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Question: Consider this reaction : 2 NO<sub>2</sub> (g) + O<sub>3</sub> (g) ¾® N<sub>2</sub>O<sub>5</sub> (g)...

Consider this reaction :

2 NO2 (g) + O3 (g) ¾® N2O5 (g) + O2 (g)

The reaction of NO2 (g) and O3 (g) represented is first order in NO2 (g) and in O3 (g) Which of the following mechanism is/are consistent with the rate law ?

MechanismI : NO2 (g) + O3 (g) ¾® NO3 (g) + O2 (g) (slow)

NO3 (g) + NO2 (g) ¾® N2O5 (g) (fast)

MechanismII : O3 (g) ⟶\longrightarrowO2 (g) + O (fast)

NO2 (g) + O ¾® NO3 (g) (slow)

NO3 (g) + NO2 (g) ¾® N2O5 (g) (fast)

A

I only

B

II only

C

Both I & II

D

Neither I nor II

Answer

I only

Explanation

Solution

As per mechanism (I), rate = k [NO2] [O2]

slowest step is the r.d.s.

As per mechanism (II), rate = k [NO2] [O]

Keq = [O2][O][O3]\frac { \left[ \mathrm { O } _ { 2 } \right] [ \mathrm { O } ] } { \left[ \mathrm { O } _ { 3 } \right] }

or [O] = Keq [O3][O2]\frac { \left[ \mathrm { O } _ { 3 } \right] } { \left[ \mathrm { O } _ { 2 } \right] }

\ as per mechanism (II), rate = k Keq [NO2] [O3] [O2]–1