Reaction A arrow B follows first order kinetics. The time taken for...

For first order reaction, (\mathrm{k}=\frac{2.303}{\mathrm{t}} \log \frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}} \ldots \ldots .(1)) Where $k$ is rate constant when 0.8 mol of $A$ produces 0.6 mol of $B$ then, $0.8-0.6=0.2$ moles of A is left . final moles ($\mathrm{A}=\mathrm{a}$-\mathrm{x}=0.2) initial moles (\mathrm{a}=0.8) ($$\begin{aligned} & \mathrm{t}=\frac{2.303}{\mathrm{k}} \log \frac{0.8}{\mathrm{o} .2} \\ & \mathrm{k}=\frac{2.303}{1} \log [4] \ldots \ldots . . (2) \\ & \mathrm{k}=2.303 \times 0.6020 \\ & \mathrm{k}=1.3865 \ldots \ldots . .(3) \end{aligned}$$) In second case, 0.9 moles of A produces 0.675 moles of $B$ so remaining moles of $A$ are $0.9-0.675=0.225$ final moles of ($\mathrm{A}=\mathrm{a}$-\mathrm{x}=0.225) initial mole $=0.9$ By using first order equation, ($$\begin{aligned} \mathrm{t} & =\frac{2.303}{\mathrm{k}} \log \frac{0.9}{\mathrm{o} .225} \\ \mathrm{t} & =\frac{2.303}{1.3865} \log (4) \ldots . . (4) \\ \mathrm{t} & =1.661 \times 0.6020 \\ \mathrm{t} & =0.999=1 \mathrm{hr} \end{aligned}$$)