Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Slide 17Slide 18Slide 19Slide 20Slide 21Slide 22Slide 23Slide 24Slide 25Slide 26Slide 27Slide 28H2O(l) --> H2O(s)Normal freezing point of H2O = 273.15KThe change in enthalpy is the enthalpy of freezing - enthalpy change associated when one mole of liquid water freezes at 1atm and 273.15KH = -6007JEntropy changeS=HT=-6007 J273.15K= -21.99 J/KG = H - TS = -6007J - (273.15K)(-21.99J/K) = 0J=> at 273.15K and 1 atm, liquid and solid water are at equilibriumThe Gibbs Free Energy Function and Phase TransitionsAs water is cooled by 10K below 273.15K to 263.15KAssume that H and S do not change with TG = H - TS = -6007J - (263.15K)(-21.99J/K) = -220JSince G < 0 water spontaneously freezes At 10K above the freezing point, 283.15KG = H - TS = -6007J - (283.15K)(-21.99J/K) = 219JG > 0 water does not spontaneously freeze at 283.15K, 10K above the normal freezing point of waterAbove 273.15K, the reverse process is spontaneous - ice melts to liquid water.Plot of H and TS vs T for the freezing of water. At 273.15 K the system is at equilibrium, G = 0freezing spontaneousFreezing not spontaneous; melting spontaneousH2O(l) --> H2O(s)For phase transitions, at the transition temperature, the system is at equilibrium between the two phases.Above or below the transition temperature, the phase that is the more stable phase is determined by thermodynamics - G of the phase transition.Reaction Free EnergyGr = nGm(products) - nGm(reactants) Standard Reaction Free EnergyGor = nGom (products) - nGom(reactants)Difference in free energy of the pure products in their standard states and pure reactants in their standard states.Standard Free-Energy of FormationThe standard free energy of formation, Gfo, of a substance is the standard reaction free energy per mole for the formation of a compound from its elements in their most stable form1/2N2 (g) + 3/2 H2 (g) --> NH3(g) Gfo = -16 kJO2 (g) --> O2 (g) Gfo = 0Gfo is an indication of a compounds stability relative to its elementsThermodynamically stable compound Gfo < 0 Thermodynamically unstable compound Gfo > 0Standard Reaction Free EnergyFor a reaction: aA + bB --> cC + dD Gro = c Gfo (C) + d Gfo(D) - a Gfo(A) -b Gfo(B)The standard free energy of formation of elements in their most stable form at 298.15K is zero.Problem: Calculate the standard free-energy change for the following reaction at 298K:N2(g) + 3H2 (g) --> 2NH3(g)Given that Gfo[NH3(g)] = -16.66 kJ/mol.What is the Go for the reverse reaction?Since the reactants N2(g) and H2(g) are in their standard state at 298 K their standard free energies of formation are defined to be zero. Gro = 2 Gfo[NH3(g)] - Gfo[N2(g)] - 3 Gfo[H2(g)] = 2 x -16.66 = -33.32 kJ/molFor the reverse reaction: 2NH3(g) --> N2(g) + 3H2 (g) Gro = +33.32 kJ/molC3H8(g) + 5O2(g) --> 3CO2 (g) + 4H2O(l) Ho = - 2220 kJa) Without using the information from the tables of thermodynamic quantities predict whether Go for this reaction will be less negative or more negative than Ho.b) Use data from App. D to calculate Gro for this reaction. Whether Gro is more negative or less negative than Ho depends on the sign of So since Gro = Ho - T SoThe reaction involves 6 moles of gaseous reactants and produces 3 moles of gaseous product => So < 0 (since fewer moles of gaseous products than reactants)Since So < 0 and Ho < 0 and Gro = Ho - T So=> Gro is less negative than HoGro = 3Gfo(CO2 (g))+4Gfo(H2O(l)-Gfo(C3H8(g))-5Gfo(O2(g)) = 3(-394.4) + 4(-237.13) - (-23.47) - 5(0) = -2108 kJEffect of temperature on GoValues of Gro calculated using the tabulated values of Gfo apply only at 298.15K.For other temperatures the relation:Gro = Ho - T Socan be used, assuming Ho and So do not vary significantly with temperature.T =HoSoWhen Gro = 0, Use this expression to determine temperature above or below which reaction becomes spontaneous.Problem: The normal boiling point is the temperature at which a pure liquid is in equilibrium with its vapor at 1 atm.a) write the chemical equation that defines the normal boiling point of liquid CCl4b) what is the value of Go at equilibrium?c) Use thermodynamic data to estimate the boiling point of CCl4.a) CCl4(l) CCl4(g) b) At equilibrium G = 0. In any equilibrium for a normal boiling point, both the liquid and gas are in their standard states. Hence, for this process G is Go.c)T = HoSowhere for this reaction T is the boiling pointNote: To determine the boiling point accurately we need the values of Ho and So for the vaporization process at the boiling point of CCl4Ho = (1mol)(-106.7 kJ/mol) - (1mol)(-139.3kJ/mol) = 32.6 kJSo = (1mol)(309.4 J/mol-K) -(1mol)(214.4 J/mol-K) = 95.0 J/KT = HoSo = 32.6 kJ0.095 kJ/K = 343 KThe Gibbs Function and the Equilibrium ConstantG = Gro + R T ln QQ - reaction quotientUnder standard conditions the concentrations of all reactants and products are set to 1(standard pressure of gases = 1 atmStandard pressure of solutions = 1 M)Under standard conditions, ln Q = 0=> G = GroAt equilibrium G = 0 and Q = KAt equilibriumGro = - R T ln K or K = e- Gro /RTGro < 0 => K > 1Gro > 0 => K < 1G = Gro + R T ln Q = - RT ln K + RT ln QG = R T ln QKwhen Q = K => G = 0; equilibriumIf the reaction mixture has too much N2 or H2 relative to NH3, Q < K, and reaction moves spontaneously in the direction to form NH3 till equilibrium is established; Q = K. If there is too much NH3 in the mixture, Q > K, decomposition of NH3 is spontaneous, till equilibrium is established; Q = K Gforward < 0Gforward > 0Temperature Dependence of KThe temperature dependence of K can be used to determine Ho and So of a reactionln K = -Gr R T = - Ho So R T R +If the equilibrium constant of a reaction is known at one temperature and the value of Ho is known, then the equilibrium constant at another temperature can be calculatedln K1 = - Ho So R T1 R + ln K2 = - Ho So R T2 R + If Ho is negative (exothermic), an increase in T reduces K; if it is positive (endothermic), an increase in T increases Kln K2 - ln K1 = lnK1=Ho R 1 T2 1 T1 ( )-K2For the equilibrium