Overall Order of Reaction for Multistep Reaction Mechanism

To find the overall order of the given multistep reaction, we analyze the mechanism by identifying the rate-determining step (RDS), which is the slowest step in the mechanism.

Step 1: Write the rate law based on the slow step The rate-determining step is: $\text{N}_2\text{O}_2(\text{g}) + \text{H}_2(\text{g}) \overset{k_2}{\rightarrow} \text{N}_2\text{O}(\text{g}) + \text{H}_2\text{O}(\text{g}) \quad (\text{slow})$

The rate of the reaction is determined by the rate of this slow elementary step: $\text{Rate} = k_2 [\text{N}_2\text{O}_2][\text{H}_2]$

Step 2: Express intermediate concentrations in terms of reactants Here, $\text{N}_2\text{O}_2$ is a reaction intermediate and should be eliminated from the rate law. From the fast equilibrium step: $2\text{NO}(\text{g}) \underset{k_{-1}}{\overset{k_1}{\rightleftharpoons}} \text{N}_2\text{O}_2(\text{g}) \quad (\text{fast equilibrium})$

The equilibrium constant $K_{\text{eq}}$ for this step is given by: $K_{\text{eq}} = \frac{k_1}{k_{-1}} = \frac{[\text{N}_2\text{O}_2]}{[\text{NO}]^2}$

Solving for $[\text{N}_2\text{O}_2]$: $[\text{N}_2\text{O}_2] = K_{\text{eq}} [\text{NO}]^2$

Step 3: Substitute into the rate law Substituting the expression for $[\text{N}_2\text{O}_2]$ into the rate equation: $\text{Rate} = k_2 K_{\text{eq}} [\text{NO}]^2 [\text{H}_2] = k_{\text{eff}} [\text{NO}]^2 [\text{H}_2]^1$

where $k_{\text{eff}} = k_2 K_{\text{eq}}$.

Step 4: Determine the overall order of the reaction

• Order of reaction with respect to $\text{NO} = 2$
• Order of reaction with respect to $\text{H}_2 = 1$

The overall order of the reaction is: $\text{Overall Order} = 2 + 1 = 3$