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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 and b) Use tabulated values of free energies of formation, Gfo. formation, Gorxn = Gorxn = Gffo (products) -G o ((products) products) Gffo (reactants) G o ((reactants) reactants) Gibbs Free Energy, G Gibbs Free Energy, G 2 Ho - T So Go = Gibbs free energy change = total energy change for system - energy lost in disordering the system If reaction is exothermic ( Ho negative) and negative) entropy increases ( So is +), then Go must be negative and reaction productfavored. If reaction is endothermic ( Ho is +), and is 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 change 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(+) endo(+) decrease(-) + React-favored exo(-) exo(-) decrease(-) ? T dependent endo(+) 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 -1238 Use standard molar entropies to calculate Sorxn = -97.4 J/K or -0.0974 kJ/K -97.4 Gorxn = -1238 kJ - (298 K)(-0.0974 J/K) -1238 = -1209 kJ Reaction is product-favored in spite of negative Sorxn. Reaction is “enthalpy driven” Page 1 Ho - T So exo(-) exo(-) 5 3 increase(+) ? T dependent Calculating Calculating Calculating Gorxn Gorxn NH4NO3(s) + heat ---> NH 4NO3 (aq) aq) Is the dissolution of ammonium nitrate productfavored? If so, is it enthalpy- or entropy-driven? 6 Calculating Calculating Calculating Gorxn Gorxn Gorxn = Gorxn = NH4NO3(s) + heat ---> NH 4NO3 (aq) aq) From tables of thermodynamic data we find Horxn = +25.7 kJ +25.7 Sorxn = +108.7 J/K or +0.1087 kJ/K +108.7 Gorxn = +25.7 kJ - (298 K)(+0.1087 J/K) +25.7 = -6.7 kJ Reaction is product-favored in spite of negative Horxn. Reaction is “entropy driven” 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 +467.9 +560.3 Gorxn = +300.8 kJ +300.8 Reaction is reactant-favored at 298 K At what T does Gorxn just change from being At just (+) to being (-)? When Gorxn = 0 = Horxn - T Sorxn When T= Calculating Calculating Calculating 7 Gffo (products) -G o ((products) products) Gorxn Gorxn 8 Gffo (reactants) G o ((reactants) 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] -394.4 Note that free energy of formation of an element in its standard state is 0. Gorxn = -394.4 kJ -394.4 Reaction is product-favored as expected. 10 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 +467.9 +560.3 Gorxn = +300.8 kJ +300.8 Reaction is reactant-favored at 298 K At what T does Gorxn just change from being At just (+) to being (-)? When Gorxn = 0 = Horxn - T Sorxn When Thermodynamics and Keq More thermo? thermo? Later! Hrxn 467.9 kJ = = 835.1 K 0.5603 kJ/K Srxn Page 2 9 Keq is related to reaction favorability. is favorability. When Gorxn < 0, reaction moves When 0, energetically “downhill” Gorxn is the change in free energy as is 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 so 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 -4.8 Here Grxn is less than Here is Gorxn , so the state so 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 --->2 Gorxn = +4.8 kJ +4.8 Here Gorxn is greater Here is than Grxn , so the than so state with both reactants and products present is more stable than complete conversion. 16 Gorxn = - RT lnK RT Calculate K for the reaction N2O4 --->2 NO2 Gorxn = +4.8 kJ --->2 +4.8 Gorxn = +4800 J = - (8.31 J/K)(298 K) ln K +4800 4800 J lnK = = -1.94 (8.31 J/K)(298K) K = 0.14 When Gorxn > 0, then K < 1 When 0, Page 3 Thermodynamics and Keq Keq is related to reaction favorability and so is to Gorxn. to The larger the value of Gorxn the larger the The the value of K. Gorxn = - RT lnK RT where R = 8.31 J/K•mol J/K•mol 15 ...
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