Case Study Question 2 on Chemical Kinetics – Chapter 4 CBSE Class 12 Chemistry

Read the given passage and answer the questions that follow.

Question. Temperature has a marked effect on the rate of reaction. For most of the reactions,the rate of reaction becomes nearly double for every 10 degree rise in temperature. The effect of temperature is usually expressed in terms of temperature coefficient. The quantitative dependence of reaction rate on temperature was first explained by Swante Arrhenius. Arrhenius proposed a simple equation known as Arrhenius equation

k = Ae^–Ea/RT

This equation provides a relationship between the rate constant (k) of a reaction and the temperature of the system. A is the Arrhenius factor or pre-exponential factor, Ea is the activation energy and R is the gas constant.

1. Define ‘activation energy’ of a reaction. [CBSE (AI) 2011]

Ans. The energy required to form the intermediate called activated complex is known as activation energy. Activation energy = Threshold energy – Average energy of the reactants

2. How does a catalyst affect the rate of a reaction? Explain with respect to the Arrhenius equation.

Ans. A catalyst decreases the activation energy. According to Arrhenius equation, lower the activation energy, greater will be the rate constant and thus the rate of reaction increases.

3. Can a reaction have zero activation energy? Justify.

Ans. No, Ea ≠ 0.

If Ea = 0, then according to Arrhenius equation,

k = Ae^–Ea/RT

k = Ae⁰ = A,

i.e., Rate constant = Collision frequency

This means every collision results into a chemical reaction which cannot be true.

4. The plot of log k vs X is linear with slope = – E_a/2.303R. What is X?

Ans. 1/T

5. What is the fraction of molecules having energy greater than activation energy, E_a?

Ans. e ^–Ea/RT at temperature T.