In the reversible reaction

$2N{O}_2\rightleftharpoons N_2{O}_4$

The rate of dis appearance of $N{O}_2$  is equal to

For the given reversible reaction,

$2N{O}_2\underset{k_2}{\overset{k_1}{\rightleftharpoons }}{N}_2{O}_4$

The rate of disapperane of ${\text{NO}}_2$ =?

Let the reaction be,

$2N{O}_2\underset{k_2}{\overset{k_1}{\rightleftharpoons }}{N}_2{O}_4$

For the forward reaction rate $\left(r_f\right)=-\frac{1}{2}\frac{d\left[N{O}_2\right]}{dt}=\frac{d\left[N_2{O}_4\right]}{dt}=K_1{\left[N{O}_2\right]}^2$

For backward reaction

Rate $=-\frac{d\left[N_2{O}_4\right]}{dt}=\frac{1}{2}\frac{d\left[N{O}_2\right]}{dt}=K_2\left[N_2{O}_4\right].$

Rate of disappearance of $N{O}_2=$ (Rate of disappearance of $N{O}_2-$  Rate of appearance of $N{O}_2$ )

$=2K_1{\left[N{O}_2\right]}^2-2K_2\left[N_2{O}_4\right].$