Chemistry - Chemical Kinetics Question with Solution | TestHub

Given, ${\mathrm{t}}_{1/2}=36\min ,\mathrm{t}=2\mathrm{hrs}=120\min ,[\mathrm{Ao}]=\log$ For first order reaction, $\mathrm{R}=\frac{0.693}{{\mathrm{t}}_{1/2}}$ $\begin{array}{rl}\therefore k& =\frac{0.693}{36}=0.01925\\ t& =\frac{2303}{k}\log \frac{\left[{\mathrm{A}}_0\right]}{[\mathrm{A}]}\end{array}$ where, $\left[{\mathrm{A}}_0\right]=$ initial amount $\begin{array}{rl}& [\mathrm{A}]=\text{ final amount }\\ & 120=\frac{2303}{0.01925}\log \frac{10}{[\mathrm{A}]}\\ & \log 10-\log [\mathrm{A}]=1\\ & 1-\log [\mathrm{A}]=1\\ & \log [\mathrm{A}]=0\\ & [\mathrm{A}]=1\mathrm{gm}\end{array}$