Calculation of reaction enthalpy 7 (thermo-19) Adopted, one decides itself for the not health variation, that the complete daily energy intake by means of drinking of alcohol (40% alcohol) implemented is. The following assumptions and specifications shall apply: Drinking alcohol would be C2H5OH (ethanol), the density of pure alcohol would be 0.9 g/ml Assumption: The alcohol is inhaled in the body, i.e. injected with O2. Note 1: Calculate first the reaction enthalpy including state of matter Note 2: Calculate from ΔHoR and the amount of ethanol per litre of alcohol the amount needed to realize the energy supply Note 3: Daily energy consumption is 9200 kJ Note 4: for the professionals: the bases of calculation is not fully, in fact, [translation unavailable] the gibbs free energy of reaction at body temperature calculated are. See details of the following exercise thermo-108. But you can see that the result would only change by about 10 percent.

  Exercise: thermo-19    

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Adopted, one decides itself for the not health variation, that the complete daily energy intake by means of drinking of alcohol (40% alcohol) implemented is. The following assumptions and specifications shall apply:

1. Drinking alcohol would be C₂H₅OH (ethanol), the density of pure alcohol would be 0.9 g/ml
2. Assumption: The alcohol is inhaled in the body, i.e. injected with O₂.
3. Note 1: Calculate first the reaction enthalpy including state of matter
4. Note 2: Calculate from ΔHoR and the amount of ethanol per litre of alcohol the amount needed to realize the energy supply
5. Note 3: Daily energy consumption is 9200 kJ
6. Note 4: for the professionals: the bases of calculation is not fully, in fact, [translation unavailable] the gibbs free energy of reaction at body temperature calculated are. See details of the following exercise thermo-108. But you can see that the result would only change by about 10 percent.

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Solution:

1. reaction equation:
       1·C₂H₅OH + 3·O₂ ⇄ 2·CO₂(g) + 3·H₂O(l)
2. ΔHoR = ∑ΔHof(Produkte) – ∑ΔHof(Edukte)
   ΔHoR = (2· ΔHof(CO₂(g)) + 3 ΔHof(H₂O(l))) – ( ΔHof(C₂H₅OH) + 3· ΔHof(O₂))
   ΔHoR = (2·-393 + 3·-286) – (-278 + 3·0) = -1366 kJ/mol
   Thus [translation unavailable] one well, you know,, that per Mol [translation unavailable] 1366 kJ released are
3. Next question: How many grams of ethanol are in alcohol?
   1 liter of alcohol, 40%, i.e. 400 ml of pure ethanol. Thus, at a density of 0.9 g/ml, 360 grams of ethanol are produced.
4. M(C₂H₅OH) = 46 g/mol, so that
   n= m/M = 360 / 46 = 7.8 mol
5. Thus, 7.8 mol to -1366 kJ per bottle of alcohol is contained, i.e. -10690 kJ.
6. So for a daily requirement of 9200 kJ it takes about 860 ml.

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