1. Criteria for Spontaneity isolated dS > 0 Closed dS ≥ dq/T or dq – TdS ≤ 0 Cst V,T: dU – TdS ≤ 0 let A = U - TS dA ≤ 0 Cst P,T: dH – TdS ≤ 0 let G = H - TS dG ≤ 0

2. Free Energy A = U - TS Helmholtz Free Energy A is minimized for any irrev process in a closed system at cst V & T. dA = dU – TdS = dq + dw – TdS if rev then dw = dw max & dS = dq/T dA = dw max A called ‘work’ function It is equivalent to the maximum work a system can perform

3. Free Energy G = H - TS Gibbs Free Energy G is minimized for any irrev process in a closed system at cst P & T. dG = dH – TdS = dU + PdV + V dP – TdS cst P & 1 st Law dq + dw + PdV – TdS if rev then dw = dw rev & dS = dq/T dG = dw rev – dw exp = dw nonexp dG is often referred to as the maximum ‘useful’ work dG = dH – TdS – SdT Because absolute S is never known (3 rd Law ignores nuclear spin contributions and isotopic mixing at T = 0 K) the change in G with change is T is undefined (like dividing by 0 in math). Therefore the last term in this equation is simply ignored.

4. Measuring  G 2.  Isothermal processes:  G =  H – T  S 1.Heating and Cooling:  G is undefined for temperature changes.  G can only be determined for constant T processes.  However, the change in  G for a process with change in T can be determined.  (more on this later) 3.  Phase changes at nmp or nbp:  G =  H – T  S sub  S tr =  H tr /T   G =  H tr – T  H tr /T = 0  At the nmp or nbp two phases are in equilibrium and  G tr = 0

5. What is  G for conversion of supercooled water at -10˚C to ice at -10˚C? (n = 1) -10˚C 0˚C  H = 1  75.4  10 + 1  -6007 + 1  38.07  -10 = -6312  S = 75.4  ln(273/263) – 6007/273 + 38.07  ln(263/273) = -20.61  G = -6312 – 263(-20.61) = -892 J You can’t measure  G for constant P heating/cooling!  S =  T1 T2 C P dT/T  S =  H tr /T  S = ∫ dq rev /T

6. Measuring  G 4. Chemical Reactions:  G 298  =  i i  Gº f,m,i  (  G 298 can be determined from thermodynamic table data) or….  G 298 =  H 298 – 298  S 298 5. Chemical Reactions (not at 298) :  Find  H T and  S T using  C P values in thermodynamic tables  Then….  G T =  H T – T  S T  Hº T =  H o 298 +   C P º dT  Sº T =  S o 298 +   C P º dT/T

7. Free Energy of Mixing IG’s  G =  H – T  S = 0 - T  S = -T  S  S = -n a R ln  a - n b R ln  b  S = n a R ln(V/V a ) + n b R ln(V/V b )  G = -n a RT ln(V/V a ) - n b RT ln(V/V b )  G = n a RTln  a + n b RTln  b

8. Compound/ element  H˚ f kJ mol -1  G˚ f kJ mol -1 S˚ J mol -1 k -1 C P ˚ J mol -1 k -1 H 2 O (l) -285.83-237.12969.9175.21 H 2(g) 00130.68428.824 O 2(g) 00205.13829.355 H 2(g) + ½ O 2(g) → H 2 O (l)  S˚ f = 0.06991 -.130684 – ½ 0.205138 = -0.163343  G˚ f = -285.83 – 298(-0.163343) = -237.15  G˚ f  -285.83 – 298(0.06991) = -306.66  f Spontaneity in chemical systems at cst T & P involves a balance between the natural tendency to minimize enthalpy and to maximize entropy. Increasing T gives entropy a higher stake in that balance. This is why solids melt, proteins unfold, and DNA unwinds at higher values of T.

9. (if closed, rev & w exp ) dU = dq rev - PdV dq rev /T = dS & dq rev = TdS dU = TdS - PdV dU = dq + dw Since U is a state function, this also applies to irreversible processes. Note that in this expression U is a function of S and V or U(S, V). S and V are called the natural variables of U. In theory, if we know how U varies with S and V, then we can determine the other thermodynamic variables of the system.

10. Hydrophobic Interactions C 3 H 8(aq)  C 3 H 8(l)  H  298 ~ +8 kJ mol -1  S  298 ~ +80 J mol -1 Hydrophobic effects are due to the entropy increase of the solvent which must be more ordered when it is in the presence of nonpolar solutes compared to when it is surrounded by other water molecules. 1.Is this process spontaneous? A = yes, b = no 2.What is the sign of  G? For this process? 3.Does entropy increase during this process? (more disorder) 4.Does enthalpy decrease during this process? (stronger bonding) Yes No  G  298 = 8 - 298  0.080 = -16 kJ mol -1 Yes (-)

11. 1. dU = dq + dw (closed sys, exp work only) 2. H = U + PV 3. A = U - TS 4. G = H - TS 5. C V = (dU/dT) V = T(dS/dT) V 6. C P = (dH/dT) P = T(dS/dT) P 1. dU = TdS - PdV (derived from 1 st Law) 2.dH = 3.dA = 4.dG = Gibbs Equations TdS + VdP -SdT - PdV -SdT + VdP

12. 1. dU = TdS - PdV U (S,V) natural variables 2. dH = TdS + VdP H (S,P) 3. dA = -SdT – PdV A (T,V) 4. dG = -SdT + VdP G (T,P) e.g. dG = (dG/dT) P dT + (dG/dP) T dP (dG/dT) P = -S find (dG/dP) T = (dG/dP) T = V (dU/dS) V = (dU/dV) S = (dH/dS) P = (dH/dP) S = (dA/dT) V = (dA/dV) T = T -P T V -S -P

13. 1. dU = TdS - PdV (derive from 1st) 2. dH = TdS + VdP 3. dA = -SdT - PdV 4. dG = -SdT + VdP (dT/dP) S = (dV/dS) P (dS/dV) T = (dP/dT) V (dS/dP) T = -(dV/dT) P Maxwell Relations Euler Reciprocity Relations Given z = z (x,y) dz = (dz/dx) Y dx + (dz/dy) x dy let M = (dz/dx) Y & N = (dz/dy) x then..... (dM/dy) x = (dN/dx) y dU = TdS – PdV ……. dH: dA: dG: (dT/dV) S = -(dP/dS) V

14. dU = TdS - PdV (derive from 1st) dH = TdS + VdP dA = -SdT – PdV dG = -SdT + VdP C V = (dU/dT) V = T(dS/dT) V (since dU = dq V ) C P = (dH/dT) P = T(dS/dT) P (since dH = dq P )  = 1/V (dV/dT) P  = -1/V (dV/dP) T  = (dP/dT) V Starting Points

15. (dH/dT) P = C P (dU/dT) V = C V (dS/dT) P = C P /T (dS/dT) V = C V /T some of the relationships are fairly easy.... (dS/dT) P (dS/dP) T (dS/dT) V (dS/dV) T (dH/dT) P (dH/dP) T (dG/dT) P (dG/dP) T (dU/dV) T (dU/dT) V (dA/dT) V (dA/dV) T desired relationships.......... C V = (dU/dT) V = T(dS/dT) V & C P = (dH/dT) P = T(dS/dT) P  = 1/V (dV/dT) P  = -1/V (dV/dP) T  = (dP/dT) V Starting PointsdU = TdS - PdV dH = TdS + VdP dA = -SdT – PdV dG = -SdT + VdP

16. C V = (dU/dT) V = T(dS/dT) V & C P = (dH/dT) P = T(dS/dT) P  = 1/V (dV/dT) P  = -1/V (dV/dP) T  = (dP/dT) V Starting PointsdU = TdS - PdV dH = TdS + VdP dA = -SdT – PdV dG = -SdT + VdP (dS/dT) P (dS/dP) T (dS/dT) V (dS/dV) T (dH/dT) P (dH/dP) T (dG/dT) P (dG/dP) T (dU/dV) T (dU/dT) V (dA/dT) V (dA/dV) T desired relationships.......... dH = TdS + VdP ÷ by dP at cst T (dH/dP) T = T(dS/dP) T + V from Maxwell (dG)... (dS/dP) T = -(dV/dT) P = - V ....... (dH/dP) T = -TV  + V Check Maxwell relations no result ….

17. (dH/dP) T = -TV  + V Water at 303 Kn = 1 density = 0.995 g/ml = 995 kg/m 3.  = 3.04 x 10 -4 K -1. What is  H if the pressure is increased by 100 atm. (1 atm. = 101325 Pa)  H = (-303 0.018/995 3.04 x 10-4 + 0.018/995) 100 101325 = 166 J Squeezing the liquid does not strengthen bonding What is  H if the pressure is decreased from 1 atm to 0.1 atm?  H = (-303 0.018/995 3.04 x 10-4 + 0.018/995) -0.9 101325 = -1.5 J

18. 1. dU = TdS - PdV (derive from 1st) 2. dH = TdS + VdP 3. dA = -SdT - PdV 4. dG = -SdT + VdP (dT/dV) S = -(dP/dS) V (dT/dP) S = (dV/dS) P (dS/dV) T = (dP/dT) V (dS/dP) T = -(dV/dT) P Maxwell Relations (dG/dT) P = -S (dG/dP) T = V (dU/dS) V = T (dU/dV) S = -P (dH/dS) P = T (dH/dP) S = V (dA/dT) V = -S (dA/dV) T = -P C V = (dU/dT) V = T(dS/dT) V & C P = (dH/dT) P = T(dS/dT) P  = 1/V (dV/dT) P  = -1/V (dV/dP) T  = (dP/dT) V

19. dU = TdS - PdV  dV T (dU/dV) T = T(dS/dV) T - P apply Euler/Maxwell from dA = -SdT - PdV Internal P = (dU/dV) T for an IG show that (dU/dV) T = 0 (dU/dV) T = T(dP/dT) V – P gas (dU/dV) T = T(  ) - P sol/liq  = 1/V (dV/dT) P  = -1/V (dV/dP) T  = (dP/dT) V For water at 298 K Internal P = 298 2.07 x 10 -4 /4.57 x 10 -10 – 101325 = 1.35 x 10 8 Pa or 1330 atm

20. The effect of T on Gibbs energy (dG/dT) P = -S or (d  G/dT) P = -  S {d(G/T)/dT} P = -H/T 2 or {d(  G/T)/dT} P = -  H/T 2 {d(G/T)/d(1/T)} P = H or {d(  G/T)/d(1/T)} P =  H d(G/T)/dT = G d(1/T)/dT + 1/T dG/dT = -G/T 2 – S/T = -(G + TS)/T 2 = -(H – TS + TS)/T 2 = -H/T 2 d(G/T) = -H dT/T 2 & dT/T 2 = d∫ dT/T 2 = -d(1/T) d(G/T) = H d(1/T)

21. Compound HHGG SS Cp  kJ mol -1 J mol -1 K -1 CO (g)-110.525-137.168197.729.116 CO 2 (g)-393.509-394.359213.737.11 O 2 (g)00205.129.355 {d(  G/T)/d(1/T)} P =  H What is  G 400 for the reaction: CO (g) + ½O 2(g) → CO 2(g)  H 400 =  H 298 +  C P  T -284 = -283 - 0.0067 (202) =  S 400 =  S 298 +  C P ln (T 2 /T 1 ) -88.5 = -86.5 - 6.8 (0.2944) =  G 400 =  H 400 - T  S 400 -248.6 = -284 - 400 (-0.0885)  G 298 = -257.2 kJ (table value) -257.2 kJ (  H 298 – T  S 298 )

22. Compound HHGG SS Cp  kJ mol -1 J mol -1 K -1 CO (g)-110.525-137.168197.67429.116 CO 2 (g)-393.509-394.359213.7437.11 O 2 (g)00205.13829.355 {d(  G/T)/d(1/T)} P =  H What is  G 400 for the reaction: CO (g) + ½O 2(g) → CO 2(g)  G/400 - (-257.2/298) = -283.0 (1/400 – 1/298)  G/400 + 0.8631 = 0.2422  G 400 = -248.4 kJ mol -1 At what T is the reaction at equilibrium?  - (-257.2/298) = -283.0 (1/T – 1/298) and T = 3244 K  G T ~  H - T  S 0 ~ -284 - T (-0.0865) & T = 3283K assumes  H and  S are constant

23. The effect of P on Gibbs Free Energy (dG/dP) T  G = G P2 – G P1 ~ V (P 2 – P 1 ) solids/liquids (since V is not affected very much by P)  G = G P2 – G P1 = nRT/P dP = nRT ln (P f /P i ) IG dG = -S dT + V dP (dG/dP) T = V dG = V dP  G = ∫ V dP Applied to process: e.g. phase change or chemical reaction  G) =  G P2 –  G P1 ~  V (P 2 – P 1 ) for a reaction involving gases you can assume  V = is entirely due to gas volumes.

24. Compound Density g/cm 3 Molar volume HHGG SS Cp  kJ mol -1 J mol -1 K -1 C(graphite)2.2675.29 x 10 -6 005.748.527 C(diamond)3.5153.41 x 10 -6 1.8972.92.3776.115 Estimate the Pressure at which graphite will change into diamond? d  G =  V m (P – P°) & 0 – 2900 = ( 3.41 x 10 -6 - 5.29 x 10 -6 ) (P – 101325) P = 1523 atm  G) =  G P2 –  G P1 ~  V (P 2 – P 1 ) Why is graphite more stable than diamond at 1 atm. P? What does the ° mean in  G°? What is  G at P ≠ P° for C dia → C gr ? Look at process qualitatively using sign change for ↑ P + + (  G P -  G P° ) =  V ( P - P°) What is  G° for C dia → C gr ? +  G P is smaller (-), 0 or + value Process is less likely at higher P

25. Compound HHGG SS Cp  kJ mol -1 J mol -1 K -1 CO (g)-110.525-137.168197.67429.116 CO 2 (g)-393.509-394.359213.7437.11 O 2 (g)00205.13829.355  G) =  G P2 –  G P1 ~  V (P 2 – P 1 ) CO (g) + ½O 2(g) → CO 2(g) Will the reaction become more or less favorable at higher pressures? Will the reaction become more or less favorable at higher T? Increasing T favors reactants …… Increasing P favors products. Putting the same amount of starting material in smaller flask will influence the equilibrium!

26. dG = -SdT + VdP This Gibb’s equation assumes that there is no change in the amount of substance in the closed system. However, if you open the lid and throw in some substance, G will change. It will also change if there is a chemical reaction or phase changes that how much of a particular substance is present. This concept is handled by expanding the Gibb’s equation above to …… dG = -S dT + V dP +  i (dG/dn i ) T,P,n´ Chemical potential (  i ) = (dG/dn i ) T,P,n´ G m (P) = Gº m + RT ln (P/Pº) ideal gas or …  i (P) =  i (P˚) + RT ln(P i /P˚)

27. Chemical Potential (  ) Is G an extensive or an intensive property? More ‘stuff’ = more free energy G m (the molar free energy) is an intensive property. G nini Chemical potential (  i ) = (dG/dn i ) T,P,n ´ For a pure substance (  i ) = G m.

28. However  i depends on other molecules in system. Just as the volume of ethanol changes when mixed with water … the free energy of one mole of a substance has a different impact in a mixture Chemical potential (  i ) = (dG/dn i ) T,P,n ´ For a mixture substance (  i ) ≠ G m, and  i changes with added moles of i. G nini

29. Chemical Potential (  ) How does the chemical potential of a pure gas change with pressure? G m (P) - Gº m = RT ln (P/Pº) 1 mole of ideal gas  i (P) =  i (P˚) + RT ln(P i /P˚) However, since an ideal gas does not interacts with other components in a mixture the chemical potential of that gas also applies to partial pressures in an ideal gas mixture.

30. What can cause a change in n i ? In closed system (no added material). dG = -SdT + VdP +  i (dG/dn i ) T,P,n´ dn i 1) Chemical Reactions … Chapter 5 2) Phase changes … Chapter 6 dG = -SdT + VdP +  i  i dn i The natural tendency for systems is to achieve their lowest potential energy in a given force field. In a gravitational field water achieves this by flowing downhill and pooling in valleys. In a chemical system, molecules do this by reacting or changing phases to minimize G (at constant T and P). Thus Free energy is the potential energy of chemical systems, and thus the name chemical potential. Molecules will spontaneously ‘flow’ from higher to lower .