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.
• B.
• C.
• D.

Answer 1

The correct option is A

Rate = $k\left[H_2{O}_2\right]=\frac{0.693}{25}\times 0.05$

Rate of formation of $H_2{O}_2=\frac{1}{2}\times \frac{0.693}{25}\times 0.05$

$=6.93\times {10}^{-4}molmi{n}^{-1}$