Calculate ΔHrxn for the following reaction: 5C(s)+6H2(g)→C5H12(l) Use the following reactions and given ΔH values: C5H12(l)+8O2(g)→5CO2(g)+6H2O(g),ΔHC (s)+O2(g)→CO2(g),ΔH2H2(g)+O2(g)→2H2 O(g),ΔH===−3244.8kJ−393.5kJ−483.5kJ

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

the answer is the

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