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 $0.05M$, the rate of formation of $O_2$ will be :

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

Answer 1

Answer: (B)

$t_{3/4}=2\times t_{1/2}=50\min$

i.e. $t_{1/2}=25\min$

$k=\frac{0.693}{t_{1/2}}=\frac{0.693}{25}{\min }^{-1}$

Rate of $H_2{O}_2$ decomposition $=k[H_2{O}_2]$

$=\frac{0.693}{25}\times 0.05=-\frac{d\left[H_2{O}_2\right]}{dt}$

$H_2{O}_2⟶H_{2}O+\frac{1}{2}{O}_2$

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

$\Rightarrow \frac{d\left[O_2\right]}{dt}=6.93\times 10^{-4}\text{mol}{\min }^{-1}$