Propanol, C3H7OH, is an organic alcohol that is used to manufacture pesticides and
plays a role in the production of many pharmaceuticals. Consider the combustion reaction of propanol: propanol (l). + oxygen (g) → carbon dioxide (g) + water (l).
A) Determine the molar enthalpy of combustion using the bond energy method.
B) The published (accepted) enthalpy of the combustion of propanol is -2021 kJ/mol. Justify why the calculated enthalpy determined using the bond energy method differs.
C) Using the published (accepted) enthalpy value, create a thermochemical equation (including the energy value) to represent the combustion of propanol.
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2020-07-23 2:07 pm
回答 (1)
2020-07-23 3:53 pm
✔ 最佳答案
A)Refer to: http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html
E(C-H) = 411 kJ/mol
E(C-C) = 346 kJ/mol
E(C-O) = 358 kJ/mol
E(O-H) = 459 kJ/mol
E(O=O) = 494 kJ/mol
E(C=O) = 799 kJ/mol
(Bond energy values from different sources may be slightly differenet).
Refer to the figure below which shows the equation for the molar enthalpy combustion of propanol.
Energy absorbed for bond breaking when burning 1 mole of propanol
= 2 E(C-C) + 7E(C-H) + E(C-0) + E(O-H) + 4.5 E(O=O)
= 2(346) + 7(411) + 358 + 459 + 4.5(494) kJ
= 6609 kJ
Energy released for bond formation when burning 1 mole of propanol
= 3 × [2 E(C=O)] + 4 × [2 E(O-H)]
= 6(799) + 8(459) kJ
= 8466 kJ
Molar enthalpy of combustion of propanol using the bond energy method
= 6609 - 8466 kJ/mol
= -1857 kJ/mol
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B)
When using bond energy method, all the reactants and products involved should be gaseous. Now, propanol and water are liquid. The enthalpy of evaporation of propanol and the enthalpy of condensation of water are not considered. This makes the difference.
Also, all bond enthalpies are average values.
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C)
C₃H₇OH(ℓ) + 4.5O₂(g) → 3CO₂(g) + 4H₂O(ℓ) ΔH = -2021 kJ
收錄日期: 2021-05-01 09:37:58
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