$A$ first order reaction takes $40 \ min$ for $30 \%$ decomposition. Calculate $t_{1/2}$.

(N/A) For a first order reaction,the rate constant $k$ is given by:
$k = \frac{2.303}{t} \log \frac{[R]_0}{[R]}$
Given $t = 40 \ min$ and $[R] = [R]_0 - 0.30[R]_0 = 0.70[R]_0$,
$k = \frac{2.303}{40} \log \frac{100}{70} = \frac{2.303}{40} \log(1.4286)$
$k = \frac{2.303}{40} \times 0.1549 = 8.918 \times 10^{-3} \ min^{-1}$
Now,the half-life $t_{1/2}$ is calculated as:
$t_{1/2} = \frac{0.693}{k} = \frac{0.693}{8.918 \times 10^{-3}} \ min$
$t_{1/2} \approx 77.7 \ min$