In a 1^{st} order reaction,75% of the reactan | vedclass

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(B) For a $1^{st}$ order reaction,the rate constant $k$ is given by: $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$ Given: $t = 1.388 \, h = 1.388 	i

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$2.8 	imes 10^{-4} \, s^{-1}$

Vedclass · Answer

(B) For a $1^{st}$ order reaction,the rate constant $k$ is given by: $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$ Given: $t = 1.388 \, h = 1.388 	imes 3600 \, s = 4996.8 \, s$. If $75\%$ is consumed,then $[A]_t = [A]_0 - 0.75[A]_0 = 0.25[A]_0$. $k = \frac{2.303}{4996.8} \log \frac{[A]_0}{0.25[A]_0} = \frac{2.303}{4996.8} \log(4) = \frac{2.303 	imes 0.602}{4996.8} \approx 2.775 	imes 10^{-4} \, s^{-1} \approx 2.8 	imes 10^{-4} \, s^{-1}$.

$S.No.$	Time $/$ $s$	Total pressure $/$ $atm$
$1.$	$0$	$0.5$
$2.$	$100$	$0.512$

In a $1^{st}$ order reaction,$75\%$ of the reactant is consumed in $1.388 \, h$. Find the rate constant of the reaction.

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