Chapter 11 Spontaneous Change and Equilibrium 11-1 Enthalpy and Spontaneous Change 11-2 Entropy 11-3 Absolute Entropies and Chemical Reactions 11-4 The Second Law of Thermodynamics 11-5 The Gibbs Function 11-6 The Gibbs Function and Chemical Reactions 11-7 The Gibbs Function and the Equilibrium Constant 11-8 The Temperature Dependence of Equilibrium Constants 11/23/2018 OFP Chapter 11 16 1

Chapter 11 Spontaneous Change and Equilibrium Second Law of Thermodynamics In a real spontaneous process the Entropy of the universe (meaning the system plus its surroundings) must increase. Suniverse > 0 (spontaneous process) Suniverse = Ssystem + Ssurroundings > 0 if Suniverse = 0, then everything is in equilibrium The 2nd Law of Thermodynamics profoundly affects how we look at nature and processes 11/23/2018 OFP Chapter 11 16 3

(conservation of energy) Summarize a few Concepts 1st Law of Thermodynamics In any process, the total energy of the universe remains unchanged: energy is conserved A process and its reverse are equally allowed Eforward = – Ereverse (conservation of energy) 2nd Law of Thermodynamics S, the entropy of a universe, increases in only one of the two directions of a reaction Processes that decrease ΔS are impossible. Or improbable beyond conception ΔSuniv > 0 Spontaneous ΔSuniv = 0 Equilibrium ΔSuniv < 0 Non-spontaneous 11/23/2018 OFP Chapter 11

(spontaneous process) Gibbs Function How are Enthalpy and Entropy related? G = H - T•S G has several names Gibbs function Gibbs free energy Free Enthalpy For the change in the Gibbs Energy of system, at constant Temperature and Pressure ΔGsys = ΔHsys - T·ΔSsys ΔGsystem = Δ Hsys - TΔSsystem = < 0 (spontaneous process) 11/23/2018 OFP Chapter 11

For a change at constant temperature and pressure ΔGsys < 0 Spontaneous ΔGsys = 0 Equilibrium ΔGsys > 0 Non-spontaneous, but the reverse is spontaneous From Earlier ΔSuniv > 0 Spontaneous ΔSuniv = 0 Equilibrium ΔSuniv < 0 Non-spontaneous 11/23/2018 OFP Chapter 11

Typical example using Gibbs Free energy Benzene, C6H6, boils at 80.1°C and ΔHvap = 30.8 kJ a) Calculate ΔSvap for 1 mole of benzene B) at 60°C and pressure = 1 atm does benzene boil? 11/23/2018 OFP Chapter 11

(equilibrium process) Benzene, C6H6, boils at 80.1°C. ΔHvap = 30.8 kJ a) Calculate ΔSvap for 1 mole of benzene B) at 60°C and pressure = 1 atm does benzene boil? ΔGsys < 0 Spontaneous ΔGsys = 0 Equilibrium ΔGsys > 0 Non-spontaneous ΔGsystem = Δ Hsys - TΔSsystem = 0 (equilibrium process) 11/23/2018 OFP Chapter 11

The Gibbs Function and Chemical Reactions ΔG = ΔH - T·ΔS ΔGf° is the standard molar Gibbs function of formation Because G is a State Property, DGrxno for a general reaction a A + b B → c C + d D 11/23/2018 OFP Chapter 11

Example 11-7 3NO(g) → N2O(g) + NO2(g) Calculate ΔG° for the following reaction at 298.15K. Use Appendix D for additional information needed. 3NO(g) → N2O(g) + NO2(g) Solution From Appendix D ΔGf°(N2O) = 104.18 kJ mol-1 ΔGf°(NO2) = 51.29 ΔGf° (NO) = 86.55 ΔG°= 1(104.18)+1(51.29)-3(86.55) ΔG°= − 104.18 kJ therefore, spontaneous 11/23/2018 OFP Chapter 11

Effects of Temperature on ΔG° For temperatures other than 298K ΔG = ΔH - TΔS Typically ΔH and ΔS are almost constant over a broad range For above reaction, as Temperature increases ΔG becomes more positive, i.e., less negative. Next Slide ΔG has a strong temperature dependence 3NO (g) → N2O (g) + NO2 (g) 11/23/2018 OFP Chapter 11

G has a strong temperature dependence 3NO(g) → N2O(g) + NO2(g) G has a strong temperature dependence 11/23/2018 OFP Chapter 11

For temperatures other than 298K or 25C ΔG = ΔH - T·ΔS 11/23/2018 OFP Chapter 11

For temperatures other than 298K or 25C ΔG = ΔH - T·ΔS B A C D 11/23/2018 OFP Chapter 11

For temperatures other than 298K or 25C ΔG = ΔH - T·ΔS Case C ΔH° > 0 ΔS° < 0 ΔG = ΔH - T·ΔS ΔG = (+) - T·(-) = positive  ΔG > 0 or non-spontaneous at all Temp. B A Case B ΔH° < 0 ΔS° > 0 ΔG = ΔH - T·ΔS ΔG = (-) - T·(+) = negative  ΔG < 0 or spontaneous at all temp. C D 11/23/2018 OFP Chapter 11

For temperatures other than 298K or 25C ΔG = ΔH - T·ΔS The spontaneity of a reaction at different temperatures depends on the signs of Δ H° & ΔS°. When both signs are the same, the temperature determines the spontaneity. For the next two cases, it is more interesting because there is a temperature at which ΔG = 0 Then above or below a temperature the ΔG is either positive or negative 11/23/2018 OFP Chapter 11

11/23/2018 OFP Chapter 11

For temperatures other than 298K or 25C ΔG = ΔH - T·ΔS Case D ΔH° < 0 ΔS° < 0 ΔG = ΔH - T·ΔS at a low Temp ΔG = (-) - T·(-) = negative  ΔG < 0 or spontaneous at low Temp. B A Case A ΔH° > 0 ΔS° > 0 ΔG = ΔH - T·ΔS ΔG = (+) - T·(+) at a High Temp  ΔG < 0 or spontaneous at high Temp. C D 11/23/2018 OFP Chapter 11

The Gibbs Function and the Equilibrium Constant What about non-standard states, other than 1 atm or a conc. [X] = 1 mol/L? ΔG = ΔG° + RT ln Q Where Q is the reaction quotient 11/23/2018 OFP Chapter 11

Where Q is the reaction quotient ΔG = ΔG° + RT ln Q Where Q is the reaction quotient a A + b B ↔ c C + d D If Q > K the rxn shifts towards the reactant side The amount of products are too high relative to the amounts of reactants present, and the reaction shifts in reverse (to the left) to achieve equilibrium If Q = K equilibrium If Q < K the rxn shifts toward the product side The amounts of reactants are too high relative to the amounts of products present, and the reaction proceeds in the forward direction (to the right) toward equilibrium compare 11/23/2018 OFP Chapter 11

- ΔG ° = RT ln K ΔG = ΔG° + RT ln Q a A + b B ↔ c C + d D Where Q is the reaction quotient a A + b B ↔ c C + d D If Q < K the rxn shifts towards the product side If Q = K equilibrium If Q > K the rxn shifts toward the reactant side At Equilibrium conditions, ΔG = 0 - ΔG ° = RT ln K NOTE: we can now calculate equilibrium constants (K) for reactions from standard ΔGf functions of formation 11/23/2018 OFP Chapter 11

Where Q is the reaction quotient ΔG = ΔG° + RT ln Q Where Q is the reaction quotient a A + b B ↔ c C + d D 11/23/2018 OFP Chapter 11

Example 11-9 Found ΔG°= - 104.18 kJ mol -1 The ΔGr° for the following reaction at 298.15K was obtained in example 11-7. Calculate the equilibrium constant for this reaction at 25C. 3NO(g) ↔ N2O(g) + NO2(g) Strategy Use - ΔG ° = RT ln K Found ΔG°= - 104.18 kJ mol -1 from example 11-7 11/23/2018 OFP Chapter 11

Example 11-9 3NO(g) ↔ N2O(g) + NO2(g) Solution - ΔG ° = RT ln K Use - ΔG ° = RT ln K Rearrange Use ΔGr°= - 104.18 kJ mol-1 from Ex. 11-7 11/23/2018 OFP Chapter 11

Chapter 11 Spontaneous Change and Equilibrium Examples / Exercises 11-2, 11-3, 11-4, 11-5, 11-6 11-7, 11-8, 11-9, 11-10, 11-11, 11-12 Problems WebCT 8abc 30abcd 44a 50abc 11/23/2018 OFP Chapter 11 16 4