Question

The possible mechanism for the reaction:

$2H_{2+}2NO\to N_2+2H_{2}O$ is

$\left(i\right)2NO\rightleftharpoons N_2{O}_2\left(ii\right)N_2{O}_2+H_2\to N_{2}O+H_{2}O\text{(slow) (}iii\text{)}{N}_{2}O+H_2\to N_2{H}_{2}O\text{(fast)}$ What is (i) the rate law for the reaction (ii) the order of the reaction?

Answer 1

As know the slowest step is the rate determining step therefore,

The rate law for the above mechanism can be written as:

Rate = $k_2\left[N_2{O}_2\right]\left[H_2\right]$ ....(i)

As $N_2{O}_2$ is not the reactant therefore its concentration can be determine as follows:

For the first reaction

K1 = $\left[N_2{O}_2\right]/[NO{]}_2$

This implies, $\left[N_2{O}_2\right]=K_1[NO{]}_2$

On substituting the above value in equation (i), we get

Rate = $K_2{K}_1\left[NO{]}_2\right[H_2]$

Let $K_2\times K_1$ = K

Where k is the new rate constant

Hence, rate law for the above mechanism is Rate = $K\left[NO{]}_2\right[H_2]$

From the above rate law it is quite clear that order to reaction is (2+1) = 3