JEE Advanced 2024 — Paper 1 — CHEM-III solution

JEE Advanced2024Paper 1CHEM-III

Consider the following reaction,

$2 H_{2} (g) + 2 NO (g) \to N_{2} (g) + 2 H_{2} O (g)$

which follows the mechanism given below:

$2 NO (g) k_{- 1} ⇌ k_{1} N_{2} O_{2} (g)$ (fast equilibrium)

$N_{2} O_{2} (g) + H_{2} (g) k_{2} N_{2} O (g) + H_{2} O (g)$ (slow reaction)

$N_{2} O (g) + H_{2} (g) k_{3} N_{2} (g) + H_{2} O (g)$ (fast reaction)

The order of the reaction is_____?

Your answer

Solution

The slowest (rate-determining) step is:

$N_{2} O_{2} + H_{2} k_{2} N_{2} O + H_{2} O$

Rate: $r = k_{2} [N_{2} O_{2}] [H_{2}]$ ... (1)

$N_{2} O_{2}$ is an intermediate; eliminate using the fast pre-equilibrium:

$K_{e q} = \frac{k _{1}}{k _{- 1}} = \frac{[ N _{2} O _{2} ]}{[ NO ] ^{2}}$ so $[N_{2} O_{2}] = \frac{k _{1}}{k _{- 1}} [NO]^{2}$ ... (2)

Substituting (2) into (1):

$r = \frac{k _{2} k _{1}}{k _{- 1}} [NO]^{2} [H_{2}]$

Overall order $= 2 + 1 = 3$ .

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