Question

For the reaction,
$N_2{O}_5\to 2N{O}_2+\frac{1}{2}{O}_2.$
$Given:-\frac{d\left[N_2{O}_5\right]}{dt}=K_1\left[N_2{O}_5\right]$

$\frac{d\left[N{O}_2\right]}{dt}=K_2\left[N_2{O}_5\right]$

$And\frac{d\left[O_2\right]}{dt}=K_3\left[N_2{O}_5\right]$

The relation between $K_1,K_2$ and $K_3$ is

• A. $2K_1=K_2=4K_3$
• B. $K_1=K_2=K_3$
• C. $2K_1=4K_2=K_3$
• D. $K_1=2K_2=3K_3$

Answer 1

The correct option is A $2K_1=K_2=4K_3$

The given reaction is,

$N_2{O}_5$ → $2N{O}_2$ + $O_2$

Thus, rate of reaction = $\frac{-d\left[N_2{O}_5\right]}{dt}$= $\frac{1}{2}$ $\frac{d\left[N{O}_2\right]}{dt}$ = 2$\frac{d\left[O_2\right]}{dt}$

Substituting the given values of $\frac{-d\left[N_2{O}_5\right]}{dt}$ , $\frac{d\left[N{O}_2\right]}{dt}$ and $\frac{d\left[O_2\right]}{dt}$ in the above equation :-

$K_1$ [$N_2{O}_5$] = $\frac{1}{2}$ $K_2$ [$N_2{O}_5$] = 2 $K_3$[$N_2{O}_5$] ---(i)

Multiply equation (i) by 2 :-

2 $K_1$ [$N_2{O}_5$] = $K_2$ [$N_2{O}_5$] = 4 $K_3$[$N_2{O}_5$]

⇒2 $K_1$ = $K_2$ = 4 $K_3$

Hence, option A is correct.