Entropy, Free Energy, and EquilibriumSlide 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 27Entropy, Free Energy, and EquilibriumChapter 18Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.Spontaneous Physical and Chemical Processes• A waterfall runs downhill• A lump of sugar dissolves in a cup of coffee• At 1 atm, water freezes below 0 0C and ice melts above 0 0C• Heat flows from a hotter object to a colder object• A gas expands in an evacuated bulb• Iron exposed to oxygen and water forms rustspontaneousnonspontaneous18.2spontaneousnonspontaneous18.2Does a decrease in enthalpy mean a reaction proceeds spontaneously?CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) H0 = -890.4 kJH+ (aq) + OH- (aq) H2O (l) H0 = -56.2 kJH2O (s) H2O (l) H0 = 6.01 kJNH4NO3 (s) NH4+(aq) + NO3- (aq) H0 = 25 kJH2OSpontaneous reactions18.2Entropy (S) is a measure of the randomness or disorder of a system.orderSdisorderSS = Sf - SiIf the change from initial to final results in an increase in randomnessSf > SiS > 0For any substance, the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid < Sliquid << SgasH2O (s) H2O (l) S > 018.3W = 1W = 4W = 6W = number of microstatesS = k ln WS = Sf - SiS = k lnWfWiWf > Wi then S > 0Wf < Wi then S < 0Entropy18.3Processes that lead to an increase in entropy (S > 0)18.2How does the entropy of a system change for each of the following processes?(a) Condensing water vaporRandomness decreasesEntropy decreases (S < 0)(b) Forming sucrose crystals from a supersaturated solutionRandomness decreasesEntropy decreases (S < 0)(c) Heating hydrogen gas from 600C to 800CRandomness increasesEntropy increases (S > 0)(d) Subliming dry iceRandomness increasesEntropy increases (S > 0)18.3EntropyState functions are properties that are determined by the state of the system, regardless of how that condition was achieved.Potential energy of hiker 1 and hiker 2 is the same even though they took different paths.energy, enthalpy, pressure, volume, temperature, entropy18.3First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed.Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process.Suniv = Ssys + Ssurr > 0Spontaneous process:Suniv = Ssys + Ssurr = 0Equilibrium process:18.4Entropy Changes in the System (Ssys)aA + bB cC + dDS0rxndS0(D)cS0(C)=[ + ]-bS0(B)aS0(A)[ + ]S0rxnnS0(products)=mS0(reactants)-The standard entropy of reaction (S0 ) is the entropy change for a reaction carried out at 1 atm and 250C.rxn18.4What is the standard entropy change for the following reaction at 250C? 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 197.9 J/K•molS0(O2) = 205.0 J/K•molS0(CO2) = 213.6 J/K•molS0rxn= 2 x S0(CO2) – [2 x S0(CO) + S0 (O2)] S0rxn= 427.2 – [395.8 + 205.0] = -173.6 J/K•molEntropy Changes in the System (Ssys)18.4When gases are produced (or consumed)• If a reaction produces more gas molecules than it consumes, S0 > 0.•If the total number of gas molecules diminishes, S0 < 0.•If there is no net change in the total number of gas molecules, then S0 may be positive or negative BUT S0 will be a small number.What is the sign of the entropy change for the following reaction? 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down, S is negative.Entropy Changes in the Surroundings (Ssurr)Exothermic ProcessSsurr > 0Endothermic ProcessSsurr < 018.4Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature.18.3S = k ln WW = 1S = 0Suniv = Ssys + Ssurr > 0Spontaneous process:Suniv = Ssys + Ssurr = 0Equilibrium process:Gibbs Free EnergyFor a constant-temperature process:G = Hsys -TSsysGibbs free energy (G)G < 0 The reaction is spontaneous in the forward direction.G > 0 The reaction is nonspontaneous as written. The reaction is spontaneous in the reverse direction.G = 0 The reaction is at equilibrium.18.518.5aA + bB cC + dDG0rxndG0 (D)fcG0 (C)f=[ + ]-bG0 (B)faG0 (A)f[ + ]G0rxnnG0 (products)f=mG0 (reactants)f-The standard free-energy of reaction (G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions.rxnStandard free energy of formation (G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states.fG0 of any element in its stable form is zero.f2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)G0rxnnG0 (products)f=mG0 (reactants)f-What is the standard free-energy change for the following reaction at 25 0C?G0rxn6G0 (H2O)f12G0 (CO2)f=[ + ]- 2G0 (C6H6)f[ ]G0rxn=[ 12x–394.4 + 6x–237.2 ] – [ 2x124.5 ] = -6405 kJIs the reaction spontaneous at 25 0C?G0 = -6405 kJ< 0spontaneous18.5G = H - TS18.5CaCO3 (s) CaO (s) + CO2 (g)H0 = 177.8 kJS0 = 160.5 J/KG0 = H0 – TS0At 25 0C, G0 = 130.0 kJG0 = 0 at 835 0C18.5Temperature and Spontaneity of Chemical ReactionsEquilibrium Pressure of CO2Gibbs Free Energy and Phase TransitionsH2O (l) H2O (g)G0 = 0 = H0 – TS0S = TH = 40.79 kJ373 K= 109 J/K18.5Efficiency = X 100%Th - TcTcChemistry In Action: The Efficiency of Heat EnginesA Simple Heat EngineGibbs Free Energy and Chemical EquilibriumG = G0 + RT lnQR is the gas constant (8.314 J/K•mol)T is the absolute temperature (K)Q is the reaction quotientAt EquilibriumG = 0Q = K0 = G0 + RT lnKG0 =  RT lnK18.618.6Free Energy Versus Extent of ReactionG0 < 0 G0 > 0G0 =  RT lnK18.618.7ATP + H2O + Alanine + Glycine ADP + H3PO4 + AlanylglycineAlanine + Glycine AlanylglycineG0 = +29 kJG0 = -2 kJK < 1K > 118.7The Structure of ATP and ADP in Ionized FormsHigh EntropyLow EntropyTS = H - GChemistry In Action: The Thermodynamics of a Rubber