Reaction $\mathrm{A} \rightarrow \mathrm{B}$ 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 conversion of 0.9 mole of A to produce 0.675 mole of $\mathrm{B}$ ?

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Aicharya · Accepted Answer

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, $\begin{aligned} \mathrm{t} & =\frac{2.303}{\mathrm{k}} \log \frac{0.9}{\mathrm{o} .225} \ \mathrm{t} & =\frac{2.303}{1.3865} \log (4) \ldots . . (4) \ \mathrm{t} & =1.661 \times 0.6020 \ \mathrm{t} & =0.999=1 \mathrm{hr} \end{aligned}$

Reaction $\mathrm{A} \rightarrow \mathrm{B}$ follows first order kinetics. The time taken for 0.8 mole of…

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