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
\(-\frac{d\left|H_{d} O_{2}\right|}{d t}=\frac{d\left|H_{2} O\right|}{d t}=\frac{2 d\left|O_{2}\right|}{d t}\)
\(\frac{2 d\left[O_{2}\right]}{d t}-\frac{d\left[H_{2} O_{2}\right]}{d t}\)
\(\frac{d\left[O_{2}\right]}{d t}=\frac{1}{2} \times \frac{d\left[H_{2} O_{2}\right]}{d t}\)
\(=\frac{1.386 \times 10^{-3}}{2}=6.93 \times 10^{-4}\,\, \mathrm{mol}\, \mathrm{min}^{-1}\)
(નજીકના પૂર્ણાંકમાં રાઉન્ડ ઑફ) (ધારી લો : $\ln 10=2.303, \ln 2=0.693$)
( $R =$ મોલર વાયુ અચળાંક $= 8.314\,JK^{-1}\,mol^{-1}$ )