1.Consider the Following Reaction: CO(G) + H2O(G) CO2(G) + H2(G)

Chem 1BName:______

Chapter 13 Exercises

Exercises #1

1.Consider the following reaction: CO(g) + H2O(g)  CO2(g) + H2(g)

(a) For a given equation, there is only one value of the equilibrium constant K at a given temperature. In a system where the number of moles of products is equal to those of reactants such as the above, carrying out the reaction from either the reactants or the products using the same initial concentrations of the starting material will yield the same molar quantities of components at equilibrium. This, in turn, will yield the same value of K.

(b) Equilibrium is a dynamic state. Some CO will become CO2 and the reverse reaction will occur at the same rate at equilibrium. Therefore, Carbon-14 will appear in both CO and CO2.

2.(A) Write the expressions for the equilibrium constant Kc for the following equilibrium systems.

(i) CH4(g) + H2O(g)  CO(g) + 3 H2(g)

(ii) N2(g) + 3 H2(g)  2 NH3(g)

(iii) (NH4)2CO3(s)  CO2(g) + 2 NH3(g) + H2O(g)

(iv) NH3(g) + HCl(g)  NH4Cl(s)

(v) HNO2(aq) + H2O(l)  H3O+(aq) + NO2-(aq)

(B) For equations (i) – (iv), write the expressions for the equilibrium constant Kp.

3.(a) For the reaction N2(g) + 3H2(g)  2NH3(g), Kc = 0.0602 and Kp = 3.57 x 10-5

(b) For the reaction NH3(aq)  ½ N2(g) + 3/2 H2(g), Kc = 4.08 and Kp = 167

(c) If [N2] is increased, net forward reaction will occur until a new equilibrium position is attained, and [NH3] will increase, but [H2] will decrease because it is consumed in the forward reaction.

4.At constant temperature the values of Kc and Kp are fixed (constant), but the chemical compositions of the equilibrium mixture can be different.

5.Equilibrium constants are values obtained using concentrations (or partial pressures) of components in the mixture at equilibrium that are input into the expression for K for that reaction. Whereas the reaction quotient (Qc or Qp) for a give reaction is calculate using the same formula for calculating K, but the concentrations (or partial pressures) may or may not be the equilibrium values. Reaction quotients are calculated in order to determine whether or not a given reaction mixture is at equilibrium.

6.For Expt.#1, Kc = 0.0603; Expt.#2, Kc = 0.0602, and Expt.#3, Kc = 0.0602.

Therefore, the value of an equilibrium constant for a given equation at a given temperature does not, depend on how the reaction is carried out.

7.Answers: [N2] = 0.0944 M; [H2] = 0.2832 M; [NH3] = 0.0112 M; Kc = 0.0585; Kp = 3.47 x 10–5

Exercises #2

1.(a) N2O(g) + ½ O2(g)  2 NO(g);Kc = 1.7 x 10-12

(b) 4 NO(g)  2 N2O(g) + O2(g);Kc = 3.4 x 1023

2.For the reaction: CO2(g) + 3 H2(g)  CH3OH(g) + H2O(g); Kc = 1.4 x 102

3.Answer: [H2] = [I2] = 0.21 M; [HI] = 1.6 M

4.Answer: [COCl2] = 0.492 M; [CO] = [Cl2] = 8.1 x 10-3 M

5.Answer: Qp = 0.0084 > Kp, and net reaction is to the left.

6.For the reaction: N2(g) + O2(g)  2NO(g);Ho = 180 kJ

(a) [NO] stays the same because for this system equilibrium is not affected by pressure change that is caused by a change in volume.

(b) This is an endothermic reaction and the reaction favors high temperature. Therefore, more NO will be formed if overheating occurs in the engine block.

7.For the reaction: CH4(g) + H2O(g)  CO(g) + 3H2(g);Ho = 206 kJ

(a) Removing CO causes equilibrium to shift forward, producing more H2 and increasing [H2]

(b) Removing water vapor causes equilibrium to shift left, consuming H2 and decreasing its [H2].

(d) Adding more CO cause equilibrium to shift left, consuming more H2 and decreasing [H2].

(e) Adding argon gas will not alter the partial pressures of the other gases in the equilibrium mixture. Therefore, the equilibrium state of the system is not affected.

(f) If pressure drops due to increase in volume, the system will react in the direction that increases the number of moles of the gas. In this case, equilibrium will shift right, producing more H2 gas.

(g) A catalyst increases the rate of forward and reversed reactions to the same magnitude, which will not alter the overall concentrations of the equilibrium mixture. Therefore, [H2] will remain the same.

(h) This is an endothermic reaction, which favors high temperature. Increasing the temperature shifts the equilibrium to the right, producing more H2 gas.

Exercise #3

1.For: 2 NOBr(g)  2 NO(g) + Br2(g), Kc = 0.014

2.For: PCl5(g)  PCl3(g) + Cl2(g), Kc = 0.573; Kp = 23.5

3.For: PCl5(g)  PCl3(g) + Cl2(g), Kp = 1.42 and PPCl5 = 2.76; PPCl3 = 2.24, and PCl2 = 1.74 atm.

4.(a) For: NH4(CO2NH2)(s)  2 NH3(g) + CO2(g), Kp = 2.3 x 10-4 at 25oC.

(b) Changes to the amount of NH3 in the mixture due to the following:

(i) Adding CO2 causes equilibrium to shift left, consuming NH3 and amount of NH3 decreases.

(ii) Adding more solid NH4(CO2NH2) will not alter the amount of the gases. There will be no shift in equilibrium and the amount of NH3 will not change.

(iii) Removing CO2 causes pressure drop and equilibrium shifts right to produce more NH3 and CO2.

(iv) Adding N2 gas will not alter the partial pressure of NH3 or CO2. Equilibrium is not disturbed and no change is amount of NH3.

(v) Increasing the volume causes pressure to drop. Equilibrium shifts right to produce more of the gases (NH3 and CO2.

(vi) The decomposition of solids to gases is an endothermic process – high temperature is needed to break bonds. Therefore, increasing the temperature causes more of the solid to decompose, producing more NH3 (and CO2).

5.For: NH4HS(s)  NH3(g) + H2S(g), where Kc = 7.20 x 10-2 at 25oC, at equilibrium,

At equilibrium the partial pressures of NH3 and H2S are equal, such that PNH3 = PH2S = 0.268 atm.