Decomposition of ${\mathrm{H}}_2{\mathrm{O}}_2$ follows a first order reaction. In fifty minutes, the concentration of ${\mathrm{H}}_2{\mathrm{O}}_2$ decreases from 0.5 M to 0.125 Min one such decomposition. When the concentration of ${\mathrm{H}}_2{\mathrm{O}}_2$ reached 0.05 M, the rate of formation of ${\mathrm{O}}_2$ will be

The decomposition reaction is ${\mathrm{H}}_2{\mathrm{O}}_2\to {\mathrm{H}}_2\mathrm{O}+\frac{1}{2}{\mathrm{O}}_2$

The decreasing concentration from 0.5 M to 0.125 M implies two half-lives: 

$0.5\mathrm{M}\stackrel{ }{\stackrel{t_{1/2}}{\to }}0.25\mathrm{M}\stackrel{ }{\stackrel{t_{1/2}}{\to }}0.125\mathrm{M}$ 

Hence, $2t_{1/2}=50\min$ i.e.  $t_{1/2}=25\min$

The rate constant of the reaction is

$k=\frac{0.693}{t_{1/2}}=\frac{0.693}{25\min }2.772\times {10}^{-2}{\min }^{-1}$

The rate expression is $\frac{1}{(1/2)}\frac{\mathrm{d}\left[{\mathrm{O}}_2\right]}{\mathrm{d}t}=k\left[{\mathrm{H}}_2{\mathrm{O}}_2\right]$

When $\left[{\mathrm{H}}_2{\mathrm{O}}_2\right]=0.05\mathrm{M}$, the rate of formation of ${\mathrm{O}}_2$ will be

$\frac{\mathrm{d}\left[{\mathrm{O}}_2\right]}{\mathrm{d}t}=\frac{1}{2}\left(2.772\times {10}^{-2}\stackrel{-1}{min} \right)\left(0.05{\mathrm{molL}}^{-1}\right)=6.93\times {10}^{-4}\mathrm{mol} {\mathrm{L}}^{-1}{\min }^{-1}$.