You visited us 1 times! Enjoying our articles? Unlock Full Access!

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.

Open in App

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.

Suggest Corrections

Similar questions

2 N_{2} O_{5} (g) \to 4 N O_{2} (g) + O_{2} (g)

N_{2} O_{5}_{a} N O_{2} + N O_{3}

N O_{3} + N O_{2}_{- a} N_{2} O_{5}

N O_{2} + N O_{3}_{b} N O_{2} + O_{2} + N O

N O + N O_{3}_{c} 2 N O_{2}

2 N O_{2 (g)} + O_{3 (g)} \to N_{2} O_{5 (g)} + O_{2 (g)}

N O_{2 (g)}

O_{3 (g)}

N O_{2} + O_{3} \to N O_{3} + O_{2}

N O_{3} + N O_{2} \to N_{2} O_{5}

O_{3} ⇌ O_{2} + O

N O_{2} + O \to N O_{3}

N O_{3} + N O_{2} \to N_{2} O_{5}

N_{2} O_{5}

N_{2} O_{5} \to 2 N O_{2} + \frac{1}{2} O_{2}

- \frac{d [N_{2} O_{5}]}{d t} = K [N_{2} O_{5}]

N_{2} O_{5} k ⇌ N O_{2} + N O_{3}

N O_{2} + N O_{3} k_{1} - ---- \to N O_{2} + N O + O_{2}

N O + N O_{3} k_{2} - ---- \to 2 N O_{2}

2 N O + O_{2} k^{'} \to 2 N O_{2}

N O + O_{2}_{e q} N O_{3}

N O_{3} + N O k_{1} - \to N O_{2} + N O_{2}

2 N O + O_{2} k^{'} \to 2 N O_{2}

N O + O_{2}_{e q} N O_{3}

N O_{3} + N O k_{1} - \to N O_{2} + N O_{2}

Join BYJU'S Learning Program