Reaction A → B follows first order kinetics. The time taken for 0.8 mole of A to produce 0.6 mole of B is 1 hour. What is the time taken for…

For first order reaction, $\mathrm{k}=\frac{2.303}{\mathrm{t}} \log \frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}} \ldots \ldots .(1)$k=t2.303​loga−xa​…….(1) Where $k$k is rate constant when 0.8 mol of $A$A produces 0.6 mol of $B$B then, $0.8-0.6=0.2$0.8−0.6=0.2 moles of A is left . final moles $\mathrm{A}=\mathrm{a}-\mathrm{x}=0.2$A=a−x=0.2 initial moles $\mathrm{a}=0.8$a=0.8 $\begin{aligned} & \mathrm{t}=\frac{2.303}{\mathrm{k}} \log \frac{0.8}{\mathrm{o} .2} \\ & \mathrm{k}=\frac{2.303}{1} \log [4] \ldots \ldots . . (2) \\ & \mathrm{k}=2.303 \times 0.6020 \\ & \mathrm{k}=1.3865 \ldots \ldots . .(3) \end{aligned}$​t=k2.303​logo.20.8​k=12.303​log[4]……..(2)k=2.303×0.6020k=1.3865……..(3)​ In second case, 0.9 moles of A produces 0.675 moles of $B$B so remaining moles of $A$A are $0.9-0.675=0.225$0.9−0.675=0.225 final moles of $\mathrm{A}=\mathrm{a}-\mathrm{x}=0.225$A=a−x=0.225 initial mole $=0.9$=0.9 By using first order equation, $\begin{aligned} \mathrm{t} & =\frac{2.303}{\mathrm{k}} \log \frac{0.9}{\mathrm{o} .225} \\ \mathrm{t} & =\frac{2.303}{1.3865} \log (4) \ldots . . (4) \\ \mathrm{t} & =1.661 \times 0.6020 \\ \mathrm{t} & =0.999=1 \mathrm{hr} \end{aligned}$tttt​=k2.303​logo.2250.9​=1.38652.303​log(4)…..(4)=1.661×0.6020=0.999=1hr​