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A catalyst is added to a first-order reaction at 500 K, causing its rate to become 2.718 times the original rate. The activation energy with the catalyst present is 4.15 kJ/mol. What was the activation energy before the catalyst was introduced? (Given: R = 8.3 J/K/mol, ln(2.718) = 1)
- 4.15 kJ/mol
- 2.08 kJ/mol
- 2.718 kJ/mol
- 8.3 kJ/mol
Correct answer: 8.3 kJ/mol
Solution
The rate increases by a factor of 2.718, so ln(k2/k1) = 1. Using Arrhenius: Ea(without catalyst) - Ea(with catalyst) = RT*1 = 8.3*500/1000 kJ = 4.15 kJ, giving Ea(without) = 4.15 + 4.15 = 8.3 kJ/mol.
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