4.66 - Thermodynamics study questions - 4

Q466-04 Hex-1-ene gas, C6H12, burns in oxygen to produce carbon dioxide and water vapour. Write an equation to represent this reaction.
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 C6H12(g) + 9O2(g) 6CO2(g) + 6H2O(g)

Use the data below to calculate the values of ΔHc and ΔSc for the combustion of hex-1-ene.
 Substance O2(g) C6H12(g) CO2(g) H2O(g) Standard enthalpy of formation, ΔHf/kJmol-1 0.0 -43 -394 -242 Entropy, S/JK-1 mol-1 205 385 214 189

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 Reactants enthalpy of formation 1 x C6H12(g) = -43 kJ Products enthalpy of formation 6 x CO2(g) = 6 x -394 = -2364 kJ 6 x H2O(g) = 6 x -242 = -1452 kJ Total = -3816 kJ The reactants are 'unformed' so change the sign of the reactants and add it to the products: Reaction enthalpy = Products ΔHf - ΔReactants Hf = -3816 - (-43) = -3773 kJ
 Reactants entropy 1 x C6H12(g) = 1 x 385 = 385 JK-1 9 x O2(g) = 9 x 205 = 1845 JK-1 Total = 2230 JK-1 Products entropy 6 x CO2(g) = 6 x 214 = 1284 JK-1 6 x H2O(g) = 6 x 189 = 1134 JK-1 Total = 2418 JK-1 There is an increase in entropy by 2418 - 2230 = +188 JK-1 ΔSc = +188 JK-1

Calculate the standard free energy change for the combustion of hex-1-ene.
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Using the Gibbs free energy equation:
 ΔG = ΔH - TΔS

At 298 K: ΔG = -3773 - 298(0.188) = -3773 - 56 = -3829 kJ

State and explain whether or not the combustion of hex-1-ene is spontaneous at 25ºC
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 The final value for Gibbs free energy is negative at 298K, so the reaction is spontaneous.