Chemistry

Chemistry

Problem

Combustion of hydrocarbons is an important source of thermal energy. Consider the combustion of ethane at 298 K298\,\text{K} and 1 atm1\,\text{atm}: C₂H₆(g) + 72\tfrac{7}{2} O₂(g) → 2 CO₂(g) + 3 H₂O(l) Selected standard enthalpies of formation, ΔH_f° (kJ mol⁻¹): • C₂H₆(g) = −84.7  • CO₂(g) = −393.5  • H₂O(l) = −285.8  • O₂(g) = 0.0 Use the data to analyze the thermochemistry of the reaction. (a) Without performing a full numerical calculation, predict the sign of ΔH° for the combustion of ethane. Cite specific features of the ΔH_f° data that support your prediction. (b) Describe, in molecular-level terms, why combustion reactions of hydrocarbons such as ethane generally release heat to the surroundings.
Intuitive explanation

Combustion reactions release energy, and you can find exactly how much using Hess's Law with standard enthalpies of formation. The key insight is that ΔHrxn°=∑ΔHf°(products)−∑ΔHf°(reactants)\Delta H_{\text{rxn}}° = \sum \Delta H_f°(\text{products}) - \sum \Delta H_f°(\text{reactants})—you're essentially calculating the energy difference between breaking apart the reactants and forming the products. Remember that ΔHf°\Delta H_f° for elements in their standard state (like O₂) is zero, so it doesn't contribute.

Intuitive explanation

Combustion reactions release energy, and you can find exactly how much using Hess's Law with standard enthalpies of formation. The key insight is that ΔHrxn°=∑ΔHf°(products)−∑ΔHf°(reactants)\Delta H_{\text{rxn}}° = \sum \Delta H_f°(\text{products}) - \sum \Delta H_f°(\text{reactants})—you're essentially calculating the energy difference between breaking apart the reactants and forming the products. Remember that ΔHf°\Delta H_f° for elements in their standard state (like O₂) is zero, so it doesn't contribute.

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