A hypothetical reaction follows the mechanism: Step 1: (fast, equilibrium constant ); Step 2: (slow, — Chemical Kinetics Chemistry Question
Question
A hypothetical reaction $2A + B \rightarrow \text{Product}$ follows the mechanism: Step 1: $2A \rightleftharpoons X$ (fast, equilibrium constant $K_{eq}$); Step 2: $X + B \rightarrow Y$ (slow, rate constant $k_2$); Step 3: $Y + B \rightarrow \text{Product}$ (fast). The overall rate law expression for the reaction is:
Answer: C
💡 Solution & Explanation
The slowest step determines the overall rate: $\text{Rate} = k_2[X][B]$. Since $X$ is an intermediate formed in the fast equilibrium step, $K_{eq} = \frac{[X]}{[A]^2} \Rightarrow [X] = K_{eq}[A]^2$. Substituting this into the rate equation gives $\text{Rate} = k_2 K_{eq}[A]^2[B]$.
💬Ask on WhatsApp →
Still have doubts about this question?
Send it to our AI chemistry tutor on WhatsApp — gets answered in minutes