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 ______?

We need to determine the overall rate law and the total order of the reaction for the given chemical equation based on its multi-step reaction mechanism.

Reaction Mechanism Analysis:

1. Step 1: Fast Equilibrium

   $$2\text{NO}_{(g)} \underset{k_{-1}}{\overset{k_1}{\rightleftharpoons}} \text{N}_2\text{O}_{2(g)} \quad (\text{Fast equilibrium})$$

   Since this step reaches a rapid equilibrium, we can express its equilibrium constant ($$K_{eq}$$) in terms of the rate constants of the forward ($$k_1$$) and reverse ($$k_{-1}$$) reactions:

   $$K_{eq} = \frac{k_1}{k_{-1}} = \frac{[\text{N}_2\text{O}_2]}{[\text{NO}]^2}$$

   Rearranging this equation gives the concentration of the intermediate $$\text{N}_2\text{O}_2$$:

   $$[\text{N}_2\text{O}_2] = \frac{k_1}{k_{-1}} [\text{NO}]^2$$



2. Step 2: Rate-Determining Step (RDS)

   $$\text{N}_2\text{O}_{2(g)} + \text{H}_{2(g)} \overset{k_2}{\rightarrow} \text{N}_2\text{O}_{(g)} + \text{H}_2\text{O}_{(g)} \quad (\text{Slow reaction})$$

   The slowest step in a reaction mechanism determines the overall rate of the chemical reaction. Therefore, the rate law expression is derived directly from this step:

   $$\text{Rate} = k_2 [\text{N}_2\text{O}_2] [\text{H}_2]$$

Substituting the Intermediate Concentration:

A final rate law must only contain the concentrations of the starting reactants, not unstable intermediates like $$\text{N}_2\text{O}_2$$. Substituting our expression for $$[\text{N}_2\text{O}_2]$$ into the rate law gives:

$$\text{Rate} = k_2 \left( \frac{k_1}{k_{-1}} [\text{NO}]^2 \right) [\text{H}_2]$$

Combining all individual constants into a single effective rate constant ($$k = \frac{k_1 k_2}{k_{-1}}$$) yields the final overall rate law:

$$\text{Rate} = k [\text{NO}]^2 [\text{H}_2]^1$$

Determining the Reaction Order:

• Order with respect to $$\text{NO}$$: 2
• Order with respect to $$\text{H}_2$$: 1

$$\text{Overall Order of Reaction} = 2 + 1 = 3$$

Conclusion:

By applying the steady-state equilibrium approximation to the fast step, the overall reaction mechanism follows third-order kinetics.

Answer: 3