is essentially pure acetylene, the heat produced by combustion of one mole of acetylene in such a torch is likely not equal to the enthalpy of combustion of acetylene listed in Table 5.2. Considering the conditions for which the tabulated data are reported, suggest an explanation. 45. How much heat is produced by burning 4.00 moles of acetylene under standard state conditions? 46. How much heat is produced by combustion of 125 g of methanol under standard state conditions? 47. How many moles of isooctane must be burned to produce 100 kJ of heat under standard state conditions? 48. What mass of carbon monoxide must be burned to produce 175 kJ of heat under standard state conditions? 49. When 2.50 g of methane burns in oxygen, 125 kJ of heat is produced. What is the enthalpy of combustion per mole of methane under these conditions? 50. How much heat is produced when 100 mL of 0.250 M HCl (density, 1.00 g/mL) and 200 mL of 0.150 M NaOH (density, 1.00 g/mL) are mixed? If both solutions are at the same temperature and the specific heat of the products is 4.19 J/g °C, how much will the temperature increase? What assumption did you make in your calculation? 51. A sample of 0.562 g of carbon is burned in oxygen in a bomb calorimeter, producing carbon dioxide. Assume both the reactants and products are under standard state conditions, and that the heat released is directly proportional to the enthalpy of combustion of graphite. The temperature of the calorimeter increases from 26.74 °C to 27.93 °C. What is the heat capacity of the calorimeter and its contents? 52. Before the introduction of chlorofluorocarbons, sulfur dioxide (enthalpy of vaporization, 6.00 kcal/mol) was used in household refrigerators. What mass of SO 2 must be evaporated to remove as much heat as evaporation of 1.00 kg of CCl 2 F 2 (enthalpy of vaporization is 17.4 kJ/mol)? The vaporization reactions for SO 2 and CCl 2 F 2 are and respectively. 53. Homes may be heated by pumping hot water through radiators. What mass of water will provide the same amount of heat when cooled from 95.0 to 35.0 °C, as the heat provided when 100 g of steam is cooled from 110 °C to 100 °C. 54. Which of the enthalpies of combustion in Table 5.2 the table are also standard enthalpies of formation? 55. Does the standard enthalpy of formation of H 2 O( g ) differ from Δ H ° for the reaction 56. Joseph Priestly prepared oxygen in 1774 by heating red mercury(II) oxide with sunlight focused through a lens. How much heat is required to decompose exactly 1 mole of red HgO( s ) to Hg( l ) and O 2 ( g ) under standard conditions? 57. How many kilojoules of heat will be released when exactly 1 mole of manganese, Mn, is burned to form Mn 3 O 4 ( s ) at standard state conditions? 58. How many kilojoules of heat will be released when exactly 1 mole of iron, Fe, is burned to form Fe 2 O 3 ( s ) at standard state conditions? 59. The following sequence of reactions occurs in the commercial production of aqueous nitric acid: Determine the total energy change for the production of one mole of aqueous nitric acid by this process. 60. Both graphite and diamond burn. For the conversion of graphite to diamond: Which produces more heat, the combustion of graphite or the combustion of diamond? 61. From the molar heats of formation in Appendix G, determine how much heat is required to evaporate one mole of water: 62. Which produces more heat? or for the phase change 63. Calculate for the process from the following information: 64. Calculate for the process from the following information: 65. Calculate Δ H for the process from the following information: 66. Calculate for the process from the following information: 67. Calculate the standard molar enthalpy of formation of NO( g ) from the following data: 68. Using the data in Appendix G, calculate the standard enthalpy change for each of the following reactions: 69. Using the data in Appendix G, calculate the standard enthalpy change for each of the following reactions: 70. The following reactions can be used to prepare samples of metals. Determine the enthalpy change under standard state conditions for each. 71. The decomposition of hydrogen peroxide, H 2 O 2 , has been used to provide thrust in the control jets of various space vehicles. Using the data in Appendix G, determine how much heat is produced by the decomposition of exactly 1 mole of H 2 O 2 under standard conditions. 72. Calculate the enthalpy of combustion of propane, C 3 H 8 ( g ), for the formation of H 2 O( g ) and CO 2 ( g ). The enthalpy of formation of propane is −104 kJ/mol. 73. Calculate the enthalpy of combustion of butane, C 4 H 10 ( g ) for the formation of H 2 O( g ) and CO 2 ( g ). The enthalpy of formation of butane is −126 kJ/mol. 74. Both propane and butane are used as gaseous fuels. Which compound produces more heat per gram when burned? 75. The white pigment TiO 2 is prepared by the reaction of titanium tetrachloride, TiCl 4 , with water vapor in the gas phase: How much heat is evolved in the production of exactly 1 mole of TiO 2 ( s ) under standard state conditions? 76. Water gas, a mixture of H 2 and CO, is an important industrial fuel produced by the reaction of steam with red hot coke, essentially pure carbon: Under the conditions of the reaction, methanol forms as a gas. Calculate for this reaction and for the condensation of gaseous methanol to liquid methanol. (c) Calculate the heat of combustion of 1 mole of liquid methanol to H 2 O( g ) and CO 2 ( g ). 77. In the early days of automobiles, illumination at night was provided by burning acetylene, C 2 H 2 . Though no longer used as auto headlamps, acetylene is still used as a source of light by some cave explorers. The acetylene is (was) prepared in the lamp by the reaction of water with calcium carbide, CaC 2 : Calculate the standard enthalpy of the reaction. The of CaC 2 is −15.14 kcal/mol. 78. From the data in Table 5.2, determine which of the following fuels produces the greatest amount of heat per gram when burned under standard conditions: CO( g ), CH 4 ( g ), or C 2 H 2 ( g ). 79. The enthalpy of combustion of hard coal averages −35 kJ/g, that of gasoline, 1.28 10 5 kJ/gal. How many kilograms of hard coal provide the same amount of heat as is available from 1.0 gallon of gasoline? Assume that the density of gasoline is 0.692 g/mL (the same as the density of isooctane). 80. Ethanol, C 2 H 5 OH, is used as a fuel for motor vehicles, particularly in Brazil. Write the balanced equation for the combustion of ethanol to CO 2 ( g ) and H 2 O( g ), and, using the data in Appendix G, calculate the enthalpy of combustion of 1 mole of ethanol. The density of ethanol is 0.7893 g/mL. Calculate the enthalpy of combustion of exactly 1 L of ethanol. Assuming that an automobile’s mileage is directly proportional to the heat of combustion of the fuel, calculate how much farther an automobile could be expected to travel on 1 L of gasoline than on 1 L of ethanol. Assume that gasoline has the heat of combustion and the density of n–octane, C 8 H 18 density = 0.7025 g/mL). 81. Among the substances that react with oxygen and that have been considered as potential rocket fuels are diborane [B 2 H 6 , produces B 2 O 3 ( s ) and H 2 O( g )], methane [CH 4 , produces CO 2 ( g ) and H 2 O( g )], and hydrazine [N 2 H 4 , produces N 2 ( g ) and H 2 O( g )]. On the basis of the heat released by 1.00 g of each substance in its reaction with oxygen, which of these compounds offers the best possibility as a rocket fuel? The 82. How much heat is produced when 1.25 g of chromium metal reacts with oxygen gas under standard conditions? 83. Ethylene, C 2 H 4 , a byproduct from the fractional distillation of petroleum, is fourth among the 50 chemical compounds produced commercially in the largest quantities. About 80% of synthetic ethanol is manufactured from ethylene by its reaction with water in the presence of a suitable catalyst. Using the data in the table in Appendix G, calculate Δ H ° for the reaction. 84. The oxidation of the sugar glucose, C 6 H 12 O 6 , is described by the following equation: The metabolism of glucose gives the same products, although the glucose reacts with oxygen in a series of steps in the body. How much heat in kilojoules can be produced by the metabolism of 1.0 g of glucose? How many Calories can be produced by the metabolism of 1.0 g of glucose? 85. Propane, C 3 H 8 , is a hydrocarbon that is commonly used as a fuel. Write a balanced equation for the complete combustion of propane gas. Calculate the volume of air at 25 °C and 1.00 atmosphere that is needed to completely combust 25.0 grams of propane. Assume that air is 21.0 percent O 2 by volume. (Hint: We will see how to do this calculation in a later chapter on gases—for now use the information that 1.00 L of air at 25 °C and 1.00 atm contains 0.275 g of O 2 .) The heat of combustion of propane is −2,219.2 kJ/mol. Calculate the heat of formation, of propane given that of H 2 O( l ) = −285.8 kJ/mol and of CO 2 ( g ) = −393.5 kJ/mol. Assuming that all of the heat released in burning 25.0 grams of propane is transferred to 4.00 kilograms of water, calculate the increase in temperature of the water. 86. During a recent winter month in Sheboygan, Wisconsin, it was necessary to obtain 3500 kWh of heat provided by a natural gas furnace with 89% efficiency to keep a small house warm (the efficiency of a gas furnace is the percent of the heat produced by combustion that is transferred into the house). (f) How many kilowatt–hours (1 kWh = 3.6 10 6 J) of electricity would be required to provide the heat necessary to heat the house? Note electricity is 100% efficient in producing heat inside a house. (g) Although electricity is 100% efficient in producing heat inside a house, production and distribution of electricity is not 100% efficient. The efficiency of production and distribution of electricity produced in a coal-fired power plant is about 40%. A certain type of coal provides 2.26 kWh per pound upon combustion. What mass of this coal in kilograms will be required to produce the electrical energy necessary to heat the house if the efficiency of generation and distribution is 40%? |
Courses/Chandler_Gilbert_Community_College/Fundamental_Organic_ala_Mech/05%3A_Unit_2/5.04%3A_Apply_to_Acids_and_Bases/5.4.05%3A_Acids_and_Bases_-_The_Lewis_Definition | The Lewis definition of acids and bases is more encompassing than the Brønsted–Lowry definition because it’s not limited to substances that donate or accept just protons. A Lewis acid is a substance that accepts an electron pair, and a Lewis base is a substance that donates an electron pair. The donated electron pair is shared between the acid and the base in a covalent bond. Lewis Acids and the Curved Arrow Formalism The fact that a Lewis acid is able to accept an electron pair means that it must have either a vacant, low-energy orbital or a polar bond to hydrogen so that it can donate H + (which has an empty 1 s orbital). Thus, the Lewis definition of acidity includes many species in addition to H + . For example, various metal cations, such as Mg 2 + , are Lewis acids because they accept a pair of electrons when they form a bond to a base. We’ll also see in later chapters that certain metabolic reactions begin with an acid–base reaction between Mg 2 + as a Lewis acid and an organic diphosphate or triphosphate ion as the Lewis base. In the same way, compounds of group 3A elements, such as BF 3 and AlCl 3 , are Lewis acids because they have unfilled valence orbitals and can accept electron pairs from Lewis bases, as shown in Figure \(\PageIndex{1}\). Similarly, many transition-metal compounds, such as TiCl 4 , FeCl 3 , ZnCl 2 , and SnCl 4 , are Lewis acids. Look closely at the acid–base reaction in Figure \(\PageIndex{1}\), and notice how it's shown. Dimethyl ether, the Lewis base, donates an electron pair to a vacant valence orbital of the boron atom in BF 3 , a Lewis acid. The direction of electron-pair flow from base to acid is shown using a curved arrow, just as the direction of electron flow from one resonance structure to another was shown using curved arrows in Section 2.6. We’ll use this curved-arrow notation throughout the remainder of this text to indicate electron flow during reactions, so get used to seeing it. Some further examples of Lewis acids follow: Lewis Bases The Lewis definition of a base—a compound with a pair of nonbonding electrons that it can use to bond to a Lewis acid—is similar to the Brønsted–Lowry definition. Thus, H 2 O, with its two pairs of nonbonding electrons on oxygen, acts as a Lewis base by donating an electron pair to an H + in forming the hydronium ion, H 3 O + . In a more general sense, most oxygen- and nitrogen-containing organic compounds can act as Lewis bases because they have lone pairs of electrons. A divalent oxygen compound has two lone pairs of electrons, and a trivalent nitrogen compound has one lone pair. Note in the following examples that some compounds can act as both acids and bases, just as water can. Alcohols and carboxylic acids, for instance, act as acids when they donate an H + but as bases when their oxygen atom accepts an H + . Note also that some Lewis bases, such as carboxylic acids, esters, and amides, have more than one atom with a lone pair of electrons and can therefore react at more than one site. Acetic acid, for example, can be protonated either on the doubly bonded oxygen atom or on the singly bonded oxygen atom. The reaction normally occurs only once in such instances, and the more stable of the two possible protonation products are formed. For acetic acid, protonation by reaction with sulfuric acid occurs on the doubly bonded oxygen because that product is stabilized by two resonance forms. Worked Example \(\PageIndex{1}\): Using Curved Arrows to Show Electron Flow Using curved arrows, show how acetaldehyde, CH 3 CHO, can act as a Lewis base. Strategy A Lewis base donates an electron pair to a Lewis acid. We therefore need to locate the electron lone pairs on acetaldehyde and use a curved arrow to show the movement of a pair toward the H atom of the acid. Solution Exercise \(\PageIndex{1}\) Using curved arrows, show how the species in part (a) can act as Lewis bases in their reactions with HCl, and show how the species in part (b) can act as Lewis acids in their reaction with OH – . CH 3 CH 2 OH, HN(CH 3 ) 2 , P(CH 3 ) 3 H 3 C + , B(CH 3 ) 3 , MgBr 2 Answer Exercise \(\PageIndex{2}\) Imidazole, which forms part of amino acid histidine, can act as both an acid and a base. Look at the electrostatic potential map of imidazole, and identify the most acidic hydrogen atom and the most basic nitrogen atom. Draw structures for the resonance forms of the products that result when imidazole is protonated by an acid and deprotonated by a base. Answer |
Courses/Portland_Community_College/CH106%3A_Allied_Health_Chemistry_III/04%3A_Aldehydes_and_Ketones/4.02%3A_Aldehydes_and_Ketones-_Structure_and_Names | Learning Objectives Identify the general structure for an aldehyde and a ketone. Use common names to name aldehydes and ketones. Use the IUPAC system to name aldehydes and ketones. The next functional group we consider, the carbonyl group, has a carbon-to-oxygen double bond. Carbonyl groups define two related families of organic compounds: the aldehydes and the ketones. The carbonyl group is ubiquitous in biological compounds. It is found in carbohydrates, fats, proteins, nucleic acids, hormones, and vitamins—organic compounds critical to living systems. In a ketone, two carbon groups are attached to the carbonyl carbon atom. The following general formulas, in which R represents an alkyl group and Ar stands for an aryl group, represent ketones. In an aldehyde, at least one of the attached groups must be a hydrogen atom. The following compounds are aldehydes: In condensed formulas, we use CHO to identify an aldehyde rather than CO H , which might be confused with an alcohol. This follows the general rule that in condensed structural formulas H comes after the atom it is attached to (usually C, N, or O). The carbon-to-oxygen double bond is not shown but understood to be present. Because they contain the same functional group, aldehydes and ketones share many common properties, but they still differ enough to warrant their classification into two families. Naming Aldehydes and Ketones Both common and International Union of Pure and Applied Chemistry (IUPAC) names are frequently used for aldehydes and ketones, with common names predominating for the lower homologs. The common names of aldehydes are taken from the names of the acids into which the aldehydes can be converted by oxidation . The stems for the common names of the first four aldehydes are as follows: 1 carbon atom: form - 2 carbon atoms: acet - 3 carbon atoms: propion - 4 carbon atoms: butyr - Because the carbonyl group in a ketone must be attached to two carbon groups, the simplest ketone has three carbon atoms. It is widely known as acetone , a unique name unrelated to other common names for ketones. Generally, the common names of ketones consist of the names of the groups attached to the carbonyl group, followed by the word ketone . (Note the similarity to the naming of ethers.) Another name for acetone, then, is dimethyl ketone. The ketone with four carbon atoms is ethyl methyl ketone. Example \(\PageIndex{1}\) Classify each compound as an aldehyde or a ketone. Give the common name for each ketone. Solution This compound has the carbonyl group on an end carbon atom, so it is an aldehyde. This compound has the carbonyl group on an interior carbon atom, so it is a ketone. Both alkyl groups are propyl groups. The name is therefore dipropyl ketone. This compound has the carbonyl group between two alkyl groups, so it is a ketone. One alkyl group has three carbon atoms and is attached by the middle carbon atom; it is an isopropyl group. A group with one carbon atom is a methyl group. The name is therefore isopropyl methyl ketone. Exercise \(\PageIndex{1}\) Classify each compound as an aldehyde or a ketone. Give the common name for each ketone. Here are some simple IUPAC rules for naming aldehydes and ketones: The stem names of aldehydes and ketones are derived from those of the parent alkanes, defined by the longest continuous chain (LCC) of carbon atoms that contains the functional group. For an aldehyde, drop the - e from the alkane name and add the ending - al . Methanal is the IUPAC name for formaldehyde, and ethanal is the name for acetaldehyde. For a ketone, drop the - e from the alkane name and add the ending - one . Propanone is the IUPAC name for acetone, and butanone is the name for ethyl methyl ketone. To indicate the position of a substituent on an aldehyde, the carbonyl carbon atom is always considered to be C1; it is unnecessary to designate this group by number. To indicate the position of a substituent on a ketone, number the chain in the manner that gives the carbonyl carbon atom the lowest possible number. In cyclic ketones, it is understood that the carbonyl carbon atom is C1. Example \(\PageIndex{2}\) Give the IUPAC name for each compound. Solution There are five carbon atoms in the LCC . The methyl group (CH 3 ) is a substituent on the second carbon atom of the chain; the aldehyde carbon atom is always C1. The name is derived from pentane. Dropping the - e and adding the ending - al gives pentanal. The methyl group on the second carbon atom makes the name 2-methylpentanal. There are five carbon atoms in the LCC. The carbonyl carbon atom is C3, and there are methyl groups on C2 and C4. The IUPAC name is 2,4-dimethyl-3-pentanone. There are six carbon atoms in the ring. The compound is cyclohexanone. No number is needed to indicate the position of the carbonyl group because all six carbon atoms are equivalent. Exercise Give the IUPAC name for each compound. Example \(\PageIndex{3}\) Draw the structure for each compound. 7-chlorooctanal 4-methyl–3-hexanone Solution The octan - part of the name tells us that the LCC has eight carbon atoms. There is a chlorine (Cl) atom on the seventh carbon atom; numbering from the carbonyl group and counting the carbonyl carbon atom as C1, we place the Cl atom on the seventh carbon atom. The hexan - part of the name tells us that the LCC has six carbon atoms. The 3 means that the carbonyl carbon atom is C3 in this chain, and the 4 tells us that there is a methyl (CH 3 ) group at C4: Exercise \(\PageIndex{3}\) Draw the structure for each compound. 5-bromo-3-iodoheptanal 5-bromo-4-ethyl-2-heptanone Summary The common names of aldehydes are taken from the names of the corresponding carboxylic acids: formaldehyde, acetaldehyde, and so on. The common names of ketones, like those of ethers, consist of the names of the groups attached to the carbonyl group, followed by the word ketone . Stem names of aldehydes and ketones are derived from those of the parent alkanes, using an - al ending for an aldehydes and an - one ending for a ketone. |
Courses/Taft_College/CHEM_1510%3A_Introductory_College_Chemistry/16%3A_Acids_and_Bases/16.08%3A_The_pH_and_pOH_Scales_-_Ways_to_Express_Acidity_and_Basicity | Learning Objectives Define pH and pOH. Determine the pH of acidic and basic solutions. Determine the hydronium ion concentration and pOH from pH. As we have seen, \([H_3O^+]\) and \([OH^−]\) values can be markedly different from one aqueous solution to another. So chemists defined a new scale that succinctly indicates the concentrations of either of these two ions. \(pH\) is a logarithmic function of \([H_3O^+]\): \[pH = −\log[H_3O^+] \label{pH} \] \(pH\) is usually (but not always) between 0 and 14. Knowing the dependence of \(pH\) on \([H_3O^+]\), we can summarize as follows: If pH < 7, then the solution is acidic. If pH = 7, then the solution is neutral. If pH > 7, then the solution is basic. This is known as the \(pH\) scale. The pH scale is the range of value s from 0 to 14 that describes the acidity or basicity of a solution. You can use \(pH\) to make a quick determination whether a given aqueous solution is acidic, basic, or neutral. Figure \(\PageIndex{1}\) illus trates this relationship, along with some examples of various solutions. Because hydrogen ion concentrations are generally less than one (for example \(1.3 \times 10^{-3}\,M\)), th e log of the number will be a negative number. To make pH even easier to work with, pH is defined as the negative log of \([H_3O^+]\) , which will give a positive value for pH. Example \(\PageIndex{1}\) Label each solution as acidic, basic, or neutral based only on the stated \(pH\). milk of magnesia, pH = 10.5 pure water, pH = 7 wine, pH = 3.0 Answer With a pH greater than 7, milk of magnesia is basic. (Milk of magnesia is largely Mg(OH) 2 .) Pure water, with a pH of 7, is neutral. With a pH of less than 7, wine is acidic. Exercise \(\PageIndex{1}\) Identify each substance as acidic, basic, or neutral based only on the stated \(pH\). human blood with \(pH\) = 7.4 household ammonia with \(pH\) = 11.0 cherries with \(pH\) = 3.6 Answer a basic Answer b basic Answer c acidic Calculating pH from Hydronium Concentration The pH of solutions can be determined by using logarithms as illustrated in the next example for stomach acid. Stomach acid is a solution of \(HCl\) with a hydronium ion concentration of \(1.2 \times 10^{−3}\; M\), what is the \(pH\) of the solution? \[ \begin{align} \mathrm{pH} &= \mathrm{-\log [H_3O^+]} \nonumber \\ &=-\log(1.2 \times 10^{−3}) \nonumber \\ &=−(−2.92)=2.92 \nonumber \end{align} \nonumber \] Logarithms To get the log value on your calculator, enter the number (in this case, the hydronium ion concentration) first, then press the LOG key. If the number is 1.0 x 10 -5 (for [H 3 O + ] = 1.0 x 10 -5 M) you should get an answer of "-5". If you get a different answer, or an error, try pressing the LOG key before you enter the number. Example \(\PageIndex{2}\): Converting Ph to Hydronium Concentration Find the pH, given the \([H_3O^+]\) of the following: 1 ×10 -3 M 2.5 ×10 -11 M 4.7 ×10 -9 M Solution Steps for Problem Solving Unnamed: 1 Identify the "given" information and what the problem is asking you to "find." Given: [H3O+] =1 × 10−3 M [H3O+] =2.5 ×10-11 M [H3O+] = 4.7 ×10-9 M Find: ? pH Plan the problem. Need to use the expression for pH (Equation \ref{pH}). pH = - log [H3O+] Calculate. Now substitute the known quantity into the equation and solve. pH = - log [1 × 10−3 ] = 3.0 (1 decimal places since 1 has 1 significant figure) pH = - log [2.5 ×10-11] = 10.60 (2 decimal places since 2.5 has 2 significant figures) pH = - log [4.7 ×10-9] = 8.30 (2 decimal places since 4.7 has 2 significant figures) The other issue that concerns us here is significant figures. Because the number(s) before the decimal point in a logarithm relate to the power on 10, the number of digits after the decimal point is what determines the number of significant figures in the final answer: Exercise \(\PageIndex{2}\) Find the pH, given [H 3 O + ] of the following: 5.8 ×10 -4 M 1.0×10 -7 Answer a 3.22 Answer b 7.00 Calculating Hydronium Concentration from pH Sometimes you need to work "backwards"—you know the pH of a solution and need to find \([H_3O^+]\), or even the concentration of the acid solution. How do you do that? To convert pH into \([H_3O^+]\) we solve Equation \ref{pH} for \([H_3O^+]\). This involves taking the antilog (or inverse log) of the negative value of pH . \[[\ce{H3O^{+}}] = \text{antilog} (-pH) \nonumber \] or \[[\ce{H_3O^+}] = 10^{-pH} \label{ph1} \] As mentioned above, different calculators work slightly differently—make sure you can do the following calculations using your calculator. Calculator Skills We have a solution with a pH = 8.3. What is [H 3 O + ] ? With some calculators you will do things in the following order: Enter 8.3 as a negative number (use the key with both the +/- signs, not the subtraction key). Use your calculator's 2nd or Shift or INV (inverse) key to type in the symbol found above the LOG key. The shifted function should be 10 x . You should get the answer 5.0 × 10 -9 . Other calculators require you to enter keys in the order they appear in the equation. Use the Shift or second function to key in the 10 x function. Use the +/- key to type in a negative number, then type in 8.3. You should get the answer 5.0 × 10 -9 . If neither of these methods work, try rearranging the order in which you type in the keys. Don't give up—you must master your calculator! Example \(\PageIndex{3}\): Calculating Hydronium Concentration from pH Find the hydronium ion concentration in a solution with a pH of 12.6. Is this solution an acid or a base? How do you know? Solution Steps for Problem Solving Unnamed: 1 Identify the "given" information and what the problem is asking you to "find." Given: pH = 12.6 Find: [H3O+] = ? M Plan the problem. Need to use the expression for [H3O+] (Equation \ref{ph1}). [H3O+] = antilog (-pH) or [H3O+] = 10-pH Calculate. Now substitute the known quantity into the equation and solve. [H3O+] = antilog (12.60) = 2.5 x 10-13 M (2 significant figures since 4.7 has 12.60 2 decimal places) or [H3O+] = 10-12.60 = 2.5 x 10-13 M (2 significant figures since 4.7 has 12.60 2 decimal places) The other issue that concerns us here is significant figures. Because the number(s) before the decimal point in a logarithm relate to the power on 10, the number of digits after the decimal point is what determines the number of significant figures in the final answer: Exercise \(\PageIndex{3}\) If moist soil has a pH of 7.84, what is [H 3 O + ] of the soil solution? Answer 1.5 x 10 -8 M The pOH scale As with the hydrogen-ion concentration, the concentration of the hydroxide ion can be expressed logarithmically by the pOH. The pOH of a solution is the negative logarithm of the hydroxide-ion concentration. \[\text{pOH} = -\text{log} \left[ \ce{OH^-} \right] \nonumber \] The relation between the hydronium and hydroxide ion concentrations expressed as p-functions is easily derived from the \(K_w\) expression: \[K_\ce{w}=\ce{[H_3O^+][OH^- ]} \label{\(\PageIndex{6}\)} \] \[-\log K_\ce{w}=\mathrm{-\log([H_3O^+][OH^−])=-\log[H_3O^+] + -\log[OH^-]}\label{\(\PageIndex{7}\)} \] \[\mathrm{p\mathit{K}_w=pH + pOH} \label{\(\PageIndex{8}\)} \] At 25 °C, the value of \(K_w\) is \(1.0 \times 10^{−14}\), and so: \[\mathrm{14.00=pH + pOH} \label{\(\PageIndex{9}\)} \] The hydronium ion molarity in pure water (or any neutral solution) is \( 1.0 \times 10^{-7}\; M\) at 25 °C. The pH and pOH of a neutral solution at this temperature are therefore: \[\mathrm{pH=-\log[H_3O^+]=-\log(1.0\times 10^{−7}) = 7.00} \label{\(\PageIndex{1}\)0} \] \[\mathrm{pOH=-\log[OH^−]=-\log(1.0\times 10^{−7}) = 7.00} \label{\(\PageIndex{1}\)1} \] And so, at this temperature, acidic solutions are those with hydronium ion molarities greater than \( 1.0 \times 10^{-7}\; M\) and hydroxide ion molarities less than \( 1.0 \times 10^{-7}\; M\) (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 \( 1.0 \times 10^{-7}\; M\) and hydroxide ion molarities greater than \( 1.0 \times 10^{-7}\; M\) (corresponding to pH values greater than 7.00 and pOH values less than 7.00). Example \(\PageIndex{4}\): Find the pOH of a solution with a pH of 4.42. Solution Steps for Problem Solving Unnamed: 1 Identify the "given" information and what the problem is asking you to "find." Given: pH =4.42 Find: ? pOH Plan the problem. Need to use the expression pOH = 14 - pH Calculate. Now substitute the known quantity into the equation and solve. pOH=14−4.42=9.58 Think about your result. The pH is that of an acidic solution, and the resulting pOH is the difference after subtracting from 14. The answer has two significant figures because the given pH has two decimal places. Exercise \(\PageIndex{4}\) The pH of a solution is 8.22. What is the pOH? Answer 5.78 The diagram below shows all of the interrelationships between [H3O+][H3O+], [OH−][OH−], pH, and pOH. Contributions & Attributions Peggy Lawson (Oxbow Prairie Heights School). Funded by Saskatchewan Educational Technology Consortium. Template:OpenStax |
Bookshelves/General_Chemistry/Chemistry_2e_(OpenStax)/18%3A_Representative_Metals_Metalloids_and_Nonmetals/18.06%3A_Occurrence_Preparation_and_Properties_of_Carbonates | Learning Objectives By the end of this section, you will be able to: 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, 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: \[\begin{align*} \ce{Na_2 O(s) + CO_2(g) &-> Na_2CO_3(s) } \\[4pt][4pt] \ce{Ca( OH )_2(s) + CO_2(g) &-> CaCO_3(s) + H2O(l)} \end{align*} \] 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: \[\begin{align*} \ce{Ca^{2+}(aq) + CO_3^{2-}(aq) &-> CaCO_3(s)} \\[4pt][4pt] \ce{Pb^{2+}(aq) + CO_3^{2-}(aq) &-> PbCO_3(s)} \end{align*} \] 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: \[\ce{OH^{-}(aq) + CO_2(aq) -> HCO_3^{-}(aq)} \nonumber \] 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: \[\ce{CaCO_3(s) + CO_2(aq) + H2O(l) -> Ca^{2+}(aq) + 2 HCO_3^{-}(aq)} \nonumber \] 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 : \[\ce{2 Na_3(CO3)(HCO3)(H2O)2(s) -> 3 Na_2CO_3(s) + 5 H2O(l) + CO_2(g)} \nonumber \] Carbonates are moderately strong bases. Aqueous solutions are basic because the carbonate ion accepts hydrogen ion from water in this reversible reaction: \[\ce{CO_3^{2-}(aq) + H2O(l) <=> HCO_3^{-}(aq) + OH^{-}(aq)} \nonumber \] 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: \[\ce{CaCO_3(s) + 2 HCl(aq) -> CaCl_2(aq) + CO_2(g) + H2O(l)} \nonumber \] 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: \[\ce{KHCO_3(aq) + KOH(aq) -> K_2 CO_3(aq) + H2O(l)} \nonumber \] 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: \[\ce{HC_4 H_4 O_6^{-}(aq) + HCO_3^{-}(aq) -> C_4 H_4 O_6^{2-}(aq) + CO_2(g) + H2O(l)} \nonumber \] Dough will trap the carbon dioxide, causing it to expand during baking, producing the characteristic texture of baked goods. |
Courses/Ursinus_College/CHEM322%3A_Inorganic_Chemistry/02%3A_Molecular_Structure/2.02%3A_Lewis_Structures_and_Molecular_Shape/2.2.03%3A_Resonance | Resonance structures are a set of two or more Lewis Structures that collectively describe the electronic bonding a single polyatomic species including fractional bonds and fractional charges. Resonance structures are capable of describing delocalized electrons that cannot be expressed by a single Lewis formula with an integer number of covalent bonds. Sometimes One Lewis Structure Is Not Enough Sometimes, even when formal charges are considered, the bonding in some molecules or ions cannot be described by a single Lewis structure. Resonance is a way of describing delocalized electrons within certain molecules or polyatomic ions where the bonding cannot be expressed by a single Lewis formula. A molecule or ion with such delocalized electrons is represented by several contributing structures (also called resonance structures or canonical forms). Such is the case for ozone (\(\ce{O3}\)), an allotrope of oxygen with a V-shaped structure and an O–O–O angle of 117.5°. Ozone (\(O_3\)) 1. We know that ozone has a V-shaped structure, so one O atom is central: 2. Each O atom has 6 valence electrons, for a total of 18 valence electrons. 3. Assigning one bonding pair of electrons to each oxygen–oxygen bond gives with 14 electrons left over. 4. If we place three lone pairs of electrons on each terminal oxygen, we obtain and have 2 electrons left over. 5. At this point, both terminal oxygen atoms have octets of electrons. We therefore place the last 2 electrons on the central atom: 6. The central oxygen has only 6 electrons. We must convert one lone pair on a terminal oxygen atom to a bonding pair of electrons—but which one? Depending on which one we choose, we obtain either Which is correct? In fact, neither is correct. Both predict one O–O single bond and one O=O double bond. As you will learn, if the bonds were of different types (one single and one double, for example), they would have different lengths. It turns out, however, that both O–O bond distances are identical, 127.2 pm, which is shorter than a typical O–O single bond (148 pm) and longer than the O=O double bond in O 2 (120.7 pm). Equivalent Lewis dot structures, such as those of ozone, are called resonance structures . The position of the atoms is the same in the various resonance structures of a compound, but the position of the electrons is different. Double-headed arrows link the different resonance structures of a compound: The double-headed arrow indicates that the actual electronic structure is an average of those shown, not that the molecule oscillates between the two structures. When it is possible to write more than one equivalent resonance structure for a molecule or ion, the actual structure is the average of the resonance structures. The Carbonate (\(CO_3^{2−} \)) Ion Like ozone, the electronic structure of the carbonate ion cannot be described by a single Lewis electron structure. Unlike O 3 , though, the actual structure of CO 3 2− is an average of three resonance structures. 1. Because carbon is the least electronegative element, we place it in the central position: 2. Carbon has 4 valence electrons, each oxygen has 6 valence electrons, and there are 2 more for the −2 charge. This gives 4 + (3 × 6) + 2 = 24 valence electrons. 3. Six electrons are used to form three bonding pairs between the oxygen atoms and the carbon: 4. We divide the remaining 18 electrons equally among the three oxygen atoms by placing three lone pairs on each and indicating the −2 charge: 5. No electrons are left for the central atom. 6. At this point, the carbon atom has only 6 valence electrons, so we must take one lone pair from an oxygen and use it to form a carbon–oxygen double bond. In this case, however, there are three possible choices: As with ozone, none of these structures describes the bonding exactly. Each predicts one carbon–oxygen double bond and two carbon–oxygen single bonds, but experimentally all C–O bond lengths are identical. We can write resonance structures (in this case, three of them) for the carbonate ion: The actual structure is an average of these three resonance structures. The Nitrate (\(NO_3^-\)) ion 1. Count up the valence electrons: (1*5) + (3*6) + 1(ion) = 24 electrons 2. Draw the bond connectivities: 3. Add octet electrons to the atoms bonded to the center atom: 4. Place any leftover electrons (24-24 = 0 ) on the center atom: 5. Does the central atom have an octet? NO , it has 6 electrons Add a multiple bond (first try a double bond) to see if the central atom can achieve an octet: 6. Does the central atom have an octet? YES Are there possible resonance structures? YES Note: We would expect that the bond lengths in the \(\ce{NO_3^{-}}\) ion to be somewhat shorter than a single bond. Warning For second period elements you cannot exced the octet, even if additional double or triple bonds would reduce the formal charges in the structure. Example \(\PageIndex{1}\): Benzene Benzene is a common organic solvent that was previously used in gasoline; it is no longer used for this purpose, however, because it is now known to be a carcinogen. The benzene molecule (\(\ce{C6H6}\)) consists of a regular hexagon of carbon atoms, each of which is also bonded to a hydrogen atom. Use resonance structures to describe the bonding in benzene. Given: molecular formula and molecular geometry Asked for: resonance structures Strategy: Draw a structure for benzene illustrating the bonded atoms. Then calculate the number of valence electrons used in this drawing. Subtract this number from the total number of valence electrons in benzene and then locate the remaining electrons such that each atom in the structure reaches an octet. Draw the resonance structures for benzene. Solution: A Each hydrogen atom contributes 1 valence electron, and each carbon atom contributes 4 valence electrons, for a total of (6 × 1) + (6 × 4) = 30 valence electrons. If we place a single bonding electron pair between each pair of carbon atoms and between each carbon and a hydrogen atom, we obtain the following: Each carbon atom in this structure has only 6 electrons and has a formal charge of +1, but we have used only 24 of the 30 valence electrons. B If the 6 remaining electrons are uniformly distributed pairwise on alternate carbon atoms, we obtain the following: Three carbon atoms now have an octet configuration and a formal charge of −1, while three carbon atoms have only 6 electrons and a formal charge of +1. We can convert each lone pair to a bonding electron pair, which gives each atom an octet of electrons and a formal charge of 0, by making three C=C double bonds. C There are, however, two ways to do this: Each structure has alternating double and single bonds, but experimentation shows that each carbon–carbon bond in benzene is identical, with bond lengths (139.9 pm) intermediate between those typically found for a C–C single bond (154 pm) and a C=C double bond (134 pm). We can describe the bonding in benzene using the two resonance structures, but the actual electronic structure is an average of the two. The existence of multiple resonance structures for aromatic hydrocarbons like benzene is often indicated by drawing either a circle or dashed lines inside the hexagon: Exercise \(\PageIndex{1}\): Nitrate Ion The sodium salt of nitrite is used to relieve muscle spasms. Draw two resonance structures for the nitrite ion (NO 2 − ). Answer Resonance structures are particularly common in oxoanions of the p -block elements, such as sulfate and phosphate, and in aromatic hydrocarbons, such as benzene and naphthalene. How do resonance structures represent the true structure? If several reasonable resonance forms for a molecule exists, the "actual electronic structure" of the molecule will probably be intermediate between all the forms that you can draw. The classic example is benzene in Example \(\PageIndex{1}\). One would expect the double bonds to be shorter than the single bonds, but if once overlays the two structures, you see that one structure has a single bond where the other structure has a double bond. The best measurements that we can make of benzene do not show two bond lengths - instead, they show that the bond length is intermediate between the two resonance structures. Resonance structures is a mechanism that allows us to use all of the possible resonance structures to try to predict what the actual form of the molecule would be. Single bonds, double bonds, triple bonds, +1 charges, -1 charges, these are our limitations in explaining the structures, and the true forms can be in between - a carbon-carbon bond could be mostly single bond with a little bit of double bond character and a partial negative charge, for example. Summary Some molecules have two or more chemically equivalent Lewis electron structures, called resonance structures. Resonance is a mental exercise and method within the Valence Bond Theory of bonding that describes the delocalization of electrons within molecules. These structures are written with a double-headed arrow between them, indicating that none of the Lewis structures accurately describes the bonding but that the actual structure is an average of the individual resonance structures. Resonance structures are used when one Lewis structure for a single molecule cannot fully describe the bonding that takes place between neighboring atoms relative to the empirical data for the actual bond lengths between those atoms. The net sum of valid resonance structures is defined as a resonance hybrid, which represents the overall delocalization of electrons within the molecule. A molecule that has several resonance structures is more stable than one with fewer. Some resonance structures are more favorable than others. |
Courses/University_of_Illinois_Springfield/CHE_124%3A_General_Chemistry_for_the_Health_Professions_(Morsch_and_Andrews)/07%3A_Energy_and_Chemical_Processes/7.1%3A_Energy_and_Its_Units | Skills to Develop To define energy and he a t. Energy is the ability to do work. You can understand what this means by thinking about yourself when you feel “energetic.” You feel ready to go—to jump up and get something done. When you have a lot of energy, you can perform a lot of work. By contrast, if you do not feel energetic, you have very little desire to do much of anything. This description is not only applicable to you but also to all physical and chemical processes. The quantity of work that can be done is related to the quantity of energy available to do it. Energy can be transferred from one object to another if the objects have different temperatures. The transfer of energy due to temperature differences is called heat. For example, if you hold an ice cube in your hand, the ice cube slowly melts as energy in the form of heat is transferred from your hand to the ice. As your hand loses energy, it starts to feel cold. Because of their interrelationships, energy, work, and heat have the same units. The SI unit of energy, work, and heat is the joule (J). A joule is a tiny amount of energy. For example, it takes about 4 J to warm 1 mL of H 2 O by 1°C. Many processes occur with energy changes in thousands of joules, so the kilojoule (kJ) is also common. Another unit of energy, used widely in the health professions and everyday life, is the calorie (cal). The calorie was initially defined as the amount of energy needed to warm 1 g of H 2 O by 1°C, but in modern times, the calorie is related directly to the joule, as follows: \[1\; cal = 4.184\; J\] We can use this relationship to convert quantities of energy, work, or heat from one unit to another. Although the joule is the proper SI unit for energy, we will use the calorie or the kilocalorie (or Calorie) in this chapter because they are widely used by health professionals. The calorie is used in nutrition to express the energy content of foods. However, because a calorie is a rather small quantity, nutritional energies are usually expressed in kilocalories (kcal), also called Calories (capitalized; Cal). For example, a candy bar may provide 120 Cal (nutritional calories) of energy, which is equal to 120,000 cal. Figure \(\PageIndex{1}\) shows an example. Proteins and carbohydrates supply 4 kcal/g, while fat supplies 9 kcal/g. Figure \(\PageIndex{1}\): Nutritional Energy. A sample nutrition facts label, with instructions from the U.S. Food and Drug Administration. Image used with permission from Wikipedia. Example \(\PageIndex{1}\) The energy content of a single serving of bread is 70.0 Cal. What is the energy content in calories? In joules? SOLUTION This is a simple conversion-factor problem. Using the relationship 1 Cal = 1,000 cal, we can answer the first question with a one-step conversion: \(\mathrm{70.0 \: Cal\times\dfrac{1,000\: cal}{1\: Cal}= 70,000\: cal}\) Then we convert calories into joules \(\mathrm{70,000 \: cal \times \dfrac{4.184\: J}{1\: cal}=293,000\: J}\) and then kilojoules \(\mathrm{293,000\: J\times\dfrac{1\: kJ}{1,000\: J}=293\: kJ}\) The energy content of bread comes mostly from carbohydrates. Exercise \(\PageIndex{1}\) The energy content of one cup of honey is 1,030 Cal. What is its energy content in calories and joules? To Your Health: Energy Expenditures Most health professionals agree that exercise is a valuable component of a healthy lifestyle. Exercise not only strengthens the body and develops muscle tone but also expends energy. After obtaining energy from the foods we eat, we need to expend that energy somehow, or our bodies will store it in unhealthy ways (e.g., fat). Like the energy content in food, the energy expenditures of exercise are also reported in kilocalories, usually kilocalories per hour of exercise. These expenditures vary widely, from about 440 kcal/h for walking at a speed of 4 mph to 1,870 kcal/h for mountain biking at 20 mph. Table \(\PageIndex{1}\) lists the energy expenditure for a variety of exercises. Exercise Energy Expended (kcal/h) aerobics, low-level 325 basketball 940 bike riding, 20 mph 830 golfing, with cart 220 golfing, carrying clubs 425 jogging, 7.5 mph 950 racquetball 740 skiing, downhill 520 soccer 680 walking upstairs 1200 yoga 280 Because some forms of exercise use more energy than others, anyone considering a specific exercise regimen should consult with his or her physician first. Summary Energy is the ability to do work. Heat is the transfer of energy due to temperature differences. Energy and heat are expressed in units of joules. Concept Review Exercises What is the relationship between energy and heat? What units are used to express energy and heat? Answers Heat is the exchange of energy from one part of the universe to another. Heat and energy have the same units. Joules and calories are the units of energy and heat. Exercises Define energy . What is heat? What is the relationship between a calorie and a joule? Which unit is larger? What is the relationship between a calorie and a kilocalorie? Which unit is larger? Express 1,265 cal in kilocalories and in joules. Express 9,043.3 J in calories and in kilocalories. One kilocalorie equals how many kilojoules? One kilojoule equals how many kilocalories? Many nutrition experts say that an average person needs 2,000 Cal per day from his or her diet. How many joules is this? Baby formula typically has 20.0 Cal per ounce. How many ounces of formula should a baby drink per day if the RDI is 850 Cal? Answers Energy is the ability to do work. 1 cal = 4.184 J; the calorie is larger. 1.265 kcal; 5,293 J 1 kcal = 4.184 kJ 8.4 × 10 6 J |
Courses/San_Diego_Miramar_College/Chem_103%3A_Fundamentals_of_GOB_Chemistry_(Garces)/08%3A_Gases/8.02%3A_Kinetic_Molecular_Theory-_A_Model_for_Gases | Learning Objectives State the major concepts behind the kinetic theory of gases. Relate the general properties of gases to the kinetic theory. Gases were among the first substances studied in terms of the modern scientific method, which was developed in the 1600s. It did not take long to recognize that gases all shared certain physical behaviors, suggesting that all gases could be described by one all-encompassing theory. Today, that theory is the kinetic theory of gases . It is based on the following statements: Gases consist of tiny particles of matter that are in constant motion. Gas particles are constantly colliding with each other and the walls of a container. These collisions are elastic; that is, there is no net loss of energy from the collisions. Gas particles are separated by large distances, with the size of a gas particle tiny compared to the distances that separate them. There are no interactive forces (i.e., attraction or repulsion) between the particles of a gas. The average speed of gas particles is dependent on the temperature of the gas. Figure \(\PageIndex{1}\) shows a representation of how we mentally picture the gas phase. This model of gases explains some of the physical properties of gases. Because most of a gas is empty space, a gas has a low density and can expand or contract under the appropriate influence. The fact that gas particles are in constant motion means that two or more gases will always mix, as the particles from the individual gases move and collide with each other. An ideal gas is a gas that exactly follows the statements of the kinetic theory. Unfortunately, real gases are not ideal. Many gases deviate slightly from agreeing perfectly with the kinetic theory of gases. However, most gases adhere to the statements so well that the kinetic theory of gases is well accepted by the scientific community. The physical behavior of gases is explained by the kinetic theory of gases. An ideal gas adheres exactly to the kinetic theory of gases. |
Ancillary_Materials/Worksheets/Worksheets%3A_Inorganic_Chemistry/Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry/05%3A_Stereochemistry/5.11%3A_Carbohydrates_in_Cyclic_Form | Carbohydrates are complicated molecules. In solution, they slowly change into different isomers. In water, most carbohydrates can change from one form to another quickly; they are described as being in equilibrium with different structures. That means they can change back and forth. The most prevalent form for most carbohydrates is a ring. One of the oxygens farther along the chain can reach around and bond to the carbon in the C=O at the head of the chain. When that happens, there are two possible orientations of the oxygen at the head of the chain. The two stereoisomers that result are diastereomers. Some of the chiral centers are the same, but one is different. Figure SC11.1. D-threose and its two possible five-membered-ring forms. For example, D-threose can form two different five-membered rings (that is, rings made from a circle of five atoms). In carbohydrates, a five-membered ring with an oxygen in it is called a furanose (another common form, a six-membered ring, is called a pyranose). There are two diastereomers formed. they are sometimes called anomers. That means they differ in 3D space at the anomeric center; the anomeric center is the C=O carbon to which an oxygen binds to form the ring. These drawings, by the way, are called Haworth projections. They are commonly used in biochemistry to depict the cyclic forms of sugars. They seem to suggest a funny see-saw shape for the carbon atoms, but the idea here is not to convey the shape of the molecule perfectly. Just like in Fischer projections, the main concern is to quickly convey stereochemical relationships. In a Haworth projection, you can easily see whether two groups are cis to each other or trans to each other. Note that a Haworth projection of an odd-numbered ring is always drawn with the ring oxygen in the back. We are looking at the ring from the edge and slightly above, so the "back" is the upper edge of the polygon. It's just like if you were looking at a swimming pool, and the far corner of the pool appeared in your field of vision above the near corner. For an even-numbered ring, the same idea applies, but the ring oxygen is always drawn on the top edge and to the right. If you want to get an idea of the true shape of the molecule, you can see it in the ball-and-stick models below. Figure SC11.2. Ball-and-stick model of β-D-threofuranose. Figure SC11.3. Ball-and-stick model of α-D-threofuranose. Go to Animation SC11.1. A three-dimensional model of β-D-threofuranose. Go to Animation SC11.2. Another three-dimensional model of α-D-threofuranose. Carbohydrates are normally found in their cyclic forms, not their chain forms. For every ring a carbohydrate can form, there are always two diastereomeric forms called anomers. The two terms, α and β, that described the two different stereoisomers of the cyclic form of threose refer to the position of the OH group on the newly formed chiral center (the old C=O was trigonal planar, not tetrahedral, so it wasn't chiral). The terms are defined by looking at the Haworth projection of the molecule. If the new OH is in the upper position, it is termed β (think butterfly); it is is in the lower position, it is termed α (think ant). Carbohydrates do not always make five-membered rings. Sometimes they make six-membered rings instead. A five-membered ring with an oxygen in it is called a furanose. A six-membered ring with an oxygen in it is called a pyranose. Ribose is a five-carbon sugar that could make either a five-membered or a six-membered ring, although we usually see it in its five-membered ring. Figure SC11.4. Ribose in equilibrium between its chain form and some of its ring forms. The presence of different diastereomers that can change back and forth complicates things. If you wanted to measure the optical activity of a sugar, you would probably open a bottle of the solid carbohydrate, dissolve some up in water and put the sample in a polarimeter. However, the optical rotation would change over time as the pure solid slowly turned into other isomers. This process is called mutarotation (which means changing optical rotation). For this reason, the time at which a measurement was performed is often reported with optical rotation values of carbohydrates. For example, the optical rotation value for D-threose may be listed as [α] D -12.3 (20 min, c = 4, water). Carbohydrates are important partly because they are structural units in biology. Carbohydrates are incorporated into DNA and RNA as well as important enzymatic cofactors such as ATP, NADH and Acetyl Coenzyme A. Very frequently, the unique pieces attached to the carbohydrates to make these different molecules are attached via the anomeric carbon. Specific OH groups on carbohydrates are frequently substituted with other groups to make more complex molecules. One more piece of terminology may be useful to review here. There are terms used in ring structures to describe whether two groups are attached to the same face or on opposite faces. For example, in β-D-threofuranose, the second hydroxy group is cis to the one on the anomeric center, but in α-D-threofuranose, the second hydroxy group is trans to the one on the anomeric center. Two groups on the same face of a ring are described as being cis to each other. Two groups on the opposite faces of a ring are described as being trans to each other. The terms cis and trans can be applied to any rings, not just carbohydrates. Problem SC11.1. In the following carbohydrate, glucose, show the cyclic structures that would form if the indicated oxygen bonded to the carbonyl carbon (the C=O in the CHO group at the top). You should move a proton from one oxygen to another so that all the atoms have the usual number of bonds. Problem SC11.2 . β-deoxyribofuranose (below) is a building block of DNA. a) What is the relationship between the two hydroxy (OH) groups that are directly attached to the 5-membered ring? b) In the other possible 5-membered ring form of deoxyribose, what is the relationship between the two hydroxy (OH) groups that are directly attached to the 5-membered ring? |
Bookshelves/Analytical_Chemistry/Instrumental_Analysis_(LibreTexts)/25%3A_Voltammetry/25.06%3A_Stripping_Methods | Another important voltammetric technique is stripping voltammetry, which consists of three related techniques: anodic stripping voltammetry, cathodic stripping voltammetry, and adsorptive stripping voltammetry. Because anodic stripping voltammetry is the more widely used of these techniques, we will consider it in greatest detail. Anodic stripping voltammetry consists of two steps (Figure \(\PageIndex{1}\)). The first step is a controlled potential electrolysis in which we hold the working electrode—usually a hanging mercury drop or a mercury film electrode—at a cathodic potential sufficient to deposit the metal ion on the electrode. For example, when analyzing Cu 2 + the deposition reaction is \[\mathrm{Cu}^{2+}+2 e^{-} \rightleftharpoons \mathrm{Cu}(\mathrm{Hg}) \label{sv1} \] where Cu(Hg) indicates that the copper is amalgamated with the mercury. This step serves as a means of concentrating the analyte by transferring it from the larger volume of the solution to the smaller volume of the electrode. During most of the electrolysis we stir the solution to increase the rate of deposition. Near the end of the deposition time we stop the stirring—eliminating convection as a mode of mass transport—and allow the solution to become quiescent. Typical deposition times of 1–30 min are common, with analytes at lower concentrations requiring longer times. In the second step, we scan the potential anodically—that is, toward a more positive potential. When the working electrode’s potential is sufficiently positive, the analyte is stripped from the electrode, returning to solution in its oxidized form. \[\mathrm{Cu}(\mathrm{Hg})\rightleftharpoons \text{ Cu}^{2+}+2 e^{-} \label{sv2} \] Monitoring the current during the stripping step gives a peak-shaped voltammogram, as shown in Figure \(\PageIndex{1}\). The peak current is proportional to the analyte’s concentration in the solution. Because we are concentrating the analyte in the electrode, detection limits are much smaller than other electrochemical techniques. An improvement of three orders of magnitude—the equivalent of parts per billion instead of parts per million—is routine. Applications Anodic stripping voltammetry is very sensitive to experimental conditions, which we must carefully control to obtain results that are accurate and precise. Key variables include the area of the mercury film or the size of the hanging Hg drop, the deposition time, the rest time, the rate of stirring, and the scan rate during the stripping step. Anodic stripping voltammetry is particularly useful for metals that form amalgams with mercury, several examples of which are listed in Table \(\PageIndex{1}\). anodic stripping voltammetry cathodic stripping voltammetry adsorptive stripping voltammetry Bi3+ Br– bilirubin Cd2+ Cl– codeine Cu2+ I– cocaine Ga3+ mercaptans (RSH) digitoxin In3+ S2– dopamine Pb2+ SCN– heme Tl+ NaN monesin Sn2+ NaN testosterone Zn2+ NaN NaN Source: Compiled from Peterson, W. M.; Wong, R. V. Am. Lab. November 1981, 116–128; Wang, J. Am. Lab. May 1985, 41–50. Source: Compiled from Peterson, W. M.; Wong, R. V. Am. Lab. November 1981, 116–128; Wang, J. Am. Lab. May 1985, 41–50. Source: Compiled from Peterson, W. M.; Wong, R. V. Am. Lab. November 1981, 116–128; Wang, J. Am. Lab. May 1985, 41–50. The experimental design for cathodic stripping voltammetry is similar to anodic stripping voltammetry with two exceptions. First, the deposition step involves the oxidation of the Hg electrode to \(\text{Hg}_2^{2+}\), which then reacts with the analyte to form an insoluble film at the surface of the electrode. For example, when Cl – is the analyte the deposition step is \[2 \mathrm{Hg}(l)+2 \mathrm{Cl}^{-}(a q) \rightleftharpoons \text{ Hg}_{2} \mathrm{Cl}_{2}(s)+2 e^{-} \label{sv3} \] Second, stripping is accomplished by scanning cathodically toward a more negative potential, reducing \(\text{Hg}_2^{2+}\) back to Hg and returning the analyte to solution. \[\mathrm{Hg}_{2} \mathrm{Cl}_{2}(s)+2 e^{-}\rightleftharpoons 2 \mathrm{Hg}( l)+2 \mathrm{Cl}^{-}(a q) \label{sv4} \] Table \(\PageIndex{1}\) lists several analytes analyzed successfully by cathodic stripping voltammetry. In adsorptive stripping voltammetry, the deposition step occurs without electrolysis. Instead, the analyte adsorbs to the electrode’s surface. During deposition we maintain the electrode at a potential that enhances adsorption. For example, we can adsorb a neutral molecule on a Hg drop if we apply a potential of –0.4 V versus the SCE, a potential where the surface charge of mercury is approximately zero. When deposition is complete, we scan the potential in an anodic or a cathodic direction, depending on whether we are oxidizing or reducing the analyte. Examples of compounds that have been analyzed by absorptive stripping voltammetry also are listed in Table \(\PageIndex{1}\). |
Courses/Lumen_Learning/Book%3A_Western_Civilization_(Lumen)/Ch._12_The_Rise_of_Nation-States/13.16%3A_The_Sun-King_and_Authoritarianism | Learning Objective Describe Louis XIV’s views on royal power and how he expanded his own authority Key Points At the time of King Louis XIII’s death in 1643, Louis XIV was only five years old. His mother, Anne of Austria, was named regent, but she entrusted the government to the chief minister, Cardinal Mazarin. Mazarin’s policies paved the way for the authoritarian reign of Louis XIV. Louis began his personal reign with administrative and fiscal reforms. National debt was quickly reduced through more efficient taxation, although reforms imposing taxes on the aristocracy were late and of limited outcome. Louis and his administration also bolstered French commerce and trade by establishing new industries in France and instituted reforms in military administration that curbed the independent spirit of the nobility by imposing order at court and in the army. Louis also attempted uniform regulation of civil procedure throughout legally irregular France by issuing a comprehensive legal code, the “Grande Ordonnance de Procédure Civile” of 1667, also known as the Code Louis. One of his most infamous decrees was the Code Noir, which sanctioned slavery in French colonies. Louis also attached nobles to his court at Versailles and thus achieved increased control over the French aristocracy. An elaborate court ritual by which the king observed the aristocracy and distributed his favors was created to ensure the aristocracy remained under his scrutiny. Following consistent efforts to limit religious tolerance, Louis XIV issued the Edict of Fontainebleau, which revoked the Edict of Nantes and repealed all the privileges that arose therefrom. Terms cuius regio, eius religio A Latin phrase that literally means “Whose realm, his religion,” meaning that the religion of the ruler was to dictate the religion of those ruled. At the Peace of Augsburg of 1555, which ended a period of armed conflict between Roman Catholic and Protestant forces within the Holy Roman Empire, the rulers of the German-speaking states and Charles V, the emperor, agreed to accept this principle. Declaration of the Clergy of France A four-article document of the 1681 Assembly of the French clergy promulgated in 1682, which codified the principles of Gallicanism into a system for the first time in an official and definitive formula. Gallicanism The belief that popular civil authority—often represented by the monarchs’ authority or the state’s authority—over the Catholic Church is comparable to the pope’s. Edict of Fontainebleau A 1685 edict issued by Louis XIV of France, also known as the Revocation of the Edict of Nantes. The Edict of Nantes (1598) had granted the Huguenots the right to practice their religion without persecution from the state. Code Noir A decree originally passed by France’s King Louis XIV in 1685 that defined the conditions of slavery in the French colonial empire, restricted the activities of free black persons, forbade the exercise of any religion other than Roman Catholicism, and ordered all Jews out of France’s colonies. The Sun King At the time of King Louis XIII’s death in 1643, Louis XIV was only five years old. His mother, Anne of Austria, was named regent in spite of her late husband’s wishes. Anne assumed the regency but entrusted the government to the chief minister, Cardinal Mazarin, who helped her expand the limited power her husband had left her. He functioned essentially as the co-ruler of France alongside Queen Anne during her regency, and until his death effectively directed French policy alongside the monarch. In 1651, when Louis XIV officially came of age, Anne’s regency legally ended. However, she kept much power and influence over her son until the death of Mazarin. On the death of Mazarin in 1661, Louis assumed personal control of the reins of government and astonished his court by declaring that he would rule without a chief minister. Louis XIV, King of France, in 1661 by Charles Le Brun. After Mazarin’s death in 1661, Louis assumed personal control of the reins of government and astonished his court by declaring that he would rule without a chief minister: “Up to this moment I have been pleased to entrust the government of my affairs to the late Cardinal. It is now time that I govern them myself. You [he was talking to the secretaries and ministers of state] will assist me with your counsels when I ask for them. I request and order you to seal no orders except by my command … I order you not to sign anything, not even a passport … without my command; to render account to me personally each day and to favor no one.” Reforms Louis began his personal reign with administrative and fiscal reforms. In 1661, the treasury verged on bankruptcy. To rectify the situation, Louis chose Jean-Baptiste Colbert as Controller-General of Finances in 1665. Colbert reduced the national debt through more efficient taxation. Excellent results were achieved, and the deficit of 1661 turned into a surplus in 1666. However, to support the reorganized and enlarged army, the panoply of Versailles, and the growing civil administration, the king needed a good deal of money, but methods of collecting taxes were costly and inefficient. The main weakness of the existing system arose from an old bargain between the French crown and nobility: the king might raise without consent if only he refrained from taxing the nobles. Only towards the close of his reign, under extreme stress of war, was Louis able, for the first time in French history, to impose direct taxes on the aristocracy. This was a step toward equality before the law and toward sound public finance, but so many concessions and exemptions were won by nobles and bourgeois that the reform lost much of its value. Louis and Colbert also had wide-ranging plans to bolster French commerce and trade. Colbert’s administration established new industries, encouraged domestic manufacturers and inventors, and invited manufacturers and artisans from all over Europe to France. This aimed to decrease foreign imports while increasing French exports, hence reducing the net outflow of precious metals from France. Louis also instituted reforms in military administration and, with the help of his trusted experts, curbed the independent spirit of the nobility by imposing order at court and in the army. Gone were the days when generals protracted war at the frontiers while bickering over precedence and ignoring orders from the capital and the larger politico-diplomatic picture. The old military aristocracy ceased to have a monopoly over senior military positions and rank. Louis also attempted uniform regulation of civil procedure throughout legally irregular France by issuing a comprehensive legal code, the “Grande Ordonnance de Procédure Civile” of 1667, also known as the Code Louis. Among other things, it prescribed baptismal, marriage, and death records in the state’s registers, not the church’s, and also strictly regulated the right of the Parlements to remonstrate. The Code Louis played an important part in French legal history as the basis for the Napoleonic code, itself the origin of many modern legal codes.One of Louis’s most infamous decrees was the Grande Ordonnance sur les Colonies of 1685, also known as the Code Noir (“black code”). It sanctioned slavery and limited the ownership of slaves in the colonies to Roman Catholics only. It also required slaves to be baptized. Centralization of Power Louis initially supported traditional Gallicanism, which limited papal authority in France, and convened an Assembly of the French clergy in November 1681. Before its dissolution eight months later, the assembly had accepted the Declaration of the Clergy of France, which increased royal authority at the expense of papal power. Without royal approval, bishops could not leave France and appeals could not be made to the pope. Additionally, government officials could not be excommunicated for acts committed in pursuance of their duties. Although the king could not make ecclesiastical law, all papal regulations without royal assent were invalid in France. Unsurprisingly, the pope repudiated the declaration. Louis also attached nobles to his court at Versailles and thus achieved increased control over the French aristocracy. Apartments were built to house those willing to pay court to the king. However, the pensions and privileges necessary to live in a style appropriate to their rank were only possible by waiting constantly on Louis. For this purpose, an elaborate court ritual was created where the king became the center of attention and was observed throughout the day by the public. With his excellent memory, Louis could see who attended him at court and who was absent, facilitating the subsequent distribution of favors and positions. Another tool Louis used to control his nobility was censorship, which often involved opening letters to discern their author’s opinion of the government and king. Moreover, by entertaining, impressing, and domesticating nobles with extravagant luxury and other distractions, Louis not only cultivated public opinion of himself, but also ensured the aristocracy remained under his scrutiny. This, along with the prohibition of private armies, prevented the aristocracy from passing time on their own estates and in their regional power-bases, from which they historically had waged local wars and plotted resistance to royal authority. Louis thus compelled and seduced the old military aristocracy (the “nobility of the sword”) into becoming his ceremonial courtiers, further weakening their power. In their place, he raised commoners or the more recently ennobled bureaucratic aristocracy as presumably easier to control. Religion Finally, Louis dramatically limited religious tolerance in France, as he saw the persistence of Protestantism as a disgraceful reminder of royal powerlessness. Responding to petitions, Louis initially excluded Protestants from office, constrained the meeting of synods, closed churches outside Edict of Nantes-stipulated areas, banned Protestant outdoor preachers, and prohibited domestic Protestant migration. He also disallowed Protestant-Catholic intermarriages where third parties objected, encouraged missions to the Protestants, and rewarded converts to Catholicism. In 1681, Louis dramatically increased his persecution of Protestants. The principle of cuius regio, eius religio generally had also meant that subjects who refused to convert could emigrate, but Louis banned emigration and effectively insisted that all Protestants must be converted. In 1685, he issued the Edict of Fontainebleau, which cited the redundancy of privileges for Protestants given their scarcity after the extensive conversions. It revoked the Edict of Nantes and repealed all the privileges that arose therefrom. By his edict, Louis no longer tolerated Protestant groups, pastors, or churches to exist in France. No further churches were to be constructed, and those already existing were to be demolished. Pastors could choose either exile or a secular life. Those Protestants who had resisted conversion were now to be baptized forcibly into the established church. 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 |
Courses/Indiana_Tech/EWC%3A_CHEM_1020_-_General_Chemistry_I_(Budhi)/01%3A_Matter_Measurement_and_Problem_Solving/1.9%3A_Matter_Measurement_and_Problem_Solving_(Exercises) | Template:HideTOC These are homework exercises to accompany the Textmap created for Chemistry: A Molecular Approach by Nivaldo Tro. Complementary General Chemistry question banks can be found for other Textmaps and can be accessed here . In addition to these publicly available questions, access to private problems bank for use in exams and homework is available to faculty only on an individual basis; please contact Delmar Larsen for an account with access permission. Q1.35 A household receives a $125 electricity bill. The cost of electricity is $0.150 /kWh. How much energy, in joules (J), did the household use? Strategy Step 1 : You must first multiply the total cost of the bill ($125) by 1 kWh and then divide that answer by the cost of each kW ($0.150 kWh). You should end up with an equation that looks like this: \[($125)\times \dfrac{1\;kWh}{0.150\;kWh}\nonumber \] Step 2 : Once you do this math, then you should receive the answer of 833.33 kWh. This is the number of kWh used during the billing cycle. Step 3 : Next, you will then take your solution found during step 1 and multiply it by \(\dfrac{1}{2.78\times 10^{-7}}\) in order to convert kWh to kWh/J. You should have the equation: \[(833.33)\times \dfrac{1}{2.78\cdot 10^{-7}}\nonumber \] Step 4 : After completing step 3, you should receive an answer of \[3.0 \times 10^{9} J\nonumber \] Step 5 : YOU ARE FINISHED! Q1.36 Determine whether the mixtures are Homogeneous or Heterogeneous. State whether it is a compound or pure substance? Mud and Water Salt and water Chicken noodle soup Gold \(H_{2}O\) Strategy Homogeneous: mixtures are uniformly distributed, easily dissolved, of a single phase, also they are easily dispersed through the membrane. Heterogeneous: mixtures are unequally distributed, do not dissolve in water, and could be of different phases. A pure substance: has a definite and constant composition, it can be a compound or and element. A element: is composed of a single atom. A compound is composed of two or more elements. For example, \(H_{2}O\) is a ratio of two hydrogen atoms and one oxygen atom to form \(H_{2}O\). So with this strategy you can solve the following question. Mud and Water: Heterogeneous mixture Salt and water: Homogeneous mixture Chicken noodle soup: Heterogeneous mixture Gold: (element) \(H_{2}O\): Compound Q1.36 Classify each substance as a pure substance or a mixture. If it is a pure substance, classify it as an element or a compound. If it is a mixture, classify it as homogeneous or heterogeneous. Plain yogurt Chicken noodle soup Titanium Table salt What we know: A pure substance is made up of only one type of particle. A mixture is a substance composed of two or more particles. An element is a substance that cannot be chemically broken down into simpler substances. A compound is a substance composed of two or more elements. A heterogeneous mixture is one in which the composition varies from one region of the mixture to another. A homogeneous mixture is one with the same composition throughout. Solution mixture, homogeneous mixture, heterogeneous pure substance, element pure substance, compound Q1.41 Classify Each statement as an Observation, Theory, or Law. All matter that exists is made up of Submicroscopic particles known as atoms. If an iron rod is placed in a closed container and it rusts, the mass of the container and its contents do not change. Matter is neither created nor destroyed When a candle is burned, heat is released Strategy You need to understand what an Observation, Theory or law is before you classify it in the question. Observation : An outcome of events that is being viewed Theory : Preserved explanation for observations and laws Law : Summarizes past observations and predicts future ones Now that we have defined them, we need to apply it to the question (AnswerS IN BOLD) This statement is proposing all matter is made up of atoms so it would become a theory because it is an idea that studies propose is true. This statement is saying, without matter being added to the container, As you See it rust, the mass doesn't change so this is assumed by Observation. This statement is called the Law of conservation of mass, so because this is true, the statement is a Law. If you are looking at a candle and see the flame, then you should understand that heat is being released. You tell this from Observation . Q1.45 When 12 g of sodium reacts with 23.5 g of chlorine to form sodium chloride, how many grams of sodium chloride is formed? (Assume that sodium chloride is the only product). Solution: What we know: We have a chemical reaction between sodium and chlorine to form only sodium chloride. We are also given the masses of the reactants, 12 g of sodium and 23.5 g of chlorine. We also know that the Law of Conservation of Mass tells us that matter is neither created nor destroyed in a chemical reaction. What we're asked for: The mass, in grams, of sodium chloride formed from the reaction. Strategy: A. Find the sum of the masses of the reactants from the chemical equation. Solution 12 g + 23.5 g = 35.5 g 35.5 g of sodium chloride is formed from the reaction. Q1.55 According to Dalton's Atomic Theory, what are compounds made up of? Strategy Find the summary of Dalton's Atomic Theory Read through it Locate where it talks about what compounds are made of Write your answer Solution Compounds are made up of two or more different atoms. Q1.64 Isotopes are atoms that have the same number of protons but differ in the number of neutrons. Write the chemical symbols for each of the following isotopes: the chlorine isotope with 18 neutrons the chlorine isotope with 20 neutrons the calcium isotope with 20 neutrons the neon isotope with 11 neutrons Strategy : Determine the given element's atomic number from the periodic table. Add the number of neutrons given in the question to the atomic number to calculate the mass number of the isotope. The mass number will be represented in the upper left corner of the chemical symbol, while the atomic number will be represented in the lower left corner. Helpful Hints: A chemical symbol is written as \[_{Z}^{A}\textrm{X}\nonumber \] . The A stands for the number of protons and neutrons, also known as the mass number. The Z stands for the number of protons, also known as the atomic number. Subtracting Z from A will give you the number of neutrons. Example: \[_{19}^{39}\textrm{K}\nonumber \] --> 39-19= 20 neutrons Therefore, adding the given number of neutrons to the atomic number, Z , gives you the mass number. Solution : \[_{17}^{35}\textrm{Cl}\nonumber \] \[_{17}^{37}\textrm{Cl}\nonumber \] \[_{20}^{40}\textrm{Ca}\nonumber \] \[_{10}^{21}\textrm{Ne}\nonumber \] Q1.65 Give the number of protons and neutrons in each isotope listed \[_{19}^{41}\textrm{K}\nonumber \] \[_{4}^{11}\textrm{Be}\nonumber \] \[_{22}^{45}\textrm{Ti}\nonumber \] \[_{8}^{16}\textrm{O}\nonumber \] Strategy Take the larger number and subtract the smaller number from it. The smaller number is the atomic number (z), which indicates the number of protons the element has. The larger number indicates the number of protons and neutron present in the element. 41 K has an atomic mass of 41. The atomic mass is the number of protons and neutrons present in the element. To get the number of neutrons, subtract the atomic number (19) from the atomic mass (41). When done so you find that K 41 has 19 protons and 22 neutrons. Keep in mind the atomic number is the number of protons present in the element, so the only thing you are trying to find through subtraction is the number of neutrons. For 11 Be, the same thing occurs. Take the atomic number and subtract it from the atomic mass. The atomic mass is 11 and the atomic number is 4, so, 11 minus for equals 7. The answer for part b. is 4 protons and 7 neutrons. For 45 Ti, you subtract 22 from 45, which gives you 22 protons and 23 neutrons. For 16 O, the same method is applied. Subtract 8 from 16 and you get 8 protons and 8 electrons. Q1.66 Question: Find the amount of protons and neutrons in the following atoms: \[_{6}^{14}\textrm{C}\nonumber \] \[_{8}^{18}\textrm{O}\nonumber \] \[_{25}^{55}\textrm{Mn}\nonumber \] \[_{30}^{64}\textrm{Zn}\nonumber \] Strategy First identify the subscript, the bottom number on the left hand side of the element's symbol. This is the atomic number and indicates how many protons the atom's nucleus contains. All isotopes of the same element contain the same number of protons. Next, identify the superscript, the top number left hand side of the element symbol. This is the atomic mass number, the total combined number of protons and neutrons. Different isotopes of a given element differ in the number of neutrons contained in their nuclei. The number of protons can be found simply by identifying the atomic number. To find the number of neutrons in the atom, the atomic number is subtracted from the atomic mass number. Solution A The atomic number of carbon is 6, the atomic number of oxygen is 8, the atomic number of manganese is 25, and the atomic number of zinc is 30. Therefore, we now know the number of protons contained within each atom since this number is represented by the atomic number. B The atomic mass number of the carbon isotope is 14, the mass number of the oxygen isotope is 18, the mass number of the manganese isotope is 55, and the mass number of the zinc isotope is 64. This is the combined number of protons and neutrons in each respective isotope. C Carbon isotope: \[14-6=8\, neutrons\nonumber \] Oxygen isotope: \[18-8=10\, neutrons\nonumber \] Manganese isotope: \[55-25=30\, neutrons\nonumber \] Zinc isotope: \[64-30=34\, neutrons\nonumber \] Q1.64 Write the isotopic symbols in the fo rm \(\ce{^A_ZX}\) for each isotope. The oxygen isotope with 9 neutrons. The neon isotope with 10 neutrons. The chlorine isotope with 18 neutrons. The carbon isotope with 8 neutrons. Strategy Identify each element’s atomic number (number of protons). Find the atomic mass of each element by adding the number of protons and the number of neutrons. Properly write the isotopic symbol of each element by placing the atomic mass as the superscript and the atomic number (number of protons) as the subscript. Solution Atomic Numbers (Number of Protons): Oxygen: 8 protons Neon: 10 protons Chlorine: 17 protons Carbon: 6 protons Atomic Mass: Oxygen: 8+9=17 Neon: 10+10=20 Chlorine: 17+18=35 Carbon: 6+8=14 Isotopic Symbols: Oxygen: 17 8 O Neon: 20 10 Ne Chlorine: 35 17 Cl Carbon: 14 6 C Q1.69 Question: Determine the number of protons and electrons in each of the following ions. Strategy First locate each element on the periodic table. Identify the atomic number associated with each element, or the number in between the element's symbol and scientific name. The atomic number is the number of protons as well as the number of electrons an element possesses in it's neutral state. Next, identify the superscript of each element, or the number to the top right of the element's symbol. This number should include a "+" or "-" sign next to it, indicating the element's charge. A "+" indicates a positively charged ion and a "-" indicates a negatively charged ion. An element becomes positively charged when it loses electrons, and negative when it gains electrons. To solve for each element's total number of electrons, take that element's atomic number, and inversely add or subtract the number in the superscript according to the element's charge. If an element has a "+" sign, it is losing the amount of electrons indicated by the superscript number, so subtract this number from the element's atomic number. The same goes if an element has a "-" sign, instead adding the number in the superscript to the element's atomic number. Solution (1) N 3- A. Nitrogen's atomic number is 7, so in a neutral state, Nitrogen has 7 protons and electrons. B. Nitrogen's superscript is "3-" indicating that it is gaining 3 electrons C. Because it gains 3 electrons, N 3- now has a total of 10 electrons. (2) Be 2 + A. Beryllium's atomic number is 4, so Beryllium possesses 4 protons and electrons B. The superscript "2+" indicates beryllium is losing 2 electrons C. 2 electrons are subtracted from Beryllium's total 4, so it now has 2 electrons. (3) K + A. Potassium as a neutral element possesses 19 protons and electrons, indicated by it's atomic number of 19. B. The superscript "+" indicates Potassium will lose 1 electron from it's total. C. 19 total electrons minus 1 makes K + number of electrons to 18. (4) Se 2 - A. Selenium's atomic number is 34, indicating 34 protons and electrons at ground state. B. It' superscript "2-" indicates that it will be gaining 2 electrons to it's total. C. 34 electrons plus 2 makes selenium's total number of electrons 34. Q 1.73 The atomic mass for Helium is 4.003 amu and the mass spectrum for Helium shows a spike at this mass. Meanwhile, the atomic mass for Bromine is 79.904 amu; however, the mass spectrum for Bromine does not show a peak at that mass. What causes the difference? Solution The best way to start this problem is to recall what the atomic mass means in the context of the problem. This is the standard atomic mass, which is the weighted average of all the isotopes. "Weighted" in this situation means that the abundance of each isotope factors in to how it factors in to the overall the standard atomic mass for that particular element. Next step is to look at Helium. Helium has two isotopes and is stable at either; however, in atmospheric values, Helium is found most abundantly as 4 He (99.999863%) and much less commonly found as 3 He (0.000137%). Consequently, 3 He does not have a strong influence on the standard atomic mass of Helium or much expression on the mass spectrum which measures relative abundance. Examining Bromine reveals two major isotopes for that element as well. Both 79 Br as well as 81 Br are rather prevalent ( 79 Br with 50.69% and 81 Br with 49.31%). This means that their relative abundance is high enough that they would make a significant impact on both the mass spectrum as well as the standard atomic weight of Bromine. To connect the dots, the dichotomy between the two elements comes from the abundance of their isotopes. Bromine's two isotopes almost equally comprise the atomic mass so the atomic mass they produce does not match up with an actual relatively abundant isotope of Br so therefore the line for that particular atomic mass is absent on the mass spectrum for Br. Helium, meanwhile, is primarily influenced by the 3 He because of its incredibly high abundance and therefore the mass spectrum and the standard atomic mass are very similar. Q1.74 The atomic mass of chlorine is 35.453 amu. There are two stable isotopes for chlorine. Would either of these isotopes have a mass of 35.453 amu? Why or why not? Important things to consider The atomic mass is the sum of the protons and neutrons. Isotopes vary in the amount of neutrons of the same element. Each isotope will have a different mass number depending upon how many neutrons are present. amu = atomic mass unit unit used for indicating mass on an atomic scale Answer No, because isotopes are variants of an element that differ in the number of neutrons present, but they have the same number of protons (atomic number the same). |
Courses/Furman_University/CHM101%3A_Chemistry_and_Global_Awareness_(Gordon)/01%3A_Introduction_to_Chemistry | What is chemistry? Simply put, chemistry is the study of the interactions of matter with other matter and with energy. This seems straightforward enough. However, the definition of chemistry includes a wide range of topics that must be understood to gain a mastery of the topic or even take additional courses in chemistry. Get ready for a fantastic journey through a world of wonder, delight, and knowledge. One of the themes of this book is “chemistry is everywhere,” and indeed it is; you would not be alive if it weren’t for chemistry because your body is a big chemical machine. If you don’t believe it, don’t worry. Every chapter in this book contains examples that will show you how chemistry is, in fact, everywhere. So enjoy the ride—and enjoy chemistry. 1.1: What is Chemistry? Chemistry is the study of matter—what it consists of, what its properties are, and how it changes. Being able to describe the ingredients in a cake and how they change when the cake is baked is called chemistry. Matter is anything that has mass and takes up space—that is, anything that is physically real. Some things are easily identified as matter—this book, for example. 1.2: Using 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 in turn lead to additional observations, hypotheses, and experiments in repeated cycles 1.3: Pure and Applied Research In science, we usually talk about two types of research: pure and applied. Pure research focuses on answering basic questions such as, "how do gases behave?" Applied research would be involved in the process of developing specific preparation for a gas in order for it to be produced and delivered efficiently and economically. This division sounds like it would be easy to make, but sometimes we cannot draw a clear line between what is "pure" and what is "applied". 1.4: Classification and Properties of Matter The properties that chemists use to describe matter fall into two general categories. Physical properties are characteristics that describe matter. They include characteristics such as size, shape, color, and mass. Chemical properties are characteristics that describe how matter changes its chemical structure or composition. An example of a chemical property is flammability—a material’s ability to burn—because burning (also known as combustion) changes the chemical composition of a material. 1.5: Applying Properties - MSDS/SDS A Material Safety Data Sheet (MSDS) is designed to provide both workers and emergency personnel with the proper procedures for handling or working with a particular substance. MSDS's include information such as physical data (melting point, boiling point, flash point etc.), toxicity, health effects, first aid, reactivity, storage, disposal, protective equipment, and spill/leak procedures. 1.6: All About the Elements We will delve more into the initial construction and modifications of today's periodic table. For now, you need to be aware that the periodic table has columns (known as families/groups). Horizontal rows are called periods. On the table below, the element symbols are either one of two letters. Previously, some of the newly discovered elements used three letter symbols. These three letter symbols corresponded to a Latin numbering system until the elements were given official names. |
Bookshelves/Environmental_Chemistry/Green_Chemistry_and_the_Ten_Commandments_of_Sustainability_(Manahan)/07%3A_Chemistry_of_Life_and_Green_Chemistry/7.01%3A_New_Page | Biochemistry is the science of chemical processes that occur in living organisms. 1 By its nature biochemistry is a green chemical and biological science. This is because over eons of evolution organisms have evolved that carry out biochemical processes sustainably. Because the enzymes that carry out biochemical processes can only function under mild conditions, particularly of temperature, biochemical processes take place under safe conditions, avoiding the high temperatures, high pressures, and corrosive and reactive chemicals that often characterize synthetic chemical operations. Therefore, it is appropriate to refer to green biochemistry . The ability of organisms to carry out chemical processes is truly amazing, even more so when one considers that many of them occur in single-celled organisms. Photosynthetic cyanobacteria consisting of individual cells less than a micrometer (μm) in size can make all the complex biochemicals they need to exist and reproduce using sunlight for energy and simple inorganic substances such as CO 2 , K + ion, NO 3 - ion and HPO 4 2- ion for raw materials. Beginning soon after conditions on Earth became hospitable to life, these photosynthetic bacteria produced the oxygen that now composes about 20% of Earth’s atmosphere. Fossilized stromatolites (bodies of sedimentary materials bound together by films produced by microorganisms) produced by cyanobacteria have been demonstrated dating back 2.8 billion years, and this remarkable microorganism that converts atmospheric carbon dioxide to biomass and atmospheric N 2 to chemically fixed N may have been on Earth as long as 3.5 billion years ago. It is fascinating to view single live cells of animal-like protozoa through an optical microscope. An ameba appears as a body of cellular protoplasm and moves by oozing about like a living blob of jelly. Examination of Euglena protozoa may show a cell several μm in size with many features including a cell nucleus that serves to direct metabolism and reproduction, green chloroplasts for photosynthetic production of biomass, a red eye-spot sensitive to light, a mouth-like contractile vacuole by which the cell expels excess water, and a thin tail-like structure(flagella) that moves rapidly and propels the cell. More detailed examination by electron microscope of such cells and those that make up more complex organisms reveals many more cell parts that are involved with biochemical function. At least a rudimentary knowledge of biochemistry is needed to understand green chemistry, environmental chemistry, and sustainability science and technology. One reason is the ability of organisms to synthesize a vast variety of substances. The most obvious of these is biomass made by the photosynthetic fixation of carbon dioxide and that forms the basis of nature’s food webs.Organisms make many of the materials upon which humans rely. In addition to food, one such material is the lignocellulose that composes most of plant biomass such as wood used for construction, paper-making, and fuel. Very complex molecules are made by organisms, for example, human insulin produced by genetically engineered organisms. Organisms make materials under very mild conditions compared to those used in the anthrosphere. An important example is chemically fixed nitrogen from the atmosphere which is produced synthetically in the anthrosphere as ammonia (NH 3 ) at high temperatures and pressures whereas Rhizobium bacteria attached to the roots of soybeans and other legume plants fix nitrogen in the mild conditions of the soil environment. Increasingly as supplies of petroleum and other non-renewable raw materials become more scarce, humans are turning to microorganisms and plants to make essential materials. Another major reason for considering biochemistry as part of green chemistry and sustainability is the protection of organisms from products and processes in the anthrosphere. It is essential to know the potential toxic effects of various materials, a subject addressed by toxicological chemistry . 2 One of the fundamental goals of green chemistry is to minimize the production and use of products that may have adverse environmental effects. Sustainability of the entire planet requires that humans not disperse into the environment substances that may undergo bioaccumulation and be toxic to humans and other organisms. Biochemical processes not only are profoundly influenced by chemical species in the environment, they largely determine the nature of these species, their degradation, and even their syntheses, particularly in the aquatic and soil environments. The study of such phenomena forms the basis of environmental biochemistry. This chapter is designed to give an overview of biochemistry and how it relates to green chemistry and sustainability science and technology. A glance at the structural formulas of some of the biochemicals shown in this chapter gives a hint of the complexity of biochemistry. This chapter is designed to provide a basic understanding of this complex science with enough detail for it to be meaningful but to avoid overwhelming the reader. It begins with an overview of the four major classes of biochemicals— proteins, carbohydrates, lipids, and nucleic acids. Many of the compounds in these classes are polymers with molecular masses of the order of a million or even larger. Proteins and nucleic acids consist of macromolecules, lipids are usually relatively small molecules, carbohydrates range from small sugar molecules to high-molar-mass macromolecules such as those in cellulose. The behavior of a substance in a biological system depends to a large extent upon whether the substance is hydrophilic (“water-loving”) or hydrophobic (“water-hating”). Some important toxic substances are hydrophobic, a characteristic that enables them to traverse cell membranes readily and to bioaccumulate in lipid (fat) tissue. Many hydrocarbons and organohalide compounds synthesized from hydrocarbons are hydrophobic. Part of the detoxification process carried on by living organisms is to render such molecules hydrophilic, therefore water-soluble and readily eliminated from the body. |
Courses/Riverland_Community_College/CHEM_1000_-_Introduction_to_Chemistry_(Riverland)/16%3A_Solutions | Template:HideTOC Solutions play a very important role in many biological, laboratory, and industrial applications of chemistry. Of particular importance are solutions involving substances dissolved in water, or aqueous solutions. Solutions represent equilibrium systems, and the lessons learned in the last chapter will be of particular importance again. Quantitative measurements of solutions are another key component of this chapter. Solutions can involve all physical states—gases dissolved in gases (the air around us), solids dissolved in solids (metal alloys), and liquids dissolved in solids (amalgams—liquid mercury dissolved in another metal such as silver, tin or copper). This chapter is almost exclusively concerned with aqueous solutions, substances dissolved in water. 16.1: Water - Some Unique Properties Earth is the only known body in our solar system that has liquid water existing freely on its surface. That is a good thing because life on Earth would not be possible without the presence of liquid water. 16.2: Solutions - Homogeneous Mixtures There are two types of mixtures: mixtures in which the substances are evenly mixed together (called a homogenous mixture, or solution) and a mixture in which the substances are not evenly mixed (called a heterogeneous mixture). When a solution, or homogenous mixture, is said to have uniform properties throughout, the definition is referring to properties at the particle level. 16.3: The Dissolution Process When a solute dissolves, its individual particles are surrounded by solvent molecules and are separated from each other. 16.4: Solutions of Solids Dissolved in Water- How to Make Rock Candy Solutions can be formed in a variety of combinations using solids, liquids, and gases. We also know that solutions have constant composition and we can also vary this composition up to a point to maintain the homogeneous nature of the solution. Reasons for why solutions form will be explored in this section, along with a discussion of why water is used most frequently to dissolve substances of various types. 16.5: Solutions of Gases in Water Several factors affect the solubility of a given substance in a given solvent. Temperature is one such factor, with gas solubility typically decreasing as temperature increases. This is one of the major impacts resulting from the thermal pollution of natural bodies of water. 16.6: Solution Concentration- Mass Percent To define a solution precisely, we need to state its concentration: how much solute is dissolved in a certain amount of solvent. Words such as "dilute" or "concentrated" are used to describe solutions that have a little or a lot of dissolved solute, respectively. However "dilute" and "concentrated" are relative terms, and have meanings dependent on various factors. The mass/mass percent (% m/m) is defined as the mass of a solute divided by the mass of a solution times 100. 16.7: Solution Concentration- Molarity Another way of expressing concentration is to give the number of moles of solute per unit volume of solution. Of all the quantitative measures of concentration, molarity is the one used most frequently by chemists. Molarity is defined as the number of moles of solute per liter of solution. The symbol for molarity is MM or moles/liter. Chemists also use square brackets to indicate a reference to the molarity of a substance. 16.8: Solution Dilution We are often concerned with how much solute is dissolved in a given amount of solution. We will begin our discussion of solution concentration with two related and relative terms—dilute and concentrated. 16.9: Freezing Point Depression and Boiling Point Elevation Freezing point depression and boiling point elevation are "colligative properties" that depend on the concentration of solute in a solvent, but not on the type of solute. What this means for the example above is that people in colder climates don't necessarily need salt to get the same effect on the roads—any solute will work. The higher the concentration of solute, the more these colligative properties will change. 16.10: Osmosis and Diffusion Fish cells, like all cells, have semipermeable membranes. Eventually, the concentration of "stuff" on either side of them will even out. A fish that lives in salt water will have somewhat salty water inside itself. Put it in freshwater, and the freshwater will, through osmosis, enter the fish, causing its cells to swell, and the fish will die. What will happen to a freshwater fish in the ocean? Template:HideTOC |
Courses/Sacramento_City_College/SCC%3A_Chem_309_-_General_Organic_and_Biochemistry_(Bennett)/Text/10%3A_Organic_Functional_Groups_-_Introduction_to_Acid-Base_Chemistry/10.04%3A_Physical_Properties_of_Esters | Learning Objectives Compare the boiling points of esters with alcohols of similar molar mass. Compare the solubilities of esters in water with the solubilities of comparable alkanes and alcohols in water. Ester molecules are polar but have no hydrogen atom attached directly to an oxygen atom. They are therefore incapable of engaging in intermolecular hydrogen bonding with one another and thus have considerably lower boiling points than their isomeric carboxylic acids counterparts. Because ester molecules can engage in hydrogen bonding with water molecules, however, esters of low molar mass are somewhat soluble in water. Borderline solubility occurs in those molecules that have three to five carbon atoms. Table \(\PageIndex{1}\) lists the physical properties of some common esters. Esters are common solvents. Ethyl acetate is used to extract organic solutes from aqueous solutions—for example, to remove caffeine from coffee. It also is used to remove nail polish and paint. Cellulose nitrate is dissolved in ethyl acetate and butyl acetate to form lacquers. The solvent evaporates as the lacquer “dries,” leaving a thin film on the surface. High boiling esters are used as softeners (plasticizers) for brittle plastics. Condensed Structural Formula Name Molar Mass Melting Point (°C) Boiling Point (°C) Aroma HCOOCH3 methyl formate 60 −99 32 NaN HCOOCH2CH3 ethyl formate 74 −80 54 rum CH3COOCH3 methyl acetate 74 −98 57 NaN CH3COOCH2CH3 ethyl acetate 88 −84 77 NaN CH3CH2CH2COOCH3 methyl butyrate 102 −85 102 apple CH3CH2CH2COOCH2CH3 ethyl butyrate 116 −101 121 pineapple CH3COO(CH2)4CH3 pentyl acetate 130 −71 148 pear CH3COOCH2CH2CH(CH3)2 isopentyl acetate 130 −79 142 banana CH3COOCH2C6H5 benzyl acetate 150 −51 215 jasmine CH3CH2CH2COO(CH2)4CH3 pentyl butyrate 158 −73 185 apricot CH3COO(CH2)7CH3 octyl acetate 172 −39 210 orange Summary Esters have polar bonds but do not engage in hydrogen bonding and are therefore intermediate in boiling points between the nonpolar alkanes and the alcohols, which engage in hydrogen bonding. Ester molecules can engage in hydrogen bonding with water, so esters of low molar mass are therefore somewhat soluble in water. |
Courses/Brevard_College/CHE_104%3A_Principles_of_Chemistry_II/04%3A_Thermochemistry_and_Thermodynamics/4.18%3A_Spontaneous_and_Nonspontaneous_Reactions | Nitroglycerin is a tricky substance. An active ingredient in dynamite (where it is stabilized), "raw" nitroglycerin is very unstable. Physical shock will cause the material to explode. The reaction is shown below. \[4 \ce{C_3H_5(ONO_2)_3} \rightarrow 12 \ce{CO_2} + 10 \ce{H_2O} + 6 \ce{N_2} + \ce{O_2}\nonumber \] The explosion of nitroglycerin releases large volumes of gases and is very exothermic. Spontaneous Reactions Reactions are favorable when they result in a decrease in enthalpy and an increase in entropy of the system. When both of these conditions are met, the reaction occurs naturally. A spontaneous reaction is a reaction that favors the formation of products at the conditions under which the reaction is occurring. A roaring bonfire is an example of a spontaneous reaction, since it is exothermic (there is a decrease in the energy of the system as energy is released to the surroundings as heat). The products of a fire are composed partly of gases such as carbon dioxide and water vapor. The entropy of the system increases during a combustion reaction. The combination of energy decrease and entropy increase dictates that combustion reactions are spontaneous reactions. A nonspontaneous reaction is a reaction that does not favor the formation of products at the given set of conditions. In order for a reaction to be nonspontaneous, it must be endothermic, accompanied by a decrease in entropy, or both. Our atmosphere is composed primarily of a mixture of nitrogen and oxygen gases. One could write an equation showing these gases undergoing a chemical reaction to form nitrogen monoxide: \[\ce{N_2} \left( g \right) + \ce{O_2} \left( g \right) \rightarrow 2 \ce{NO} \left( g \right)\nonumber \] Fortunately, this reaction is nonspontaneous at normal temperatures and pressures. It is a highly endothermic reaction with a slightly positive entropy change \(\left( \Delta S \right)\). Nitrogen monoxide is capable of being produced at very high temperatures and has been observed to form as a result of lightning strikes. One must be careful not to confuse the term spontaneous with the notion that a reaction occurs rapidly. A spontaneous reaction is one in which product formation is favored, even if the reaction is extremely slow. A piece of paper will not suddenly burst into flames, although its combustion is a spontaneous reaction. What is missing is the required activation energy to get the reaction started. If the paper were to be heated to a high enough temperature, it would begin to burn, at which point the reaction would proceed spontaneously until completion. In a reversible reaction, one reaction direction may be favored over the other. Carbonic acid is present in carbonated beverages. It decomposes spontaneously to carbon dioxide and water, according to the following reaction. \[\ce{H_2CO_3} \left( aq \right) \rightleftharpoons \ce{CO_2} \left( g \right) + \ce{H_2O} \left( l \right)\nonumber \] If you were to start with pure carbonic acid in water and allow the system to come to equilibrium, more than \(99\%\) of the carbonic acid would be converted into carbon dioxide and water. The forward reaction is spontaneous because the products of the forward reaction are favored at equilibrium. In the reverse reaction, carbon dioxide and water are the reactants, and carbonic acid is the product. When carbon dioxide is bubbled into water, less than \(1\%\) is converted to carbonic acid when the reaction reaches equilibrium. The reverse reaction, as written above, is not spontaneous. Summary Spontaneous and nonspontaneous reactions are defined. Examples of both types of reactions are given. |
Bookshelves/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/18%3A_Reactions_of_Aromatic_Compounds/18.11%3A__NAS_Reactions_-_the_Elimination-Addition_(Benzyne)_Mechanism | Elimination-Addition Mechanism of Nucleophilic Aromatic Substitution via Benzyne (Arynes) The reactivities of aryl halides, such as the halobenzenes, are exceedingly low toward nucleophilic reagents that normally effect displacements with alkyl halides and activated aryl halides. Substitutions do occur under forcing conditions of either high temperatures or very strong bases. For example, chlorobenzene reacts with sodium hydroxide solution at temperatures around \(340^\text{o}\) and this reaction was once an important commercial process for the production of benzenol (phenol): In addition, aryl chlorides, bromides, and iodides can be converted to areneamines \(\ce{ArNH_2}\) by the conjugate bases of amines. In fact, the reaction of potassium amide with bromobenzene is extremely rapid, even at temperatures as low as \(-33^\text{o}\) with liquid ammonia as solvent: However, displacement reactions of this type differ from the previously discussed displacements of activated aryl halides in that rearrangement often occurs. That is, the entering group does not always occupy the same position on the ring as that vacated by the halogen substituent . For example, the hydrolysis of 4-chloromethylbenzene at \(340^\text{o}\) gives an equimolar mixture of 3- and 4-methylbenzenols: Even more striking is the exclusive formation of 3-methoxybenzenamine in the amination of 2-chloromethoxybenzene. Notice that this result is a violation of the principle of least structural change (Section 1-1H): The mechanism of this type of reaction has been studied extensively, and much evidence has accumulated in support of a stepwise process, which proceeds first by base-catalyzed elimination of hydrogen halide \(\left( \ce{HX} \right)\) from the aryl halide - as illustrated below for the amination of bromobenzene: Elimination The product of the elimination reaction is a highly reactive intermediate \(9\) called benzyne , or dehydrobenzene , which differs from benzene in having two less hydrogen and an extra bond between two ortho carbons. Benzyne reacts rapidly with any available nucleophile, in this case the solvent, ammonia, to give an addition product: Addition The rearrangements in these reactions result from the reaction of the nucleophile at one or the other of the carbons of the extra bond in the intermediate. With benzyne the symmetry is such that no rearrangement would be detected. With substituted benzynes isomeric products may result. Thus 4-methylbenzyne, \(10\), from the reaction of hydroxide ion with 4-chloro-1-methylbenzene gives both 3- and 4-methylbenzenols: In the foregoing benzyne reactions the base that produces the benzyne in the elimination step is derived from the nucleophile that adds in the addition step. This need not always be so, depending on the reaction conditions. In fact, the synthetic utility of aryne reactions depends in large part of the success with which the aryne can be generated by one reagent but captured by another. One such method will be discussed in Section 14-10C and involves organometallic compounds derived from aryl halides. Another method is to generate the aryne by thermal decomposition of a 1,2-disubstituted arene compound such as \(11\), in which both substituents are leaving groups - one leaving with an electron pair, the other leaving without: When \(11\) decomposes in the presence of an added nucleophile, the benzyne intermediate is trapped by the nucleophile as it is formed. Or, if a conjugated diene is present, benzyne will react with it by a [4 + 2] cycloaddition. In the absence of other compounds with which it can react, benzyne will undergo [2 + 2] cycloaddition to itself: Exercise 25. When p -chlorotoluene is reacted with NaOH, two products are seen. While when m -chlorotoluene is reacted with NaOH, three products are seen. Explain this. Answer 25. You need to look at the benzyne intermediates. The para substituted only allows for two products, while the para produces two different alkynes which give three different products. |
Bookshelves/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/02%3A_Atoms_Molecules_and_Ions/2.2%3A_Evolution_of_Atomic_Theory | Learning Objectives Outline milestones in the development of modern atomic theory Summarize and interpret the results of the experiments of Thomson, Millikan, and Rutherford Describe the three subatomic particles that compose atoms Introduce the term isotopes In the two centuries since Dalton developed his ideas, scientists have made significant progress in furthering our understanding of atomic theory. Much of this came from the results of several seminal experiments that revealed the details of the internal structure of atoms. Here, we will discuss some of those key developments, with an emphasis on application of the scientific method, as well as understanding how the experimental evidence was analyzed. While the historical persons and dates behind these experiments can be quite interesting, it is most important to understand the concepts resulting from their work. Atomic Theory after the Nineteenth Century If matter were composed of atoms, what were atoms composed of? Were they the smallest particles, or was there something smaller? In the late 1800s, a number of scientists interested in questions like these investigated the electrical discharges that could be produced in low-pressure gases, with the most significant discovery made by English physicist J. J. Thomson using a cathode ray tube. This apparatus consisted of a sealed glass tube from which almost all the air had been removed; the tube contained two metal electrodes. When high voltage was applied across the electrodes, a visible beam called a cathode ray appeared between them. This beam was deflected toward the positive charge and away from the negative charge, and was produced in the same way with identical properties when different metals were used for the electrodes. In similar experiments, the ray was simultaneously deflected by an applied magnetic field, and measurements of the extent of deflection and the magnetic field strength allowed Thomson to calculate the charge-to-mass ratio of the cathode ray particles. The results of these measurements indicated that these particles were much lighter than atoms (Figure \(\PageIndex{1}\)). Based on his observations, here is what Thomson proposed and why: The particles are attracted by positive (+) charges and repelled by negative (−) charges, so they must be negatively charged (like charges repel and unlike charges attract); they are less massive than atoms and indistinguishable, regardless of the source material, so they must be fundamental, subatomic constituents of all atoms. Although controversial at the time, Thomson’s idea was gradually accepted, and his cathode ray particle is what we now call an electron , a negatively charged, subatomic particle with a mass more than one thousand-times less that of an atom. The term “electron” was coined in 1891 by Irish physicist George Stoney, from “ electr ic i on .” In 1909, more information about the electron was uncovered by American physicist Robert A. Millikan via his “oil drop” experiments. Millikan created microscopic oil droplets, which could be electrically charged by friction as they formed or by using X-rays. These droplets initially fell due to gravity, but their downward progress could be slowed or even reversed by an electric field lower in the apparatus. By adjusting the electric field strength and making careful measurements and appropriate calculations, Millikan was able to determine the charge on individual drops (Figure \(\PageIndex{2}\)). Looking at the charge data that Millikan gathered, you may have recognized that the charge of an oil droplet is always a multiple of a specific charge, 1.6 \(\times\) 10 −19 C. Millikan concluded that this value must therefore be a fundamental charge—the charge of a single electron—with his measured charges due to an excess of one electron (1 times 1.6 \(\times\) 10 −19 C), two electrons (2 times 1.6 \(\times\) 10 −19 C), three electrons (3 times 1.6 \(\times\) 10 −19 C), and so on, on a given oil droplet. Since the charge of an electron was now known due to Millikan’s research, and the charge-to-mass ratio was already known due to Thomson’s research (1.759 \(\times\) 10 11 C/kg), it only required a simple calculation to determine the mass of the electron as well. \[\mathrm{Mass\: of\: electron=1.602\times 10^{-19}\:\cancel{C}\times \dfrac{1\: kg}{1.759\times 10^{11}\:\cancel{C}}=9.107\times 10^{-31}\:kg} \tag{2.3.1} \] Scientists had now established that the atom was not indivisible as Dalton had believed, and due to the work of Thomson, Millikan, and others, the charge and mass of the negative, subatomic particles—the electrons—were known. However, the positively charged part of an atom was not yet well understood. In 1904, Thomson proposed the “plum pudding” model of atoms, which described a positively charged mass with an equal amount of negative charge in the form of electrons embedded in it, since all atoms are electrically neutral. A competing model had been proposed in 1903 by Hantaro Nagaoka , who postulated a Saturn-like atom, consisting of a positively charged sphere surrounded by a halo of electrons (Figure \(\PageIndex{3}\)). The next major development in understanding the atom came from Ernest Rutherford , a physicist from New Zealand who largely spent his scientific career in Canada and England. He performed a series of experiments using a beam of high-speed, positively charged alpha particles (α particles) that were produced by the radioactive decay of radium; α particles consist of two protons and two neutrons (you will learn more about radioactive decay in the chapter on nuclear chemistry). Rutherford and his colleagues Hans Geiger (later famous for the Geiger counter) and Ernest Marsden aimed a beam of α particles, the source of which was embedded in a lead block to absorb most of the radiation, at a very thin piece of gold foil and examined the resultant scattering of the α particles using a luminescent screen that glowed briefly where hit by an α particle. What did they discover? Most particles passed right through the foil without being deflected at all. However, some were diverted slightly, and a very small number were deflected almost straight back toward the source (Figure \(\PageIndex{4}\)). Rutherford described finding these results: “It was quite the most incredible event that has ever happened to me in my life. It was almost as incredible as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you” 1 (p. 68). Here is what Rutherford deduced: Because most of the fast-moving α particles passed through the gold atoms undeflected, they must have traveled through essentially empty space inside the atom. Alpha particles are positively charged, so deflections arose when they encountered another positive charge (like charges repel each other). Since like charges repel one another, the few positively charged α particles that changed paths abruptly must have hit, or closely approached, another body that also had a highly concentrated, positive charge. Since the deflections occurred a small fraction of the time, this charge only occupied a small amount of the space in the gold foil. Analyzing a series of such experiments in detail, Rutherford drew two conclusions: The volume occupied by an atom must consist of a large amount of empty space. A small, relatively heavy, positively charged body, the nucleus , must be at the center of each atom. This analysis led Rutherford to propose a model in which an atom consists of a very small, positively charged nucleus, in which most of the mass of the atom is concentrated, surrounded by the negatively charged electrons, so that the atom is electrically neutral (Figure \(\PageIndex{5}\)). After many more experiments, Rutherford also discovered that the nuclei of other elements contain the hydrogen nucleus as a “building block,” and he named this more fundamental particle the proton , the positively charged, subatomic particle found in the nucleus. With one addition, which you will learn next, this nuclear model of the atom, proposed over a century ago, is still used today. Another important finding was the discovery of isotopes. During the early 1900s, scientists identified several substances that appeared to be new elements, isolating them from radioactive ores. For example, a “new element” produced by the radioactive decay of thorium was initially given the name mesothorium. However, a more detailed analysis showed that mesothorium was chemically identical to radium (another decay product), despite having a different atomic mass. This result, along with similar findings for other elements, led the English chemist Frederick Soddy to realize that an element could have types of atoms with different masses that were chemically indistinguishable. These different types are called isotopes —atoms of the same element that differ in mass. Soddy was awarded the Nobel Prize in Chemistry in 1921 for this discovery. One puzzle remained: The nucleus was known to contain almost all of the mass of an atom, with the number of protons only providing half, or less, of that mass. Different proposals were made to explain what constituted the remaining mass, including the existence of neutral particles in the nucleus. As you might expect, detecting uncharged particles is very challenging, and it was not until 1932 that James Chadwick found evidence of neutrons , uncharged, subatomic particles with a mass approximately the same as that of protons. The existence of the neutron also explained isotopes: They differ in mass because they have different numbers of neutrons, but they are chemically identical because they have the same number of protons. This will be explained in more detail later in this chapter. Summary Although no one has actually seen the inside of an atom, experiments have demonstrated much about atomic structure. Thomson’s cathode ray tube showed that atoms contain small, negatively charged particles called electrons. Millikan discovered that there is a fundamental electric charge—the charge of an electron. Rutherford’s gold foil experiment showed that atoms have a small, dense, positively charged nucleus; the positively charged particles within the nucleus are called protons. Chadwick discovered that the nucleus also contains neutral particles called neutrons. Soddy demonstrated that atoms of the same element can differ in mass; these are called isotopes. Footnotes Ernest Rutherford, “The Development of the Theory of Atomic Structure,” ed. J. A. Ratcliffe, in Background to Modern Science , eds. Joseph Needham and Walter Pagel, (Cambridge, UK: Cambridge University Press, 1938), 61–74. Accessed September 22, 2014, https://ia600508.us.archive.org/3/it...e032734mbp.pdf . Glossary alpha particle (α particle) positively charged particle consisting of two protons and two neutrons electron negatively charged, subatomic particle of relatively low mass located outside the nucleus isotopes atoms that contain the same number of protons but different numbers of neutrons neutron uncharged, subatomic particle located in the nucleus proton positively charged, subatomic particle located in the nucleus nucleus massive, positively charged center of an atom made up of protons and neutrons |
Courses/Duke_University/CHEM_310L%3A_Physical_Chemistry_I_Laboratory/CHEM310L_-_Physical_Chemistry_I_Lab_Manual/01%3A_Orientation_to_this_course | 1.1: Pre-lab orientation assignment 1.2: Introductory Details Experiments have been set up for students working in small groups. Be sure to have a work plan that makes efficient use of your time. Lengthy sample preparation procedures should be started early in the six hours assigned to a given experiment. 1.3: Rubrics The points available for the semester and that will be used to calculate your final grade are described here. TAs will use rubrics similar to the ones given here to score your assignments. Rubrics are subject to minor changes during the semester. 1.4: Statements |
Courses/SUNY_Oneonta/Chem_322_Lecture_Content/06%3A_Nucleophilic_Acyl_Substitution_Reactions/6.10%3A_Nucleophilic_Acyl_Substitution_Reactions_in_the_Laboratory | All of the biological nucleophilic acyl substitution reactions we have seen so far have counterparts in laboratory organic synthesis. Mechanistically, one of the biggest differences between the biological and the lab versions is that the lab reactions usually are run with a strong acid or base as a catalyst, whereas biological reactions are of course taking place at physiological \(pH.\) When proposing mechanisms, then, care must be taken to draw intermediates in their reasonable protonation states: for example, a hydronium ion (\(H_3O^+\)) intermediate is reasonable to propose in an acidic reaction, but a hydroxide (\(OH^-\)) intermediate is not. Ester reactions - bananas, soap and biodiesel Acid-catalyzed synthesis of flavor compounds such as isopentyl acetate (an ester with the flavor of banana) is simple to carry out in the lab. In this esterification reaction, acetic acid is combined with isopentyl alcohol along with a catalytic amount of sulfuric acid. Acid-catalyzed esterification (laboratory reaction): Mechanism: The carbonyl oxygen of acetic acid is first protonated (step 1), which draws electron density away from the carbon and increases its electrophilicity. In step 2, the alcohol nucleophile attacks: notice that under acidic conditions, the nucleophile is not deprotonated simultaneously as it attacks (as we would show in a biochemical mechanism), and the tetrahedral intermediate is a cation rather than an anion. In step 3, a proton is transferred from one oxygen atom to another, creating a good leaving group (water) which is expelled in step 4. Finally (step 5), the carbonyl oxygen on the ester is deprotonated, regenerating the catalytic acid. This reaction is highly reversible, because carboxylic acids are approximately as reactive as esters. In order to obtain good yields of the ester, an excess of acetic acid can be used, which by Le Chatelier's principle (see your General Chemistry textbook for a review) shifts the equilibrium toward the ester product. Saponification is a common term for base-induced hydrolysis of an ester. For example, methyl benzoate will hydrolyze to benzoate and methanol when added to water with a catalytic amount of sodium hydroxide. Mechanism of base-catalyzed ester hydrolysis (saponification): Addition of the base provides hydroxide ion to act as a nucleophile (hydroxide is of course a better nucleophile than water) in step 1. The tetrahedral intermediate (anionic in this case, because the reaction conditions are basic) then collapses in step 2, and the alkoxide (\(CH_3O^-\)) leaves. We are not used to seeing alkoxides or hydroxides as leaving groups in biochemical reactions, because they are strong bases - but in a basic solution, this is a reasonable chemical step. Step 3 is simply an acid-base reaction between the carboxylic acid and the alkoxide. Note that this is referred to as base-induced rather than base-catalyzed because hydroxide is not regenerated, and thus a full molar equivalent of base must be used. The saponification process derives its name from the ancient craft of soap-making, in which the ester groups of triacylglycerols in animal fats are hydrolized under basic conditions to glycerol and fatty acyl anions (see section 2.5A for a reminder of how fatty acyl anions work as soap). We learned earlier about transesterification reactions in the context of the chemical mechanism of aspirin. Transesterification also plays a key role in a technology that is already an important component in the overall effort to develop environmentally friendly, renewable energy sources: biodeisel. You may have heard stories about people running their cars on biodeisel from used french fry oil. To make biodeisel, triacylglycerols in fats and oils can be transesterified with methanol or ethanol under basic conditions. The fatty acyl methyl and ethyl ester products are viable motor fuels. Exercise \(\PageIndex{1}\) Draw structures of the carboxylic acid and alcohol starting materials that could be used to synthesize the fragrant fruit esters shown in section 6.2. Exercise \(\PageIndex{2}\) What would happen if you tried to synthesize isopentyl acetate (banana oil) with basic rather than acidic conditions? Would this work? Exercise \(\PageIndex{3}\) Consider the reverse direction of the acid-catalyzed esterification reaction. What would you call this reaction in organic chemistry terms? Exercise \(\PageIndex{4}\) An alternative way to synthesize esters is to start with a carboxylate and an alkyl halide. Draw a mechanism for such a synthesis of methyl benzoate - what type of reaction mechanism is this? Acid chlorides and acid anhydrides In the cell, acyl phosphates and thioesters are the most reactive of the carboxylic acid derivatives. In the organic synthesis lab, their counterparts are acid chlorides and acid anhydrides, respectively. Of the two, acid chlorides are the more reactive, because the chloride ion is a weaker base and better leaving group than the carboxylate ion (the \(pK_a\) of \(HCl\) is -7, while that of carboxylic acids is about 4.5: remember, a stronger conjugate acid means a weaker conjugate base). Acid chlorides can be prepared from carboxylic acids using \(SOCl_2\): Acid anhydrides can be prepared from carboxylic acids and an acid chloride under basic conditions: Acetic anhydride is often used to prepare acetate esters and amides from alcohols and amines, respectively. The synthesis of aspirin and acetaminophen are examples: A carboxylic acid cannot be directly converted into an amide because the amine nucleophile would simply act as a base and deprotonate the carboxylic acid: Instead, the carboxylic acid is first converted to an acid chloride (in other words, the carboxylic acid is activated), then the acid chloride is combined with an amine to make the amide. This sequence of reactions is a direct parallel to the biochemical glutamine and asparagine synthase reactions we saw earlier in the chapter ( section 11.5), except that the activated form of carboxylic acid is an acid chloride instead of an acyl phosphate or acyl-AMP. Exercise \(\PageIndex{5}\) For the preparation of the amide below, sh ow a starting carboxylate and amine and the intermediate acid chloride species. Polyesters and polyamides If you have ever had the misfortune of undergoing surgery or having to be stitched up after a bad cut, it is likely that you benefited from our increasing understanding of polymers and carboxylic ester chemistry. Polyglycolic acid is a material commonly used to make dissolving sutures. It is a polyester - a polymer linked together by ester groups - and is formed from successive acyl substitution reactions between the alcohol group on one end of a glycolic acid monomer and the carboxylic acid group on a second: The resulting polymer - in which each strand is generally several hundred to a few thousand monomers long - is strong, flexible, and not irritating to body tissues. It is not, however, permanent: the ester groups are reactive to gradual, spontaneous hydrolysis at physiological \(pH\), which means that the threads will dissolve naturally over several weeks, eliminating the need for them to be cut out by a doctor. Exercise \(\PageIndex{6}\) Dacron, a polyester used in clothing fiber, is made of alternating dimethyl terephthalate and ethylene glycol monomers. Draw the structure of a Dacron tetramer (in other words, four monomers linked together). Water is a side product of glycolic acid polymerization. What is the equivalent side product in Dacron production? Exercise \(\PageIndex{7}\) Nylon 6,6 is a widely used polyamide composed of alternating monomers. Nylon 6,6 has the structure shown below -the region within the parentheses is the repeating unit, with 'n' indicating a large number of repeats. Identify the two monomeric compounds used to make the polymer. The Gabriel synthesis of primary amines The Gabriel synthesis, named after the 19 th -century German chemist Siegmund Gabriel, is a useful way to convert alkyl halides to amines and another example of \(S_N2\) and acyl substitution steps in the laboratory. The nitrogen in the newly introduced amine group comes from phthalimide. In the first step of the reaction, phthalimide is deprotonated by hydroxide, then in step 2 it acts as a nucleophile to displace a halide in an \(S_N2\) reaction (phthalimide is not a very powerful nucleophile, so this reaction works only with unhindered primary or methyl halides). Step 3 is simply a pair of hydrolytic acyl substitution steps to release the primary amine, with an aromatic dicarboxylate by-product. Exercise \(\PageIndex{8}\) Phthalimide contains an ' imide ' functional group, and has a \(pK_a\) of approximately 10. What makes the imide group so much more acidic than an amide, which has a \(pK_a\) of approximately 17? As an alternative procedure, release of the amine in step 3 can be carried out with hydrazine (\(H_2NNH_2\)) instead of hydroxide. Again, this occurs through two nucleophilic acyl substitution reactions. In 2000, chemists at MIT synthesizing a porphyrin-containing molecule introduced two amine groups using the Gabriel synthesis with hydrazine. Porphyrins, which include the 'heme' in our red blood cells, are an important family of biomolecules with a variety of biochemical function (J. Org. Chem. 2000, 65, 5298). |
Bookshelves/General_Chemistry/ChemPRIME_(Moore_et_al.)/08%3A_Properties_of_Organic_Compounds/8.23%3A_Synthetic_Macromolecules-_Some_Applied_Organic_Chemistry | Other important classes of substances containing very large molecules are the plastics and artificial fibers, which are such a conspicuous, though not always a positive, feature of modern life. Most of these materials are made in the same basic way. The starting materials or monomers are relatively simple molecules—usually carbon compounds derived from petroleum—which can be persuaded to link up with each other in order to form a long chain of repeating units called a polymer . If we think of the monomer as a bead, then the polymer corresponds to a string of beads. Polymers are usually classified into two types: addition polymers and condensation polymers , according to the kind of reaction by which they are made. |
Bookshelves/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/01%3A_General_Techniques/1.05%3A_Filtering_Methods | There are many methods used to separate a mixture containing a solid and liquid. If the solid settles well, the liquid can sometimes be poured off (decanted). If the solid has very small sized particles or forms a cloudy mixture, the mixture can sometimes be centrifuged or passed through a filter pipette (on the microscale, < 5 mL). 1.5A: Overview of Methods The most common methods of solid-liquid separation in the organic lab are gravity and suction filtration. Gravity filtration refers to pouring a solid-liquid mixture through a funnel containing a filter paper, allowing the liquid to seep through while trapping the solid on the paper. Suction filtration is a similar process with the difference being the application of a vacuum beneath the funnel in order to pull liquid through the filter paper with suction. 1.5B: Decanting When there is a need to separate a solid-liquid mixture, on occasion it is possible to pour off the liquid while leaving the solid behind. This process is called decanting, and is the simplest separation method. 1.5C: Gravity Filtration Gravity filtration is generally used when the filtrate (liquid that has passed through the filter paper) will be retained, while the solid on the filter paper will be discarded. 1.5D: Suction Filtration Suction filtration (vacuum filtration) is the standard technique used for separating a solid-liquid mixture when the goal is to retain the solid (for example in crystallization). Similar to gravity filtration, a solid-liquid mixture is poured onto a filter paper, with the main difference being that the process is aided by suction beneath the funnel. 1.5E: Hot Filtration A hot filtration is generally used in some crystallization, when a solid contains impurities that are insoluble in the crystallization solvent. It is also necessary in crystallization when charcoal is used to remove highly colored impurities from a solid, as charcoal is so fine that it cannot be removed by decanting. 1.5F: Pipette Filtration (Microscale) For the separation of small volumes (< 10mL ) of solid-liquid mixtures, pipette filters are ideal as filter papers absorb a significant amount of material. Pipette filtration may also be used if small amounts of solid are noticed in NMR or GC samples, as both instruments require analysis of liquids without suspended solids. 1.5G: Centrifugation Centrifugation is used for the separation of solid-liquid mixtures that are stubborn to settle or difficult to otherwise filter. It uses centrifugal force by rapidly spinning samples so that the solid is forced to the bottom of the tube. In this section is shown centrifugation of a suspension of yellow lead(II) iodide in water (Figure 1.90b). |
Courses/SUNY_Potsdam/Book%3A_Organic_Chemistry_I_(Walker)/05%3A_Physical_Organic_Chemistry/5.06%3A_Reactive_intermediates | In chemistry, a reactive intermediate or an intermediate is a short-lived, high-energy, highly reactive molecule. When generated in a chemical reaction, it will quickly convert into a more stable molecule. Only in exceptional cases can these compounds be isolated and stored, e.g. low temperatures, matrix isolation. When their existence is indicated, reactive intermediates can help explain how a chemical reaction takes place. Most chemical reactions take more than one elementary step to complete, and a reactive intermediate is a high-energy, yet stable, product that exists only in one of the intermediate steps. The series of steps together make a reaction mechanism. A reactive intermediate differs from a reactant or product or a simple reaction intermediate only in that it cannot usually be isolated but is sometimes observable only through fast spectroscopic methods. It is stable in the sense that an elementary reaction forms the reactive intermediate and the elementary reaction in the next step is needed to destroy it. When a reactive intermediate is not observable, its existence must be inferred through experimentation. This usually involves changing reaction conditions such as temperature or concentration and applying the techniques of chemical kinetics, chemical thermodynamics, or spectroscopy. We will often refer to certain reactive intermediates based on carbon, viz., carbocations, radicals, carbanions and carbenes. Common features Reactive intermediates have several features in common: low concentration with respect to reaction substrate and final reaction product often generated on chemical decomposition of a chemical compound it is often possible to prove the existence of this species by spectroscopic means cage effects have to be taken into account often stabilization by conjugation or resonance often difficult to distinguish from a transition state prove existence by means of chemical trapping References ^ Carey, Francis A.; Sundberg, Richard J.; (1984). Advanced Organic Chemistry Part A Structure and Mechanisms (2nd ed.). New York N.Y.: Plenum Press. ISBN 0-306-41198-9 . ^ March Jerry; (1885). Advanced Organic Chemistry reactions, mechanisms and structure (3rd ed.). New York: John Wiley & Sons, inc. ISBN 0-471-85472-7 ^ Gilchrist, T. L. (1966). Carbenes nitrenes and arynes . Springer US. ISBN 9780306500268 . ^ Moss, Robert A.; Platz, Matthew S.; Jones, Jr., Maitland (2004). Reactive intermediate chemistry . Hoboken, N.J.: Wiley-Interscience. ISBN 9780471721499 . Carbocations (R+) A carbocation is an ion with a positively-charged carbon atom . Among the simplest examples are methenium CH 3 + , methanium CH 5 + , and ethanium C 2 H 7 + . Until the early 1970s, all carbocations were called carbonium ions.[1] In present-day chemistry, a carbocation is any positively charged carbon atom, classified in two main categories according to the valence of the charged carbon: +3 in carbenium ions (protonated carbenes), +5 or +6 in the carbonium ions (protonated alkanes, named by analogy to ammonium). These are much less common. Structure and properties The charged carbon atom in a carbocation is a “sextet”, i.e. it has only six electrons in its outer valence shell instead of the eight valence electrons that ensures maximum stability (octet rule). Therefore, carbocations are often reactive, seeking to fill the octet of valence electrons as well as regain a neutral charge. One could reasonably assume a carbocation to have $$sp^3$$ hybridization with an empty $$sp_3$$ orbital giving positive charge. However, the reactivity of a carbocation more closely resembles $$sp^2$$ hybridization with atrigonal planar molecular geometry. An example is the methyl cation, $$CH_3^+$$. Order of stability of examples of tertiary (3 o ), secondary (2 o ), and primary (1 o ) alkyl carbenium ions , as well as the methyl cation (far right). The methyl group is so unstable it is only observed in the gas phase. Carbocations are often the target of nucleophilic attack by nucleophiles such as water or halide ions. Carbocations typically undergo rearrangement reactions from less stable structures to equally stable or more stable ones with rate constants in excess of 10 9 /sec. This fact complicates synthetic pathways to many compounds. For example, when 3-pentanol is heated with aqueous HCl, the initially formed 3-pentyl carbocation rearranges to a statistical mixture of the 3-pentyl and 2-pentyl. These cations react with chloride ion to produce about 1/3 3-chloropentane and 2/3 2-chloropentane. A carbocation may be stabilized by resonance by a carbon-carbon double bond next to the ionized carbon. Such cations as allyl cation CH 2 =CH–CH 2 + and benzyl cation C 6 H 5 –CH 2 + are more stable than most other carbocations. Molecules that can form allyl or benzyl carbocations are especially reactive. These carbocations where the C+ is adjacent to another carbon atom that has a double or triple bond have extra stability because of the overlap of the empty p orbital of the carbocation with the p orbitals of the π bond. This overlap of the orbitals allows the charge to be shared between multiple atoms – delocalization of the charge – and, therefore, stabilizes the carbocation. References Gold Book definition for carbocation Radicals A radicals is a seven electron intermediate that adopts a flat, sp 2 structure despite the fact that it has four electron groups; the lone electron resides in a half-filled p-orbital. This sp 2 structure allows radicals to delocalize the single electron through resonance. We will study radical reactions in detail in the second semester. Being short of the octet, radicals are electrophilic, and therefore they are stabilized by alkyl groups. Thus the order for stability is the same as for carbocations, namely tertiary > secondary > primary > methyl . Carbanions A carbanion is an eight electron intermediate with an sp 3 structure as shown in A. Despite its full octet, it is very reactive due to the fact that carbon is not very electronegative. Although it is sp 3 , it can participate in resonance because it can easily re-hybridize to an sp 2 structure (see B), which allows overlap. Carbanions are electron-rich and nucleophilic, so in fact they are destabilized by alkyl groups. This means that the order for stability is the opposite of that for carbocations, namely methyl > primary > secondary > tertiary . Carbenes Carbenes are the least obvious of the four common intermediates; in most cases they have a six-electron sp 2 structure that has a lone pair but no overall charge. Although they are short of a full octet, they also have a reactive lone pair, so (depending on structure) carbenes can be either electrophilic or nucleophilic, or sometimes both – they just like to react with almost anything! We will learn about carbene reactions in section 10.7. CC licensed content, Shared previously Reactive_intermediate. Authored by : Wikipedia contributors. Provided by : Wikimedia Foundation. Located at : https://en.wikipedia.org/wiki/Reactive_intermediate . Project : Wikipedia. License : CC BY-SA: Attribution-ShareAlike Carbocations. Authored by : Libretexts contributors. Provided by : UC Davis. Located at : https://chem.libretexts.org/Textbook_Maps/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Fundamentals/Reactive_Intermediates/Carbocations . Project : Libretexts. License : CC BY-NC: Attribution-NonCommercial |
Courses/Palomar_College/PC%3A_CHEM100_-_Fundamentals_of_Chemistry/15%3A_Chemical_Bonding/15.4%3A_Lewis_Structures%3A_Counting_Valence_Electrons | Learning Objectives Draw Lewis structures for covalent compounds. The following procedure can be used to construct Lewis electron structures for more complex molecules and ions: How-to: Constructing Lewis electron structures 1. Determine the total number of valence electrons in the molecule or ion. Add together the valence electrons from each atom. (Recall that the number of valence electrons is indicated by the position of the element in the periodic table.) If the species is a polyatomic ion, remember to add or subtract the number of electrons necessary to give the total charge on the ion. For CO 3 2− , for example, we add two electrons to the total because of the −2 charge. 2. Arrange the atoms to show specific connections. When there is a central atom, it is usually the least electronegative element in the compound. Chemists usually list this central atom first in the chemical formula (as in CCl 4 and CO 3 2− , which both have C as the central atom), which is another clue to the compound’s structure. Hydrogen and the halogens are almost always connected to only one other atom, so they are usually terminal rather than central. 3. Place a bonding pair of electrons between each pair of adjacent atoms to give a single bond. In H 2 O, for example, there is a bonding pair of electrons between oxygen and each hydrogen. 4. Beginning with the terminal atoms, add enough electrons to each atom to give each atom an octet (two for hydrogen). These electrons will usually be lone pairs. 5. If any electrons are left over, place them on the central atom. We will explain later that some atoms are able to accommodate more than eight electrons. 6. If the central atom has fewer electrons than an octet, use lone pairs from terminal atoms to form multiple (double or triple) bonds to the central atom to achieve an octet. This will not change the number of electrons on the terminal atoms. 7. Final check Always make sure all valence electrons are accounted for and each atom has an octet of electrons except for hydrogen (with two electrons). The central atom is usually the least electronegative element in the molecule or ion; hydrogen and the halogens are usually terminal. Now let’s apply this procedure to some particular compounds, beginning with one we have already discussed. Example \(\PageIndex{1}\): Water Write the Lewis Structure for H 2 O. Solution Steps for Writing Lewis Structures Example \(\PageIndex{1}\) 1. Determine the total number of valence electrons in the molecule or ion. Each H atom (group 1) has 1 valence electron, and the O atom (group 16) has 6 valence electrons, for a total of 8 valence electrons. 2. Arrange the atoms to show specific connections. H O H Because H atoms are almost always terminal, the arrangement within the molecule must be HOH. 3. Place a bonding pair of electrons between each pair of adjacent atoms to give a single bond. 4. Beginning with the terminal atoms, add enough electrons to each atom to give each atom an octet (two for hydrogen). Placing one bonding pair of electrons between the O atom and each H atom gives H -O- H with 4 electrons left over. Each H atom has a full valence shell of 2 electrons. 5. If any electrons are left over, place them on the central atom. Adding the remaining 4 electrons to the oxygen (as two lone pairs) gives the following structure: 6. If the central atom has fewer electrons than an octet, use lone pairs from terminal atoms to form multiple (double or triple) bonds to the central atom to achieve an octet. Not necessary 7. Final check The Lewis structure gives oxygen an octet and each hydrogen two electrons, Example \(\PageIndex{2}\) Write the Lewis structure for the \(CH_2O\) molecule Solution Steps for Writing Lewis Structures Example \(\PageIndex{2}\) 1. Determine the total number of valence electrons in the molecule or ion. Each hydrogen atom (group 1) has one valence electron, carbon (group 14) has 4 valence electrons, and oxygen (group 16) has 6 valence electrons, for a total of [(2)(1) + 4 + 6] = 12 valence electrons. 2. Arrange the atoms to show specific connections. Because carbon is less electronegative than oxygen and hydrogen is normally terminal, C must be the central atom. 3. Place a bonding pair of electrons between each pair of adjacent atoms to give a single bond. Placing a bonding pair of electrons between each pair of bonded atoms gives the following: Six electrons are used, and 6 are left over. 4. Beginning with the terminal atoms, add enough electrons to each atom to give each atom an octet (two for hydrogen). Adding all 6 remaining electrons to oxygen (as three lone pairs) gives the following: Although oxygen now has an octet and each hydrogen has 2 electrons, carbon has only 6 electrons. 5. If any electrons are left over, place them on the central atom. Not necessary There are no electrons left to place on the central atom. 6. If the central atom has fewer electrons than an octet, use lone pairs from terminal atoms to form multiple (double or triple) bonds to the central atom to achieve an octet. To give carbon an octet of electrons, we use one of the lone pairs of electrons on oxygen to form a carbon–oxygen double bond: 7. Final check Both the oxygen and the carbon now have an octet of electrons, so this is an acceptable Lewis electron structure. The O has two bonding pairs and two lone pairs, and C has four bonding pairs. This is the structure of formaldehyde, which is used in embalming fluid. Exercise \(\PageIndex{1}\) Write Lewis electron structures for CO 2 and SCl 2 , a vile-smelling, unstable red liquid that is used in the manufacture of rubber. Answer CO 2 . Answer SCl 2 . The United States Supreme Court has the unenviable task of deciding what the law is. This responsibility can be a major challenge when there is no clear principle involved or where there is a new situation not encountered before. Chemistry faces the same challenge in extending basic concepts to fit a new situation. Drawing of Lewis structures for polyatomic ions uses the same approach, but tweaks the process a little to fit a somewhat different set of circumstances. Writing Lewis Structures for Polyatomic Ions Recall that a polyatomic ion is a group of atoms that are covalently bonded together and which carry an overall electrical charge. The ammonium ion, \(\ce{NH_4^+}\), is formed when a hydrogen ion \(\left( \ce{H^+} \right)\) attaches to the lone pair of an ammonia \(\left( \ce{NH_3} \right)\) molecule in a coordinate covalent bond. When drawing the Lewis structure of a polyatomic ion, the charge of the ion is reflected in the number of total valence electrons in the structure. In the case of the ammonium ion: \(1 \: \ce{N}\) atom \(= 5\) valence electrons \(4 \: \ce{H}\) atoms \(= 4 \times 1 = 4\) valence electrons subtract 1 electron for the \(1+\)charge of the ion total of 8 valence electrons in the ion It is customary to put the Lewis structure of a polyatomic ion into a large set of brackets, with the charge of the ion as a superscript outside the brackets. Exercise \(\PageIndex{2}\) Draw the Lewis electron dot structure for the sulfate ion. Answer Exceptions to the Octet Rule As important and useful as the octet rule is in chemical bonding, there are some well-known violations. This does not mean that the octet rule is useless—quite the contrary. As with many rules, there are exceptions, or violations. There are three violations to the octet rule. Odd-electron molecules represent the first violation to the octet rule. Although they are few, some stable compounds have an odd number of electrons in their valence shells. With an odd number of electrons, at least one atom in the molecule will have to violate the octet rule. Examples of stable odd-electron molecules are NO, NO 2 , and ClO 2 . The Lewis electron dot diagram for NO is as follows: Although the O atom has an octet of electrons, the N atom has only seven electrons in its valence shell. Although NO is a stable compound, it is very chemically reactive, as are most other odd-electron compounds. Electron-deficient molecules represent the second violation to the octet rule. These stable compounds have less than eight electrons around an atom in the molecule. The most common examples are the covalent compounds of beryllium and boron. For example, beryllium can form two covalent bonds, resulting in only four electrons in its valence shell: Boron commonly makes only three covalent bonds, resulting in only six valence electrons around the B atom. A well-known example is BF 3 : The third violation to the octet rule is found in those compounds with more than eight electrons assigned to their valence shell. These are called expanded valence shell molecules. Such compounds are formed only by central atoms in the third row of the periodic table or beyond that have empty d orbitals in their valence shells that can participate in covalent bonding. One such compound is PF 5 . The only reasonable Lewis electron dot diagram for this compound has the P atom making five covalent bonds: Formally, the P atom has 10 electrons in its valence shell. Example \(\PageIndex{3}\): Octet Violations Identify each violation to the octet rule by drawing a Lewis electron dot diagram. ClO SF 6 Solution a. With one Cl atom and one O atom, this molecule has 6 + 7 = 13 valence electrons, so it is an odd-electron molecule. A Lewis electron dot diagram for this molecule is as follows: b. In SF 6 , the central S atom makes six covalent bonds to the six surrounding F atoms, so it is an expanded valence shell molecule. Its Lewis electron dot diagram is as follows: Exercise \(\PageIndex{3}\): Xenon Difluoride Identify the violation to the octet rule in XeF 2 by drawing a Lewis electron dot diagram. Answer: The Xe atom has an expanded valence shell with more than eight electrons around it. Summary Lewis dot symbols provide a simple rationalization of why elements form compounds with the observed stoichiometries. A plot of the overall energy of a covalent bond as a function of internuclear distance is identical to a plot of an ionic pair because both result from attractive and repulsive forces between charged entities. In Lewis electron structures, we encounter bonding pairs , which are shared by two atoms, and lone pairs , which are not shared between atoms. Lewis structures for polyatomic ions follow the same rules as those for other covalent compounds. There are three violations to the octet rule: odd-electron molecules, electron-deficient molecules, and expanded valence shell molecules |
Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Chemistry_(Blinder)/01%3A_Chapters/1.03%3A_Quantum_Mechanics_of_Some_Simple_Systems | The simple quantum-mechanical problem we have just solved can provide an instructive application to chemistry: the free-electron model (FEM) for delocalized \(\pi\)-electrons. The simplest case is the 1,3-butadiene molecule \[\rho =2\psi _{1}^2+2\psi _{2}^2\label{28}\] A chemical interpretation of this picture might be that, since the \(\pi\)-electron density is concentrated between carbon atoms 1 and 2, and between 3 and 4, the predominant structure of butadiene has double bonds between these two pairs of atoms. Each double bond consists of a \(\pi\)-bond, in addition to the underlying \(\sigma\)-bond. However, this is not the complete story, because we must also take account of the residual \(\pi\)-electron density between carbons 2 and 3. In the terminology of valence-bond theory, butadiene would be described as a resonance hybrid with the contributing structures CH 2 =CH-CH=CH 2 (the predominant structure) and ºCH 2 - CH=CH-CH 2 º (a secondary contribution). The reality of the latter structure is suggested by the ability of butadiene to undergo 1,4-addition reactions. The free-electron model can also be applied to the electronic spectrum of b utadiene and other linear polyenes . The lowest unoccupied molecular orbital ( LUMO ) in butadiene corresponds to the n=3 particle-in-a-box state. Neglecting electron-electron interaction, the longest-wavelength (lowest-energy) electronic transition should occur from n=2, the highest occupied molecular orbital (HOMO). The energy difference is given by \[\Delta E=E_{3}-E_{2}=(3^2-2^2)\dfrac{h^2}{8mL^2}\label{29}\] Here m represents the mass of an electron (not a butadiene molecule!), 9.1x10 -31 Kg, and L is the effective length of the box, 4x1.40x10 -10 m. By the Bohr frequency condition \[\Delta E=h\upsilon =\dfrac{hc}{\lambda }\label{30}\] The wavelength is predicted to be 207 nm. This compares well with the experimental maximum of the first electronic absorption band, \(\lambda_{max} \approx\) 210 nm, in the ultraviolet region. We might therefore be emboldened to apply the model to predict absorption spectra in higher polyenes CH 2 =(CH-CH=) n- 1 CH 2 . For the molecule with 2 n carbon atoms ( n double bonds), the HOMO → LUMO transition corresponds to n → n + 1, thus \[\dfrac{hc}{\lambda} \approx \begin{bmatrix}(n+1)^2-n^2\end{bmatrix}\dfrac{h^2}{8m(2nL_{CC})^2}\label{31}\] A useful constant in this computation is the Compton wavelength \[\dfrac{h}{mc}= 2.426 \times 10^{-12} m.\] For n =3, hexatriene, the predicted wavelength is 332 nm, while experiment gives \(\lambda _{max}\approx \) 250 nm. For n =4, octatetraene, FEM predicts 460 nm, while \(\lambda _{max}\approx\) 300 nm. Clearly the model has been pushed beyond range of quantitate validity, although the trend of increasing absorption band wavelength with increasing n is correctly predicted. Incidentally, a compound should be colored if its absorption includes any part of the visible range 400-700 nm. Retinol (vitamin A), which contains a polyene chain with n =5, has a pale yellow color. This is its structure: |
Courses/Saint_Francis_University/Chem_114%3A_Human_Chemistry_II_(Hargittai)/18%3A_Amino_Acids_and_Proteins/18.05%3A_Handedness | Learning Objectives Explain how a peptide is formed from individual amino acids. Explain why the sequence of amino acids in a protein is important. Two or more amino acids can join together into chains called peptides. In an earlier chapter, we discussed the reaction between ammonia and a carboxylic acid to form an amide. In a similar reaction, the amino group on one amino acid molecule reacts with the carboxyl group on another, releasing a molecule of water and forming an amide linkage: An amide bond joining two amino acid units is called a peptide bond . Note that the product molecule still has a reactive amino group on the left and a reactive carboxyl group on the right. These can react with additional amino acids to lengthen the peptide. The process can continue until thousands of units have joined, resulting in large proteins. A chain consisting of only two amino acid units is called a dipeptide ; a chain consisting of three is a tripeptide . By convention, peptide and protein structures are depicted with the amino acid whose amino group is free (the amino-terminal or N-terminal end) on the left and the amino acid with a free carboxyl group (the carboxyl-terminal or C-terminal end) to the right. Individual amino acids joined in a chain are called residues . The general term peptide refers to an amino acid chain of unspecified length. However, chains of about 50 amino acids or more are usually called proteins or polypeptides. In its physiologically active form, a protein may be composed of one or more polypeptide chains. Note: Order is Important For peptides and proteins to be physiologically active, it is not enough that they incorporate certain amounts of specific amino acids. The order, or sequence , in which the amino acids are connected is also of critical importance. Bradykinin is a nine-amino acid peptide (Figure \(\PageIndex{1}\)) produced in the blood that has the following amino acid sequence: arg-pro-pro-gly-phe-ser-pro-phe-arg This peptide lowers blood pressure, stimulates smooth muscle tissue, increases capillary permeability, and causes pain. When the order of amino acids in bradykinin is reversed, arg-phe-pro-ser-phe-gly-pro-pro-arg the peptide resulting from this synthesis shows none of the activity of bradykinin. Just as millions of different words are spelled with our 26-letter English alphabet, millions of different proteins are made with the 20 common amino acids. However, just as the English alphabet can be used to write gibberish, amino acids can be put together in the wrong sequence to produce nonfunctional proteins. Although the correct sequence is ordinarily of utmost importance, it is not always absolutely required. Just as you can sometimes make sense of incorrectly spelled English words, a protein with a small percentage of “incorrect” amino acids may continue to function. However, it rarely functions as well as a protein having the correct sequence. There are also instances in which seemingly minor errors of sequence have disastrous effects. For example, in some people, every molecule of hemoglobin (a protein in the blood that transports oxygen) has a single incorrect amino acid unit out of about 300 (a single valine replaces a glutamic acid). That “minor” error is responsible for sickle cell anemia, an inherited condition that usually is fatal. Summary The amino group of one amino acid can react with the carboxyl group on another amino acid to form a peptide bond that links the two amino acids together. Additional amino acids can be added on through the formation of addition peptide (amide) bonds. A sequence of amino acids in a peptide or protein is written with the N-terminal amino acid first and the C-terminal amino acid at the end (writing left to right). Concept Review Exercises Distinguish between the N-terminal amino acid and the C-terminal amino acid of a peptide or protein. Describe the difference between an amino acid and a peptide. Amino acid units in a protein are connected by peptide bonds. What is another name for the functional group linking the amino acids? Answers The N-terminal end is the end of a peptide or protein whose amino group is free (not involved in the formation of a peptide bond), while the C-terminal end has a free carboxyl group. A peptide is composed of two or more amino acids. Amino acids are the building blocks of peptides. amide bond Exercises Draw the structure for each peptide. gly-val val-gly Draw the structure for cys-val-ala. Identify the C- and N-terminal amino acids for the peptide lys-val-phe-gly-arg-cys. Identify the C- and N-terminal amino acids for the peptide asp-arg-val-tyr-ile-his-pro-phe. Answers C-terminal amino acid: cys; N-terminal amino acid: lys |
Courses/Widener_University/Widener_University%3A_Chem_135/10%3A_Liquids_and_Solids | The great distances between atoms and molecules in a gaseous phase, and the corresponding absence of any significant interactions between them, allows for simple descriptions of many physical properties that are the same for all gases, regardless of their chemical identities. As described in the final module of the chapter on gases, this situation changes at high pressures and low temperatures—conditions that permit the atoms and molecules to interact to a much greater extent. In the liquid and solid states, these interactions are of considerable strength and play an important role in determining a number of physical properties that do depend on the chemical identity of the substance. In this chapter, the nature of these interactions and their effects on various physical properties of liquid and solid phases will be examined. 10.1: Prelude to Liquids and Solids In the liquid and solid states, these interactions are of considerable strength and play an important role in determining a number of physical properties that do depend on the chemical identity of the substance. In this chapter, the nature of these interactions and their effects on various physical properties of liquid and solid phases will be examined. 10.2: Intermolecular Forces The physical properties of condensed matter (liquids and solids) can be explained in terms of the kinetic molecular theory. In a liquid, intermolecular attractive forces hold the molecules in contact, although they still have sufficient kinetic energy to move past each other. Intermolecular attractive forces, collectively referred to as van der Waals forces, are responsible for the behavior of liquids and solids and are electrostatic in nature. 10.3: Properties of Liquids The intermolecular forces between molecules in the liquid state vary depending upon their chemical identities and result in corresponding variations in various physical properties. Cohesive forces between like molecules are responsible for a liquid’s viscosity (resistance to flow) and surface tension. Adhesive forces between the molecules of a liquid and different molecules composing a surface in contact with the liquid are responsible for surface wetting and capillary rise. 10.4: Phase Transitions 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. 10.E: Liquids and Solids (Exercises) End of chapter homework problems for Chapter \(\PageIndex{1}\). |
End of preview. Expand
in Data Studio
README.md exists but content is empty.
- Downloads last month
- 73
Size of downloaded dataset files:
259 MB
Size of the auto-converted Parquet files:
259 MB
Number of rows:
66,861