JEE Advanced 2024 Paper 1 · Q09 · Rate Law from Mechanism (Steady-State / Pre-equilibrium)

Question

Consider the following reaction,
$$2H_2(g) + 2NO(g) 	o N_2(g) + 2H_2O(g)$$ which follows the mechanism given below:
$2NO(g) \underset{k_{-1}}{\overset{k_1}{ightleftharpoons}} N_2O_2(g)$ \quad (fast equilibrium)
$N_2O_2(g) + H_2(g) \xrightarrow{k_2} N_2O(g) + H_2O(g)$ \quad (slow reaction)
$N_2O(g) + H_2(g) \xrightarrow{k_3} N_2(g) + H_2O(g)$ \quad (fast reaction)
The order of the reaction is ______.

SolveKar AI · Accepted Answer

Slow (rate-determining) step: $r = k_2 [N_2O_2][H_2]$.
The first step is a fast pre-equilibrium: $K_{eq} = k_1/k_{-1} = [N_2O_2]/[NO]^2$, so $[N_2O_2] = (k_1/k_{-1})[NO]^2$.
Substituting:
$$r = k_2 \cdot \frac{k_1}{k_{-1}} [NO]^2 [H_2] = k_{obs}[NO]^2[H_2]$$
Overall order $= 2 + 1 = \mathbf{3}$.

JEE Advanced 2024 Paper 1 · Q09 · Rate Law from Mechanism (Steady-State / Pre-equilibrium)

Solution