JEE Mains · Chemistry · STD 12 - 3. Chemical kinetics
Decomposition of a hydrocarbon follows the equation \(k = (5.5 \times 10^{11}\,\text{s}^{-1})\,e^{\frac{-28000\,\text{K}}{T}}\). The activation energy of reaction is __________ kJ mol\(^{-1}\). (Nearest Integer) Given : R \(= 8.3\) J K\(^{-1}\) mol\(^{-1}\)
- A 230
- B 232
- C 234
- D 236
Answer & Solution
Correct Answer
(B) 232
Step-by-step Solution
Detailed explanation
According to the Arrhenius equation, \(k = A e^{-\dfrac{E_a}{RT}}\). The given equation is \(k = (5.5 \times 10^{11} \text{ s}^{-1}) e^{-\dfrac{28000}{T}}\). Comparing the exponent of \(e\) in both equations: \(\dfrac{E_a}{RT} = \dfrac{28000}{T}\) \(E_a = 28000 \times R\)…
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