**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(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.