Question

In the reversible reaction, $ 2NO_2 \xrightleftharpoons[k_2]{k_1}N_2O_4 $ ,
the rate of disappearance of $ NO_2 $ is equal to

• A. $ \frac{2k_{1}}{k_{2}}\left[NO_{2}\right]^{2} $
• B. $ 2k_{1} \left[NO_{2}\right]-2k_{2}\left[N_{2}O_{4}\right] $
• C. $ 2k_{1} \left[NO_{2}\right]^{2}-2k_{2}\left[N_{2}O_{4}\right] $
• D. $ (2k_1 - k_2) [NO_2] $

Answer 1

Answer: (C)

$2NO_2 \xrightleftharpoons[k_2]{k_1} N_2O_4$
The rate of disappearance of $NO_2 = \frac{-1}{2} \frac{d[NO_2]}{dt}$
$\Rightarrow \frac{-1}{2} \frac{d[NO_2]}{dt} = k_1[NO_2]^2 - k_2[N_2O_4]$
$\frac{-dNO_2}{dt} = 2k_1[NO_2]^2 - 2k_2[N_2O_4]$