(N/A) The integrated rate equations for a first-order reaction are as follows:
$1. \ln [R] = -k(t) + \ln [R]_0$
$2. \log [R] = -\frac{k}{2.303}(t) + \log [R]_0$
These equations follow the straight-line form $y = mx + c$. Thus,the plots of $\ln [R]$ versus $t$ and $\log [R]$ versus $t$ are straight lines with a negative slope and a positive intercept on the $Y$-axis.
- For the plot of $\ln [R]$ vs $t$: The slope is $-k$ and the intercept is $\ln [R]_0$.
- For the plot of $\log [R]$ vs $t$: The slope is $-\frac{k}{2.303}$ and the intercept is $\log [R]_0$.
Additionally,based on the integrated rate equation $\log \frac{[R]_0}{[R]} = \frac{k}{2.303}(t)$,the plot of $\log \frac{[R]_0}{[R]}$ versus $t$ is a straight line passing through the origin $(0, 0)$ with a slope of $\frac{k}{2.303}$.