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Half life

A decompositi...

Question

A decomposition reaction has following mechanism:
$2 N_{2} O_{5} ⟶ 4 N O_{2} + O_{2}$ (overal)
$N_{2} O_{5} ⇌ N O_{2} + N O_{3}$ (fast decomposition)
$N O_{2} + N O_{3} ⟶ N O + N O_{2} + O_{2}$ (slow)
$N O + N O_{3} ⟶ 2 N O_{2}$ (fast)
Determine rate law and show the mechanism corresponds to first order reaction.

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Solution

The rate will be determined from the slowest step i.e, RDS step, hence
$R a t e = K [{N O}_{2}] [{N O}_{3}] (1)$
From the fast decomposition reaction we can write
$K e q = \frac{[{N O}_{2}] [{N O}_{3}]}{[N_{2} O_{5}]} (2)$
From $(1)$ & $(2)$ we get
$R a t e = K K e q [N_{2} O_{5}]$
Hence mechanism corresponds to first order reaction.

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