Answer:
ΔHrxn = -173.2 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 = -3244.8kJ (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 = + 3244.8 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 = + 3244.8 - 1967.5 - 1450.5
ΔHrxn = -173.2 kJ
Explanation:
Using Hess' Law,
Overall reaction:
5C(s) + 6H₂(g) → C₅H₁₂(l)
I. C₅H₁₂(l) + 8O₂(g) → 5CO₂(g) + 6H₂O(g)
ΔH = −3244.8kJ
II. C(s) + O₂(g) → CO₂(g)
ΔH = -393.5 kJ
III. 2H₂(g) + O₂(g) → 2H₂O(g)
ΔH = -483.5 kJ
Multiplying, equation II by 5 and equation III by 3 and their enthalpies respectively. Also, invert equation I and its enthalpy changes. We get:
IV. 5CO₂(g) + 6H₂O(g) → C₅H₁₂(l) + 8O₂(g)
ΔH = +3244.8kJ
V. 5C(s) + 5O₂(g) → 5CO₂(g) ΔH = -1967.5 kJ
VI. 6H₂(g) + 3O₂(g) → 6H₂O(g) ΔH = -1450.5 kJ
Adding up equations IV, V and VI as well as their enthalpies. We get:
5C(s) + 6H₂(g) → C₅H₁₂(l)
ΔHrxn = + 3244.8 - 1967.5 - 1450.5
ΔHrxn = -173.2 kJ
