Chemistry - Chemical Kinetics Question with Solution | TestHub

for 1st order reaction $$ \begin{aligned} & \text { rate constant }(\mathrm{k})=\frac{2.303}{\ell} \log \frac{\mathrm{a}_{\mathrm{o}}}{\mathrm{a}_{\mathrm{t}}} \\ & \mathrm{a}_{\mathrm{o}}=\text { Initial amount }=20 \mathrm{~m} \mathrm{~mol} \\ & \text { at }=\text { final amount }=10 \mathrm{~m} \mathrm{~mol} \\ & \mathrm{k}=\frac{2.303}{1.151} \cdot \log \frac{20}{10} \\ & =\frac{2.303}{1.151} \times \log 2 \\ & =\frac{2.303 \times .0 .3010}{1.151} \\ & \mathrm{k}=0.60 \mathrm{~min}^{-1} \\ & \end{aligned} II ${}^{\text{nd }}$ Method As concentration decreases from $20\mathrm{m}\mathrm{mol}$ to $10\mathrm{m}\mathrm{mol}$ in $1.151\min$ ${\ell }_{1/2}=1.151\min =\frac{\ell \mathrm{n}2}{\mathrm{K}}$ $\mathrm{k}=\frac{\ln 2}{1.151}=\frac{0.693}{1.151}$ $\mathrm{k}=0.60{\min }^{-1}$