Q66 / 160 · JIPMER 2012 · Chemistry

In the first order reaction, $75 \%$ of the reactant gets disappeared in 1.386 hours. The rate constant of the reaction is

  • A
    $3.0 \times 10^{-3} \mathrm{~s}^{-1}$
  • Correct answer
    B
    $2.8 \times 10^{-4} \mathrm{~s}^{-1}$
  • C
    $17.2 \times 10^{-3} \mathrm{~s}^{-1}$
  • D
    $1.8 \times 10^{-3} \mathrm{~s}^{-1}$

Solution

For a first order reaction, $k=\frac{2.303}{1.386 \times 60 \times 60} \log \frac{a}{a-0.75 a}$ $=\frac{2.303}{1.386 \times 60 \times 60} \log 4$ $=2.8 \times 10^{-4} \mathrm{~s}^{-1}$

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

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