In a first order reaction, the concentration of reactant is reduced to $$\frac{1}{8}$$  of the initial concentration in 75 minutes at $$298\,K.$$  What is the half-life period of the reaction in minutes?

Solution :
$$\eqalign{ & {\text{Let}}\,\,a = 1,a - x = \frac{1}{8},t = 75\,\min . \cr & k = \frac{{2.303}}{t}\log \frac{a}{{a - x}} \cr & \,\,\,\, = \frac{{2.303}}{{75}}\log \frac{1}{{\frac{1}{8}}} \cr & \,\,\,\, = \frac{{2.303 \times 0.903}}{{75}}\,\min {.^{ - 1}} \cr & {\text{For first order reaction,}} \cr & {t_{\frac{1}{2}}} = \frac{{0.693}}{k} \cr & \,\,\,\,\,\,\, = \frac{{0.693 \times 75}}{{2.303 \times 0.903}} \cr & \,\,\,\,\,\,\, = 25\,\min . \cr} $$