For a first order reaction $2{\mathrm{N}}_2{\mathrm{O}}_5\stackrel{k}{⟶}4{\mathrm{NO}}_2+{\mathrm{O}}_2$, the half-life period is given by

 

For $2{\mathrm{N}}_2{\mathrm{O}}_5\to 4{\mathrm{NO}}_2+{\mathrm{O}}_2$, we have

$-\frac{1}{2}\frac{\mathrm{d}\left[{\mathrm{N}}_2{\mathrm{O}}_5\right]}{\mathrm{d}t}=k\left[{\mathrm{N}}_2{\mathrm{O}}_5\right]$ or $-\frac{\mathrm{d}\left[{\mathrm{N}}_2{\mathrm{O}}_5\right]}{\left[{\mathrm{N}}_2{\mathrm{O}}_5\right]}=2k\mathrm{d}t$

On integrating, we get

$\ln \frac{{\left[{\mathrm{N}}_2{\mathrm{O}}_5\right]}_0}{{\left[{\mathrm{N}}_2{\mathrm{O}}_5\right]}_t}=2\mathrm{kt}.$   For a half-life  ${\left[{\mathrm{N}}_2{\mathrm{O}}_5\right]}_t=(1/2){\left[{\mathrm{N}}_2{\mathrm{O}}_5\right]}_0$ Hence $t_{1/2}=\frac{\ln 2}{2k}=\frac{0.693}{2k}$