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System strives for minimum Free Energy

System strives for minimum Free Energy. The standard free-energy of reaction ( D G 0 ) is the free-energy change for a reaction when it occurs under standard-state conditions. rxn. a A + b B c C + d D. -. [. +. ]. [. +. ]. =. -. m D G 0 (reactants). S. S. =. f.

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System strives for minimum Free Energy

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  1. System strives for minimum Free Energy

  2. The standard free-energy of reaction (DG0 ) is the free-energy change for a reaction when it occurs under standard-state conditions. rxn aA + bB cC + dD - [ + ] [ + ] = - mDG0 (reactants) S S = f Standard free energy of formation (DG0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states. DG0 DG0 rxn rxn f DG0 of any element in its stable form is zero. f dDG0 (D) nDG0 (products) cDG0 (C) aDG0 (A) bDG0 (B) f f f f f

  3. - mDG0 (reactants) S S = f 2C6H6(l) + 15O2(g) 12CO2(g) + 6H2O (l) DG0 DG0 DG0 - [ ] [ + ] = rxn rxn rxn [ 12x–394.4 + 6x–237.2 ] – [ 2x124.5 ] = -6405 kJ = Is the reaction spontaneous at 25 0C? 12DG0 (CO2) 2DG0 (C6H6) f f 6DG0 (H2O) f nDG0 (products) f What is the standard free-energy change for the following reaction at 25 0C? DG0 = -6405 kJ < 0 spontaneous 18.4

  4. DG = DH - TDS DG gives direction of reaction If A  B DG < 0 then B  A DG > 0 A  B DG = 0 DG does not indicate the rate of the process.

  5. CaCO3(s) CaO (s) + CO2(g) Temperature and Spontaneity of Chemical Reactions DH0 = 177.8 kJ DS0 = 160.5 J/K DG0 = DH0 – TDS0 At 25 0C, DG0 = 130.0 kJ When does reaction switch from being spontaneous to non-spontaneous? DG0 = 0 = DHsys -TDSsys DHsys / DSsys = T 177.8 kJ / 0.1605 kJ/K = T T = 835 0C 18.4 Watch units !!!

  6. H2O (l) H2O (g) 40.79 kJ = DS = 373 K DH T Gibbs Free Energy and Phase Transitions DG0 = 0 = DH0 – TDS0 = 109 J/K 18.4

  7. Gibbs Free Energy and Chemical Equilibrium DG = DG0 + RT lnQ R is the gas constant (8.314 J/K•mol) T is the absolute temperature (K) Q is the reaction quotient = [Products]0 or P°Products [Reactants]0 P°Reactants Can correct Gibbs Free Energy change for non-standard state 18.4

  8. - mDG0 (reactants) S S = f DG0 DG0 DG0 rxn rxn rxn [ 2(–16.66 kJ/mole) – 0 – 0 = - 33.32 kJ = 2DG0 (NH3) f nDG0 (products) f What is the standard free-energy change for the following reaction at 25 0C? N2(g) + 3H2(g) 2 NH3(g) - 3DG0 (H2) DG0 (N2) - = f f What is the standard free-energy change for the following mixture of gases at 25 0C: 0.10 atm N2, 0.10 atm H2and 0.10 atm NH3 ? DG = DG0 + RT lnQ = DG0 + RT ln PNH32 PN2PH23 = -33.32 kJ + (0.008314 kJ /molK)(298K) ln 0.102 (0.10)(0.10)3 = -33.32 kJ + 11.4 kJ = -21.92 kJ

  9. Gibbs Free Energy and Chemical Equilibrium DG = DG0 + RT lnQ R is the gas constant (8.314 J/K•mol) T is the absolute temperature (K) Q is the reaction quotient At Equilibrium Q = K DG = 0 0 = DG0 + RT lnK DG0 = -RT lnK 18.4

  10. -DG0/RT Kp= e What is Kp for the following reaction at 25 0C? N2(g) + 3H2(g) 2 NH3(g) - (-33.32 kJ)/(0.008314 kJ /molK)(298K) Kp= e 13.4 Kp= e Kp=6.6 x 105

  11. DG0 < 0 DG0 > 0 18.4

  12. DG0 = -RT lnK 18.4

  13. DG0 = -RT lnK Consider the reaction: NH3(aq) + H2O (l) NH4+(aq) + OH-(aq) Kb= 1.8 x 10-5 What is DG0 for the reaction at 25 0C? DG0 = -RT lnK = -(0.008314 kJ /molK)(298K) ln(1.8 x 10-5) DG0 = 27.08 kJ /mol K < 1, reinforces that this is a weak base and left side is favored.

  14. How does K change with Temperature? At equilibrium: -RT lnK =DG0 = DH0 – TDS0 RT lnK + DH0 = TDS0 constant R lnK + DH0/T = DS0 Compare 2 Temperatures: R lnK1+ DH0/T1 = R lnK2+ DH0/T2 R(lnK2 - lnK1) = DH0/T1 - DH0/T2 ln(K2 / K1) = DH0/R(1/T1 - 1/T2) Similar to Clausius-Clapeyron Equation for vapor pressure and rate constant change equation

  15. DH0 rxn -DG0/RT Kp= e What is Kp for the following reaction at 25 0C? H2(g) + I2(g) 2 HI (g) DG0 = 2.6 kJ - (2.6 kJ)/(0.008314 kJ /molK)(298K) -1.05 = e = e Kp= 0.350 What is Kp at 50 0C? = 2(25.9 kJ/mole) – 0 – 0 = 51.8 kJ ln(K2 / K1) = DH0/R(1/T1 + 1/T2) ln(K2 / 0.350 = (51.8 kJ )/(0.008314 kJ /molK)(1/298K + 1/323K) lnK2 - ln0.350 = (6.23 x 103)(2.60 x 10-4) lnK2 = 1.62 - 1.05 = 0.60 K2 = 1.82

  16. DS0 DH0 rxn rxn = 2 x DH0(NO2) – [2 x DH0(NO) + DH0(O2)] = 2 x S0(NO2) – [2 x S0(NO) – S0(O2)] Calculate DG0for the following reaction at 25 0C? 2 NO (g) + O2(g) 2 NO2(g) Given: f DG0 = DH0 – TDS0 f f f = 2(33.84 kJ) - 2(90.37 kJ) - 0 = -113.06 kJ = 2(240.45J/molK) - 2(210.62 J/molK) - 205.0 j/molK = -145.34 J/K

  17. DG0 rxn = 2 x DG0(NO2) – [2 x DG0(NO) + DG0(O2)] DG0 = DH0 – TDS0 = -113.06 kJ - (298 K)(- 0.14534 kJ/K = -113.06 kJ + 43.31 kJ = -69.75 kJ Can also calculate with DG0 ’s from tabulated data f f f f = 2(51.84 kJ/mol) - 2(86.71 kJ/mol) - 0 = 103.68 kJ - 173.42 kJ = -69.74 kJ Get same answer either way DG0, DH0 and DS0 are all state functions

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