1. Education
You can opt-out at any time. Please refer to our privacy policy for contact information.

Discuss in my forum

Activation Energy Example Problem

Calculate Activation Energy from Reaction Rate Constants

By

Reactions usually proceed more quickly at higher temperatures.

Reactions usually proceed more quickly at higher temperatures.

Wataru Yanagida, Getty Images
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) = Ea/R x (1/T1 - 1/T2)

where
Ea 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 Ea

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

or in kJ/mol, (divide by 1000)

Ea = 45.9 kJ/mol

Answer:

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

©2014 About.com. All rights reserved.