Find the time period of a 1^{st} order reacti | vedclass

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(D) For a $1^{st}$ order reaction,the rate equation is given by: $t = \frac{2.303}{k} \log \frac{[A]_0}{[A]_t}$ Given that the reaction is $\frac{2}{3

Vedclass · Accepted Answer

$2.5 	imes 10^{3} \, s$

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(D) For a $1^{st}$ order reaction,the rate equation is given by: $t = \frac{2.303}{k} \log \frac{[A]_0}{[A]_t}$ Given that the reaction is $\frac{2}{3}$ complete,the amount reacted $x = \frac{2}{3} a$,where $a$ is the initial concentration. Therefore,the remaining concentration $[A]_t = a - \frac{2}{3} a = \frac{a}{3}$. Substituting the values into the equation: $t = \frac{2.303}{4.3 	imes 10^{-4}} \log \frac{a}{a/3}$ $t = \frac{2.303}{4.3 	imes 10^{-4}} \log 3$ Using $\log 3 \approx 0.4771$: $t = \frac{2.303 	imes 0.4771}{4.3 	imes 10^{-4}} \approx \frac{1.0988}{4.3 	imes 10^{-4}} \approx 2555 \, s$ Rounding to the given options,$t = 2.5 	imes 10^{3} \, s$.

Find the time period of a $1^{st}$ order reaction when the reaction is $\frac{2}{3} ^{rd}$ complete. If the value of the rate constant is $4.3 \times 10^{-4} \, s^{-1}$.

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