A hypothetical chemical reaction obeys the following sequence: (i) (fast equilibrium, constant ); (i — Chemical Kinetics Chemistry Question
Question
A hypothetical chemical reaction $X_2 + Y_2 \rightarrow 2XY$ obeys the following sequence: (i) $X_2 \rightleftharpoons 2X$ (fast equilibrium, constant $K$); (ii) $X + Y_2 \rightarrow XY + Y$ (slow, constant $k_2$); (iii) $X + Y \rightarrow XY$ (fast). The overall apparent order of this complete reaction is:
Answer: B
💡 Solution & Explanation
The rate of the overall reaction is dictated exclusively by the slow step: $\text{Rate} = k_2[X][Y_2]$. Since $X$ is an intermediate from the fast equilibrium, $K = \frac{[X]^2}{[X_2]} \Rightarrow [X] = K^{1/2} [X_2]^{1/2}$. Substituting this back gives $\text{Rate} = k_2 K^{1/2} [X_2]^{1/2} [Y_2]$. The total order is the sum of the exponents: $1/2 + 1 = 1.5$.
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