Consider the following reaction,

$$2H_2(g) + 2NO(g) \rightarrow N_2(g) + 2H_2O(g)$$

which follows the mechanism given below:

$$2NO(g) \underset{k_{-1}}{\overset{k_1}{\rightleftharpoons}} N_2O_2(g)$$     (fast equilibrium)

$$N_2O_2(g) + H_2(g) \overset{k_2}{\rightarrow} N_2O(g) + H_2O(g)$$     (slow reaction)

$$N_2O(g) + H_2(g) \overset{k_3}{\rightarrow} N_2(g) + H_2O(g)$$     (fast reaction)

The order of the reaction is ______?

The overall rate of a multi-step reaction is governed by the slow (rate-determining) step. In the given mechanism, the slow step is

$$N_2O_2(g)+H_2(g)\;\xrightarrow{k_2}\;N_2O(g)+H_2O(g)$$

Hence the instantaneous rate is

$$\text{rate}=k_2\,[N_2O_2]\,[H_2]\;-(1)$$

The species $$N_2O_2$$ is an intermediate; its concentration must be expressed in terms of the reactants that appear in the overall balanced equation. This is done using the preceding fast equilibrium

$$2NO(g)\;\rightleftharpoons\;N_2O_2(g)$$

For this equilibrium, the equilibrium constant $$K$$ is

$$K=\frac{[N_2O_2]}{[NO]^2}\;-(2)$$

Rearranging $$-(2)$$ gives

$$[N_2O_2]=K\,[NO]^2\;-(3)$$

Substituting $$-(3)$$ into the rate expression $$-(1)$$:

$$\text{rate}=k_2\,K\,[NO]^2\,[H_2]$$

Since $$k_2K$$ is a constant, we can write

$$\text{rate}=k_{\text{obs}}\,[NO]^2\,[H_2]$$

Thus, the reaction is

• second order with respect to $$NO$$
• first order with respect to $$H_2$$

Total (overall) order = $$2+1=3$$.

Therefore, the order of the reaction is 3.