The first order decomposition of H_2O_2 in an | vedclass

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(B) Given,rate constant $(k)$ for first order $= 0.2303 \ min^{-1}$.For a first order reaction,the time $(t)$ is given by:$t = \frac{2.303}{k} \log \l

Vedclass · Accepted Answer

$10$

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(B) Given,rate constant $(k)$ for first order $= 0.2303 \ min^{-1}$.For a first order reaction,the time $(t)$ is given by:$t = \frac{2.303}{k} \log \left( \frac{a}{a-x} ight)$where $a$ is the initial concentration and $x$ is the amount reacted.Let the initial concentration $(a) = 1$.Given that the fraction completed is $9/10$,so $x = 0.9$.The remaining concentration $(a-x) = 1 - 0.9 = 0.1$.Substituting the values into the equation:$t = \frac{2.303}{0.2303} \log \left( \frac{1}{0.1} ight)$$t = 10 \log(10)$Since $\log(10) = 1$,we get $t = 10 	imes 1 = 10 \ min$.Therefore,the correct option is $(b)$.

Experiment	$[A]$ $(mol \ L^{-1})$	$[B]$ $(mol \ L^{-1})$	Initial rate $(mol \ L^{-1} \ min^{-1})$
$I$	$0.10$	$0.20$	$6.93 \times 10^{-3}$
$II$	$0.10$	$0.25$	$6.93 \times 10^{-3}$
$III$	$0.20$	$0.30$	$1.386 \times 10^{-2}$

The first order decomposition of $H_2O_2$ in an appropriate medium is characterised by a rate constant of $0.2303 \ min^{-1}$. What is the time (in $min$) required to complete $9/10$ fraction of the reaction?

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