Question

State and explain the law of mass action. Apply it to the following equilibria: (i) \(\mathrm{H}_{2(\mathrm{~g})}+\mathrm{F}_{2(\mathrm{~g})} \rightleftarrows 2 \mathrm{HF}_{(\mathrm{g})}\) (ii) \(\mathrm{NH}_{4} \mathrm{HS}_{(\mathrm{S})} \rightleftarrows \mathrm{NH}_{3(\mathrm{~g})}+\mathrm{H}_{2} \mathrm{~S}_{(\mathrm{g})}\) (iii) \(\mathrm{PCl}_{5(s)} \rightleftarrows \mathrm{PCl}_{3(t)}+\mathrm{Cl}_{2(\mathrm{~g})}\)

Short answer

The law of mass action states that the ratio of the concentration of products raised to their respective stoichiometric coefficients over the concentration of reactants raised to their respective stoichiometric coefficients is constant at a given temperature. For the 3 given equilibrium reactions, the equilibrium constant (K) expressions are: (i) H2 + F2 ↔ 2HF: \(K = \frac{[\mathrm{HF}]^2}{[\mathrm{H}_2][\mathrm{F}_2]}\) (ii) NH4HS(s) ↔ NH3(g) + H2S(g): \(K' = [\mathrm{NH}_3] [\mathrm{H}_2\mathrm{S}]\) (iii) PCl5(s) ↔ PCl3(l) + Cl2(g): \(K' = [\mathrm{Cl}_2]\)

Step-by-step solution

(i) Law of Mass Action for H2 + F2 ↔ 2HF

First, write the balanced chemical equation for the reaction: \(\mathrm{H}_{2}(\mathrm{g}) + \mathrm{F}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{HF}_{(\mathrm{g})}\) Now, apply the law of mass action to determine the equilibrium expression, K: \(K = \frac{[\mathrm{HF}]^2}{[\mathrm{H}_2][\mathrm{F}_2]}\) where [HF], [H2], and [F2] are the equilibrium concentrations of HF, H2, and F2, respectively.

(ii) Law of Mass Action for NH4HS(s) ↔ NH3(g) + H2S(g)

First, write the balanced chemical equation for the reaction: \(\mathrm{NH}_{4} \mathrm{HS}_{(\mathrm{s})} \rightleftarrows \mathrm{NH}_{3(\mathrm{g})} + \mathrm{H}_{2} \mathrm{S}_{(\mathrm{g})}\) Now, apply the law of mass action to determine the equilibrium expression, K: \(K = \frac{[\mathrm{NH}_3][\mathrm{H}_2\mathrm{S}]}{[\mathrm{NH}_{4}\mathrm{HS}]}\) However, since the reaction has a solid reactant, its concentration remains constant and is incorporated into the equilibrium constant. Thus, the equilibrium expression becomes: \(K' = [\mathrm{NH}_3] [\mathrm{H}_2\mathrm{S}]\)

(iii) Law of Mass Action for PCl5(s) ↔ PCl3(l) + Cl2(g)

First, write the balanced chemical equation for the reaction: \(\mathrm{PCl}_{5(s)} \rightleftarrows \mathrm{PCl}_{3(l)} + \mathrm{Cl}_{2(\mathrm{g})}\) Now, apply the law of mass action to determine the equilibrium expression, K: \(K = \frac{[\mathrm{PCl}_3][\mathrm{Cl}_2]}{[\mathrm{PCl}_5]}\) Since the reaction has a solid product, its concentration can be assumed to be constant, and we can redefine the equilibrium constant: \(K' = [\mathrm{Cl}_2]\)