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The Gibbs free energy change is calculated from the expression Δ G = Δ H - T(Δ S) For the combustion of ethene (assuming it takes place at 25oC): C2H4 + 3O2 --> 2CO2 + 2H2O you need to find the enthalpy and entropy changes, which are Δ H (combustion) = - 1400 kJ/mol Δ S (combustion) = - 1102 J/mol/K Substituting into the first equation, remembering to divide the entropy value by 1000 because it's in J per mol per kelvin, not kJ, and converting the 25 degrees C to kelvin, we get: Δ G = -1314.35 kJ http://www.docbrown.info/page07/delta3SGc.htm

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7y ago
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11y ago

2C2H6 + 7O2 ----> 4CO2 + 6H2O

C - C ----> 347 * 2 = 694

C - H ----> 435 * 12 = 5220

O - O ----> 497 * 7 = 3479

total = 9393 KJ/Mol

C- O -----> 803 * 8 = 646

H -O ----> 464 * 12= 5568

total = 11992 enthlphy change =

9393 - 11992

=-2599 KJ/mol

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13y ago

ΔG°rxn = (2 mol)(-394.4 kJ/mol) + (2 mol)(-228.6 kJ/mol) - (1 mol)(68.15 kJ/mol) - (3 mol)(0 kJ/mol)

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Q: How can you calculate the enthalpy change for combustion of ethane?
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