Activation Energy Example Problem

This example problem demonstrates how to determine the activation energy of a reaction from reaction rate constants at different temperatures.

Problem:

A second-order reaction was observed. The reaction rate constant at 3 °C was found to be 8.9 x 10^-3 L/mol and 7.1 x 10^-2 L/mol at 35 °C. What is the activation energy of this reaction?

Solution

Activation energy is the amount of energy required to initiate a chemical reaction. The activation energy can be determined from reaction rate constants at different temperatures by the equation

ln(k₂/k₁) = E_a/R x (1/T₁ - 1/T₂)

where
E_a is the activation energy of the reaction in J/mol
R is the ideal gas constant = 8.3145 J/K·mol
T₁ and T₂ are absolute temperatures
k₁ and k₂ are the reaction rate constants at T₁ and T₂

Step 1 - Convert °C to K for temperatures

T = °C + 273.15
T₁ = 3 + 273.15
T₁ = 276.15 K

T₂ = 35 + 273.15
T₂ = 308.15 K

Step 2 - Find E_a

ln(k₂/k₁) = E_a/R x (1/T₁ - 1/T₂)
ln(7.1 x 10^-2/8.9 x 10^-3) = E_a/8.3145 J/K·mol x (1/276.15 K - 1/308.15 K)
ln(7.98) = E_a/8.3145 J/K·mol x 3.76 x 10^-4 K^-1
2.077 = E_a(4.52 x 10^-5 mol/J)
E_a = 4.59 x 10⁴ J/mol

or in kJ/mol, (divide by 1000)

E_a = 45.9 kJ/mol

Answer:

The activation energy for this reaction is 4.59 x 10⁴ J/mol or 45.9 kJ/mol.