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Difficulty: HardActivation Energy and Energy Profile Diagrams

Consider the exothermic reaction 2NO2(g)⇌N2O4(g)2\text{NO}_2\text{(g)} \rightleftharpoons \text{N}_2\text{O}_4\text{(g)}, where the potential energy of the reactants is 180 kJ mol−1180\text{ kJ mol}^{-1} and that of the products is 70 kJ mol−170\text{ kJ mol}^{-1}. If the uncatalyzed forward activation energy is 90 kJ mol−190\text{ kJ mol}^{-1} and a catalyst lowers the activation energy of the forward reaction by 35 kJ mol−135\text{ kJ mol}^{-1}, what is the activation energy for the catalyzed reverse reaction?

  1. 165 kJ mol−1165\text{ kJ mol}^{-1}Answer
  2. B
    200 kJ mol−1200\text{ kJ mol}^{-1}
  3. C
    55 kJ mol−155\text{ kJ mol}^{-1}
  4. D
    145 kJ mol−1145\text{ kJ mol}^{-1}

Answer

The activation energy for the catalyzed reverse reaction is 165 kJ mol−1165\text{ kJ mol}^{-1}.
A catalyst lowers the energy barrier peak for both forward and reverse paths by the same magnitude (35 kJ mol−135\text{ kJ mol}^{-1}). Starting with reactants at 180 kJ mol−1180\text{ kJ mol}^{-1} and a catalyzed forward activation energy of 55 kJ mol−155\text{ kJ mol}^{-1}, the transition state peak is located at 235 kJ mol−1235\text{ kJ mol}^{-1}. Since the products lie at 70 kJ mol−170\text{ kJ mol}^{-1}, the energy required for the reverse reaction to reach the transition state peak is 235−70=165 kJ mol−1235 - 70 = 165\text{ kJ mol}^{-1}.

Step-by-Step Solution

1
Calculate the uncatalyzed transition state peak energy
Epeak, uncatalyzed=180 kJ mol−1+90 kJ mol−1=270 kJ mol−1E_{\text{peak, uncatalyzed}} = 180\text{ kJ mol}^{-1} + 90\text{ kJ mol}^{-1} = 270\text{ kJ mol}^{-1}
The peak energy equals the reactant energy plus the forward activation energy.
2
Calculate the catalyzed transition state peak energy
Epeak, catalyzed=270 kJ mol−1−35 kJ mol−1=235 kJ mol−1E_{\text{peak, catalyzed}} = 270\text{ kJ mol}^{-1} - 35\text{ kJ mol}^{-1} = 235\text{ kJ mol}^{-1}
A catalyst lowers the transition state energy barrier by 35 kJ mol−135\text{ kJ mol}^{-1}.
3
Determine the catalyzed reverse activation energy
Ea,reverse, catalyzed=235 kJ mol−1−70 kJ mol−1=165 kJ mol−1E_{a, \text{reverse, catalyzed}} = 235\text{ kJ mol}^{-1} - 70\text{ kJ mol}^{-1} = 165\text{ kJ mol}^{-1}
The reverse activation energy is the difference between the catalyzed peak energy and the energy level of the products.

Key Concept

Activation Energy and Catalyzed Energy Profiles
Estimated Time:2m 0s
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Topics

Activation Energy and Energy Profile Diagrams