Question

The enthalpy of combustion for octane (C8H18(l)), a key component of gasoline, is -5074 kJ/mol. The reaction equation is: C8H18(l) + 12.5O₂(g) → 8CO₂(g) + 9H₂O(g). What is the enthalpy change for this reaction?

1) -16CO₂(g) - 18H₂O(g) + 2C8H18(l) + 25O₂(g)
2) -8CO₂(g) - 9H₂O(g) + C8H18(l) + 12.5O₂(g)
3) -8CO₂(g) - 9H₂O(g) + 2C8H18(l) + 25O₂(g)
4) -16CO₂(g) - 18H₂O(g) + C8H18(l) + 12.5O₂(g)

Answer 1

The correct enthalpy change for the reaction in question is option 3, which reflects the doubled and reversed stoichiometry of the provided combustion reaction for octane, leading to an enthalpy change of -10148 kJ/mol.

Step-by-step explanation:

The enthalpy change for the combustion of octane (C8H18(l)) is given as -5074 kJ/mol for the following balanced chemical equation:

C8H18(l) + 12.5O2(g) → 8CO2(g) + 9H2O(g)

When examining the provided reaction options, it’s important to identify the one that reflects the correct stoichiometry when comparing it with the given enthalpy of combustion. Using stoichiometric analysis, if we reverse and double the given reaction, we would form:

2C8H18(l) + 25O2(g) → 16CO2(g) + 18H2O(g)

This is because when you reverse the reaction, the sign of the enthalpy change also reverses, and when you double the amounts of reactants and products, the enthalpy change doubles. Therefore, if the enthalpy change for the initial reaction is -5074 kJ/mol, for the doubled and reversed reaction, the enthalpy change would be +2(-5074 kJ/mol) = -10148 kJ/mol, which is the doubled amount because we have two moles of C8H18. Hence, the correct answer is the one which matches this stoichiometry which is option 3: -8CO2(g) - 9H2O(g) + 2C8H18(l) + 25O2(g).