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Enthalpy of formation

I’m not sure how to start this question. Or where to even finish it.

Calculate the standard enthalpy of formation of ethanol, C2H6O, given the following standard enthalpies of combustion:
C = -393.5 KJmol-1
H2 = -285.9 KJmol-1
C2H6O = -1336.7 KJmol-1
Can you construct Hess Cycles? And if you have the enthalpies of combustion, how would you use these in a Hess Cycle?
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Cwarne35
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My issue is the oxygen, it’s just confused me, I’m struggling to work out what goes at the top of the cycle and what goes at the bottom, the rest I should be fine with.
The formation equation would go at the top, so
2C(s) + 3H2(g) + 1/2 O2 (g) = C2H6O
and at the bottom would be the elements used to form C2H6O (in their standard states) The arrows would point upwards and the oxygen will have an enthalpy of formation of 0 as it is already an element in its standard state. I hope this helps.
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Cwarne35
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Thanks that’s helped a lot!
Original post by KirstyPayton23
The formation equation would go at the top, so
2C(s) + 3H2(g) + 1/2 O2 (g) = C2H6O


The above is correct.

Original post by KirstyPayton23

and at the bottom would be the elements used to form C2H6O (in their standard states) The arrows would point upwards and the oxygen will have an enthalpy of formation of 0 as it is already an element in its standard state. I hope this helps.


This bit is mostly incorrect, sorry. At the bottom of the Hess's cycle / triangle is the combustion products of the elements or of C2H6O (both are the same), i.e. carbon dioxide and water.

Oxygen isn't considered to combust on it's own, hence it's enthalpy of combustion is omitted. Consider it to be zero if that helps.
The oxygen combines with the carbon and hydrogen to be part of their combustion. In fact extra oxygen is necessary for the complete combustion of the carbon and hydrogen in the equation. No need to worry about the extra oxygen - the enthalpy of combustion value assumes that there is sufficient oxygen available.

The arrows are pointing down towards the combustion products.
(edited 6 years ago)
Sorry misread that combustion data was being used.
Reply 7
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Cwarne35
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Sooooo am I correct in thinking 2CO2 + 3H2O goes at the bottom?
Original post by Cwarne35
Sooooo am I correct in thinking 2CO2 + 3H2O goes at the bottom?


Yes, that's correct.

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