Answer:
ΔHcomb = -3849 kJ/mol
Step-by-step explanation:
For the alkane C₆H₁₄ the balance reaction of combustion is:
C₆H₁₄+ ¹⁹/₂ O₂(g) → 6CO₂(g) + 7H₂O(g)
It is possible to obtain ΔHcomb for this reaction from the standard enthalpies of formation using:
ΔHcomb = ΔH°products - ΔH°reactants.
For the combustion of the alkane C₆H₁₄:
ΔHcomb = 6ΔH°CO₂ + 7ΔH°H₂O - (ΔH°C₆H₁₄ + ¹⁹/₂ΔH°O₂)
Where:
ΔH°CO₂: -393,5 kJ/mol
ΔH°H₂O: -241,8 kJ/mol
ΔH°C₆H₁₄: -204,6 kJ/mol
ΔH°O₂: 0 kJ/mol
Replacing:
ΔHcomb = 6×-393,5 kJ/mol + 7×-241,8 kJ/mol - (-204,6 kJ/mol + ¹⁹/₂×0 kJ/mol)
ΔHcomb = -3849 kJ/mol
I hope it helps!