1. H 2 O(l) --> H 2 O(s) Normal freezing point of H 2 O = 273.15K The change in enthalpy is the enthalpy of freezing - enthalpy change associated when one mole of liquid water freezes at 1atm and 273.15K  H = -6007J Entropy change  S= HH T = -6007 J 273.15K = -21.99 J/K  G =  H - T  S = -6007J - (273.15K)(-21.99J/K) = 0J => at 273.15K and 1 atm, liquid and solid water are at equilibrium The Gibbs Free Energy Function and Phase Transitions

2. As water is cooled by 10K below 273.15K to 263.15K Assume that  H and  S do not change with T  G =  H - T  S = -6007J - (263.15K)(-21.99J/K) = -220J Since  G < 0 water spontaneously freezes At 10K above the freezing point, 283.15K  G =  H - T  S = -6007J - (283.15K)(-21.99J/K) = 219J  G > 0 water does not spontaneously freeze at 283.15K, 10K above the normal freezing point of water Above 273.15K, the reverse process is spontaneous - ice melts to liquid water.

3. Plot of  H and T  S vs T for the freezing of water. At 273.15 K the system is at equilibrium,  G = 0 freezing spontaneous Freezing not spontaneous; melting spontaneous H 2 O(l) --> H 2 O(s)

4. For phase transitions, at the transition temperature, the system is at equilibrium between the two phases. Above or below the transition temperature, the phase that is the more stable phase is determined by thermodynamics -  G of the phase transition.

5. Reaction Free Energy  G r =  nG m (products) -  nG m (reactants) Standard Reaction Free Energy  G o r =  nG o m (products) -  nG o m (reactants) Difference in free energy of the pure products in their standard states and pure reactants in their standard states.

6. Standard Free-Energy of Formation The standard free energy of formation,  G f o, of a substance is the standard reaction free energy per mole for the formation of a compound from its elements in their most stable form 1/2N 2 (g) + 3/2 H 2 (g) --> NH 3 (g)  G f o = -16 kJ O 2 (g) --> O 2 (g)  G f o = 0

7.  G f o is an indication of a compounds stability relative to its elements Thermodynamically stable compound  G f o < 0 Thermodynamically unstable compound  G f o > 0

8. Standard Reaction Free Energy For a reaction: aA + bB --> cC + dD  G r o = c  G f o (C) + d  G f o (D) - a  G f o (A) -b  G f o (B) The standard free energy of formation of elements in their most stable form at 298.15K is zero.

9. Problem: Calculate the standard free-energy change for the following reaction at 298K: N 2 (g) + 3H 2 (g) --> 2NH 3 (g) Given that  G f o [NH 3 (g)] = -16.66 kJ/mol. What is the  G o for the reverse reaction? Since the reactants N 2 (g) and H 2 (g) are in their standard state at 298 K their standard free energies of formation are defined to be zero.  G r o = 2  G f o [NH 3 (g)] -  G f o [N 2 (g)] - 3  G f o [H 2 (g)] = 2 x -16.66 = -33.32 kJ/mol For the reverse reaction: 2NH 3 (g) --> N 2 (g) + 3H 2 (g)  G r o = +33.32 kJ/mol

10. C 3 H 8 (g) + 5O 2 (g) --> 3CO 2 (g) + 4H 2 O(l)  H o = - 2220 kJ a) Without using the information from the tables of thermodynamic quantities predict whether  G o for this reaction will be less negative or more negative than  H o. b) Use data from App. D to calculate  G r o for this reaction. Whether  G r o is more negative or less negative than  H o depends on the sign of  S o since  G r o =  H o - T  S o The reaction involves 6 moles of gaseous reactants and produces 3 moles of gaseous product =>  S o < 0 (since fewer moles of gaseous products than reactants) Since  S o < 0 and  H o < 0 and  G r o =  H o - T  S o =>  G r o is less negative than  H o

11.  G r o = 3  G f o (CO 2 (g))+4  G f o (H 2 O(l)-  G f o (C 3 H 8 (g))-5  G f o (O 2 (g)) = 3(-394.4) + 4(-237.13) - (-23.47) - 5(0) = -2108 kJ

12. Effect of temperature on  G o Values of  G r o calculated using the tabulated values of  G f o apply only at 298.15K. For other temperatures the relation:  G r o =  H o - T  S o can be used, assuming  H o and  S o do not vary significantly with temperature. T = HoHo SoSo When  G r o = 0, Use this expression to determine temperature above or below which reaction becomes spontaneous.

15. Problem: The normal boiling point is the temperature at which a pure liquid is in equilibrium with its vapor at 1 atm. a) write the chemical equation that defines the normal boiling point of liquid CCl 4 b) what is the value of  G o at equilibrium? c) Use thermodynamic data to estimate the boiling point of CCl 4. a) CCl 4 (l) CCl 4 (g) b) At equilibrium  G = 0. In any equilibrium for a normal boiling point, both the liquid and gas are in their standard states. Hence, for this process  G is  G o.

16. c) T = HoHo SoSo where for this reaction T is the boiling point Note: To determine the boiling point accurately we need the values of  H o and  S o for the vaporization process at the boiling point of CCl 4  H o = (1mol)(-106.7 kJ/mol) - (1mol)(-139.3kJ/mol) = 32.6 kJ  S o = (1mol)(309.4 J/mol-K) -(1mol)(214.4 J/mol-K) = 95.0 J/K T = HoHo SoSo = 32.6 kJ 0.095 kJ/K = 343 K

17. The Gibbs Function and the Equilibrium Constant  G =  G r o + R T ln Q Q - reaction quotient Under standard conditions the concentrations of all reactants and products are set to 1 (standard pressure of gases = 1 atm Standard pressure of solutions = 1 M) Under standard conditions, ln Q = 0 =>  G =  G r o

18. At equilibrium  G = 0 and Q = K At equilibrium  G r o = - R T ln K or K = e -  G r o /RT  G r o K > 1  G r o > 0 => K < 1  G =  G r o + R T ln Q = - RT ln K + RT ln Q  G = R T ln Q K when Q = K =>  G = 0; equilibrium

19. If the reaction mixture has too much N 2 or H 2 relative to NH 3, Q K, decomposition of NH 3 is spontaneous, till equilibrium is established; Q = K  G forward < 0  G forward > 0

20. Temperature Dependence of K The temperature dependence of K can be used to determine  H o and  S o of a reaction ln K = -Gr-Gr R T = -  H o SoSo R TR +

21. If the equilibrium constant of a reaction is known at one temperature and the value of  H o is known, then the equilibrium constant at another temperature can be calculated ln K 1 = -  H o SoSo R T 1 R + ln K 2 = -  H o SoSo R T 2 R + If  H o is negative (exothermic), an increase in T reduces K; if it is positive (endothermic), an increase in T increases K ln K 2 - ln K 1 =ln K1K1 = HoHo R 1 T 2 1 T 1 () - K2K2

22. For the equilibrium between a pure liquid and its vapor, the equilibrium constant is simply the equilibrium vapor pressure (since pure liquids do not appear in the equilibrium expression) LiquidGasK = P gas Hence ln p1p1 =  H vap R 1 T 2 1 T 1 () - p2p2 Clausius-Clapeyron equation The Clausius-Clapeyron equation indicates the variation of vapor pressure of a liquid with temperature

23. Driving Non-spontaneous Reactions The magnitude of  G is a useful tool for evaluating reactions. For a spontaneous process at constant pressure and temperature, the change in free energy for a process equals the maximum amount of work that can be done by the system on its surroundings w max =  G r For non-spontaneous reactions (  G r > 0), the magnitude of  G r is a measure of the minimum amount of work that must be done on the system to cause the process to occur. A reaction for which  G r is large and negative (like the combustion of gasoline), is much more capable of doing work on the surroundings than a reaction for which  G r is small and negative (like the melting of ice).

24. Many chemical reactions, including a large number that are central to biological systems, are non-spontaneous. For example: Cu can be extracted from the mineral chalcolite which contains Cu 2 S. The decomposition of Cu 2 S is non-spontaneous Cu 2 S --> 2Cu(s) + S(s)  G r o = +86.2 kJ Since this reaction is not spontaneous Cu cannot be directly obtained from the above reaction. Instead, work needs to be done on this reaction, and this is done by coupling this non-spontaneous reaction with a spontaneous reaction so that the overall reaction is spontaneous.

25. Consider the following reaction: S(s) + O 2 (g) --> SO 2 (g)  G r o = -300.4kJ Coupling this reaction with the extraction of Cu from Cu 2 S Cu 2 S --> 2Cu(s) + S(s)  G r o = +86.2 kJ S(s) + O 2 (g) --> SO 2 (g)  G r o = -300.4kJ Net:Cu 2 S(s) + O 2 (g) --> 2Cu(s) + SO 2 (g)  G r o = -214.2kJ

26. Biological systems employ the same principle by using spontaneous reactions to drive non-spontaneous reactions. Many biochemical reactions are not spontaneous. These reactions are made to occur by coupling them with spontaneous reactions which release heat. The metabolism of food is the usual source of free energy needed to do the work to maintain biological systems. C 6 H 12 O 6 (s) + 6O 2 (g) --> 6CO 2 (g) + 6H 2 O(l)  H o = -2803 kJ

27. A means of transporting the energy released by glucose metabolism to the reactions that require energy is needed. The free energy released by the metabolism of glucose is used to convert lower-energy ADP (adenine diphosphate) to higher energy ATP (adenine triphosphate). The energy stored in the ATP molecule is then available to a biochemical reaction that requires energy and in the process ATP is converted to ADP.

28. C 6 H 12 O 6 6CO 2 (g) + 6H 2 O(l) ADP ATP Free energy released by oxidation of glucose converts ADP to ATP Free energy released by ATP converts simple molecules to more complex molecules