In a first-order reaction, half of the reaction is completed in 100 s. The time required for 99% of the reaction to be completed is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 664.64 s
For a first-order reaction the half-life gives the rate constant, and the integrated rate equation then gives the time for any extent of reaction.
Rate constant from the half-life:
\( \displaystyle k=\frac{0.693}{t_{1/2}}=\frac{0.693}{100\ \mathrm{s}}=6.93\times10^{-3}\ \mathrm{s^{-1}} \)
Concentration left after 99% reaction: [R] = 0.01[R]0, so [R]0/[R] = 100.
Time for 99% reaction:
\( \displaystyle t=\frac{2.303}{k}\log\frac{[\ce{R}]_0}{[\ce{R}]} \)
\( \displaystyle t=\frac{2.303}{6.93\times10^{-3}\ \mathrm{s^{-1}}}\times\log 100=\frac{2.303\times2}{6.93\times10^{-3}}\ \mathrm{s} \)
\( \displaystyle t=664.6\ \mathrm{s} \)