Chapter 17 Free Energy and Thermodynamics

1. Chemistry II Chapter 17Free Energy and Thermodynamics

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 ΔEuniverse = 0 = ΔEsystem + ΔEsurroundings

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 High PE → Low PE The direction of spontaneity can be determined by comparing the potential energy of the system at the start and the end. 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 • entropyis 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. These microstates all have the same macrostate So there are 6 different particle arrangements that result in the same macrostate This macrostate can be achieved through several different arrangements of the particles Macrostates → Microstates

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, ΔSuniv > 0 • Spontaneous processes increase the entropy of the universe Δ S = Sfinal – Sinitial • 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 2nd 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 ΔSuniv = 0, • for irreversible (spontaneous) process Δ Suniv > 0 ΔSuniverse = ΔSsystem + Δ Ssurroundings • if the entropy of the system decreases, then the entropy of the surroundings must increase by a larger amount • when ΔSsystem is negative, ΔSsurroundings is positive • e.g. when water freezes • the increase in ΔSsurroundings often comes from the heat released in an exothermic reaction

26. Temperature Dependence of ΔSsurroundings • 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. Temperature Dependence of ΔSsurroundings • Example: 2N2 (g) + O2 (g) → 2N2O (g) ΔH = +163.2 kJ/mol (i) Calculate the entropy change in the surroundings at 25 ºC ΔSsurr. = -ΔHsys. / 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 -ΔSsys (iii) Determine the sign of Δ Suniverse , is reaction spontaneous? Δ Suniverse = ΔSsystem + Δ Ssurroundings -ve + -ve = -ve Not spontaneous

28. Gibbs Free Energy, ΔG • Enthalpy (ΔHsys) and entropy (ΔSsurr) are related by: ΔSsurr. = -ΔHsys. / T • Also: Δ Suniverse = ΔSsystem + Δ Ssurroundings • Combining these: ΔSuniverse = ΔSsystem-ΔHsys. / T • Now ΔSuniv can be calculated with the ‘system’ only • Multiplying by –T: -TΔSuniverse = -TΔSsystem-ΔHsys. • This is now called the Gibbs free Energy expression (remember we multiplied by a –ve)

29. Gibbs Free Energy, ΔG • maximum amount of energy from the system available to do work on the surroundings G = H – T∙S ΔGsys = Δ Hsys – T Δ Ssys The change in Gibbs free energy for a process at constant T and P is ~ -ΔSuniv Δ Gsys = – T Δ Suniverse • 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)

30. Gibbs Free Energy, ΔG • process will be spontaneous when ΔG is negative ΔGsys = Δ Hsys – T Δ Ssys • 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

31. Which of the following is spontaneous? ΔHº TΔSº TΔSº ΔHº ΔHº TΔSº ΔHº TΔSº

32. ΔG, ΔH, and ΔS

33. T, ΔH, ΔS Δ G Ex. 17.3a – The reaction CCl4(g) C(s, graphite) + 2 Cl2(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. Given: Find: ΔH = +95.7 kJ, Δ S = 142.2 J/K, T = 298 K ΔG, kJ Concept Plan: Relationships: Solution: Since ΔG is +, the reaction is not spontaneous at this temperature. To make it spontaneous, we need to increase the temperature. Answer:

34. ΔG, ΔH, ΔS T Ex. 17.3a – The reaction CCl4(g) C(s, graphite) + 2 Cl2(g) has ΔH = +95.7 kJ and ΔS = +142.2 J/K. Calculate the minimum temperature it will be spontaneous. Given: Find: ΔH = +95.7 kJ, ΔS = 142.2 J/K, ΔG < 0 T, K Concept Plan: Relationships: Solution: The temperature must be higher than 673K for the reaction to be spontaneous Answer:

36. Entropy • entropyis 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,

37. The 3rd Law of ThermodynamicsAbsolute Entropy • the absolute entropy of a substance is the amount of energy it has due to dispersion of energy through its particles • the 3rd 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

38. 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

40. Relative Standard EntropiesStates • 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

41. Relative Standard EntropiesMolar Mass • the larger the molar mass, the larger the entropy

42. Relative Standard EntropiesAllotropes • the less constrained the structure of an allotrope is, the larger its entropy

43. Relative Standard EntropiesMolecular 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

44. Relative Standard EntropiesDissolution • dissolved solids generally have larger entropy • distributing particles throughout the mixture

45. Standard Entropy of Reaction, ΔSrxn • Calculated similar to ΔHrxn • Where n = stoichiometric coefficient

46. SNH3, SO2, SNO, SH2O, ΔS Ex. 17.4 –Calculate ΔS for the reaction4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(l) Given: Find: standard entropies from Appendix IIB ΔS, J/K Concept Plan: Relationships: Solution: Check: ΔS is +, as you would expect for a reaction with more gas product molecules than reactant molecules

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

48. T, ΔH, ΔS ΔG Ex. 17.6 – The reaction SO2(g) + ½ O2(g) SO3(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. Given: Find: ΔH = -98.9 kJ, ΔS = -94.0 J/K, T = 398 K ΔG, kJ Concept Plan: Relationships: Solution: Since ΔG is -, the reaction is spontaneous at this temperature, though less so than at 25C Answer:

49. Calculating ΔGrxn with ΔG f • at 25C: ΔGorxn = ΣnpΔGof(products) - ΣnrΔGof(reactants) • Like ΔHf°, ΔGf° = 0 for an element in its standard state: • Example: ΔGf° (O2(g)) = 0