Question

Write the equilibrium expressions \((K)\) for the following reactions: (a) \(\mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{CO}(g)+3 \mathrm{H}_{2}(g)\) (b) \(4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \rightleftharpoons 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)\) (c) \(\mathrm{BaCO}_{3}(s) \rightleftharpoons \mathrm{BaO}(s)+\mathrm{CO}_{2}(g)\) (d) \(\mathrm{NH}_{3}(g)+\mathrm{HCl}(g) \rightleftharpoons \mathrm{NH}_{4} \mathrm{Cl}(s)\)

Short answer

In summary, we calculated the equilibrium expressions (K) for the four given reactions as follows: (a) \(K = \frac{[\mathrm{CO}(g)][\mathrm{H}_2(g)]^3}{[\mathrm{CH}_4(g)]}\) (b) \(K = \frac{[\mathrm{NO}(g)]^4[\mathrm{H}_2\mathrm{O}(g)]^6}{[\mathrm{NH}_3(g)]^4[\mathrm{O}_2(g)]^5}\) (c) \(K = [\mathrm{CO}_2(g)]\) (d) \(K = \frac{1}{[\mathrm{NH}_{3}(g)][\mathrm{HCl}(g)]}\)

Step-by-step solution

(a) Write the equilibrium expression for the reaction of CH4 and H2O

For the reaction \(\mathrm{CH}_4(g) + \mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{CO}(g) + 3\mathrm{H}_2(g)\), the equilibrium expression is: \(K = \frac{[\mathrm{CO}(g)][\mathrm{H}_2(g)]^3}{[\mathrm{CH}_4(g)]}\) H2O is a liquid and not included in the equilibrium expression.

(b) Write the equilibrium expression for the reaction of NH3 and O2

For the reaction \(4 \mathrm{NH}_3(g) + 5 \mathrm{O}_2(g) \rightleftharpoons 4 \mathrm{NO}(g) + 6 \mathrm{H}_2\mathrm{O}(g)\), the equilibrium expression is: \(K = \frac{[\mathrm{NO}(g)]^4[\mathrm{H}_2\mathrm{O}(g)]^6}{[\mathrm{NH}_3(g)]^4[\mathrm{O}_2(g)]^5}\)

(c) Write the equilibrium expression for the reaction of BaCO3

For the reaction \(\mathrm{BaCO}_{3}(s) \rightleftharpoons \mathrm{BaO}(s) + \mathrm{CO}_2(g)\), the equilibrium expression is: \(K = [\mathrm{CO}_2(g)]\) Both BaCO3 and BaO are solids and are not included in the equilibrium expression.

(d) Write the equilibrium expression for the reaction of NH3 and HCl

For the reaction \(\mathrm{NH}_{3}(g) + \mathrm{HCl}(g) \rightleftharpoons \mathrm{NH}_{4}\mathrm{Cl}(s)\), the equilibrium expression is: \(K = \frac{1}{[\mathrm{NH}_{3}(g)][\mathrm{HCl}(g)]}\) Since NH4Cl is a solid, it is not included in the equilibrium expression. Remember that the concentration of a pure solid is considered to be constant, so K is actually equal to the reciprocal of the product of the reactant concentrations.