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Unformatted text preview: S117 Homework 7 Out of 85 Points Chapter 6 (to turn in) 12. (out of 5 points) Equilibrium is not static, but dynamic. This means that when a reaction is at an equilibrium position, both the forward and reverse reaction are occurring; they're simply occurring at the same rate so that there is no net change. Therefore, even at equilibrium, some of the radioactive 14C atoms in the CO will react and end up in CO2. 20. (out of 5 points) K with no subscript is shorthand for Kc. , where the n is the number of moles of gas in the products minus the number of moles of gas in the reactants. Therefore, when the number of moles of gas is the same in the reactants as in the products, , so , so . The only reaction in problem 19 for which this is true is reaction d. 24. (out of 5 points) The equilibrium constant in terms of partial pressures for this reaction is , where division by 1atm was done because equilibrium constants deal with activities rather than absolute pressures. (From hereafter, units will simply not be shown in equilibrium constants.) At the other given partial pressures, . Since , this system is not at equilibrium. 26. (out of 5 points) The balanced reaction is . Since this problem is at constant volume, the number of moles is directly proportional to the partial pressure. Since we're not given the volume or the number of moles, the temperature is useless, but we can work through this problem without it. We can use an ICE table. Initial 1.00atm 2.00atm 0atm Change -x -3x +2x Equilibrium 1.00atm-x 2.00atm-3x 2x We are given that the total pressure at equilibrium is 2.00atm. Mathematically, this means that , so . Therefore, the partial pressure of hydrogen gas at equilibrium is . Also, . This is not even close to the equilibrium constant calculated in problem 84, but that is okay since the temperature may be different. 34. (out of 10 points) The equilibrium constant is more is added is . The concentration of F2 after . After more F2 is added, . Since , the reaction must shift to the right to reach equilibrium, which is exactly what Le Chtelier's principle predicts. Let's set up an ICE table to calculate the equilibrium concentrations after the F2 is added. Initial Change Equilibrium Now Let's manipulate this to solve for x. , so . Therefore, , and 38. (out of 5 points) ICE tables are useful here. a. Initial Change Equilibrium Now . First guessing , 2.0mol/2.0L=1.0M -2x 1.0M-2x 0M +2x 2x 0M +x x 0.0500M -x 0.0500M-x , so , so , . 0.0500M -x 0.0500M-x 0.400M +2x 0.400M+2x . . We'll use successive approximation, so . Now using . Since this . Therefore, , and . in the right side of the equation, yields the same value of x, we can see that our final x is , b. Initial Change Equilibrium Now 0M +2x 2x 2M -2x 2M-2x 1M -x 1M-x . Using successive approximation here is quite difficult because only a very good first approximation will work; otherwise, each approximation gets worse and worse. This will almost always happen when x is expected to be large compared to the concentrations with which it's being added/subtracted. In this case, equilibrium lies far to the left, so it is expected that x will be close to 1, so it is large. Instead of trying and failing at successive approximation, the solutions to this equation were solved on a computer algebra system to yield . Therefore, , , and . On an exam, you of course will not have a computer algebra system to solve cubic equations exactly for you, so that sounds like trouble. Successive approximations don't work, and it's not feasible to solve this analytically. There is yet another method. In this problem, we start with all products and no reactants; yet we know that at equilibrium, we'll have mostly reactants and few products since K is small. Let's assume for the moment that all of the products react to form the reactants. Initial Change Final 0M +2M 2M 2M -2M 0M 1M -1M 0M Now we can solve this as a regular ICE table, taking the final conditions above as the initial conditions of our ICE table. Initial Change Equilibrium 2M -2x 2.0M-2x 0M +2x 2x 0M +x x Now this problem is very similar to part a., and we can use successive approximation. The results are , , and . This is the general method to use when x is expected to be large in the given initial conditions. c. Initial Change Equilibrium Now successive approximation. Therefore, . d. Initial Change Equilibrium 0M +2x 2x 3M -2x 3M-2x 1M -x 1M-x 1M -2x 1M-2x , so , , and 1M +2x 1M+2x 0M +x x using Now . As in part b., successive approximation will fail here because x will be nearly 1M. Rather than solving a cubic equation, let's assume that all of the Cl2 reacts. (In this case, Cl2 is a limiting reagent, so not all of the NO can react.) Initial Change Final 0M +2M 2M 3M -2M 1M 1M -1M 0M We can use these conditions to set up an ICE table and solve it with successive approximation. Initial Change Equilibrium Now for x in terms of x. 2M -2x 2.0M-2x 1M +2x 1M+2x 0M +x x . To use successive approximation, we can solve . First assuming that and going through successive , approximations (actually, the first approximation is good enough), we find that so , , and . e. It is not immediately obvious in which direction the reaction must move to reach equilibrium, but the reaction quotient will tell us this. . Since , the reaction must shift to the left, but we have yet another situation where x will be large, so let all of the products react. Initial 2M 2M 1M Change +2M -2M -1M Final 4M 0M 0M Proceeding as in parts b. and d., we would solve an ICE table, which will not be shown for the third time. We find that , , and . f. Again, it is not immediately obvious in which direction the reaction must move to reach equilibrium, but the reaction quotient will tell us this. . Since , the reaction must shift to the left. Let's do exactly what we did in parts b., d., and e. Initial Change Final 1M +1M 2M 1M -1M 0M 1M -0.5M 0.5M We can use these conditions to set up an ICE table and solve it with successive approximation. Initial Change Equilibrium Now . Assuming Therefore, 2M -2x 2.0M-2x 0M +2x 2x 0.5M +x 0.5M+x . Solving for the x in the [NO] factor, and doing successive approximation, we find that , , and . 48. (out of 5 points) a. Water does not appear in this reaction, but it does change the concentrations of the reactants. Doubling the volume by adding water will halve the concentrations of everything. Now . Since , the reaction will shift left to reach equilibrium. b. Without knowing whether Fe3+ or FeSCN2+ will react with NO3-, it is impossible to tell what will happen to the equilibrium position. If neither reacts, then equilibrium will shift to the left since the formation of AgSCN will pull SCN- out of solution. If they do react, then the result depends on the equilibrium constants of the reactions and . c. Assuming that the only species which reacts with Na+ or OH- is Fe3+ (forming Fe(OH)3), adding NaOH will shift the equilibrium position to the left since reactant will be removed. d. Again, not knowing whether NO3- will react with FeSCN2+, this question is impossible to answer. Assuming that the only effect of adding Fe(NO)3 is to increase the concentration of Fe3+, then the equilibrium position will shift right. 56. (out of 5 points) As the temperature is raised (in other words, as energy is put into the system as heat), the equilibrium constant becomes smaller and smaller, meaning that reactants are favored more and more. If reactants are favored more as heat is input, then the equilibrium position is shifting left when heat is input. One way of imagining this is to think of heat as a component of the reaction. Since its addition shifts the equilibrium position left, it must be a "product" of the reaction, making the reaction exothermic. 58. (out of 10 points) a. Adding the first and the reverse of the third reactions gives , which is equivalent to constant is . The equilibrium constant is then . , the reaction we want. Since we've essentially subtracted the third reaction, its new equilibrium b. Add the first and second reactions and subtract the third reaction to get , which is equivalent to . c. Doubling the second reaction and subtracting the third reaction from it gives , which is equivalent to , the reaction we want. Doubling the second reaction means that the coefficients of everything double. The effect of this is that the equilibrium constant gets squared. Therefore, . 70. (out of 5 points) Naphthalene is a hydrocarbon, so it consists only of carbon and hydrogen atoms. From the given information, its molar mass is . Since naphthalene is 93.71% carbon by mass, of its molar mass comes from carbon. This corresponds to 10 moles of carbon atoms per mole of naphthalene (since the molar mass of carbon is approximately ). Therefore, the remaining of naphthalene come from hydrogen which corresponds to 8 moles of hydrogen atoms per mole of naphthalene. Therefore, naphthalene's molecular formula is C10H8. The equilibrium expression for this reaction is . Recall that equilibrium expressions deal with activities rather than concentrations, so it can be written as , so the concentration of naphthalene is . It is in a 5.0-L container, so . The total amount of naphthalene is , so the fraction which sublimed is . , the reaction we want. Therefore, Chapter 7 (to turn in) 42. (out of 10 points) The relevant reaction is given by an ICE table. Initial Change Equilibrium 0.1M -x 0.1M-x , so the acid dissociation expression is , so monochloroacetic acid is a weak acid. Let's use 0M +x x 0M +x x Therefore, approximation gives 55. (out of 5 points) The relevant reaction is , so , so , so . Successive . , so . This is a different sort of problem than usual since we know the equilibrium concentration (since we are given the equilibrium pH) rather than the initial concentration of something. However, an ICE table will still help us. The ICE table is Initial Change Equilibrium Now concentration of HCOOH was 0.024M. x -10-2.7M x-10-2.7M , so 0M +10-2.7M 10-2.7M 0M +10-2.7M 10-2.7M , so the initial Chapter 8 (to turn in) 24. (out of 5 points) HONH2 is a weak base with conjugate acid HONH3+. For HONH2, as given on page A-25. We can, as usual, use an ICE table. Note that HONH3Cl can be assumed to dissociate completely into Cl- and HONH3+. Initial Change Equilibrium Now , so is give 0.1M -x 0.1M-x 0.1M +x 0.1M+x . Using successive approximation, . We have seen an expression for the autoionization of water; it . Here, we can substitute in our hydroxide concentration to . Now 0M +x x . There is another way to do this problem, but we haven't seen it in class. If you're curious, here it is. HONH2 is a weak base with conjugate acid HONH3+. For HONH2, as given on page A-25, so dissociate completely into Cl and
- . HONH3Cl can be assumed to HONH3+, so the Henderson-Hasselbalch equation can be used. . 26. (out of 5 points) NaOH will dissociate essentially completely into Na+ and OH-, so adding 0.020mol OH- will react with 0.020mol HONH3+, bringing the concentration of HONH3+ down to and thereby raising the concentration of HONH2 up to Initial Change Equilibrium 0.120M -x 0.120M-x 0.080M +x 0.080M+x . 0M +x x , so Using successive approximation and solving for x gives us . Now . HCl will dissociate essentially completely into H+ and Cl-, so adding 0.020mol H+ will react with 0.020mol HONH2, bringing the concentration of HONH2 down to and thereby raising the concentration of HONH3+ up to Initial Change Equilibrium 0.080M -x 0.080M-x . 0.120M +x 0.120M+x 0M +x x , so Using successive approximation and solving for x gives us . Now . ...
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This note was uploaded on 01/17/2012 for the course S 117 taught by Professor Stephenjacobson during the Fall '11 term at Indiana.
- Fall '11