Discussion Forum

Decomposition of H₂O₂  follows a first-order reaction. In 50 min, the concentration of H₂O₂ - decreases from 0.5 to 0.25 in one such decomposition. When the concentration of H₂O₂   reaches 0.05  M, the rate of formation of O₂ will be 

Answer:

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