Chapter 16 Thermodynamics: Entropy, Free Energy, and Equilibrium spontaneous nonspontaneous In this chapter we will determine the direction of a chemical reaction and calculate equilibrium constant using thermodynamic values: Entropy and Enthalpy.

Spontaneous Processes Spontaneous Process: A process that, once started, proceeds on its own without a continuous external influence.

Universe: System + Surroundings The system is what you observe; surroundings are everything else.

Thermodynamics State functions are properties that are determined by the state of the system, regardless of how that condition was achieved. energy , pressure, volume, temperature Potential energy of hiker 1 and hiker 2 is the same even though they took different paths.

Laws of Thermodynamics 1) Energy is neither created nor destroyed. 2) In any spontaneous process the total Entropy of a system and its surrounding always increases. We will prove later: ∆G = ∆Hsys – T∆Ssys ∆G < 0 Reaction is spontaneous in forward direction 3) The entropy of a perfect crystalline substance is zero at the absolute zero ( K=0)

Enthalpy, Entropy, and Spontaneous Processes State Function: A function or property whose value depends only on the present state, or condition, of the system, not on the path used to arrive at that state. Enthalpy Change (DH): The heat change in a reaction or process at constant pressure. Entropy (DS): The amount of Molecular randomness change in a reaction or process at constant pressure.

Enthalpy Change Exothermic: CO2(g) + 2H2O(l) CH4(g) + 2O2(g) = -890.3 kJ DH° Endothermic: H2O(l) H2O(s) = +6.01 kJ DHfusion H2O(g) H2O(l) = +40.7 kJ DHvap All are spontaneous processes. Enthalpy alone isn’t enough to determine spontaneity. 2NO2(g) N2O4(g) = +57.1 kJ DH° Na1+(aq) + Cl1-(aq) NaCl(s) H2O = +3.88 kJ DH°

Entropy Change DS = Sfinal - Sinitial

Entropy

Entropy

The sign of entropy change, DS, associated with the boiling of water is _______. Positive Negative Zero

Correct Answer: Positive Negative Zero Vaporization of a liquid to a gas leads to a large increase in volume and hence entropy; DS must be positive.

Entropy and Temperature 02

Third Law of Thermodynamics The entropy of a perfect crystalline substance is zero at the absolute zero ( K=0) Ssolid < Sliquid < Sgas

Entropy Changes in the System (DSsys) When gases are produced (or consumed) If a reaction produces more gas molecules than it consumes, DS0 > 0. If the total number of gas molecules diminishes, DS0 < 0. If there is no net change in the total number of gas molecules, then DS0 may be positive or negative . 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, DS is negative.

Standard Molar Entropies and Standard Entropies of Reaction Standard Molar Entropy (S°): The entropy of 1 mole of a pure substance at 1 atm pressure and a specified temperature. Calculated by using S = k ln W , W = Accessible microstates of translational, vibrational and rotational motions

DH° = DH°f (Products) – DH°f (Reactants) Review of Ch 8: ∆G = ∆H – T∆S see page 301 of your book We will show the proof of this formula later in this chapter Calculating DH° for a reaction: DH° = DH°f (Products) – DH°f (Reactants) For a balanced equation, each heat of formation must be multiplied by the stoichiometric coefficient. aA + bB cC + dD DH° = [cDH°f (C) + dDH°f (D)] – [aDH°f (A) + bDH°f (B)] See Appendix B, end of your book

Entropy Changes in the System (DSsys) The standard entropy change of reaction (DS0 ) is the entropy change for a reaction carried out at 1 atm and 250C. rxn aA + bB cC + dD DS0 rxn dS0(D) cS0(C) = [ + ] - bS0(B) aS0(A) DS0 rxn nS0 (products) = S mS0 (reactants) - What is the standard entropy change for the following reaction at 250C? 2CO (g) + O2 (g) 2CO2 (g) S0(CO) = 197.6 J/K•mol S0(CO2) = 213.6 J/K•mol S0(O2) = 205.0 J/K•mol DS0 rxn = 2 x S0(CO2) – [2 x S0(CO) + S0 (O2)] DS0 rxn = 427.2 – [395.2 + 205.0] = -173.0J/K•mol

Calculating ∆S for a Reaction ∆So =  So (products) -  So (reactants) Consider 2 H2(g) + O2(g) ---> 2 H2O(liq) ∆So = 2 So (H2O) - [2 So (H2) + So (O2)] ∆So = 2 mol (69.9 J/K•mol) - [2 mol (130.6 J/K•mol) + 1 mol (205.0 J/K•mol)] ∆So = -326.4 J/K Note that there is a decrease in Entropy because 3 mol of gas give 2 mol of liquid.

Calculate DS° for the equation below using the standard entropy data given: 2 NO(g) + O2(g)  2 NO2(g) DS° values (J/mol-K): NO2(g) = 240, NO(g) = 211, O2(g) = 205. +176 J/mol +147 J/mol 147 J/mol 76 J/mol

Correct Answer:   ( ) ( )  DS° = 2(240)  [2(211) + 205] +176 J/K +147 J/K 147 J/K 76 J/K  ( )  ( ) D S  = D S  products  D S  reactants DS° = 2(240)  [2(211) + 205] DS° = 480  [627] DS° =  147

Thermite Reaction

Spontaneous Processes and Entropy 07 Consider the gas phase reaction of A2 molecules (red) with B atoms (blue). (a) Write a balanced equation for the reaction. (b) Predict the sign of ∆S for the reaction.

Entropy Changes in the Surroundings (DSsurr) Second Law Of Thermodynamic: In any spontaneous process the total Entropy of a system and its surrounding always increases. Entropy Changes in the Surroundings (DSsurr) Exothermic Process DSsurr > 0 Endothermic Process DSsurr < 0

Entropy and the Second Law of Thermodynamics Dssurr a ( - DH) T 1 Dssurr a ( ) Dssurr = T - DH See example of tossing a rock into a calm waters vs. rough waters Page 661 of your book

2nd Law of Thermodynamics 01 The total entropy increases in a spontaneous process and remains unchanged in an equilibrium process. Spontaneous: ∆Stotal = ∆Ssys + ∆Ssur > 0 Equilibrium: ∆Stotal = ∆Ssys + ∆Ssur = 0 The system is what you observe; surroundings are everything else.

2nd Law of Thermodynamics Calculate change of entropy of surrounding in the following reaction: 2 H2(g) + O2(g) ---> 2 H2O(liq) ∆H ° = -571.7 kJ D S surr = -D H sys T ∆Sosurroundings = +1917 J/K

Spontaneous Reactions 01 The 2nd law tells us a process will be spontaneous if ∆Stotal > 0 which requires a knowledge of ∆Ssurr. spontaneous: ∆Stotal = ∆Ssys + ∆Ssur > 0 D S sur = -D H sys T ∆Stotal = ∆Ssys + ( ) > 0 -D H sys T -T (∆Stotal = ∆Ssys + ( ) ) < 0 -D H sys T

∆G = ∆Hsys – T∆Ssys Gibbs Free Energy 02 The expression –T∆Stotal is equated as Gibbs free energy change (∆G)*, or simply free energy change: ∆G = ∆Hsys – T∆Ssys ∆G < 0 Reaction is spontaneous in forward direction. ∆G = 0 Reaction is at equilibrium. ∆G > 0 Reaction is spontaneous in reverse direction. -T . ∆Stotal = -T. ∆Ssys + ΔHsys < 0 –T∆Stotal = ΔG * Driving force of a reaction, Maximum work you can get from a system.

Calculate ∆Go rxn for the following: C2H2(g) + 5/2 O2(g) --> 2 CO2(g) + H2O(g) Use enthalpies of formation to calculate ∆Horxn = -1256 kJ Use standard molar entropies to calculate ∆Sorxn ( see page 658, appendix A-10) ∆Sorxn = -97.4 J/K or -0.0974 kJ/K ∆Gorxn = -1256 kJ - (298 K)(-0.0974 kJ/K) = -1227 kJ Reaction is product-favored in spite of negative ∆Sorxn. Reaction is “enthalpy driven”

Calculate DG° for the equation below using the thermodynamic data given: N2(g) + 3 H2(g)  2 NH3(g) DHf° (NH3) = 46 kJ; S° values (J/mol-K) are: NH3 = 192.5, N2 = 191.5, H2 = 130.6. +33 kJ +66 kJ 66 kJ 33 kJ

Correct Answer: DH° = 92 kJ DS° = 198.3J/mol . K DG° = DH°  TDS° DG° = (92) + (59.2) = 32.9 KJ

Gibbs Free Energy 03 Using ∆G = ∆H – T∆S, we can predict the sign of ∆G from the sign of ∆H and ∆S. 1) If both ∆H and ∆S are positive, ∆G will be negative only when the temperature value is large. Therefore, the reaction is spontaneous only at high temperature. 2) If ∆H is positive and ∆S is negative, ∆G will always be positive. Therefore, the reaction is not spontaneous

Gibbs Free Energy: 04 ∆G = ∆H – T∆S 3) If ∆H is negative and ∆S is positive, ∆G will always be negative. Therefore, the reaction is spontaneous 4) If both ∆H and ∆S are negative, ∆G will be negative only when the temperature value is small. Therefore, the reaction is spontaneous only at Low temperatures.

Gibbs Free Energy 04

Gibbs Free Energy 06 What are the signs (+, –, or 0) of ∆H, ∆S, and ∆G for the following spontaneous reaction of A atoms (red) and B atoms (blue)?

∆G = ∆Hsys – T∆Ssys Gibbs Free Energy 02 The expression –T∆Stotal is equated as Gibbs free energy change (∆G)*, or simply free energy change: ∆G = ∆Hsys – T∆Ssys ∆G < 0 Reaction is spontaneous in forward direction. ∆G = 0 Reaction is at equilibrium. ∆G > 0 Reaction is spontaneous in reverse direction. -T . ∆Stotal = -T. ∆Ssys + ΔHsys < 0 –T∆Stotal = ΔG * Driving force of a reaction, Maximum work you can get from a system.

Standard Free-Energy Changes for Reactions Calculate the standard free-energy change at 25 °C for the Haber synthesis of ammonia using the given values for the standard enthalpy and standard entropy changes: 2NH3(g) N2(g) + 3H2(g) DH° = -92.2 kJ DS° = -198.7 J/K Negative value for DG° means it’s spontaneous. Spontaneous doesn’t mean fast, however. Rate of reaction is kinetics. DG° = DH° - TDS° -92.2 kJ - 298 K K -198.7 J 1000 J 1 kJ = x x = -33.0 kJ

Fe2O3(s) + 3 H2(g)  2 Fe(s) + 3 H2O(g) Gibbs Free Energy 05 Iron metal can be produced by reducing iron(III) oxide with hydrogen: Fe2O3(s) + 3 H2(g)  2 Fe(s) + 3 H2O(g) ∆H° = +98.8 kJ; ∆S° = +141.5 J/K Is this reaction spontaneous at 25°C? At what temperature will the reaction become spontaneous?

Decomposition of CaCO3 has a H° = 178 Decomposition of CaCO3 has a H° = 178.3 kJ/mol and S° = 159 J/mol  K. At what temperature does this become spontaneous? 121°C 395°C 848°C 1121°C

Correct Answer: T = H°/S° T = 178.3 kJ/mol/0.159 kJ/mol  K T (°C) = 1121 – 273 = 848

Standard Free Energies of Formation DG° = DG°f (products) - DG°f (reactants) cC + dD aA + bB DG° = [cDG°f (C) + dDG°f (D)] - [aDG°f (A) + bDG°f (B)] Products Reactants

Standard free energy of formation (DG0f) is the free-energy change that occurs when1 mole of the compound is formed from its elements in their standard ( 1 atm) states. Figure: Table 17-03 Title: Caption:

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 DG0 rxn dDG0 (D) f cDG0 (C) = [ + ] - bDG0 (B) aDG0 (A) -- DG0 rxn nDG0 (products) f = S mDG0 (reactants) - DG0 of any element in its stable form is zero. f

Calculate DG° for the equation below using the Gibbs free energy data given: 2 SO2(g) + O2(g)  2 SO3(g) DGf° values (kJ): SO2(g) = 300.4, SO3(g) = 370.4 +70 kJ +140 kJ 140 kJ 70 kJ

Correct Answer: ( ) ( )    G =  G products  G reactants ° ° ° +70 kJ +140 kJ 140 kJ 70 kJ ( ) ( )  G ° =   G °  products   G ° reactants f f DG° = (2  370.4)  [(2  300.4) + 0] DG° = 740.8  [600.8] = 140

Gibbs Free Energy 07

Calculation of Nonstandard ∆G The sign of ∆G° tells the direction of spontaneous reaction when both reactants and products are present at standard state conditions. Under nonstandard( condition where pressure is not 1atm or concentrations of solutions are not 1M) conditions, ∆G˚ becomes ∆G. ∆G = ∆G˚ + RT lnQ The reaction quotient is obtained in the same way as an equilibrium expression.

Free Energy Changes and the Reaction Mixture DG = DG° + RT ln Q Calculate DG for the formation of ethylene (C2H4) from carbon and hydrogen at 25 °C when the partial pressures are 100 atm H2 and 0.10 atm C2H4. C2H4 P H2 2 C2H4(g) 2C(s) + 2H2(g) Qp = Calculate ln Qp: = -11.51 1002 0.10 ln

Free Energy Changes and the Reaction Mixture Calculate DG: DG = DG° + RT ln Q K mol J 1000 J 1 kJ = 68.1 kJ/mol + 8.314 (298 K)(-11.51) ( see page A-13 of your book, page 667) DG = 39.6 kJ/mol Since the value of DG is positive, it is nonspontaneous in the forward direction.

Free Energy and Chemical Equilibrium 05 At equilibrium ∆G = 0 and Q = K ∆G = ∆G˚ + RT lnQ ∆G˚rxn = –RT ln K

Free Energy and Chemical Equilibrium Calculate Kp at 25 °C for the following reaction: CaO(s) + CO2(g) CaCO3(s) Calculate DG°: DG° = [DG°f (CaO(s)) + DG°f (CO2(g))] - [DG°f (CaCO3(s))] Please see appendix B: = [(1 mol)(-604.0 kJ/mol) + (1 mol)(-394.4 kJ/mol)] - [(1 mol)(-1128.8 kJ/mol)] DG° = +130.4 kJ/mol

Free Energy and Chemical Equilibrium Calculate ln K: DG° = -RT ln K -130.4 kJ/mol RT -DG° ln K = = K mol J 1000 J 1 kJ 8.314 (298 K) ln K = -52.63 Calculate K: -52.63 K = e = 1.4 x 10-23

Figure: 19-17a, b Title: Potential energy and free energy. Caption: An analogy between the gravitational potential-energy change of a boulder rolling down a hill (a) and the free-energy change in a spontaneous reaction (b). The equilibrium position in (a) is given by the minimum gravitational potential energy available to the system. The equilibrium position in (b) is given by the minimum free energy available to the system.

Free Energy and Chemical Equilibrium 04 ∆G˚rxn = –RT ln K

Suppose G° is a large, positive value Suppose G° is a large, positive value. What then will be the value of the equilibrium constant, K? K = 0 K = 1 0 < K < 1 K > 1

Correct Answer: G° = RTlnK K = 0 K = 1 0 < K < 1 K > 1 Thus, large positive values of DG° lead to large negative values of lnK. The value of K itself, then, is very small. K = 0 K = 1 0 < K < 1 K > 1

More thermo? You betcha!

Thermodynamics and Keq ∆Gorxn = - RT lnK Calculate K for the reaction N2O4 --->2 NO2 ∆Gorxn = +4.8 kJ/mole ( see page A-11) ΔG° = RT lnK ∆Gorxn = +4800 J = - (8.31 J/mol.K)(298 K) ln K K = 0.14 When ∆Gorxn > 0, then K < 1