d
\(H_{2} O_{2}(a q) \rightarrow H_{2} O(a q)+\frac{1}{2} O_{2}(g)\)

For a first order reaction

\(k=\frac{2.303}{t} \log \frac{a}{(a-x)}\)

Given \(a=0.5,(a-x)=0.125,\,t=50\,min\)

\(k=\frac{2.303}{50} \log \frac{0.5}{0.125}\)

\(=2.78 \times 10^{-2} \,\mathrm{min}^{-1}\)

\(r=k\left[H_{2} O_{2}\right]=2.78 \times 10^{-2} \times 0.05\)

\(=1.386 \times 10^{-3} \,\,\mathrm{mol} \,\mathrm{min}^{-1}\)

Now

\(-\frac{d\left|H_{d} O_{2}\right|}{d t}=\frac{d\left|H_{2} O\right|}{d t}=\frac{2 d\left|O_{2}\right|}{d t}\)

\(\frac{2 d\left[O_{2}\right]}{d t}-\frac{d\left[H_{2} O_{2}\right]}{d t}\)

\(\frac{d\left[O_{2}\right]}{d t}=\frac{1}{2} \times \frac{d\left[H_{2} O_{2}\right]}{d t}\)

\(=\frac{1.386 \times 10^{-3}}{2}=6.93 \times 10^{-4}\,\, \mathrm{mol}\, \mathrm{min}^{-1}\)