The rate equation for a first-order reaction is given by R = R 0 e - kt . A straight line with positive…

First-order reaction is the rate of reaction is proportional to the first power of the concentration of the reactant $\mathrm{R}$.

$\mathrm{R}\to \mathrm{P}$

Rate $=-\frac{\mathrm{d}\left[\mathrm{R}\right]}{\mathrm{dt}}=\mathrm{kR}$

Rate equation, $\left[\mathrm{R}\right]={\left[\mathrm{R}\right]}_0{\mathrm{e}}^{-\mathrm{kt}}$

Comparing with $\mathrm{y}=\mathrm{mx}+\mathrm{c}$, if we plot $\mathrm{lnR}$ against $\mathrm{t}$, we get a straight line with a slope $=-\mathrm{k}$ and intercept $=\ln \left[{\mathrm{R}}_0\right]$.

Also,

$$\begin{gathered} \mathrm{k}=\frac{2.303}{\mathrm{t}}\log \frac{\left[{\mathrm{R}}_0\right]}{\left[\mathrm{R}\right]} \\ \Rightarrow \log \frac{\left[{\mathrm{R}}_0\right]}{\left[\mathrm{R}\right]}=\frac{\mathrm{kt}}{2.303} \end{gathered}$$

If we plot a graph between $\log \frac{\left[{\mathrm{R}}_0\right]}{\left[\mathrm{R}\right]}$ Vs $\mathrm{t}$, slope $=\frac{\mathrm{k}}{2.303}$.