Mechanism of a hypothetical reaction $X_2 + Y_2 \rightarrow 2XY$ is given below:
(i) $X_2 \rightleftharpoons X

Question

Mechanism of a hypothetical reaction $X_2 + Y_2 \rightarrow 2XY$ is given below:
(i) $X_2 \rightleftharpoons X + X$ (fast)
(ii) $X + Y_2 \rightarrow XY + Y$ (slow)
(iii) $X + Y \rightarrow XY$ (fast)
The overall order of the reaction will be:

Accepted Answer

The overall rate of a complex reaction is governed by the slowest step, which is the rate-determining step (RDS).
From the given mechanism, the slowest step is:
$X + Y_2 ightarrow XY + Y$
Therefore, the rate law based on the slowest step is:
$	ext{Rate} = k[X][Y_2]$
Since $X$ is an intermediate formed in a fast equilibrium step, its concentration must be expressed in terms of the reactants using the equilibrium step:
$X_2 ightleftharpoons 2X$
The equilibrium constant $K_c$ is:
$K_c = \frac{[X]^2}{[X_2]}$
$\Rightarrow [X]^2 = K_c[X_2]$
$\Rightarrow [X] = K_c^{1/2}[X_2]^{1/2}$
Substituting $[X]$ back into the rate equation gives:
$	ext{Rate} = k \cdot (K_c^{1/2}[X_2]^{1/2}) \cdot [Y_2]$
$	ext{Rate} = k' [X_2]^{1/2}[Y_2]^1$ (where $k' = k \cdot K_c^{1/2}$)
The overall order of the reaction is the sum of the powers of the concentration terms in the rate law:
$	ext{Overall order} = \frac{1}{2} + 1 = 1.5$

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