Question

The rate of first-order reaction is $0.04 {~mol} {~L}^{-1} {~s}^{-1}$ at $10 {~s}$ and $0.03 {~mol} {~L}^{-1} {~s}^{-1}$ at $20 {~s}$ after initiation of the reaction. The half-life period of the reaction is

(a) $44.1 {s}$

(b) $54.1 {s}$

(c) $24.1 {s}$

(d) $34.1 {s}$

Answer 1

Correct Option : (c)

Explanation—

For a first order reaction, $A \rightarrow$ Products and for concentration of the reactant at two different times,

$k=\frac{2.303}{t_2-t_1} \log \frac{[A]_1}{[A]_2}$

$\therefore \quad k=\frac{2.303}{t_2-t_1} \log \frac{(\text { rate })_1}{(\text { rate })_2} \quad(\because \text { rate } \propto[A])$

$k=\frac{2.303}{(20-10)} \log \left(\frac{0.04}{0.03}\right)=0.0287 {sec}^{-1}$

$t_{1 / 2}=\frac{0.693}{k}=\frac{0.693}{0.0287 {sec}^{-1}}=24.14 {sec}$