Question

A chemical reaction procceds into the following steps,
Step I, $2A\rightleftharpoons X\left(fast\right)$
Step II, $X+B\to Y\left(slow\right)$
Step III, $Y+BA\rightleftharpoons X\left[Product\right]\left(fast\right)$
The rate law for the overall reaction is:

• A. Rate $=k[A{]}^2$
• B. Rate $=k[B{]}^2$
• C. Rate $=k\left[A\right]\left[B\right]$
• D. Rate $=k\left[A{]}^2\right[B]$

Answer 1

The correct option is D Rate $=k\left[A{]}^2\right[B]$

Step $I$

$2A\leftrightharpoons X$

Rate constant for forward reaction $r=k[A{]}^2$ (EQ 1)

Rate constant for backward reaction$r=k\left[X\right]$ (EQ 2]

Equating 1 & 2

$[A{]}^2=[X]$

For step $II$

$X+Y\leftrightharpoons Y$

$r=\left[X\right]\left[B\right]$

Putting the value of $\left[X\right]=[A{]}^2$

$r=[A{]}^{2}B$ is the correct answer because it is the slowest step which is rate determining.