Slope of the graph between $\log \frac{[\mathrm{A}]_0}{[\mathrm{~A}]_{\mathrm{t}}}$ (y axis) and time ( $x$…
Slope of the graph between $\log \frac{[\mathrm{A}]_0}{[\mathrm{~A}]_{\mathrm{t}}}$ (y axis) and time ( $x$ axis) for first order reaction is equal to:
$\pm \frac{k}{2.303}$
$\mathrm{k}$
$-\mathrm{k}$
$-\frac{2.303}{k}$
Solution
The integrated rate law for the first order reaction is
$\begin{gathered}
k=\frac{2.303}{t} \log _{10} \frac{[\mathrm{A}]_0}{[\mathrm{~A}]_{\mathrm{t}}} \\
\log _{10} \frac{[\mathrm{A}]_0}{[\mathrm{~A}]_{\mathrm{t}}}=\frac{\mathrm{k}}{2.303} \mathrm{t} \\
\downarrow \\
\mathrm{y} \quad \mathrm{m} x
\end{gathered}$
The graph of $\log _{10} \frac{[\mathrm{A}]_0}{[\mathrm{~A}]_{\mathrm{t}}}$ versus time $(\mathrm{t})$ is a straight line passing through origin with slope $(\mathrm{m})=+\frac{\mathrm{k}}{2.303}$