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: CH4(g) + 2O2(g) CO2(g) + 2H2O(l) DH° = -890.3 kJ Endothermic: H2O(s) H2O(l) H2O(l) H2O(g) N2O4(g) NaCl(s) H2O DHfusion = +6.01 kJ DHvap = +40.7 kJ 2NO2(g) DH° = +57.1 kJ Na1+(aq) + Cl1-(aq) DH° = +3.88 kJ Entropy Change DS = Sfinal - Sinitial Entropy Entropy The sign of entropy change, DS, associated with the boiling of water is _______. 1. Positive 2. Negative 3. Zero Correct Answer: 1. Positive 2. Negative 3. 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 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) See Appendix B, end of your book • 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)] Entropy Changes in the System (DSsys) 0 ) is the The standard entropy change of reaction (DS rxn entropy change for a reaction carried out at 1 atm and 250C. aA + bB DS0rxn = cC + dD [ cS0(C) + dS0(D) ] - [ aS0(A) + bS0(B) ] DS0rxn = S 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(O2) = 205.0 J/K•mol S0(CO2) = 213.6 J/K•mol DS0rxn = 2 x S0(CO2) – [2 x S0(CO) + S0 (O2)] DS0rxn = 427.2 – [395.2 + 205.0] = -173.0J/K•mol Calculating ∆S for a Reaction ∆So = S So (products) - S 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. 1. +176 J/mol 2. +147 J/mol 3. 147 J/mol 4. 76 J/mol Correct Answer: 1. +176 J/K 2. +147 J/K 3. 147 J/K 4. 76 J/K DS = S DS (products) S DS (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. 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) Dssurr a ( Dssurr = 1 ) T - DH T 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 DSsurr = DS D H sys o surr oundings T = - (-571.7 kJ)(1000 J/kJ) 298.15 K ∆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 DS sur = •∆Stotal = ∆Ssys + ( D H sys T D H sys ) > 0 T D H sys -T (∆Stotal = ∆Ssys + ( T ) ) <0 Gibbs Free Energy • 02 The expression –T∆Stotal is equated as Gibbs free energy change (∆G)*, or simply free energy change: • -T • • ∆G • . ∆S total = -T. ∆Ssys + ΔHsys < 0 –T∆Stotal = ΔG = ∆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. * 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; 1. +33 kJ 2. +66 kJ 3. 66 kJ 4. 33 kJ Correct Answer: 1. +33 kJ 2. +66 kJ 3. 66 kJ 4. 33 kJ DH° = 92 kJ DS° = 198.3J/mol . K DG° = DH° TDS° DG° = (92) (298)(0.1983) 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) 2) 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. If ∆H is positive and ∆S is negative, ∆G will always be positive. Therefore, the reaction is not spontaneous Gibbs Free Energy: ∆G = ∆H – T∆S 04 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)? Gibbs Free Energy • 02 The expression –T∆Stotal is equated as Gibbs free energy change (∆G)*, or simply free energy change: • -T • • ∆G • . ∆S total = -T. ∆Ssys + ΔHsys < 0 –T∆Stotal = ΔG = ∆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. * 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: N2(g) + 3H2(g) 2NH3(g) DH° = -92.2 kJ DS° = -198.7 J/K DG° = DH° - TDS° = -92.2 kJ - 298 K x -198.7 J K x 1 kJ 1000 J = -33.0 kJ 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 1. Is this reaction spontaneous at 25°C? 2. At what temperature will the reaction become spontaneous? Decomposition of CaCO3 has a DH° = 178.3 kJ/mol and DS° = 159 J/mol K. At what temperature does this become spontaneous? 1. 2. 3. 4. 121°C 395°C 848°C 1121°C Correct Answer: 1. 2. 3. 4. 121°C 395°C 848°C 1121°C T = DH°/DS° T = 178.3 kJ/mol/0.159 kJ/mol K T = 1121 K T (°C) = 1121 – 273 = 848 Standard Free Energies of Formation DG° = DG°f (products) - DG°f (reactants) aA + bB cC + dD 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. The standard free-energy of reaction (DG0rxn) is the freeenergy change for a reaction when it occurs under standardstate conditions. aA + bB cC + dD 0 DGrxn = [cDG0f (C) + dDG0f (D) ] - [aDG0f (A) + bDG0f (B) ] 0 DGrxn = S nDG0f (products) - S mDG0f (reactants) DG0f of any element in its stable form is zero. -- 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 1. +70 kJ 2. +140 kJ 3. 140 kJ 4. 70 kJ Correct Answer: 1. +70 kJ 2. +140 kJ 3. 140 kJ 4. 70 kJ DG ° = DGf (products) DGf (reactants) ° ° 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. 2C(s) + 2H2(g) C2H4(g) Qp = PC H 2 4 2 PH Calculate ln Qp: 0.10 ln 1002 = -11.51 2 Free Energy Changes and the Reaction Mixture Calculate DG: DG = DG° + RT ln Q = 68.1 kJ/mol + 8.314 ( see page A-13 of your book, page 667) DG = 39.6 kJ/mol J K mol 1 kJ 1000 J (298 K)(-11.51) Free Energy and Chemical Equilibrium • At equilibrium ∆G = 0 and Q = K • ∆G = ∆G˚ + RT lnQ ∆G˚rxn = –RT ln K 05 Free Energy and Chemical Equilibrium Calculate Kp at 25 °C for the following reaction: CaCO3(s) CaO(s) + CO2(g) 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 -DG° ln K = = RT 8.314 J 1 kJ K mol 1000 J ln K = -52.63 Calculate K: -52.63 K=e = 1.4 x 10-23 (298 K) Free Energy and Chemical Equilibrium ∆G˚rxn = –RT ln K 04 Suppose DG° is a large, positive value. What then will be the value of the equilibrium constant, K? 1. 2. 3. 4. K=0 K=1 0<K<1 K>1 Correct Answer: DG° = RTlnK 1. 2. 3. 4. 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. 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 ln K = - 4800 J/mole = - 1.94 (8.31 J/m.K)(298 K) K = 0.14 When ∆Gorxn > 0, then K < 1