Thermodynamics of simple solutions

A solution is a homogeneous mixture, i.e., a solution is a one-phase system containing more than one component.

The phase may be solid, liquid, or gas.

1

Thermodynamics of simple solutions Some ways of specifying solution composition (1) Mole fraction

xi ≡

ni ntot

(2) Molar concentration (molarity) n ci ≡ i V (3) Molality n ni mi ≡ i = wA n A M A Mass of solvent (kg)

Moles of solvent x molar mass of solvent (kg/mol) 2

1

Thermodynamics of simple solutions Mixing quantities For an extensive thermodynamic quantity X the change in X on mixing is given by

∆ mix X = X − X * where X represents the value of X for the mixture/solution, and X* represents the value of X for the pure components of the solution. Ex.:

∆ mixV = V − V ∗ ∆ mix G = G − G ∗

Similarly for ∆ mixU , ∆ mix H , ∆ mix S etc.

3

Thermodynamics of simple solutions Liquid mixtures and vapour pressure Consider a pure liquid A which has been placed in an initially evacuated container, and allowed to attain equilibrium with its vapour. Pure vapour A

µ A∗,vap

µ A∗,liq Pure liquid A

For phase equilibrium µ ∗A,liq = µ A∗,vap

4

2

Thermodynamics of simple solutions Liquid mixtures and vapour pressure Pure vapour A

µ A∗,vap

µ

∗,liq A

=µ

∗, vap A

µ A∗,liq

Suppose the vapour pressure of pure liquid A is pA*.

Pure liquid A

From eqn 5 the chemical potential of A in the vapour phase:  ∗  po    p∗  = µ Ao,vap + RT ln  A o  p  

µ A∗,vap = µ Ao,vap + RT ln  p A

∴µ

∗,liq A

16 5

Thermodynamics of simple solutions Liquid mixtures and vapour pressure Suppose we now repeat the experiment, but suppose that this time we place a sample of a solution which contains both A and a second component B in the originally evacuated container. Vapour mixture A+B

µ Avap

µ Aliq Liquid solution A+B

For phase equilibrium µ Aliq = µ Avap

Suppose the partial vapour pressure of component A is pA.

6

3

Thermodynamics of simple solutions Liquid mixtures and vapour pressure Vapour mixture A+B

µ Avap

µ =µ liq A

vap A

µ Aliq Liquid solution A+B

Suppose the partial vapour pressure of component A is pA.

From eqn 6 the chemical potential of A in the vapour phase µ Avap = µ Ao,vap + RT ln  p A µ Aliq

 p   p  = µ Ao,vap + RT ln  A o  p   o

17 7

Thermodynamics of simple solutions Liquid mixtures and vapour pressure o, vap :− Eqn 17 – eqn 16, and cancelling out µi

µ −µ liq A

∗,liq A

 p ∗A   pA  = RT ln  − RT ln  p o  p o    p  µ =µ = RT ln  A ∗  pA   µ =µ

i.e.,

 

µ Aliq = µ A∗,liq + RT ln  p A

 p  ∗ A

∗ ,liq A

o , vap A

 p∗  + RT ln  A o  p  

liq A

o , vap A

p  + RT ln  A o  p  

17

An assumption we have made throughout is that the vapour phase8 is ideal.

4

Thermodynamics of simple solutions Raoult’s Law In a series of experiments on mixtures of closely related liquids (like benzene and methylbenzene), the French chemist Raoult found that the ratio of the partial vapour pressure of each component to its vapour pressure as a pure liquid, pA/pA*, is approximately equal to the mole fraction of A in the liquid mixture. This observation is now called Raoult’s Law. Equilibrium partial vapour pressure of A in the vapour phase in equilibrium with the liquid phase.

p A = x A p ∗A

Vapour pressure of the pure liquid A.

Mole fraction of A in the liquid phase

9

Thermodynamics of simple solutions Raoult’s Law For a liquid mixture consisting of two components A and B, Raoult’s Law can be plotted on a diagram of partial pressure vs mole fraction: ptot = p A + pB = xA p∗A + xB pB∗ Vapour pressure

pB∗

•

p∗A

p A = x A p∗A

•

pB = xB pB∗

0 1

xA xB

1 0

10

5

Thermodynamics of simple solutions Ideal solutions Mixtures which obey Raoult’s Law throughout the composition range from pure A to pure B are called ideal solutions. Ideal mixtures or solutions are formed when the molecules of the solution components are so similar to one another that molecules of one component can replace molecules of the other component in the solution without changing the solution’s energy or spatial structure. So, for a mixture or solution of two species A and B:11

Thermodynamics of simple solutions Ideal solutions (i)A molecules must be of approximately the same size and shape as B molecules (avoiding a change in spatial structure of the solution). (ii)Molecular interaction energies must be essentially the same for A↔A, A↔B, and B↔B interactions (avoiding a change in the system’s energy). Ex.: The best examples of ideal solutions are isotopic mixtures: 12CH I and 13CH I. Other pairs of substances which form nearly ideal 3 3 solutions are: C2H5Cl / C2H5Br, n-C7H16 / n-C8H18, and C(CH3)4/Si(CH3)4. 12

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Thermodynamics of simple solutions Ideal solutions A short while ago we saw (eqn 17) that, for any liquid mixture, assuming an ideal vapour   µ Aliq = µ A∗,liq + RT ln  p A ∗  pA   For an ideal solution, we can substitute from Raoult’s Law

pA = xA p ∗A

to obtain

µ Aliq = µ A∗,liq + RT ln x A

18

Equation 18 constitutes a thermodynamic definition of an 13 ideal solution.

Thermodynamics of simple solutions Properties of ideal solutions 1. An ideal solution obeys Raoult’s Law for the entire range of component mole fractions. 2. For the process in which an ideal solution is formed from its pure components: n1A1 + n 2 A 2 + ⋅⋅⋅ + n i Ai + ⋅⋅⋅⋅ → ideal solution (n1 , n 2 , ....... n i , ........)

∆ mix G = Gmixture − G pure components = n1Gm ,1 + n2Gm,2 + ⋅⋅⋅ + n j Gm , j + ⋅⋅⋅ − n1Gm∗ ,1 − n2Gm∗ ,2 − ⋅⋅⋅ − n j Gm∗ , j − ⋅⋅⋅ = n1µ1 + n2 µ 2 + ⋅⋅⋅ + n j µ j + ⋅⋅⋅ − n1µ1∗ − n2 µ 2∗ − ⋅⋅⋅ − n j µ ∗j − ⋅⋅⋅ = n1 ( µ1 − µ1∗ ) + n2 ( µ2 − µ 2∗ ) + ⋅⋅⋅ + n j ( µ j − µ ∗j ) + ⋅⋅⋅ = ∑ ni ( µi − µi∗ )

19

14

i

7

Thermodynamics of simple solutions Properties of ideal solutions (contd.) : ∆mixG For any component of an ideal solution we have that, by definition

µi = µi∗ + RT ln xi

Substituting into the previous equation ∆ mix G = Gmixture − G pure components = ∑ ni ( µi − µi∗ ) i

∆ mix G = ∑ ni RT ln xi

20

i

We see that ∆mixG is a function of no. of moles and mole fractions only, and does not depend on the chemical nature of the components. Since 15 xi ≤ 1, ∆mixG ≤ 0 (as expected).

Thermodynamics of simple solutions Properties of ideal solutions (contd.) : ∆mixS From the Gibbs equation for the Gibbs function, we have that dG = Vdp − SdT  ∂G    = −S  ∂T  p

 ∂G  ∴S = −   ∂T  p

 ∂∆ G  ∆ mix S = −  mix  = − R ∑ ni ln xi  ∂T  p ,ni i

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We see that ∆mixS is a function of no. of moles and mole fractions only, and does not depend on the chemical nature of the components. Since 16 xi ≤ 1, ∆mixS ≥ 0 (as expected).

8

Thermodynamics of simple solutions Properties of ideal solutions (contd.) : ∆mixH Since, for any chemical process occurring at constant T ∆G = ∆H − T ⋅ ∆S

we have that ∆ mix G = ∆ mix H − T ⋅ ∆ mix S ∆ mix H = ∆ mix G + T ⋅ ∆ mix S = RT ∑ ni ln xi − RT ∑ ni ln xi i

i

∆ mix H = 0

22

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Thermodynamics of simple solutions Properties of ideal solutions (contd.) : ∆mixV  ∂G  From the Gibbs dG = Vdp − SdT ⇒ ∴   =V equation for G ∂ p  T Rewriting in terms of partial molar quantities, we have that, for component i of a liquid mixture:  ∂Gi    = Vi  ∂p T

or

 ∂µi    = Vi  ∂p T

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For component i of a liquid mixture we have that (eqn 17):

 

µi = µi∗ + RT ln  pi

 p  ∗ i

24 18

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Thermodynamics of simple solutions Properties of ideal solutions (contd.) : ∆mixV Differentiating with respect to total pressure p at constant temperature T:   pi    ∂ ln  p∗    ∂µi   ∂µi∗     + RT    = ∂p    ∂p T  ∂p T    T   pi    ∂ ln  p∗     Vi = Vi * + RT  ∂ p      

For an ideal liquid solution, by Raoult’s Law:

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Thermodynamics of simple solutions Properties of ideal solutions (contd.) : ∆mixV  ∂ ln xi  Vi = Vi * + RT    ∂p T

The mole fraction xi of components of the liquid mixture are not a function of pressure  ∂ ln xi  ∴  =0  ∂p T

∴Vi = Vi ∗

25

Before mixing, the combined volume of the pure liquids:

V pure liquids = nAVA∗ + nBVB∗ + ⋅⋅⋅⋅

20

10

Thermodynamics of simple solutions Properties of ideal solutions (contd.) : ∆mixV After mixing, the volume of the solution is given by:

Vmixture = n AVA + nBVB + ⋅⋅⋅⋅ The volume change accompanying the mixing process: ∆ mixV = Vmixture − V pure components

∆ mixV = nA (VA − VA∗ ) + nB (VB − VB∗ ) + ⋅⋅⋅⋅ ∆ mixV = 0

21

Thermodynamics of simple solutions Properties of ideal solutions (contd.) : ∆mixU For the mixing process occurring at constant pressure, from the definition of enthalpy ∆ mix H = ∆ mixU + ∆ mix ( pV ) ∆ mix H = ∆ mixU + p ( ∆ mixV )

∆ mixU = ∆ mix H − p ( ∆ mixV ) = 0 − p ( 0)

∆ mixU = 0 22

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Thermodynamics of simple solutions Properties of ideal solutions (contd.) : Summary Collecting together all the properties we have derived for an ideal solution: ∆ mixU = 0

∆ mixV = 0 ∆ mix H = 0 ∆ mix S = − R ∑ ni ln xi i

∆ mix G = ∑ ni RT ln xi i

Do these results make sense in terms of our molecular interpretation of an ideal mixture?

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Thermodynamics of simple solutions Properties of ideal solutions (contd.) : Summary (a)The fact that ∆mixU = 0 accords well with the view that molecular interactions in solution are indistinguishable from interactions in the pure separated liquids. Under these circumstances one would not expect to get an energy change when replacing a molecule of one constituent with a molecule of another constituent. (b)The fact that ∆mixV = 0 accords well with the view that the molecules of the various components in solution are very similar in shape and size. This, together with indistinguishable intermolecular forces, would be expected to lead to no volume change on mixing. (c)From the relations obtained for ∆mixH, ∆mixS, and ∆mixG it is apparent that the formation of a ideal solution is an entropy 24 driven process.

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Thermodynamics of simple solutions Solutes and Solvent Sometimes it is convenient to call one component of a solution the solvent and the other component/s of the solution the solutes. • The solvent is invariably the major component of the solution. We shall denote the solvent by the letter “A”. • The solute/s is/are the minor component/s of the solution. We shall denote the solute/s by the letter “i”. When two liquids are miscible in all proportions we usually do not distinguish between solutes and solvent, for obvious reasons. 25

Thermodynamics of simple solutions Deviations from Raoult’s Law Solutions that obey Raoult’s Law are called ideal solutions. Solutions which deviate from Raoult’s Law are termed nonideal or real solutions. Nonideal solutions are formed when the molecules of the solution components differ from one another in shape and size and in their intermolecular forces. There are two kinds of deviations from Raoult’s Law, i.e., positive and negative deviations from Raoult’s Law. 26

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Thermodynamics of simple solutions Positive deviations from Raoult’s Law Ex.: a mixture of acetone and carbon disulphide ∗ • pCS

2

p ∗Acetone •

0

xCS2 →

1

27

Thermodynamics of simple solutions Positive deviations from Raoult’s Law Ex.: a mixture of acetone and carbon disulphide ∗ • pCS

2

p ∗Acetone •

0

xCS2 →

1

28

14

Thermodynamics of simple solutions Positive deviations from Raoult’s Law Ex.: A mixture of acetone and carbon disulphide

ptot p ∗Acetone •

∗ • pCS

2

pCS2 p Acetone

0

xCS2 →

1

29

Thermodynamics of simple solutions Detailed examination of experimental data shows that, at the extremes of the composition range, the partial vapour pressure of the minor component (the solute) is directly proportional to its mole fraction:

ptot p∗Acetone •

∗ • pCS

2

pCS2 p Acetone

0

xCS2 →

1

30

15

Thermodynamics of simple solutions In sufficiently dilute solutions the minor component/s of a solution (the solute/s) obey Henry’s Law: pi = kixi . ptot p ∗Acetone •

∗ • pCS

2

pCS2 p Acetone

Henry’s Law obeyed

Henry’s Law obeyed

0

xCS2 →

1

31

Thermodynamics of simple solutions In sufficiently dilute solutions the major component of a solution (the solvent) obeys Raoult’s Law: pi = pi*xi . ptot p∗Acetone • Raoult’s Law obeyed

∗ • pCS

2

Raoult’s Law obeyed

pCS2 p Acetone

Henry’s Law obeyed

Henry’s Law obeyed

0

xCS2 →

1

32

16

Thermodynamics of simple solutions Positive deviations from Raoult’s Law Can we explain why a mixture of acetone and carbon disulphide displays positive deviations from Raoult’s Law? To answer this question we must consider the intermolecular forces acting within the two pure liquids and within a mixture of the two components. Carbon disulphide Acetone Mixture H3C

Acetone molecules will interfere with interactions between CS2 molecules, and viceversa.

O

H3C Dipole-dipole & London (dispersion) forces

S] [S

C

[S S]

London (dispersion) forces

Mixture will be destabilized relative to the pure liquids

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Thermodynamics of simple solutions Negative deviations from Raoult’s Law Ex.: A mixture of acetone and chloroform

p ∗Acetone • ∗

•pCHCl

3

0

xCHCl3 →

1

34

17

Thermodynamics of simple solutions Negative deviations from Raoult’s Law Ex.: A mixture of acetone and chloroform

p ∗Acetone • ∗ • pCHCl

3

0

xCHCl3 →

1

35

Thermodynamics of simple solutions Negative deviations from Raoult’s Law Ex.: A mixture of acetone and chloroform

p ∗Acetone • ∗ • pCHCl

ptot

3

p Acetone

pCHCl3

0

xCHCl3 →

1

36

18

Thermodynamics of simple solutions Once again, experimental data show that, at the extremes of the composition range, the partial vapour pressure of the minor component (the solute) is directly proportional to its mole fraction.

p ∗Acetone • ∗ • pCHCl

ptot

3

p Acetone

pCHCl3

0

xCHCl3 →

1

37

Thermodynamics of simple solutions Once again, in sufficiently dilute solution, the minor component/s of the solution (the solute/s) obeys Henry’s Law: pi = kixi . p ∗Acetone • ∗ • pCHCl

ptot

3

p Acetone

pCHCl3 Henry’s Law obeyed

0

xCHCl3 →

Henry’s Law obeyed

1

38

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Thermodynamics of simple solutions In sufficiently dilute solutions the major component of a solution (the solvent) obeys Raoult’s Law: pi = pi*xi . Raoult’s Law p ∗Acetone • obeyed ∗ Raoult’s Law • pCHCl obeyed

ptot

3

p Acetone

pCHCl3 Henry’s Law obeyed

0

Henry’s Law obeyed

xCHCl3 →

1

39

Thermodynamics of simple solutions Negative deviations from Raoult’s Law Can we explain why a mixture of acetone and chloroform displays negative deviations from Raoult’s Law? Again, we must consider the intermolecular forces acting within the two pure liquids and within a mixture of the two components. Acetone H3C

O

H3C Dipole-dipole & London (dispersion) forces

Mixture H3C H3C

Cl

Cl

O⋅⋅ ⋅H

Cl H-bonding & London forces

Mixture will be stabilized relative to the pure liquids

Chloroform Cl H

Cl Cl

London (dispersion) forces 40

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Thermodynamics of simple solutions Ideally dilute solutions Any solution that is sufficiently dilute so that the solvent obeys Raoult’s Law and the solute/s obey Henry’s Law is called an ideally dilute solution. Notes: 1. An ideally dilute solution is a very different thing from an ideal solution. 2. All non-electrolyte solutions display ideally dilute solution behaviour when they are sufficiently dilute.

If we look at the vapour pressure behaviour of the mixture of acetone and carbon disuphide we can now identify the composition ranges over which the solution formed is 41 ideally dilute.

Thermodynamics of simple solutions Ideally dilute solutions ptot p ∗Acetone • Raoult’s Law obeyed

∗ • pCS

2

Raoult’s Law obeyed

pCS2

p Acetone

Henry’s Law obeyed

Henry’s Law obeyed

0

xCS2 →

1 42

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