# 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

#### Practise This Question

Fractional distillation is the separation of a mixture into its component parts or fractions. This is the process of separation of chemical compounds due to difference in their boiling point. The mixture is heated to a temperature at which one or more fractions (component parts) will vaporize. It is a special type of distillation.

Air is a homogeneous mixture of 78.09% nitrogen, 20.95% oxygen, 0.93% argon, 0.039% carbon dioxide, and small amounts of other gases such as helium, krypton, nitrogen dioxide etc.

Consider three gases oxygen, nitrogen and argon whose boiling points are -183, -196, -186 degree celsius, they are to be separated from air using fractional distillation.

Air is compressed by increasing the pressure and then cooled by decreasing the temperature in order to get liquid air. This liquid air is allowed to warm-up slowly in a fractional distillation column, where gases get separated at different heights depending upon their boiling points. As the height of tower increases the temperature increases.

Statement: The temperature is ____ at the base of the fractional distillation column.