Question

The half life decomposition of $N_2{O}_5$ in a first order reaction represented by
$N_2{O}_5\to N_2{O}_4+\frac{1}{2}{O}_2$
After 20 min the volume of $O_2$ produced is 10 ml and at the end of the reaction 40 ml. The value of rate constant is: $\left[Given2.303\log \left(\frac{4}{3}\right)=0.28\right]$

• A. $0.14{min}^{-1}$
• B. $0.014{min}^{-1}$
• C. $1.4{min}^{-1}$
• D. $14{min}^{-1}$

Answer 1

The correct option is B $0.014{min}^{-1}$

For first order reaction:

$KT=ln\left(\frac{a}{a-x}\right)$

So for given reaction : $N_2{O}_5\to N_2{O}_4+1/2O_2$

$K\times 20=ln\left(\frac{40-0}{40-10}\right)$

$K\times 20=ln\frac{3}{4}$

Given $ln\frac{3}{4}=0.28$

So,

$K\times 20=0.28$

$K=0.014mi{n}^{-1}$

Option B is correct.