Question

For the reaction;
$N_2{O}_5⟶2{NO}_2+\frac{1}{2}{O}_2$,
given:
$-\frac{d\left[N_2{O}_5\right]}{dt}=K_1\left[N_2{O}_5\right]$
$-\frac{d\left[{NO}_2\right]}{dt}=K_2\left[{NO}_2\right]$,
and $-\frac{d\left[O_2\right]}{dt}=K_3\left[O_2\right]$
The relation between $K_1,K_2,K_3$ 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:-

$N_2{O}_5⟶2N{O}_2+\frac{1}{2}{O}_2$

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

Putting the value from $\left(ii\right)$ & $\left(iii\right)$ in $\left(i\right)$, we get:-

$K_1\left[N_2{O}_5\right]=\frac{1}{2}{K}_2\left[N_2{O}_5\right]$

$\Rightarrow K_1=\frac{K_2}{2}\Rightarrow K_2=2K_1$ $-\left(A\right)$

and from $(iii), (iv) and (i):-

$\Rightarrow \frac{1}{2}{K}_2\left[N_2{O}_5\right]=2K_3\left[N_2{O}_5\right]$

$\Rightarrow K_2=4K_3$ $-\left(B\right)$

From $\left(A\right)$ and $\left(B\right)$, we get:-

$2K_1=K_2=4K_3$