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Difficulty: HardActivation Energy and Energy Profile Diagrams

An exothermic reversible reaction P(g)+Q(g)⇌R(g)P_{(g)} + Q_{(g)} \rightleftharpoons R_{(g)} has an enthalpy change (ΔH\Delta H) of −40 kJ mol−1-40\text{ kJ mol}^{-1} and an uncatalyzed forward activation energy of 65 kJ mol−165\text{ kJ mol}^{-1}. In the presence of a catalyst, the activation energy for the forward reaction is lowered by 25 kJ mol−125\text{ kJ mol}^{-1}. What is the activation energy for the reverse catalyzed reaction?

  1. A
    50 kJ mol−150\text{ kJ mol}^{-1}
  2. 80 kJ mol−180\text{ kJ mol}^{-1}Answer
  3. C
    105 kJ mol−1105\text{ kJ mol}^{-1}
  4. D
    40 kJ mol−140\text{ kJ mol}^{-1}

Answer

The activation energy for the reverse catalyzed reaction is 80 kJ mol−180\text{ kJ mol}^{-1}.
In an exothermic reaction with ΔH=−40 kJ mol−1\Delta H = -40\text{ kJ mol}^{-1}, the products reside at a lower energy level than the reactants. The catalyzed forward activation energy is 65−25=40 kJ mol−165 - 25 = 40\text{ kJ mol}^{-1}. To go from products back to the activated complex, reactants must gain the 40 kJ mol−140\text{ kJ mol}^{-1} lost during the forward reaction plus the 40 kJ mol−140\text{ kJ mol}^{-1} required to reach the transition state. Thus, the reverse catalyzed activation energy is 40−(−40)=80 kJ mol−140 - (-40) = 80\text{ kJ mol}^{-1}.

Step-by-Step Solution

1
Calculate the forward catalyzed activation energy
Ea,f(cat)=65 kJ mol−1−25 kJ mol−1=40 kJ mol−1E_{a,\text{f(cat)}} = 65\text{ kJ mol}^{-1} - 25\text{ kJ mol}^{-1} = 40\text{ kJ mol}^{-1}
A catalyst lowers the forward activation energy by the given reduction amount.
2
Relate reverse activation energy to forward activation energy and reaction enthalpy
Ea,r=Ea,f−ΔHE_{a,\text{r}} = E_{a,\text{f}} - \Delta H
For any reaction step, the energy barrier in the reverse direction equals the forward energy barrier minus the enthalpy change.
3
Calculate the reverse catalyzed activation energy using the catalyzed forward value and enthalpy change
Ea,r(cat)=40 kJ mol−1−(−40 kJ mol−1)=80 kJ mol−1E_{a,\text{r(cat)}} = 40\text{ kJ mol}^{-1} - (-40\text{ kJ mol}^{-1}) = 80\text{ kJ mol}^{-1}
Substituting Ea,f(cat)=40 kJ mol−1E_{a,\text{f(cat)}} = 40\text{ kJ mol}^{-1} and ΔH=−40 kJ mol−1\Delta H = -40\text{ kJ mol}^{-1} yields the energy required to convert products back to the transition state under catalysis.

Key Concept

Relationship between forward activation energy, reverse activation energy, enthalpy change, and catalytic lowering in reaction profile diagrams
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Topics

Activation Energy and Energy Profile Diagrams