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The half life period of a first order reaction is 1 $\min 40$ secs. Calculate its rate constant.
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Verified Answer
The correct answer is:
$6.93 \times 10^{-3} \mathrm{sec}^{-1}$
For $1^{\text {st }}$ order reaction.
$\mathrm{t}_{1 / 2} \text { (halflife time) }=\frac{0.693}{\mathrm{~K}}$
Hence $\mathrm{K}=\frac{0.693}{\mathrm{t}^{1 / 2}}=\frac{0.693}{(60+40) \mathrm{sec}}=0.693 \times$ $10^{-2} \mathrm{sec}^{-1}=6.93 \times 10^{-3} \mathrm{sec}^{-1}$
$\mathrm{t}_{1 / 2} \text { (halflife time) }=\frac{0.693}{\mathrm{~K}}$
Hence $\mathrm{K}=\frac{0.693}{\mathrm{t}^{1 / 2}}=\frac{0.693}{(60+40) \mathrm{sec}}=0.693 \times$ $10^{-2} \mathrm{sec}^{-1}=6.93 \times 10^{-3} \mathrm{sec}^{-1}$
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