Question

The half-life period of a first order reaction is $1$ hour. What is the time/in hour taken for $87.5\%$ completion of the reaction?

Answer 1

Answer: 3

For first order reaction

$\frac{t_{1/2}}{t_{x\%}}=\frac{0.3}{\log \left(\frac{100}{100-x}\right)}\Rightarrow \frac{t_{1/2}}{t_{87.5\%}}=\frac{0.3}{\log \left(\frac{100}{100-87.5}\right)}=0.3$

$\Rightarrow t_{87.5}=\frac{t_{1/2}}{0.3}=\frac{1}{0.3}=3.33hr\approx 3hr$