Question

A first order reaction is completed $90\%$ in 40 minutes. Calculate its half-life period. ($\log 10=1$).

Answer 1

Let initial conc. of reactant $\left(a\right)100,t=40$ minutes.
$90\%$ reaction gets completed in 40 minutes i.e.,
$x=90$
$k=\frac{2.303}{t}\log \frac{a}{a-x}$
$=\frac{2.303}{40}\log \frac{100}{100-90}=\frac{2.303}{40}\log 10$
$=\frac{2.303}{40}\times 1=5.757\times {10}^{-2}mi{n}^{-1}$.
But $t_{1/2}=\frac{0.693}{k}$
$t_{1/2}=\frac{0.693}{5.757\times {10}^{-2}}=10.3min$