Question

For the reaction, A ⇌ B, E_a = 50 kJ mol⁻¹ and ΔH = −20 kJ mol⁻¹. When a catalyst is added, E_a decreases by 10 kJ mol⁻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

Hint:

The backward activation energy is calculated using E a,b=Ea,f−ΔH.

Answer 1

The Correct Option is A

To solve the problem of determining the activation energy for the backward reaction in the presence of a catalyst, let's follow these steps:

1. Initially, we have the activation energy (\(E_a\)) for the forward reaction, which is 50 kJ mol⁻¹. The enthalpy change (\(\Delta H\)) for the reaction is -20 kJ mol⁻¹.
2. From the relationship between \(E_a\), \(\Delta H\), and the activation energy for the backward reaction (\(E_{a,\text{backward}}\)), we use the equation: 
   
   \(E_{a,\text{backward}} = E_a + |\Delta H|\)  
   
   Since \(\Delta H\) is negative, we take its absolute value.
3. Substituting the given values: 
   
   \(E_{a,\text{backward}} = 50 \, \text{kJ mol}^{-1} + |{-20 \, \text{kJ mol}^{-1}}| = 50 \, \text{kJ mol}^{-1} + 20 \, \text{kJ mol}^{-1} = 70 \, \text{kJ mol}^{-1}\)
4. When a catalyst is added, it reduces the activation energy of both the forward and backward reactions by the same amount. Here, the reduction is 10 kJ mol⁻¹.
5. Therefore, the activation energy for the backward reaction in the presence of the catalyst is: 
   
   \(E_{a,\text{backward, catalyst}} = 70 \, \text{kJ mol}^{-1} - 10 \, \text{kJ mol}^{-1} = 60 \, \text{kJ mol}^{-1}\)

Hence, the correct answer is 60 kJ mol⁻¹. This means that when the catalyst is added, it decreases the activation energy for the backward reaction to 60 kJ mol⁻¹. This is the correct option among the given choices.