The time taken for the completion of 90% of a first-order reaction is ‘t’ min. What is the time (in seconds) taken for the completion of 99% of the reaction?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 120 t
For a first-order reaction, the time to reach a given fraction of completion follows from the integrated rate equation:
\( \displaystyle t=\frac{2.303}{k}\log\frac{[R]_0}{[R]} \)
90% completion: [R] = 0.1[R]0, so \( \displaystyle t_{90\%}=\frac{2.303}{k}\log 10=\frac{2.303}{k} \)
99% completion: [R] = 0.01[R]0, so \( \displaystyle t_{99\%}=\frac{2.303}{k}\log 100=2\times\frac{2.303}{k} \)
Ratio: \( \displaystyle t_{99\%}=2\,t_{90\%}=2t\ \mathrm{min} \)
In seconds: \( \displaystyle 2t\ \mathrm{min}\times60\ \mathrm{s\,min^{-1}}=120\,t\ \mathrm{s} \)