TS EAMCET · Chemistry · Chemical Kinetics
The specific rate constant of decomposition of a compound is given by \(\ln k=5.0-\frac{12000}{T}\). The activation energy of decomposition for this compound at \(300 \mathrm{~K}\) is
- A \(24 \mathrm{ kcal} \mathrm{mol}^{-1}\)
- B \(12 \mathrm{kcal} \mathrm{mol}^{-1}\)
- C \(24 \mathrm{cal} \mathrm{mol}^{-1}\)
- D \(12 \mathrm{cal} \mathrm{mol}^{-1}\)
Answer & Solution
Correct Answer
(A) \(24 \mathrm{ kcal} \mathrm{mol}^{-1}\)
Step-by-step Solution
Detailed explanation
Given, \(\text { In } k=5.0-\frac{12000}{T}\) Also, \(\because \quad \text { In } k=\log A-\frac{E_a}{R T} \text { and } k=A e^{-E_a / R T}\) or, \(\quad \log _e K=\log _e A=\frac{E_a}{R T}\) where, \(A=\) pre-exponential constant i.e. frequency factor \(E_a=\) activation energy…
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