Q26 / 120 · VITEEE 2008 · Chemistry

The half life period of a first order reaction is 1 $\min 40$ secs. Calculate its rate constant.

  • A
    $6.93 \times 10^{-3} \mathrm{~min}^{-1}$
  • Correct answer
    B
    $6.93 \times 10^{-3} \mathrm{sec}^{-1}$
  • C
    $6.93 \times 10^{-3} \mathrm{sec}$
  • D
    $6.93 \times 10^{3} \mathrm{sec}$

Solution

For $1^{\text {st }}$ order reaction. $\mathrm{t}_{1 / 2} \text { (halflife time) }=\frac{0.693}{\mathrm{~K}}$ Hence $\mathrm{K}=\frac{0.693}{\mathrm{t}^{1 / 2}}=\frac{0.693}{(60+40) \mathrm{sec}}=0.693 \times$ $10^{-2} \mathrm{sec}^{-1}=6.93 \times 10^{-3} \mathrm{sec}^{-1}$

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

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