**The activation energy of a chemical reaction can be calculated using the Arrhenius equation:**

k = A * e^(-Ea/RT)

Where: k = rate constant A = pre-exponential factor or frequency factor Ea = activation energy R = ideal gas constant T = temperature in Kelvin

Taking the natural logarithm of both sides of the equation gives:

ln(k) = ln(A) – (Ea/RT)

This equation has a linear form, with a slope of -Ea/R and a y-intercept of ln(A). Therefore, plotting ln(k) vs. 1/T should give a straight line, from which the activation energy can be determined by the slope of the line.

## Solved Example

**Question:** The rate of a reaction quadruples when the temperature changes from 290 to 330K. Find the energy of activation of the reaction assuming that it does not change with temperature.

**Solution:**

Given:

T1 = 290K

T2 = 330K

K2 =4K1

From the Arrhenius equation, we obtain

log k2/k1= Ea/2.303R [T2-T1/T1T2]

by substituting all the values we get,

Ea = 1,103276.8/40

=27,581.9 J/mol

Ea = 27.5819 kJ/mol