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Chapter 17 Free Energy and Thermodynamics

Chapter 17 Free Energy and Thermodynamics. Goals. Entropy (S, S) and spontaneity. Free energy; G ,  G o. G , K, product- or reactant-favored. Review : H (Enthalpy) and the 1st Law of Thermodynamics Chemical Equilibria (ch. 14, etc). First Law of Thermodynamics.

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Chapter 17 Free Energy and Thermodynamics

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  1. Chapter 17Free Energy and Thermodynamics

  2. Goals Entropy (S, S) and spontaneity Free energy; G, Go G, K, product- or reactant-favored Review: H (Enthalpy)and the 1st Law of Thermodynamics Chemical Equilibria (ch. 14, etc)

  3. First Law of Thermodynamics First Law of Thermodynamics: Energy cannot be created or destroyed the total energy of the universe cannot change it can be transfered from one place to another DEuniverse = 0 = DEsystem + DEsurroundings system = reactants & products surroundings = everything else (the transfer of energy from one to the other does not change the energy of the universe)

  4. First Law of Thermodynamics For an exothermic reaction, heat from the system goes into the surroundings • two ways energy can be “lost” from a system, • converted to heat, q • used to do work, w Energy conservation requires that the energy change in the system = heat exchanged + work done on the system. • DE = q + w (E = internal energy change) • DE = DH – PDV (at const. P, qp = DH, enthalpy change) • State functions (H, P, V). qandw are not. • internal energy change (DE) independent of how done

  5. Enthalpy, H • related to (includes) 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(DHf°prod) - S(DHf°react)

  6. Thermodynamics and Spontaneity • thermodynamics predicts whether a process will proceed (occur) under the given conditions • spontaneous process • nonspontaneousprocess does not occur under specific conditions. • spontaneity is determined by comparing the free energy (G) 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 (rate); this is kinetics

  7. Spontaneous Nonspontaneous ice melts @ 25oC water freezes @ 25oC water freezes @ -10 oC ice melts @ -10oC ball rolls downhill ball rolls uphill 2Na(s) + 2H2O(l) H2(g) + 2NaOH(aq)  H2(g) + 2NaOH(aq) 2Na(s) + 2H2O(l)

  8. Diamond → Graphite kinetics: how fast Spontaneity: direction & extent kinetics Graphite is thermodynamically more stable than diamond, so the conversion of diamond into graphite is spontaneous – but it’s kinetically too slow (inert) it will never happen in many, many generations.

  9. 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 relate to the randomness/orderliness of a system • DS • The enthalpy factor is generally more important than the entropy factor

  10. Entropy, S Entropy is usually described as a measure of the randomness or disorder; the greater the disorder of a system, the greater its S. The greater the order  the smaller its S. • Entropyis a thermodynamic function that increases as the number of energetically equivalent ways of arranging the components increases. • S generally in J/K (joules/K) • S = k lnW • k = Boltzmann Constant (R/NA) = 1.38  10-23 J/K • W is the number of energetically equivalent ways, (microstates). It is unitless.

  11. S = k ln W Entropy & Microstates, W Energetically Equivalent States for the Expansion of a Gas (4 gas molecules) 1 microstate 1 microstate S = k ln Wf - k ln Wi if Wf > Wi , S > 0 & entropy increases. 6 microstates (most probable distribution)

  12. Changes in Entropy, DS • entropy change is favorable when the result is a more random system (State C: higher entropy). • DS is positive (DS > 0) • Some changes that increase the entropy are: • rxns where products are in a more disordered state. • (solid > liquid > gas) less order • (solid< liquid < gas) larger S (disorder) • reactions which have larger numbers of product molecules than reactant molecules. • increase in temperature (more movement) • solids dissociating into ions upon dissolving

  13. Changes in Entropy in a System(melting) Particles fixed in space Particles can occupy many positions

  14. Changes in Entropy in a System(vaporization) Particles occupy more space (larger volume)

  15. Changes in Entropy in a System (solution process) Structure of solute and solvent disrupted (also more solute particles)

  16. Predict entropy change for a process/reaction For which process/reaction is S negative? Freezing ethanol  entropy dec Mixing CCl4 with C6H6  entropy dec Condensing bromine vapor 2O3(g)  3O2(g) 4Fe(s) + 3O2(g)  2Fe2O3(s)  entropy dec 2H2O2(aq)  2H2O(l) + O2(g) 2Li(s) + 2H2O(l)  2LiOH(aq) + H2(g) 2NH3(g)  N2(g) + 3H2(g)

  17. The 2ndLaw of Thermodynamics • The entropy of the universe increases in a spontaneous process. • DSuniverse = DSsystem + DSsurroundings> 0 • DSuniverse = DSsystem + DSsurroundings = 0 (equilibrium) If DSsystem >> 0, DSsurroundings < 0 for DSuniverse > 0! • If DSsystem < 0, DSsurroundings >> 0 for DSuniverse > 0! • the increase in DSsurroundings often comes from the heat released in an exothermic reaction, DHsystem< 0.

  18. The 3rd Law of Thermodynamics -allows determination of entropy of substances. (W = 1, there is only one way to arrange the particles to form a perfect crystal) S = k ln W = k ln 1 = 0 • the 3rd Law states that for a perfect crystal at absolute zero, the absolute entropy = 0 J/mol∙K • S = Sf – Si; where Si = 0 @ 0 K • the absolute entropy of a substance is always (+) positive at the new T S = k ln W

  19. Standard Entropies S° 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 Values can be used to calculate the standard entropy change for a reaction, Sorxn (= Sosys)

  20. Trends: Standard EntropiesMolar Mass • For monatomic species, the larger the molar mass, the larger the entropy • available energy states more closely spaced, allowing more dispersal of energy through the states

  21. Trends: Standard EntropiesStates • the standard entropy of a substance in the gas phase is greater than the standard entropy of the same substance in the solid or liquid phase at a particular temperature

  22. Trends: Standard EntropiesAllotropes -different forms of an element • the more highly ordered form has the smaller entropy

  23. Trends: Standard EntropiesMolecular Complexity (inc # of atoms) larger, more complex molecules generally have larger entropy • more available energy states, allowing more dispersal of energy through the states

  24. Trends: Standard EntropiesDissolution • dissolved solids generally have larger entropy • distributing particles throughout the mixture

  25. Q. Arrange the following in order of increasing entropy @ 25oC! (lowest to highest) Ne(g), SO2(g), Na(s), NaCl(s) and H2(g) Na(s) < NaCl(s) < H2(g) < Ne(g) < SO2(g) Q .Which has the larger entropy in each pair? • Li(s) or Li(l) • b) C2H5OH(l) or CH3OCH3 (l) • c) Ar(g) or Xe(g) • d) O2(g) or O3(g)

  26. SNH3, SO2, SNO, SH2O, DS Calculate DS for the reaction4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g) 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

  27. Calculating ∆So for a Reaction ∆So =  So (products) -  So (reactants) Consider 2 H2(g) + O2(g)  2 H2O(liq) @ 25oC ∆So = 2 So (H2O) - [2 So (H2) + So (O2)] ∆So = 2 mol (69.9 J/K•mol) - [2 mol (130.6 J/K•mol) + 1 mol (205.0 J/K•mol)] ∆So = -326.4 J/K Note that there is a decrease in S because 3 mol of gas give 2 mol of liquid.

  28. Temperature Dependence of DSsurroundings DHsystem< 0 (exothermic), it adds heat to the surroundings, increasing the entropy of the surroundings (DSsurroundings > 0 ) DHsystem> 0 (endothermic), it takes heat from the surroundings, decreasing the entropy of the surroundings (DSsurroundings < 0 )

  29. Calculating ∆So for the surroundings 2 H2(g) + O2(g)  2 H2O(liq) @ 25 oC ∆Sosystem = -326.4 J/K Can calculate ∆Hosystem = ∆Horxn = -571.7 kJ (also from tabulated data) ∆Sosurroundings = +1917 J/K

  30. 2 H2(g) + O2(g)  2 H2O(liq) @ 25oC ∆Sosystem = -326.4 J/K ∆Sosurroundings = +1917 J/K ∆Souniverse = +1591 J/K The entropy of the universe is increasing, so the reaction is spontaneous ( product- favored). (see slide # 18) DSuniverse = DSsystem + DSsurroundings Given DSosurr, DSosys and T, determine DSounivand predict if the reaction will be spontaneous.

  31. T, DH DS 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 surroundings should increase (inc in # gas mol)

  32. Spontaneous or Not? DSuniverse = DSsystem – DHsystem/T originally: DSuniverse = DSsystem + DSsurroundings but

  33. Without doing any calculations, determine the sign of Ssys and Ssurr for each reaction. Predict under what temperatures (all T, low T, or high T) the reaction will be spontaneous. DSuniverse = DSsystem + (-DHsys/T) Ssurr= (-DHsys/T) 2CO(g) + O2(g)  2CO2(g) DHrxn = -566.0 kJ DSsystem = (-); 3 mol gas form 2 mol gas DSsurr = (+); spontaneous @ low T 2NO2(g)  O2(g) + 2NO(g) DHrxn = +113.1 kJ DSsystem = (+); 2 mol gas form 3 mol gas DSsurr = (-); spontaneous @ high T

  34. Without doing any calculations, determine the sign of Ssys and  Ssurr for each reaction. Predict under what temperatures (all T, low T, or high T) the reaction will be spontaneous. DSuniverse = DSsystem + (-DHsys/T) 2H2(g) + O2(g)  2H2O(g) DHrxn = -483.6 kJ DSsystem = (-); 3 mol gas form 2 mol gas DSsurr = (+) ; spontaneous @ low T CO2(g)  C(s) + O2(g) DHrxn = +393.5 kJ DSsystem = (-); complicated gas forms a solid & gas DSsurr = (-); nonspontaneous @ all T Ssurr= (-DHsys/T)

  35. At what temperature is the change in entropy for the reaction equal to the change in entropy for the surroundings, if Horxn = -127 kJ and Sorxn = 314 J/K. Plan: set Sorxn = Sosurr and solve for T; convert kJ to J rxn implies system!!! Ans: T = +404 K

  36. Gibbs Free Energy, G = H -TS ∆Suniv = ∆Ssurr + ∆Ssys Multiply through by -T -T∆Suniv = ∆Hsys - T∆Ssys -T∆Suniv = change in Gibbs free energy for the system = ∆Gsystem hence, ∆G = ∆H -T∆S Under standard conditions — ∆Gosys = ∆Hosys - T∆Sosys J. Willard Gibbs1839-1903

  37. ∆G = ∆H - T∆S Gibbsfree energychange =total energy change for system - energy lost in disordering the system If the reaction is • exothermic (negative ∆H) • and entropy increases (positive ∆So) • then∆G must be NEGATIVE • the reaction is spontaneous (and product-favored) at ALL temperatures.

  38. ∆G = ∆H - T∆S • DG will be positive (∆G > 0) when DH is positive (endothermic) and DS is negative (more ordered). So the change in free energy will be positive at all temperatures. • The reaction will therefore be nonspontaneous at ALL temperatures • When DG = 0 the reaction is at equilibrium

  39. DG = DH – TDSSpontaneous or Not? A decrease in Gibbs free energy (DG < 0) corresponds to a spontaneous process An increase in Gibbs free energy (DG > 0) corresponds to a nonspontaneous process

  40. Calculating ∆Go: ∆Go= ∆H -T∆So Combustion of acetylene @ 25 oC C2H2(g) + 5/2 O2(g)  2 CO2(g) + H2O(g) Use enthalpies of formation to calculate ∆Horxn = -1238 kJ Use standard molar entropies to calculate ∆Sorxn = -97.4 J/K or -0.0974 kJ/K ∆Gorxn = -1238 kJ - (298 K)(-0.0974 J/K) = -1209 kJ (spontaneous) Reaction is product-favored in spite of negative ∆Sorxn. Reaction is “enthalpy driven”

  41. Calculating ∆Go: ∆Go= ∆H -T∆So Is the dissolution of ammonium nitrate product-favored? If so, is it enthalpy- or entropy-driven? From tables of thermodynamic data we find ∆Horxn = +25.7 kJ (endothermic) ∆Sorxn = +108.7 J/K or +0.1087 kJ/K (disorder) ∆Gorxn = +25.7 kJ - (298 K)(+0.1087 kJ/K) = -6.7 kJ (spontaneous) Reaction is product-favored in spite of positive ∆Horxn. Reaction is “entropy driven” NH4NO3(s) + heat NH4NO3(aq)

  42. T, DH, DS DG 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:

  43. DG, DH, DS T The reaction CCl4(g) C(s, graphite) + 2 Cl2(g) has DH = +95.7 kJ and DS = +142.2 J/K. Calculate the minimum T @ which 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 (i.e. 674 K) Answer:

  44. Gibbs Free Energy, G Calculating ∆Go(two ways) a) Determine ∆Horxn and ∆Sorxn and use Gibbs equation (at various temps). b) Use tabulated values offree energies of formation, ∆Gfo@ 25oC ∆Gorxn =  ∆Gfo (products) -  ∆Gfo (reactants)

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

  46. T, DH, DS DG 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 (PRACTICE PROBLEM) Concept Plan: Relationships: Solution: Since DG is -ve, the rxn is spontaneous at this temperature, but less spontaneous than at 25C (-127 kJ) Answer:

  47. ∆G is the change in free energy at non-standard conditions. ∆G is related to ∆G˚ ∆G = ∆G˚ + RT ln Q where Q = reaction quotient When Q < K or Q > K, the reaction may be spontaneous or nonspontaneous. When Q = K reaction is at equilibrium When ∆G = 0 reaction is at equilibrium Therefore, ∆G˚ = - RT ln K ∆G, ∆G˚, and Keq

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