Calculate ΔHrxn for the following reaction:
5C(s)+6H2(g)→C5H12(l)
Use the following reactions and given ΔH′s.
C5H12(l)+8O2(g)→5CO2(g)+6H2O(g), ΔH= -3244.8 kJ
C(s)+O2(g)→CO2(g), ΔH= -393.5 kJ
2H2(g)+O2(g)→2H2O(g), ΔH= -483.5 kJ
Calculate ΔHrxn for the following reaction:?
2016-05-29 5:41 pm
回答 (1)
2016-05-29 5:56 pm
The three given thermochemical equations can be rewritten as:
5C(s) + 5O₂(g) → 5CO₂(g) ...... ΔH = 5(-393.5 kJ)
6H₂(g) + 3O₂(g) → 6H₂O(g) ...... ΔH = 3(-483.5 kJ)
5CO₂(g) + 6H₂O(g) → C₅H₁₂(l) + 8O₂(g) ...... ΔH = -(-3244.8 kJ)
Add the above three thermochemical equations, and cancel 8O₂(g), 5CO₂(g) and 6H₂O(g) on the both sides. We get
5C(s) + 6H₂(g) →C₅H₁₂(l) ...... ΔH = ΔHrxn
ΔHrxn = 5(-393.5 kJ) + 3(-483.5 kJ) - (-3244.8 kJ) = -173.2 kJ
5C(s) + 5O₂(g) → 5CO₂(g) ...... ΔH = 5(-393.5 kJ)
6H₂(g) + 3O₂(g) → 6H₂O(g) ...... ΔH = 3(-483.5 kJ)
5CO₂(g) + 6H₂O(g) → C₅H₁₂(l) + 8O₂(g) ...... ΔH = -(-3244.8 kJ)
Add the above three thermochemical equations, and cancel 8O₂(g), 5CO₂(g) and 6H₂O(g) on the both sides. We get
5C(s) + 6H₂(g) →C₅H₁₂(l) ...... ΔH = ΔHrxn
ΔHrxn = 5(-393.5 kJ) + 3(-483.5 kJ) - (-3244.8 kJ) = -173.2 kJ
收錄日期: 2021-04-18 14:54:06
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