Question

Decomposition of $H_2{O}_2$ follows a first order reaction. In fifty minutes the concentration of $H_2{O}_2$ decreases from 0.5 to 0.125 M in one such decomposition. When the concentration of $H_2{O}_2$ reaches 0.05 M, the rate of formation of $O_2$, will be:

• A. $2.78\times {10}^{-4}molmi{n}^{-1}$
• B. $2.66Lmi{n}^{-1}$ at STP
• C. $1.34\times {10}^{-2}molmi{n}^{-1}$
• D. $6.93\times {10}^{-4}molmi{n}^{-1}$

Answer 1

The correct option is D $6.93\times {10}^{-4}molmi{n}^{-1}$

$2H_2{O}_2\to 2H_{2}O+O_2$

In 50 minutes, the concentration is reduced to one fourth
$(\because \frac{0.5}{0.125}=4)$

Hence 2 half life periods correxponds to 50 minutes.
$t_{1/2}=25$ min

The rate constant $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{25}$

Rate of decomposition of $H_2{O}_2=k\left[H_2{O}_2\right]=\frac{0.693}{25}\times 0.05=1.39\times {10}^{-3}mol/min$

Rate of formation of oxygen is one half the rate of decomposition of $H_2{O}_2$.

It is $\frac{1}{2}\times 1.39\times {10}^{-3}=6.93\times {10}^{-4}mol/min$