A hypothetical reaction obeys the following mechanism: (i) (fast equilibrium, ), (ii) (slow, ), (iii — Chemical Kinetics Chemistry Question
Question
A hypothetical reaction $X_2 + Y_2 \rightarrow 2XY$ obeys the following mechanism: (i) $X_2 \rightleftharpoons 2X$ (fast equilibrium, $K_{eq}$), (ii) $X + Y_2 \rightarrow XY + Y$ (slow, $k_2$), (iii) $X + Y \rightarrow XY$ (fast). What is the overall order of the reaction?
Answer: B
💡 Solution & Explanation
The slow step determines the rate: $r = k_2[X][Y_2]$. From the fast equilibrium, $K_{eq} = [X]^2 / [X_2] \Rightarrow [X] = K_{eq}^{1/2}[X_2]^{1/2}$. Substituting this into the rate equation gives $r = k_2 K_{eq}^{1/2} [X_2]^{1/2} [Y_2]^1$. The overall order is $0.5 + 1 = 1.5$.
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