Kinetics to Equilibrium - Solutions

Chem Worksheet - Week 8 Kinetics to Equilibrium - Solutions Key Questions 1. Consider the following reversible reaction, which is believed to proceed by a one-step mechanism in each direction: 2NO2 kf N2O4 kr a. Write the rate expression for each direction. ratef = kf[NO2]2 b. rater = kr[N2O4] At equilibrium, the net rate of the reaction is zero, because the rate of the forward reaction is exactly equal to that of the reverse reaction. Set your two rate expressions equal to each other, and solve the equality for kf/kr. This ratio of constants is itself a constant, which we call the equilibrium constant, Kc. kf[NO2]2 = kr[N2O4] 2. Write the expressions for the equilibrium constant, Kc for the following reactions. a. N2(g) + 3H2(g) º 2NH3(g) b. Cu(NO3)2(aq) + Zn(s) º Cu(s) + Zn(NO3)2(aq) [Hint: Write the net ionic equation first.] Cu2+(aq) + Zn(s) º Cu(s) + Zn2+(aq) Note that the spectator ions do not appear in the Kc expression c. CaO(s) + CO2(g) º CaCO3(s) Note that the solids are omitted from the Kc expression. d. CaCO3(s) º CaO(s) + CO2(g) How does this compare to the expression you wrote for the reverse reaction in part c? Kc = [CO2] This is the inverse of the expression for the reverse reaction. e. HF(aq) + H2O(l) º H3O+(aq) + F–(aq) [Hint: H2O(l) is a solvent.] 3. Consider the following reaction H2(g) + I2(g) º 2 HI(g) At 425oC, an equilibrium mixture has the following concentrations [HI] = 1.01 x 10-2 mol/L [H2] = 1.25 x 10-3 mol/L [I2] = 1.49 x 10-3 mol/L a. What is the value of Kc? b. In general, does an equilibrium mixture of H2(g), I2(g), and HI(g) at 425 oC contain mostly reactants or mostly products? Kc is much greater than 1, so products are favored over reactants. c. In another experiment at 425 oC, an equilibrium mixture is found to contain 0.25 mol/L of H2(g) and 0.50 mol/L of I2(g). What is the concentration of HI(g) in this mixture? [HI]2 = (54.8)(0.125 mol2/L2) = 6.85 mol2/L2 [HI] = 2.61725 mol/L = 2.62 mol/L 4. Consider the equilibrium HF(aq) + H2O(l) º H3O+(aq) + F–(aq) and its Kc expression as you defined it in Key Question 2.e. A solution of hydrofluoric acid is prepared by dissolving 0.100 mol HF in enough water to make a liter of solution. Once equilibrium is established, the concentration of hydronium ion is found to be [H3O+] = 7.91 x 10–3 M. a. What is the concentration of undissociated HF in the solution? To answer this question, fill in the values in the following table. The given information is shown in bold. All other entries are inferred from the given data and the stoichiometry of the equilibrium equation. Start Change Equilibrium [HF] [H3O+] [F–] 0.100 M ~0 M 0M –7.91 x 10–3 M +7.91 x 10–3 M +7.91 x 10–3 M 0.092 M 7.91 x 10–3 M 7.91 x 10–3 M b. What is the value of Kc for HF? c. Strong acids are completely dissociated in water, and weak acids are partially dissociated in water. Kc is much less than 1, so undissociated HF predominates at equilibrium. Thus, HF is a weak acid, based on Kc = 6.80 x 10–4. d. What is the concentration of undissociated HF in a hydrofluoric acid solution in which [H3O+] = [F–] = 0.0100 M? 5. A mixture of 0.297 mol of NO and 0.297 mol of H2 is placed in an evacuated 1.00-L vessel at 300 K. The following equilibrium is established: 2 NO(g) + 2 H2(g) º N2(g) + 2 H2O(g) At equilibrium, [N2] = 0.120 M. a. Calculate the equilibrium concentrations of NO, H2, and H2O. [Hint: Make a table similar to that shown in question 4, above.] The given information is shown in bold. All other entries are inferred from the given data and the stoichiometry of the equilibrium equation. [NO] [H2] [N2] [H2O] Start 0.297 M 0.297 M 0M 0M Change –0.240 M –0.240 M +0.120 M +0.240 M Equilibrium 0.057 M 0.057 M 0.120 M 0.240 M b. Calculate Kc. c. In general, does an equilibrium mixture of NO, H2, N2, and H2O at 300 K favor products or reactants? Because Kc >> 1, products are favored at equilibrium. 6. At 1000 K, the value of Kc is 4.08 x 10–3 mol/L for the equilibrium 2 SO3(g) º 2 SO2(g) + O2(g) What is the value of Kp? [R = 0.0821 L@atm/K@mol] For this reaction, Δn = 3 – 2 = 1. Therefore, Kp = KcRT. Kp = (4.08 x 10–3 mol/L)(0.0821 L@atm/K@mol)(1000K) = 0.335 atm 7. At 800K, the value of Kp is 4.55 atm–1 for the equilibrium CaO(s) + CO2(g) º CaCO3(s) What is the value of Kc? For this reaction, counting only moles of gas species, Δn = 0 – 1 = –1. Kp = Kc(RT)–1 Y Kc = KpRT = (4.55 atm–1)(0.0821 L@atm/K@mol)(800K) = 300 L/mol