Question

Decomposition of $H_2O_2$ follows a first order reaction. In fifty minutes the concentration of $H_2O_2$ decreases from $0.5$ to $0.125 \,M$ in one such decomposition. When the concentration of $H_2O_2$ reaches $0.05 \,M$, the rate of formation of $O_2$ will be :

• A. $6.93 \times 10^{-2} \,mol \, min^{-1}$
• B. $6.93 \times 10^{-4} \,mol \, min^{-1}$
• C. $2.66 \,L \, min^{-1} $ at $STP$
• D. $1.34 \times 10^{-2} mol \, min^{-1}$

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 $H_2O_2$ decomposition $= k[H_{2}O_{2}] $

$= \frac{0.693}{25} \times0.05 = - \frac{d\left[H_{2}O_{2}\right]}{dt} $

$H _{2} O _{2} \longrightarrow H _{2} O +\frac{1}{2} O _{2}$

$ - \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 \times10^{-4} \text{mol} \min^{-1} $