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Using the balanced equation and the data in the table below, calculate the theoretical enthalpy of combustion. Note: you will n...Asked by Theory
Using the balanced equation and the data in the table below, calculate the theoretical enthalpy of combustion.
Note: you will need to include the enthalpy of vaporisation for the liquid components which are also given.
CH3OH(l) + 1.5O2(g) → CO2(g) + 2H2O(l)
Average Bond Enthalpies KJmol-1
C-H 412
C-C 348
C-O 358
O=O 496
C=O 743
O-H 463
Enthalpy of vaporisation KJmol-1
Methanol 35
Water 41
Note: you will need to include the enthalpy of vaporisation for the liquid components which are also given.
CH3OH(l) + 1.5O2(g) → CO2(g) + 2H2O(l)
Average Bond Enthalpies KJmol-1
C-H 412
C-C 348
C-O 358
O=O 496
C=O 743
O-H 463
Enthalpy of vaporisation KJmol-1
Methanol 35
Water 41
Answers
Answered by
Anonymous
This is a somewhat tedious problem. Tell me what you've done with it, what you don't understand and how I can help OTHER THAN WORKING THE PROBLEM COMPLETELY FOR YOU.
Answered by
Theory
a) I think this is the way to find the average bond enthalpy:
3 x 412 = 1236 kJ
1.5 x 496 = 744 kJ
1 x 463 = 463 kJ
358 kJ
1486 kJ
Formed 2 x 743 = 1486 kJ
4 x 463 = 1852 kJ
I did not want to do anything else before checking this was right.
The note says to include the enthalpy of vaporisation for the liquid components which are also given. I did not. I do not know what to do.
3 x 412 = 1236 kJ
1.5 x 496 = 744 kJ
1 x 463 = 463 kJ
358 kJ
1486 kJ
Formed 2 x 743 = 1486 kJ
4 x 463 = 1852 kJ
I did not want to do anything else before checking this was right.
The note says to include the enthalpy of vaporisation for the liquid components which are also given. I did not. I do not know what to do.
Answered by
DrBob222
See your other post above,
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