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Difficulty: HardExothermic and Endothermic Reactions
Consider the gas-phase combustion of methane represented by the balanced equation:
CH4(g)+2O2(g)→CO2(g)+2H2O(g)\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(g)

Given the following average bond dissociation energies:
- C–H=414 kJ mol−1\text{C--H} = 414\text{ kJ mol}^{-1}
- O=O=498 kJ mol−1\text{O=O} = 498\text{ kJ mol}^{-1}
- C=O=803 kJ mol−1\text{C=O} = 803\text{ kJ mol}^{-1}
- O–H=464 kJ mol−1\text{O--H} = 464\text{ kJ mol}^{-1}

What is the overall enthalpy change (ΔH\Delta H) for this reaction, and how is the process classified thermodynamically?

  1. −810 kJ mol−1-810\text{ kJ mol}^{-1}, and the reaction is exothermicAnswer
  2. B
    +810 kJ mol−1+810\text{ kJ mol}^{-1}, and the reaction is endothermic
  3. C
    −810 kJ mol−1-810\text{ kJ mol}^{-1}, but the heat released is doubled upon adding a catalyst
  4. D
    −1308 kJ mol−1-1308\text{ kJ mol}^{-1}, and the reaction is exothermic

Answer

−810 kJ mol−1-810\text{ kJ mol}^{-1}, and the reaction is exothermic
The correct answer is obtained by summing the energies needed to break reactant bonds (4×C–H+2×O=O=2652 kJ4 \times \text{C--H} + 2 \times \text{O=O} = 2652\text{ kJ}) and subtracting the energy released from forming product bonds (2×C=O+4×O–H=3462 kJ2 \times \text{C=O} + 4 \times \text{O--H} = 3462\text{ kJ}). The result ΔH=−810 kJ mol−1\Delta H = -810\text{ kJ mol}^{-1} shows that heat is liberated to the surroundings, defining an exothermic reaction.

Step-by-Step Solution

1
Calculate the total energy required to break all reactant bonds.
Energy absorbed =4(C–H)+2(O=O)=4(414)+2(498)=1656+996=2652 kJ mol−1= 4(\text{C--H}) + 2(\text{O=O}) = 4(414) + 2(498) = 1656 + 996 = 2652\text{ kJ mol}^{-1}.
Bond breaking is an endothermic process requiring energy input.
2
Calculate the total energy released when forming all product bonds.
Energy released =2(C=O)+4(O–H)=2(803)+4(464)=1606+1856=3462 kJ mol−1= 2(\text{C=O}) + 4(\text{O--H}) = 2(803) + 4(464) = 1606 + 1856 = 3462\text{ kJ mol}^{-1}.
Bond formation is an exothermic process that releases energy.
3
Determine the net enthalpy change (ΔH\Delta H) and thermodynamic classification.
ΔH=Energy absorbed−Energy released=2652−3462=−810 kJ mol−1\Delta H = \text{Energy absorbed} - \text{Energy released} = 2652 - 3462 = -810\text{ kJ mol}^{-1}. Since ΔH<0\Delta H < 0, the reaction is exothermic.
A negative enthalpy change indicates that more energy is released in bond formation than absorbed during bond breaking.

Key Concept

Bond Energy and Enthalpy Change of Reaction
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Topics

Exothermic and Endothermic Reactions