Exercises
Exercise 1. The rate of a forward reaction is 2.6 x 10-4 s-1 and the rate of the reverse reaction is 1.9 x 10-3 s-1. Calculate Kc and draw a reaction diagram.
\(\displaystyle K_c\;=\;\frac{k_f}{k_r}\;=\;\frac{2.6\times 10^{-4}\;s^{-1}}{1.9\times 10^{-3}\;s^{-1}}\;=\;0.14\)
Back to the Equilibrium Constant Study Guide
Exercise 2. Write Kp for the following reaction.
NH4Cl (s) ⇄ NH3 (g) + HCl (g)
\(\displaystyle K_p\;=\;[P_{NH_3}][P_{HCl}]\)
Back to the Equilibrium Constant Study Guide
Exercise 3. Write Kc for the following reactions.
a) CO2 (g) ⇄ CO2 (s)
\(\displaystyle K_c\;=\;\frac{1}{[CO_2]}\)
b) H2O (l) + SO3 (g) ⇄ H2SO4 (aq)
\(\displaystyle K_c\;=\;\frac{[H_2SO_4]}{[SO_3]}\)
c) N2 (g) + H2O (g) ⇄ NO (g) + H2 (g)
Balance the equation
N2 (g) + 2 H2O (g) ⇄ 2 NO (g) + 2 H2 (g)
\(\displaystyle K_c\;=\;\frac{[NO]^2[H_2]^2}{[N_2][H_2O]^2}\)
Back to the Equilibrium Constant Study Guide
Exercise 4. Chlorine gas is prepared by the following 2-step reaction.
ClF (g) + F2 (g) ⇄ ClF3 (g)
a) Write the overall reaction.
Here we will use the reaction quotient, Q.
Cl2 (g) + F2 (g) ⇄ 2 ClF (g) \(\displaystyle \;\;\;Q_1\;=\;\frac{[ClF]^2}{[Cl_2][F_2]}\)
2 ClF (g) + 2 F2 (g) ⇄ 2 ClF3 (g) \(\displaystyle \;\;\;Q_2\;=\;\frac{[ClF_3]^2}{[ClF]^2[F_2]^2]}\)
______________________________________________________________
Cl2 (g) + 3 F2 (g) ⇄ 2 ClF3 (g) \(\displaystyle \;\;\;Q_{overall}\;=\;\frac{[ClF_3]^2}{Cl_2][F_2]^3}\)
b) Show that the overall Kc is equal to the product of Kc‘s for each individual step.
Q1 x Q2 = Qoverall
\(\displaystyle \require{cancel}\frac{\cancel{[ClF]^2}}{[Cl_2][F_2]}\times \frac{[ClF_3]^2}{\cancel{[ClF]^2}[F_2]^2}\;=\;\frac{[ClF_3]^2}{[Cl_2][F_2]^3}\)
Back to the Equilibrium Constant Study Guide
Exercise 5. Write the equilibrium constant expression and indicate if products or reactants are favored at equilibrium for each of the following reactions.
\(\displaystyle K_c\;=\;\frac{[CO][H_2]^3}{[CH_4][H_2O]}\)
There will be more reactants than products at equilibrium as Kc < 1.
b) CO (g) + Cl2 (g) ⇄ COCl2 (g) Kp = 3.7 x 10-2 at 727oC
\(\displaystyle K_c\;=\;\frac{[COCl_2]}{[CO][Cl_2]}\)
There will be more reactants at equilibrium as K < 1.
c) S2 (g) + C (s) ⇄ CS2 (g) Kc = 28.6 at 227oC.
\(\displaystyle K_c\;=\;\frac{[CS_2]}{[S_2]}\)
There will be more products at equilibrium because K > 1.