Half-life for a first order reaction is 6.93 | vedclass

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Question

(D) For a first order reaction,the rate constant $k$ is given by $k = \frac{0.693}{t_{1/2}}$.Substituting the given value,$k = \frac{0.693}{6.93} = 0.

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(D) For a first order reaction,the rate constant $k$ is given by $k = \frac{0.693}{t_{1/2}}$.Substituting the given value,$k = \frac{0.693}{6.93} = 0.1 \ hour^{-1}$.For $80 \%$ completion,the remaining concentration $[A]_t = [A]_0 - 0.80[A]_0 = 0.20[A]_0$.The time $t$ is calculated using the formula $t = \frac{2.303}{k} \log_{10} \frac{[A]_0}{[A]_t}$.$t = \frac{2.303}{0.1} \log_{10} \frac{100}{20} = 23.03 	imes \log_{10} 5$.Using $\log_{10} 5 \approx 0.699$,we get $t = 23.03 	imes 0.699 \approx 16.10 \ hours$.

Half-life for a first order reaction is $6.93 \ hour$. What is the time required for $80 \%$ completion of the reaction (in $hours$)?

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