If k is the rate constant and t is the time,t | vedclass

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(B) For a first-order reaction,the concentration of the reactant at time $t$ is given by the integrated rate law: $[A]_t = [A]_0 e^{-kt}$.Here,$[A]_0$

Vedclass · Accepted Answer

$1 - e^{-kt}$

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(B) For a first-order reaction,the concentration of the reactant at time $t$ is given by the integrated rate law: $[A]_t = [A]_0 e^{-kt}$.Here,$[A]_0$ is the initial concentration and $[A]_t$ is the concentration at time $t$.The amount of substance dissociated is $[A]_0 - [A]_t = [A]_0 - [A]_0 e^{-kt} = [A]_0(1 - e^{-kt})$.The degree of dissociation $(\alpha)$ is defined as the ratio of the amount dissociated to the initial amount:$\alpha = \frac{[A]_0(1 - e^{-kt})}{[A]_0} = 1 - e^{-kt}$.

$t \ (sec)$	$P \ (mm \ of \ Hg)$
$0$	$35.0$
$360$	$54.0$
$720$	$63.0$

If $k$ is the rate constant and $t$ is the time,then the degree of dissociation for a substance undergoing decay with first-order kinetics is:

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