Q29 / 90 · KVPY 2016 (SB/SX) · Chemistry

It takes $1 \mathrm{~h}$ for a first order reaction to go to $50 \%$ completion. The total time required for the same reaction to reach $87.5 \%$ completion will be

  • A
    $1.75 \mathrm{~h}$
  • B
    $6.00 \mathrm{~h}$
  • C
    $3.50 \mathrm{~h}$
  • Correct answer
    D
    $3.00 \mathrm{~h}$

Solution

$\mathrm{t}_{1 / 2}=1 \mathrm{hr}$ $\mathrm{t}_{87.5}=\frac{2.303}{\mathrm{k}} \log \frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}$ $=\frac{2.303}{\frac{0.693}{1}} \log \frac{\mathrm{a}}{\mathrm{a}-.875}$ $=\frac{2.303}{0.693} \log 8$ $=\frac{2.303}{0.693} \times 3 \times .3010$ $=3$

Hence, the correct answer is option: 'D'.

More Chemical Kinetics Questions