Chemistry - Chemical Kinetics Question with Solution | TestHub

For first order reaction, $\mathrm{k}=\frac{2.303}{\mathrm{t}}\log \frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}\dots \dots .(1)$ 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 $\mathrm{a}=0.8$ $\begin{array}{rl}& \mathrm{t}=\frac{2.303}{\mathrm{k}}\log \frac{0.8}{\mathrm{o}.2}\\ & \mathrm{k}=\frac{2.303}{1}\log [4]\dots \dots ..(2)\\ & \mathrm{k}=2.303\times 0.6020\\ & \mathrm{k}=1.3865\dots \dots ..(3)\end{array}$ 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, $\begin{array}{rl}\mathrm{t}& =\frac{2.303}{\mathrm{k}}\log \frac{0.9}{\mathrm{o}.225}\\ \mathrm{t}& =\frac{2.303}{1.3865}\log (4)\dots ..(4)\\ \mathrm{t}& =1.661\times 0.6020\\ \mathrm{t}& =0.999=1\mathrm{hr}\end{array}$