chapter 5 chemical potential phase transitions mixtures

Chemical Thermodynamics : Georg Duesberg Chemical Thermodynamics : Georg Duesberg

Chapter 1 : Slide 1

Chapter 5

Chemical Potential Phase Transitions

Mixtures

Chapter11111 1 : Slide 1

Chemical Thermodynamics : Georg Duesberg

Chemical potential

ii

i nPTnG ,,⎟⎟⎠

⎞⎜⎜⎝

⎛

∂

∂=µ

Where µ = chemical potential (kJ/mol) ΔG = free energy (kJ) ni = moles of component (i)

iiii STHG ⋅−=≡µ

For a single component system containing 1 mole of substance:

µ=mG

Chemical Thermodynamics : Georg Duesberg Chemical Thermodynamics : Georg Duesberg

Properties of the Gibbs energy

G = H - TS

dG = dH –TdS - SdT

dG = dU + pdV + Vdp –TdS - SdT dU = dqrev +dwrev = TdS –pdV (for closed systems and no non-volume work) Explanation: Both are not state functions but the sum dq + dw can be treated to be reversible because they only depend on the state functions

dG = TdS – pdV + pdV + Vdp –TdS - SdT

dG = Vdp - SdT

G = f ( p, T )

dH = dU +pdV + Vdp

H = U + pV


4

Properties of the Gibbs energy dG = Vdp - SdT

STG

p

−=⎟⎠

⎞⎜⎝

⎛∂

∂V

pG

T

=⎟⎟⎠

⎞⎜⎜⎝

⎛

∂

∂

V is positive so G is increasing with increasing p

G

T (constant p)

Slope = -S G

P (constant T)

Slope = V

S is positive (-S is negative) so G is decreasing with increasing T


5

Dependence of G on p

It would be useful to determine the Gibbs energy at one pressure knowing its value at a different pressure.

dG = Vdp - SdT

We set dT = 0 (we make sure that the temperature is steady) and integrate:

∫+=f

i

)()( if

p

pVdppGpG

∫∫ =f

i

f

i

p

p

G

GVdpGd


∫=−=Δf

i

)()( if

p

pVdppGpGG


6

Dependence of G on p in Liquids and Solids

∫+=f

i

)()( if

p

pVdppGpG

Only slight changes of volume with pressure mean that we can effectively treat V as a constant, and take it out of the integral.

)()()()( iifif pGppVpGpG ≈−+=

pVpGpG Δ+= )()( if

Often V Δp is very small and may be neglected i.e. G for solids and liquids under normal conditions is independent of p. (For Water 0.02 kJ for going from 1 to 10 bar) Exception: Geography , very high pressures…


7

Dependence of G on p: Ideal Gases.

∫+=f

i

)()( if

p

pVdppGpG

For gases V cannot be considered a constant with respect to pressure. For a perfect gas we may use:

i

fiif ln)( )()( f

i ppnRTpG

pdpnRTpGpG

p

p+=+= ∫

We can set pi to equal the standard pressure, pθ ( = 1 bar). Then the Gibbs energy at a pressure p is related to its standard Gibbs energy, Gθ, by:


Dependence of G on p

θθ

ppnRTGpG f

f ln)( +=

Example: Going from 1 to 10 bar: dG ≈ 6 kJ (R = 8.31 JK-1, T = 298 K, ln 10= 2.3) Ø  Can derive that, for a gas: Ø  ΔGm = RT ln(pf/pi)

Ø  For p -> 0 Gm -> - ∞

Variation of G with pressure

VpG

T

=⎟⎟⎠

⎞⎜⎜⎝

⎛

∂

∂


Fugacity (lat.) = “urge to flee”

f= same units as pressure

The fugacity is a parameter which enables us to apply the perfect gas expression to real gases…. In the thermo dynamical treatment of the Chemical Potential

Chemical potential of real gases


11

Dependence of G on p of real Gases

θθ

pfRTGpG ln)( mfm +=

For real gases we modify the expression for a perfect gas and replace the true pressure by a new parameter, f, which we call the fugacity.

The fugacity is a parameter we have simply invented to enable us to apply the perfect gas expression to real gases.


Dependence of G on p Real Gases.

We may show that the ratio of fugacity to pressure is called the fugacity coefficient:

γ=pf

Where γ is the fugacity coefficient

γ is related to the compression factor Z and the Viral coefficients:

dppZp

∫−

=0

1lnγ

1

2lnffRTG =ΔWe may then write

..)'' 2+=×= PCpBepfViral coefficients


A real gas is in its standard state when its fugacity is equal to 1 bar and it is behaving as if it were an ideal gas at some specified temperature (the diagram shown below is exaggerated to make this point):

In the limit of zero pressure real gases behave more and more like ideal gases and the fugacities of real gases approach their partial pressures: lim fi --> Pi

Pi --> 0

ideal gas fi = Pi

real gas

fugacity

pressure

f = 1 bar

P = 1 bar

hypothetical standard state

not the standard state

The standard state of a pure liquid or solid at a fugacity of 1 bar and some specified temperature. Since the molar volumes of solids and liquids are generally small and relatively insensitive to pressure, the activities of solids and liquids at pressures that are not too far removed from 1 bar remain close to unity.

Pressure region Z f α γ

I (very Low) ≈1 P ≈0 ≈0

II (moderate) <1 <P >0 >0

III (high) >1 >P <0 <0


15

Dependence of G on T

STG

p

−=⎟⎠

⎞⎜⎝

⎛∂

∂

Using the same procedure as for the dependence of G on p we get:

TSdGd ∫∫ −=

To go any further we need S as a function of T

G = H - TS

G depennce stronger in the case for gases than for fluids or solids, because S is much higher in the gaseous state


16

Dependence of G on T

STH

TG

−=

Let G/T = x STHx −=

2TH

Tx

p

−=⎟⎠

⎞⎜⎝

⎛∂

∂

2

)/(TH

TTG

p

−=⎟⎠

⎞⎜⎝

⎛∂

∂

This is the Gibbs-Helmholtz Equation Important when dealing with chemical reaction and phase changes.

G = H - TS

Put back G/T = x


17

Gibbs-Helmholtz Equation

2

)/(TH

TTG

p

Δ−=⎟

⎠

⎞⎜⎝

⎛∂

Δ∂ STG

p

Δ−=⎟⎠

⎞⎜⎝

⎛∂

Δ∂

Two expressions:

Changes in entropy or, more commonly, changes in enthalpy can be used to show how changes in the Gibbs energy vary with temperature. For a spontaneous (ΔG < 0) exothermic reaction (ΔH < 0) the change in Gibbs energy increases with increasing temperature.

ΔG/T

T (constant p)

Slope = -ΔH/T2 = positive for exothermic reaction

Very negative

Less negative

STG

p

−=⎟⎠

⎞⎜⎝

⎛∂

∂from

18

Variation of G with temperature

ΔGm = -SmΔT Can help us to understand why transitions occur

The transition temperature is the temperature when the molar Gibbs energy of the two phases are equal.

The two phases are in EQUILIBIRIUM at this temperature

STG

p

−=⎟⎠

⎞⎜⎝

⎛∂

∂

19

Variation of G with pressure

19

VpG

T

=⎟⎟⎠

⎞⎜⎜⎝

⎛

∂

∂



µ

T

High P

low P

Tf Tb Tf Tb

Variation of G with temperature and pressure

Freezing and boiling point raise with pressure – much prounced Effect for boiling point – anomaly: water

21

Phase Equilibria Phase transitions Changes in phase without a change in chemical composition. Gibbs Energy is at the centre of the discussion of transitions

If two phases are in equilibrium as depicted here, along the phase transition line, then both phases have the same chemical potential. If a pressure is applied, which shifts the system out of equilibrium then the temperature will change (as a result of some particles migrating from one phase to the other) until equilibrium is re-established. The slope of the phase boundary is dp/dT.

22

Derivation of Clapeyron equation dGm(1) = dGm(2)

with dGm = Vmdp – SmdT Vm(1)dp – Sm(1)dT = Vm(2)dp – Sm(2)dT {Vm(2) – Vm(1)}dp = {Sm(2) – Sm(1)}dT

ΔtrsV dp = ΔtrsS dT Or

dp/dT = ΔtrsS/ ΔtrsV (Clapeyron equation) with ΔtrsS=ΔtrsH/T

dp/dT = ΔtrsH/(T ΔtrsV)

23

Location of phase boundaries

Ø  Clapeyron equation

Ø  Clausius-Clapeyron

TVTHptrs

trs ΔΔΔ

=Δ

( )

constant11lnln

ln

1212

2

+⎟⎟⎠

⎞⎜⎜⎝

⎛−

Δ−=

ΔΔ

=Δ

TTRH

pp

TRTH

p

vap

vap

Constant is

ΔvapS/R

24

Phase diagrams

Ø Map showing conditions of T and p at which various phases are thermodynamically stable

Ø At any point on the phase boundaries, the phases are in dynamic equilibrium

25

Characteristic points

Ø When vapour pressure is equal to external pressure bubbles form: boiling point

Normal bp: 1 atm, Standard bp: 1 bar Ø When a liquid is heated in a closed vessel the liquid density eventually

becomes equal to the vapour density: a supercritical fluid is formed.

26

Water Ice I structure

Ø  Solid-liquid boundary slopes to the left with increasing pressure

Ø volume decreases when ice melts, liquid is denser that solid at 273 K

27

first-order phase transition. Remember that at the transition point, we are changing the enthalpy of the system but not its temperature. Thus, the heat capacity Cp at the transition point is infinite. In other words, as we heat the system that is at the transition point, no temperature change happens because all the heat is going into the phase transition.

Duesberg, Chemical Thermodynamics

Chapter 7 : Slide 28

Chapter 6

Chemical Potential in Mixtures

Partial molar quantities

Thermodynamics of mixing The chemical potentials of liquids

Concentration Units

•  There are three major concentration units in use in thermodynamic descriptions of solutions.

•  These are – molarity – molality – mole fraction

•  Letting J stand for one component in a solution (the solute), these are represented by

•  [J] = nJ/V (V typically in liters) •  bJ = nJ/msolvent (msolvent typically in kg) •  xJ = nJ/n (n = total number of moles of all species

present in sample)



PARTIAL MOLAR QUANTITIES

Recall our use of partial pressures:

p = xA p + xB p + …

pA = xA p is the partial pressure

We can define other partial quantities…

For binary mixtures (A+B): xA + xB= 1




The contribution of one mole of a substance to the volume of a mixture is called the partial molar volume of that component.

( )...,,, BA nnTpfV =

AnTpAA n

VV≠

⎟⎟⎠

⎞⎜⎜⎝

⎛

∂

∂=

,,

...+⎟⎟⎠

⎞⎜⎜⎝

⎛

∂

∂+⎟⎟

⎠

⎞⎜⎜⎝

⎛

∂

∂= B

BA

A

dnnVdn

nVdV

At constant T and p

In a system that contains at least two substances, the total value of any extensive property of the system is the sum of the contribution of each substance to that property.



Partial molar volume

Very Large Mixture of A and B

Add nA of A to mixture

When you add nA of A to a large mixture of A and B, the composition remains essentially unchanged. In this case: Va can be considered constant and the volume change of the mixture is nAVA. Likewise for addition of B.

constnVV

AnTpAA ≈⎟⎟

⎠

⎞⎜⎜⎝

⎛

∂

∂=

≠,,

The total change in volume is nAVA + nBVB . (Composition is essentially unchanged).

...++= BBAA nVnVVScoop out of the reservoir a sample containing nA of A and nB of B its volume is nAVA + nBVB . Because V is a state function:


Volume Vs. Composition ¡  The partial molar volume of a

substance l  slope of the variation of the total

sample volume plotted against composition.

¡  partial molar volumes vary with solution composition

In general, partial molar quantities vary with the composition, as shown by the different slopes at the compositions a and b. Note that the partial molar volume at b is negative: the overall volume of the sample decreases as a is added. (e.g. MgSO4 in water)

AnTpAA n

VV≠

⎟⎟⎠

⎞⎜⎜⎝

⎛

∂

∂=

,,



Partial molar volume

A different answer is obtained if we add 1 mol of water to a large volume of ethanol.

3

,OHOH cm18

2

2=⎟

⎟⎠

⎞⎜⎜⎝

⎛

∂

∂=

TpnVVThe change in

volume is 18cm3

3

OH)CHCH(,,OHOH cm14

232

2=⎟

⎟⎠

⎞⎜⎜⎝

⎛

∂

∂=

nTpnVVThe change in

volume is 14cm3

Example: What is the change in volume of adding 1 mol of water to a large volume of water?

35

Introduction to mixtures

Ø Homogeneous mixtures of a solvent (major component) and solute (minor component).

Ø  Introduce partial molar property: contribution that a substance makes to overall property.

V = nAVA + nBVB



VA is not generally a constant: It is a function of composition, because the environment and therefore their interaction of the molecules changes :


with the composition :

Exercise Calculate the density of a mixture of 20 g of water and 100g of ethanol.

•  First calculate the mole fractions. –  20 g H2O = 1.11 mol; –  100 g EtOH = 2.17 mol –  xH2O = 0.34; xEtOH = 0.66

•  Then interpolate from the mixing curve –  VH2O = 17.1 cm3 mol-1; –  VEtOH = 57.4 cm3 mol-1

•  Then plug the moles and partial molar volume –  (1.11 mol)(17.1 cm3/mol) + (2.17 mol)

(57.4 cm3/mol) =19.0 cm3 + 125 cm3 = 144 cm3

•  Finally, the total mass is divided by the total volume: 120 g/144 cm3 = 0.83 g/ cm3

Task: You mix 300 ml alcohol with a bottle of juice (water, 0.7 l) - Will you really end up with for your x-mas party supply (1l)? -Of course not… - How much do you have to mix to get 1 l of the same strength?

38

The chemical potential, µ

• We can extend the concept of partial molar properties to state functions, such as Gibbs energy, G.

•  This is so important that it is given a special name and symbol, the chemical potential, µ.

G = nAGA + nBGB

G = nAµA + nBµB



The Thermodynamic Criterion of Equilibrium

At equilibrium, the chemical potential of a substance is the same throughout a sample, regardless of how many phases are present.

µ1 µ2

Consider this single substance system:

dG = -µ1 dn dG = +µ2 dn

Total dG = 0 only if µ1 = µ2



Chemical Potential

1,,1

µ=⎟⎟⎠

⎞⎜⎜⎝

⎛

∂

∂

jnpTnG

Partial molar Gibbs energy Chemical potential for component 1

...2,,2

1,,1

+⎟⎟⎠

⎞⎜⎜⎝

⎛

∂

∂+⎟⎟

⎠

⎞⎜⎜⎝

⎛

∂

∂= dn

nGdn

nGdG

jj npTnpT

If T and p are kept constant (i.e. dT and dp = 0) then

Or: ...2211 ++= nnG µµ

For a single component system containing 1 mole of substance:

µ=mG

...2211 ++= dndndG µµ

Chemical Potential

dG is negative

Phase β Phase α Phase γ

dG = 0

Phase β Phase α Phase γ

γβα µµµ iii >>

γβα µµµ iii ==

dniα

dnjα

dniβ

dnjβ

dniγ

dnjγ



The Partial Molar Gibbs Energy

Ap,T,nAA n

Gµ≠

⎟⎟⎠

⎞⎜⎜⎝

⎛

∂

∂=

The partial molar Gibbs energy is called the “chemical potential”

...+⎟⎟⎠

⎞⎜⎜⎝

⎛

∂

∂+⎟⎟

⎠

⎞⎜⎜⎝

⎛

∂

∂= B

BA

A

dnnGdn

nGdG

G = nAµA + nBµB

At constant T and p

...++= BBAA dndndG µµ(At equilibrium dG = 0)

The implication of G = f ( p,T, ni ) is that we now write:

∑++−= iidnVdpSdTdG µ

...++++−= AAAA dndnVdpSdTdG µµ

Fundamental Equation of Chemical Thermodynamics.

( )...,,, BA nnTpfG =



The chemical potential of a substance is the slope of the total Gibbs energy of a mixture with respect to the amount of substance of interest. In general, the chemical potential varies with composition, as shown for the two values at a and b. In this case, both chemical potentials are positive.

∑= iidndG µ

In case of constant temperature and pressure reduces to:

∑++−= iidnVdpSdTdG µ

Chemical Potential = Partial Molar Gibbs Energy



The Wider Significance of the Chemical Potential

∑ ⎟⎟⎠

⎞⎜⎜⎝

⎛

∂

∂++−= i

npTi

dnnGVdpSdTdG

j,,

∑ ⎟⎟⎠

⎞⎜⎜⎝

⎛

∂

∂++= i

npSi

dnnHVdpTdSdH

j,,

∑ ⎟⎟⎠

⎞⎜⎜⎝

⎛

∂

∂+−−= i

nVTi

dnnApdVSdTdA

j,,

∑ ⎟⎟⎠

⎞⎜⎜⎝

⎛

∂

∂+−= i

nVSi

dnnUpdVTdSdU

j,,

µi

jnpTinG

,,⎟⎟⎠

⎞⎜⎜⎝

⎛

∂

∂

jnpSinH

,,⎟⎟⎠

⎞⎜⎜⎝

⎛

∂

∂

jnVTinA

,,⎟⎟⎠

⎞⎜⎜⎝

⎛

∂

∂

jnVSinU

,,⎟⎟⎠

⎞⎜⎜⎝

⎛

∂

∂

All extensive thermodynamic properties change with composition!

chem "chemical work"j jj

dw dnµ− =∑Example: E-chem cell:

Chemical Potential

∑== iiadd dndGdw µmax,In case of constant temperature and pressure also:

j jj

j jj

j jj

j jj

dU TdS PdV dn

dH TdS VdP dn

dA SdT PdV dn

dG SdT VdP dn

µ

µ

µ

µ

= − +

= + +

= − − +

= − + +

∑

∑

∑

∑

, , , ,

, , , ,

j i j i

j i j i

ii iS V n S P n

i iT V n T P n

U Hn n

A Gn n

µ≠ ≠

≠ ≠

⎛ ⎞ ⎛ ⎞∂ ∂= =⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞∂ ∂= =⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠



2211 nnG µµ +=

22112211 µµµµ dndndndndG +++=

The Gibbs-Duhem Equation

by differentiating:

At equilibrium: dG = 0 0 = µ1dn1 + µ2dn2

22110 µµ dndn +=∴

21

21or µµ d

nnd −=

The Gibbs-Duhem Equation. A similar expression may be deduced for all partial molar quantities (If one is changed the other changes as well…)

...2211 ++= dndndG µµ

(For binary systems)

∑= iidndG µand recall

iidnor µ∑=0



The Gibbs-Duhem Equation

0j jj

SdT VdP n dµ− + =∑

48

The chemical potential of perfect gases in a mixture

Recall that Gm(pf) = Gm(pi) + RT ln (pf/pi)

At standard pressure Gm(p) = Gm° + RT ln (p/p°)

Therefore, for a mixture of gases

µJ = µJ° + RT ln (pJ/p°) More simply (at p° = 1 bar)

µJ = µJ° + RT ln pJ

System is at equilibrium when µ for each substance has the same value in every phase

49

Gas mixtures

Compare ΔGmix = nRT {xAln xA+ xB ln xB}

ΔG = ΔH – TΔS

Therefore ΔH = 0

ΔSmix = − nR {xAln xA+ xB ln xB}

Perfect gases mix spontaneously in all proportions

50

Ideal Liquid Solutions

pJ = xJpJ*

Due to effect of solute on entropy of solution

Raoult’s Law

51

Real Solutions

52

Chemical potential of a solvent

Ø At equilibrium µA(g) = µA(l) µA(l)= µA°(g) + RT ln pA

µA(l)= µA°(g) + RT ln xApA* µA(l)= µA°(g) + RT ln pA* + RT

ln xA └────────────────┘

µA*

Ø µA(l)= µA*+ RT ln xA

53

Is solution formation spontaneous?

G = nAµA + nBµB Can show that ΔGmix = nRT {xAln xA+ xB ln xB} and

ΔH = 0

ΔSmix = −nR {xAln xA+ xB ln xB}

54

Ideal-dilute solutions Ø Raoult’s law generally describes well solvent vapour

pressure when solution is dilute, but not the solute vapour pressure

Ø  Experimentally found (by Henry) that vp of solute is proportional to its mole fraction, but proportionality constant is not the vp of pure solute.

Henry’s Law

pB = xBKB

55

Gas solubility

KH/(kPa m3 mol−1)

Ammonia, NH3 5.69

Carbon dioxide, CO2 2.937

Helium, He 282.7

Hydrogen, H2 121.2

Methane, CH4 67.4

Nitrogen, N2 155

Oxygen, O2 74.68

Henry’s law constants for gases dissolved in water at 25°C

Concentration of 4 mg/L of oxygen is required to support aquatic life, what partial pressure of oxygen can achieve this?

56

Application-diving

Table 1 Increasing severity of nitrogen narcosis symptoms with depth in feet and

pressures in atmospheres.

Depth P Total P N2 Symptoms

100 4.0 3.0 Reasoning measurably slowed.

150 5.5 4.3 Joviality; reflexes slowed; idea fixation.

200 7.1 5.5 Euphoria; impaired concentration; drowsiness.

250 8.3 6.4 Mental confusion; inaccurate observations.

300 10. 7.9 Stupefaction; loss of perceptual faculties.

Gas narcosis caused by nitrogen in normal air dissolving into nervous tissue during dives of more than 120 feet [35 m]

Pain due to expanding or contracting trapped gases, potentially leading to Barotrauma. Can occur either during ascent or descent, but are potentially most severe when gases are expanding. Decompression sickness due to evolution of inert gas bubbles.

57

Real Solutions-Activities µJ = µJ° + RT ln aJ

Substance Standard state Activity, a

Solid Pure solid, 1 bar 1

Liquid Pure liquid, 1 bar 1

Gas Pure gas, 1 bar p/po

Solute Molar concentration of

1 mol dm−3 [J]/co

po = 1 bar (= 105 Pa), co = 1 mol dm−3.

* Activities are for perfect gases and ideal-dilute solutions; all activities are dimensionless.

58

Colligative properties Properties of solutions that are a result of changes in the disorder of the solvent, and rely only on the number of solute particles present

Lowering of vp of pure liquid is one colligative property

Freezing point depression

Boiling point elevation

Osmotic pressure

59

Colligative properties Ø Chemical potential of a solution

(but not vapour or solid) decreases by a factor (RTlnxA) in the presence of solute

Ø Molecular interpretation is based on an enhanced molecular randomness of the solution

Ø Get empirical relationship for FP and BP (related to enthalpies of transition)

mKTmKT

bb

ff

=Δ

=Δ

60

Cryoscopic and ebullioscopic constants

Solvent Kf/(K kg mol−1) Kb/(K kg mol−1)

Acetic acid 3.90 3.07

Benzene 5.12 2.53

Camphor 40

Carbon disulfide 3.8 2.37

Naphthalene 6.94 5.8

Phenol 7.27 3.04

Tetrachloromethane 30 4.95

Water 1.86 0.51

61

Osmotic pressure

Van’t Hoff equation

MRT=Π

62

Phase diagrams of mixtures

We will focus on two-component systems (F = 4 ─ P), at constant pressure of 1 atm (F’ = 3 ─ P), depicted as temperature-composition diagrams.

63

Fractional Distillation-volatile liquids

Important in oil refining

64

Exceptions-azeotropes Azeotrope: boiling without changing High-boiling and Low-boiling

Favourable interactions between components reduce vp of mixture Trichloromethane/propanone HCl/water (max at 80% water, 108.6°C)

Unfavourable interactions between components increase vp of mixture Ethanol/water (min at 4% water, 78°C)

65

Liquid-Liquid (partially miscible)

Ø Hexane/nitrobenzene as example

Ø Relative abundances in 2 phases given by Lever Rule n’l’ = n’’l’’

Ø Upper critical Temperature is limit at which phase separation occurs. In thermodynamic terms the Gibbs energy of mixing becomes negative above this temperature

66

Other examples

Water/triethylamine Weak complex at low temperature disrupted at higher T.

Nicotine/water Weak complex at low temperature disrupted at higher T. Thermal motion homogenizes mixture again at higher T.

67

Liquid-solid phase diagrams

chapter 5 chemical potential phase transitions mixtures

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