Question

In the reversible reaction,
$2 NO _{2} \underset{k_{2}}{\stackrel{k_{1}}{\rightleftharpoons}} N _{2} O _{4}$
the rate of disappearance of $NO _{2}$ is equal to

• A. $\frac{2 k_{1}}{k_{2}}\left[ NO _{2}\right]^{2}$
• B. $2 k_{1}\left[ NO _{2}\right]^{2}-2 k_{2}\left[ N _{2} O _{4}\right]$
• C. $2 k_{1}\left[ NO _{2}\right]^{2}-k_{2}\left[ N _{2} O _{4}\right]$
• D. $\left(2 k_{1}-k_{2}\right)\left[ NO _{2}\right]$

Answer 1

Answer: (B)

$2 NO _{2} \underset{ k _{2}}{\stackrel{k_{1}}{\rightleftharpoons}} N _{2} O _{4}$

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., $-\frac{d\left[ NO _{2}\right]}{d t}=2 k_{1}\left[ NO _{2}\right]^{2}-2 k_{2}\left[ N _{2} O _{4}\right]$