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

Free Energy

A = U - TS Helmholtz Free Energy

A is minimized for any irrev processin a closed system at cst V & T.

dA = dU – TdS = dq + dw – TdS if rev then dw = dwmax & dS = dq/T

dA = dwmax

A called ‘work’ functionIt is equivalent to the maximum work a system can perform

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 & 1st Law

dq + dw + PdV – TdS if rev then dw = dwrev & dS = dq/T

dG = dwrev – dwexp = dwnonexp

dG is often referred to as the maximum ‘useful’ work

dG = dH – TdS – SdTBecause absolute S is never known (3rd 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.

Measuring DG

2. Isothermal processes: DG = DH – T DS

1. Heating and Cooling: DG is undefined for temperature changes. DG can only be determined for constant T processes. However, the change in DG for a process with change in T can be determined. (more on this later)

3. Phase changes at nmp or nbp: DG = DH – T DS sub DStr = DHtr/T DG = DHtr – T DHtr/T = 0 At the nmp or nbp two phases are in equilibrium and DGtr = 0

What is DG for conversion of supercooled water at -10˚C to ice at -10˚C? (n = 1)

-10˚C

-10˚C

0˚C

0˚C

DH = 1 75.4 10 + 1 -6007 + 1 38.07 -10 = -6312

DS = 75.4 ln(273/263) – 6007/273 + 38.07 ln(263/273) = -20.61

DG = -6312 – 263(-20.61) = -892 J

You can’t measure DG for constant P heating/cooling!

DS = T1T2 CP dT/T

DS = DHtr/T

DS = ∫ dqrev/T

Measuring DG

4. Chemical Reactions: DG298 = Si ni DGºf,m,i

(DG298 can be determined from thermodynamic table data) or….

DG298 = DH298 – 298 • DS298

5. Chemical Reactions (not at 298): Find DHT and DST using DCP values in thermodynamic tables Then…. DGT = DHT – TDST

DHºT = DHo298 + DCPº dT DSºT = DSo

298 + DCPº dT/T

Free Energy of Mixing IG’sDG = DH – TDS = 0 - TDS = -TDS

DS = -naR lnca - nbR lncb

DS = naR ln(V/Va) + nbR ln(V/Vb)

DG = -naRT ln(V/Va) - nbRT ln(V/Vb)

DG = naRTlnca + nbRTlncb

Compound/element

DH˚fkJ mol-1

DG˚fkJ mol-1

S˚J mol-1 k-1

CP˚J mol-1 k-1

H2O(l) -285.83 -237.129 69.91 75.21

H2(g) 0 0 130.684 28.824

O2(g) 0 0 205.138 29.355

H2(g) + ½ O2(g) → H2O(l)

DS˚f = 0.06991 - .130684 – ½ 0.205138 = -0.163343

DG˚f = -285.83 – 298(-0.163343) = -237.15

DG˚f -285.83 – 298(0.06991) = -306.66

D

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.

(if closed, rev & wexp)

dU = dqrev - PdV dqrev/T = dS & dqrev = 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.

Hydrophobic Interactions

C3H8(aq) C3H8(l)

DH298 ~ +8 kJ mol-1 DS298 ~ +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 = no2. What is the sign of DG? For this process? 3. Does entropy increase during this process? (more disorder)4. Does enthalpy decrease during this process? (stronger bonding)

YesNo

DG298 = 8 - 298 0.080 = -16 kJ mol-1

Yes (-)

1. dU = dq + dw (closed sys, exp work only)

2. H = U + PV3. A = U - TS4. G = H - TS5. CV = (dU/dT)V = T(dS/dT)V 6. CP = (dH/dT)P = T(dS/dT)P

1. dU = TdS - PdV (derived from 1st Law)

2. dH = 3. dA = 4. dG =

Gibbs Equations

TdS + VdP-SdT - PdV-SdT + VdP

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

1. dU = TdS - PdV (derive from 1st)2. dH = TdS + VdP3. dA = -SdT - PdV4. 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 dylet 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

dU = TdS - PdV (derive from 1st)dH = TdS + VdPdA = -SdT – PdVdG = -SdT + VdPCV = (dU/dT)V = T(dS/dT)V (since dU = dqV)

CP = (dH/dT)P = T(dS/dT)P (since dH = dqP)

a = 1/V (dV/dT)P k = -1/V (dV/dP)T

/a k = (dP/dT)V

Starting Points

(dH/dT)P = CP

(dU/dT)V = CV

(dS/dT)P = CP/T

(dS/dT)V = CV/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 ..........

CV = (dU/dT)V = T(dS/dT)V & CP = (dH/dT)P = T(dS/dT)P

a = 1/V (dV/dT)P k = -1/V (dV/dP)T /a k = (dP/dT)V

Starting Points dU = TdS - PdVdH = TdS

+ VdPdA = -SdT

– PdVdG = -SdT

+ VdP

CV = (dU/dT)V = T(dS/dT)V & CP = (dH/dT)P = T(dS/dT)P

a = 1/V (dV/dT)P k = -1/V (dV/dP)T /a k = (dP/dT)V

Starting Points dU = TdS - PdVdH = TdS

+ VdPdA = -SdT

– PdVdG = -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 = - Va .......

(dH/dP)T = -TVa + V

Check Maxwell relations no result ….

(dH/dP)T = -TVa + V

Water at 303 K n = 1 density = 0.995 g/ml = 995 kg/m3.a = 3.04 x 10-4 K-1.

What is DH if the pressure is increased by 100 atm. (1 atm. = 101325 Pa)

DH = (-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 DH if the pressure is decreased from 1 atm to 0.1 atm?

DH = (-303 • 0.018/995 • 3.04 x 10-4 + 0.018/995) • -0.9 • 101325 =

-1.5 J

1. dU = TdS - PdV (derive from 1st)2. dH = TdS + VdP3. dA = -SdT - PdV4. 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

CV = (dU/dT)V = T(dS/dT)V & CP = (dH/dT)P = T(dS/dT)P

a = 1/V (dV/dT)P k = -1/V (dV/dP)T /a k = (dP/dT)V

dU = TdS - PdV dVT

(dU/dV)T = T(dS/dV)T - Papply 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( /a k) - P sol/liq

a = 1/V (dV/dT)P k = -1/V (dV/dP)T /a k = (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 108 Pa or 1330 atm

The effect of T on Gibbs energy(dG/dT)P = -S or (dDG/dT)P = -DS

{d(G/T)/dT}P = -H/T2 or {d(DG/T)/dT}P = -DH/T2

{d(G/T)/d(1/T)}P = H or {d(DG/T)/d(1/T)}P = DH

d(G/T)/dT = G d(1/T)/dT + 1/T dG/dT = -G/T2 – S/T

= -(G + TS)/T2

= -(H – TS + TS)/T2 = -H/T2

d(G/T) = -H dT/T2 & dT/T2 = d∫ dT/T2 = -d(1/T) d(G/T) = H d(1/T)

Compound

DH° DG° S° Cp°

kJ mol-1 kJ mol-1 J mol-1 K-1 J mol-1 K-1

CO (g) -110.525 -137.168 197.7 29.116

CO2 (g) -393.509 -394.359 213.7 37.11

O2 (g) 0 0 205.1 29.355

{d(DG/T)/d(1/T)}P = DH

What is DG400 for the reaction: CO(g) + ½O2(g) → CO2(g)

DH400 = DH298 + DCP • DT -284 = -283 - 0.0067 (202) =

DS400 = DS298 + DCP • ln (T2/T1)

-88.5 = -86.5 - 6.8 (0.2944) =

DG400 = DH400 - TDS400

-248.6 = -284 - 400 (-0.0885)

DG298 = -257.2 kJ (table value) -257.2 kJ (DH298 – TDS298)

Compound

DH° DG° S° Cp°

kJ mol-1 kJ mol-1 J mol-1 K-1 J mol-1 K-1

CO (g) -110.525 -137.168 197.674 29.116

CO2 (g) -393.509 -394.359 213.74 37.11

O2 (g) 0 0 205.138 29.355

{d(DG/T)/d(1/T)}P = DH

What is DG400 for the reaction: CO(g) + ½O2(g) → CO2(g)

DG/400 - (-257.2/298) = -283.0 (1/400 – 1/298)

DG/400 + 0.8631 = 0.2422 DG400 = -248.4 kJ mol-1

At what T is the reaction at equilibrium?

0 - (-257.2/298) = -283.0 (1/T – 1/298) and T = 3244 K

DGT ~ DH - TDS

0 ~ -284 - T (-0.0865) & T = 3283K assumes DH and DS are constant

The effect of P on Gibbs Free Energy (dG/dP)T

DG = GP2 – GP1 ~ V (P2 – P1) solids/liquids (since V is not affected very much by P)

DG = GP2 – GP1 = nRT/P dP = nRT ln (Pf/Pi) IG

dG = -S dT + V dP (dG/dP)T = V

dG = V dP DG = ∫ V dP

Applied to process: e.g. phase change or chemical reaction

(D DG) = DGP2 – DGP1 ~ DV (P2 – P1) for a reaction involving gases you can assume DV = is entirely due to gas volumes.

CompoundDensity g/cm3 Molar volume

DH° DG° S° Cp°

kJ mol-1 kJ mol-1 J mol-1 K-1 J mol-1 K-1

C(graphite) 2.267 5.29 x 10 -6 0 0 5.74 8.527C(diamond) 3.515 3.41 x 10 -6 1.897 2.9 2.377 6.115

Estimate the Pressure at which graphite will change into diamond?

dDG = DVm (P – P°) & 0 – 2900 = (3.41 x 10 -6 - 5.29 x 10 -6) (P – 101325)

P = 1523 atm

(D DG) = DGP2 – DGP1 ~ DV (P2 – P1)Why is graphite more stable than diamond at 1 atm. P?

What does the ° mean in DG°?

What is DG at P ≠ P° for Cdia → Cgr? Look at process qualitatively using sign change for ↑ P + +

(DGP - DGP°) = DV ( P - P°)

What is DG° for Cdia → Cgr?

+ DGP is smaller (-), 0 or + value Process is less likely at higher P

Compound

DH° DG° S° Cp°

kJ mol-1 kJ mol-1 J mol-1 K-1 J mol-1 K-1

CO (g) -110.525 -137.168 197.674 29.116

CO2 (g) -393.509 -394.359 213.74 37.11

O2 (g) 0 0 205.138 29.355

(D DG) = DGP2 – DGP1 ~ DV (P2 – P1)

CO(g) + ½O2(g) → CO2(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!

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 + Si (dG/dni)T,P,n´

Chemical potential (mi) = (dG/dni)T,P,n´

Gm(P) = Gºm + RT ln (P/Pº) ideal gas or … mi(P) = mi(P˚) + RT ln(Pi/P˚)

Chemical Potential (m)

Is G an extensive or an intensive property?

More ‘stuff’ = more free energy

Gm (the molar free energy) is an intensive property.

G

ni

Chemical potential (mi) = (dG/dni)T,P,n´

For a pure substance (mi) = Gm.

However mi 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 (mi) = (dG/dni)T,P,n´

For a mixture substance (mi) ≠ Gm, and mi changes with added moles of i.

G

ni

Chemical Potential (m)

How does the chemical potential of a pure gas change with pressure?

Gm(P) - Gºm = RT ln (P/Pº) 1 mole of ideal gas

mi(P) = mi(P˚) + RT ln(Pi/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.

What can cause a change in ni? In closed system (no added material).

dG = -SdT + VdP + Si (dG/dni)T,P,n´ dni

1) Chemical Reactions … Chapter 5

2) Phase changes … Chapter 6

dG = -SdT + VdP + Si mi dni

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 minimizeG (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 m.