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Calculate the Gibbs free energy change at standard conditions for the combustion of propane:C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l) Given the standard enthalpy of formation data: ΔH°f(CO2(g)) = -393.5 kJ/mol ΔH°f(H2O(l)) = -285.8 kJ/mol ΔH°f(C3H8(g)) = -103.85 kJ/mol and the standard entropy data: S°(CO2(g)) = 213.6 J/(mol•K) S°(H2O(l)) = 69.9 J/(mol•K) S°(C3H8

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To calculate the Gibbs free energy change  G  at standard conditions for the combustion of propane, we first need to find the standard enthalpy change  H  and the standard entropy change  S  for the reaction.1. Calculate the standard enthalpy change  H  for the reaction:H = [ Hf products ] - [ Hf reactants ]H = [3Hf CO2 g   + 4Hf H2O l  ] - [Hf C3H8 g   + 5Hf O2 g  ]Since the standard enthalpy of formation for O2 g  is zero  it's an element in its standard state , we can simplify the equation:H = [3 -393.5  + 4 -285.8 ] - [-103.85]H =  -1180.5 - 1143.2  + 103.85H = -2219.85 kJ/mol2. Calculate the standard entropy change  S  for the reaction:S = [ S products ] - [ S reactants ]S = [3S CO2 g   + 4S H2O l  ] - [S C3H8 g   + 5S O2 g  ]The standard entropy for O2 g  is 205.0 J/ molK . Now we can plug in the values:S = [3 213.6  + 4 69.9 ] - [269.9 + 5 205.0 ]S =  640.8 + 279.6  -  269.9 + 1025.0 S = 920.4 - 1294.9S = -374.5 J/ molK 3. Calculate the Gibbs free energy change  G  at standard conditions  298 K :G = H - TSG = -2219.85 kJ/mol -  298 K  -0.3745 kJ/ molK  Note that we need to convert S to kJ/ molK  by dividing by 1000:G = -2219.85 kJ/mol -  298 K  -0.0003745 kJ/ molK  G = -2219.85 kJ/mol + 111.621 kJ/molG = -2108.229 kJ/molSo, the Gibbs free energy change at standard conditions for the combustion of propane is -2108.229 kJ/mol.

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