The rate equation for a first order reaction is given by $[R]=[R]_0 e^{-k t}$. A straight line with positive…
The rate equation for a first order reaction is given by $[R]=[R]_0 e^{-k t}$. A straight line with positive slope is obtained by plotting. $[R]_0=$ initial concentration of reactant, $[R]=$ concentration of reactant at time, $t$
$\log \frac{[R]_0}{[R]}$ vs time
$[R]$ vs time
$\log [R]$ vs time
$\log \frac{[R]}{[R]_0}$ vs time
Solution
For the equation $[R]=[R]_0 e^{-k t}$
Taking log we get
$
\begin{array}{rlrl}
& \ln [R] & =-k t+\operatorname{in}[R]_0 \\
& \therefore & \ln \frac{[R]}{[R]_0} & =-k t \\
& & & \\
k & =\frac{1}{t} \text { in } \frac{[R]_0}{[R]} \\
k & =\frac{2.303}{t} \log \frac{[R]_0}{[R]}
\end{array}
$
When graph is ploted of $\log \frac{[R]_0}{[R]}$ against $t$, we get a straight line with positive slope $=\frac{k}{2.303}$