更新1:
694 + 1486 +3304 - 2475 - 5944 = -1721 I got it now, I was adding instead of subtracting
更新2:
nevermind the under part
更新3:
The correct answer is -1339
CH3(C=O)CH3(g) + 4 O2(g) → 3 CO2(g) + 3 H2O(g) I need help determine the approximate enthalpy (in kj) please help?
回答 (1)
2017-11-01 11:08 am
Refer to: http://www.kentchemistry.com/links/Kinetics/BondEnergy.htm
The average bond energy in kJ/mol:
C-H: 413
C-C: 348
O-H: 463
C=O: 799
O=O: 495
Energy needed to break the bonds of CH₃COCH₃ = (413×6 + 348×2 + 799) kJ = 3973 kJ
Energy needed to break the bonds of O₂ = 495 × 4 kJ = 1980 kJ
Energy released in forming the bonds of CO₂ = 799 × 2 × 3 kJ = 4794 kJ
Energy released in forming the bonds of H₂O = 463 × 2 × 3 kJ = 2778 kJ
Approximate ΔH = (3973 + 1980 - 4794 - 2778) kJ = -1619 kJ
The average bond energy in kJ/mol:
C-H: 413
C-C: 348
O-H: 463
C=O: 799
O=O: 495
Energy needed to break the bonds of CH₃COCH₃ = (413×6 + 348×2 + 799) kJ = 3973 kJ
Energy needed to break the bonds of O₂ = 495 × 4 kJ = 1980 kJ
Energy released in forming the bonds of CO₂ = 799 × 2 × 3 kJ = 4794 kJ
Energy released in forming the bonds of H₂O = 463 × 2 × 3 kJ = 2778 kJ
Approximate ΔH = (3973 + 1980 - 4794 - 2778) kJ = -1619 kJ
收錄日期: 2021-04-18 18:00:37
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