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:
- A \(\pm \frac{k}{2.303}\)
- B \(\mathrm{k}\)
- C \(-\mathrm{k}\)
- D \(-\frac{2.303}{k}\)
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 \((\mathrm{m})=+\frac{\mathrm{k}}{2.303}\)
\(
\begin{gathered}
k=\frac{2.303}{t} \log _{10} \frac{[\mathrm{A}]_0}{[\mathrm{~A}]_{\mathrm{t}}} \\
\log _{10} \frac{[\mathrm{A}]_0}{[\mathrm{~A}]_{\mathrm{t}}}=\frac{\mathrm{k}}{2.303} \mathrm{t} \\
\downarrow \\
\mathrm{y} \quad \mathrm{m} x
\end{gathered}
\)
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 \((\mathrm{m})=+\frac{\mathrm{k}}{2.303}\)
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