The combustion of heptane, C7H16, occurs via the reaction
C7H16(g)+1102(g)→7CO2(g) + 8H2O(g)
with heat of formation values given by the following table:
Calculate the enthalpy for the combustion of 1 mole of heptane.
The Correct Answer and Explanation is :
To calculate the enthalpy change (ΔH) for the combustion of 1 mole of heptane (C₇H₁₆), we use the standard enthalpies of formation (ΔH_f°) of the reactants and products. The balanced chemical equation for the combustion is:
C₇H₁₆(g) + 11O₂(g) → 7CO₂(g) + 8H₂O(g)
The standard enthalpy change of the reaction can be determined using Hess’s law, which states:
ΔH°_reaction = Σ(ΔH_f° of products) – Σ(ΔH_f° of reactants)
Given the standard enthalpies of formation:
- ΔH_f°[C₇H₁₆(g)] = -187.9 kJ/mol
- ΔH_f°[CO₂(g)] = -393.5 kJ/mol
- ΔH_f°[H₂O(g)] = -241.8 kJ/mol
- ΔH_f°[O₂(g)] = 0 kJ/mol (standard state)
Calculating the total enthalpy of the products:
- 7 moles of CO₂: 7 × (-393.5 kJ/mol) = -2,754.5 kJ
- 8 moles of H₂O: 8 × (-241.8 kJ/mol) = -1,934.4 kJ
Total enthalpy of products = -2,754.5 kJ + -1,934.4 kJ = -4,688.9 kJ
Calculating the total enthalpy of the reactants:
- 1 mole of C₇H₁₆: 1 × (-187.9 kJ/mol) = -187.9 kJ
- 11 moles of O₂: 11 × 0 kJ/mol = 0 kJ
Total enthalpy of reactants = -187.9 kJ
Applying Hess’s law:
ΔH°_reaction = (-4,688.9 kJ) – (-187.9 kJ) = -4,501 kJ
Therefore, the enthalpy change for the combustion of 1 mole of heptane is -4,501 kJ.
This negative value indicates that the combustion of heptane is an exothermic process, releasing 4,501 kJ of energy per mole. Combustion reactions of hydrocarbons like heptane are highly exothermic due to the formation of strong bonds in the products (CO₂ and H₂O) compared to the reactants. The energy released during such reactions is harnessed in various applications, including internal combustion engines and heating systems.
It’s important to note that the enthalpy change calculated here assumes standard conditions (298 K and 1 atm pressure) and that all reactants and products are in their standard states. Variations in temperature, pressure, or physical states can affect the enthalpy change of the reaction.
For a visual representation of the balanced combustion equation and further explanation, you may refer to the following video: