# Decomposition of

Question:

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 $0.5$ to $0.125 \mathrm{M}$ in one such decomposition. When the concentration of $\mathrm{H}_{2} \mathrm{O}_{2}$ reaches $0.05 \mathrm{M}$, the rate of formation of $\mathrm{O}_{2}$ will be :-

1. $1.34 \times 10^{-2} \mathrm{~mol} \mathrm{~min}-1$

2. $6.93 \times 10^{-2} \mathrm{~mol} \mathrm{~min}{ }^{-1}$

3. $6.93 \times 10^{-4} \mathrm{~mol} \mathrm{~min}^{-1}$

4. $2.66 \mathrm{~L} \mathrm{~min}^{-1}$ at STP

Correct Option: , 3

Solution:

$\mathrm{H}_{2} \mathrm{O}_{2(\mathrm{aq})} \longrightarrow \mathrm{H}_{2} \mathrm{O}_{(\mathrm{aq})}+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g})$

$k=\frac{1}{t} \ln \left(\frac{a_{0}}{a_{1}}\right)$

$=\frac{1}{50} \ln \left(\frac{0.5}{0.125}\right)$

$=\frac{1}{50} \operatorname{\ell n} 4 \min ^{-1}$