Question

Mechanism of a hypothetical reaction $X_2+Y_2\to 2XY$ is given below
$I)X_2\rightleftharpoons X+X(fast\left)II\right)X+Y_2\to XY+Y\left(slow\right)III)X+Y\to XY(fast)$
The overall order of the reaction will be

• A. 1
• B. 2
• C. 0
• D. 1.5

Answer 1

The correct option is D

1.5

We know that, slowest step is the rate determining step.
$\therefore$ Rater $\left(r\right)=K_1\left[X\right]\left[Y_2\right]$ ….(i)
Now, from equation. (i), i.e.
$X_2\to 2X$ [fast]
$K_{eq}=\frac{[X{]}^2}{\left[X_2\right]}$
$\left[X\right]={\left\{K_{eq}\right[X_2\left]\right\}}^{\frac{1}{2}}$ .....(ii)
Now, substitute the value of [X] from equation. (ii) in equation.(i), we get
$Rate\left(r\right)=K_1\times \left(K_{eq}{)}^{\frac{1}{2}}\right[X_2{]}^{\frac{1}{2}}\left[Y_2\right]=K\left[X_2{]}^{\frac{1}{2}}\right[Y_2]$
$\therefore$ Order of reaction $=\frac{1}{2}+1=\frac{3}{2}=1.5$