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Difficulty: MediumActivation Energy and Energy Profile Diagrams

An uncatalyzed endothermic reaction has a forward activation energy of 120 kJ mol−1120\text{ kJ mol}^{-1} and an overall enthalpy change (ΔH\Delta H) of +50 kJ mol−1+50\text{ kJ mol}^{-1}. In the presence of a positive catalyst, the activation energy of the forward reaction is reduced by 40 kJ mol−140\text{ kJ mol}^{-1}. What is the activation energy for the reverse reaction in the presence of the catalyst?

  1. 30 kJ mol−130\text{ kJ mol}^{-1}Answer
  2. B
    70 kJ mol−170\text{ kJ mol}^{-1}
  3. C
    110 kJ mol−1110\text{ kJ mol}^{-1}
  4. D
    130 kJ mol−1130\text{ kJ mol}^{-1}

Answer

The activation energy for the catalyzed reverse reaction is 30 kJ mol−130\text{ kJ mol}^{-1}.
For an endothermic reaction, the activation energy of the reverse reaction is equal to the forward activation energy minus the enthalpy change (Ea,reverse=Ea,forward−ΔHE_{a,\text{reverse}} = E_{a,\text{forward}} - \Delta H). First, determine the forward activation energy under catalyzed conditions: 120 kJ mol−1−40 kJ mol−1=80 kJ mol−1120\text{ kJ mol}^{-1} - 40\text{ kJ mol}^{-1} = 80\text{ kJ mol}^{-1}. Then, subtract the enthalpy change: 80 kJ mol−1−50 kJ mol−1=30 kJ mol−180\text{ kJ mol}^{-1} - 50\text{ kJ mol}^{-1} = 30\text{ kJ mol}^{-1}.

Step-by-Step Solution

1
Calculate the forward activation energy of the catalyzed reaction.
Ea,forward, catalyzed=120 kJ mol−1−40 kJ mol−1=80 kJ mol−1E_{a,\text{forward, catalyzed}} = 120\text{ kJ mol}^{-1} - 40\text{ kJ mol}^{-1} = 80\text{ kJ mol}^{-1}.
A positive catalyst lowers the activation energy barrier.
2
Relate forward activation energy, reverse activation energy, and enthalpy change.
Ea,reverse=Ea,forward−ΔHE_{a,\text{reverse}} = E_{a,\text{forward}} - \Delta H.
For an endothermic reaction, the energy of products is higher than reactants by ΔH\Delta H.
3
Substitute the catalyzed forward activation energy and enthalpy change into the relation.
Ea,reverse, catalyzed=80 kJ mol−1−50 kJ mol−1=30 kJ mol−1E_{a,\text{reverse, catalyzed}} = 80\text{ kJ mol}^{-1} - 50\text{ kJ mol}^{-1} = 30\text{ kJ mol}^{-1}.
Subtracting ΔH\Delta H from the catalyzed forward activation energy gives the required reverse activation energy.

Key Concept

Activation energy relationship for forward and reverse reactions with catalysts
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Topics

Activation Energy and Energy Profile Diagrams