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Activation Energy Example Problem

Calculate Activation Energy from Reaction Rate Constants

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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.
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