1. Entropy, Free Energy, and Equilibrium Chapter 18 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

2. 열역학 자연의 방향 - 자발적 과정 엔트로피 깁스 자유에너지 화학평형

3. Spontaneous Physical and Chemical Processes A waterfall runs downhill A lump of sugar dissolves in a cup of coffee At 1 atm, water freezes below 0 0 C and ice melts above 0 0 C Heat flows from a hotter object to a colder object A gas expands in an evacuated bulb Iron exposed to oxygen and water forms rust spontaneous nonspontaneous

4. spontaneous nonspontaneous

5. Does a decrease in enthalpy mean a reaction proceeds spontaneously? CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (l)  H 0 = -890.4 kJ H + (aq) + OH - (aq) H 2 O (l)  H 0 = -56.2 kJ H 2 O (s) H 2 O (l)  H 0 = 6.01 kJ NH 4 NO 3 (s) NH 4 + (aq) + NO 3 - (aq)  H 0 = 25 kJ H2OH2O Spontaneous reactions

6. Entropy (S) is a measure of the randomness or disorder of a system. orderS disorder S  S = S f - S i If the change from initial to final results in an increase in randomness S f > S i  S > 0 For any substance, the solid state is more ordered than the liquid state and the liquid state is more ordered than gas state S solid < S liquid << S gas H 2 O (s) H 2 O (l)  S > 0

7. W = 1 W = 4 W = 6 W = number of microstates S = k ln W  S = S f - S i  S = k ln WfWf WiWi W f > W i then  S > 0 W f < W i then  S < 0 Entropy

8. S = k ln WΔS = S f - S i ΔS = k ln WfWf WiWi Entropy 이상 기체의 경우 등온에서 부피가 두 배 늘어나면 ? W i = a N 이라고 하면, W f = (2a) N 따라서 ΔS = W f - W i = kN ln(2a/a) = kN ln(2) = nR ln 2 ΔS = Δq rev T 2. 엔트로피의 고전 열역학적 정의 1. 엔트로피의 통계 열역학적 정의

9. Processes that lead to an increase in entropy (  S > 0)

10. How does the entropy of a system change for each of the following processes? (a) Condensing water vapor (b) Forming sucrose crystals from a supersaturated solution (c) Heating hydrogen gas from 60 0 C to 80 0 C (d) Subliming dry ice Randomness decreases Entropy decreases (  S < 0) Randomness decreases Entropy decreases (  S < 0) Randomness increases Entropy increases (  S > 0) Randomness increases Entropy increases (  S > 0)

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

12. First Law of Thermodynamics Energy can be converted from one form to another but energy cannot be created or destroyed. Second Law of Thermodynamics The entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process.  S univ =  S sys +  S surr > 0 Spontaneous process:  S univ =  S sys +  S surr = 0 Equilibrium process:

13. Entropy Changes in the System (  S sys ) aA + bB cC + dD S0S0 rxn dS 0 (D) cS 0 (C) = [+] - bS 0 (B) aS 0 (A) [+] S0S0 rxn nS 0 (products) =  mS 0 (reactants)  - The standard entropy of reaction (  S 0 ) is the entropy change for a reaction carried out at 1 atm and 25 0 C. rxn What is the standard entropy change for the following reaction at 25 0 C? 2CO (g) + O 2 (g) 2CO 2 (g) S 0 (CO) = 197.9 J/K mol S 0 (O 2 ) = 205.0 J/K mol S 0 (CO 2 ) = 213.6 J/K mol S0S0 rxn = 2 x S 0 (CO 2 ) – [2 x S 0 (CO) + S 0 (O 2 )] S0S0 rxn = 427.2 – [395.8 + 205.0] = -173.6 J/K mol

14. Entropy Changes in the System (  S sys ) When gases are produced (or consumed) If a reaction produces more gas molecules than it consumes,  S 0 > 0. If the total number of gas molecules diminishes,  S 0 < 0. If there is no net change in the total number of gas molecules, then  S 0 may be positive or negative BUT  S 0 will be a small number. What is the sign of the entropy change for the following reaction? 2Zn (s) + O 2 (g) 2ZnO (s) The total number of gas molecules goes down,  S is negative.

15. Entropy Changes in the Surroundings (  S surr ) Exothermic Process  S surr > 0 Endothermic Process  S surr < 0

16. Entropy Changes in the Surroundings (  S surr )  S univ =  S surr +  S sys = q(1/T’ - 1/T) = q(T-T’)/TT’ > 0  S surr = q/T’ When T’ < T When T < T’  S sys = -q/T  S surr = -q/T’  S sys = q/T  S univ =  S surr +  S sys = q(-1/T’ + 1/T) = q(T’-T)/TT’ > 0

17. Third Law of Thermodynamics The entropy of a perfect crystalline substance is zero at the absolute zero of temperature. S = k ln W W = 1 S = 0 ΔS = Δq rev T

18.  S univ =  S sys +  S surr > 0 Spontaneous process:  S univ =  S sys +  S surr = 0 Equilibrium process: Gibbs Free Energy For a constant-T, P process:  G =  H sys - T  S sys  S univ =  S sys -  H sys /T > 0  S surr = -q/T = -  H sys /T 일정한 압력에서 계에 주어진 열은 계의 엔탈피 변화와 같다 T  S sys -  H sys > 0  H sys - T  S sys < 0

19.  S univ =  S sys +  S surr > 0 Spontaneous process:  S univ =  S sys +  S surr = 0 Equilibrium process: Gibbs Free Energy For a constant-temperature process:  G =  H sys -T  S sys Gibbs free energy (G)  G < 0 The reaction is spontaneous in the forward direction.  G > 0 The reaction is nonspontaneous as written. The reaction is spontaneous in the reverse direction.  G = 0 The reaction is at equilibrium.

20. aA + bB cC + dD G0G0 rxn d  G 0 (D) f c  G 0 (C) f = [+] - b  G 0 (B) f a  G 0 (A) f [+] G0G0 rxn n  G 0 (products) f =  m  G 0 (reactants) f  - The standard free-energy of reaction (  G 0 ) is the free- energy change for a reaction when it occurs under standard- state conditions. rxn Standard free energy of formation (  G 0 ) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states. f  G 0 of any element in its stable form is zero. f

21. 2C 6 H 6 (l) + 15O 2 (g) 12CO 2 (g) + 6H 2 O (l) G0G0 rxn n  G 0 (products) f =  m  G 0 (reactants) f  - What is the standard free-energy change for the following reaction at 25 0 C? G0G0 rxn 6  G 0 (H 2 O) f 12  G 0 (CO 2 ) f = [+] - 2  G 0 (C 6 H 6 ) f [] G0G0 rxn = [ 12x–394.4 + 6x–237.2 ] – [ 2x124.5 ] = -6405 kJ Is the reaction spontaneous at 25 0 C?  G 0 = -6405 kJ < 0 spontaneous

22.  G =  H - T  S

23. But many spontaneous reactions or processes are endothermic... Ba(OH) 2 ∙8H 2 O(s) + 2NH 4 Cl(s)  BaCl 2 ∙ 2H 2 O (s) + 2NH 3 (aq) + 8 H 2 O (l)  S = 368 J/mol K,  H = +63.6 kJ/mol Ba(OH) 2 ∙8H 2 O(s) + 2NH 4 SCN (s)  Ba 2+ (aq) + 2SCN - (aq) + 2NH 3 (aq) + 10 H 2 O (l)  S = 495 J/mol K,  H = +102.2 kJ/mol Entropy Driven Spontaneous Reactions Ba(OH) 2 ∙8H 2 O(s) + 2NH 4 NO 3 (s)  Ba(NO 3 ) 2 (s) + 2NH 3 (aq) + 10 H 2 O (l)  S = 406 J/mol K,  H = +62.3 kJ/mol

24. CaCO 3 (s) CaO (s) + CO 2 (g)  H 0 = 177.8 kJ  S 0 = 160.5 J/K  G 0 =  H 0 – T  S 0 At 25 0 C,  G 0 = 130.0 kJ Temperature and Spontaneity of Chemical Reactions  G 0 = 0 at 835 0 C 835 0 C 1 atm

25.  G 0 = 0 at 835 0 C 835 0 C 1 atm

26. Gibbs Free Energy and Phase Transitions H 2 O (l) H 2 O (g)  G 0 = 0=  H 0 – T  S 0  S = T HH = 40.79 kJ 373 K = 109 J/K

27. Efficiency = X 100% T h - T c ThTh Chemistry In Action: The Efficiency of Heat Engines

28. Gibbs Free Energy and Chemical Equilibrium  G =  G 0 + RT lnQ R is the gas constant (8.314 J/K mol) T is the absolute temperature (K) Q is the reaction quotient At Equilibrium  G = 0 Q = K 0 =  G 0 + RT lnK  G 0 =  RT lnK

29. Free Energy Versus Extent of Reaction  G 0 < 0  G 0 > 0

30.  G 0 =  RT lnK

31. ATP + H 2 O + Alanine + Glycine ADP + H 3 PO 4 + Alanylglycine Alanine + Glycine Alanylglycine  G 0 = +29 kJ  G 0 = -2 kJ K < 1 K > 1

33. High Entropy Low Entropy T  S =  H -  G Chemistry In Action: The Thermodynamics of a Rubber Band

34. Rubber-band 실험 고무줄을 입술에 대고 힘차게 당겨 고무줄의 온도 변화를 살펴 보아라. 1. 실험에서 고무줄의 온도는 어떻게 되는가 ? 2. 이 과정에서의  G 의 부호는 어떻게 될까 ? 그 이유를 설명하라. 3. 이 과정에서의  H 의 부호는 어떻게 될까 ? 그 이유를 설명하라. 4. 이 과정에서의  S 의 부호는 어떻게 될까 ? 그 이유를 설명하라.

35. Rubber-band 실험 온도 ΔGΔGΔHΔHΔSΔS 당길 때 ++–– 놓을 때 ––++ Rubber Band (Contracted) →←→← Rubber Band (Stretched)  G =  H - T  S

36. magnetic refrigeration and vapor cycle refrigeration. H = externally applied magnetic field; Q = heat quantity; P = pressure; ΔTad = adiabatic temperature variation From wikipedia