Question

A first order reaction takes 40 min for $30\%$ decomposition. Calculate $t_{1/2}$ for this reaction. (Given, $\log \left(1.428\right)=.1548)$

Answer 1

After $40min$ rest compound is $70\%$, applying this first order kinetics

$k=\frac{2.303}{t}\log \left(\frac{100}{70}\right)$ or $k=\frac{2.303}{40}\log \left(1.428\right)$ ,As given $\log 1.428=.1548$

So $k=\frac{2.303}{40}\times .1548$, or $k=8.9\times {10}^{-3}$

Now-

$k=\frac{.693}{T_{half}}$ or $T_{half}=\frac{.693}{9\times {10}^{-3}}$ or $T_{half}=77min.$