# Activation Energy

In order to start a reaction, molecule requires energy. This can be understood by a simple example –  molecules need some kinetic energy or velocity to collide with other molecules to start a reaction. No reaction will take place, if collision don’t happen, or molecules don’t have enough kinetic energy. The energy needs to initiate the reaction is known as Activation energy.

“Activation Energy is the minimum amount of energy that is needed to start a chemical reaction.”

Activation Energy Formula

$\dpi{120}&space;\large&space;lnK=lnA-\frac{E_{a}}{RT}$

$\dpi{120}&space;\large&space;logK=logA-\frac{E_{a}}{2.303RT}$

Where

k = rate constant

A = frequency factor

Ea = activation energy

R = gas constant

T =  absolute temperature

If we know the rate constant k1 and k2 at T1 and T2 the activation energy formula is

$\dpi{120}&space;\large&space;log\frac{K_{2}}{K_{1}}=&space;-\frac{E_{_{a}}}{2.303&space;R}\left&space;(&space;\frac{T_{2}-T_{1}}{T_{1}T_{2}}&space;\right&space;)$

Solved Examples

Question 1: The rate constant for the reaction

2N2O5(g) → 4NO2(g) + O2(g) is 5.0 ×× 10-4S-1. Frequency factor is 2.812 ×× 1013S-1. Find the activation energy of the reaction at 45oC.

Solution:

T = 273 + 45oC

$\dpi{120}&space;\large&space;logK=logA-\frac{E_{a}}{2.303&space;RT}$

Therefore,replace all the values and rearrange the equation to get Ea value.

log (5.0 ×× 10-4) = log (2.812 ×× 1013S-1) + Ea2.303 ×8.314 ×318Ea2.303 ×8.314 ×318

Ea = 102000J

Ea = 102KJ

Question 2: The rate of a reaction increases when the temperature changes from 293 to 313K. Find the energy of activation of the reaction assuming that it does not change with temperature.

Solution:

The T1and T2 are 293K and 313K

Rearrange the following equation to get Ea.

$\dpi{120}&space;\large&space;log\frac{K_{2}}{K_{1}}=&space;-\frac{E_{_{a}}}{2.303&space;R}\left&space;(&space;\frac{T_{2}-T_{1}}{T_{1}T_{2}}&space;\right&space;)$

So that Ea = 2.303R [T1T2T2−T1T1T2T2−T1 ×× logK2K1K2K1]

T1T2T2−T1T1T2T2−T1 = 4585.45K

logK2K1K2K1 = log4141 = log4 = 0.6021

Ea = 2.303 ×× 8.314JK-1mol-1 ×× 4585.45K ×× 0.6021

Ea = 52848 Jmol-1

Ea = 52.8 KJ mol-1