AP EAMCET · Chemistry · Chemical Kinetics
The slope of the graph drawn between \(\ln k\) and \(\frac{1}{T}\) as per Arrhenius equation gives the value \((R=\) gas constant, \(E_\alpha=\) Activation energy)
- A \(\frac{R}{E_a}\)
- B \(\frac{E_a}{R}\)
- C \(\frac{-E_a}{R}\)
- D \(\frac{-R}{E_a}\)
Answer & Solution
Correct Answer
(D) \(\frac{-R}{E_a}\)
Step-by-step Solution
Detailed explanation
Arrhenius equation, \(k=A e^{-E_a / \mathrm{RT}}\) On taking log on both sides, we get \(\ln k=\ln \mathrm{A}-\frac{E_a}{R T}\) A graph between in \(k\) and \(1 / \mathrm{T}\) is a straight line with \(-\frac{E_a}{R}\) slope.
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