1 Q about Enthalpy

這是一條舊題目，而且是一條陳舊而『錯誤』的題目。這題目流傳至今居然還是繼續沒有更改而繼續存在，可堪一嘆！

題目錯誤的原因，是題目給予的standard enthalpy change of formation of H2O(g)，但在 standard enthalpy change of combustion of ethene 中，state of water under standard conditions should be liquid, i.e. H2O(l)。但是若果你堅持『正確』的話， H2O(l) 與 H2O(g) 不能在計算中互相消去。計不到答案是題目的錯誤，不是我能力的問題。

以下所列的流傳數十年的計法，是將錯就錯變成答案的計法。若果你老師逼你要計到答案的話，你可以照抄。但紅色的 state symbol (g) 其實應該是 (l)。記著，若果你堅持要『計算正確』的話，你會計算不到答案的。（苦笑）

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Given :
Standard enthalpy change of formation of CO2(g):
C(graphite) + O2(g) → CO2(g) aaΔH = -394 kJ mol-1
Standard enthalpy change of formation of H2O(g):
H2(g) + 0.5O2(g) → H2O(g) aaΔH = -242 kJ mol-1
Standard enthalpy change of combustion of ethane:
CH3­H3(g) + 3.5O2(g) → 2CO2(g) + 3H2O(g)aaΔH = -1560 kJ mol-1
Standard enthalpy change of reduction of ethane to ethane by hydrogen:
CH2CH2­(g) + H2(g) → CH3CH3(g)aaΔH = -(-138) = +138 kJ mol-1

The solution :

CH3CH3(g) → CH2CH2­(g) + H2(g) aaΔH = -(-138) = +138 kJ mol-1

2CO2(g) + 3H2O(g) → CH3­H3(g) + 3.5O2(g) aaΔH = -(-1560) = +1560 kJ mol-1

3H2(g) + 1.5O2(g) → 3H2O(g) aaΔH = 3(-242) = -726 kJ mol-1

2C(graphite) + 2O2(g) → 2CO2(g) aaΔH = 2(-394) = -788 kJ mol-1

Add the above four thermochemical equations together, we get
2C(graphite) + 2H2(g) → CH2CH2(g)
Standard enthalpy change of formation ΔH of ethane, ΔHf[CH2CH2]
= 138 + 1560 - 726 - 788 = +184 kJ mol-1

With reference to the following energy cycle:

圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Mar08/Crazythermalchem1.jpg

By Hess's law:
△Hf + △H1 + △H2 = △H3 + △H4
△Hf - 138 - 1560 = -242 x 3 - 394 x 2
△Hf = 184 kJ/mol