Answer:
Option A
Explanation:
For the first-order reaction,
$k=\frac{2.303}{t}\log\frac{a}{a-x}$
Given, t= 50min a= 0.5 M
a-x= 0.125 M
Therefore,
$k=\frac{2.303}{50}\log\frac{0.5}{0.125}$$=0.0277min^{-1}$
Now, as per reaction
$2H_2O_2\rightarrow2H_2O+O_2 $
$-\frac{1}{2}\frac{\text{d}[H_2O_2]}{\text{d}t}=\frac{1}{2}\frac{\text{d}[H_2O]}{\text{d}t}=\frac{\text{d}[O_2]}{\text{d}t}$
Rate of reaction
$-\frac{\text{d}[H_2O_2]}{\text{d}t}=k[H_2O_2]$
Therefore,
$-\frac{\text{d}[O_2]}{\text{d}t}=\frac{1}{2}\frac{\text{d}H_2O_2}{\text{d}t}=\frac{1}{2}k[H_2O_2]$....(i)
When concentration of of $H_2O_2$ reaches 0.05 M
$\frac{\text{d}[O_2]}{\text{d}t}=\frac{1}{2}\times .0277\times .05 $
or $\frac{\text{d}[O_2]}{\text{d}t}=6.93\times 10^-4 mol$ $min^{-1}$
