The half-life of 1 g of a radioactive sample | vedclass

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(C) For a first-order reaction,the rate constant $k$ is given by $k = \frac{0.693}{t_{1/2}} = \frac{0.693}{9} \ h^{-1}$.Using the integrated rate law:

Vedclass · Accepted Answer

$20.9$

Vedclass · Answer

(C) For a first-order reaction,the rate constant $k$ is given by $k = \frac{0.693}{t_{1/2}} = \frac{0.693}{9} \ h^{-1}$.Using the integrated rate law: $k = \frac{2.303}{t} \log \left( \frac{[A]_0}{[A]_t} ight)$.Substituting the values: $\frac{0.693}{9} = \frac{2.303}{t} \log \left( \frac{1}{0.2} ight)$.$\frac{0.693}{9} = \frac{2.303}{t} \log(5)$.Since $\log(5) \approx 0.699$,we have $\frac{0.693}{9} = \frac{2.303 	imes 0.699}{t}$.$t = \frac{2.303 	imes 0.699 	imes 9}{0.693} \approx 20.9 \ hours$.

The half-life of $1 \ g$ of a radioactive sample is $9 \ hours$. The radioactive decay obeys first-order kinetics. The time required for the original sample to reduce to $0.2 \ g$ is .......... $hours$.

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