Thermodynamics: Entropy, Free Energy, and Equilibrium

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1 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.

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

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

4 Thermodynamics State functionsare 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 1and hiker 2 is the same even though they took different paths.

5 *

6 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= H sys T S sys G < 0 Reaction is spontaneous in forward direction 3) The entropy of a perfect crystalline substance is zero at the absolute zero ( K=0)

7 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 ( H): The heat change in a reaction * or process at constant pressure. Entropy ( S): The amount of Molecular randomness change in a reaction or process at constant pressure.

8 Exothermic: Enthalpy Change CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(l) H = kj Endothermic: H 2 O(s) H 2 O(l) * H fusion = kj H 2 O(l) H 2 O(g) H vap = kj N 2 O 4 (g) 2NO 2 (g) H = kj NaCl(s) H 2O Na 1+ (aq) + Cl 1- (aq) H = kj

9 Entropy Change S= S final - S initial *

10 Entropy *

11 Entropy *

12 The sign of entropy change, S, associated with the boiling of water is. 1. Positive 2. Negative 3. Zero *

13 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; S must be positive.

14 Entropy and Temperature 02 *

15 Third Law of Thermodynamics The entropy of a perfect crystalline substance is zero at the absolute zero ( K=0) * S solid < S liquid < S gas

16 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. 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, Sis negative.

17 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 lnw, W = Accessible microstates of translational, vibrational and rotational motions

18 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 H for a reaction: H = H f (Products) H 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 H = [c H f (C) + d H f (D)] [a H f (A) + b H f (B)]

19 Entropy Changes in the System ( S sys ) The standard entropy change of reaction( S rxn 0 ) is the entropy change for a reaction carried out at 1 atm and 25 0 C. S 0 rxn S 0 rxn aa + bb cc + dd = [ cs 0 (C) + ds 0 (D)] - [ as 0 (A) + bs 0 (B)] = Σ ΣmS 0 (reactants) ns 0 (products) 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) = J/K mol S 0 (O 2 ) = J/K mol - S 0 (CO 2 ) = J/K mol S 0 rxn= 2 x S 0 (CO 2 ) [2 x S 0 (CO) + S 0 (O 2 )] S 0 rxn= [ ] = J/K mol

20 Calculating S for a Reaction S o = ΣS o (products) -ΣS o (reactants) Consider 2 H 2 (g) + O 2 (g) ---> 2 H 2 O(liq) S o = 2 S o (H 2 O) - [2 S o (H 2 ) + S o (O 2 )] S o = 2 mol (69.9 J/K mol) - [2 mol (130.7 J/K mol) + 1 mol (205.3 J/K mol)] S o = J/K Note that there is a decrease in Entropy because 3 mol of gas give 2 mol of liquid.

21 Calculate S for the equation below using the standard entropy data given: 2 NO(g)+ O 2 (g) 2 NO 2 (g) S values (J/mol-K): NO 2 (g)= 240, NO(g)= 211, O 2 (g)= J/mol J/mol J/mol J/mol

22 Correct Answer: J/K J/K J/K J/K S = Σ S ( ) Σ products S( reactants) 쨠 S = 2(240) [2(211) + 205] S = 480 [627] S = 147

23 Thermite Reaction

24 Spontaneous Processes and Entropy 07 Consider the gas phase reaction of A 2 molecules (red) with B atoms (blue). (a) Write a balanced equation for the reaction. (b) Predict the sign of Sfor the reaction. 췠 ѥ

25 Entropy Changes in the Surroundings ( S surr ) Exothermic Process S surr > 0 Endothermic Process S surr < 0

26 2nd Law of Thermodynamics 01 The total entropy increases in a spontaneous process and remains unchanged in an equilibrium process. 漠 ѯ Spontaneous: S total = S sys + S sur > 0 Equilibrium: S total = S sys + S sur = 0 The system is what you observe; surroundings are everything else.

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

28 2nd Law of Thermodynamics Calculate change of entropy of surrounding in the following reaction: 2 H 2 (g) + O 2 (g)---> 2 H 2 O(liq) H = kj S o system = J/K S surr = H sys T 漠 ѯ S o surroundings = - ( kj)(1000 J/kJ) K S o surroundings = J/K

29 Spontaneous Reactions 01 The 2nd law tells us a process will be spontaneous if S total > 0 which requires a knowledge of S surr. 쳐ѯ spontaneous: S total = S sys + S sur > 0 S total = S sys + ( Hsys ) > 0 T S sur = H sys T -T ( S total = S sys + ( ) ) < 0 H sys T

30 Gibbs Free Energy 02 The expression T S total is equated as Gibbs free energy change ( G), or simply free energy change: -T. S total = -T. S sys + H sys < 0 T S total = G G= H sys T S sys G < 0 Reaction is spontaneous in forward direction. G = 0 Reaction is at equilibrium. G > 0 Reaction is spontaneous in reverse direction.

31 Calculate G o rxn for the following: C 2 H 2 (g) + 5/2 O 2 (g) --> 2 CO 2 (g) + H 2 O(g) Use enthalpies of formation to calculate H o rxn = kj Use standard molar entropies to calculate S o rxn( see page 658) S o rxn = J/K or kj/k G o rxn= -1238kJ - (298K)( kJ/K) = -1209kJ Reactionisproduct-favoredinspiteofnegative S o rxn. Reaction is enthalpy driven

32 Calculate G for the equation below using the thermodynamic data given: N 2 (g)+ 3 H 2 (g) 2 NH 3 (g) H (NH 3 ) = 46 kj; kj kj kj kj

33 Correct Answer: kj kj kj kj H = 92 kj S = 198.7J/mol.K G = H T S G = ( 92) (298)( ) 쳐ѯ G = ( 92) + (59.2) = 33 KJ

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

35 Gibbs Free Energy: G= H T S 04 3) If His negative and Sis positive, Gwill always be negative. Therefore, the reaction is spontaneous SѨ 4) If both Hand Sare negative, Gwill be negative only when the temperature value is small. Therefore, the reaction is spontaneous only at Low temperatures.

36 Gibbs Free Energy 04 SѨ

37 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)? SѨ

38 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: N 2 (g) + 3H 2 (g) 2NH 3 (g) SѨ H = kj S = J/K G = H -T S J = kj -298 K x x K 1 kj 1000 J = kj

39 Gibbs Free Energy 05 Iron metal can be produced by reducing iron(iii) oxide with hydrogen: Fe 2 O 3 (s) + 3 H 2 (g) 2 SѨ Fe(s) + 3 H 2 O(g) H = kj; S = J/K 1. Is this reaction spontaneous at 25 C? 2. At what temperature will the reaction become spontaneous?

40 Decomposition of CaCO 3 has a H = kj/mol and S = 159 J/mol K. At what temperature does this become spontaneous? C C C C SѨ

41 Correct Answer: C C C C T = H / S T= kj/mol/0.159 kj/mol K T= 1121 K SѨ T( C) = = 848

42 Standard Free Energies of Formation G = G f (products) - G f (reactants) aa + bb cc + dd SѨ G = [c G f (C) + d G f (D)] -[a G f (A) + b G f (B)] Products Reactants

43 Standard free energy of formation ( G 0 f) is the free-energy change that occurs when1 moleof the compound is formed f from its elements in their standard( 1 atm) states.

44 The standard free-energy of reaction ( G 0 rxn)is the freeenergy change for a reaction when it occurs under standardstate conditions. G 0 rxn G 0 rxn aa + bb cc + dd = [ c G 0 f(c) + d G 0 (D)] - [ a G 0 f(a) + b G 0 f(b) ] SѨ G 0 of any element in its stable f f = Σn G 0 f(products) - Σm G 0 f (reactants) form is zero. --

45 Calculate G for the equation below using the Gibbs free energy data given: 2 SO 2 (g)+ O 2 (g) 2 SO 3 (g) G values (kj): SO 2 (g)= 300.4, SO 3 (g)= kj kj kj kj SѨ

46 kj kj kj kj Correct Answer: G = ( ) ( ) ѫ G products f G reactants G = ( ) [( ) + 0] G = [ 600.8] = 140 f

47 Gibbs Free Energy 07 ѫ

48 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 ѫ concentrations of solutions are not 1M) conditions, G becomes G. G= G + RTlnQ The reaction quotientis obtained in the same way as an equilibrium expression.

49 Free Energy Changes and the Reaction Mixture G= G + RTln Q Calculate Gfor the formation of ethylene (C 2 H 4 ) from carbon and hydrogen at 25 C when the partial pressures are 100 atm H 2 and 0.10 atm C 2 H 4. 2C(s) + 2H 2 (g) C 2 H ѫ 4 (g) Q p = P C2 H 4 P 2 H 2 Calculate ln Q p : ln =

50 Free Energy Changes and the Reaction Mixture Calculate G: G= G + RTln Q = 68.1 kj/mol + ( see page A-13 of your book) J ѫ K mol 1 kj 1000 J (298 K)(-11.51) G= 39.6 kj/mol

51 Free Energy and Chemical Equilibrium 05 At equilibrium G= 0 and Q= K G= G + RTlnQ ѫ G rxn = RT ln K

52 Free Energy and Chemical Equilibrium Calculate K p at 25 C for the following reaction: CaCO 3 (s) CaO(s) + CO 2 (g) Calculate G : G = [ G f (CaO(s)) + G f (CO 2 (g))]-[ G f (CaCO 3 (s))] = [(1 mol)( kj/mol) + (1 mol)( kj/mol)] -[(1 mol)( kj/mol)] G = kj/mol

53 Free Energy and Chemical Equilibrium Calculate ln K: G = -RTln K ln K= - G RT = ѫ kj/mol J 1 kj K mol 1000 J (298 K) ln K= Calculate K: K= e = 1.4 x 10-23

54 ѫ

55 ѫ

56 Free Energy and Chemical Equilibrium 04 G rxn = RT ln K ѫ

57 Suppose G is a large, positive value. What then will be the value of the equilibrium constant, K? 1. K= 0 2. K= < K< 1 4. K> 1 ѫ

58 Correct Answer: 1. K= 0 2. K= < K< 1 4. K> 1 G = RTlnK Thus, large positive values of G lead to large negative values of lnk. The value ѫ of Kitself, then, is very small.

59 More thermo? You betcha! ѫ

60 Thermodynamics and K eq G o rxn= -RT lnk Calculate K for the reaction N 2 O 4 --->2 NO 2 G = RT lnk G o rxn= +4.8 kj/mole ( see page A-11) G o rxn= J = -(8.31 J/mol.K)(298 K) ln K ѫ lnk = J/mole (8.31 J/m.K)(298K) =-1.94 K = 0.14 When G o rxn> 0, then K < 1

Entropy, Free Energy, and Equilibrium

Entropy, Free Energy, and Equilibrium Chapter 17 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1 Spontaneous Physical and Chemical Processes A waterfall runs