Question

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

• A. 1
• B. 2
• C. \N
• D. 1.5

Answer 1

The correct option is D 1.5

The solution of this question is given by assuming step (i) to be reversible which is not given in question
Overall rate = Rate of slowest step (ii)
= $k\left[X\right][Y_2$] ...(1)
k = rate constant of step (ii)
Assuming step (i) to be reversible, its equilibrium constant,
$k_{eq}=\frac{[X{]}^2}{\left[X_2\right]}\Rightarrow \left[X\right]=k_{eq}{}^{\frac{1}{2}}[X_2{]}^{\frac{1}{2}}...(2)$
Putting (2) in (1),
$\text{Rate}=k{k}_{eq}{}^{\frac{1}{2}}\left[X_2{]}^{\frac{1}{2}}\right[Y_2]$
Overall order = $\frac{1}{2}+1=\frac{3}{2}$