 Download Download Presentation Chapter 17 Free Energy and Thermodynamics

# Chapter 17 Free Energy and Thermodynamics

Download Presentation ## Chapter 17 Free Energy and Thermodynamics

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
##### Presentation Transcript

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

2. 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 • DEuniverse = 0 = DEsystem + DEsurroundings Tro, Chemistry: A Molecular Approach

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 • DE = q + w • DE = DH + PDV • DE is a state function • internal energy change independent of how done Tro, Chemistry: A Molecular Approach

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 Tro, Chemistry: A Molecular Approach

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

6. 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 Tro, Chemistry: A Molecular Approach

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. Tro, Chemistry: A Molecular Approach

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 Tro, Chemistry: A Molecular Approach

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

10. 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). Tro, Chemistry: A Molecular Approach

11. 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. • DH • The entropy factors relates to the randomness/orderliness of a system • DS • The enthalpy factor is generally more important than the entropy factor Tro, Chemistry: A Molecular Approach

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

13. Entropy • entropyis 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 Tro, Chemistry: A Molecular Approach

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

15. 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 Tro, Chemistry: A Molecular Approach

16. 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 Tro, Chemistry: A Molecular Approach

17. Changes in Entropy, DS • entropy change is favorable when the result is a more random system. • DS 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 Tro, Chemistry: A Molecular Approach

18. Increases in Entropy Tro, Chemistry: A Molecular Approach

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

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 Tro, Chemistry: A Molecular Approach

21. Entropy Change and State Change Tro, Chemistry: A Molecular Approach

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 Tro, Chemistry: A Molecular Approach

23. Temperature Dependence of DSsurroundings • 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 Tro, Chemistry: A Molecular Approach

24. Gibbs Free Energy, DG • maximum amount of energy from the system available to do work on the surroundings G = H – T∙S DGsys = DHsys – TDSsys DGsys = – TDSuniverse DGreaction = SnDGprod – SnDGreact • when DG < 0, there is a decrease in free energy of the system that is released into the surroundings; therefore a process will be spontaneous when DG is negative Tro, Chemistry: A Molecular Approach

25. T, DH DS Ex. 17.2a – The reaction C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(g) has DHrxn = -2044 kJ at 25°C. Calculate the entropy change of the surroundings. Given: Find: DHsystem = -2044 kJ, T = 298 K DSsurroundings, J/K Concept Plan: Relationships: Solution: Check: combustion is largely exothermic, so the entropy of the surrounding should increase significantly

26. Free Energy Change and Spontaneity Tro, Chemistry: A Molecular Approach

27. Gibbs Free Energy, DG • process will be spontaneous when DG is negative • DG will be negative when • DH is negative and DS is positive • exothermic and more random • DH is negative and large and DS is negative but small • DH is positive but small and DS is positive and large • or high temperature • DG will be positive when DH is + and DS is − • never spontaneous at any temperature • when DG = 0 the reaction is at equilibrium Tro, Chemistry: A Molecular Approach

28. DG, DH, and DS Tro, Chemistry: A Molecular Approach

29. T, DH, DS DG Ex. 17.3a – The reaction CCl4(g) C(s, graphite) + 2 Cl2(g) has DH = +95.7 kJ and DS = +142.2 J/K at 25°C. Calculate DG and determine if it is spontaneous. Given: Find: DH = +95.7 kJ, DS = 142.2 J/K, T = 298 K DG, kJ Concept Plan: Relationships: Solution: Since DG is +, the reaction is not spontaneous at this temperature. To make it spontaneous, we need to increase the temperature. Answer:

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

31. 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 + Tro, Chemistry: A Molecular Approach

32. 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 Tro, Chemistry: A Molecular Approach

33. 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 Tro, Chemistry: A Molecular Approach

34. Relative Standard EntropiesMolar Mass • the larger the molar mass, the larger the entropy • available energy states more closely spaced, allowing more dispersal of energy through the states Tro, Chemistry: A Molecular Approach

35. Relative Standard EntropiesAllotropes • the less constrained the structure of an allotrope is, the larger its entropy Tro, Chemistry: A Molecular Approach

36. Relative Standard EntropiesMolecular Complexity • larger, more complex molecules generally have larger entropy • more available energy states, allowing more dispersal of energy through the states Tro, Chemistry: A Molecular Approach

37. Relative Standard EntropiesDissolution • dissolved solids generally have larger entropy • distributing particles throughout the mixture Tro, Chemistry: A Molecular Approach

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

39. Calculating DG • at 25C: DGoreaction = SnGof(products) - SnGof(reactants) • at temperatures other than 25C: • assuming the change in DHoreaction and DSoreaction is negligible DGreaction = DHreaction – TDSreaction Tro, Chemistry: A Molecular Approach

40. DGf of prod & react DG Ex. 17.7 –Calculate DG at 25C for the reactionCH4(g) + 8 O2(g) CO2(g) + 2 H2O(g) + 4 O3(g) Given: Find: standard free energies of formation from Appendix IIB DG, kJ Concept Plan: Relationships: Solution:

41. T, DH, DS DG Ex. 17.6 – The reaction SO2(g) + ½ O2(g) SO3(g) has DH = -98.9 kJ and DS = -94.0 J/K at 25°C. Calculate DG at 125C and determine if it is spontaneous. Given: Find: DH = -98.9 kJ, DS = -94.0 J/K, T = 398 K DG, kJ Concept Plan: Relationships: Solution: Since DG is -, the reaction is spontaneous at this temperature, though less so than at 25C Answer:

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

43. 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 Tro, Chemistry: A Molecular Approach

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

45. DG under Nonstandard Conditions • DG = DG 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, DG = DG + RTlnQ • Q is the reaction quotient • at equilibrium DG = 0 • DG = ─RTlnK Tro, Chemistry: A Molecular Approach

46. Tro, Chemistry: A Molecular Approach

47. PNH32 PN21 x PH23 (2.0 atm)2 (33.0 atm)1 (99.0)3 Q = = = 1.2 x 10-7 DG = DG° + RTlnQ DG = +46400 J + (8.314 J/K)(700 K)(ln 1.2 x 10-7) DG = -46300 J = -46 kJ Example - DG • Calculate DG at 427°C for the reaction below if the PN2 = 33.0 atm, PH2= 99.0 atm, and PNH3= 2.0 atm N2(g) + 3 H2(g) ® 2 NH3(g) DH° = [ 2(-46.19)] - [0 +3( 0)] = -92.38 kJ = -92380 J DS° = [2 (192.5)] - [(191.50) + 3(130.58)] = -198.2 J/K DG° = -92380 J - (700 K)(-198.2 J/K) DG° = +46400 J