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Thermodynamics

Production of quicklime. Thermodynamics. Liquid benzene. ⇅. Solid benzene. Chapter 19. CaCO 3 (s) ⇌ CaO + CO 2. S. order. disorder. S. q rev T.  S =. Entropy (S) - measure of randomness or disorder of a system.

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Thermodynamics

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  1. Production of quicklime Thermodynamics Liquid benzene ⇅ Solid benzene Chapter 19 CaCO3 (s) ⇌ CaO + CO2

  2. S order disorder S qrev T S = Entropy (S) - measure of randomness or disorder of a system State functions - properties that are determined by the state of the system, regardless of how that condition was achieved. DS = Sfinal - Sinitial For an isothermal process: at constant T

  3. First Law of Thermodynamics Energy can be converted from one form to another but energy cannot be created or destroyed. Second Law of Thermodynamics The entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process. DSuniv = DSsys + DSsurr≥ 0 Third Law of Thermodynamics The entropy of a pure crystalline substance at absolute zero is 0.

  4. Entropy and Physical States • Entropy increases with the freedom of motion of molecules S(g) >> S(l) > S(s) • Generally, when a solid is dissolved in a solvent, entropy increases. Fig 19.10 Dissolving an ionic solid in water

  5. Entropy Changes Fig 19.11 • In general, entropy increases when: • Gases are formed from liquids and solids • Liquids or solutions are formed from solids • Number of gas molecules increases • Number of moles increases

  6. Table 19.2 Standard Molar Entropies • Standard entropies tend to increase with increasing molar mass and complexity Fig 19.15 Molar entropies

  7. Entropy Changes • Entropy changes for a reaction can be estimated in a manner analogous to that by which H is estimated: S = nS(products) — mS(reactants) where n and m are the coefficients in the balanced chemical equation.

  8. Sample Exercise 19.5 Calculating ΔS from Tabulated Entropies Calculate ΔSº for the synthesis of ammonia from N2(g) and H2(g) at 298 K: N2(g) + 3 H2(g) → 2 NH3(g) ΔS° = 2S°(NH3) − [S°(N2) + 3S°(H2)] ΔS° = [(2 mol)(192.5 J/mol-K)] − [(1 mol)(191.5 J/mol-K) + (3 mol)(130.6 J/mol-K)] ΔS° = −198.3 J/K

  9. qsys T ΔHsys T Ssurr = Ssurr = Entropy Changes in Surroundings • Heat that flows into or out of the system changes the entropy of the surroundings. • For an isothermal process: • At constant pressure, qsys is simply H for the system.

  10. Spontaneous Physical and Chemical Processes Cannot use entropy as the sole criterion for spontaneous change!

  11. CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) DH° = −890.4 kJ H+ (aq) + OH- (aq) H2O (l) DH° = −56.2 kJ H2O (s) H2O (l) DH° = 6.01 kJ H2O NH4NO3 (s) NH4+(aq) + NO3- (aq) DH° = 25 kJ Does a decrease in enthalpy mean a reaction proceeds spontaneously? i.e., Are all spontaneous processes exothermic? Spontaneous reactions Conclusion:ΔH° is not a reliable indicator of spontaneity

  12. Entropy Change in the Universe • The universe is composed of the system and the surroundings. • Therefore, Suniverse = Ssystem + Ssurroundings • For spontaneous processes Suniverse > 0

  13. qsystem T Hsystem T Entropy Change in the Universe • Since Ssurroundings = and qsystem = Hsystem • This becomes: Suniverse = Ssystem + • Multiplying both sides by T, we get: TSuniverse = Hsystem  TSsystem

  14. TSuniverse = Hsystem  TSsystem • TΔSuniverse ≡ the Gibbs (free) energy, G • When Suniverse > 0, G < 0 • Therefore, when G < 0, a process is spontaneous. Gibbs (Free) Energy

  15. Gibbs Energy For a constant-pressure & constant temperature process: Gibbs energy (G) DG = DHsys - TDSsys DG < 0 The reaction is spontaneous in the forward direction DG > 0 The reaction is nonspontaneous as written. The reaction is spontaneous in the reverse direction DG = 0 The reaction is at equilibrium

  16. Fig 19.17 Potential Energy and Free Energy

  17. Fig 19.18 Free Energy and Equilibrium

  18. Standard free-energy of reaction (DGorxn) - free-energy change for a reaction when it occurs under standard-state conditions aA + bB cC + dD - [ + ] [ + ] = - mDGo (reactants) S S = f Standard free energy of formation(DGo) - the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states DGo DGo rxn rxn f DGo of any element in its stable form is zero f bDGo (B) dDGo (D) aDGo (A) cDGo (C) nDGo (products) f f f f f

  19. Sample Exercise 19.7 Calculating Standard Free-Energy Change from Free Energies of Formation (a) Use data from Appendix C to calculate the standard free-energy change for the following reaction at 298 K: P4(g) + 6 Cl2(g) → 4 PCl3(g) (b) What is ΔG° for the reverse of the above reaction? 4 PCl3(g) → P4(g) + 6 Cl2(g) ΔG° = +1102.8 kJ

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