What is a first-order reaction? Determine the integrated rate equation for a first-order reaction $R \to P$.

(N/A) First-order reaction: $A$ reaction in which the rate of reaction is proportional to the first power of the concentration of the reactant $R$ is called a first-order reaction.
The rate of a first-order reaction $\propto [R]^1$.
For the reaction $R \to P$,the differential rate expression is:
$Rate = -\frac{d[R]}{dt} = k[R]$
$\therefore \frac{d[R]}{[R]} = -k dt \dots (i)$
Integrating this equation on both sides:
$\int \frac{d[R]}{[R]} = -\int k dt$
$\ln [R] = -kt + I \dots (ii)$
Here,$I$ is the constant of integration.
When $t = 0$,$[R] = [R]_0$,where $[R]_0$ is the initial concentration of the reactant. Substituting these values into equation $(ii)$:
$\ln [R]_0 = -k(0) + I \implies I = \ln [R]_0 \dots (iii)$
Substituting $I = \ln [R]_0$ into equation $(ii)$:
$\ln [R] = -kt + \ln [R]_0 \dots (iv)$
Rearranging the terms:
$kt = \ln [R]_0 - \ln [R]$
$kt = \ln \frac{[R]_0}{[R]}$
$k = \frac{1}{t} \ln \frac{[R]_0}{[R]} \dots (v)$
Converting to base $10$ logarithm:
$k = \frac{2.303}{t} \log \frac{[R]_0}{[R]} \dots (vi)$
Taking the antilog of equation $(iv)$,we get:
$[R] = [R]_0 e^{-kt} \dots (vii)$