Question

In the reaction $2N_2 O_5\rightarrow 4NO_2 + O_2 $ the rate is expressed as
(i)$\frac{d\left[N_{2}O_{5}\right]}{dt} k_{1}\left[N_{2}O_{5}\right] $
(ii)$\frac{d\left[NO_{2}\right]}{dt}=k_{2}\left[N_{2}O_{5}\right] $0
(iii)$\frac{d\left[O_{2}\right]}{dt}=k_{3}\left[N_{2}O_{5}\right]$
Relation between $ k_1 k_2$ and $ 3 $ is

• A. $2k_1 = 4k_2 = k_3 $
• B. $2k_1 =k_2 = 4k_3 $
• C. $2k_1 = k_2 = 2k_3 $
• D. $k_1 = 4k_2 = 2k_3 $

Answer 1

Answer: (B)

Rate $= \frac{1}{2} \frac{d\left[N_{2}O_{5}\right]}{dt} =\frac{1}{4} \frac{d\left[NO_{2}\right]}{dt} =\frac{d\left[O_{2}\right]}{dt} $
Rate $=\frac{1}{2}k_{1}\left[N_{2}O_{2}\right]=\frac{1}{4}k_{2}\left[N_{2}O_{5}\right]=\frac{1}{4}\left[N_{2}O_{5}\right]=K_{3}\left[N_{2}O_{5}\right]$
On dividing all the terms by $ [N_2O_5] $ and multiplying by $ 4 $ we get, $2k_1 = k_2 = 4k_3 $