For the reaction $2N_2O_5 \rightarrow 4NO_2 + O_2$, the rate of reaction is given by dividing the rate of disappearance or appearance of a species by its respective stoichiometric coefficient:

$\text{Rate} = -\frac{1}{2}\frac{d[N_2O_5]}{dt} = \frac{1}{4}\frac{d[NO_2]}{dt} = \frac{d[O_2]}{dt}$

We are given the individual rate expressions: $-\frac{d[N_2O_5]}{dt} = k[N_2O_5]$ $\frac{d[NO_2]}{dt} = k'[N_2O_5]$ $\frac{d[O_2]}{dt} = k''[N_2O_5]$

Substituting these into the overall rate expression, we get: $\frac{1}{2}(k[N_2O_5]) = \frac{1}{4}(k'[N_2O_5]) = k''[N_2O_5]$

Dividing the entire equation by $[N_2O_5]$: $\frac{k}{2} = \frac{k'}{4} = k''$

From this equality, we can deduce: $\frac{k'}{4} = \frac{k}{2} \implies k' = 2k$ $k'' = \frac{k}{2}$