For the reaction: (g) + (g) ⇌ 2(g). The rate law expression is: — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

Reaction: $\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g)$ For a reversible reaction at equilibrium, the net rate of change of any species is the difference between the forward and backward contributions. Consider the net rate of disappearance of HI (i.e., the backward reaction dominates when written from the HI side): \begin{itemize} \item Backward reaction: $2\text{HI} \to \text{H}_2 + \text{I}_2$ with rate constant $K_1$; rate $= K_1[\text{HI}]^2$ \item Forward reaction: $\text{H}_2 + \text{I}_2 \to 2\text{HI}$ with rate constant $K_{-1}$; rate $= K_{-1}[\text{H}_2][\text{I}_2]$ \end{itemize} Net rate of \emph{disappearance} of HI (with stoichiometric factor $\frac{1}{2}$): \[ -\frac{1}{2}\frac{d[\text{HI}]}{dt} = K_1[\text{HI}]^2 - K_{-1}[\text{H}_2][\text{I}_2] \] At equilibrium: $-\frac{d[\text{HI}]}{dt} = 0 \Rightarrow K_1[\text{HI}]^2 = K_{-1}[\text{H}_2][\text{I}_2] \Rightarrow \dfrac{K_{-1}}{K_1} = K_c$ \textbf{Answer: A}