Calculation of reaction enthalpy 5 (thermo-17) To sweat and cut, you need fuel gas. You usually add ethane (C2H2). Why not the cheaper methane (CH4)? Note: For both gases, formulate the combustion reaction and determine the reaction enthalpy per (!) molecule C2H2 and CH4 respectively. ΔHof(CH4) = -75 kJ/mol ΔHof(C2H2) = 227 kJ/mol ΔHof(H2O) = -285 kJ/mol ΔHof(CO2) = -393 kJ/mol

  Exercise: thermo-17    

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To sweat and cut, you need fuel gas. You usually add ethane (C₂H₂). Why not the cheaper methane (CH₄)?
Note: For both gases, formulate the combustion reaction and determine the reaction enthalpy per (!) molecule C₂H₂ and CH₄ respectively.
ΔHof(CH₄) = -75 kJ/mol
ΔHof(C₂H₂) = 227 kJ/mol
ΔHof(H₂O) = -285 kJ/mol
ΔHof(CO₂) = -393 kJ/mol

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Solution:

1. Incineration of ethylene:
   1·C₂H₂(g) + 5/2·O₂(g) → 1·H₂O(g) + 2·CO₂(g)
   ΔHoR = ∑ΔHof(Produkte) – ∑ΔHof(Edukte)
   ΔHoR = (1·-285+2·-393)-(1·227+5/2·0) = -1298 kJ/mol
2. Burning of methane:
   1·CH₄(g) + 2·O₂(g) → 2·H₂O(g) + 1·CO₂(g)
   ΔHoR = ∑ΔHof(Produkte) – ∑ΔHof(Edukte)
   ΔHoR = (2·-285+1·-393)-(1·-75+2·0) = -880 kJ/mol

The calculation shows that for every turnover of one mol of ethine, about 1.5 times more energy It's released more than methane.

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