Question

For the reaction, \( A \rightleftharpoons B \), \( E_a = 50 \, \text{kJ/mol} \) and \( \Delta H = -20 \, \text{kJ/mol} \). When a catalyst is added, \( E_a \) decreases by \( 10 \, \text{kJ/mol} \). What is the \( E_a \) for the backward reaction in the presence of the 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.

Answer 2

For a reaction \( A \rightleftharpoons B \), the activation energy for the forward reaction is \( E_a = 50 \, \text{kJ/mol} \). The enthalpy change \( \Delta H \) for the reaction is \( -20 \, \text{kJ/mol} \), meaning the backward reaction has an enthalpy change of \( +20 \, \text{kJ/mol} \). The activation energy for the backward reaction without the catalyst is given by: \[ E_a^{\text{backward}} = E_a^{\text{forward}} + \Delta H \] \[ E_a^{\text{backward}} = 50 + 20 = 70 \, \text{kJ/mol} \] When a catalyst is added, the activation energy decreases by 10 kJ/mol, so the activation energy for the backward reaction with the catalyst becomes: \[ E_a^{\text{backward (catalyst)}} = 70 - 10 = 60 \, \text{kJ/mol} \] Thus, the activation energy for the backward reaction in the presence of a catalyst is 60 kJ/mol.