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Mechanism of a hypothetical reaction $X_2 + Y_2 \rightarrow 2XY$ is given below:
(i) $X_2 \rightarrow X + X$ (fast)
(ii) $X + Y_2 \rightleftharpoons XY + Y$ (slow)
(iii) $X + Y \rightarrow XY$ (fast). The overall order of the reaction will be
(i) $X_2 \rightarrow X + X$ (fast)
(ii) $X + Y_2 \rightleftharpoons XY + Y$ (slow)
(iii) $X + Y \rightarrow XY$ (fast). The overall order of the reaction will be
Explanation
Substitute the pre-equilibrium $[X] \propto [X_2]^{1/2}$ into the slow step rate.
Detailed Solution
The solution is given by assuming step (i) to be reversible, which is not stated in the question.
Overall rate = rate of slowest step (ii) $= k[X][Y_2]$ ...(1), where k = rate constant of step (ii)
Assuming step (i) to be reversible, its equilibrium constant $k_{eq} = \frac{[X]^2}{[X_2]} \Rightarrow [X] = k_{eq}^{1/2}[X_2]^{1/2}$ ...(2)
Put (2) in (1): Rate $= kk_{eq}^{1/2}[X_2]^{1/2}[Y_2]$
Overall order $= \frac{1}{2} + 1 = \frac{3}{2}$
Note: the source answer assumes step (i) is a fast reversible equilibrium, which the question does not state.
Overall rate = rate of slowest step (ii) $= k[X][Y_2]$ ...(1), where k = rate constant of step (ii)
Assuming step (i) to be reversible, its equilibrium constant $k_{eq} = \frac{[X]^2}{[X_2]} \Rightarrow [X] = k_{eq}^{1/2}[X_2]^{1/2}$ ...(2)
Put (2) in (1): Rate $= kk_{eq}^{1/2}[X_2]^{1/2}[Y_2]$
Overall order $= \frac{1}{2} + 1 = \frac{3}{2}$
Note: the source answer assumes step (i) is a fast reversible equilibrium, which the question does not state.