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

### Activation Energy 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.

If less energy is available, a chemical reaction is unable to proceed. The activation energy can be determined from reaction rate constants at different temperatures by the equation

ln(k_{2}/k_{1}) = E_{a}/R x (1/T_{1} - 1/T_{2})

where

E_{a} is the activation energy of the reaction in J/mol

R is the ideal gas constant = 8.3145 J/K·mol

T_{1} and T_{2} are absolute temperatures

k_{1} and k_{2} are the reaction rate constants at T_{1} and T_{2}**Step 1** - Convert °C to K for temperatures

T = °C + 273.15

T_{1} = 3 + 273.15

T_{1} = 276.15 K

T_{2} = 35 + 273.15

T_{2} = 308.15 K**Step 2** - Find E_{a}

ln(k_{2}/k_{1}) = E_{a}/R x (1/T_{1} - 1/T_{2})

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 10^{4} 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 10^{4} J/mol or 45.9 kJ/mol.

### Using a Graph to Find Activation Energy from Rate Constant

Another way to calculate the activation energy of a reaction is to graph ln k (the rate constant) versus 1/T (the inverse of the temperature in Kelvin).

The plot will form a straight line where:

m = - E_{a}/R

where m is the slope of the line, Ea is the activation energy, and R is the ideal gas constant of 8.314 J/mol-K. If you took temperature measurements in Celsius or Fahrenheit, remember to convert them to Kelvin before calculating 1/T and plotting the graph!

If you were to make a plot of the energy of the reaction versus the reaction coordinate, the difference between the energy of the reactants and the products would be ΔH, while the excess energy (the part of the curve above that of the products) would be the activation energy.

### Who Discovered Activation Energy?

Swedish scientist Svante Arrhenius proposed the term "activation energy" in 1880 to define the minimum energy needed for the chemical reactants to interact and form products. In a diagram, activation energy is graphed as the height of an energy barrier between to minimum points of potential energy. The minimum points are the energies of the stable reactants and products.