Activation energy is often studied under physical chemistry and it is a very important concept related to chemical kinetics in JEE. The term Activation Energy was first used by a Swedish scientist named Svante Arrhenius in the year 1889. Today, we learn about this concept as it helps us understand the energy requirements for a chemical reaction, which further gives us control over our actions and the environment.
Download Complete Chapter Notes of Chemical Kinetics and Nuclear Chemistry
Almost all the chemical reactions require some activation energy to go forward. Therefore, it is important for students to know what activation energy is. We shall learn more about this concept in this lesson.
Table of Content
- Activation Energy Definition
- Factors Affecting Activation Energy
- Activation Energy Examples
- Activated Complex
- Activation Energy Formula
Activation Energy Definition
Activation energy is defined as the minimum amount of extra energy required by a reacting molecule to get converted into a product. It can also be described as the minimum amount of energy needed to activate or energise molecules or atoms so that they can undergo a chemical reaction or transformation.
What Is the SI Unit of Activation Energy?
Factors Affecting Activation Energy
Activation energy depends on two factors.
1. Nature of Reactants
In the case of ionic reactant, the value of (Ea) will be low because there is an attraction between reacting species. While in the case of covalent reactant the value of Ea will be high because energy is required to break the older bonds.
Also Read: Chemical Kinetics
2. Effect of Catalyst
Positive catalyst provides such an alternate path in which the value of Ea will be low, while the negative catalyst provides such an alternate path in which the value of Ea will be high.
Activation Energy Examples
Activation energy for Forward reaction (Ea)f
Activation energy for Forward reaction (Ea)b
(Ea)f < (Ea)b
(ROR)f > (ROR)b
△H = (Ea)f – (Ea)b
△H = -ve value
(Ea)f = △H Only one can be possible for Exothermic Reaction
(Ea)f > △H
(Ea)f < △H
Read More: Exothermic Reaction
(Ea)f > (Ea)b
(ROR)f < (ROR)b
△H = (Ea)b – (Ea)f
△H = +ve value
(Ea)f > △H (Always )…………universal
Read More: Endothermic Reaction
A catalyst is a chemical substance that either increases or decreases the rate of a chemical reaction. In the case of activation energy, a catalyst lowers it. However, the energies of the original reactants remain the same. A catalyst only alters the activation energy.
Types of Catalysts
A catalyst that helps to increase the rate of reaction or which support the reaction to carry out quickly is called a positive catalyst. Such catalyst decreases activation energy by accepting a smaller path, so the rate of reaction is increased.
Negative Catalysts (Inhibitors)
A catalyst that decreases or retards or helps in slowing down the rate of reaction is called a negative catalyst.
It is because a negative catalyst increases activation energy by taking a longer alternative path.
Points to Remember
Forward Reaction: Those reactions in which the product formed are produced from reactants
When reactant molecules collide with each other at its highest energy point, an intermediate is formed which remains in equilibrium with the main reactant. If this intermediate complex has an energy equal to or greater than the Threshold Energy, it will be converted into a product.
Activation Energy Formula
The formula used to find the value of Activation Energy, Ea is;
K = Ae-Ea/RT
K = Rate Constant
A = Arrhenius Constant
Ea = Activation Energy
R = Gas constant = 8.34J/K/mol
= 2 cal/K/mol
= 0.0821 lit atm/K/mol
K = Ae-Ea/RT
Taking log on both sides
ln K = ln A – (Ea /RT)ln e
2.303 log K = 2.303 log A – Ea/RT
log K = log A – Ea /2.303RT
Solved Examples Related to Activation Energy
1. The rate constant of a 1st order reaction increases from 3 × 10-2 to 8 × 10-2 when the temperature changes from 310K to 330K. Calculate the activation energy (Ea).
Given k2 = 8 × 10-2, k1 = 3 × 10-2, T1 = 310K, T2 = 330K
Ea = 41,613.62 J/mole
= 41.614 kJ/mol
2. The rate constant of the first-order reaction raises from 3 X 10-2 to 5×10-2 when the temperature changes from 300K to 310K. Calculate the activation energy.
Given k2 = 3 × 10-2, k1 = 5 × 10-2, T1 = 300K, T2 = 310K
Ea = 39,495.7 J/mole
= 39.495 kJ/mol
3. The first order reaction has rate constant of 2.0×10-2 and 6.0×10-2 at 00C and 300C. Calculate the activation energy of the reaction.
Given k2 = 6 × 10-2, k1 = 2 × 10-2, T1 = 273K, T2 = 303K
Ea = 25,188.74 J/mole
= 25.188 kJ/mol