Calculate ΔHrxn for the reaction: 5C(s)+6H2(g)→C5H12(l) given these reactions and their ΔH′s: C5H12(l)+8O2(g)→5CO2(g)+6H2O(g)ΔH=−3271.5kJ C(s)+O2(g)→CO2(g)ΔH=−393.5kJ 2H2(g)+O2(g)→2H2O(g)ΔH=−483.5kJ

By Hess' Law, when the reaction is summed, the enthalpy of the global reaction will be the sum of the enthalpy of the steps reaction. If the reaction is multiplied by a number, the enthalpy will be multiplied by it too, and if the reaction is inverted, the enthalpy will have it's signal changed.

C₅H₁₂(l) + 8O₂(g) → 5CO₂(g) + 6H₂O(g) ΔH = -3271.5 kJ (Must be inverted)

C(s) + O₂(g) → CO₂(g) ΔH = -393.5 kJ (x5)

2H₂(g) + O₂(g) → 2H₂O(g) ΔH = -483.5 kJ (x3)

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5CO₂(g) + 6H₂O(g) → C₅H₁₂(l) + 8O₂(g) ΔH = + 3271.5 kJ

5C(s) + 5O₂(g) → 5CO₂(g) ΔH = -1967.5 kJ

6H₂(g) + 3O₂(g) → 6H₂O(g) ΔH = -1450.5 kJ

The bold substances have the same amount on both sides of the reaction, so they will be eliminated in the global reaction.

5C(s) + 6H₂(g) → C₅H₁₂(l)

ΔHrxn = + 3271.5 - 1967.5 - 1450.5

ΔHrxn = -146.5 kJ