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TOPIC 3

IDEAL SOLUTIONS, FUGACITY, ACTIVITY, AND STANDARD STATES

1

I. PARTIAL AND APPARENT MOLAR PROPERTIES

2

MOLAR VS. PARTIAL MOLAR QUANTITIES

Molar values of state functions are defined as follows: U =U / n S = S/n V =V /n etc. These are useful only in the case of singlecomponent systems and dependent only on pressure and temperature, not composition. Partial molar quantities are defined according to:

U Ui = ni T ,P ,n j

S Si = n i T ,P ,n j

PHYSICAL INTERPRETATION OF PARTIAL MOLAR VOLUMES

The partial molar volume of component i in a system is equal to the infinitesimal increase or decrease in the volume, divided by the infinitesimal number of moles of the substance which is added, while maintaining T, P and quantities of all other components constant. Another way to visualize this is to think of the change in volume on adding a small amount of component i to an infinite total volume of the system. Note: partial molar quantities can be positive or negative! 4

V Vi = n i T ,P ,n j

These are dependent on T, P, and composition.

3

SUMMING PARTIAL MOLAR QUANTITIES

The total value for a state function of a system is obtained by summing the partial molar volumes of its components according to: n

We can also express the summations in terms of molar state functions and mole fractions:

U=

X U

i i =1

n

i

H = X i Hi

i=1

n

V = X iVi

i=1

n

U = n1U 1 + n2U 2 + n3U 3 + n4U 4 + K + nnU n = niU i

H = ni H i

i=1 n

Xi =

ni

V = niVi

i= 1

n

i=1

n

i= 1

n

i

We can manipulate partial molar quantities in a manner identical to the way we manipulate total quantities. As with total state functions, we cannot know absolute values, only differences (except for V and S)!

5

In the case of the volume of a two-component system, e.g., NaCl-H2O, we can write:

V = X NaCl NaCl + X H 2OVH 2O V

6

1

Schematic plot of the molar volume of aqueous NaCl solutions as a function of mole fraction of NaCl .

HOW TO DETERMINE PARTIAL MOLAR VOLUME

Refer to the previous diagram. Triangles A and B are similar, so it is true that:

X NaCl /(V solution - x ) = X H 2O /( y - Vsolution )

V

y

X NaCl y - X NaClV solution = X H 2 OV solution - X H 2O x

A

x

(X

H 2O

+ X NaCl ) Vsolution = X NaCl y + X H 2O x

so

B

XNaCl

XH

Vsolution

2

O

but

X NaCl + X H 2 O = 1

VNaCl = y

V solution = X NaCl y + X H 2O x

VH 2O = x

comparison with shows that

V = X NaClVNaCl + X H2OVH2O

0.0

0.2

0.4

0.6

0.8

1.0

7

X NaCl

So the partial molar volumes can be determined from the intercepts of a line tangent to the plot of volume vs. mole fraction. 8

G Gi = µi n i T , P,n j

Absolute Molar Gibbs Free Energy

PARTIAL MOLAR FREE ENERGY THE CHEMICAL POTENTIAL

Chemical potential

Schematic plot of chemical potential vs. mole fraction for a binary system

µBo

o + X Aµ A o X BµB

µA

o

G mixing

The previous relationships also apply:

µB

G = ni µ i

i= 1

n

G = X i µi

i=1

n

It can also be shown that:

U µi = n i A H = = S V ,n j n i T ,V ,n j ni , S ,P ,n j

µA

0.0 0.2 0.4 0.6 0.8 1.0

10

9

XB

COMPOSITIONAL CHANGES

The Master equations that we developed previously for one-component systems can now be written as:

dU = TdS - PdV + µ i dni

i =1 n n

AN ADDITIONAL REQUIREMENT FOR EQUILIBRIUM

Consider a system with components i, j, k, l, ... distributed among phases , , , , ... At equilibrium it must be true that: µi = µi = µ i = µ i = ... µj = µj = µ j = µ j = ... µk = µk = µk = µk = ... µl = µl = µ l = µ l = ... etc.

12

dH = TdS + VdP + µ i dni

i =1

dG = - SdT + VdP + µ i dni

i =1 n

n

dA = - SdT - PdV + µ i dni

i =1

11

2

Chemical potentials represent the slope of the Gibbs free energy surface in compositional space. Thus, a component will move from a phase in which it has a high chemical potential, to one in which it has a low chemical potential, until its chemical potential in all phases is the same. Specific example: Consider a silicate melt in equilibrium with forsterite (Mg 2 SiO4), and enstatite (Mg 2 Si2O6 ). At equilibrium the following must be true: µ Mgmelt = µ Mg Fo = µ Mg En µSi melt = µ SiFo = µ SiEn µO melt = µ OFo = µOEn

13

GIBBS-DUHEM EQUATION

For a homogenous phase of two components, A and B, the Master Equation becomes:

dG = - SdT + VdP + µ Adn A + µ B dnB

If we now specify equilibrium at constant T and P

dG = µ Adn A + µ B dnB = 0

Now, we have shown above that

G = µ A nA + µ B nB

Differentiating this we obtain

dG = µ AdnA + µ B dnB + nA dµ A + nB dµ B

14

GIBBS-DUHEM EQUATION CONTINUED

At equilibrium

dG = µ A dnA + µ B dnB + nAdµ A + nB dµ B = 0

GIBBS-DUHEM EQUATION CONTINUED

Starting with the expression:

n Adµ A + n B dµ B = 0

Substituting the previous expression

dG = µ Adn A + µ B dnB = 0

If we divide through both sides by dXA we get

µ µ n A A + nB B =0 X X A T , P ,nB A T , P ,nB

we obtain

n Adµ A + n B dµ B = 0

And now dividing by nA + n B we get:

µ µ X A A + (1 - X A ) B =0 X X A T , P ,nB A T , P ,nB

15 16

In the general case we get the Gibbs-Duhem equation n

n dµ = 0

i =1 i i

APPARENT MOLAR QUANTITIES

Although in principle, partial molar quantities can be measured from intercepts of lines tangent to a plot of state functions vs. mole fraction as outlined previously, they are not determined this way in practice. In practice, apparent molar quantities are determined. For a state function like volume, the apparent molar volume, V, is given by V = (V - n 1V1 °)/n 2 where n 1 , and n2 are the number of moles of solvent and solute, respectively, and V 1 ° is the molar volume of pure solvent.

17

Total volume of a solution as a function of solute concentration. Illustrates the difference between partial and apparent molar volume.

Volume cm3 (V)

Volume of pure solvent

n2 V

V2

Volume attributed to solute

2

n1 V1

o

Volume attributed to solvent

0

n2

Moles of solute n2

18

3

Comparison of apparent molar and partial molar volumes

The apparent molar volume is the volume that would be attributed to one mole of solute in solution if it is assumed that the solvent contributes the same volume it has in the pure state. Starting with the definition of apparent molar volume: V = (V - n 1V1 °)/n 2 we can rearrange to get V = n1 V1° + n 2 V and dividing by (n1 + n2 ), V = X1 V1 ° + X2 V Thus, the volume of solution can be calculated knowing V instead of the partial molar volume.

19

V 2o V1o V1

V

V2

V

2

0.0

0.2

0.4

0.6

0.8

1.0

20

X2

PARTIAL MOLAR VOLUMES FROM APPARENT MOLAR VOLUMES

V ( n V o + n2V ) V2 = = 1 1 n n2 2 T ,P , n1 T , P ,n1

V2 = n2 V n 2 + V n1

or

V2 = m V + V m n1

II. IDEAL SOLUTIONS

If V measurements are fit by an equation of the type: V = a + bm + cm 2 then we have V 2 = m(b + 2cm) + a + bm + cm2 or V2 = a + 2bm + 3cm2

21 22

THERMODYNAMICS OF IDEAL SOLUTIONS

An ideal solution is one that satisfies the following equation: µi - µ i° = RT ln Xi where µ i is the chemical potential of some component i in a solution and µ i° is the chemical potential of that component in the pure form. Recall that G = X i µi

i

G ideals o l' n = X iµi + RT X i ln X i

o i i

G mech mix = X i µi

i

o

Gidealsol'n = Gmech mix + RT X i ln X i

i

Gideal mix = Gideal s o l' n - Gmech mix = RT X i ln X i

i

substituting we get

Gidealsol' n = X i ( µio + RT ln X i )

i

23

These equations tell us that the free energy of an ideal solution is the sum of two terms: the free energy of a mechanical mixture, and a free energy of ideal mixing.

24

4

ENTHALPY AND VOLUME OF AN IDEAL SOLUTION

H ideal s o l'n = H mech mix =

ENTROPY OF MIXING

Gideal mix = RT X i ln X i

i

X H

i i i

o i

= XiH i

i

o

Videal s o l'n = Vmech mix = X iVi =

o

XV

i

o

If i is a pure substance.

i i

but we have and so

Gideal mix = Hideal mix -TS ideal mix Hideal mix = 0

i

There is no volume or enthalpy change upon ideal mixing. In other words:

Gideal mix = - TSideal mix = RT X i ln X i Sideal mix = - R X i ln X i

i

Videal mix = H ideal mix = 0

However, there is a change in entropy upon ideal mixing. Because the solution is more disordered, 25 entropy increases!

Thus, the only contribution to Gideal mix is an entropy contribution!

26

µB o

Molar Gibbs Free Energy

G mech mix

o µ + = XA A

o X Bµ B

Molar Volume

µA o

V sol'n

=

o X AV A

+ XB VB

o

VB o

Gideal sol'n

on soluti ideal

V Ao

0.0

0.2

0.4

0.6

0.8

1.0

27

0.0

0.2

0.4

0.6

0.8

1.0

28

XB

XB

Molar Enthalpy

H sol'n

HAo

H = XA A lution eal so id

o

+ X BHB

o

HB o

III. FUGACITY AND ACTIVITY

0.0

0.2

0.4

0.6

0.8

1.0

XB

29 30

5

FUGACITY

Starting with dG = VdP - SdT at constant T this becomes dG = VdP For an ideal gas dG = (RT/P)dP = RT dln P This is true for ideal gases only, but it would be nice to have a similar form for real fluids. dG = RT dln where is the fugacity = /P 1 as P 0 is the fugacity coefficient = P Fugacity may be thought of as a thermodynamic pressure; it has units of pressure.

31

MEASUREMENT OF FUGACITY

G =V P T (RT ln f ) RT ( ln f ) = =V P P T T V ( ln f ) = P T RT

V ln f = P RT

V ln f - ln P = P - ln P RT

32

V 1 ln f = ln = P P - P P RT 1 V ln = - P RT P

V 1 ln = - P RT P 0

P

Z ( ln f ) = P T P

ln f = Z P P = Z ln P

ln f - ln P = Z ln P - ln P f = ln = ( Z - 1) ln P = Z - 1 P ln P P

( )

Alternatively, we can begin again with: V ( ln f ) = P T RT But we now define the compressibility factor Z: PV Z= RT

ln =

P 0

P

0

Z - 1 P P

P Z -1 ln = dP 0 P

33 34

The above equation is the basis of the experimental determination of fugacities from P-V-T data. We can substitute into the integral (Z-1)/P calculated from any equation of state, or we can integrate graphically.

CALCULATION OF EQUILIBRIUM BOUNDARIES INVOLVING FLUIDS

A number of mineral reactions involve only solid minerals and either H 2 O or CO2, e.g. KAl3 Si3O10 (OH)2 KAlSi3 O8 + Al2 O3 + H 2 O or CaCO3 + SiO2 CaSiO3 + CO2 or in general A(s) B(s) + C(fluid) rV = VB + VC - VA = (VB - VA) + V C rV = sV + Vfluid

36

(Z-1)/P

area = ln

0

P

P

35

6

VdP = VdP + V

r s

fluid

dP

The pressure integral for the solids is then evaluated using the constant sV approximation and that for the fluid is evaluated using fugacities.

rV° = V°Kspar + V°cor + V°H 2 O - V°musc rV° = (V°Kspar + V° cor - V° musc) + V°H2 O rV° = sV° + V°H2 O

P

f P=1 rVdP = sV ( P - 1) + RT ln P f P=1

P

V

r P r

o

P P f o dP = sV odP + VH 2O dP = sV o ( P - 1) + RT ln H2O , P fH O, 1bar P Pr r 2

For the muscovite breakdown reaction above, we can start with the equation:

o o o o r GT , P = r GTr , Pr + r G dT + r G dP T P = P P T = T r r Tr Pr T P o = r GT r, Pr - r S odT + rV odP Tr Pr

37

T o o 0 = r GTr - r STr (T - Tr ) + r aT - Tr - T ln T r 2 2 rb r c (T + Tr - 2TTr ) 2 2 o + (2TTr - T - Tr ) + + sV ( P - 1) 2 2TTr2 fH O , P 2 + RT ln f H 2O, 1bar

A function of pressure and temperature!

38

T

P

FUGACITIES IN GASEOUS SOLUTIONS

Starting with the following, in terms of partial molar volumes: Vi ( ln f i ) = P T RT We obtain the expression for the fugacity coefficient of a component in a solution:

P 1 V ln i = i - P P 0 RT

39

ALL CONSTITUENTS HAVE A FUGACITY

The expression dG = RT dln may be integrated between two states 1 and 2 to give: G2 - G1 = RT ln ( 2 /1 ) This equation applies to a pure one-component system. For a solution we must use chemical potentials and we write: µi&quot; - µi' = RT ln (i&quot;/ i') This equation makes no stipulation as to the state of component i, and can therefore refer to solid, liquid or gas.

40

· Solids and liquids therefore are also associated with a fugacity. In some cases, this fugacity can be thought of as a vapor pressure. Fugacity can also be thought of as an escaping tendency. · However, in some cases, a vapor phase may not exist, but a fugacity always exists. One must realize that the fugacity is a thermodynamic model parameter, not always an approximation to a real pressure. · Fugacities of solid phases or individual components of solid solutions are not generally known. · Fugacities are absolute physical properties.

41

ACTIVITIES

The absolute values of the fugacities of solids and liquids cannot always be determined, but their ratios can be. Consider µ i&quot; - µi' = RT ln (i&quot;/ i') If we let one of these states be a reference state, this can be rewritten: µi - µ i° = RT ln ( i/i°) We now define the activity of constituent i to be ai = i/i° Thus µi - µ i° = RT ln ai

42

7

DALTON'S LAW

Dalton (1811) discovered that, at low total pressures, a mixture of gases exerts a pressure equal to the sum of the pressures that each constituent gas would exert if each alone occupied the same volume. Strictly true only for ideal gases, but is approximately true at low total pressure where real gases approach ideality. For each gas we have: P1 V = n 1 RT P2 V = n 2 RT etc. For the gas mixture we have: PtotalV = ni RT

i

43

If we divide the expression for each constituent by the expression for the mixture we obtain:

P n 1 = 1 = X1 P ni

i

P2 n = 2 = X2 P ni

i

etc.

or P1 = X 1 · Ptotal P2 = X 2 · Ptotal etc. Ptotal = P1 + P 2 + P3 + ... P1 , P 2 , etc. are called the partial pressures.

44

HENRY'S LAW

Henry (1803) was studying the solubility of gases in liquids. He found that the amount of gas dissolved in a liquid in contact with it was directly proportional to the pressure on the gas, i.e., Pi = Kh,i· Xi Kh,i is a constant called the Henry's Law constant. In practice, this law hold's only at relatively low values of P i.

RAOULT'S LAW

Raoult (1887) studied vapor-liquid systems in which two or more liquid components were mixed in known proportions and the liquid was equilibrated with its own vapor. The composition of the vapor was then determined. The total vapor pressure of the system was low, so the vapor behaved ideally and conformed to Dalton's law. In such systems, the partial pressures of the gaseous components were found to be a linear function of the their mole fraction in the liquid.

46

45

Thus, for a binary system A-B, he obtained: PA = XA·PA º and PB = X B ·PB º where P Aº and PB º are the vapor pressures of pure components A and B, respectively.

P A 10

8

o

P

tota l

P

6

PB

4

o

2

0 0.0 0.2 0.4 0.6 0.8 1.0

The only way that such a simple relationship as Raoult's law can hold is if the intermolecular forces between A-A, B-B, and A-B are identical. Solutions in which this is the case are called ideal solutions. The most general way of expressing Raoult's Law is: Pi = Xi·Piº Very few systems follow Raoult's Law over the entire range of composition from Xi = 0 to Xi = 1. However, Raoult's Law often applies to the solvent in dilute solutions, whereas the solute in dilute solutions follows Henry's Law.

47 48

A

XB

B

8

Partial pressure in the mixture acetone-chloroform at 35.2°C. This mixture exhibits negative deviations from Raoult's Law

Partial pressure in the mixture carbon disulfide-acetone at 35.2°C. This mixture exhibits positive deviations from Raoult's Law

49

50

THE GIBBS-DUHEM EQUATION REVISITED

Previously we derived the Gibbs-Duhem equation for a binary solution:

(1 - X B ) µA

X B T ,P , nA

µ + XB B =0 X B T , P ,n A

This equation shows that the slopes of tangents to curves of chemical potential vs. mole fraction for binary solutions are not independent of one another. µ A For example, if XB = 0, and X has a finite

B

T ,P , nA

This can be rearranged to give:

µ A X B T , P , nA XB =- µ B (1 - X B ) X B T , P , nA

51

µ =. value, then B X B T , P ,nA

If XA = 0.5, then

µ A µ = - B etc. X X B T , P ,n A B T , P ,n A

52

Chemical potentials in solutions of carbon disulfide and acetone.

THE DUHEM-MARGULES EQUATION

Starting with the Gibbs-Duhem equation:

(1 - X B ) µ A

Gibbs- Duhem Equation

X B T , P, n A

µ + X B B =0 X B T ,P , n A

µ A X B T , P ,nA XB =- µ B (1 - X B ) X B T , P ,n

A

If we recall that dµi = RT dln i the substitution and obtain:

we can make

(1 - X B ) ln f A

X B T ,P , n A

ln f B = -X B X B T ,P ,n A

53

ln f A X B T ,P , n XB A =- ln fB (1- X B ) X B T ,P , n

A

54

9

When the vapors are nearly perfect gases, we may substitute partial pressures for fugacities to obtain the approximate relation:

ln PA X B T , P ,n A XB =- ln PB (1 - X B ) X B T , P ,n

A

Application of the Duhem -Margules equation. Partial pressure is plotted on the Y-axis.

Duhem -Margules Equation

Realizing that dX B = -dX A, and that d ln P = d P/P we can rewrite this as:

PA PA X A T , P ,nB XA = PB P B XB X B T , P ,n

A

PA PA X A T , P ,nB XA = PB P B XB X B T , P ,n

A

55

56

57

58

THE LEWIS FUGACITY RULE

This is a variation on Raoult's Law:

IDEAL MIXING AND ACTIVITY

If we compare the definition of the activity: ai = i/i° and a rearrangement of the Lewis Fugacity Rule: Xi = fimixture /fipure we see that for solutions that obey the Lewis Fugacity Rule ai = Xi We can also now write: µi - µ i° = RT ln Xi which is considered another form of Raoult's Law.

60

fimixture = Xi· fipure

This states that the fugacity of a constituent in a mixture is equal to it's mole fraction times its fugacity in the pure state. Many substances that do not obey Raoult's Law do in fact obey the Lewis Fugacity Rule.

fA

pure

Fugacity

p f B ure

0.0

0.2

0.4

0.6

0.8

1.0

A

XB

B

59

10

Activity relations for an ideal binary system.

1.0

NON-IDEAL MIXING

aA aB

Activity

0.0

0.2

0.4

0.6

0.8

1.0

XB

It turns out that these relations hold not only for liquid and gaseous solutions, but also for solid solutions.

· As already discussed, most real solutions do not conform to Raoult's Law over the entire compositional range. · However, whether solid, liquid or gas, in many solutions, the component in excess (solvent) follows Raoult's Law and the minor component (solute) follows Henry's Law over a limited range at low mole fractions.

61 62

Positive deviation from Raoult's Law

1.0

ANOTHER NIFTY APPLICATION OF THE GIBBS-DUHEM EQUATION

Task: Prove that, if the solute in a binary solution obeys Henry's Law, then the solvent obeys Raoult's Law. Starting with the Gibbs-Duhem equation for a binary system

n Ad µ A + n B d µ B = 0

0.8

aA

aB

Raoultian activity

0.6

0.4

and dividing through both sides by nA + n B we get

0.2

X Adµ A + X B dµ B = 0

X Ad µ A = - X B d µ B

0.0 0.0 0.2 0.4

XB

0.6

0.8

1.0

63

At low pressures we have

dµi = RT dln Pi

64

So we can now write: X A d ln PA = - X B d ln PB Now if component A is the solute and obeys Henry's Law we have: PA = Kh,A · XA Taking the natural logarithm of both sides we have: ln PA = ln Kh,A + ln XA Now differentiating we get: d ln P A = d ln XA so now X Ad ln X A = - X B d ln PB

XA dX A = - X B d ln PB XA

Now

-dXB = dXA

so

- dX B = - X B d ln PB dX B = d ln PB d ln X B = d ln P B XB

Now we integrate from X B = 1 to X B = XB

ln

where P B° is the partial pressure of B when XB = 1, i.e., the partial pressure of pure B.

ln X B = ln PB PBo XB = PB o PB

66

XB P = ln B 1 Po B

dX A = - X B d ln P B

65

PB = X B PBo

Raoult's Law!

11

ACTIVITY COEFFICIENTS

The ideal solution is useful as a model with which real solutions are compared. This comparison is effected by taking the ratio of the activity of the real solution relative to that of the ideal solution. This ratio is called the activity coefficient. Deviations from Raoult's Law are expressed by the Raoultian activity coefficient R ai = R,iXi

Deviations from Henry's Law are expressed by the Henryian activity coefficient H ai = H,iXi Activity coefficients, because they are ratios of activities, are unitless. A major difference between the two types of activity coefficients is that: R 1 as X 1, but H 1 as X 0 Thus, H is usually more useful for solutes in dilute solutions.

68

67

STANDARD STATES

Because the activity is the ratio of two fugacities, i.e., ai = i/i° the value of the activity depends on the reference state chosen for i°. This state we usually refer to as the standard state. The choice of the standard state is completely arbitrary. The standard state need not be a real state. It is only necessary that we be able to calculate or measure the ratio of the fugacity of the constituent in the real state to that in the standard state. 70

IV. STANDARD STATES

69

A STANDARD STATE HAS FOUR ATTRIBUTES

· · · · Temperature Pressure Composition A particular, well-defined state (e.g., ideal gas, ideal solution, solid, liquid, etc.)

STANDARD STATES FOR GASES

Single Ideal Gas Starting with the relationship: µ2 - µ1 = RT ln (P2 /P1 ) we can assign our standard state to be the ideal gas at 1 bar and any temperature. In this case we can write: µ - µ° = RT ln P and µ - µ° is the difference in chemical potential between an ideal gas at T and P, and an ideal gas at T and 1 bar.

If desirable, we can permit T or P to be variable, i.e., on a sliding scale.

71

72

12

Ideal mixture of ideal gases For such a mixture we can write: µ1 - µ1 ° = RT ln [(X 1 P)/(X1 P)°] If we choose our standard state to be the pure ideal gas 1 at any temperature and 1 bar, then X1° = 1 and P° = 1, so µ1 - µ1 ° = RT ln (X1 P) = RT ln P 1 Non-ideal gases For non-ideal gases we would write: µ1 - µ1 ° = RT ln (f1 /f1°) but recall that lim (fi/Pi) Pi 0 = 1. So if we chose our standard state to be the pure, ideal gas at any temperature and P = 1 bar, we get

µ1 - µ1 ° = RT ln f1 This equation is frequently written, but rarely understood. It only has meaning if the standard state is specified to be the pure ideal gas at any temperature and 1 bar. This is the most commonly chosen standard state for gases and supercritical fluids. However, there is no reason why this particular standard state has to be chosen. We could just as easily choose: 1) the pure ideal gas at any T and 10 bars; 2) a pure real gas at 25°C and 1 bar; 3) a specific mixture of gases at any T and bars; or 4) any other welldefined standard state.

74

73

LIQUIDS AND SOLIDS

The following equation applies to liquids and solids as well: µi - µ i° = RT ln (fi /fi°) Fixed pressure standard state The standard state is chosen to be the pure phase at the temperature of interest and 1 bar. Then fi° = 1, so a i = fi. In this case, it is necessary to know fi at each and every set of P-T conditions of interest. Variable pressure standard state The standard state is the pure phase at the pressure and temperature of interest.

75

Under these conditions, fi = fi°, so a i = 1. The only way the activity of a solid deviates from unity under this standard state is when the solid is not pure, but is a solid solution. It may seem that the second standard state is easier to deal with in terms of pressure corrections. However, with the first standard state, the pressure correction is applied to fi, whereas in the second standard state, the correction is applied to µ i°. In either case, volume data for the constituent are required to make the correction.

76

AQUEOUS SOLUTIONS

The activities of solutes in dilute solutions are more closely approximated with Henry's Law than Raoult's Law. Thus, a somewhat different standard state is applied. We start with the equation expressing the difference in chemical potentials between two solutions with different molalities: µi&quot; - µ i' = RT ln (H&quot;m&quot;/ H'm') If we let one solution be the standard state, we can write: µi - µ i° = RT ln ( Hm)/(Hm)°

77

We then define the standard state to be the hypothetically ideal one-molal solution at the temperature and pressure of interest. Under these conditions H = 1 and m = 1, so ( Hm)° = 1, and we write: µi - µ i° = RT ln H m This somewhat strange standard state is necessary, because if we let the standard state be the infinitely dilute solution, we would have H = 1 and m = 0, so ( Hm)° = 0, which would result in an undefined value of: µi - µ i° = RT ln ( Hm)/ ( Hm)°

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