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Question: Consider the reaction : \(2N_{2}O_{4}\) \(\rightleftharpoons\) \(\rightleftharpoons\) \(4NO_{2}\) I...

Consider the reaction : 2N2O42N_{2}O_{4} ⇌\rightleftharpoons ⇌\rightleftharpoons 4NO24NO_{2}

If −d[N2O4]dt=k- \frac{d\lbrack N_{2}O_{4}\rbrack}{dt} = k and d[NO2]dt=k′\frac{d\lbrack NO_{2}\rbrack}{dt} = k' then

A

2k′=k2k' = k

B

k′=2kk' = 2k

C

k′=kk' = k

D

k=14k′k = \frac{1}{4}k'

Answer

k′=2kk' = 2k

Explanation

Solution

2N2O42N_{2}O_{4} 4NO24NO_{2}

Rate =−12d[N2O4]dt=+14d[NO2]dt=12k=14k′= - \frac{1}{2}\frac{d\lbrack N_{2}O_{4}\rbrack}{dt} = + \frac{1}{4}\frac{d\lbrack NO_{2}\rbrack}{dt} = \frac{1}{2}k = \frac{1}{4}k'

Or k′=2kk' = 2k