1. Chapter 17 Free Energy and Thermodynamics 2008, Prentice Hall Chemistry: A Molecular Approach, 1 st Ed. Nivaldo Tro Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA

2. Tro, Chemistry: A Molecular Approach2 First Law of Thermodynamics you can’t win! First Law of Thermodynamics: Energy cannot be Created or Destroyed the total energy of the universe cannot change though you can transfer it from one place to another   E universe = 0 =  E system +  E surroundings

3. Tro, Chemistry: A Molecular Approach3 First Law of Thermodynamics Conservation of Energy For an exothermic reaction, “lost” heat from the system goes into the surroundings two ways energy “lost” from a system, converted to heat, q used to do work, w Energy conservation requires that the energy change in the system equal the heat released + work done  E = q + w  E =  H + P  V  E is a state function internal energy change independent of how done

4. Tro, Chemistry: A Molecular Approach4 Energy Tax you can’t break even! to recharge a battery with 100 kJ of useful energy will require more than 100 kJ every energy transition results in a “loss” of energy conversion of energy to heat which is “lost” by heating up the surroundings

5. Tro, Chemistry: A Molecular Approach5 Heat Tax fewer steps generally results in a lower total heat tax

6. Tro, Chemistry: A Molecular Approach6 Thermodynamics and Spontaneity thermodynamics predicts whether a process will proceed under the given conditions spontaneous process  nonspontaneous processes require energy input to go spontaneity is determined by comparing the free energy of the system before the reaction with the free energy of the system after reaction. if the system after reaction has less free energy than before the reaction, the reaction is thermodynamically favorable. spontaneity ≠ fast or slow

7. Tro, Chemistry: A Molecular Approach7 Comparing Potential Energy The direction of spontaneity can be determined by comparing the potential energy of the system at the start and the end.

8. Tro, Chemistry: A Molecular Approach8 Reversibility of Process any spontaneous process is irreversible it will proceed in only one direction a reversible process will proceed back and forth between the two end conditions equilibrium results in no change in free energy if a process is spontaneous in one direction, it must be nonspontaneous in the opposite direction

9. Tro, Chemistry: A Molecular Approach9 Thermodynamics vs. Kinetics

10. Tro, Chemistry: A Molecular Approach10 Diamond → Graphite Graphite is more stable than diamond, so the conversion of diamond into graphite is spontaneous – but don’t worry, it’s so slow that your ring won’t turn into pencil lead in your lifetime (or through many of your generations).

11. Tro, Chemistry: A Molecular Approach11 Factors Affecting Whether a Reaction Is Spontaneous The two factors that determine the thermodynamic favorability are the enthalpy and the entropy. The enthalpy is a comparison of the bond energy of the reactants to the products. bond energy = amount needed to break a bond.  H The entropy factors relates to the randomness/orderliness of a system  S The enthalpy factor is generally more important than the entropy factor

12. Tro, Chemistry: A Molecular Approach12 Enthalpy related to the internal energy  H generally kJ/mol stronger bonds = more stable molecules if products more stable than reactants, energy released exothermic  H = negative if reactants more stable than products, energy absorbed endothermic  H = positive The enthalpy is favorable for exothermic reactions and unfavorable for endothermic reactions. Hess’ Law  H° rxn =  (  H° prod ) -  (  H° react )

14. Tro, Chemistry: A Molecular Approach14 Entropy entropy is a thermodynamic function that increases as the number of energetically equivalent ways of arranging the components increases, S  S generally J/mol S = k ln W k = Boltzmann Constant = 1.38 x 10 -23 J/K W is the number of energetically equivalent ways, unitless Random systems require less energy than ordered systems

15. Tro, Chemistry: A Molecular Approach15 W Energetically Equivalent States for the Expansion of a Gas

16. Tro, Chemistry: A Molecular Approach16 Macrostates → Microstates This macrostate can be achieved through several different arrangements of the particles These microstates all have the same macrostate So there are 6 different particle arrangements that result in the same macrostate

17. Tro, Chemistry: A Molecular Approach17 Macrostates and Probability There is only one possible arrangement that gives State A and one that gives State C There are 6 possible arrangements that give State B Therefore State B has higher entropy than either State A or State B The macrostate with the highest entropy also has the greatest dispersal of energy

18. Tro, Chemistry: A Molecular Approach18 Changes in Entropy,  S entropy change is favorable when the result is a more random system.  S is positive Some changes that increase the entropy are: reactions whose products are in a more disordered state.  (solid > liquid > gas) reactions which have larger numbers of product molecules than reactant molecules. increase in temperature solids dissociating into ions upon dissolving

19. Tro, Chemistry: A Molecular Approach19 Increases in Entropy

20. Tro, Chemistry: A Molecular Approach20 The 2 nd Law of Thermodynamics the total entropy change of the universe must be positive for a process to be spontaneous for reversible process  S univ = 0, for irreversible (spontaneous) process  S univ > 0  S universe =  S system +  S surroundings if the entropy of the system decreases, then the entropy of the surroundings must increase by a larger amount when  S system is negative,  S surroundings is positive the increase in  S surroundings often comes from the heat released in an exothermic reaction

21. Tro, Chemistry: A Molecular Approach21 Entropy Change in State Change when materials change state, the number of macrostates it can have changes as well for entropy: solid < liquid < gas because the degrees of freedom of motion increases solid → liquid → gas

22. Tro, Chemistry: A Molecular Approach22 Entropy Change and State Change

23. Tro, Chemistry: A Molecular Approach23 Heat Flow, Entropy, and the 2 nd Law Heat must flow from water to ice in order for the entropy of the universe to increase

24. Tro, Chemistry: A Molecular Approach24 Temperature Dependence of  S surroundings when a system process is exothermic, it adds heat to the surroundings, increasing the entropy of the surroundings when a system process is endothermic, it takes heat from the surroundings, decreasing the entropy of the surroundings the amount the entropy of the surroundings changes depends on the temperature it is at originally the higher the original temperature, the less effect addition or removal of heat has

25. Tro, Chemistry: A Molecular Approach25 Gibbs Free Energy,  G maximum amount of energy from the system available to do work on the surroundings G = H – T∙S  G sys =  H sys – T  S sys  G sys = – T  S universe  G reaction =  n  G prod –  n  G react when  G < 0, there is a decrease in free energy of the system that is released into the surroundings; therefore a process will be spontaneous when  G is negative

26. Ex. 17.2a – The reaction C 3 H 8(g) + 5 O 2(g)  3 CO 2(g) + 4 H 2 O (g) has  H rxn = -2044 kJ at 25°C. Calculate the entropy change of the surroundings. combustion is largely exothermic, so the entropy of the surrounding should increase significantly  H system = -2044 kJ, T = 298 K  S surroundings, J/K Check: Solution: Concept Plan: Relationships: Given: Find: SST,  H

27. Tro, Chemistry: A Molecular Approach27 Free Energy Change and Spontaneity

28. Tro, Chemistry: A Molecular Approach28 Gibbs Free Energy,  G process will be spontaneous when  G is negative   G will be negative when  H is negative and  S is positive  exothermic and more random  H is negative and large and  S is negative but small  H is positive but small and  S is positive and large  or high temperature  G will be positive when  H is + and  S is − never spontaneous at any temperature when  G = 0 the reaction is at equilibrium

29. Tro, Chemistry: A Molecular Approach29  G,  H, and  S

30. Ex. 17.3a – The reaction CCl 4(g)  C (s, graphite) + 2 Cl 2(g) has  H = +95.7 kJ and  S = +142.2 J/K at 25°C. Calculate  G and determine if it is spontaneous. Since  G is +, the reaction is not spontaneous at this temperature. To make it spontaneous, we need to increase the temperature.  H = +95.7 kJ,  S = 142.2 J/K, T = 298 K  G, kJ Answer: Solution: Concept Plan: Relationships: Given: Find: GGT,  H,  S

31. Ex. 17.3a – The reaction CCl 4(g)  C (s, graphite) + 2 Cl 2(g) has  H = +95.7 kJ and  S = +142.2 J/K. Calculate the minimum temperature it will be spontaneous. The temperature must be higher than 673K for the reaction to be spontaneous  H = +95.7 kJ,  S = 142.2 J/K,  G < 0  Answer: Solution: Concept Plan: Relationships: Given: Find: T  G,  H,  S

32. Tro, Chemistry: A Molecular Approach32 The 3 rd Law of Thermodynamics Absolute Entropy the absolute entropy of a substance is the amount of energy it has due to dispersion of energy through its particles the 3 rd Law states that for a perfect crystal at absolute zero, the absolute entropy = 0 J/mol∙K therefore, every substance that is not a perfect crystal at absolute zero has some energy from entropy therefore, the absolute entropy of substances is always +

33. Tro, Chemistry: A Molecular Approach33 Standard Entropies S° extensive entropies for 1 mole at 298 K for a particular state, a particular allotrope, particular molecular complexity, a particular molar mass, and a particular degree of dissolution

35. Tro, Chemistry: A Molecular Approach35 Relative Standard Entropies States the gas state has a larger entropy than the liquid state at a particular temperature the liquid state has a larger entropy than the solid state at a particular temperature Substance S°, (J/mol∙K) H 2 O (g)70.0 H 2 O (l)188.8

36. Tro, Chemistry: A Molecular Approach36 Relative Standard Entropies Molar Mass the larger the molar mass, the larger the entropy available energy states more closely spaced, allowing more dispersal of energy through the states

37. Tro, Chemistry: A Molecular Approach37 Relative Standard Entropies Allotropes the less constrained the structure of an allotrope is, the larger its entropy

38. Tro, Chemistry: A Molecular Approach38 Relative Standard Entropies Molecular Complexity larger, more complex molecules generally have larger entropy more available energy states, allowing more dispersal of energy through the states Substance Molar Mass S°, (J/mol∙K) Ar (g)39.948154.8 NO (g)30.006210.8

39. Tro, Chemistry: A Molecular Approach39 Relative Standard Entropies Dissolution dissolved solids generally have larger entropy distributing particles throughout the mixture Substance S°, (J/mol∙K) KClO 3 (s)143.1 KClO 3 (aq)265.7

40. Ex. 17.4 –Calculate  S  for the reaction 4 NH 3(g) + 5 O 2(g)  4 NO (g) + 6 H 2 O (l)  S is +, as you would expect for a reaction with more gas product molecules than reactant molecules standard entropies from Appendix IIB  S, J/K Check: Solution: Concept Plan: Relationships: Given: Find: SS S  NH3, S  O2, S  NO, S  H2O, Substance S , J/mol  K NH 3 (g)192.8 O2(g)O2(g)205.2 NO(g)210.8 H 2 O(g)188.8

41. Tro, Chemistry: A Molecular Approach41 Calculating  G  at 25  C:  G o reaction =  nG o f (products) -  nG o f (reactants) at temperatures other than 25  C: assuming the change in  H o reaction and  S o reaction is negligible  G  reaction =  H  reaction – T  S  reaction

42. 42

43. Ex. 17.7 –Calculate  G  at 25  C for the reaction CH 4(g) + 8 O 2(g)  CO 2(g) + 2 H 2 O (g) + 4 O 3(g) standard free energies of formation from Appendix IIB  G , kJ Solution: Concept Plan: Relationships: Given: Find: GG  G  f of prod & react Substance  G  f, kJ/mol CH 4 (g)-50.5 O2(g)O2(g)0.0 CO 2 (g)-394.4 H 2 O(g)-228.6 O3(g)O3(g)163.2

44. Ex. 17.6 – The reaction SO 2(g) + ½ O 2(g)  SO 3(g) has  H  = -98.9 kJ and  S  = -94.0 J/K at 25°C. Calculate  G  at 125  C and determine if it is spontaneous. Since  G is -, the reaction is spontaneous at this temperature, though less so than at 25  C  H  = -98.9 kJ,  S  = -94.0 J/K, T = 398 K  G , kJ Answer: Solution: Concept Plan: Relationships: Given: Find: GG T,  H ,  S 

45. Tro, Chemistry: A Molecular Approach45  G Relationships if a reaction can be expressed as a series of reactions, the sum of the  G values of the individual reaction is the  G of the total reaction  G is a state function if a reaction is reversed, the sign of its  G value reverses if the amounts of materials is multiplied by a factor, the value of the  G is multiplied by the same factor the value of  G of a reaction is extensive

46. Tro, Chemistry: A Molecular Approach46 Free Energy and Reversible Reactions the change in free energy is a theoretical limit as to the amount of work that can be done if the reaction achieves its theoretical limit, it is a reversible reaction

47. Tro, Chemistry: A Molecular Approach47 Real Reactions in a real reaction, some of the free energy is “lost” as heat if not most therefore, real reactions are irreversible

48. Tro, Chemistry: A Molecular Approach48  G under Nonstandard Conditions   G =  G  only when the reactants and products are in their standard states  there normal state at that temperature  partial pressure of gas = 1 atm  concentration = 1 M  under nonstandard conditions,  G =  G  + RTlnQ  Q is the reaction quotient  at equilibrium  G = 0   G  = ─RTlnK

49. Tro, Chemistry: A Molecular Approach49

50. 50 Example -  G Calculate  G at 427°C for the reaction below if the P N2 = 33.0 atm, P H2 = 99.0 atm, and P NH3 = 2.0 atm N 2 (g) + 3 H 2 (g)  2 NH 3 (g) Q = P NH3 2 P N2 1 x P H2 3 (2.0 atm) 2 (33.0 atm) 1 (99.0) 3 == 1.2 x 10 -7  G =  G° + RTlnQ  G = +46400 J + (8.314 J/K)(700 K)(ln 1.2 x 10 -7 )  G = -46300 J = -46 kJ  H° = [ 2(-46.19)] - [0 +3( 0)] = -92.38 kJ = -92380 J  S° = [2 (192.5)] - [(191.50) + 3(130.58)] = -198.2 J/K  G° = -92380 J - (700 K)(-198.2 J/K)  G° = +46400 J

51. Tro, Chemistry: A Molecular Approach51

52. 52 Example - K Estimate the equilibrium constant and position of equilibrium for the following reaction at 427°C N 2 (g) + 3 H 2 (g)  2 NH 3 (g)  G° = -RT lnK +46400 J = -(8.314 J/K)(700 K) lnK lnK = -7.97 K = e -7.97 = 3.45 x 10 -4 since K is << 1, the position of equilibrium favors reactants  H° = [ 2(-46.19)] - [0 +3( 0)] = -92.38 kJ = -92380 J  S° = [2 (192.5)] - [(191.50) + 3(130.58)] = -198.2 J/K  G° = -92380 J - (700 K)(-198.2 J/K)  G° = +46400 J

53. Tro, Chemistry: A Molecular Approach53 Temperature Dependence of K for an exothermic reaction, increasing the temperature decreases the value of the equilibrium constant for an endothermic reaction, increasing the temperature increases the value of the equilibrium constant