The thermal decomposition of a molecule follo | vedclass

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Question

(C) For a first-order reaction,the rate constant $K$ is given by $K = \frac{0.693}{t_{1/2}}$.Given $t_{1/2} = 120 	ext{ minutes}$,so $K = \frac{0.693

Vedclass · Accepted Answer

$398.8$

Vedclass · Answer

(C) For a first-order reaction,the rate constant $K$ is given by $K = \frac{0.693}{t_{1/2}}$.Given $t_{1/2} = 120 	ext{ minutes}$,so $K = \frac{0.693}{120} 	ext{ min}^{-1}$.For $90\%$ decomposition,the remaining amount is $100 - 90 = 10\%$.The time $t$ required is given by $t = \frac{2.303}{K} \log \left( \frac{a}{a-x} ight)$.Substituting the values: $t = \frac{2.303}{(0.693/120)} \log \left( \frac{100}{10} ight)$.Since $\log(10) = 1$,we get $t = \frac{2.303 	imes 120}{0.693} \approx 398.8 	ext{ minutes}$.

$S. No.$	$Time$ $(s)$	$Total$ $pressure$ $(atm)$
$1$	$0$	$0.5$
$2$	$100$	$0.6$

The thermal decomposition of a molecule follows a first-order reaction. The molecule decomposes $50\%$ in $120 \text{ minutes}$. How many minutes will it take for $90\%$ decomposition?

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$S. No.$ $Time$ $(s)$ $Total$ $pressure$ $(atm)$
$1$ $0$ $0.5$
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