Question
Define isomorphism. Derive integrated law expression for first order reaction.

Answer

isomorphism:
Two or more substances having the same crystal structures are called isomorphous substances and the phenomenon is called an isomorphism. e.g. $NaF$ and $MgO, NaNO_3$ and $CaCO_3.$
Consider the first$-$order reaction,
$A \rightarrow$ product
The differential rate law is given by
rate $=-\frac{ d [ A ]}{ dt }= k [ A ] \ldots(1)$
where, $[A]$ is the concentration of reactant at time $t.$ Rearranging Eq. $(1)$
$\frac{d[A]}{[A]}=-k d t...(2)$
Let $[A]_0$ be the initial concentration of the reactant $A$ at time $t = 0.$
Suppose $[A]_t$ is the concentration of $A$ at time $= t$
The equation $(2)$ is integrated between limits $[A] = [A]_0$ at $t = 0$ and $[A] = [A]_t$ at $t = t$
$\int_{[A]_0}^{[A]_t} \frac{d[A]}{[A]}=-k \int_0^t d t$
On integration,
$\ln [A]_{[A]_0}^{[A]_t}=-k t_0^t$
Substitution of limits gives
$\ln [A]_t-\ln [A]_0=-k t$
or $\ln \frac{[ A ]_{ t }}{[ A ]_0}=- kt...(3)$
or $k=\frac{1}{t} \ln \frac{[ A ]_0}{[ A ]_{ t }}$
Converting ln to $log_{10},$ we write
$k =\frac{2.303}{ t } \log _{10} \frac{[ A ]_0}{[ A ]_{ t }}...(4)$
Eq. $(4)$ gives the integrated rate law for the first$-$order reactions.

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