The reaction, $$A \to B$$  follows first order kinetics. The time taken for $$0.8\,mole$$   of $$A$$ to produce $$0.6\,mole$$   of $$B$$ is $$1h.$$  What is the time taken for the conversion of $$0.9\,mole$$   of $$A$$ to $$0.675\,mole$$   of $$B?$$

Solution :
$$\eqalign{ & {\text{Rate constant of first order reaction}} \cr & k = \frac{{2.303}}{t}{\text{lo}}{{\text{g}}_{10}}\frac{{{{\left( A \right)}_0}}}{{{{\left( A \right)}_t}}} \cr & {\text{or}}\,\,k = \frac{{2.303}}{1} \times {\text{lo}}{{\text{g}}_{10}}\frac{{0.8}}{{0.2}}\,\,\,...{\text{(i)}} \cr & \left( {{\text{because }}0.6{\text{ }}mole{\text{ of }}B{\text{ is formed}}} \right) \cr} $$
Suppose $${t_1}$$  hour are required for changing the concentration of $$A$$ from $$0.9\,mole$$   to $$0.675\,mole$$   of $$B.$$
Remaining mole of $$A = 0.9 - 0.675 = 0.225$$
$$\eqalign{ & \therefore \,\,k = \frac{{2.303}}{{{t_1}}}{\text{lo}}{{\text{g}}_{10}}\frac{{0.9}}{{0.225}}\,\,\,...{\text{(ii)}} \cr & {\text{From Eqs}}{\text{. (i) and (ii)}} \cr & \frac{{2.303}}{1}{\text{lo}}{{\text{g}}_{10}}\frac{{0.8}}{{0.2}} = \frac{{2.303}}{{{t_1}}}{\text{lo}}{{\text{g}}_{10}}\frac{{0.9}}{{0.225}} \cr & 2.303\,{\text{lo}}{{\text{g}}_{10}}\,4 = \frac{{2.303}}{{{t_1}}}{\text{lo}}{{\text{g}}_{10}}4 \cr & {t_1} = 1\,h \cr} $$