A first order reaction is 60% complete in 20 | vedclass

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(C) For a first order reaction, the rate constant $k$ is given by $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.For $60\%$ completion, $[A]_t = [A]_0

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(C) For a first order reaction, the rate constant $k$ is given by $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.For $60\%$ completion, $[A]_t = [A]_0 - 0.60[A]_0 = 0.40[A]_0$ and $t = 20 	ext{ min}$.$k = \frac{2.303}{20} \log \frac{[A]_0}{0.40[A]_0} = \frac{2.303}{20} \log(2.5) \approx \frac{2.303 	imes 0.3979}{20} \approx 0.0458 	ext{ min}^{-1}$.For $84\%$ completion, $[A]_t = [A]_0 - 0.84[A]_0 = 0.16[A]_0$.$t = \frac{2.303}{k} \log \frac{[A]_0}{0.16[A]_0} = \frac{2.303}{0.0458} \log(6.25)$.Since $\log(6.25) = 2 \log(2.5) \approx 2 	imes 0.3979 = 0.7958$.$t = \frac{2.303 	imes 0.7958}{0.0458} \approx 40 	ext{ min}$.

$A$ first order reaction is $60\%$ complete in $20 \text{ minutes}$. How long will the reaction take to complete $84\%$ (in $\text{ min}$)?

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