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Activation Energy Formula Online - Tutorpace

# Activation Energy Formula

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Every molecule required some minimum amount of energy to react; there is some formation of bonds or
breaking of bond in the chemical reaction which also require some minimum amount of energy. The energy
can be in the form of Potential energy or Kinetic energy or both.

This minimum energy which requires to starting the reaction is called the activation energy. We represent it
by “Ea”.

We use the Arrhenius equation to derive formula for activation energy.

K = A e^-Ea/RT

Thus, ln K = -Ea/RT + ln A

Where,  K = Rate constant

A = Frequency factor

R = gas constant

T = absolute temperature

If we have rate constant K1 and K2 at Temperature T1 and T2, the activation energy formula will be:

Log(K2/K1) = - Ea/2.303R (T2 – T1)/T1T2

Example: The rate constant for a reaction is 6.0 × 10^-4S^-1. Frequency factor is 3.621 × 10^13 S^-1.
Calculate the activation energy of the reaction at 45oC.

Solution: Given that,

Rate constant, K = 6.0 × 10^-4S^-1

Frequency factor, A = 3.621 × 10^13 S^-1

Temperature, T = 45

So absolute temperature = 273 + 45 = 318

Now we have the formula for activation energy:

Log K = Log A – Ea/2.303RT

Log(6.0 × 10^-4) = Log (3.621 × 10^13) – Ea/2.303(8.314)(318)

Therefore, Ea = 111657 J = 112 KJ

Example: The rate of a reaction doubled when the temperature changes from 295 to 308K. Calculate the
energy of activation of the reaction assuming that it does not change with temperature.

Solution: Given, K2 = 2K1

And T1 = 295,  T2 = 308

Now, use the formula

Log(K2/K1) = - Ea/2.303R (T2 – T1)/T1T2

Log(2K2/K1) = -Ea/2.303R (308-295)/308*295

Log 2 = -Ea/2.303*8.314 (13)/90860

Therefore, Ea = 40,285 Jmol^-1 = 40 KJ mole^-1