For a first order reaction 2N2O5⟶k4NO2+O2, the half-life period is given by - Sri Chaitanya Infinity Learn Best Online Courses for NCERT Solutions, CBSE, ICSE, JEE, NEET, Olympiad and Class 6 to 12

A
$2.303 / k$
B
$0.693 / k$
C
$2.303 / 2 k$
D
$0.693 / 2 k$

Solution:

For $2 N_{2} O_{5} \to 4 {NO}_{2} + O_{2}$ , we have

$- \frac{1}{2} \frac{d [N_{2} O_{5}]}{d t} = k [N_{2} O_{5}]$ or $- \frac{d [N_{2} O_{5}]}{[N_{2} O_{5}]} = 2 k d t$

On integrating, we get

$\ln \frac{{[N_{2} O_{5}]}_{0}}{{[N_{2} O_{5}]}_{t}} = 2 kt .$ For a half-life ${[N_{2} O_{5}]}_{t} = (1 / 2) {[N_{2} O_{5}]}_{0}$ Hence $t_{1 / 2} = \frac{\ln 2}{2 k} = \frac{0.693}{2 k}$

For a first order reaction $2 N_{2} O_{5} \overset{k}{⟶} 4 {NO}_{2} + O_{2}$ , the half-life period is given by

Solution:

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