Question

A chemical reaction proceeds into the following steps
Step I: $$\text{2A}\rightleftharpoons \text{ X}$$ (fast)
Step II: $$\text{X + B}\to \text{ Y}$$ (slow)
Step III: $$\text{Y + B }\rightleftharpoons \text{ Product}$$ (fast)
The rate low for the overall reaction *is:

Answer 1

Hint:First identify the rate determining step and write its rate law expression. If any concentration term for an intermediate is present, then use suitable expression from the equilibrium step to express the concentration of the intermediate.

Complete answer:
Consider the step I: $$\text{2A}\rightleftharpoons \text{ X}$$ (fast)
Write the expression for the rate of the forward reaction
  $$\text{r = k}{\left[\text{A}\right]}^2$$ … …(1)
Write the expression for the rate of the reverse reaction
  $$\text{r = k'}\left[\text{X}\right]$$ … …(2)
For a reaction at equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction. So equation (1) is equal to the equation (2).
 $$\text{k}{\left[\text{A}\right]}^2\text{ = k'}\left[\text{X}\right]$$
$$\left[\text{X}\right]={\displaystyle \frac{\text{k}}{\text{k'}}}{\left[\text{A}\right]}^2$$
$$\left[\text{X}\right]=\text{K}{\left[\text{A}\right]}^2$$ … …(3)
Here $$\text{K}={\displaystyle \frac{\text{k}}{\text{k'}}}$$ is the equilibrium constant for the step I.
Now consider step II: It is slow step or rate determining step.
$$\text{X + B}\to \text{ Y}$$
Write the expression for the rate of the reaction for the step II:
$\text{r=k’’[X][B]}$ … ….(4)
Substitute equation (3) into equation (4)
${\text{r=k’’K[A]}}^{\text{2}}\text{[B]}$
${\text{r=k’’’[A]}}^{\text{2}}\text{[B]}$
Here $\text{k’’’=k’’K}$ is the rate constant for the overall reaction.

Hence, the rate law expression for the overall reaction is ${\text{r=k’’’[A]}}^{\text{2}}\text{[B]}$.

Note:
When several steps are present in the reaction, then the rate of the overall reaction is equal to the rate of the rate determining step. The rate determining step is the slowest step of the reaction.