Chapter 15 - Spontaneity, Entropy, and Free Energy 1 Spontaneous Processes 2 The Isothermal Expansion and Compression of an Ideal Gas 3 The Definition of Entropy 4 Entropy and Physical Changes 5 Entropy and the Second Law of Thermodynamics 6 The Effect of Temperature on Spontaneity 7 Free Energy 8 Entropy Changes in Chemical Reactions 9 Free Energy and Chemical Reactions 10 The Dependence of Free Energy on Pressure 11 Free Energy and Equilibrium 12 Free Energy and Work (skip) 13 Reversible and Irreversible Processes: A Summary (skip) 14 Adiabatic Processes (skip)

Spontaneous Processes Kinetics vs Thermo Thermodynamics predicts direction and “driving force”. Kinetics predicts speed (rate). Spontaneous Processes Occur on some timescale (maybe slowly) without outside intervention (examples: a battery will discharge, a hot cup of coffee will cool to ambient temperature). All spontaneous processes proceed toward “states” (macrostates) with the greatest number of accessible microstates.

Microstates and Macrostates: An available microstate describes a specific detailed microscopic configuration (molecular rotations, translations, vibrations, electronic configuration) that a system can visit in the course of its fluctuations. A macrostate describes macroscopic properties such as temperature and pressure. For a gas at constant T: the number of available microstates increases with volume. For gas, liquid or solid, the number of available microstates increases with T (the number of available vibrational microstates, electronic microstates, etc. increases with T). When you heat anything, you increase the number of available microstates. When a liquid vaporizes, the number of available microstates increases. When a liquid freezes, the number of available microstates decreases.

What is a spontaneous process? probable not probable This is not a spontaneous process. The reverse process (going from right to left) is spontaneous. a) A gas will spontaneously expand to fill the available space. b) There is a ‘driving’ force that causes a gas to spontaneously expand to fill a vacuum. c) The entropy of the universe increases with a gas expands to fill a vacuum. a=b=c

Why is this not a spontaneous process? probable not probable There are more available microstates on the left hand side than on the right hand side. If a system gains degrees of freedom (more constituents, more room to move, more available quantum states, more available rotational, vibrational, translational or electronic states), then it gains entropy. A spontaneous process increases entropy (but you must consider both the system and the surroundings)

probable not probable The probability of finding both molecules on the left side is ¼ (this is one available microstate out of four possible microstates that will give this arrangement) The probability of finding one molecule on the each side is ½. (this are two possible microstates out of four possible microstates that will give this arrangement) The probability of finding both molecules on the right side is ¼ (this is one microstate out of four possible microstates) The probability of occurrence of a particular arrangement (state) depends on the number of ways (microstates) in which that arrangement can be achieved. All microstates are equally probable.

You will see that arrangement III is most probable. There are three possible arrangements of four molecules in two chambers. The arrangement with the greatest number of microstates is most probable. Label the molecules a,b,c,d and count the microstates. You will see that arrangement III is most probable.

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2 molecules Probability of finding 2 molecules on the same side is 1/4

Definitions of Entropy “S” Entropy is related to probability If a system has several available macrostates, it will spontaneously proceed to the one with the largest number of available microstates. The macrostate with the greatest probability (largest number of available microstates) has the highest entropy. When you heat something you increase its entropy. S = kB ln Ω Joules/Kelvin Kb = Boltzmann’s constant, the gas constant per molecule (R/NA) Ω = the number of available microstates of a given state ∆S = q/T J / mol-K

Ludwig Boltzmann (1844-1906) Highlights Moments in a Life Established the logarithmic connection between entropy and probability in his kinetic theory of gases. The Boltzmann constant (k or kB) is the physical constant relating temperature to energy. Moments in a Life Suffered from bipolar disorder and depression Ironically, in Max Planck’s Nobel Prize speech in 1918, it was pointed out that Boltzmann never introduced the constant k, Planck did. 5/19/2018 Zumdahl Chapter 10

Twice the number of microstates One He in the gas phase expands from volume V1 to 2V1 Ω2 = 2Ω1 Twice the number of microstates If 1 mole of He (instead of 2 He) and the gas expands from V1 to V2 The change in entropy of a gas is dependent on the change in volume of the gas 5/19/2018 Zumdahl Chapter 10

The isothermal expansion of an ideal gas. Isothermal – system and surroundings maintain constant temperature. ΔE = 0 = q + w then q = – w Consider only reversible and irreversible processes For a reversible, cyclic process both the system and the surroundings are returned exactly to their original positions. Cyclic expansion-compression process “work is converted to heat” Work → Heat 5/19/2018 Zumdahl Chapter 10

The isothermal expansion of an ideal gas. ∆E=0 (energy of a perfect gas depends only on T) ∆E= w + q w = -q This important relationship entropy (determined by number of available microscopic states) is related to a macroscopic properties of heat and temperature.

↓ ΔS(A) < 0 (cools) ΔS(B) > 0 (heats) |ΔS(A)| > |ΔS(B)| Brick A (warm) Brick B (warm) w(A) q(A) ΔE(A) ΔH(A) ΔS(A) W(B) Q(B) ΔE(B) ΔH(B) ΔS(B) ↓ Brick A (cold) Brick B (hot) ΔS(A) < 0 (cools) ΔS(B) > 0 (heats) |ΔS(A)| > |ΔS(B)| ΔS(uni) = ΔS(A) + ΔS(B) ΔS(uni) < 0 This is not a spontaneous process.

Entropy and Physical Change Temperature Dependence of Entropy: Cp and Cv are is the heat capacities of the system. ΔS(T1 to T2) here should be written ΔSsys(T1 to T2)

Example Calculate the change in entropy that occurs when a sample containing 1.00 mol of water (ice) is heated from – 20 °C to +20°C at 1 atm pressure. The molar heat capacities of H2O (s) and H2O (l) are 38.1 J K-1mol-1 and 75.3 J K-1mol-1 respectively and the enthalpy of fusion (melting) is 6.01 kJ mol-1 at 0°C. Solution ΔS from 253K to 273K = n Cp ln(T2/T1) = (1.00)(38.1)ln (273/253) = 2.90 J/K ΔS phase change from liq to gas = qrev/T = ΔHfus/T = (6010/273) = 22.0 J/K Δfrom 273K to 293K = n Cp ln(T2/T1) = (1.00)(75.3)ln (293/273) = 5.3 J/K Total ΔS = ΔS1 + ΔS2 + ΔS3

First Law of Thermodynamics The change in the internal energy of a system is equal to the work done on it plus the heat transferred to it. The Law of Conservation of Energy E = q + w Second Law of Thermodynamics For a spontaneous process the Entropy of the universe (meaning the system plus its surroundings) increases. Suniverse > 0 Third Law of Thermodynamics In any thermodynamic process involving only pure phases at equilibrium, the entropy change,  S, approaches zero at absolute zero temperature; also the entropy of a crystalline substance approaches zero at 0K. S = 0 at 0 K 5/19/2018 Zumdahl Chapter 10 9

0 =ΔEforward + ΔEreverse (Energy is conserved in both directions) 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 by the first law 0 =ΔEforward + ΔEreverse (Energy is conserved in both directions) 2nd Law of Thermodynamics Processes that increase ΔSuniverse are spontaneous. ΔSuniv > 0 Spontaneous Forward ΔSuniv = 0 At Equilibrium ΔSuniv < 0 Spontaneous Reverse 5/19/2018 Zumdahl Chapter 10

Suniverse = Ssystem + Ssurroundings The sign of ΔSsur depends on the direction of the heat flow. The magnitude of ΔSsur depends on the temperature This is ΔH of the system. If the reaction is exothermic, ΔH has a negative sign and ΔSsurr is positive If the reaction is endothermic, ΔH has a positive sign and ΔSsurr is negative

Ssystem + Ssurroundings = Suniverse

Ssolid < S liquid < Sgas Summary of Entropy Entropy is a quantitative measure of the number of microstates available to the molecules in a system. It is a measure of the number of ways in which energy or molecules can be arranged. Entropy is the degree of randomness or disorder in a system The Entropy of all substances is positive Ssolid < S liquid < Sgas ΔSsys is the Entropy Change of the system ΔSsur is the Entropy Change of the surroundings ΔSuni is the Entropy Change of the universe S has the units J K-1mol-1

Josiah Willard Gibbs (1839-1903) Highlights Devised much of the theoretical foundation for chemical thermodynamics. Established the concept free energy Moments in a Life 1863 Yale awarded him the first American Ph.D. in engineering Book: Equilibrium of Heterogeneous Substances, deemed one of the greatest scientific achievements of the 19th century. Will never be famous like Michael Jackson. 5/19/2018 Zumdahl Chapter 10

Gibbs Free Energy ΔG = ΔHsys - TΔSsys Allows us to focus on the system only, without considering the surroundings. ΔG = - TΔSuni G is called Gibbs Function, or Gibbs Free Energy, or Free Energy.

Free Energy ΔG < 0 Spontaneous ΔG = 0 Equilibrium ΔG > 0 Spontaneous Reverse Entropy ΔSuniv > 0 Spontaneous Forward ΔSuniv = 0 Equilibrium ΔSuniv < 0 Spontaneous Reverse

Benzene, C6H6, boils at 80.1°C (at 1 atm) and ΔHvap = 30.8 kJ Gibbs Free energy Benzene, C6H6, boils at 80.1°C (at 1 atm) and ΔHvap = 30.8 kJ a) Calculate ΔSvap for 1 mole of benzene at 60°C and pressure = 1 atm.

Benzene, C6H6, boils at 80°C at 1 atm. ΔHovap = 30.8 kJ a) Calculate ΔSvap for 1 mole of benzene. Start with ΔGvap=ΔHvap-TΔSvap at the boiling point, ΔGvap = 0 so ΔHvap = TbΔSvap

Benzene, C6H6, boils at 80.1°C (at 1 atm) and ΔHvap = 30.8 kJ Gibbs Free energy Benzene, C6H6, boils at 80.1°C (at 1 atm) and ΔHvap = 30.8 kJ b) Does benzene spontaneously boil at 60°C?

Effects of Temperature on ΔG° ΔG° = ΔH° - TΔS° Typically ΔH and ΔS are almost constant over a broad range For the reaction above, as Temperature increases ΔG becomes more positive, i.e., less negative. 3NO (g) → N2O (g) + NO2 (g)

Effects of Temperature on ΔG

For temperatures other than 298K or 25C ΔG = ΔH - T·ΔS B A D C

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. D C

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. D C

ΔSrxn° = ΣS°products – ΣS°reactants Entropies of Reaction ΔSrxn° = ΣS°products – ΣS°reactants ΔSrxn° is the sum of products minus the sum of the reactants, for one mole of reaction (that is what ° means) For a general reaction a A + b B → c C + d D Appendix 4 tabulates standard molar entropy values, S° in units JK-1mol-1

2H2S(g) + 3O2(g) → 2SO2(g) + 2H2O(g) Example Calculate ΔSr° at 298.15 K for the reaction 2H2S(g) + 3O2(g) → 2SO2(g) + 2H2O(g) (b) Calculate ΔS° of the system when 26.71 g of H2S(g) reacts with excess O2(g) to give SO2(g) and H2O(g) and no other products at 298.15K

2H2S(g) + 3O2(g) → 2SO2(g) + 2H2O(g) Calculate ΔSr° at 298.15 K for the reaction 2H2S(g) + 3O2(g) → 2SO2(g) + 2H2O(g) Solution (a) Look up each S° of formation [Note this is for “one mole of the reaction” as written: i.e. 2 moles of H2S, 3 moles of O2, etc] ΔSrxn°= 2S°(SO2(g) ) + 2S°(H2O(g)) -2S°(H2S(g) ) - 3S°(O2(g)) ΔSrxn°= 2(248) + 2(189) -2(206) - 3(205) = – 153 JK-1mol-1

2H2S(g) + 3O2(g) → 2SO2(g) +2H2O(g) (a) ΔSrxn= -153 JK-1mol-1 (b) Calculate ΔS° when 26.7 g of H2S(g) reacts with excess O2(g) to give SO2(g) and H2O(g) and no other products at 298.15K Solution:

Free Energy and Chemical Reactions ΔG = ΔH - T·ΔS ΔGf° is the standard molar Gibbs function of formation Because G is a State Property, for a general reaction a A + b B → c C + d D

Calculate ΔG° for the following reaction at 298. 15K Calculate ΔG° for the following reaction at 298.15K. Use Appendix 4 for additional information needed. 3NO(g) → N2O(g) + NO2(g) Solution From Appendix 4 ΔGf°(N2O) = 104 kJ mol-1 ΔGf°(NO2) = 52 ΔGf° (NO) = 87 ΔG°= 1(104) + 1(52) – 3(87) ΔG°= − 105 kJ therefore, spontaneous 5/19/2018 Zumdahl Chapter 10

Where Q is the reaction quotient The Dependence of Free Energy on Pressure 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 5/19/2018 Zumdahl Chapter 10

At Equilibrium conditions, ΔG = 0 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 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 5/19/2018 Zumdahl Chapter 10

Use ΔG°= - 105 kJ mol -1 (from previous) Calculate the equilibrium constant for this reaction at 25C. 3NO(g) ↔ N2O(g) + NO2(g) Strategy Use - ΔG ° = RT ln K Use ΔG°= - 105 kJ mol -1 (from previous)

3NO(g) ↔ N2O(g) + NO2(g) Solution ΔGrxn°= – 105 kJ mol-1 Use – ΔG ° = RT ln K Rearrange ΔGrxn°= – 105 kJ mol-1

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 5/19/2018 Zumdahl Chapter 10

The Temperature Dependence of Equilibrium Constants Where does this come from? Recall ΔG = ΔH - T·ΔS Divide by RT, then multiply by -1 5/19/2018 Zumdahl Chapter 10

A plot of y = mx + b or ln K vs. 1/T Notice that this is y = mx + b the equation for a straight line A plot of y = mx + b or ln K vs. 1/T 5/19/2018 Zumdahl Chapter 10

If we have two different Temperatures and K’s (equilibrium constants) or Now given ΔH and T at one temperature, we can calculate K at another temperature, assuming that ΔH and ΔS are constant over the temperature range 5/19/2018 Zumdahl Chapter 10

ΔT = − i m K J.H. van’t Hoff (1852-1901) van’t Hoff Factor (i) The Person Behind the Science J.H. van’t Hoff (1852-1901) Highlights Discovery of the laws of chemical dynamics and osmotic pressure in solutions his work led to Arrhenius's theory of electrolytic dissociation or ionization The Van't Hoff equation in chemical thermodynamics relates the change in temperature to the change in the equilibrium constant given the enthalpy change. Moments in a Life 1901 awarded first Noble Prize in Chemistry van’t Hoff Factor (i) ΔT = − i m K 5/19/2018 Zumdahl Chapter 10

The reaction 2 Al3Cl9 (g) → 3 Al2Cl6 (g) Has an equilibrium constant of 8.8X103 at 443K and a ΔHr°= 39.8 kJmol-1 at 443K. Estimate the equilibrium constant at a temperature of 600K. 5/19/2018 Zumdahl Chapter 10

P = mg/A V Infinite-step expansion 2 step expansion 6 step expansion Expansion (V2 > V1): Work flows out of the system and the Work sign is negative 5/19/2018 Zumdahl Chapter 10

Compression (V2 < V1): Work is put into system, P = mg/A V Compression (V2 < V1): Work is put into system, ln(V2/V1) is negative and the Work is positive