Question

In the reversible reaction, $2 NO _{2} \ce{->[{k_{1}}][{K_{2}}]} N _{2} O _{4}$ the rate of disappearance of $N O_{2}$ is equal to

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

Answer 1

Answer: (C)

$2k_{1}\left[NO_{2}\right]^{2} - k_{2}\left[N_{2}O_{4}\right]$

Answer 2

$2NO_{2} \ce{->[{k_1}][{k_2}]} N_{2}O_{4}$
Rate = $-\frac{1}{2} \frac{d\left[NO_{2}\right]}{dt} = k_{1}\left[NO_{2}\right]^{2} - k_{2}\left[N_{2}O_{4}\right]$
$\therefore$ Rate of disappearance of $NO_{2}$ i.e.,
$-\frac{d\left[NO_{2}\right]}{dt} = 2k_{1}\left[NO_{2}\right]^{2} - k_{2}\left[N_{2}O_{4}\right]$