For first order reaction the slope of the graph of $\log _{10}[\mathrm{~A}]_{\mathrm{t}} \mathrm{Vs}$. time…
For first order reaction the slope of the graph of $\log _{10}[\mathrm{~A}]_{\mathrm{t}} \mathrm{Vs}$. time is equal to
$\mathrm{k}$
$-\mathrm{k} / 2.303$
$-\mathrm{k}$
$\mathrm{k} / 2.303$
Solution
From first order rate equation,
$\log [A]=\log \left[A_0\right]-\frac{k t}{2.303}$
On compairing this equation with equation of straight line.
$\begin{aligned}
& y=m x+c \text { and } m=-\frac{k}{2.303} \\
& \therefore \text { slope }(X)=-\frac{K}{2.303}
\end{aligned}$