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Chemical Thermodynamics

Chemical Thermodynamics. BLB 12 th Chapter 19. Chemical Reactions. Will the reaction occur, i.e. is it spontaneous? Ch. 5, 19 How fast will the reaction occur? Ch. 14 How far will the reaction proceed? Ch. 15. energy heat work pathway state function system surroundings. Review terms.

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Chemical Thermodynamics

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  1. Chemical Thermodynamics BLB 12th Chapter 19

  2. Chemical Reactions • Will the reaction occur, i.e. is it spontaneous? Ch. 5, 19 • How fast will the reaction occur? Ch. 14 • How far will the reaction proceed? Ch. 15

  3. energy heat work pathway state function system surroundings Review terms

  4. exothermic endothermic enthalpy enthalpy change standard state std. enthalpy of formation 1st Law of Thermodynamics Review terms, cont.

  5. 19.1 Spontaneous Processes • Spontaneous – proceeds on its own without any outside assistance • product-favored; K > 1 • not necessarily fast • the direction a process will take if left alone and given enough time. • Nonspontaneous • opposite direction of spontaneous • reactant-favored; K < 1 • not necessarily slow

  6. Spontaneity and Energy • Examples of spontaneous systems: • Brick falling • Ball rolling downhill • Hot objects cooling • Combustion reactions • Are all spontaneous processes accompanied by a loss of heat, that is, exothermic? • Spontaneity is temperature-dependent.

  7. Reversible & Irreversible Systems • Reversible – a change in a system for which the system can restored by exactly reversing the change – a system at equilibrium ex. melting ice at 0°C • Irreversible – a process that cannot be reversed to restore the system and surroundings to their original states – a spontaneous process ex. melting ice at 25°C • See p. 788-790 (last paragraph of section)

  8. 19.2 Entropy and the 2nd Law of Thermodynamics • Entropy, S – measure of randomness • State function • Temperature-dependent • A random (or dispersed) system is favored due to probability. “Entropy Is Simple – If We Avoid the Briar Patches” Frank Lambert, Occidental College, ret. http://entropysimple.oxy.edu

  9. Entropy Change • ΔS = Sfinal − Sinitial (a state function) (isothermal) as for phase changes. • ΔS > 0 is favorable

  10. Calculating ΔS for Changes of State

  11. Problem 26 The freezing point of Ga is 29.8°C and the enthalpy of fusion is 5.59 kJ/mol. • Is ΔS + or − for Ga(l) → Ga(s) at the freezing point? • Calculate the value of ΔS when 60.0 g of Ga(l) solidifies.

  12. System & Surroundings Dividing the universe: • System – dispersal of matter by reaction: reactants → products • Surroundings – dispersal of energy as heat

  13. 2nd Law of Thermodynamics • The entropy of the universe increases for any spontaneous process. • ΔSuniv > 0 • ΔSuniv = ΔSsys + ΔSsurr • For a “reversible” process: ΔSuniv = 0. • For an irreversible process: • Net entropy increase ►spontaneous • Net entropy decrease ► nonspontaneous

  14. 19.3 The Molecular Interpretation of Entropy • Molecules have degrees of freedom based upon their motion • Translational • Vibrational • Rotational • Motion of water (Figure 19.8, p. 796) • Lowering the temperature decreases the entropy.

  15. Boltzman & Microstates • S = klnW (W = # of microstates) • If # microstates ↑, then entropy ↑. • Increasing volume, temperature, # of molecules increases the # of microstates.

  16. Examples of systems that have increased entropy Entropy increases for: • Changes of state: solid → liquid → gas (T) • Expansion of a gas (V) • Dissolution: solid → solution (V) • Production of more moles in a chemical reaction (# of particles) • Ionic solids: lower ionic charge S° (J/mol·K) Na2CO3 136 MgCO3 66

  17. Changes of State

  18. # microstates ↑, system entropy ↑

  19. Dissolution

  20. Expansion of a Gas

  21. 2 NO(g) + O2(g) → 2 NO2(g) S° + or −?

  22. Problem 22 TNT (trinitrotoluene) Detonation 4 C3H5N3O9(l) → 6 N2(g) + 12 CO2(g) + 10 H2O(g) + O2(g) Spontaneous? Sign of q? Can the sign of w be determined? Can the sign of ΔE be determined?

  23. 3rd Law of Thermodynamics • The entropy, S, of a pure crystalline substance at absolute zero (0 K) is zero.

  24. 19.4 Entropy Changes in Chemical Reactions Standard molar entropy values, S° (J/mol·K): • increase in value as temperature increases from 0 K • have been determined for common substances (Appendix C, pp. 1059-1061) • increase with molar mass • increase with # of atoms in molecule

  25. Calculating ΔS°sys ΔS°sys = ∑nS°(products) - ∑mS°(reactants) (where n and m are coefficients in the chemical equation)

  26. 2 NO(g) + O2(g) → 2 NO2(g) S° =

  27. Problem 54(c)

  28. Problem

  29. Entropy Changes in the Surroundings • Heat flow affects surroundings. • As T increases, ΔH becomes less important. • As T decreases, ΔH becomes more important.

  30. Calculating ΔS°univ • ΔSuniv = ΔSsys + ΔSsurr by obtaining ΔSsys and ΔSsurr • If ΔSuniv> 0, the reaction is spontaneous. But there is a better way – one in which only the system is involved.

  31. 19.5 Gibbs Free Energy • The spontaneity of a reaction involves both enthalpy (energy) and entropy (matter). • Gibbs Free Energy, ΔG makes use of ΔHsys and ΔSsys to predict spontaneity. • ΔGsys represents the total energy change for a system. • G = H – TS or ΔG = ΔH – TΔS • or, under standard conditions: ΔG° = ΔH° – TΔS°

  32. Gibbs Free Energy • If: • ΔG < 0, forward reaction is spontaneous • ΔG = 0, reaction is at equilibrium • ΔG > 0, forward reaction is nonspontaneous • In any spontaneous process at constant temperature and pressure, the free energy always decreases. • ΔG is a state function. • ΔGf° of elements in their standard state is zero.

  33. Calculating ΔG°sys ΔG°sys = ΔH°sys − TΔS°sys or ΔG°sys = ∑nΔG°f(products) - ∑mΔG°f(reactants) (where n and m are coefficients in the chemical equation)

  34. 19.6 Free Energy and Temperature

  35. Problem 60(b) & 83

  36. Driving force of a reaction For a reaction where ΔG < 0: • Enthalpy-driven – if ΔH < 0 and ΔS < 0; at low temp. • Entropy-driven – if ΔH > 0 and ΔS > 0; at high temp. • “cross-over point” is where ΔG = 0

  37. Problem

  38. 19.7 Free Energy and K • If conditions are non-standard: ΔG = ΔG° + RT lnQR = 8.3145 J/mol·K • If at equilibrium: ΔG = ΔG° + RT lnQ = 0 ΔG° = −RT lnK

  39. Problem

  40. Problem 81

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