Thermodynamics Chapter 19 Liquid benzene Production of quicklime Solid benzene ⇅ CaCO 3 (s) ⇌ CaO + CO 2

Entropy (S) - measure of randomness or disorder of a system orderS disorder S  S = S final - S initial For an isothermal process:  S = q rev T State functions - properties that are determined by the state of the system, regardless of how that condition was achieved. at constant T

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.  S univ =  S sys +  S surr ≥ 0 Third Law of Thermodynamics The entropy of a pure crystalline substance at absolute zero is 0.

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 Dissolving an ionic solid in water

 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 Fig 19.11

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

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.

Sample Exercise 19.5 Calculating ΔS from Tabulated Entropies Calculate ΔSº for the synthesis of ammonia from N 2 (g) and H 2 (g) at 298 K: N 2 (g) + 3 H 2 (g) → 2 NH 3 (g) ΔS° = 2S°(NH 3 ) − [S°(N 2 ) + 3S°(H 2 )] 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

Entropy Changes in Surroundings Heat that flows into or out of the system changes the entropy of the surroundings. For an isothermal process:  S surr =  q sys T At constant pressure, q sys is simply  H  for the system.  S surr =  ΔH sys T

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

Does a decrease in enthalpy mean a reaction proceeds spontaneously? i.e., Are all spontaneous processes exothermic? CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (l)  H° = −890.4 kJ H + (aq) + OH - (aq) H 2 O (l)  H° = −56.2 kJ H 2 O (s) H 2 O (l)  H° = 6.01 kJ NH 4 NO 3 (s) NH 4 + (aq) + NO 3 - (aq)  H° = 25 kJ H2OH2O Spontaneous reactions Conclusion: ΔH° is not a reliable indicator of spontaneity

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

Entropy Change in the Universe Since  S surroundings = and q system =  H system This becomes:  S universe =  S system + Multiplying both sides by  T, we get:  T  S universe =  H system  T  S system  H system T  q system T

Gibbs (Free) Energy  T  S universe =  H system  T  S system  TΔS universe ≡ the Gibbs (free) energy,  G When  S universe > 0,  G < 0 Therefore, when  G < 0, a process is spontaneous.

Gibbs Energy For a constant-pressure & constant temperature process:  G =  H sys - T  S sys Gibbs energy (G)  G < 0 The reaction is spontaneous in the forward direction  G > 0 The reaction is nonspontaneous as written. The reaction is spontaneous in the reverse direction  G = 0 The reaction is at equilibrium

Fig Potential Energy and Free Energy

Fig Free Energy and Equilibrium

aA + bB cC + dD GoGo rxn d  G o (D) f c  G o (C) f = [+] - b  G o (B) f a  G o (A) f [+] GoGo rxn n  G o (products) f =  m  G o (reactants) f  - Standard free-energy of reaction (  G o rxn ) - free-energy change for a reaction when it occurs under standard-state conditions Standard free energy of formation (  G o ) - the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states f  G o of any element in its stable form is zero f

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: P 4 (g) + 6 Cl 2 (g) → 4 PCl 3 (g) (b) What is ΔG° for the reverse of the above reaction? 4 PCl 3 (g) → P 4 (g) + 6 Cl 2 (g) ΔG° = kJ