Engineering Board Exam Reviewer · Chemistry · hard

The rate constant of a reaction doubles when the temperature increases from 300 K to 310 K. What is the activation energy? (R = 8.314 J/mol· K)

  1. 0.0006 kJ/mol
  2. 123.4 kJ/mol
  3. 26.8 kJ/mol
  4. 53.6 kJ/mol

Answer

D. 53.6 kJ/mol

Solution

(k2/k1) = (Ea/R)(1/T1 - 1/T2) Ea = 8.314 2/(1/300 - 1/310) = 53594 J/mol = 53.6 kJ/mol.

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