Using the equations below, determine the molar enthalpy change (ΔH°) for the combustion of octane (C₈H₁₈) in kJ/mol. 1. 2 C₈H₁₈ (l) + 15 O₂ (g) → 12 CO₂ (g) + 6 H₂O (g) ΔH° = -6271 kJ/mol2. 2 H₂ (g) + O₂ (g) → 2 H₂O (g) ΔH° = -483.6 kJ/mol3. C (s) + O₂ (g) → CO₂ (g) ΔH° = -393.5 kJ/mol

• A. -5471 kJ/mol
• B. -2418 kJ/mol
• C. -6271 kJ/mol
• D. -3135.5 kJ/mol

Answer: (D)

Verified step by step guidance

1. Identify the target reaction for which you need to find the molar enthalpy change (ΔH°). In this case, it is the combustion of octane: 2 C₈H₁₈ (l) + 15 O₂ (g) → 12 CO₂ (g) + 6 H₂O (g).
2. Recognize that the given ΔH° value for the combustion of octane is for 2 moles of C₈H₁₈. To find the molar enthalpy change, you need to divide the given ΔH° by 2, since the reaction involves 2 moles of octane.
3. Understand that the enthalpy change for a reaction can be calculated using Hess's Law, which states that the total enthalpy change is the sum of the enthalpy changes for each step of the reaction. However, in this problem, the enthalpy change for the combustion of octane is directly given.
4. Use the given enthalpy change for the combustion of octane: ΔH° = -6271 kJ/mol for 2 moles of octane. Divide this value by 2 to find the molar enthalpy change for 1 mole of octane.
5. Express the final step mathematically: ΔH° (molar) = \( \frac{-6271 \text{ kJ/mol}}{2} \). This will give you the molar enthalpy change for the combustion of octane in kJ/mol.