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 $0.125M$ 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. $6.93\times {10}^{-4}mol.mi{n}^{-1}$
• B. $2.66L.mi{n}^{-1}$ at STP
• C. $1.34\times {10}^{-2}mol.mi{n}^{-1}$
• D. $6.93\times {10}^{-1}mol.mi{n}^{-1}$

Answer 1

The correct option is A $6.93\times {10}^{-4}mol.mi{n}^{-1}$

Concentration of $H_2{O}_2$ decreases from $0.5mto0.125mi.e.1/4^{th}in50min$

Thus, half life of $\left[H_2{O}_2\right]$ is:

$t_{1/2}=\frac{50}{2}$

$t_{1/2}=25min$

Rate constant for first order reaction:

$k=\frac{0.693}{t_{1/2}}$

$k=\frac{0.693}{25}$

Decomposition of $H_2{O}_2$ takes place as follows -

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

From this, we can write -

$d\left[O_2\right]/dt=1/2d\left[H_2{O}_2\right]/dt$

$d\left[O_2\right]/dt=1/2\times 0.693/25\times 0.05$

$d\left[O_2\right]/dt=1/2\times k\times \left[H_2{O}_2\right]$

$d\left[O_2\right]/dt=6.93\times {10}^{-4}molmin$

Hence, the rate of formation of $O_2$ will be $6.93\times {10}^{-4}molmin$

Option A is correct.