Question

For the reaction, A⇌B, E_a = 50kJ mol^-1 and ΔH = -20 kJ mol^-1. When a catalyst is added, E_a decreases by 10kJmol^-1. What is the E_a for the backward reaction in the presence of catalyst?

• A. 60 kJ mol^-1
• B. 40 kJ mol^-1
• C. 70 kJ mol^-1
• D. 20 kJ mol^-1

Answer 1

The Correct Option is A

To calculate the activation energy (\(E_a\)) for the backward reaction in the presence of a catalyst, we use the following relationship:

\[ E_a^{\text{(forward)}} - E_a^{\text{(backward)}} = \Delta H \]

Where:

• \(E_a^{\text{(forward)}}\) is the activation energy for the forward reaction.
• \(E_a^{\text{(backward)}}\) is the activation energy for the backward reaction.
• \(\Delta H\) is the enthalpy change for the reaction, given as \(-20 \, \text{kJ/mol}\).

Step 1: Calculate \(E_a^{\text{(backward)}}\) without the catalyst:

Using the equation:

\[ E_a^{\text{(backward)}} = E_a^{\text{(forward)}} - \Delta H \]

Substituting the given values:

\[ E_a^{\text{(backward)}} = 50 \, \text{kJ/mol} - (-20 \, \text{kJ/mol}) \]

\[ E_a^{\text{(backward)}} = 50 \, \text{kJ/mol} + 20 \, \text{kJ/mol} = 70 \, \text{kJ/mol} \]

Step 2: Consider the effect of the catalyst:

The catalyst decreases \(E_a^{\text{(forward)}}\) by 10 kJ/mol, so the new activation energy for the forward reaction is:

\[ E_a^{\text{(forward)}} = 50 \, \text{kJ/mol} - 10 \, \text{kJ/mol} = 40 \, \text{kJ/mol} \]

Since the difference between the forward and backward activation energies remains the same (\(\Delta H = -20 \, \text{kJ/mol}\)), the activation energy for the backward reaction with the catalyst is:

\[ E_a^{\text{(backward)}} = 40 \, \text{kJ/mol} + 20 \, \text{kJ/mol} = 60 \, \text{kJ/mol} \]

So, the activation energy for the backward reaction in the presence of the catalyst is 60 kJ/mol, so the correct answer is (A) 60 kJ/mol.