Question

In the reversible reaction $2N{O}_2\stackrel{k_1}{\rightleftharpoons }{N}_2{O}_4$, the rate of disappearance of $N{O}_2$ is equal to:

• A. $\frac{2k_{1^2}}{k_{-1}}[N{O}_2{]}^2$
• B. $2k_1[N{O}_2{]}^2-2k_{-1}[N_2{O}_4]$
• C. $2k_1[N{O}_2{]}^2-k_{-1}[N_2{O}_4]$
• D. $(2k_1-k_{-1})\left[N{O}_2\right]$

Answer 1

Answer: (C)

$2k_1[N{O}_2{]}^2-k_{-1}[N_2{O}_4]$

$N{O}_2$ is consumed in the forward reaction and formed in the reverse reaction.

The rate of disappearance of $N{O}_2$ in the forward reaction is $-\frac{d\left[N{O}_2\right]}{dt}=2k_1[N{O}_2{]}^2$.

The rate of formation of $N{O}_2$ in the reverse reaction is $-\frac{d\left[N{O}_2\right]}{dt}=k_{-1}\left[N_2{O}_4\right]$.

The rate of disappearance of $N{O}_2$ in the reversible reaction is $-\frac{d\left[N{O}_2\right]}{dt}=2k_1[N{O}_2{]}^2-k_{-1}[N_2{O}_4]$.

Hence, option C is correct.