Question

The rate of a first-order reaction is 0.04 mol litre${}^1$sec${}^{{}^{-1}}$at 10 minute and 0.03 mol litre${}^1$ at 20 minute after initiation. Find the half-life of the reaction.

• A. 2.406 min
• B. 24.06 min
• C. 240.6 min
• D. 0.204 min

Answer 1

The correct option is B 24.06 min

The rate law expression for the first order reaction is $=K\times \left[A\right].$
Substitute values in the above expression.
$0.04=K[A{]}_{10}$......(1)
$0.03=K[A{]}_{20}$......(2)
Divide equation (1) by equation (2).
$\frac{[A{]}_{10}}{[A{]}_{20}}=\frac{0.04}{0.03}=\frac{4}{3}$
At 10 min, the expression for the rate constant will be $t=\frac{2.303}{K}\log \frac{[A{]}_{10}}{[A{]}_{20}}.$
$10=\frac{2.303}{K}\log \frac{4}{3}.$
Hence, the rate constant $K=\frac{2.303}{10}\log \frac{4}{3}=0.0288mi{n}^{-1}$
The half life period is $t_{1/2}=\frac{0.693}{K}=\frac{0.693}{0.0288}=24.06mi{n}^{-1}$