Derive the equation showing the relation between the concentration $[R]_1$ and $[R]_2$ at time $t_1$ and $t_2$ for a first-order reaction.
(N/A) The integrated rate equation for a first-order reaction is:
$\ln [R] = -kt + \ln [R]_0$ $\quad \dots (I)$
At time $t_1$,the concentration is $[R]_1$:
$\ln [R]_1 = -kt_1 + \ln [R]_0$ $\quad \dots (II)$
At time $t_2$,the concentration is $[R]_2$:
$\ln [R]_2 = -kt_2 + \ln [R]_0$ $\quad \dots (III)$
Subtracting equation $(III)$ from equation $(II)$:
$\ln [R]_1 - \ln [R]_2 = (-kt_1 + \ln [R]_0) - (-kt_2 + \ln [R]_0)$
$\ln [R]_1 - \ln [R]_2 = -kt_1 + kt_2$
$\ln \frac{[R]_1}{[R]_2} = k(t_2 - t_1)$
Converting to base $10$ logarithm:
$2.303 \log \frac{[R]_1}{[R]_2} = k(t_2 - t_1)$
$\log \frac{[R]_1}{[R]_2} = \frac{k}{2.303}(t_2 - t_1)$
$\ln [R] = -kt + \ln [R]_0$ $\quad \dots (I)$
At time $t_1$,the concentration is $[R]_1$:
$\ln [R]_1 = -kt_1 + \ln [R]_0$ $\quad \dots (II)$
At time $t_2$,the concentration is $[R]_2$:
$\ln [R]_2 = -kt_2 + \ln [R]_0$ $\quad \dots (III)$
Subtracting equation $(III)$ from equation $(II)$:
$\ln [R]_1 - \ln [R]_2 = (-kt_1 + \ln [R]_0) - (-kt_2 + \ln [R]_0)$
$\ln [R]_1 - \ln [R]_2 = -kt_1 + kt_2$
$\ln \frac{[R]_1}{[R]_2} = k(t_2 - t_1)$
Converting to base $10$ logarithm:
$2.303 \log \frac{[R]_1}{[R]_2} = k(t_2 - t_1)$
$\log \frac{[R]_1}{[R]_2} = \frac{k}{2.303}(t_2 - t_1)$
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