Ch20_DeltaG

# Ch20_DeltaG - Gibbs Free Energy G Gibbs Free Energy G Suniv...

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Unformatted text preview: Gibbs Free Energy, G Gibbs Free Energy, G Suniv = Ssurr + Suniv = Hsys T 1 Ssys + Go = Ssys = Ho -T So Gibbs Free Energy, G Gibbs Free Energy, G Go = 4 Ho - T So Two methods of calculating Go a) Determine Horxn and Determine GIbbs equation. Sorxn and use b) Use tabulated values of free energies of formation, Gfo. formation, Gorxn = Gorxn = Gffo ((products) -G o products) Gffo ((reactants) G o reactants) Gibbs Free Energy, G Gibbs Free Energy, G 2 Ho - T So Go = Gibbs f ree energy change = total energy change for system - energy lost in disordering the system energy If reaction is exothermic ( Ho negative) and entropy increases ( So is +), then Go must be negative and reaction productfavored. If reaction is endothermic ( Ho is +), and entropy decreases ( So is -), then Go must be + and reaction is reactant-favored. Multiply through by -T -T Suniv = Hsys - T Ssys -T Suniv = change in Gibbs free energy for the system = Gsystem for Under standard conditions — Go Gibbs Free Energy, G Gibbs Free Energy, G Calculating Calculating Calculating Gorxn Gorxn Ho So Go Reaction increase(+) - Prod-favored endo(+) decrease(-) + React-favored exo(-) decrease(-) ? T dependent endo(+) Combustion of acetylene C2H2(g) + 5/2 O 2(g) --> 2 CO 2(g) + H 2O(g) Use enthalpies of formation to calculate Horxn = -1238 kJ Use standard molar entropies to calculate Sorxn = -97.4 J/K or -0.0974 kJ/K Gorxn = -1238 kJ - (298 K)(-0.0974 J/K) = -1209 kJ -1209 Reaction is product-favored in spite of negative Reaction product-favored Sorxn. Reaction is “enthalpy driven” Reaction “enthalpy Page 1 Ho - T So exo(-) 5 3 increase(+) ? T dependent Calculating Calculating Calculating Gorxn Gorxn NH4NO3(s) + heat ---> NH 4NO3 (aq) Is the dissolution of ammonium nitrate productfavored? If so, is it enthalpy- or entropy-driven? 6 Calculating Calculating Calculating Gorxn Gorxn 7 Gorxn = Gorxn = NH4NO3(s) + heat ---> NH 4NO3 (aq) From tables of thermodynamic data we find Horxn = +25.7 kJ Sorxn = +108.7 J/K or +0.1087 kJ/K Gorxn = +25.7 kJ - (298 K)(+0.1087 J/K) = -6.7 kJ -6.7 Reaction is product-favored in spite of negative Reaction product-favored Horxn. Reaction is “entropy driven” Reaction “entropy Free Energy and Temperature 2 Fe2O3(s) + 3 C(s) ---> 4 Fe(s) + 3 CO 2(g) Horxn = +467.9 kJ Sorxn = +560.3 J/K Gorxn = +300.8 kJ Reaction is reactant-favored at 298 K Reaction reactant-favored At what T does Gorxn just change from being At (+) to being (-)? When Gorxn = 0 = Horxn - T Sorxn When T= Calculating Calculating Calculating Gffo ((products) -G o products) Gorxn Gorxn 8 Gffo ((reactants) G o reactants) Combustion of carbon C(graphite) + O2(g) --> CO 2(g) Gorxn = Gfo(CO2) - [ Gfo(graph) + Gfo(O2)] (graph) Gorxn = -394.4 kJ - [ 0 + 0] Note that free energy of formation of an element in its standard state is 0. Gorxn = -394.4 kJ Reaction is product-favored as expected. Reaction product-favored 10 More thermo? More thermo Later! Hrxn 467.9 kJ = = 835.1 K 0.5603 kJ/K Srxn Page 2 Free Energy and Temperature 9 2 Fe2O3(s) + 3 C(s) ---> 4 Fe(s) + 3 CO 2(g) Horxn = +467.9 kJ Sorxn = +560.3 J/K Gorxn = +300.8 kJ Reaction is reactant-favored at 298 K Reaction reactant-favored At what T does Gorxn just change from being At (+) to being (-)? When Gorxn = 0 = Horxn - T Sorxn When Thermodynamics and Keq Keq is related to reaction favorability. When Gorxn < 0, reaction moves When energetically “downhill” Gorxn is the change in free energy as reactants convert completely to products. But systems often reach a state of equilibrium in which reactants have not converted completely to products. In this case Grxn is < Gorxn , so state with In both reactants and products present is more stable than complete conversion. 12 Thermodynamics and Keq 13 Product-favored reaction. 2 NO2 ---> N 2O4 Gorxn = -4.8 kJ Here Grxn is less than Here Gorxn , so the state with both reactants and products present is more stable than complete conversion. Thermodynamics and Keq Thermodynamics and Keq 14 Reactant-favored reaction. N2O4 --->2 NO2 Gorxn = +4.8 kJ Here Gorxn is greater Here than Grxn , so the than state with both reactants and products present is more stable than complete conversion. 16 Gorxn = - RT lnK Calculate K for the reaction N2O4 --->2 NO2 Gorxn = +4.8 kJ Gorxn = +4800 J = - (8.31 J/K)(298 K) ln K 4800 J lnK = = -1.94 (8.31 J/K)(298K) K = 0.14 When Gorxn > 0, then K < 1 When Page 3 Thermodynamics and Keq Keq is related to reaction favorability and so to Gorxn. to The larger the value of Gorxn the larger the The value of K. Gorxn = - RT lnK where R = 8.31 J/K• mol 15 ...
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## This note was uploaded on 10/19/2011 for the course CHM 2210 taught by Professor Reynolds during the Fall '01 term at University of Florida.

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