Chemistry - Chemical Kinetics Question with Solution | TestHub

Given. Half-life of ${\mathrm{C}}^4,t_{1/2}=5760$ years Initial concentration of sample of ${\mathrm{C}}^4,{\mathrm{N}}_0=200\mathrm{mg}$ Final concentration of sample of ${\mathrm{C}}^4,{\mathrm{N}}_1=2\mathrm{S}\mathrm{mg}$ The given decay is radioactive and all radiouctive decay follows first order kinetics. $$t_{1/2}=\frac{0.693}{\lambda }$$ $\therefore$ $$\lambda =\frac{0.693}{t_{1/2}}=\frac{0.693}{5760}{\mathrm{yr}}^{-1}$$ We know that, for first order reaction $\begin{array}{rl}t& =\frac{2.303}{\lambda }\log \frac{N_0}{N_1}=\frac{2303}{0.693}\log \frac{[200]}{125}\\ & =\frac{2303\times 5760}{0.693}\log 8\\ & =17,28678\mathrm{yrs}\approx 17.280\mathrm{yrs}\end{array}$ Therefore, the time taken to change 200 me to $25\mathrm{mg}$ is $17,286.78\mathrm{yr}$, which is very close to option (d)