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Spontaneity, Entropy and Free Energy

Spontaneity, Entropy and Free Energy. CH 4 (g) + O 2 (g) CO 2 (g) + H 2 O(g) + energy. How much energy Energy flow Final form of energy. Spontaneous Processes and Entropy. First Law of Thermodynamics “Energy can neither be created nor destroyed"

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Spontaneity, Entropy and Free Energy

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  1. Spontaneity, Entropyand Free Energy CH4(g) + O2(g) CO2(g) + H2O(g) + energy • How much energy • Energy flow • Final form of energy

  2. Spontaneous Processes and Entropy • First Law of Thermodynamics • “Energy can neither be created nor destroyed" • The energy of the universe is constant • Spontaneous Processes • Processes that occur without outside intervention • Spontaneous processes may be fast or slow • Many forms of combustion are fast • Conversion of diamond to graphite is slow

  3. Figure 16.2Thermodynamics domain

  4. Let’s Explain Spontaneity Book Falling Nail Rusting Gas filling

  5. Plant materials burn to form CO2 and H2O

  6. Formation of Rust on Bare Steel is a Spontaneous Process

  7. Entropy (S) • A measure of the randomness or disorder • The driving force for a spontaneous process is an increase in the entropy of the universe • Entropy is a thermodynamic function describing the number of arrangements that are available to a system • Nature proceeds toward the states that have the highest probabilities of existing

  8. Niranjan’s Bedroom Before Guests Next Day

  9. A Disordered Pile of Playing Cards

  10. ENTROPY • Is a thermodynamic function that describes • the number of arrangements (positions) that are • available to a system existing in a given state. • Entropy is closely related to probablility. • Nature spontaneously proceeds toward the states • that have the highest probabilities of existing.

  11. Figure 16.3 The expansion of an ideal gas into an evacuated bulb.

  12. Figure 16.4 Possible arrangements (states) of four molecules in a two-bulbed flask.

  13. Table 16.1 The Microstates That Give a Particular Arrangement (State)

  14. Positional Entropy • The probability of occurrence of a particular state depends on the number of ways (microstates) in which that arrangement can be achieved Ssolid < Sliquid << Sgas

  15. Example • Which has higher positional S? • Solid CO2 or gaseous CO2? • N2 gas at 1 atm or N2 gas at 0.01 atm? • Predict the sign of S • Solid sugar is added to water • + because larger volume • Iodine vapor condenses on cold surface to form crystals • - because gs, less volume

  16. Second Law of Thermodynamics • "In any spontaneous process there is always an increase in the entropy of the universe" • "The entropy of the universe is increasing" • For a given change to be spontaneous, Suniverse must be positive Suniv = Ssys + Ssurr

  17. The Effect of Temperature onSpontaneity H2O(l) H2O(g) Δssys = + Δssurr = ??? Exothermic or Endothermic process? Δssurr = - Δsuniv = Δssurr Δssys + Since the signs are in opposition… What controls the situation?

  18. The Effect of Temperature onSpontaneity H2O(l) H2O(g) Entropy changes in the surroundings are primarily determined by HEAT FLOW Remember…a system tends to undergoe changes that LOWER its ENERGY Δssurr = - ΔH T

  19. Think you got it???? Try this: In the metallurgy of antimony, the pure metal is recovered via different reactions depending on the composition of the ore. For example, iron is used to reduce antimony in sulfide ores: Sb2S3(s) + 3Fe(s) 2Sb(s) + 3FeS(s) ΔH = -125kJ Carbon is used as a reducing agent for oxide ores: Sb4O6(s) + 6C(s) 4Sb(s) + 6CO(g) ΔH = 778kJ Calculate Δssurr for each of these reactions at 25oC and 1 atm.

  20. Ssurr in terms of the Change in Enthalpy for a Process Occurring at Constant Pressure. H Ssurr = - T 419 J/K -2.61 x 103 J/K

  21. Table 16.3 Interplay of ΔSsys and ΔSsurr in Determining the Sign of ΔSuniv

  22. What about Chemical Changes? • We have been working with only physical changes so far… • Compare using # of independent states N2(g) + 3H2(g)  2NH3(g) Entropy decreases (-∆S) • Compare using # molecules in higher entropy states 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g) Entropy increases (+∆S)

  23. Example • Predict the sign of ∆S for the following reactions: • CaCO3(s)  CaO(s) + CO2(g) • Increasing S, +∆S • 2SO2(g) + O2(g)  2SO3(g) • Decreasing S, -∆S

  24. Calculating Entropy Change in a Reaction • Entropy is an extensive property (a function of the number of moles) • Generally, the more complex the molecule, the higher the standard entropy value

  25. Al2O3(s) + 3H2(g)  2Al(s) + 3H2O(g) Would you expect ∆S to be LARGE or small for the above reaction? Generally, the more complex the molecule, the higher the standard entropy value Answer: 179 J/K LARGE and POSITIVE! Why???

  26. Example • Find the ∆S° at 25°C for: 2NiS(s) + 3O2(g)  2SO2(g) + 2NiO(s)

  27. G What is this???? • A thermodynamic function (yes another one!) • called free energy or Gibbs free energy.

  28. Calculating Free Energy Method #1 G = H - TS G - H ΔS + = T = Δ + ΔS = ΔSuniv surr ΔSuniv =

  29. Calculating Free Energy Method #1 For reactions at constant temp & press: G0 = H0 - TS0 H2O(s)  H2O(l) where ∆H°=6.033x10 J/mol, ∆S°=22.1 J/mol.K Since reaction’s values are in opposition, the T determines spontaneity. At -10°C, is it spontaneous? 3

  30. Example NO! the reverse is spontaneous Is it spontaneous at 0°C? NO! it is at equilibrium so no shift

  31. Example Is it spontaneous at +10°C? YES! When H and S are in opposition, the spontaneity depends on T.

  32. Calculating Free Energy: Method #2 An adaptation of Hess's Law: Cdiamond(s) + O2(g) CO2(g) G0 = -397 kJ Cgraphite(s) + O2(g) CO2(g) G0 = -394 kJ Cdiamond(s) + O2(g) CO2(g) G0 = -397 kJ CO2(g) Cgraphite(s) + O2(g) G0 = +394 kJ Cdiamond(s)  Cgraphite(s) G0 = -3 kJ

  33. Calculating Free Energy Method #3 Using standard free energy of formation (Gf0): Gf0 of an element in its standard state is zero

  34. H, S, G and Spontaneity G = H - TS H is enthalpy, T is Kelvin temperature

  35. Standard Free Energy Change • G0 is the change in free energy that will occur if the reactants in their standard states are converted to the products in their standard states • G0 cannot be measured directly • The more negative the value for G0, the farther to the right the reaction will proceed in order to achieve equilibrium • Equilibrium is the lowest possible free energy position for a reaction

  36. A system at constant temperature and pressure will proceed spontaneously in the direction that _______????? Lowers it free energy Free energy is dependent on pressure. Why? How?

  37. The Dependence of Free Energy on Pressure • Enthalpy, H, is not pressure dependent • Entropy, S • entropy depends on volume, so it also depends on pressure Slarge volume > Ssmall volume Slow pressure > Shigh pressure

  38. Free Energy and Work • The maximum possible useful work obtainable from a process at constant temperature and pressure is equal to the change in free energy wmax = ΔG • The amount of work obtained is always less than the maximum • Henry Bent's First Two Laws of Thermodynamics • First law: You can't win, you can only break even • Second law: You can't break even

  39. Henry Bent

  40. The third law of thermodynamics

  41. The third law of thermodynamics The entropy of a perfect crystal at 0 K is zero.

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