Question

Decomposition of $H_2{O}_2$ follows first order kinetics. 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. $6.93\times {10}^{-4}$ mol ${\text{min}}^{-1}$
• B. $2.66{\text{L min}}^{-1}$ at STP
• C. $1.34\times {10}^{-2}$ mol ${\text{min}}^{-1}$
• D. $6.93\times {10}^{-2}$ mol ${\text{min}}^{-1}$

Answer 1

The correct option is A $6.93\times {10}^{-4}$ mol ${\text{min}}^{-1}$

$H_2{O}_2\to H_{2}O+\frac{1}{2}{O}_2$

$2t_{\frac{1}{2}}=50\text{mins}$

$t_{\frac{1}{2}}=25\text{mins}$

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

$-\frac{d\left[H_2{O}_2\right]}{dt}=\frac{d\left[O_2\right]}{dt}\times 2$

$\therefore \frac{d\left[O_2\right]}{dt}=\frac{1}{2}\times \frac{0.693}{25}\times 0.05$

$=6.93\times {10}^{-4}$