reaction enthalpy Calculation 8 (thermo-20) Analogs exercise to thermo-18. Es It should be But It 's just … The free of charge reaction enthalpy In the case of KHeat and temperature for The Air conditioning (Incineration) of Ethanol Calculated will be.

  Aufgabe: thermo-20    

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Analogs exercise to thermo-18. Es It should be But It 's just … The free of charge reaction enthalpy In the case of KHeat and temperature for The Air conditioning (Incineration) of Ethanol Calculated will be.

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Lösung:

Table of values:

	ΔHof(kJ·mol-1)	So(J·K-1·mol-1)
EtOH(l)	-277	161
O2 (g)	0	205
CO2 (g)	-393	214
H2O (l)	-286	67

1. reaction equation:
       1·C₂H₅OH + 3·O₂ ⇄ 2·CO₂(g) + 3·H₂₀(l)
2. ΔHoR = ∑ΔHof(Produkte) – ∑ΔHof(Edukte)
   ΔHoR = (2· ΔHof(CO₂,g) + 3· ΔHof(H₂O, l)) – ( ΔHof(C₂H₅OH, l) + 3· ΔHof(O₂,g))
   ΔHoR = (2·-393 + 3·-286) – (-277 + 3·0) = -1367 kJ/mol (Analogue to the exercise thermo-107)
   ΔSoR = (2·S^o(CO₂,g) + 3·S^o(H₂O, l)) – (S^o(C₂H₅OH, l) + 3·S^o(O₂,g))
   ΔSoR = (2·214 + 3·67) – (161 + 3·205) = -147 J·mol^-1·K^-1
   ΔGoR = ΔHoR – T· ΔSoR
   ΔGoR = -1367 – (273+37)·-0.147 = -1321.43 kJ/mol

This Results is a Not much less Exergonically How The Similar calculation The exercise 107. Quintessence: The required Quantity Alcohol would be to about. 4% Greater.

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