devarshi-dt-logo

Question:

The rate of a reaction quadruples when the temperature changes from 300 to 310 K. The activation energy of this reaction is :(Assume activation energy and pre-exponential factor are independent of temperature; ln 2=0.693; R = 8.314 Jmol⁻¹K⁻¹)

26.8 kJmol⁻¹

414.4 kJmol⁻¹

107.2 kJmol⁻¹

53.6 kJmol⁻¹

Solution:

The rate of the reaction can be written as:
ln(K₂/K₁) = Ea/R[1/T₁ - 1/T₂]
ln(4/1) = Ea/8.314[1/300 - 1/310]
⇒Ea = 107.2 KJ/mol