For the reaction, $2{\mathrm{N}}_2{\mathrm{O}}_5\to 4{\mathrm{NO}}_2+{\mathrm{O}}_2$ , the rate equation can be expressed in term of $-\frac{\mathrm{d}\left[{\mathrm{N}}_2{\mathrm{O}}_5\right]}{\mathrm{d}t}=k\left[{\mathrm{N}}_2{\mathrm{O}}_5\right]$ and $\frac{\mathrm{d}\left[{\mathrm{NO}}_2\right]}{\mathrm{d}t}=k^{\mathrm{\prime }}\left[{\mathrm{N}}_2{\mathrm{O}}_5\right]$ The constant$k$ and $k\mathit{\text{'}}$ are related as 

For the reaction $2{\mathrm{N}}_2{\mathrm{O}}_5\to 4{\mathrm{NO}}_2+{\mathrm{O}}_2$, the rate of expression is

$-\frac{1}{2}\frac{\mathrm{d}\left[{\mathrm{N}}_2{\mathrm{O}}_5\right]}{\mathrm{d}t}=\frac{1}{4}\frac{\mathrm{d}\left[{\mathrm{NO}}_2\right]}{\mathrm{d}t}=k^{\prime \prime }\left[{\mathrm{N}}_2{\mathrm{O}}_5\right]$

It is given that $-\frac{\mathrm{d}\left[{\mathrm{N}}_2{\mathrm{O}}_5\right]}{\mathrm{d}t}=k\left[{\mathrm{N}}_2{\mathrm{O}}_5\right]$; Hence,  $-\frac{1}{2}\frac{\mathrm{d}\left[{\mathrm{N}}_2{\mathrm{O}}_5\right]}{\mathrm{d}t}=\frac{k}{2}\left[{\mathrm{N}}_2{\mathrm{O}}_5\right]$

    $\frac{\mathrm{d}\left[{\mathrm{NO}}_2\right]}{\mathrm{d}t}=k^{\mathrm{\prime }}\left[{\mathrm{N}}_2{\mathrm{O}}_5\right]$ , Hence, $\frac{1}{4}\frac{\mathrm{d}\left[{\mathrm{NO}}_2\right]}{\mathrm{d}t}=\frac{k^{\mathrm{\prime }}}{4}\left[{\mathrm{N}}_2{\mathrm{O}}_5\right]$

Equation ( 1) requires that $\frac{k}{2}=\frac{k^{\mathrm{\prime }}}{4}$ i.e. $2k=k^{\mathrm{\prime }}$.