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CHAPTER 15 (B&O) NON ACID/BASE CHEMICAL EQUILIBRIUM

Problems to prepare students for hourly exam II. x Non acid-base equilibrium concepts x Non acid-base equilibrium calculations x Le Chatelier’s Principle E. Tavss 2/12

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NON ACID-BASE EQUILIBRIUM CONCEPTS

Chem 162-2011 Hourly Exam II + Answers Chapter 14 - Non Acid/Base Chemical Equilibrium Non acid-base equilibrium concepts 13. 2NO(g) + Cl2(g) ඬ 2NOCl(g) Kp =1.40 x 105 at 298K What is Kp at 298 K for the reaction: NOCl(g) ඬ�NO(g) + ½ Cl2(g) A. 7.14 x 10-6 B. 3.57 x 10-6

C. 1.34 x 10-3 D. 5.34 x 10-3 E. 2.67 x 10-3 The equilibrium equation is reversed and halved. Therefore, the new K is the square root of the inverse of the

original K. 1/((1.40x105)0.5) = 2.67 x 10-3

Chem 162-2011 Hourly Exam II + Answers Chapter 14 - Non Acid/Base Chemical Equilibrium Non acid-base equilibrium concepts 10. Which of the following statements concerning the reaction quotient Q and/or the equilibrium

constant K is false? A. Q must be equal to K at equilibrium. B. The larger the value of K, the farther the forward reaction proceeds toward completion. C. As the temperature increases, the value of K increases. D. The concentrations of pure solids and pure liquids are omitted from the expression for Q, because they do not change during the course of a chemical reaction. E. By comparing the value of Q with K one can predict the direction of change for a system not initially in equilibrium. A. True. If Q = K, then the system is at equilibrium.

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B. True. The larger the value of K, the further the reaction goes to the right (which means, toward completion).

C. False. As the temperature increases in an endothermic reaction, the value of K increases; but as the temperature increases in an exothermic reaction the value of K decreases.

D. True. Whether the measurement is Q or K, the concentrations of pure solids and pure liquids are omitted from the equilibrium expression because only substances that change in concentration are included, and pure solids and pure liquids don’t change in concentration.

E. True. If Q is smaller than K then the reaction will go to the right; if Q is larger than K then the reaction will go to the left.

CHEM 162-2010 EXAM II Chapter 14 - Non Acid/Base Chemical Equilibrium Non acid-base equilibrium concepts 25. SO2Cl2(g) mo SO2(g) + Cl2(g) K = 0.070 Flask X: 0.50 mol SO2Cl2, 0.10 mol SO2, 0.20 mol Cl2 in 0.50L Flask Y: 0.50 mol SO2Cl2, 0.10 mol SO2, 0.20 mol Cl2 in 1.00L Flask Z: 0.50 mol SO2Cl2, 0.10 mol SO2, 0.20 mol Cl2 in 2.00L Which flask will result in a reaction forming more Cl2? A. Flask X only B. Flask Y only C. Flask Z only D. Flasks X and Y only E. Flasks Y and Z only Flask X: SO2Cl2(g) mo SO2(g) + Cl2(g)

Initial 1.00 0.20 0.40 Change Equilibrium Q = ([SO2][Cl2])/[SO2Cl2] Q = ([0.20][0.40])/[1.00] = 0.080. Q > K. Reaction will go to the left; [Cl2] will be decreased. Flask Y: SO2Cl2(g) mo SO2(g) + Cl2(g)

Initial 0.50 0.10 0.20 Change Equilibrium

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Q = ([SO2][Cl2])/[SO2Cl2] Q = ([0.10][0.20])/[0.50] = 0.040. Q < K. Reaction will go to the right; [Cl2] will be formed. Flask Z: SO2Cl2(g) mo SO2(g) + Cl2(g)

Initial 0.25 0.05 0.10 Change Equilibrium Q = ([SO2][Cl2])/[SO2Cl2] Q = ([0.05][0.10])/[0.25] = 0.020. Q < K. Reaction will go to the right; [Cl2] will be formed. Answer: Flasks Y and Z only CHEM 162-2010 EXAM II Chapter 14 - Non Acid/Base Chemical Equilibrium Non acid-base equilibrium concepts 16. Given: C(s) + CO2(g) mo 2CO Kp = 1.3 x 1014

COCl2(g) mo CO(g) + Cl2(g) Kp = 167 Calculate Kp for: C(s) + CO2(g) + 2Cl2(g) mo 2COCl2 A. 2.2 x 1016 B. 7.8 x 1011 C. 1.3 x 1012 D. 2.6 x 1011 E. 4.7 x 109 Use Hess’s Law. C(s) + CO2(g) mo 2CO Kp = 1.3 x 1014 For the 2nd equation, reverse it and double it; correspondingly, invert and square the Kp. 2CO(g) + 2Cl2(g) mo 2COCl2(g) Kp = (1/167)2 Add the equations and multiply the equilibrium constants. C(s) + CO2(g) + 2Cl2(g) mo 2COCl2 K = 4.66 x 109 CHEM 162-2010 EXAM II Chapter 14 - Non Acid/Base Chemical Equilibrium Non acid-base equilibrium concepts 6. Which of the following is true about a chemical reaction at equilibrium?

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X. The rate of the forward reaction equals the rate of the reverse reaction. Y. There is no observable change in any property of the system. Z. The molecules have less kinetic energy than before equilibrium was reached. A. X and Y only B. X only C. Y only D. X and Z only E. X, Y, and Z At equilibrium, the forward and reverse reactions proceed at equal rates, and the concentrations of reactants and products remain constant. Since there is no change in the concentrations of reactants and products, and change in properties (e.g., vapor pressure) will only result if the concentrations of reactants and products change, then there is no change in any property of the system. A. X and Y only.

CHEM 162-2009 FINAL EXAM CHAPTER 14 - NON ACID/BASE CHEMICAL EQUILIBRIUM NON ACID-BASE EQUILIBRIUM CONCEPTS

44. Consider the equilibrium reaction: N2O4(g) mo 2NO2(g) Which of the following correctly describes the relationship between Kc and Kp for the reaction? A. Kp = RT x Kc B. Kp = Kc C. Kp = (RT x Kc)-1 D. Kp = Kc/RT E. Kp = RT/Kc N2O(g) mo 2NO2(g) ¨n = 2 – 1 = 1 Kp = Kc(RT)¨n Kp = Kc(RT)1 Kp = Kc(RT) Kp = RT x Kc A

CHEM 162-2009 HOURLY EXAM II + ANSWERS CHAPTER 14: NON ACID/BASE CHEMICAL EQUILIBRIUM NON ACID-BASE EQUILIBRIUM CONCEPTS 11. Which statement is correct about a system at equilibrium? A. The concentrations of reactants must equal the concentrations of the products.

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B. The forward and reverse reactions occur at identical rates. C. The concentrations of reactants and products can be changed by adding a catalyst. D. The concentrations of reactants and products are not affected by a change in temperature. E. The reaction continuously oscillates between reactants and products. The two rules regarding equilibrium are (1) the forward rate of the reaction is equal to the reverse rate of the reaction, and (2) there is no change in the concentration of the reactants or products A. False. The concentrations of reactants and products are rarely equal. B. True. The forward and reverse reactions must occur at identical rates. This results in no change in the concentration of reactants or products. C. False. A catalyst doesn’t change the Free Energy of reactants and products (to be discussed in the chapter on Thermodynamics). Since the equilibrium constant is directly related to the free energy, then the equilibrium constant won’t change, and therefore the concentrations of reactants and products won’t change. D. False. The concentrations of reactants and products are affected by a change in temperature. In fact, a change in temperature is the only thing that will affect the equilibrium constant, and correspondingly, the concentrations of reactants and products. E. False. The system slowly and steadily comes to an equilibrium value, and then stays there. It doesn’t oscillate back and forth like a pendulum.

Chem 162-2008 Exam II + Answers Chapter 14 Non Acid/Base Chemical Equilibrium Non acid-base equilibrium concepts Adding equilibrium equations and multiplying K’s 3. Consider the following two reactions and their equilibrium constants. Reaction 1: NO2(g) ඬ NO(g) + O(g) K1 = 6.8 x 10-49

Reaction 2: NO2(g) + O2(g) ඬO3(g) + NO(g) K2 = 1.7 x 1033 Calculate K for the reaction: O2(g) + O(g) ඬ O3(g) (a) 1.2 x 10-15 (b) 4.0 x 10-82 (c) 8.7 x 1014 (d) 2.5 x 1081 (e) 4.3 x 1041 This is Hess’ law. Reaction 1 gets reversed; therefore, K1 gets inverted. Reaction 2 doesn’t change. The sum of reversed reaction 1 and reaction 2 equals the product of K2 and inverted K1. NO(g) + O(g) ඬ�NO2(g) 1/K1 = 1/(6.8 x 10-49)= 1.47 x 1048

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NO2(g) + O2(g) ඬ�O3(g) + NO(g) K2 = 1.7 x 1033

O2(g) + O(g) ඬ O3(g) (1/K1) x (K2) = (1.47 x 1048) x (1.7 x 1033) = 2.50 x 1081

4 Chem 162-2007 Final exam + answers Chapter 14 – Non Acid-Base Chemical Equilibrium Non Acid-Base equilibrium concepts For which of the following reactions is Kc > Kp?

A. H2O(s) ҡ H2O(g) B. H2(g) + ½ O2(g) ҡ H2O(g) C. H2(g) + Cl2(g) ҡ 2HCl(g) D. 2KClO3(s) ҡ KCl(s) + 3O2(g) E. 2O3(g) ҡ 3O2(g)

Kpinatm = KcinM(RT)¨ngas Kp = Kc when 'ngas = 0 Kp > Kc when 'ngas >0 Kc > Kp when 'ngas < 0. A. 'n = 1 – 0 = 1 B. 'n = 1 – 1.5 = -0.5 C. 'n = 2 – 2 = 0 D. 'n = 3 – 0 = 3 E. 'n = 3 – 2 = 1 Actually, the relative values of Kc and Kp also depend on the temperature. At 298K the rules I wrote apply. B

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29 Chem 162-2007 Final exam + answers Chapter 14 – Non Acid-Base Chemical Equilibrium Non Acid-Base equilibrium concepts H2(g) + I2(g) ҡ 2HI(g) K = 54.5 What is the equilibrium constant for the reaction: HI(g) ҡ ½ H2(g) + ½ I2(g)

A. 27.2 B. 0.0367 C. 7.38 D. 1.45 E. 0.135

Reversing the reaction corresponds to inverting K. Taking half the reaction corresponds to taking the square root of K. Hence, we want to take the square root of the inverse of K. (1/54.5)0.5 = 0.135

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32 Chem 162-2007 Final exam + answers Chapter 14 – Non Acid-Base Chemical Equilibrium Non Acid-Base equilibrium concepts Consider the following plot for the reaction: A(g) ҡ B(g) K = 3.0

X

Y

Z

[A] M

[B] M

Equilibrium line

Which of the following statements is true?

A. Point Z represents Q < K. B. Point X represents Q>K. C. Point Y represents Q<K D. Point X represents Q<K E. The system is at equilibrium when [B] = 4.0M and [A] = 2.0M

Visualize an example of 6 on the B axis meeting 2 on the A axis at Z A. False. Point Z represents a 3/1 ratio of [B]/[A], which is equal to K. B. False. This would be approximately 6/5 = 1.2, which is < K. C. False. This would be approximately 15/3 = 5, which is > K. D. True. This would be approximately 6/5 = 1.2, which is < K. E. False. 4.0/2.0 = 2.0, which is < K.

CHEM 162-2007 EXAM II + ANSWERS CHAPTER 14 – NON-ACID/BASE CHEMICAL EQUILIBRIUM NON-ACID BASE EQUILIBRIUM CONCEPTS

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15. When I2(s) is added to water, a saturated solution at equilibrium results. The dissolved iodine imparts a gold color to the water. The equilibrium is represented as:

I2(s) ඬ I2(aq) After equilibrium is established, some radioactive solid iodine is added to the solution. After a period of time, the solution remains saturated, and the intensity of the gold color in the solution remains the same. When the solution is tested, it is found that some of the radioactivity is now in the solution as well as in the solid. What does this experiment indicate about the equilibrium? (a) The rate of dissolving is greater than the rate of crystallizing (b) The rate of crystallizing is greater than the rate of dissolving (c) After equilibrium is established, no further solid iodine dissolves (d) The rates of dissolving and crystallizing both approach zero. (e) The iodine continues to dissolve and crystallize at equal rates (a) The rate of dissolving is greater than the rate of crystallizing False. The rate of dissolving is only greater than the rate of crystallizing at the beginning of the reaction when the solid is first added to the solvent. At equilibrium the rate of dissolving is equal to the rate of crystallizing. (b) The rate of crystallizing is greater than the rate of dissolving False. The rate of crystallizing would only be greater than the rate of dissolving in a pre-equilibrium situation, such as if the solution was cooled and the solid began precipitating out of solution. However, in this case we are at equilibrium, where, by definition, the rate of crystallizing is equal to the rate of dissolving. (c) After equilibrium is established, no further solid iodine dissolves False. This is a dynamic equilibrium in which it can microscopically be determined that even at equilibrium solid iodine is dissolving in the solvent, while dissolved iodine is crystallizing onto the solid. (d) The rates of dissolving and crystallizing both approach zero. False. If the rates of dissolving and crystallizing both approached zero, then there would be no dynamic equilibrium. However, there is always dynamic equilibrium (except for some unusual circumstances, such as at absolute zero temperature). (e) The iodine continues to dissolve and crystallize at equal rates True. The first sentence says that the pre-radioactive solution is at equilibrium. The fifth sentence suggests that the radioactive system continues to be at equilibrium. In a dynamic equilibrium, the forward rate (dissolving) and the reverse rate (crystallization/precipitation) are equal.

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NON ACID-BASE EQUILIBRIUM CALCULATIONS

25 Chem 162-2011 Final exam

Chapter 14 - Non-Acid/Base Chemical Equilibrium Non acid-base equilibrium calculations Solid ammonium chloride, NH4Cl, decomposes when heated according to the equation

NH4Cl(s) ҡ NH3(g) + HCl(g) A sample of solid ammonium chloride is heated to 400 K in an evacuated container. When the system comes to equilibrium, some NH4Cl(s) remains and the total pressure in the container is found to be 1.5 x 10-4 atm. Assuming that the total pressure is due only to NH3 and HCl, find Kp for this reaction at 400 K.

A. 7.5x10-5 B. 2.8x10-9 C. 5.6 x 10-9 D. 2.0x10-8 E. 4.0x10-8

NH4Cl(s) ҡ NH3(g) + HCl(g) Initial Y 0 0 Change -X +X +X Equilibrium Y-X +X +X

X + X = 1.5 x 10-4 2X = 1.5 x 10-4 X = 7.5 x 10-5 atm [NH3][HCl] = Kp [7.5 x 10-5][ 7.5 x 10-5] = Kp Kp = 5.625 x 10-9

C

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50 Chem 162-2011 Final exam Chapter 14 - Non-Acid/Base Chemical Equilibrium Non acid-base equilibrium calculations A reaction vessel at 27°C contains a mixture of SO2 (initial P=3.00 atm) and O2 (initial P= 1.00 atm). A reaction proceeds according to the equation below : 2SO2(g)+ O2(g) ҡ 2SO3(g) At equilibrium, the total pressure was found to be 3.75 atm. What is the equilibrium constant Kp?

A. 1.9 x 101 B. 3.0 C. 2.7 x 10-1 D. 5.3 x 10-2 E. 1.1 x 10-1

2SO2(g) + O2(g) ҡ 2SO3(g)

Initial 3.00 1.00 0 Change -2X -X +2X Equilibrium 3.00-2X 1.00-X +2X

(3.00-2X) + (1.00-X) + (2X) = 3.75 atm X = 0.25 atm 2SO2(g) + O2(g) ҡ 2SO3(g)

Initial 3.00 1.00 0 Change -2X -X +2X Equilibrium 2.50 0.75 0.50

P2so3/(P2

so2 x Po2) = (0.502)/((2.502) x 0.75) = 0.0533 = Kp

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Chem 162-2011 Hourly Exam II + Answers Chapter 14 - Non Acid/Base Chemical Equilibrium Non acid-base equilibrium calculations 16. 2SO3(g) ඬ 2SO2(g) + O2(g) Kp = 1.2 x 10-5

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A sample of SO3(g) is placed in a container, initially exerting a pressure of 0.800 atm. What is the pressure of O2(g) at equilibrium?

A. 0.0035 atm B. 0.0070 atm C. 0.012 atm D. 0.0024 atm E. 0.0096 atm 2SO3(g) ඬ 2SO2(g) + O2(g)

Initial 0.800 0 0 Change -2X +2X +X Equilibrium 0.800-2X +2X +X (Pso2

2 x Po2)/Pso32 = 1.2 x 10-5

((2X)2 x X)/(0.800-2X)2 = 1.2 x 10-5 Use the small K rule to get around a quadratic equation. ((2X)2 x X)/(0.800)2 = 1.2 x 10-5 4X3/0.64 = 1.2 x 10-5

X = 1.24 x 10-2 = Po2 Chem 162-2011 Hourly Exam II + Answers Chapter 14 - Non Acid/Base Chemical Equilibrium Non acid-base equilibrium calculations 17. Br2(g) + 3F2(g) ඬ 2BrF3(g) Kp = 5.4 x 108

Assume 0.30 atm Br2 and 0.60 atm of F2 are placed in a container. They react according to the

equation above. What is the pressure of Br2(g) at equilibrium? A. 0.10 atm B. 2.6 x 10-4 atm C. 0.40 atm D. 4.8 x 10-3 atm E. 0.028 atm Br2(g) + 3F2(g) ඬ 2BrF3(g)

Initial 0.30 0.60 0 Change -X -3X +2X Equilibrium 0.30-X 0.60-3X +2X [BrF3]2/([Br2] x [F2]3) = 5.4 x 108 [2X]2/([0.30-X] x [0.60-3X]3) = 5.4 x 108 Use a computer or a good calculator to solve for X: X = 0.199523 0.30-X = PBr2 = 0.100477 atm

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0.60-3X = PF2 = 0.00143 atm 2X = PBrF3 = 0.399046 atm Proof: [BrF3]2/([Br2] x [F2]3) = 5.4 x 108 [0.399046]2/([0.100477] x [0.00143]3) = 5.42 x 108 But since we don’t have a computer or a good calculator, we have to resort to mathematical trickery. [BrF3]2/([Br2] x [F2]3) = 5.4 x 108 [2X]2/([0.30-X] x [0.60-3X]3) = 5.4 x 108 Quadratic equation, so use large K rule. F2 is the limiting reactant. Br2(g) + 3F2(g) ඬ 2BrF3(g)

Initial 0.30 0.60 0 Change -0.20 -0.60 +0.40 Equilibrium 0.10 0 +0.40 Pbr2 at equilibrium = 0.10 atm New problem: Let’s say that the problem asked for the pressure of F2 at equilibrium, not the pressure of Br2 at equilibrium. How would we do that? Here is where we would use the large K rule followed by the small K rule. We already used the large K rule to get to the ICE table above. Now let’s use the small K rule to find the pressure of F2 at equilibrium. We went all the way to completion (to the right) with the large K rule. But we were forced to go a bit too far, because F2 cannot have a “0” value. It must have some finite value, even if the value is extremely small. So let’s go back, a little to the left, to get an equilibrium value for F2. That is, let’s now call the BrF3 the reactant, and convert some of this reactant into products. Br2(g) + 3F2(g) ඬ 2BrF3(g)

Initial 0.10 0 +0.40 Change +X +3X -2X Equilibrium 0.10 + X +3X 0.40 – 2X [BrF3]2/([Br2] x [F2]3) = 5.4 x 108 [0.40-2X]2/([0.10+X] x [3X]3) = 5.4 x 108 This gives us another quadratic equation. How do we get around the quadratic equation. Now it’s a little confusing because now the product is the reactant and the reactants are the products. The equilibrium expression and the equilibrium constant are confusing because they are based on product divided by reactant, which is now totally reversed. So to get around this confusion, let’s reverse the ICE table, and correspondingly, reverse the equilibrium equation, so that we will have the reactant (BrF3) on the left side and the products (Br2 and F2) on the right side. Since we reversed the equilibrium equation, we must now correspondingly invert the equilibrium expression and the equilibrium constant. 2BrF3(g) ඬ Br2(g) + 3F2(g)

Initial +0.40 0.10 0 Change -2X +X +3X Equilibrium 0.40 – 2X 0.10 + X +3X

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([Br2] x [F2]3)/[BrF3]2= 1/(5.4 x 108) = K = 1.85 x 10-9 ([0.10+X] x [3X]3)/[0.40-2X]2 = 1.85 x 10-9 I don’t know what we call this equation, but it is more complex than a quadratic equation. Let’s use the small K rule to get around this complex equation. Drop the “2X” from 0.40-2X, and the “X” from 0.10+X, as 2X and X must be extremely small. ([0.10] x [3X]3)/[0.40]2 = 1.85 x 10-9 X = 4.79 x 10-4 3X = 1.44 x 10-3atm = F2 pressure Chem 162-2011 Hourly Exam II + Answers Chapter 14 - Non Acid/Base Chemical Equilibrium Non acid-base equilibrium calculations 11. Solid ammonium chloride, NH4Cl, decomposes when heated according to the equation: NH4Cl(s) ඬ NH3(g) + HCl(g) A sample of solid ammonium chloride is heated to 400 K in an evacuated container. When the

system comes to equilibrium, some NH4Cl(s) remains and the total pressure in the container is found to be 1.5 x 10-4 atm. Calculate Kp for this reaction at 400 K.

A. 5.6 x 10-9 B. 2.3 x 10-8 C. 3.5 x 10-6 D. 1.5 x 10-4 E. 7.8 x 10-5 NH4Cl(s) ඬ NH3(g) + HCl(g)

Initial Y 0 0 Change -X +X +X Equilibrium Y-X X X X + X = 1.5 x 10-4 atm X = 0.75 x 10-4 atm Pnh3 x Phcl = Kp 0.75 x 10-4 x 0.75 x 10-4 = 0.56 x 10-8 = 5.6 x 10-9 CHEM162-2010 HOURLY EXAM II + ANSWERS CHAPTER 14 - NON ACID-BASE CHEMICAL EQUILIBRIUM NON-ACID-BASE EQUILIBRIUM CALCULATIONS 35. Solid ammonium chloride decomposes when heated according to the following reaction: NH4Cl(s) mo NH3(g) + HCl(g) A sample of ammonium chloride is heated in an evacuated container. At equilibrium, the total pressure in the container is 1.5x 10-4 atm. Calculate Kp for this reaction. A. 7.8x10-5 B. 2.3x10-8

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C. 3.5x10-6 D. 1.5x10-4 E. 5.6x10-9 NH4Cl(s) mo NH3(g) + HCl(g) NH4Cl(s) mo NH3(g) HCl(g)

Initial Y 0 0 Change -X +X +X Equilibrium Y-X +X +X [NH3][HCl] = Kp [X][X] = Kp But what is the value of X? X + X = 0.00015 2X = 0.00015 X = 0.000075 Kp = X2 = [0.000075]2 = 5.6 x 10-9

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CHEM-2010 FINAL EXAM + ANSWERS CHAPTER 14 - NON ACID AND BASE CHEMICAL EQUILIBRIUM NON ACID-BASE EQUILIBRIUM CALCULATIONS

21. 2BrCl(g) mo Br2(g) + Cl2(g) Kc = 32 at 500K What would happen if a 1.0L bulb were filled with 0.10mol BrCl(g), 0.50mol Br2(g), and 0.50mol Cl2(g)? A. Some BrCl will react spontaneously to form additional Br2 and Cl2. B. Some of the Br2 and Cl2 will react spontaneously to form additional BrCl. C. Nothing will happen because the initial concentrations correspond to equilibrium concentrations. D. All of the BrCl will be consumed. E. All of the Br2 will be consumed. 2 BrCl(g) mo Br2(g) + Cl2(g) 0.10 0.50 0.50 Kc = ([Br2][Cl2])/([BrCl]2) ([0.50][0.50])/([0.10]2) = 25 = Q When Q is less than Kc, then the numerator is too small, so the reaction will go to the right to make the numerator larger and to reach equilibrium. A. True B. False. The reaction will shift to the right, not to the left. C. False. The initial concentrations do not correspond to equilibrium concentrations. D. False. The reaction will shift to the right, but no component is ever totally consumed. E. False. The concentration of Br2 will increase, not decrease. CHEM 162-2010 FINAL EXAM CHAPTER 14 - NON ACID/BASE CHEMICAL EQUILIBRIUM NON ACID-BASE EQUILIBRIUM CALCULATIONS

46. Given: C(s) + 2H2(g) mo CH4(g) Kp = 0.262 at 1270K What is Kc for the above reaction? A. 20.5 B. 19.6 C. 12.9 D. 27.3 E. 16.3 Kp = Kc(RT)¨ngas 0.262 = Kc(0.08214 x 1270)-1 Kc = 27.3

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CHEM 162-2010 EXAM II Chapter 14 - Non Acid/Base Chemical Equilibrium Non acid-base equilibrium calculations 23. 2N2O5(g) mo 4NO2(g) + O2(g) 1.00 atm of N2O5 is placed in a container, and reacts to reach equilibrium in the above reaction. The total pressure at equilibrium is found to be 2.35 atm. Assume constant volume and temperature. Calculate Kp for this reaction. A. 66.0 B. 131 C. 262 D. 393 E. 472 2N2O5(g) mo 4NO2(g) + O2(g) 2N2O5(g) mo 4NO2(g) + O2(g)

Initial 1.00 0 0 Change Equilibrium 2N2O5(g) mo 4NO2(g) + O2(g)

Initial 1.00 0 0 Change -2X +4X +X Equilibrium 1.00 – 2X +4X +X Kp = ([NO2]4[O2])/[N2O5]2 ([4X]4[X])/[1.00-2X]2 = Kp Two unknowns in one equation. However, (1.00-2X) + 4X + X = 2.35 atm X = 0.45 atm 2N2O5(g) mo 4NO2(g) + O2(g)

Initial Change Equilibrium 0.10 atm 1.80 atm +0.45 atm Kp = ([NO2]4[O2])/[N2O5]2 ([1.80]4[0.45])/([0.10]2) = 472 = Kp CHEM 162-2010 EXAM II Chapter 14 - Non Acid/Base Chemical Equilibrium Non acid-base equilibrium calculations 8. 2NOCl(g) mo 2NO(g) + Cl2(g) Kp = 7.14 x 10-6

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NOCl(g) is placed in a container, exerting a pressure of 1.5 atm before any reaction. Calculate the equilibrium pressure of Cl2. A. 0.0085 atm B. 0.016 atm C. 0.024 atm D. 0.032 atm E. 0.048 atm 2NOCl(g) mo 2NO(g) + Cl2(g)

Initial 1.5 0 0 Change -2X +2X +X Equilibrium 1.5-2X +2X +X Kp = ([NO]2[Cl2])/([NOCl]2) ([2X]2[X])/([1.5-2X]2) = 7.14x10-6 This is a quadratic equation. Use the small K rule to get around the quadratic equation. ([2X]2[X])/([1.5]2) = 7.14x10-6 X = 1.59 x 10-2 atm CHEM 162-2010 EXAM II Chapter 14 - Non Acid/Base Chemical Equilibrium Non acid-base equilibrium calculations 7. A container initially contains 0.500 atm of NH3(g), which decomposes into N2(g) and H2(g). 2NH3(g) mo N2(g) + 3H2(g) If 10.0% of the NH3 reacts to reach equilibrium calculate Kp for this reaction. A. 5.2 x 10-5 B. 2.3 x 10-4 C. 8.3 x 10-4 D. 3.8 x 10-3 E. 7.6 x 10-3 2NH3(g) mo N2(g) + 3H2(g)

Initial 0.500 0 0 Change -0.050 +0.025 +0.075 Equilibrium 0.450 +0.025 +0.075 Kp = ([N2][H2]3)/([NH3]2) ([0.025][0.075]3)/([0.450]2) = 5.21 x 10-5

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CHEM 162-2009 FINAL EXAM CHAPTER 14 - NON ACID/BASE CHEMICAL EQUILIBRIUM NON ACID-BASE EQUILIBRIUM CALCULATIONS 10. Consider the reaction C(s, graphite) + CO2(g) mo 2CO(g) If the initial pressure of CO2(g) is 0.458 atm and the total pressure at equilibrium is 0.757 atm, what is Kp of the reaction? A. 1.26 B. 3.42 C. 4.65 D. 2.25 E. 5.98 C(s) + CO2(g) mo 2CO(g)

Initial 0.458 atm 0 Change -X +2X Equilibrium 0.458 – X +2X 0.458 – X + 2X = 0.757 X = 0.299 atm C(s) + CO2(g) mo 2CO(g)

Initial Change Equilibrium 0.159 atm 0.598 atm [CO]2/[CO2] = Kp ([0.598]2)/0.159 = 2.25 = Kp D CHEM 162-2009 FINAL EXAM CHAPTER 14 - NON ACID/BASE CHEMICAL EQUILIBRIUM NON ACID-BASE EQUILIBRIUM CALCULATIONS

49. 2CO2(g) mo 2CO(g) + O2(g) Kc = 2.0 x 10-7 If 2.0 mol of CO2 is placed in a 5.0 L container, and decomposes according to the above equation, calculate the concentration of O2 at equilibrium. A. 5.8 x 10-3M B. 2.7 x 10-3M C. 4.6 x 10-3M D. 2.0 x 10-3M E. 9.2 x 10-3M 2CO2(g) mo 2CO(g) + O2(g)

Initial 0.40M 0 0 Change -2X +2X +X Equilibrium 0.40-2X +2X +X

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([CO]2[O2])/[CO2]2 = 2.0 x 10-7

Use small K rule to avoid a quadratic equation. ([2X]2[X])/[0.40]2 = 2.0 x 10-7 X = 0.002M D CHEM 162-2009 HOURLY EXAM II + ANSWERS CHAPTER 14: NON ACID/BASE CHEMICAL EQUILIBRIUM NON ACID-BASE EQUILIBRIUM CALCULATIONS 1. 2SO3 mo 2SO2 + O2 Kp = 1.21 x 10-5 If 0.60 atm of SO3 and 0.30 atm of SO2 are placed in a container, what is the partial pressure of O2 at equilibrium? A. 4.8 x 10-5 atm B. 0.042 atm C. 8.4 x 10-3 atm D. 1.8 x 10-4 atm E. 7.2 x 10-4 atm 2SO3 mo 2SO2 + O2

Initial 0.60 0.30 0 Change -2X +2X +X Equilibrium 0.60 - 2X 0.30 + 2X +X ([SO2]2[O2])/[SO3]2 = Kp ([0.30+2X]2[X])/[0.60-2X]2 = Kp Use small K rule to avoid a quadratic equation: 2SO3 mo 2SO2 + O2

Initial 0.60 0.30 0 Change -2X +X +X Equilibrium 0.60 0.30 +X ([SO2]2[O2])/[SO3]2 = Kp ([0.30]2[X])/[0.60]2 = 1.21 x 10-5 X = 4.84 x 10-5 CHEM 162-2009 HOURLY EXAM II + ANSWERS CHAPTER 14: NON ACID/BASE CHEMICAL EQUILIBRIUM NON ACID-BASE EQUILIBRIUM CALCULATIONS 6. Consider the reaction: NH4CO2NH2(s) mo 2NH3(g) + CO2(g)

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When solid NH4CO2NH2 is placed in a previously evacuated container and allowed to reach equilibrium according to the reaction above, the total pressure in the container is measured to be 0.447 atm. Calculate Kp for this reaction. A. 3.31 x 10-3 B. 0.0444 C. 5.12 x 10-4 D. 0.0132 E. 0.762 NH4CO2NH2(s) mo 2NH3(g) + CO2(g)

Initial Y 0 0 Change Equilibrium NH4CO2NH2(s) mo 2NH3(g) + CO2(g)

Initial Y 0 0 Change -X +2X +X Equilibrium Y-X +2X +X 2X + X = 0.447 atm X = 0.149 atm NH4CO2NH2(s) mo 2NH3(g) + CO2(g)

Initial Y 0 0 Change -X +2X +X Equilibrium Y-X 0.298 +0.149 [NH3]2[CO2] = Kp [0.298]2[0.149] = 1.32 x 10-2 CHEM 162-2009 HOURLY EXAM II + ANSWERS CHAPTER 14: NON ACID/BASE CHEMICAL EQUILIBRIUM NON ACID-BASE EQUILIBRIUM CALCULATIONS 24. 2N2O5(g) mo 4NO2(g) + O2(g) 1.00 atm of N2O5 is placed in a container, and reacts to reach equilibrium in the above reaction. The partial pressure of NO2 at equilibrium is found to be 1.60 atm. Calculate Kp for this reaction. A. 4.1 B. 26 C. 13 D. 84

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E. 66 2N2O5 mo 4NO2 + O2

Initial 1.00 0 0 Change Equilibrium 1.60 2N2O5 mo 4NO2 + O2

Initial 1.00 0 0 Change -2X +4X +X Equilibrium 1.00 – 2X 1.60 +X 0 + 4X = 1.60 X = 0.40 2N2O5 mo 4NO2 + O2

Initial 1.00 0 0 Change -0.80 +1.60 +0.40 Equilibrium 0.20 1.60 +0.40 ([NO2]4[O2])/[N2O5]2 = Kp ([1.60]4[0.40])/[0.20]2 = 65.5

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36 Chem 162-2008 Final Exam + Answers Chapter 14 – Chemical Equilibrium Non acid-base equilibrium calculations At 830 °C, Kp = 0.50 for the reaction CaCO3(s) ҡ CaO(s) + CO2(g) A 1.00L flask containing some CaCO3(s) is evacuated and heated to 830 °C. What mass of CaO (molar mass = 56.08 g/mol) is also produced when equilibrium is established?

A. 0.310g B. 0.500g C. 0.00552g D. 5.3g E. 0.411g CaCO3(s) ҡ CaO(s) + CO2(g) Initial Y 0 0 Change -X +X +X Equilibrium Y-X +X +X

[CO2] = Kp = 0.50 [CO2 = X = 0.50atm PV = nRT 0.50atm x 1.00L = n x 0.08205Latmdeg-1mol-1 x (830 + 273K) n = 5.525 x 10-3 mol CO2 Therefore, moles of CaO formed is also 5.525 x 10-3 mol. 5.525 x 10-3mol x 56.08g/mol = 0.310 g CaO.

A

24 Chem 162-2008 Final Exam + Answers Chapter 14 – Chemical Equilibrium Non acid-base equilibrium calculations The reaction 2NO(g) + Br2(g) ҡ 2NOBr(g) has Kc = 1.3×10-2 at 1000K. What is Kp at this temperature?

A. 88 B. 108 C. 1.6×10-6 D. 1.1 E. 1.6×10-4

Kp = Kc(RT)¨ngas Kp = 0.013 x ((0.08206 x 1000)-1) Kp = 1.58 x 10-4

E

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Chem 162-2008 Exam II + Answers Chapter 14 Non Acid/Base Chemical Equilibrium Non acid-base equilibrium calculations Given initial and total pressure, find Kp 16. H2 at an initial pressure of 10.0 atm at 3000 K dissociates into H atoms. When equilibrium is

established, the final total pressure is 14.0 atm. Calculate Kp for the dissociation reaction: H2(g) ඬ 2H(g)

(a) 1.33 (b) 10.7 (c) 2.75 (d) 5.50 (e) 3.80 H2(g) ඬ 2H(g)

Initial 10.0 0 Change -X +2X Equilibrium 10.0 - X +2X 10.0 – X + 2X = 14.0 atm X = 4.0 atm H2(g) ඬ 2H(g)

Initial Change Equilibrium 6.0 atm +8.0 atm [H]2/[H2] = K [8]2/[6] = K K = 10.7 Chem 162-2008 Exam II + Answers Chapter 14 Non Acid/Base Chemical Equilibrium Non acid-base equilibrium calculations Given initial value and solubility, find K 8. 2NH3(g) ඬ N2(g) + 3H2(g) When 1.0 mol of NH3(g) is placed in a 5.0 L container, 10.0% of the NH3(g) reacts to reach

equilibrium. Calculate K for this reaction. (a) 1.4 x 10-3 (b) 3.5 x 10-5

(c) 7.0 x 10-4

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(d) 8.3 x 10-6 (e) 1.9 x 10-6 2NH3(g) ඬ N2(g) + 3H2(g)

Initial 0.2M 0 0 Change -0.02 +0.01 +0.03 Equilibrium 0.18 0.01 0.03 K = ([N2][H2]3)/[NH3]2 = ([0.01][0.03]3)/([0.18]2) = 8.3 x 10-6

Chem 162-2008 Exam II + Answers Chapter 14 Non Acid/Base Chemical Equilibrium Non acid-base equilibrium calculations Given initial value and K, find equilibrium values 4. 2NOCl(g) ඬ 2NO(g) + Cl2(g) Kp = 7.2 x 10-6 NOCl(g) is added to a container, exerting a pressure of 0.80 atm before any reaction. After equilibrium is

established, what is the pressure of Cl2 in the container? (a) 0.010 atm (b) 0.040 atm (c) 0.060 atm (d) 0.025 atm (e) 0.075 atm 2NOCl(g) ඬ 2NO(g) + Cl2(g)

Initial 0.80 atm 0 0 Change -2X +2X +X Equilibrium 0.80 – 2X +2X +X [NO]2[Cl2]/[NOCl]2 = 7.2 x 10-6 [2X]2[X]/[0.80-2X]2 = 7.2 x 10-6 Avoid a quadratic equation by using the small K rule. [2X]2[X]/[0.80]2 = 7.2 x 10-6 X = 0.0105 atm

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20 Chem 162-2007 Final exam + answers Chapter 14 – Non Acid-Base Chemical Equilibrium Non Acid-Base equilibrium calculations Common ion effect Calculate the molar solubility of Ag2SO4 in 1.00M Na2SO4(aq). Ksp of Ag2SO4 is 1.4×10-5.

A. 1.9×10-3M B. 2.3×10-3M C. 3.5×10-3M D. 4.7×10-4M E. 6.2×10-3M

Ag2SO4 mo 2Ag+ + SO4

2- Ag2SO4(s) mo 2Ag+(aq) + SO4

2-(aq)

Initial Y 0 1.00 Change -X +2X +X Equilibrium Y-X +2X 1.00 + X

[Ag+]2 x [SO42-] = Ksp

[2X]2 x [1.00 + X] = (1.4 x 10-5) Simplify due to small K rule: [2X]2 x [1.00] = (1.4 x 10-5) X = 1.88 x 10-3 M = Ag2SO4 solubility

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41 Chem 162-2007 Final exam + answers Chapter 14 – Non Acid-Base Chemical Equilibrium Non Acid-Base equilibrium calculations Consider the following equilibrium: PCl5 (g) ҡ PCl3 (g) + Cl2 (g) A flask is filled with PCl5 to a pressure of 2.00 atm at 300°C and allowed to come to equilibrium. Analysis shows the total pressure in the flask at equilibrium is 3.96 atm. Determine the value of Kp at this temperature.

A. 3.96 B. 49.0 C. 96.0 D. 0.886 E. 7.84

PCl5 (g) mo PCl3 (g) + Cl2 (g)

Initial 2.00 0 0 Change -X +X +X Equilibrium 2.00 - X +X +X

At equilibrium, PPCl5 + PCl3 + PCl2 = 3.96 atm (2.00 – X) + X + X = 3.96 atm X = 1.96 atm K = (PPCl3 x PCl2)/PPCl5 K = (X x X)/(2.00 – X) K = (1.96 x 1.96)/(2.00 – 1.96) K = 96.04

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49 Chem 162-2007 Final exam + answers Chapter 14 – Non Acid-Base Chemical Equilibrium Non acid-base equilibrium calculations The pH in a saturated solution of Mg(OH)2 is 10.56. The Ksp of Mg(OH)2 is

A. 1.3 u 10-7 B. 4.8 u 10-11 C. 1.2 u 10-11 D. 3.6 u 10-4 E. 2.4×10-11

Mg(OH)2(solid)mo Mg2+(aqueous) + 2OH-(aqueous) pH = 10.56 pOH = 14.00 – 10.56 = 3.44 [OH-] = 10-pOH = 10-3.44 = 3.63 x 10-4 [Mg2+] = ½ [OH-] = 1.82 x 10-4 Mg(OH)2 mo Mg2+ + 2OH-

Initial Change Equilibrium 1.82 x 10-4 3.63 x 10-4

[Mg2+][OH-]2 = Ksp [1.82 x 10-4] x ([3.63 x 10-4]2) = Ksp = 2.40 x 10-11

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50 Chem 162-2007 Final exam + answers Chapter 14 – Non Acid-base Chemical Equilibria Non acid-base equilibrium calculations Selective precipitation The concentration of both Ag+(aq) and Pb2+(aq) is 0.10M in a particular solution. Solid NaCl is added slowly. What is the concentration of Ag+ that remains in solution when PbCl2 just begins to precipitate?

Ksp of AgCl = 1.6×10-10; Ksp of PbCl2 =2.4×10-4

A. 3.3×10-9 M B. 4.9 × 10-3 M C. 6.7×10-9 M D. 2.0×10-10 M E. 7.8 × 10-3 M

AgCl mo Ag+ + Cl- AgCl mo Ag+ + Cl-

Initial Change Equilibrium 0.10 X

[Ag+][Cl-] = Ksp [0.10][X] = 1.6 x 10-10 X = 1.6 x 10-9 M Cl- to begin precipitation of AgCl PbCl2 mo Pb2+ + 2Cl-

PbCl2 mo Pb2+ + 2Cl-

Initial Change Equilibrium 0.10 X

[Pb2+][Cl-]2 = Ksp [0.10][X]2 = 2.4 x 10-4 X = 4.899 x 10-2M Cl- to begin precipitation of PbCl2. When PbCl2 just begins to precipitate, the [Cl-] = 4.889 x 10-2 M At 4.889 x 10-2M, the Ag+ remaining in the solution: [Ag+][ 4.889 x 10-2] = 1.6 x 10-10 [Ag+] = 3.27 x 10-9 M

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CHEM 162-2007 EXAM II + ANSWERS CHAPTER 14 – NON-ACID/BASE CHEMICAL EQUILIBRIUM NON-ACID BASE EQUILIBRIUM CALCULATIONS 6. 2SO3(g) ඬ�2SO2(g) + O2(g) Kp = 4.00 at 1100 K Some pure SO3(g) is placed in a container and reacts to reach equilibrium. At equilibrium, the pressure

of O2 = 0.10 atm. Calculate the pressure of SO3 remaining at equilibrium. (a) 0.055 atm (b) 0.015 atm (c) 0.080 atm (d) 0.032 atm (e) 0.0050 atm 2SO3(g) ඬ 2SO2(g) + O2(g)

Initial Y Change Equilibrium +0.10 2SO3(g) ඬ 2SO2(g) + O2(g)

Initial Y 0 0 Change -2X +2X +X Equilibrium Y-2X +2X +0.10 0 + X = 0.10 X = 0.10 atm 2SO3(g) ඬ 2SO2(g) + O2(g)

Initial Change Equilibrium Y-0.20 0.20 +0.10 [SO2]2[O2]/[SO3]2 = Kp [0.20]2[0.10]/[Y-0.20]2 = 4.00 ((((([0.20]2) x [0.10])/([Y-0.20]2)) = (4.00)),X) Y = 0.2316 0.2316 – 0.20 = 0.0316 atm SO3 remaining at equilibrium Proof: (([0.20]2) x [0.10])/([0.0316]2) = 4.006

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CHEM 162-2007 EXAM II + ANSWERS CHAPTER 14 – NON-ACID/BASE CHEMICAL EQUILIBRIUM NON-ACID BASE EQUILIBRIUM CALCULATIONS 9. Solid NH4CO2NH2 is placed in a container and decomposes into NH3 and CO2 to reach equilibrium. The

reaction is: NH4CO2NH2(s) ඬ 2NH3(g) + CO2(g) Kp = 0.45 At equilibrium , what is the partial pressure of CO2 in the container? (a) 0.67 atm (b) 0.74 atm (c) 0.32 atm (d) 0.48 atm (e) 0.15 atm NH4CO2NH2(s) ඬ 2NH3(g) + CO2(g)

Initial Y 0 0 Change -X +2X +X Equilibrium Y-X +2X +X [NH3]2[CO2] = Kp [2X]2[X] = 0.45 (((([2X]2) x [X]) = (0.45)),X) X = 0.483 atm CHEM 162-2007 EXAM II + ANSWERS CHAPTER 14 – NON-ACID/BASE CHEMICAL EQUILIBRIUM NON-ACID BASE EQUILIBRIUM CALCULATIONS 12. 2NO(g) + Cl2(g) ඬ 2NOCl(g) Kp = 1.6 x 108 If 3.0 atm NO(g) and 2.0 atm Cl2(g) are placed in a container and allowed to react to reach equilibrium,

what is the pressure of the Cl2(g) at equilibrium? (a) 3.0 atm (b) 2.0 atm (c) 1.0 atm (d) 0.5 atm (e) 0.1 atm 2NO(g) + Cl2(g) ඬ 2NOCl(g)

Initial 3.0 2.0 0 Change -2X -X +2X Equilibrium 3.0 - 2X 2.0 - X 0 + 2X

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[NOCl]2/[NO]2[Cl]2 = Kp [2X]2/[3.0-2X]2[2.0-X]2 = 1.6 x 108 This calculation will result in an equation more complicated than a quadratic equation (i.e., an equation with

four roots). One way of solving this is with an advanced calculator: (((([2X]2)/([3.0-2X]2) x ([2.0-X]2)) = (1.6 x 108)),X) X = 1.500 Therefore, [Cl2] = 2.0 – 1.5 = 0.5 atm Another way to solve this is to take advantage of the large K, i.e., bring the reaction to completion. NO is the limiting reactant. 2NO(g) + Cl2(g) ඬ 2NOCl(g)

Initial 3.0 2.0 0 Change -3.0 -1.5 +3.0 Equilibrium 0 0.5 3.0 [Cl2] = 0.5 atm CHEM 162-2007 EXAM II + ANSWERS CHAPTER 14 – NON-ACID/BASE CHEMICAL EQUILIBRIUM NON-ACID BASE EQUILIBRIUM CALCULATIONS 17. Suppose 2.00 mol of H2(g) and 1.00 mol of I2(g) are placed in a 1.00 L container, and they react to

form HI(g). At equilibrium, it is found that 1.80 moles of HI(g) are present in the container. Calculate K for the reaction:

H2(g) + I2(g) ඬ 2HI(g) (a) 57.5 (b) 29.5 (c) 82.2 (d) 40.1 (e) 17.8 H2(g) + I2(g) ඬ 2HI(g)

Initial 2.00 1.00 Change Equilibrium 1.80 H2(g) + I2(g) ඬ 2HI(g)

Initial 2.00 1.00 0 Change -X -X +2X Equilibrium 2.00-X 1.00-X 1.80

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0+2X = 1.80 X = 0.90M H2(g) + I2(g) ඬ 2HI(g)

Initial Change Equilibrium 1.10 0.10 1.80 [HI]2/[H2][I2] = Kc [1.80]2/[1.10][0.10] = Kc ([1.80]2)/([1.10][0.10]) = 29.45

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LE CHATELIER’S PRINCIPLE Chem 162-2011 Hourly Exam II + Answers Chapter 14 - Non Acid/Base Chemical Equilibrium Le Chatelier’s Principle 8. N2(g) + 3H2(g) ඬ 2NH3(g) 'H = -92 kJ Assume this system is at equilibrium in a 1.0 L container at 500 K. Which of the following changes

will result in a new equilibrium system with a higher value of the equilibrium constant K? X. Compress the system to 0.50 L at constant temperature. Y. Cool the system to 400 K. Z. Add a catalyst. A. Y only

B. X only C. X and Y only D. Y and Z only E. X, Y, and Z This is an exothermic reaction. N2(g) + 3H2(g) ඬ 2NH3(g) + heat X. False. Compression increases the concentration of all of the substances. According to LeChatelier, the

stress of the increased concentration will be relieved if the equilibrium is shifted to the right (fewer molecules). However, this does nothing to change the equilibrium constant. The only thing that will change the equilibrium constant is a change of temperature.

Y. True. If the system is cooled to 400 K, then heat will be removed from the system. In an exothermic reaction, if heat is removed, then the equilibrium will shift to the right to provide more heat. This provides a higher value of the equilibrium constant.

Z. False. Addition of a catalyst has no effect on changing the equilibrium reaction or the equilibrium constant.

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CHEM 162-2010 EXAM II Chapter 14 - Non Acid/Base Chemical Equilibrium Le Chatelier’s Principle 4. For a particular gas phase reaction, the value of Kp will decrease as temperature increases: A. if ¨H < 0 B. if ¨H > 0 C. if more moles of gas are on product side of reaction D. if more moles of gas are on reactant side of reaction. E. under no circumstances. According to LeChatelier, the only way that the value of Kp will decrease as temperature increases would be if the reaction is exothermic. A + B mo C + D + ¨ Addition of heat will drive the reaction to the left, thereby decreasing the equilibrium constant. An exothermic reaction means that ¨H < 0, i.e., that ¨H is negative. CHEM 162-2010 EXAM II Chapter 14 - Non Acid/Base Chemical Equilibrium Le Chatelier’s Principle 10. 2NO(g) + O2(g) mo 2NO2(g) ¨H = -114 kJ Assume that this system has reached equilibrium in a 1.00 L container at 298 K. Which of the following changes will result in a greater amount of NO2(g) at equilibrium? X. Raise temperature to 398 K Y. Lower volume to 0.50 L Z. Add a catalyst A. Y only B. X only C. X and Y only D. Y and Z only E. X, Y, and Z 2NO(g) + O2(g) mo 2NO2(g) + ¨ X. Raising the temperature will force the reaction to the left, according to LeChatelier. This decreases the amount of NO2. Y. Lowering the volume is the same as raising the pressure. According to LeChatelier, the system resists the increase of pressure. If the reaction moves to the right, then three moles of gas will become two moles of gas, resulting in a reduced pressure. Moving to the right results in an increase in NO2 concentration. Z. Adding a catalyst only speeds up the reaction. It has no effect on the concentration of reactants or products. CHEM 162-2009 FINAL EXAM

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CHAPTER 14 - NON ACID/BASE CHEMICAL EQUILIBRIUM LE CHATELIER’S PRINCIPLE 28. A 1.0 liter container holds PCl3, Cl2 and PCl5 in equilibrium at 298 K PCl3(g) + Cl2(g) mo PCl5(g) ¨H = -87.9 kJ Kp = 3.3 x 106 Which of the following changes will increase the numerical value of Kp? W. Compress the entire system to 0.50 liter. X. Expand the entire system to 2.0 liter. Y. Raise the temperature to 323 K. Z. Lower the temperature to 273 K. A. W and Y B. Z only C. Y and Z D. W and Z E. Y only The only thing that will change the equilibrium constant is a change in temperature. In this case, the reaction is exothermic, so we can write: PCl3(g) + Cl2(g) mo PCl5(g) + ¨ If we remove the heat, the reaction will shift to the right, thereby increasing the numerical value of the equilibrium constant. Lower the temperature to 273K B CHEM 162-2009 HOURLY EXAM II + ANSWERS CHAPTER 14: NON ACID/BASE CHEMICAL EQUILIBRIUM LE CHATELIER’S PRINCIPLE 14. N2O4(g) mo 2NO2(g) N2O4(g) and NO2(g) reach equilibrium in a 1.00 L container. The volume of the container is halved at constant temperature. When equilibrium is reestablished: A. The partial pressures of both N2O4 and NO2 double. B. The partial pressure of NO2 more than doubles, and the partial pressure of N2O4 less than doubles. C. The partial pressure of N2O4 more than doubles and the partial pressure of NO2 less than doubles. D. The value of K increases. E. The value of K decreases. This problem might have been better written if it would have said, “Relative to the initial equilibrium, when equilibrium is reestablished:” If the volume of the container is halved, then the pressure is doubled. This means that for an instant in time the NO2 and the N2O4 pressures are doubled. LeChatelier tries to bring the pressure back toward

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where the pressure was before it was doubled. Since conversion of two NO2 molecules to one N2O4 molecule decreases the number of molecules and therefore decreases the pressure, then the equilibrium equation shifts to the left. This results in an increase in the N2O4 pressure, to be slightly more than double, and a decrease in the NO2 pressure to be significantly less than double, resulting in the total pressure being less than double the initial. A. False. This means that there is no shift in the reaction from right to left, which would result in decreasing the added pressure. B. False. This means that there is a shift in the reaction from left to right, which would result in increased pressure, not decreased pressure. C. True. This shift from right to left decreases the total pressure, so that it is less than double the original pressure. D. False. The shift in the equilibrium equation results in a reestablishment of the original K. E. False. The shift in the equilibrium equation results in a reestablishment of the original K.

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14 Chem 162-2008 Final Exam + Answers Chapter 14 – Chemical Equilibrium Le Chatelier’s Principle For the reaction NH3(g) + HCl(g) ҡ NH4Cl(s) 'Ho = �177 kJ What conditions favor the production of ammonium chloride (NH4Cl)? A. High temperature and low pressure B. Low temperature and high pressure C. Low temperature and low pressure D. High temperature and high pressure E. Addition of an inert gas (constant volume) ¨Ho of -177 kJ means that it is an exothermic reaction. NH3(g) + HCl(g) ҡ NH4Cl(s) + ' The production of NH4Cl would be favored if the equilibrium reaction would shifto the right. According to LeChatelier, removing heat (i.e., lowering the temperature) will force the reaction to the right. Also, since there are more moles of gas on the left than on the right, increasing the pressure will result in forcing the reaction to the right (thereby lowering the pressure since a solid is formed from the gases). Hence, option “B”, lowering the temperature and increasing the pressure will force the reaction to form more NH4Cl. Addition of an inert gas will have no effect on the equilibrium reaction because the inert gas is not part of the equilibrium expression.

B

Chem 162-2008 Exam II + Answers Chapter 14 Non Acid/Base Chemical Equilibrium Le Chatelier’s Principle Le Chatelier; effect of temperature in shifting equilibrium 12. CuO(s) + H2(g) ඬ Cu(s) + H2O(g) 'H = -2.0 kJ When the substances in the equation above are at equilibrium at pressure P and temperature T, the

equilibrium can be shifted to favor the products by: (a) increasing the pressure by lowering the volume of the system. (b) decreasing the temperature. (c) increasing the pressure by adding an inert gas such as nitrogen.

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(d) allowing some gases to escape at constant P, T. (e) adding a catalyst. ¨H is negative; therefore it is an exothermic reaction, so put heat on the right hand side of the equation. CuO(s) + H2(g) ඬ Cu(s) + H2O(g) + '�(a) False. Since there is one mole of gas on the left side, and one mole of gas on the right side, increasing

the pressure will have no effect. (b) True. Decreasing the temperature in an exothermic reaction will force the equilibrium to the right. (c) False. Even if changing the pressure would shift the equilibrium (which it won’t in this case), adding

an inert gas would have no effect on the equilibrium because the inert gas is not in the equilibrium expression.

(d) False. If we assume that both gases have an equal diffusion rate, then allowing both gases to somewhat escape, at constant pressure, would keep the final pressure of each gas the same, and therefore wouldn’t put a stress on the system. Hence, the equilibrium wouldn’t shift. On the other hand, if we assume that H2 has a greater rate of diffusion than H2O, which is true, then more H2 would escape than H2O, and the equilibrium would shift to the left, not to the right.

(e) False. Adding a catalyst speeds up the forward and reverse reactions, but has no effect on shifting the equilibrium.

30 Chem 162-2007 Final exam + answers Chapter 14 – Non Acid-Base Chemical Equilibrium Le Chatelier’s Principle Which condition will increase of the value of the equilibrium constant?

A. Raising the temperature at which an exothermic reaction is conducted. B. Adding a catalyst to an endothermic reaction. C. Raising the temperature at which an endothermic reaction is conducted. D. Removing the products from the reaction mixture. E. Reducing the size of the container in a gas phase reaction.

A. False. Raising the temperature of an exothermic reaction will drive the reaction to the left, decreasing the value of the equilibrium constant. B. False. Adding a catalyst to any reaction doesn’t affect the equilibrium constant. C. True. Raising the temperature of an endothermic reaction will drive the reaction to the right, increasing the value of the equilibrium constant. D. False. Removing the products from the reaction mixture (i.e., from the reactants and/or the products) will tend to drive the reaction one way or the other, but will not affect the equilibrium constant. E. False. Reducing the size of the container will increase the pressure if the reaction mixture contains gases, which will probably shift the reaction to the left or to the right, depending on the reaction, but will not affect the equilibrium constant.

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CHEM 162-2007 EXAM II + ANSWERS CHAPTER 14 – NON-ACID/BASE CHEMICAL EQUILIBRIUM LE CHATELIER’S PRINCIPLE 3. N2O4(g) ඬ 2NO2(g) Kp = 80 at 25 oC Assume the reaction is in a 1.0 L container at equilibrium at 25oC. The volume of the container is

halved to 0.50 L with the temperature maintained at 25 oC. Which of the following would be correct when equilibrium is reestablished?

(a) The partial pressures of N2O4 and NO2 both double and Kp > 80 (b) The partial pressures of N2O4 and NO2 both double and Kp = 80 (c) The partial pressure of N2O4 more than doubles, the partial pressure of NO2 less than doubles, and Kp < 80 (d) The partial pressure of NO2 more than doubles, the partial pressure of N2O4 less than doubles and Kp > 80 (e) The partial pressure of N2O4 more than doubles, the partial pressure of NO2 less than doubles, and Kp = 80. If the volume of the container is halved (at constant temperature and moles), then the pressure is

momentarily doubled, according to the Combination Gas Law. Doubling the pressure would momentarily make the pressure of NO2 double, and the pressure of N2O4 double, which would result in a momentary increase in the equilibrium constant. According to LeChatelier, the reaction will shift in a direction to reduce this new high pressure, and thereby bring the equilibrium constant back to its original value. Since going to the left will convert two molecules (of NO2) into one molecule (of N2O4), then going to the left will reduce the pressure and bring the equilibrium constant back to its original value. Hence, the NO2 pressure will be reduced from double its value, and concurrently the N2O4 pressure will be increased to above double the value, and the equilibrium constant would go back to its original value.

[NO2]2/[N2O4] = 80 As an example, let [N2O4] = 1 atm, and [NO2] = 8.94 atm. [8.94]2/[1] = 80 Let the pressure instantaneously double. [17.88]2/[2] = 160 Then the equilibrium reaction adjusts so that the K of 80 is restored. [15.71]2/3.085 = 80 Note that the [NO2] decreases to a little less than double, while the [N2O4] increases

to more than double. (a) False, because Kp = 80 must be re-established. (b) False, because if the partial pressures of N2O4 and NO2 both double then Kp would be > 80. (c) False, because Kp of 80 must be re-established. (d) False, because Kp of 80 must be re-established. (e) True, as can be seen by the final concentrations being: [15.71]2/3.085 = 80 CHEM 162-2007 EXAM II + ANSWERS CHAPTER 14 – NON-ACID/BASE CHEMICAL EQUILIBRIUM LE CHATELIER’S PRINCIPLE

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7 . Consider the reaction N2(g) + 3H2(g) ඬ 2NH3(g) 'H = -92.6 kJ Assume that the reaction is at equilibrium in a 1.0 L container. Which of the following changes

would result in a greater amount of NH3 in the container at equilibrium? X. Add argon gas to the container with no change of volume or temperature. Y. Add a catalyst to the container. Z. Lower the temperature. (a) Z only (b) X only (c) Y only (d) X and Z only (e) X, Y, ands Z N2(g) + 3H2(g) ඬ 2NH3(g) + ¨ X. False. Adding argon doesn’t affect the equilibrium. Although it affects the total pressure, it doesn’t

affect the pressure of any of the components in the equilibrium expression. Y. False. Adding a catalyst doesn’t affect the equilibrium. Although it speeds up the reaction, it has no

effect, whatsoever, on any of the components in the equilibrium expression. Z. True. Lowering the temperature will shift the equilibrium to the right in the direction of greater

ammonia. This can be looked at according to LeChatelier, in which case we are removing a component from the right side of the equilibrium reaction, so the equilibrium reaction will shift to the right in an attempt to regain some of that lost heat. This can also be looked at as the reaction products have higher enthalpy than the reactants. Hence, forming more product will give off heat, bringing the system back to where it was originally, from the point of view of temperature.