Chapter 19: Spontaneous Change: Entropy and Gibbs Energy Chemistry 140 Fall 2002 General Chemistry Principles and Modern Applications Petrucci • Harwood • Herring 8th Edition Chapter 19: Spontaneous Change: Entropy and Gibbs Energy Philip Dutton University of Windsor, Canada N9B 3P4 Prentice-Hall © 2002

General Chemistry: Chapter 20 Chemistry 140 Fall 2002 Contents 19-1 Spontaneity: The Meaning of Spontaneous Change 19-2 The Concept of Entropy 19-3 Evaluating Entropy and Entropy Changes 19-4 Criteria for Spontaneous Change: The Second Law of Thermodynamics 19-5 Standard Gibbs Energy Change, ΔG° 19-6 Gibbs Energy Change and Equilibrium 19-7 ΔG° and Keq as Functions of Temperature 19-8 Coupled Reactions Focus On Coupled Reactions in Biological Systems Prentice-Hall © 2002 General Chemistry: Chapter 20

19-1 Spontaneity: The Meaning of Spontaneous Change Prentice-Hall © 2002 General Chemistry: Chapter 20

General Chemistry: Chapter 20 Spontaneous Process A process that occurs in a system left to itself. Once started, no external actions is necessary to make the process continue. A non-spontaneous process will not occur without external action continuously applied. 4 Fe(s) + 3 O2(g) → 2 Fe2O3(s) H2O(s)  H2O(l) 2 Fe2O3(s) → 4 Fe(s) + 3 O2(g) Prentice-Hall © 2002 General Chemistry: Chapter 20

General Chemistry: Chapter 20 Spontaneous Process Potential energy decreases. For chemical systems the internal energy U is equivalent to potential energy. Berthelot and Thomsen 1870’s Spontaneous change occurs in the direction in which the enthalpy of a system decreases. Mainly true but there are exceptions. Prentice-Hall © 2002 General Chemistry: Chapter 20

19-2 The Concept of Entropy Entropy, S. The greater the number of configurations of the microscopic particles among the energy levels in a particular system, the greater the entropy of the system. ΔS > 0 spontaneous ΔU = ΔH = 0 Prentice-Hall © 2002 General Chemistry: Chapter 20

The Boltzmann Equation for Entropy Chemistry 140 Fall 2002 The Boltzmann Equation for Entropy S = k lnW S is the entropy k = R/NA: Boltzmann constant W = number of microstates. The particular way in which particles are distributed amongst the microscopic energy levels called states available in a system with the same total energy. Ludwig Boltzmann. Associated the number of energy levels in a system with the number of ways of arranging the particles in these levels. General Chemistry: Chapter 20

Boltzmann Distribution FIG. 19-3 Chemistry 140 Fall 2002 Boltzmann Distribution FIG. 19-3 FIGURE 19-3 Energy levels for a particle in a one-dimensional box (a) The energy levels of a particle in a box become more numerous and closer together as the length of the box increases. The range of thermally accessible levels is indicated by the tinted band. The solid circles signify a system consisting of 15 particles. Each drawing represents a single microstate of the system. Can you see that as the box length increases there are many more microstates available to the particles for a given amount of thermal energy? As the number of possible microstates for a given total energy increases, so does the entropy. (b) More energy levels become accessible in a box of fixed length as the temperature is raised. Because the average energy of the particles also increases, the internal energy and entropy both increase as the temperature is raised. Prentice-Hall © 2002 General Chemistry: Chapter 20

General Chemistry: Chapter 20 Chemistry 140 Fall 2002 Entropy Change FIG. 19-4 ΔS = qrev T FIGURE 19-4 Three processes in which entropy increases Each of the processes pictured (a) the melting of a solid, (b) the evaporation of a liquid, and (c) the dissolving of a solute results in an increase in entropy. For part (c), the generalization works best for nonelectrolyte solutions, in which ion dipole forces do not exist. Prentice-Hall © 2002 General Chemistry: Chapter 20

19-3 Evaluating Entropy and Entropy Changes Chemistry 140 Fall 2002 19-3 Evaluating Entropy and Entropy Changes Phase transitions. Exchange of heat can be carried out reversibly. ΔS = ΔH Ttr H2O(s, 1 atm)  H2O(l, 1 atm) ΔHfus° = 6.02 kJ at 273.15 K ΔSfus = ΔHfus Ttr ° = 6.02 kJ mol-1 273.15 K = 22 J mol-1 K-1 Prentice-Hall © 2002 General Chemistry: Chapter 20

General Chemistry: Chapter 20 Chemistry 140 Fall 2002 Trouton’s Rule The molar boiling entropy changes are similar for many materials (e.g. toluene, 87.3, benzene, 89.5, chloroform, 87.9,…): ΔS = ΔHvap Tbp = 87 kJ mol-1 K-1 Failures of this rule are understandable: Water and hydrogen bonding. Sliquid is lower than expected so ΔS is larger than 87. Prentice-Hall © 2002 General Chemistry: Chapter 20

General Chemistry: Chapter 20 Chemistry 140 Fall 2002 Absolute Entropies Third law of thermodynamics. The entropy of a pure perfect crystal at 0 K is zero. Standard molar entropy. Tabulated in Table 17. Extensive state function, entropy can be summed in a multistep reaction ΔS = [ ∑pS°(products) - ∑rS°(reactants)]  are the stoichiometric coefficients. General Chemistry: Chapter 20

Molar entropies for the standard conditions Compound Smo, J K-1mol-1 Solids C (diamond) 2.38 C (graphite) 5.74 Na 51.2 K 64.2 Rb 69.5 Cs 85.2 NaCl 72.1 KCl 82.6 I2 116.1 General Chemistry: Chapter 20

Molar entropies for the standard conditions Compound Smo, J K-1mol-1 Liquids Hg 76.0 Br2 152.2 H2O 69.9 CH3OH 126.8 C2H5OH 160.7 C6H6 172.8 Prentice-Hall © 2002 General Chemistry: Chapter 20

Molar entropies for the standard conditions Compound Smo, J K-1mol-1 Triatomic Gases H2O 188.8 NO2 240.1 H2S 205.8 CO2 213.7 SO2 248.2 Prentice-Hall © 2002 General Chemistry: Chapter 20

Calculate DSº for the complete combustion of ethane (C2H6) with oxygen (O2). Equation: C2H6 + 7/2 O2  → 2 CO2 + 3 H2O Vapor S°, JK-1mol-1 C2H6 229.6 O2 205.1 CO2 213.7 H2O 188.8 DSº= [2(213.7) + 3(188.8)] - [229.6 + 7/2(205.1)] = 46.35 JK-1mol-1 Prentice-Hall © 2002 General Chemistry: Chapter 20

Entropy as a Function of Temperature Prentice-Hall © 2002 General Chemistry: Chapter 20

Vibrational Energy and Entropy Prentice-Hall © 2002 General Chemistry: Chapter 20

General Chemistry: Chapter 20 19-4 Criteria for Spontaneous Change: The Second Law of Thermodynamics. ΔStotal = ΔSuniverse = ΔSsystem + ΔSsurroundings The Second Law of Thermodynamics (Clausius): ΔSuniverse = ΔSsystem + ΔSsurroundings > 0 All spontaneous processes produce an increase in the entropy of the universe. (Macroscopic, non equilibrium, isolated systems – spontaneous process: the entropy increases; it is maximal in the equilibrium.) Prentice-Hall © 2002 General Chemistry: Chapter 20

Gibbs Energy and Gibbs Energy Change Chemistry 140 Fall 2002 Gibbs Energy and Gibbs Energy Change Hypothetical process: only pressure-volume work, at constant T and P. qsurroundings = -qp = -ΔHsys Make the enthalpy change reversible. large surroundings, infinitesimal change in temperature. Under these conditions we can calculate entropy. Want to evaluate the system only, not the surroundings. Prentice-Hall © 2002 General Chemistry: Chapter 20

Ice Melting Tsys= 273 K < Tsurr= 298 K and qsys = DHsys>0. P=const. Tsys= 273 K < Tsurr= 298 K and qsys = DHsys>0. Egyensúly

Ice Melting: Entalpy and Entropy The entropy change of the surroundings and system: The ice melts because the total entropy change is positive: the entropy of the universe increases. Egyensúly

Gibbs Energy and Energy Change For the universe (P and Tsurr = const.) reversible: Tsurr ΔSuniv. = Tsurr ΔSsys – ΔHsys = -(ΔHsys – TsurrΔSsys) -TΔSuniv. = ΔHsys – TΔSsys For the system (P and T = const.): G = H - TS ΔG = ΔH - TΔS ΔGsys = - TΔSuniverse General Chemistry: Chapter 20

Criteria for Spontaneous Change ΔGsys < 0 (negative), the process is spontaneous. ΔGsys = 0 (zero), the process is at equilibrium. ΔGsys > 0 (positive), the process is non-spontaneous. Prentice-Hall © 2002 General Chemistry: Chapter 20

19-5 Standard Gibbs Energy Change, ΔG° The standard Gibbs energy of formation, ΔfG°. The Gibbs energy change for a reaction in which a substance in its standard state is formed from its elements in reference forms in their standard states. The standard Gibbs energy of reaction, ΔG°. where i s are the stoichiometric factors ΔG° = [ ∑p ΔfG°(products) - ∑r ΔfG°(reactants)] Prentice-Hall © 2002 General Chemistry: Chapter 20

Standard changes of extensive quantities Reaction Gibbs energy: ΔG° = [ ∑p ΔfG°(products) - ∑r ΔfG°(reactants)] Reaction entropy change: ΔS° = [ ∑p S°(products) - ∑r S°(reactants)] Reaction enthalpy change: ΔH° = [ ∑p ΔfH°(products) - ∑r ΔfH°(reactants)] where i s are the stoichiometric factors. General Chemistry: Chapter 20

Általános Kémia, 2008 tavasz Termodinamic table P = 1 atm, T =298 K DfH° S° DfG° kJ mol-1 J mol-1 K-1 H2O(l) -285.8 69.9 -237.1 H2O(g) -241.8 188.8 -228.6 H2O(l) → H2O(g) 44.0 118.9 8.5 The enthalpy change is positive, endothermic. DG° positive, the reaction is non spontaneous at atmospheric pressure DG° = 0 if P = 23.76 mmHg - equilibrium. Equilibrium

Gibbs Energy Change and Equilibrium Prentice-Hall © 2002 General Chemistry: Chapter 20

19-6 Gibbs Energy Change and Equilibrium DG(T) = DH –TDS DS negative General Chemistry: Chapter 20

Temperature dependence of the Gibbs free energy DH and DS are constant for 298 - 400 K. At the boiling point: (DG = 0) 𝑇= Δ𝐻 Δ𝑆 Tboil = 44000/118.9 = 370 K ( 373 K) Above this temperature the DG < 0, and the water boils, evaporates completely. Δ𝐺=Δ𝐻−𝑇⋅Δ𝑆 0=Δ𝐻−𝑇⋅Δ𝑆 Egyensúly

Water boiling, P = 1 atm Egyensúly

+ Spontneous reactions DG(T) = DH –TDS DH DS DG(T) Examples - H2O(l) → H2O(g) High T helps H2O(g) → H2O(l) Low T helps 4Fe(s) +3O2(g) →2Fe2O3(s) always 2Fe2O3(s)→ 4Fe(s) +3O2(g) never T T 32 11/18/2018 Professor Gabor I. Csonka

Relationship of ΔG° to ΔG for Nonstandard Conditions N2(g) + 3 H2(g)  2 NH3(g) ΔG = ΔH - TΔS ΔG° = ΔH° - TΔS° For ideal gases ΔH = ΔH° ΔG = ΔH° - TΔS Prentice-Hall © 2002 General Chemistry: Chapter 20

Relationship Between S and S° Vf qrev = -w = RT ln Vi ΔS = qrev T = R ln Vf Vi ΔS = Sf – Si = R ln Vf Vi Pi Pf = -R ln S = S° - R ln P P° = S° 1 = S° - R ln P Prentice-Hall © 2002 General Chemistry: Chapter 20

N2(g) + 3 H2(g)  2 NH3(g) SN2 = S°N2 – R·ln PN2 SH2 = S°H2 – R·ln PH2 SNH3 = S°NH3 – R·ln PNH3 ΔSrxn = 2·(S°NH3 – R·ln PNH3) – (S°N2 – R·ln PN2) – 3· (S°H2 – R·ln PH2) PN2PH2 3 ΔSrxn = 2 S°NH3 –S°N2 –3S°H2+ R·ln PNH3 2 PN2PH2 3 ΔSrxn = ΔS°rxn + R·ln PNH3 2 Prentice-Hall © 2002 General Chemistry: Chapter 20

ΔG Under Non-standard Conditions PN2PH2 3 ΔG = ΔH° - TΔS ΔSrxn = ΔS°rxn + Rln PNH3 2 PN2PH2 3 ΔG = ΔH° - TΔS°rxn – TR ln PNH3 2 PNH3 2 ΔG = ΔG° + RT ln PN2PH2 3 ΔG = ΔG° + RT ln Q Q reaction quotient Prentice-Hall © 2002 General Chemistry: Chapter 20

ΔG and the Equilibrium Constant Keq ΔG = ΔG° + RT ln Q ΔG = ΔG° + RT ln Keq= 0 If the reaction is at equilibrium then: ΔG° = -RT ln Keq Prentice-Hall © 2002 General Chemistry: Chapter 20

Criteria for Spontaneous Change Every chemical reaction consists of both a forward and a reverse reaction. The direction of spontaneous change is the direction in which the Gibbs energy decreases. Prentice-Hall © 2002 General Chemistry: Chapter 20

Significance of the Magnitude of ΔG0 Prentice-Hall © 2002 General Chemistry: Chapter 20

The Thermodynamic Equilibrium Constant: Activities Chemistry 140 Fall 2002 The Thermodynamic Equilibrium Constant: Activities For ideal gases at 1.0 bar: S = S° - R ln P P° = S° 1 PV=nRT or P=(n/V)RT, pressure is an effective concentration Therefore, in solution: S = S° - R ln c c° = S° - R ln a Activity is the effective concentration divided by the effective concentration in the standard state. The effective concentration in the standard state for an ideal solution is c° = 1 M. Prentice-Hall © 2002 General Chemistry: Chapter 20

General Chemistry: Chapter 20 Activities For pure solids and liquids: a = 1 For ideal gases: a = P (in bars, 1 bar = 0.987 atm) For ideal solutes in aqueous solution: a = c (in mol L-1) Prentice-Hall © 2002 General Chemistry: Chapter 20

The Thermodynamic Equilibrium Constant, Keq A dimensionless equilibrium constant expressed in terms of activities. Often Keq = Kc Must be used to determine ΔG. ΔG = ΔG° + RT ln agah… aaab… Prentice-Hall © 2002 General Chemistry: Chapter 20

19-7 ΔG° and Keq as Functions of Temperature ΔG° = ΔH° -TΔS° ΔG° = -RT ln Keq ln Keq = -ΔG° RT = -ΔH° TΔS° + ln Keq = -ΔH° RT ΔS° R + ln = -ΔH° RT2 ΔS° R + RT1 - = 1 T2 T1 Keq1 Keq2 Prentice-Hall © 2002 General Chemistry: Chapter 20

Temperature Dependence of Keq ln Keq = -ΔH° RT ΔS° R + slope = -ΔH° R -ΔH° = R · slope = -8.3145 J mol-1 K-1 · 2.2·104 K = -1.8·102 kJ mol-1 Prentice-Hall © 2002 General Chemistry: Chapter 20

General Chemistry: Chapter 20 19-8 Coupled Reactions In order to drive a non-spontaneous reactions we changed the conditions (i.e. temperature or electrolysis) Another method is to couple two reactions. One with a positive ΔG and one with a negative ΔG. Overall spontaneous process. Prentice-Hall © 2002 General Chemistry: Chapter 20

General Chemistry: Chapter 20 Smelting Copper Ore Cu2O(s) → 2 Cu(s) + ½ O2(g) ΔG°673K = +125 kJ Δ Cu2O(s) → 2 Cu(s) + ½ O2(g) Non-spontaneous reaction: +125 kJ C(s) + ½ O2(g) → CO(g) Spontaneous reaction: -175 kJ Cu2O(s) + C(s) → 2 Cu(s) + CO(g) -50 kJ Spontaneous reaction! Prentice-Hall © 2002 General Chemistry: Chapter 20

Focus On Coupled Reactions in Biological Systems ADP3- + HPO42- + H+ → ATP4- + H2O ΔG° = -9.2 kJ mol-1 ΔG = ΔG° + RT ln aATPaH2O aADPaPiaH3O+ But [H3O+] = 10-7 M not 1.0 M. ΔG = -9.2 kJ mol-1 + 41.6 kJ mol-1 = +32.4 kJ mol-1 = ΔG°' The biological standard state: Prentice-Hall © 2002 General Chemistry: Chapter 20

Focus On Coupled Reactions in Biological Systems Glucose → 2 lactate + 2 H+ -218 kJ 2 ADP3- + 2 HPO42- + 2 H+ → 2 ATP4- + 2 H2O +64 kJ 2 ADP3- + 2 HPO42- + glucose → 2 ATP4- + 2 H2O + 2 lactate -153 kJ Prentice-Hall © 2002 General Chemistry: Chapter 20