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Difficulty: Very hardActivation Energy and Energy Profile Diagrams

For a reversible gas-phase reaction X(g)+Y(g)⇌Z(g)X_{(g)} + Y_{(g)} \rightleftharpoons Z_{(g)}, the total enthalpy of the reactants is +120 kJ mol−1+120\text{ kJ mol}^{-1}. The reaction is exothermic with a standard enthalpy change (ΔH\Delta H) of −45 kJ mol−1-45\text{ kJ mol}^{-1}. In the absence of a catalyst, the activation energy for the reverse reaction (Ea,revE_{a,\text{rev}}) is +185 kJ mol−1+185\text{ kJ mol}^{-1}. If a catalyst is introduced that lowers the activation energy barrier by 30 kJ mol−130\text{ kJ mol}^{-1}, what is the activation energy for the catalyzed forward reaction?

  1. +110 kJ mol−1+110\text{ kJ mol}^{-1}Answer
  2. B
    +140 kJ mol−1+140\text{ kJ mol}^{-1}
  3. C
    +155 kJ mol−1+155\text{ kJ mol}^{-1}
  4. D
    +200 kJ mol−1+200\text{ kJ mol}^{-1}

Answer

+110 kJ mol−1+110\text{ kJ mol}^{-1}
The correct answer is +110 kJ mol−1+110\text{ kJ mol}^{-1}. For an exothermic reaction with ΔH=−45 kJ mol−1\Delta H = -45\text{ kJ mol}^{-1} and reverse activation energy Ea,rev=+185 kJ mol−1E_{a,\text{rev}} = +185\text{ kJ mol}^{-1}, the uncatalyzed forward activation energy is Ea,fwd=Ea,rev+ΔH=185−45=+140 kJ mol−1E_{a,\text{fwd}} = E_{a,\text{rev}} + \Delta H = 185 - 45 = +140\text{ kJ mol}^{-1}. Adding a catalyst lowers this activation energy barrier by 30 kJ mol−130\text{ kJ mol}^{-1}, yielding 140−30=+110 kJ mol−1140 - 30 = +110\text{ kJ mol}^{-1}.

Step-by-Step Solution

1
Calculate the uncatalyzed forward activation energy (Ea,fwd, uncatalyzedE_{a,\text{fwd, uncatalyzed}})
Ea,fwd, uncatalyzed=Ea,rev+ΔH=+185 kJ mol−1+(−45 kJ mol−1)=+140 kJ mol−1E_{a,\text{fwd, uncatalyzed}} = E_{a,\text{rev}} + \Delta H = +185\text{ kJ mol}^{-1} + (-45\text{ kJ mol}^{-1}) = +140\text{ kJ mol}^{-1}
For any chemical system, the relationship between forward activation energy, reverse activation energy, and enthalpy change is ΔH=Ea,fwd−Ea,rev\Delta H = E_{a,\text{fwd}} - E_{a,\text{rev}}.
2
Apply the catalyst reduction to the forward activation energy
Ea,fwd, catalyzed=Ea,fwd, uncatalyzed−30 kJ mol−1=140−30=+110 kJ mol−1E_{a,\text{fwd, catalyzed}} = E_{a,\text{fwd, uncatalyzed}} - 30\text{ kJ mol}^{-1} = 140 - 30 = +110\text{ kJ mol}^{-1}
A catalyst lowers both forward and reverse activation energies by equal amounts, reducing the energy barrier height.

Key Concept

Relationship between forward activation energy, reverse activation energy, enthalpy change, and catalyst effect in energy profile diagrams

Alternative Method

Calculate the energy levels directly: Reactants = +120 kJ mol−1+120\text{ kJ mol}^{-1}. Products = 120+(−45)=+75 kJ mol−1120 + (-45) = +75\text{ kJ mol}^{-1}. Uncatalyzed Transition State = Products + Ea,rev=75+185=+260 kJ mol−1E_{a,\text{rev}} = 75 + 185 = +260\text{ kJ mol}^{-1}. Catalyzed Transition State = 260−30=+230 kJ mol−1260 - 30 = +230\text{ kJ mol}^{-1}. Catalyzed Ea,fwd=Catalyzed Transition State−Reactants=230−120=+110 kJ mol−1E_{a,\text{fwd}} = \text{Catalyzed Transition State} - \text{Reactants} = 230 - 120 = +110\text{ kJ mol}^{-1}.
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Topics

Activation Energy and Energy Profile Diagrams