MHT CET · Chemistry · Chemical Kinetics
What is the value of slope, if \(\log _{10} \mathrm{~K}\) (y-axis) is plotted versus \(1 / \mathrm{T}\) (x-axis) for Arrhenius equation?
- A \(\log _{10} \mathrm{~A}\)
- B \(\frac{2 \cdot 303 \mathrm{R}}{\mathrm{E}_{\mathrm{a}}}\)
- C \(\frac{-\mathrm{E}_{\mathrm{a}}}{2 \cdot 303 \mathrm{R}}\)
- D \(-\log _{10} \mathrm{~A}\)
Answer & Solution
Correct Answer
(C) \(\frac{-\mathrm{E}_{\mathrm{a}}}{2 \cdot 303 \mathrm{R}}\)
Step-by-step Solution
Detailed explanation
Arrhenius equation is \(k=A e^{-E_a / R T}\)
\(\therefore \quad \ln \mathrm{k}=-\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{RT}}+\ln \mathrm{A}\)
\(\therefore \quad \log _{10} \mathrm{k}=-\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{R}} \frac{1}{\mathrm{~T}}+\log _{10} \mathrm{~A}\)

Thus, the slope of the line is \(-\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{R}}\)
\(\therefore \quad \ln \mathrm{k}=-\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{RT}}+\ln \mathrm{A}\)
\(\therefore \quad \log _{10} \mathrm{k}=-\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{R}} \frac{1}{\mathrm{~T}}+\log _{10} \mathrm{~A}\)

Thus, the slope of the line is \(-\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{R}}\)
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