Enthalpy of combustion of ethanol (1 Viewer)

[Chimaera]

i know its usually pretty simple but i always get confused when working out bond enthalpies. would really appreicate some help in this one sum i've got to do from a worksheet:

Q. Use the mean bond enthalpy data to calculate the enthalpy of combustion of ethanol, CH3CH2OH
mean bond enthalpies:
C-H : 412
C-C : 348
O=O : 496
C=O : 743
C-O : 360
O-H : 463

 

[Casmira]

Shit ... I've never seen a question like that before. Sorry , do you go to a selective school or something? This seems bit advanced for the HSC syllabus.

 

[insert-username]

To do this question, add up the energy used to break all the broken bonds, then subtract the energy gained from creating the new bonds. So:


CH₃CH₂OH (l) + 3O₂ (g) -> 2CO₂ (g) + 3H₂O (l)

Bonds broken:

5 x C-H -> 2060 kJ
3 x O=O -> 1488 kJ
1 x C-C -> 348 kJ
1 x C-O -> 360 kJ
1 x O-H -> 463 kJ

Total: 4719 kJ


Bonds formed:

4 x C=O -> 2972 kJ
6 x O-H -> 2778 kJ

Total: 5750 kJ


Total enthalpy change = (energy lost) - (energy gained)

= 4719 - 5750

= -1031 kJ/mol


The reason this differs slightly from the "real" molar heat of combustion of ethanol (1360 kJ/mol) is due to the use of mean bond enthalpies. Just remember: (energy of total bonds broken) - (energy of total bonds formed).


I_F

 

[Riviet]

I don't think this type of question is tested in the hsc.

 

[Chimaera]

thanks for the help!

 

[insert-username]

No problemo.


I_F

 

[bmc]

i know its usually pretty simple but i always get confused when working out bond enthalpies. would really appreicate some help in this one sum i've got to do from a worksheet:

Q. Use the mean bond enthalpy data to calculate the enthalpy of combustion of ethanol, CH3CH2OH
mean bond enthalpies:
C-H : 412
C-C : 348
O=O : 496
C=O : 743
C-O : 360
O-H : 463

yes i have done hsc chemistry and it not in the syllabus. They will give u the enthalpy of the alkanol if they require it or u will b given data and have to use the delta H= -mc delta t formula.