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Chapter 19 Entropy and Free Energy

Chapter 19 Entropy and Free Energy. Objectives: Define entropy and spontaneity. Predict whether a process will be spontaneous. Describe free energy. Describe the relationship between DG, K, and product favorability. Introduction. How to predict if a reaction can occur, given enough time?

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Chapter 19 Entropy and Free Energy

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  1. Chapter 19Entropy and Free Energy Objectives: Define entropy and spontaneity. Predict whether a process will be spontaneous. Describe free energy. Describe the relationship between DG, K, and product favorability.

  2. Introduction How to predict if a reaction can occur, given enough time? THERMODYNAMICS How far will it proceed? (K) How to predict if a reaction can occur at a reasonable rate? KINETICS (Ea)

  3. Thermodynamics • Energy relationships • 1st law of thermodynamics • Energy is conserved: energy is not created nor destroyed • DE = q + w • qp = DH Enthalpy: _______ transferred in a process at constant pressure • 2nd law of thermodynamics • DS Entropy: increases in ________ processes • Direction of the process

  4. Lighter reaction: • C4H8 + 6 O2 4 CO2 + 4 H2O • Burning candle • Drop a pen • Gas expansion • Heat transfer non spontaneous spontaneous

  5. SpontaneousProcesses • Proceed on its own without outside assistance. • Occurs in a define _______ • Leads to ___________ • Maybe determined by T, P • The reverse process is ____________________

  6. Identify spontaneous processes Predict whether the following are a) spontaneous as described, b) spontaneous in the reverse direction, c) in equilibrium • When a piece of metal heated to 150oC is added to water at 40oC, the water gets hotter. • Benzene vapor, C6H6(g), at P= 1atm, condenses to liquid benzene at the normal boiling point of benzene, 80.1oC • Water at room temperature decomposes into H2(g) and O2(g) • AgCl(s) Ag+(aq) + Cl- (aq) ; K = 1.8 x 10-10

  7. Reversible vs Irreversible Processes • Sadi Carnot (French engineer -1824) • Efficiency of heat to work (steam engines) • Significant amount of heat is lost to surroundings • Rudolph Clausius (German physicist ~ 1924) • Ratio of heat in an ideal engine and temperature at which is delivered (q/T) • Entropy • Amount of work extracted from spontaneous processes depends on the manner in which the process is carried (pathway)

  8. Reversible vs Irreversible Processes • Reversible: a process which reverse direction whenever an infinitesimal change is made in some property of the system. Often at equilibrium. • Here entropy can be obtained at any T by measuring the heat required rise the temperature from 0K, with the slow addition of heat in very small amounts. • DS = qrev/T • Any spontaneous process is Irreversible, they often involve non equilibrium conditions; in order to reverse the surroundings must do some work on the system (so the surroundings change).

  9. Entropy • Associated with the ___________ in a system • Associated with the extent to which energy is ___________among the various motions of molecules of the system. • Related to heat transfer and __________.

  10. Entropy • S, is a ________ function (like internal energy, E, and enthalpy, H). • The value of S is characteristic of the state of a system (and a property of the bulk matter). • DS, change in entropy, depends only on the initial and final states of the system, and not in the path taken from one state to the other: DS = Sfinal – Sinitial • For isothermal processes: DS = qrev/T • q: heat absorbed/released • T: temperature in K Adding entropy changes = total entropy

  11. DS for phase changes • Melting of a substance at its m.p. and vaporization of a substance at its b.p. are isothermal processes. • Change can be achieved by adding heat to/from the system to/from surroundings. • qrev = DH fusion (melting) • T = 273 K (normal m.p. 1 atm and 273K). • DS fusion = qrev/T = DH fusion /T

  12. Calculate DS • Calculate DSsystem and the DSsurroundings when 1 mol of ice (~size of an ice cube) melts in your hand. • Process is not reversible (different T’s). • DS can be calculated whether rev or irrev. • DHfusion H2O = 6.01 kJ/mol (melting is endothermic processes, DH is positive). DSsystem = qrev/T = = DSsurroundings = qrev/T (heat gained = - heat lost and T~ 37oC) = = DStotal =DSsystem + DSsurroundings = Spontaneous or non spontaneous? If T’s were about the same, process would be reversible – overall DS = 0

  13. 2nd Law of Thermodynamics • Any irreversible process results in an overall (increase/decrease) ___________ in entropy, whereas a reversible process results in no overall change in entropy. • The sum of entropy of a system + entropy of the surroundings = total entropy change = DS universe. • Irreversible processes occur of their own accord are _________ (spontaneous/non spontaneous). • The total entropy of the universe increases (DSuniverse is __________ (positive/negative)) in any spontaneous process.

  14. DSsystem decreases as Fe + O2 form rust • DSsurroundings? • DStotal =DSsystem + DSsurroundings

  15. Entropy- Molecular Interpretation • Ludwig Boltzmann (1844-1906) • Molecules store energy • KMT • Higher T, higher KE and broader distribution of molecular speeds • 3 kinds of motion: translational, vibrational , rotational = motional energy of the molecule

  16. Entropy- Molecular Interpretation • Molecules moving – snapshot – microstate W= number of microstates of a system S = k ln W k= Boltzmann’s constant 1.38 x 10-23 J/K • DS = k ln Wfinal – k lnWinitial = k ln Wfin/Winit • Change leading to increase in number of microstates leads to a ________ (positive/negative) value of DS. • Related to probability.

  17. Entropy and probability Most often case is when energy is distributed over all particles and to a large number of states.

  18. Entropy and probability Probability and the locations of gas molecules. The two molecules are colored red and blue. a) Before the stopcock is opened, both molecules are in the left-hand flask. b) After the stopcock is opened, there are 4 possible arrangements of the two molecules. The greater number of possible arrangements corresponds to greater disorder in the system, in general, the probability that the molecules will stay in the original flask is (1/2)n, where n is the number of molecules.

  19. Dispersal of Energy • Dispersal of matter often contributes to energy dispersal.

  20. W increases when: • Change involves an: • Increase in _______ • Increase in __________ • Increase in _______________ • Entropy change, DS sign will be ________ • The maximum entropy will be achieved at ____________ – a state in which W has the maximum value.

  21. Explain why melting of ice is spontaneous process

  22. Explain formation of solutions of some ionic solids

  23. If energy and matter are both dispersed in a process, the process is (spontaneous/non spontaneous) _______________. • If only matter is dispersed, __________________________. • If energy is NOT dispersed, the process will (be/never be) ______________ spontaneous.

  24. Predict the sign of DS – each process occurs at constant T • Ag+(aq) + Cl- (aq) AgCl(s) • H2O(l) H2O(g) • N2(g) + O2(g) 2 NO (g) • CO(g) + 3 H2(g) CH4(g) + H2O(g)

  25. 3rd Law of Thermodynamics • Lower the temperature until there is only a single microstate: • The entropy of a pure crystalline substance at absolute zero is ________: S (0K) = ____ • The units of the lattice have no thermal motion. • S= k ln W = • As T increases, S ___________: S solid is (larger/smaller) than S liquid is (larger/smaller) than S gas • Then, all substances have ________ (positive/negative) entropy values at T > 0K.

  26. S increases slightly with T S increases a large amount with phase changes Entropy Change

  27. Standard Molar Entropy Values

  28. Standard Molar Entropy Values

  29. Thermodynamics vs Kinetics Diamond is thermodynamically (favored/not favored) ___________ to convert to graphite, and _______ (favored/not favored) kinetically.

  30. Standard Entropy • So, is the entropy gained by converting it from a perfect crystal at 0K to standard state conditions (1 bar, 1 molal solution). • Units: J/Kmol • Entropies of gases are ________ than those for liquids, entropies of liquids are __________than those for solids. • Larger molecules have a __________ entropy than smaller molecules, molecules with more complex structures have _______entropies than simpler molecules.

  31. Entropy • The entropy of liquid water is _______ than the entropy of solid water (ice) at 0˚ C. S˚(H2O sol) ____ S˚(H2O liq)

  32. Entropy of Solids • Entropy values of solids depend on: • Columbic attractions So (J/K•mol) MgO 26.9 NaF 51.5 Mg2+ & O2- Na+ & F- The larger coulombic attraction on MgO than NaF leads to a lower entropy.

  33. Which sample has the higher entropy • 1 mol NaCl(s) or 1 ml HCl(g) at 25oC b) 2 mol HCl(g) or 1 mol HCl(g) at 25oC c) 1 mol HCl(g) or 1 mol Ar(g) at 298 K d) O2 (g) or O3 (g)

  34. Entropy Change • The entropy change is the sum of the entropies of the products minus the sum of the entropies of reactants: • DS0system = S S0(products) – S S0(reactants) You will find DSo values in the Appendix L of your book.

  35. Calculate the standard entropy changes for the evaporation of 1.0 mol of liquid ethanol to ethanol vapor. C2H5OH(l)  C2H5OH(g)

  36. Calculate the standard entropy change for forming 2.0 mol of NH3(g) from N2(g) and H2(g) N2(g) + 3 H2(g)  2 NH3 (g)

  37. Using standard absolute entropies at 298K, calculate the entropy change for the system when 2.35 moles of NO(g) react at standard conditions. 2 NO(g) + O2(g)  2 NO2(g)

  38. Calculate the standard entropy change for the oxidation of ethanol vapor (CH2H5OH (g)).

  39. Show that DS0univ is positive (>0) for dissolving NaCl in water DS0univ = DS0sys + DS0surr 1) Determine DS0sys 2) Determine DS0surr NaCl(s) NaCl (aq)

  40. Classify the following as one of the four types of Table 19.2 DH0 (kJ) DS0 (J/K) CH4 (g) + 2 O2(g) 2 H2O (l) + CO2(g) -890 -242.8 2 FeO3(s) + 3 C (graphite) 4 Fe(s) + 3 CO2(g) +467 +560.7

  41. Calculate the entropy change of the UNIVERSE when 1.890 moles of CO2(g) react under standard conditions at 298.15 K. Consider the reaction6 CO2(g) + 6 H2O(l)  C6H12O6 + 6O2(g)for which DHo = 2801 kJ and DSo = -259.0 J/K at 298.15 K. • Is this reaction reactant or product favored under standard conditions?

  42. Gibbs Free Energy DSuniv = DSsurr + DSsys DSsurr = -DHsys/T DSuniv = -DHsys/T + DSsys Multiply equation by –T -T DSuniv = DHsys –TDSsys J. Willard Gibbs (1839-1903) DGsys = -T DSuniv DGsys = DHsys –TDSsys DGsys < 0, a reaction is spontaneous DGsys = 0, a reaction is at equilibrium DGsys > 0, the reaction is not spontaneous

  43. Gibbs Free Energy and Spontaneity • J. Willard Gibbs (1839-1903) • Gibbs free energy, G, “free energy”, a thermodynamic function associated with the ________________. • G = H –TS • H- Enthalpy • T- Kelvin temperature • S- Entropy • Changes during a process: DG • Use to determine whether a reaction is spontaneous. • DG is ___________related to the value of the equilibrium constant K, and hence to product favorability.

  44. “Free” Energy • DG = w max • The free energy represents the maximum energy ____________________________. • Example: C(graphite) + 2 H2 (g) CH4 (g) • DH0rx = -74.9 kJ; DS0rx = -80.7 J/K • DG0rx = DH0 – TDS0 • = -74.9 kJ – (298)(-80.7)/1000 kJ • = -74.9 kJ + 24.05 kJ • DG0rx = - 50.85 kJ • Some of the energy liberated by the reaction is needed to “order” the system. The energy left is energy available energy to do work, “free” energy. • DG < 0, the reaction is _______________.

  45. Calculate DGo for the reaction below at 25.0 C. P4(s) + 6 H2O(l) → 4 H3PO4(l) DG0rx = DH0 – TDS0

  46. Standard Molar Free Energy of Formation • The standard free energy of formation of a compound, DG0f, is the free energy change when forming one mole of the compound from the component elements, with products and reactants in their standard states. • Then, DG0f of an element in its standard states is _________.

  47. Gibbs Free Energy DG0rxn is the increase or decrease in free energy as the reactants in their standard states are converted completely to the products in their standard states. * Complete reaction is not always observed. * Reactions reach an ______________. DG0system = S G0(products) – S G0(reactants)

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