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(B) The time interval between $10:00 \ am$ and $11:30 \ am$ is $t = 90 \ min$. At $10:00 \ am$,$20 \%$ is completed,so the remaining concentration is

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(B) The time interval between $10:00 \ am$ and $11:30 \ am$ is $t = 90 \ min$. At $10:00 \ am$,$20 \%$ is completed,so the remaining concentration is $80 \%$ of initial concentration $([A]_0)$. At $11:30 \ am$,$20 \%$ is remaining,so $[A]_t = 0.20 [A]_0$. The rate constant $k$ for a first-order reaction is given by $k = \frac{2.303}{t} \log \frac{[A]_{initial}}{[A]_{final}}$. Substituting the values: $k = \frac{2.303}{90} \log \frac{0.80 [A]_0}{0.20 [A]_0} = \frac{2.303}{90} \log 4$. Using $\log 4 \approx 0.602$,$k = \frac{2.303 	imes 0.602}{90} \approx 0.0154 \ min^{-1}$. The half-life $t_{1/2} = \frac{0.693}{k} = \frac{0.693}{0.0154} \approx 45 \ min$.

$20 \%$ of a first order reaction was found to be completed at $10:00 \ am$. At $11:30 \ am$ on the same day,$20 \%$ of the reaction was found to be remaining. The half-life period in minutes of the reaction is

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