Question

Dinitrogen pentaoxide decomposes as follows:
$N_2{O}_5\to 2N{O}_2+\frac{1}{2}{O}_2$. If, $-\frac{d\left[N_2{O}_5\right]}{dt}=K^{\prime }\left[N_2{O}_5\right]$
$\frac{d\left[N{O}_2\right]}{dt}=K^{\prime \prime }\left[N_2{O}_5\right]$
$\frac{d\left[O_2\right]}{dt}=K^{\prime \prime \prime }\left[N_2{O}_5\right]$
Derive a relation in $K^{\prime },K^{\prime \prime }$ and $K^{\prime \prime \prime }$.

Answer 1

$-\frac{d\left[N_2{O}_5\right]}{dt}=+\frac{1}{2}\frac{d\left[N{O}_2\right]}{dt}=2\frac{d\left[O_2\right]}{dt}$
On substituting the values,
$K^{\prime }\left[N_2{O}_5\right]=\frac{1}{2}{k}^{\prime \prime }\left[N_2{O}_5\right]=2K^{\prime \prime \prime }\left[N_2{O}_5\right]$
or $2K^{\prime }=K^{\prime \prime }=4K^{\prime \prime \prime }$