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MHT CET · Chemistry · Chemical Kinetics

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:

  1. A \(\pm \frac{k}{2.303}\)
  2. B \(\mathrm{k}\)
  3. C \(-\mathrm{k}\)
  4. D \(-\frac{2.303}{k}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\pm \frac{k}{2.303}\)

Step-by-step Solution

Detailed explanation

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 \((\mathrm{m})=+\frac{\mathrm{k}}{2.303}\)