1. Thermodynamics: Entropy, Free Energy Direction of Chemical Reactions Chapter 20 Entropy and Free Energy - Spontaneity of Reaction 1. The Second Law of Thermodynamics: Predicting Spontaneous Change First Law and enthalpy, Second Law and entropy, Standard Entropy and Third Law 2. Calculating the Change in Entropy of a Reaction The standard entropy of reaction, entropy change and equilibrium 3. Entropy, Free Energy, and Work Free energy change and spontaneity, Standard free energy changes 4. Free Energy, Equilibrium, and Reaction Direction

2. Important criterions releasing free energy lower energythermodynamically stable Spontaneous reaction, in chemical reaction terms, is one which occurs with the system releasing free energy in some form (often, but not always, heat) and moving to a lower energy, hence more thermodynamically stable, state. sum of the internal energyproduct of the pressure and volume Enthalpy, H, is the sum of the internal energy and the product of the pressure and volume of a thermodynamic system. degree of disorder (randomness) Entropy, S, is a state function which measures the degree of disorder (randomness) of a system. The larger the disorder, the greater the value of S in units of J/K. spontaneity Free Energy is the measure of the spontaneity of a process and of the useful energy available from it. conserve energy E = q + w The First Law of Thermodynamics indicates that to conserve energy, the change in energy must equal the heat transferred plus the work performed, E = q + w. spontaneous ∆S universe > 0 (∆S universe ) The Second Law of Thermodynamics states that for a spontaneous process, ∆S universe = ∆S system + ∆S surroundings > 0 (∆S universe positive)

3. Figure 1 A spontaneous endothermic chemical reaction. water Ba(OH) 2 8H 2 O(s) + 2NH 4 NO 3 (s) Ba 2+ (aq) + 2NO 3 - (aq) + 2NH 3 (aq) + 10H 2 O(l).  H 0 rxn = + 62.3 kJ Spontaneous reaction and exothermic /endothermic Common mistake – exothermic is spontaneous.

4. The Concept of Entropy (S) Entropy refers to the state of order, measures the degree of disorder of a system.. A change in order is a change in the number of ways of arranging the particles, and it is a key factor in determining the direction of a spontaneous process. solid liquid gas more orderless order crystal + liquid ions in solution more orderless order more orderless order crystal + crystal gases + ions in solution

5. Figure 2 1 atm evacuated Spontaneous expansion of a gas stopcock closed stopcock opened 0.5 atm In thermodynamic terms, a change in order is the number of ways of arranging the particles. The change in order is likely the key factor to determine the direction of a spontaneous reaction. The number of possible arrangement is increased. The number is of ways of arranging a system is related to the entropy.

6. 1877 Ludwig Boltzman S = k ln W where S is entropy, W is the number of ways of arranging the components of a system, and k is a constant (the Boltzman constant), R/N A (R = universal gas constant, N A = Avogadro’s number. A system with relatively few equivalent ways to arrange its components (smaller W) has relatively less disorder and low entropy. A system with many equivalent ways to arrange its components (larger W) has relatively more disorder and high entropy.  S universe =  S system +  S surroundings > 0 This is the second law of thermodynamics.

7. Figure 4 Random motion in a crystal The third law of thermodynamics. A perfect crystal has zero entropy at a temperature of absolute zero. S system = 0 at 0 K

8. Predicting Relative S 0 Values of a System 1. Temperature changes 2. Physical states and phase changes 3. Dissolution of a solid or liquid 5. Atomic size or molecular complexity 4. Dissolution of a gas S 0 increases as the temperature rises. S 0 increases as a more ordered phase changes to a less ordered phase. S 0 of a dissolved solid or liquid is usually greater than the S 0 of the pure solute. However, the extent depends upon the nature of the solute and solvent. A gas becomes more ordered when it dissolves in a liquid or solid. In similar substances, increases in mass relate directly to entropy. In allotropic substances, increases in complexity (e.g. bond flexibility) relate directly to entropy.

9. Thermodynamics: Entropy, Free Energy, and the Direction of Chemical Reactions 1 Three Laws of Thermodynamics: First Law : ∆Universe = 0, Second Law: ∆S>0,  G<0, Third law : S solid =0 2 Calculating the Change in Entropy of a Reaction ∆ S 0 =  mS 0 products -  nS 0 reactants Combine standard molar entropies to find the standard entropy for a reaction 3 Entropy, Free Energy, and Work G = H-TS, ∆ G= ∆H- ∆TS, ∆G= ∆H- T∆S, ∆G 0 = ∆H 0 - T∆S 0, 4 Free Energy, Equilibrium, and Reaction Direction Lecture 2

10. Summary Spontaneous reaction happens at a specific condition without continuous input of energy. Neither first law of thermodynamic or the sign of ∆H predicts the direction of reaction. The spontaneous reaction involves a change in the universe from more to less order. Entropy is the measure of disorder and directly related to the number of ways of arrangement for components in the system. The second law of thermodynamic states that the entropy of universe increases in a spontaneous process. Entropy value are absolute since the pure single crystal has a zero entropy at absolute zero degree. Standard molar entropy is dependant on the temperature, phase change, dissolution, atomic size, and molecular complexity.

11. Sample Problem SOLUTION: Predicting Relative Entropy Values PROBLEM:Choose the member with the higher entropy in each of the following pairs, and justify your choice [assume constant temperature, except in part (e)]: (a) 1mol of SO 2 (g) or 1mol of SO 3 (g) (b) 1mol of CO 2 (s) or 1mol of CO 2 (g) (c) 3mol of oxygen gas (O 2 ) or 2mol of ozone gas (O 3 ) (d) 1mol of KBr(s) or 1mol of KBr(aq) (e) Seawater in midwinter at 2 0 C or in midsummer at 23 0 C (f) 1mol of CF 4 (g) or 1mol of CCl 4 (g) PLAN:In general less ordered systems have higher entropy than ordered systems and entropy increases with an increase in temperature. (a) 1mol of SO 3 (g) - more atoms (b) 1mol of CO 2 (g) - gas > solid (c) 3mol of O 2 (g) - larger #mols (d) 1mol of KBr(aq) - solution > solid (e) 23 0 C - higher temperature (f) CCl 4 - larger mass

12. Important criterions Spontaneous reaction, in chemical reaction terms, is one which occurs with the system releasing free energy in some form (often, but not always, heat) and moving to a lower energy, hence more thermodynamically stable, state. Entropy, S, measures the degree of disorder of a system. The larger the disorder, the greater the value of S in units of J/K. Free Energy, G, is the measure of the spontaneity of a process and of the useful energy available from it.

13. Sample Problem Calculating the Standard Entropy of Reaction,  S 0 rxn PROBLEM: Calculate  S 0 rxn for the combustion of 1mol of propane at 25 0 C. C 3 H 8 ( g ) + 5O 2 ( g ) 3CO 2 ( g ) + 4H 2 O( l ) PLAN:Use summation equations. It is obvious that entropy is being lost because the reaction goes from 6 mols of gas to 3 mols of gas. SOLUTION:Find standard entropy values in the Appendix or other table.  S = [(3 mol)(S 0 CO 2 ) + (4 mol)(S 0 H 2 O)] - [(1 mol)(S 0 C 3 H 8 ) + (5 mol)(S 0 O 2 )]  S = [(3 mol)(213.7J/mol*K) + (4 mol)(69.9J/mol*K)] - [(1 mol)(269.9J/mol*K) + (5 mol)(205.0J/mol*K)]  S = - 374 J/K When a reaction occurs, chemists expect to learn how to predict and calculate the change in entropy using the following equation: ∆S 0 =  mS 0 products -  nS 0 reactants Lecture 2

14. given condition, ∆S 0 rxn <0 In many spontaneous reaction, the system become more ordered at a given condition, ∆S 0 rxn <0. The Second Law of Thermodynamics indicated that the decrease in entropy of a system only can occur when the increase the entropy in surrounding overweigh them. The entropy change of the surrounding is related to an opposite change in the heat of the system. ∆S 0 surr  - q sys The entropy change of the surrounding is also reversely proportional to the temperature of the heat transferred. ∆S 0 surr  1/T Combining these changes, we obtain the equation: ∆S 0 surr = - Determining Reaction Spontaneity q sys T When a pressure is constant, the heat is ∆H. ( ∆H = q + ∆ PV) ∆ H sys T Lecture 2

15. Sample Problem SOLUTION: Determining Reaction Spontaneity PROBLEM: At 298K, the formation of ammonia has a negative  S 0 sys ; Calculate  S 0 rxn, and state whether the reaction occurs spontaneously at this temperature. N 2 ( g ) + 3H 2 ( g ) 2NH 3 ( g )  S 0 sys = -197 J/K PLAN:  S 0 universe must be > 0 in order for this reaction to be spontaneous, so  S 0 surroundings must be > 197 J/K. To find  S 0 surr, first find  H sys ;  H sys =  H rxn which can be calculated using  H 0 f values from tables.  S 0 universe =  S 0 surr +  S 0 sys.  H 0 rnx = [(2 mol)(  H 0 f NH 3 )] - [(1 mol)(  H 0 f N 2 ) + (3 mol)(  H 0 f H 2 )]  H 0 rnx = -91.8 kJ  S 0 surr = -  H 0 sys /T = -(-91.8x10 3 J/298K)= 308 J/K  S 0 universe =  S 0 surr +  S 0 sys = 308 J/K + (-197 J/K) = 111 J/K  S 0 universe > 0 so the reaction is spontaneous. Lecture 2

16.  G 0 system =  H 0 system - T  S 0 system  G 0 rxn =  m  G 0 products -  n  G 0 reactants Gibbs Free Energy (G)  G, the change in the free energy of a system, is a measure of the spontaneity of the process and of the useful energy available from it.  G < 0 for a spontaneous process  G > 0 for a nonspontaneous process  G = 0 for a process at equilibrium Lecture 2

17. Sample Problem SOLUTION: Calculating  G 0 from Enthalpy and Entropy Values PROBLEM: Potassium chlorate, a common oxidizing agent in fireworks and matchheads, undergoes a solid-state disproportionation reaction when heated. Note that the oxidation number of Cl in the reactant is higher in one of the products and lower in the other (disproportionation). 4KClO 3 ( s ) 3KClO 4 ( s ) + KCl( s )  +7+5 Use  H 0 f and S 0 values to calculate  G 0 sys (  G 0 rxn ) at 25 0 C for this reaction. PLAN:Use Appendix B values for thermodynamic entities; place them into the Gibbs Free Energy equation and solve.  H 0 rxn =  m  H 0 products -  n  H 0 reactants  H 0 rxn = (3 mol)(-432.8 kJ/mol) + (1 mol)(-436.7 kJ/mol) - (4 mol)(-397.7 kJ/mol)  H 0 rxn = -144 kJ Lecture 2

18. Sample Problem Calculating  G 0 from Enthalpy and Entropy Values continued  S 0 rxn =  m  S 0 products -  n  S 0 reactants  S 0 rxn = (3 mol)(151 J/mol*K) + (1 mol)(82.6 J/mol*K) - (4 mol)(143.1 J/mol*K)  S 0 rxn = -36.8 J/K  G 0 rxn =  H 0 rxn - T  S 0 rxn  G 0 rxn = -144 kJ - (298K)(-36.8 J/K)(kJ/10 3 J)  G 0 rxn = -133 kJ Lecture 2

19. Entropy, Free Energy, Work Free Gibbs energy is the measure of the spontaneity of a process. ∆G 0 sys = ∆H 0 sys - T∆S 0 sys T∆S 0 univ = - (∆H 0 sys - T∆S 0 sys ) The sign for ∆G 0 sys and ∆S 0 univ can tell us the spontaneity of a reaction. ∆G 0 sys < 0 for a spontaneous reaction; ∆G 0 sys = 0 for a reaction at equilibrium; ∆G 0 sys > 0 for a nonspontaneous reaction; ∆S 0 univ > 0 for a spontaneous reaction; ∆S 0 univ = 0 for a reaction at equilibrium; ∆S 0 univ < 0 for a nonspontaneous reaction. Lecture 2

20. The spontaneity of reaction and the sign of ∆H, ∆G, ∆S. ∆G 0 = ∆H 0 - T∆S 0 ∆H 0 ∆S 0 -T∆S 0 ∆G 0 Description -+--Spontaneous at all T +-++Non-spontaneous at all T ++-+ or -Spontaneous at high T Non-spontaneous at low T --++ or -Spontaneous at low T Non-spontaneous at high T

21. PROBLEM: Determining the effect of temperature on  G An important reaction in production of sulfuric acid is the oxidation of SO 2 (g) to SO 3 (g), 2SO 2 ( g ) + O 2 ( g ) 2SO 3 ( g ) At 298K, ∆G 0 = -141.6 kJ/mol, ∆H 0 = -198.4 kJ/mol, ∆S 0 = -187.9kJ/mol, (a). Use the data to predict if this reaction is spontaneous reaction at 25  C. (b). Predict how ∆G 0 changes when temperature increases. (c). Assume ∆H 0 and ∆S 0 are constant with increasing T, is the reaction spontaneous at 900  C? PLAN:Note the sign of ∆G 0, ∆H 0, and T∆S 0 Use the equations ∆G 0 = ∆H 0 - T∆S 0

22. SOLUTION: (a) Noting  G = -141.6 kJ, is negative, so the reaction at room temperature is spontaneous. Determining the effect of temperature on  G (b) Noting  G =  H – T  S = -198.4 - (- 0.1879T) = -198.4 + 0.1879T If T increases, the  G is less negative. Therefore, the reaction is less spontaneous. (c) Calculating  G =  H – T  S at 900  C = -198.4 - (- 0.1879T) = -198.4 + 0.1879(900+273.15) = 22.0 kJ At 900  C,  G is positive. Therefore, the reaction is non- spontaneous.

23. Free Energy, Equilibrium and Reaction Direction If Q/K < 1, then ln Q/K < 0; the reaction proceeds to the right (  G < 0) If Q/K > 1, then ln Q/K > 0; the reaction proceeds to the left (  G > 0) If Q/K = 1, then ln Q/K = 0; the reaction is at equilibrium (  G = 0)  G = RT ln Q/K = RT lnQ - RT lnK Under standard conditions (1M concentrations, 1atm for gases), Q = 1 and ln Q = 0 so  G 0 = - RT lnK

24. FORWARD REACTION REVERSE REACTION Table The Relationship Between  G 0 and K at 25 0 C  G 0 (kJ) KSignificance 200 100 50 10 1 0 -10 -50 -100 -200 9x10 -36 3x10 -18 2x10 -9 2x10 -2 7x10 -1 1 1.5 5x10 1 6x10 8 3x10 17 1x10 35 Essentially no forward reaction; reverse reaction goes to completion Forward and reverse reactions proceed to same extent Forward reaction goes to completion; essentially no reverse reaction

25. PROBLEM: Calculating  G at Nonstandard Conditions The oxidation of SO 2,2SO 2 ( g ) + O 2 ( g ) 2SO 3 ( g ) is too slow at 298K to be useful in the manufacture of sulfuric acid. To overcome this low rate, the process is conducted at an elevated temperature. (a) Calculate K at 298K and at 973K. (  G 0 298 = -141.6kJ/mol of reaction as written using  H 0 and  S 0 values at 973K.  G 0 973 = -12.12kJ/mol of reaction as written.) (b) In experiments to determine the effect of temperature on reaction spontaneity, two sealed containers are filled with 0.500atm of SO 2, 0.0100atm of O 2, and 0.100atm of SO 3 and kept at 25 0 C and at 700. 0 C. In which direction, if any, will the reaction proceed to reach equilibrium at each temperature? (c) Calculate  G for the system in part (b) at each temperature. PLAN:Use the equations and conditions.  G 0 = -RTlnK

26. SOLUTION: Calculating  G at Nonstandard Conditions (a) Calculating K at the two temperatures:  G 0 = -RTlnK so At 298, the exponent is -  G 0 /RT = - (-141.6kJ/mol)(10 3 J/kJ) (8.314J/mol*K)(298K) = 57.2 = e 57.2 = 7x10 24 At 973, the exponent is -  G 0 /RT (-12.12kJ/mol)(10 3 J/kJ) (8.314J/mol*K)(973K) = 1.50 = e 1.50 = 4.5

27. Calculating  G at Nonstandard Conditions (b) The value of Q = pSO 3 2 (pSO 2 ) 2 (pO 2 ) = (0.100) 2 (0.500) 2 (0.0100) = 4.00 Since Q is < K at both temperatures the reaction will shift right; for 298K there will be a dramatic shift while at 973K the shift will be slight. (c) The nonstandard  G is calculated using  G =  G 0 + RTLnQ  G 298 = -141.6kJ/mol + (8.314J/mol*K)(kJ/10 3 J)(298K)(Ln4.00)  G 298 = -138.2kJ/mol  G 973 = -12.12kJ/mol + (8.314J/mol*K)(kJ/10 3 J)(973K)(Ln4.00)  G 298 = -0.9kJ/mol 2SO 2 ( g ) + O 2 ( g ) 2SO 3 ( g )

28. Important equations ∆G 0 sys = ∆H 0 sys - T∆S 0 sys ∆S 0 univ = - q sys /T ∆G 0 = ∆H 0 - T∆S 0 ∆S 0 =  mS 0 products -  nS 0 reactants ∆G 0 =  mG 0 products -  nG 0 reactants ∆G 0 = - RT lnK ∆G = ∆G 0 - RT lnQ Thermodynamic