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

-146.5 kJ

Explanation:

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)

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

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