1. Chapter 17 Free Energy and Thermodynamics Chemistry II

2. First Law of Thermodynamics 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 ΔE universe = 0 = ΔE system + ΔE surroundings

3. 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 Δ E = q + w

4. 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 e.g. if a light bulb is 5 % efficient 95% is lost as heat

5. Heat Tax fewer steps generally results in a lower total heat tax

6. Spontaneous and Nonspontaneous Processes e.g. in physics - will an object fall when dropped? in chemistry - will iron rust? thermodynamics predicts whether a process will proceed under the given conditions - spontaneous process  nonspontaneous processes require energy inputs In physics sponaneity is determined by looking at the potential energy change of a system In chemistry spontaneity is determined by comparing the free energy change of the system if the system after reaction has less free energy than before the reaction, the reaction is thermodynamically favorable. spontaneity ≠ fast or slow

7. Comparing Potential Energy The direction of spontaneity can be determined by comparing the potential energy of the system at the start and the end. High PE → Low PE High ‘Free Energy’ → Low ‘Free Energy’

8. 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. Thermodynamics vs. Kinetics Don’t confuse spontaneity (direction and extent of reaction) with speed This reaction will require an energy input since ‘free energy’ of reactants < ‘free energy’ products

10. Diamond → Graphite Graphite has less FE 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. Factors Affecting Whether a Reaction Is Spontaneous The two factors that determine spontaneity (free energy) 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. Δ H The entropy factors relates to the randomness/orderliness of a system ΔS The enthalpy factor is generally more important than the entropy factor

12. Enthalpy, ΔH ΔH related to the internal energy (kJ/mol) stronger bonds = more stable molecules if products more stable than reactants, energy released Exothermic (ΔH = negative) if reactants more stable than products, energy absorbed Endothermic (ΔH = positive) Hess’ Law: Δ H° rxn = Σ(nΔ H° prod ) - Σ(nΔ H° react ) Where n = stoichiometric coefficient

13. Changes in Entropy, ΔS entropy change is favorable when the result is a more random system ΔS 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

14. Increases in Entropy Melting Evaporating Dissolution

15. Entropy entropy is a thermodynamic function that increases as the number of energetically equivalent ways of arranging the components increases, a measure of randomness, S  S generally J/K S = k ln W k = Boltzmann Constant = 1.38 x 10 -23 J/K W is the number of energetically equivalent ways, unitless A chemical system proceeds in a direction that increases the entropy of the universe Next we consider W,

16. W Energetically Equivalent States for the Expansion of a Gas in a vacuum The following are macro states

17. 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

18. Macrostates and Probability There is only one possible arrangement that gives State A and one that gives State B There are 6 possible arrangements that give State C Therefore State C has higher entropy than either State A or State B The macrostate with the highest entropy also has the greatest dispersal of energy, now imagine thousands of atoms!

19. Entropy Second law of thermodynamics: For any spontaneous process, the entropy of the universe increases, ΔS univ > 0 Spontaneous processes increase the entropy of the universe Δ S = S final – S initial In our flask analogy, ΔS > 0 Explains why heat travels from a substance at high temp. to a substance at low temp. but never the opposite

20. 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

21. Entropy Change and State Change

22. 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

23. Is Entropy Increases or Decreasing? Iodine sublimes A liquid boils A solid decomposes into two gaseous products

25. Heat Transfer and Changes in S of Surroundings the total entropy change of the universe must be positive for a process to be spontaneous for reversible process ΔS univ = 0, for irreversible (spontaneous) process Δ S univ > 0 ΔS universe = ΔS system + Δ S surroundings if the entropy of the system decreases, then the entropy of the surroundings must increase by a larger amount when ΔS system is negative, ΔS surroundings is positive e.g. when water freezes the increase in ΔS surroundings often comes from the heat released in an exothermic reaction

26. Temperature Dependence of ΔS 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

27. Ex. 17.2a – The reaction C 3 H 8(g) + 5 O 2(g)  3 CO 2(g) + 4 H 2 O (g) has DH 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 Δ H system = -2044 kJ, T = 298 K ΔS surroundings, J/K Check: Solution: Concept Plan: Relationships: Given: Find: Δ ST, Δ H

28. Temperature Dependence of ΔS surroundings Example: 2N 2 (g) + O 2 (g) → 2N 2 O (g) ΔH = +163.2 kJ/mol (i)Calculate the entropy change in the surroundings at 25 ºC ΔS surr. = -ΔH sys. / T = -163.2 kJ/298K = -0.548 kJ/K = -548 J/K (ii) Determine the sign of the entropy change 3 mols on LHS, 2 mols on RHS so -ΔS sys (iii) Determine the sign of Δ S universe, is reaction spontaneous? Δ S universe = ΔS system + Δ S surroundings -ve + -ve = -ve Not spontaneous

29. Gibbs Free Energy, ΔG Enthalpy (ΔH sys ) and entropy (ΔS surr ) are related by: ΔS surr. = -ΔH sys. / T Also: Δ S universe = ΔS system + Δ S surroundings Combining these: ΔS universe = ΔS system -ΔH sys. / T Now ΔS univ can be calculated with the ‘system’ only Multiplying by –T: -TΔS universe = -TΔS system -ΔH sys. This is now called the Gibbs free Energy expression (remember we multiplied by a –ve)

30. Gibbs Free Energy, ΔG maximum amount of energy from the system available to do work on the surroundings G = H – T∙S ΔG sys = Δ H sys – T Δ S sys The change in Gibbs free energy for a process at constant T and P is ~ -ΔS univ Δ G sys = – T Δ S universe ΔG (Gibbs free energy) being –ve is a criterion for spontaneity there is a decrease in free energy of the system that is released into the surroundings; therefore a process will be spontaneous when ΔG is negative (i.e. ΔS +ve, since we mult. by - T)

31. Free Energy Change and Spontaneity

32. Gibbs Free Energy, ΔG process will be spontaneous when ΔG is negative ΔG sys = Δ H sys – T Δ S sys  Three possibilities: (1) Δ G will be negative - spontaneous Δ H is negative and ΔS is positive  exothermic and more random ΔH is negative and large and ΔS is negative but small Δ H is positive but small and ΔS is positive  high temperature favors spontaneity, nonspon. at low T (2) ΔG will be +ve when ΔH is +ve and ΔS is −ve - nonspontaneous never spontaneous at any temperature (3) when ΔG = 0 the reaction is at equilibrium

33. Which of the following is spontaneous? ΔHºTΔSº ΔHº TΔSº ΔHº TΔSº ΔHºTΔSº

34. ΔG, ΔH, and ΔS

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

36. Ex. 17.3a – The reaction CCl 4(g)  C (s, graphite) + 2 Cl 2(g) has ΔH = +95.7 kJ and ΔS = +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 ΔH = +95.7 kJ, ΔS = 142.2 J/K, ΔG < 0 T, K Answer: Solution: Concept Plan: Relationships: Given: Find: TΔG, ΔH, ΔS

38. Entropy entropy is a thermodynamic function that increases as the number of energetically equivalent ways of arranging the components increases, a measure of randomness, S  S generally J/K S = k ln W k = Boltzmann Constant = 1.38 x 10 -23 J/K W is the number of energetically equivalent ways, unitless A chemical system proceeds in a direction that increases the entropy of the universe Next we consider W,

39. 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 +ve

40. Standard Entropies S° (J/mol.K) Extensive property (depends on amount) 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

42. 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 (l)70.0 H 2 O (g)188.8

43. Relative Standard Entropies Molar Mass the larger the molar mass, the larger the entropy

44. Relative Standard Entropies Allotropes the less constrained the structure of an allotrope is, the larger its entropy

45. Relative Standard Entropies Molecular Complexity larger, more complex molecules have larger entropy than single atoms more available energy states (translational, rotational, vibrational) 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

46. 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

47. Standard Entropy of Reaction, ΔS rxn Calculated similar to ΔH rxn Where n = stoichiometric coefficient

48. Ex. 17.4 –Calculate ΔS  for the reaction 4 NH 3(g) + 5 O 2(g)  4 NO (g) + 6 H 2 O (l) ΔS is +, as you would expect for a reaction with more gas product molecules than reactant molecules standard entropies from Appendix IIB ΔS, J/K Check: Solution: Concept Plan: Relationship s: Given: Find: ΔSΔS 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

49. Calculating ΔG  rxn At 25  C and at temperatures other than 25  C: assuming the change in ΔH o reaction and Δ S o reaction are small over a limted temp. range Δ G  rxn = Δ H  rxn – T Δ S  rxn

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

51. Calculating ΔG  rxn with ΔG  f at 25  C: ΔG o rxn = Σn p ΔG o f (products) - Σn r ΔG o f (reactants) Like ΔH f °, ΔG f ° = 0 for an element in its standard state: Example: ΔG f ° (O 2 (g)) = 0

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53. Ex. 17.7 –Calculate ΔG  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 ΔG , kJ Solution: Concept Plan: Relationship s: Given: Find: ΔGΔG ΔG  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

54. ΔG Relationships if a reaction can be expressed as a series of reactions, the sum of the ΔG values of the individual reaction is the ΔG of the total reaction ΔG is a state function if a reaction is reversed, the sign of its ΔG value reverses if the amounts of materials is multiplied by a factor, the value of the ΔG is multiplied by the same factor the value of ΔG of a reaction is extensive

55. ΔG° rxn ≠ Reaction Rate Although ΔG° rxn can be used to predict if a reaction will be spontaneous as written, it does not tell us how fast a reaction will proceed. Example: C(s, diamond) + O 2 (g) CO 2 (g)ΔG° rxn = -397 kJ<<0 But diamonds are forever….ΔG°rxn≠ rate

56. Why Free Energy is ‘Free’ If ΔG –ve: change in Gibbs free energy = max amount of available ‘free’ energy to do work, should be called ‘Gibbs available energy’ For many reactions ΔG < ΔH e.g. C(s) + 2H 2 (g) →CH 4 (g) ΔH rxn = -74.6 kJ, ΔSº rxn = -80.8 J/K, ΔGº rxn = -50.5 kJ Exothermic reaction but only half as much available to do work Entropy is –ve, so some of emitted heat must inc. entropy of the universe in order to make ΔGº rxn -ve

57. 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 theoretically a reversible reaction

58. Real Reactions in a real reaction, some of the free energy is “lost” as heat if not most therefore, real reactions are irreversible Energy is lost – when the battery does work on the motor some energy is lost in the wires as heat (resistance to e - flow)

59. ΔG under Nonstandard Conditions  Δ G = Δ 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, ΔG = ΔG  + RTlnQ  Q is the reaction quotient  at equilibrium ΔG = 0  ΔG  = ─RTlnK

60. ΔG under Nonstandard Conditions  Spilled water spontaneously evaporates even though ΔG◦ rxn for the vaporization of water is +ve..why?  We do not live at std. conditions

61. Q = P H2O Under std. conditions water does not vaporize ΔG rxn = ΔG◦ rxn = +ve Spontaneous in reverse direction Water condenses ΔG◦ rxn = +8.56 kJ/mol ΔG rxn = -ve Non-std. Conditions

62. Example – ΔG nonstandard Calculate ΔG 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 ΔG = ΔG° + RTlnQ ΔG = +46400 J + (8.314 J/K)(700 K)(ln 1.2 x 10 -7 ) ΔG = -46300 J = -46 kJ ΔH° rxm = [ 2(-46.19)] - [0 +3( 0)] = -92.38 kJ = -92380 J ΔS° rxn = [2 (192.5)] - [(191.50) + 3(130.58)] = -198.2 J/K ΔG° = -92380 J - (700 K)(-198.2 J/K) ΔG° = +46400 J

63. Free Energy and Equilibrium: Relating ΔG° rxn to K Standard free energy change of a reaction and the equilibrium constant are related K becomes larger as ΔG° becomes more –ve If ΔG° is large and –ve then reaction has a large equil. constant, if ΔG° is large and +ve reaction has a small equil. Constant, favors reactants at equilibrium ΔG = ΔG  + RTlnQ and at equilibrium Q = K, ΔG = 0 ΔG° rxn = -RT lnK

64. Free Energy and Equilibrium: Relating ΔG° rxn to K ΔG rxn = ΔG  rxn + RT lnQ and at equilibrium Q = K, ΔG rxn = 0 ΔG° rxn = -RT lnK When K < 1, lnK is –ve and ΔG° rxn is +ve When Q = 1, reaction is spontaneous in reverse direction When K > 1, lnK is +ve and ΔG° rxn is -ve When Q = 1, reaction is spontaneous in forward direction When K = 1, lnK is 0 and ΔG° rxn is 0 Reaction is at equilibrium under std. conditions

66. 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) ΔG° = -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 ΔH° = [ 2(-46.19)] - [0 +3( 0)] = -92.38 kJ = -92380 J ΔS° = [2 (192.5)] - [(191.50) + 3(130.58)] = -198.2 J/K ΔG° = -92380 J - (700 K)(-198.2 J/K) ΔG° = +46400 J

67. 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