In chapter 7, we go from the behavior of simple single substance systems to mixtures of substances. In general, these will still be non-reacting substances and we will be dealing mostly with the energetics of the mixing process itself and to a lesser extent with the types of nonideal interactions which can take place between different substances. We will also start to learn how to deal with liquid solutions, which will greatly increase our arsenal of systems that we know how to deal with.
Chapter 7.1
In order to talk about mixtures, we must define some way of specifying the values of thermodynamic parameters for components of a mixture. Such a description is provided by the concept of a partial molar quantity. The formalism goes like this. If there is some parameter, X, that we wish to describe in terms of its partial value for each component of a solution then the value of X should just be the sum of the values of each of the contributions from the components. Of course, how much a particular component contributes to X will depend on how much of it is present. Thus we write: dX





X n
1

 dn
1





X n
2

 dn
2





X n
3

 dn
3

...
This just says that any change in X will be given by how much the amounts of components 1, 2, 3… are changed (dn i
) and by the coefficient that tells us what the amount of X is per mole of substance 1, 2, 3…,




X n i

 . These coefficients are called partial molar quantities. For small changes we can often integrate this expression and obtain:
X





X n
1

 n
1




X
 n
2

 n
2




X
 n
3

 n
3

...
Lets consider an example. One common partial molar quantity is the partial molar volume. The partial molar volume of component i is just given by V i




V
 n i


T , P , n
'
. Thus it is the amount that the volume would change if we changed the number of moles of substance i on a per mole basis. The simplest case is a mixture of ideal gases, call them
A and B. Note that in this case we hold T and P constant as well as the amounts of all other substances in the mixture (n'). We know that ideal gases do not interact and that their volumes depend only on how much is present at constant T and P:
V
 n
A
RT
P
 n
B
RT
P
. Thus



V
 n
A


T , P , n '

RT
P
. So it turns out that the partial molar volume of any component in a mixture of ideal gases is RT/P. It is not always that simple, but more on that later.
OK, now we come to the really important partial molar quantity -- the partial molar Gibbs free energy. Just as for volume, we can write:

G




G
 n
1


T , P , n ' n
1




G
 n
2


T , P , n ' n
2

...
But we recognize the coefficients in this case as chemical potentials:

1





G n
1


T , P , n '
,

2





G n
2


T , P , n '
,...
Thus
G
 
1 n
1
 
2 n
2

...
and dG
 
1 dn
1
 
2 dn
2

...
At constant temperature and pressure and under conditions where the chemical potentials are constant with changing composition. In fact, we can just add these new terms to our previous expression for dG at constant pressure and constant n: dG

Vdp

SdT
 
1 dn
1
 
2 dn
2

...
which reduces to the equation just above this if the temperature and pressure are constant.
An aside…
In principle, we could have written dG from the following equation differently
G
 
1 n
1
 
2 n
2

...
and dG
 
1 dn
1
 
2 dn
2

...
 d

1 n
1
 d

2 n
2

...
Yet from the arguments made above, we can see that this equation must be the same as the previous equation for dG: dG
 
1 dn
1
 
2 dn
2

...
Thefore
0
  d

1 n
1
 d

2 n
2

...
This is an important equation called the Gibbs-Duhem equation. What it says is that any change in the chemical potential of one component of a closed system must be balanced by another. This turns out to be a rather important fact. Don't worry about this particular equation (which we will not use much). Just realize that the values of individual chemical potentials in solution are intimately dependent on one another. The equations that we will need are the ones derived previously.
Chapter 7.2
Ok, enough talk. What happens, then, to the Gibbs free energy if we take two pure ideal gases and mix them together? Consider gas A and gas B, both in separate containers at pressure P. Before we mix them, we can say that their initial Gibbs free energy total is just the sum of the individual Gibbs free energies:

G i
 n
A

A
 n
B

B
 n
A

A
 
RT ln
P
P

 n
A

A
  n
A
RT ln
P
P

 n
B

B
 
RT ln
P
P

 n
B

B
  n
B
RT ln
P
P

Now lets mix them. As you can see in the diagram, both A and B now occupy the total volume of the system.
Thus, the partial pressures of each are decreased. Therefore we have:
G f
 n
A

A
 n
B

B
 n
A



A
 
RT ln
P
P
A



 n
B



B
 
RT ln
P
P
B



 n
A

A
  n
A
RT ln
P
P
A

 n
B

B
  n
B
RT ln
P
P
B
 where P
A
+ P
B
= P. You can see that the ratio term in the logs both got smaller, thus the entire value of the equation got smaller. This means that G f
< G
I
which is what is expected for a spontaneous process. Just how big is the free energy change of mixing?
Lets calculate

G:

G mix

G f

G i
 n
A

A
  n
A
RT ln
P
A
P

 n
B

B
  n
B
RT ln
P
B
P

 n
A

A
  n
A
RT ln
P
P

 n
B

B
  n
B
RT ln
P
P


 n
A
RT
G mix
 ln
RT
P
A
P



 n
A n ln
B
RT
P
P
A ln
 n
B
P
B
P
 ln

P
B
P n
A


RT ln
P
P

 n
B
RT ln
P
P

Sometimes it is convenient to write this in terms of mole fractions:

X
A
 n
A

P
A n P n
B X
B
 
P
B n P
Which leads to:

G mix
 nRT

X
A ln X
A

X
B ln X
B

These very simple ideas are extremely important. They tell us that simply by allowing different substances to mix together, the free energy of the system will drop (assuming that no negative interactions exist between the molecules -- we will explore that later).
Remember the above results are only exact for ideal gases .
An obvious question at this point is, what happens if we mix two flasks with the same type of gas together at the same initial pressure? As you can see from the equations above, if the final mole fraction of the gas is 1.0, then the equation goes to zero. But why does it matter what the identity of the ideal gas is? This is interesting. It has to do with the fact that the driving force for mixing is actually the entropy change of the system upon mixing (see below). Mixing two buckets of balls together that are indistinguishable
(have all properties in common) does not change the amount of disorder. But if you simply paint one bucket of balls red and the other green and then mix them, they become less ordered (the red balls are all mixed up with the green ones). It is probably not obvious to you why this changes the free energy. Lets try another example and perhaps it will make sense. In the picture here we have two different sized balls.
The little ones can pass through the paddle and the big ones cannot. Because of this, the system can do work in exchange for mixing. However, if both sides had identical types of balls, it could not do work -- no change in Gibbs free energy. Anytime particles can be physically distinguished, there is the possibility of using those differences as big and little where used in this example. However, the mixing of indistinguishable particles be coupled to performing work under any conditions.

We can also calculate the entropy and enthalpy of mixing for an ideal gas. Let's start with entropy. Remember what we are doing is to let all the A molecules occupy more space and all the B molecules occupy more space at constant pressure. Thus we have:

S
A
 n
A
R ln
V f
V i
  n
A
R ln
P
A
P

S
B
 n
B
R ln
V f
V i
  n
B
R ln
P
B
P

S mix
 
S
A
 
S
B
 
R

 n
A ln
P
A
P
 n
B ln
P
B
P


  nR

X
A ln X
A

X
B ln X
B

This looks awfully familiar. If we multiplied it by -T, we would get the Gibbs free energy. Thus for mixing of a two ideal gases:

G mix

H mix



T

S
0 mix
The enthalpy being zero can be seen simply by comparison of the equation above it to

G =

H - T

S
Finally, if you want to know what happens with the gases a start at different initial pressures, see Example 7.2 in the book (you just have to consider the value of the pressures of the two gases relative to one another).
Question you should now be able to answer: at what ratio of amounts of two ideal gases is the Gibbs free energy of mixing going to be the most negative (sounds like a good test problem to me)?
Chapter 7.3
Our next job is to consider systems other than mixtures of ideal gases. The first one we will look at is ideal mixtures of liquids. Basically, an ideal solution or mixture of liquids is one where the vapor pressure of the liquid A is changed in proportion to its mole fraction when liquid A is mixed with liquid B. This will be true if the interactions between A and B molecules in the liquid is essentially the same as the interactions between A and A molecules and between B and B molecules.

Vapor pressure is the pressure due to A that would be generated if the gas molecules and the liquid molecules were allowed to come to equilibrium in a closed system. In fact, this is the key point, equilibrium. At equilibrium, the chemical potentials of the two phases of
A (liquid and gas) are equal.
First, lets consider this for a pure liquid, A (the book uses an * to denote a pure substance):

*
A
  
A

RT ln


P
P
A

*


 
*
A
So at equilibrium, we have a simple way of getting the chemical potential for A in a liquid state. Now, if we mix two liquids together, the vapors will also mix in equilibrium above the liquid and the chemical potential of the vapor of A will depend on the partial pressure of A, just as before when we mixed ideal gases. We then have:

A
  
A

RT ln
P
A
P

 
A
Next I am going to consider the difference in going from a pure liquid A to a mixture by subtracting the last two equations:

A
 
 
*
A
 
  
A

RT ln


P
P
A



  
A

RT ln


P
P
A

*



A
 
  *
A
 

RT ln
P
A
P


RT ln


P
A
*
P




A
 
 
*
A
 

RT ln


P
A
P
A
*


Notice something important -- we have done away with the standard chemical potential of the gas and have in its place as a standard state the chemical potential of the pure liquid A.

Now, if the solution or mixture of liquids is ideal, as described previously, then the vapor pressure of A above the liquid will just be proportional to the molar fraction of A in the liquid. This is an important point and is called Raoult's Law :
P
A

P
A
*
X
A
Remember that here P
A
* is the vapor pressure of A above a pure liquid A. In order to understand this at a molecular level, let's take a look at a tutorial .
Using this equation now, we can substitute:


A
A
 
 
*
A
 
   
*
A
 


RT
RT ln ln


X
A
P
A
*
 
A
P
A
*


See what we have now done -- all of the variables that depended on the vapor are gone and we are left only with parameters that are defined in the liquid phase. Also, since we have everything defined in terms of the liquid, we no longer need to worry about whether we are in equilibrium with the gas phase or not, since the properties shown are independent of the state of the vapor. This is a general expression for the chemical potential of some component A in an ideal solution of liquids.
Ok, so the world is not an ideal place. The interactions between like components of the solution are not always the same as the interactions between different components, and
Raoult's law does not work very well. In fact, there are relatively few situations for which Raoult's law is really very good. What do we do about that? As always, reality calls for a good fudge factor. We throw an extra empirical factor into Raoult's law and call it Henry's law.
P

A
K
A
X
A
Here K
A
is the effective vapor pressure of component A when it is surrounded entirely by component B. Our chemical potential would now be:

A
 
 
*
A
 

RT ln


X
A
P
K
*
A
A


We are still OK, because everything is still in terms of properties of the liquid (including the vapor pressure of the pure liquid). Note that Henry's law works well in general for very dilute solutions. Why? Because when A is in very low concentrations, it is surrounded almost entirely by B. Thus, it's enviroment is homogeneous -- it acts rather like a simple pure material again, even though it is in a mixture.

So, we use Henry's law to describe dilute solutions of liquids or (as explained below) solutes. These kind of dilute solutions are called ideal-dilute solutions .
Chapter 7.4
At this point calculating the Gibbs free energy of mixing for ideal liquid solutions is a nobrainer. Remember for ideal gases that the chemical potential was given by:

A
 
A
 
RT ln
P
P
A
 and the Gibbs free energy of mixing was

G mix
 nRT

X
A ln X
A

X
B ln X
B
 where the mole fractions are mole fractions in the gas phase.
Well for the liquid, we have

A
 
 
*
A
 

RT ln
 
A where the mole fraction is now in the liquid phase, and it should not surprise you to learn that the Gibbs free energy of mixing is just

G mix
 nRT

X
A ln X
A

X
B ln X
B

In other words, the Gibbs free energy of mixing two ideal gases is the same as the Gibbs free energy for forming an ideal solution of two liquids. For more components, you just add more terms. The entropy and enthalpy of mixing are also the same as with ideal gases.

S mix


G mix

H mix


 nR

X
A

T

S mix
0 ln X
A

X
B ln X
B

Remember, these equations only apply to ideal solutions! In real situations, the enthalpy, in particular, is usually not zero.
Let's just see what effects a mildly nonideal case would have on these (in the book, this is partly covered at the end of the chapter under activities). We can see that for ideal dilute solutions (Henry's law) instead of just X
A
in the logarithm, we would have something proportional to X
A
. As we saw before:


A
 
 
*
A
 

RT ln


X
A
K
A


P
A
* so for two components, we first consider them separately:
G i
 n

A
 n
A
A
*


A n

B n
B

B

*
B
Now for a ideal dilute mixture where there is just a wee bit of A mixed in with a lot of B:
G f
 n
A

A
 n
B

B
 n
A

*
A

RT ln


X
A
K
A
P
A
*


 n
B

*
B

RT ln
 
B
 n
A

A
*  n
A
RT ln
X
A
K
A
P
A
*
 n
B

B
*  n
B
RT ln X
B
 n
A

A
*  n
A
RT ln
K
A
P
A
*
 n
A
RT ln X
A
 n
B

B
*  n
B
RT ln X
B
Now subtract the initial from the final Gibbs free energy:

G mix

G f

G i
 n
A

A
*  n
A
RT ln
K
A
P
A
*
 n
A
RT ln X
A
 n
B

B
*  n
B
RT ln X
B
 n
A

A
*  n
B

B
*

G mix

G mix
 n
A
RT ln
K
A
P
A
*
 nRT

 X
A ln

K
A
P
A
* n
A
RT

X
A ln ln
X
A
X
A
 n
B
RT

X
B ln ln X
B
X
B

 or

G mix
 nRTX
A ln
K
A
P
A
*
 nRT

X
A ln X
A

X
B ln X
B

How this extra term partitions itself between enthalpy and entropy depends on the details of the liquid interactions and structure. However, usually the bulk of the extra term ends up making the enthalpy of mixing nonzero.
Chapter 7.5
Ok, let's go back to simple, ideal mixtures of liquids for awhile and discuss various observable effects that change upon mixing liquids. Since we are considering the effects of changing the numbers of different molecules in solution, but not interactions (ideal solutions), these effects are effects that depend strictly on numbers of molecules and are thus called colligative properties (properties of a collection of molecules). For everything we talk about below, we will assume that we have some solvent A with a little bit of some solute B in it. We will further assume that B is nonvolital (does not contribute to the vapor pressure above the liquid) and does not incorporate itself into any solid which is formed upon freezing.

The first and most obvious of these is boiling point. The boiling point of a liquid is when the vapor pressure of the liquid is equal to the vapor pressure of the surrounding atmosphere. If we add something to the liquid which decreases its mole fraction below one, we have from Raoult's law that P
A
= X
A
P
A
*. Thus, the vapor pressure of A in the mixture will be lower at any particular temperature than pure A. Thus the temperature required to achieve the point where the vapor pressure of the mixture is equal to that of the atmosphere will be higher in the presence of some solute B:
Another way to think about this, which is more general, is in terms of chemical potentials. The phase transition (boiling) will occur when the chemical potentials of the liquid and vapor phases are equal. But if we add something to the system, as we have seen before, the chemical potential of the liquid drops. Since the chemical potential of the ideal mixture is:

A
 
 
*
A
 

RT ln
 
A when X
A
is less than 1.0, the chemical potential of the liquid drops and thus T will have to be greater before it reaches the point where the chemical potential of the liquid solution is greater than the chemical potential of the pure vapor (at which point the liquid boils). Note that the chemical potential of the vapor is not directly affected by addition of
B since B is nonvolital. In the book, it shows how you can manipulate this formula to determine the change in the boiling point as a function of the amount of solute added.
The result is

T

KX
B

2
RT vap

H vap
X
B
An almost identical argument can be made for the freezing point. Again, our solute only exists in the liquid phase and so it lowers the chemical potential for the liquid by not for the solid. The liquid in the presence of the solute is stabilized relative to either the gas or the solid phases. Thus it is harder to go to the gas (higher boiling temperature, see above) and harder to go from liquid to solid (lower freezing temperature). The equation describing the decrease in the temperature of fusion (the freezing or melting temperature) upon addition of solute is essentially identical to the one for boiling point elevation above:

T

K ' X
B

2
RT fus

H fus
X
B

Note, however, that the book has chosen to define

T differently in the two cases . This is very confusing (I do not know why they do this, but we are stuck with it). You simply have to remember that the change in temperature for boiling is an increase while that for freezing is a decrease . Let's consider an example of freezing point depression in a tutorial .
The next property is maximum solubility (this is not actually a colligative property, but it fits in well here from the point of view of technique). Many times our solutes are not other liquids, but small amounts of some solid. If we add a great deal of a solid to a solvent and let them come to equilibrium, the chemical potentials of the solute in the solid and liquid should be the same. Let's call the solvent A and the solute B:

*
B
 
 
*
B
 

RT ln
 
B
Now X
B
is always less than zero, so by making the mole fraction of the solute in the solvent low, we decrease the right side of the equation. If something is a solid at room temperature, that means that the chemical potential of its solid form is less than that of the liquid form. Effectively, however, in the presents of some solvent, we are lowering the chemical potential of the liquid form of the solute (the right side of the equation).
The amount of B that can dissolve just depends on the mole fraction that makes the chemical potentials of the liquid and solid balance. The book goes through such an argument and comes to the conclusion that we can express the mole fraction of solute present at saturation as: ln X
B
 

H
R fus
1
T

1
T fus
Don't get too excited about this equation. It is darn near useless is reality because it does not take any solute/solvent interactions into account. However, it allows you to make rough guesses about how the solubility will depend on temperature for small temperature changes, and it is another good example of balancing chemical potentials. We won't use this one very much.
The final colligative property is Osmosis. This is probably the most obvious example of a colligative property and it appears in many applications. Also, the equations tend to work pretty well over limited ranges of concentrations and temperatures. The picture below tells the story:

We have a liquid partitioned in two volumes. The two volumes are separated by a membrane that will let the solvent through, but not the solute (could have done it the other way around as well). To one side we add some solute, the other side is pure solvent. The side with the solute has a lower chemical potential. Thus, solvent from the pure solvent side will migrate towards through the membrane towards the side that has a lower chemical potential. The osmotic pressure is defined as the amount of pressure (in addition to atmospheric pressure) that one would have to apply to the side that has the solute in it in order to keep more solvent from going over to that side. Again, by balancing the chemical potentials, one can show that the pressure required is just:
 
 
RT where [B] is the molarity (number of moles per liter) of solute in the solvent.
Chapter 7.6
Henry's law was one way of dealing with nonidealities, but really only works for very dilute solutions. We can more generally do the same thing by putting a fudge factor in front of the mole fraction in our chemical potential equation. Just as fugacity was a sort of effective pressure, we can also have effective mole fractions or effective concentrations (see below). For liquids we give these the symbol 'a' and call them activities. Thus, we can write:

A
 
  *
A
 

RT ln
 
A where the activity is just whatever it has to be in order to make the equation work. We can then write in terms of the activity coefficient,

, a
A
 
A
X
A which can be plugged in to give:

A
 
  *
A
 

RT ln
 
A

RT ln
 
A
This is just like fugacity. We get our ideal equation with some extra additive term that contains nasty realities in it. This term we consider to be empirical (although we will

attempt to calculate it later for ions) and we write it down in a book somewhere. If you work back through the equations that we used to derive our chemical potential for a liquid mixture, you will find that we can also express the activity as: a
A

P
P
A
*
A where the top term is the actual vapor pressure of A when mixed with something else and the bottom term is the vapor pressure of pure A. This makes the activity coefficient simple to measure experimentally for solvents.
For ideal dilute solutions, remember we had (we will now continue to call the solute B and the solvent A, for clarity):

B
 
 
*
B
 

RT ln


X
B
P
K
*
B
B

 or

B
 
  *
B
 

RT ln


K
B
P
B
*



RT ln
 
B people often lump the first two terms together and get:

B
 
 
1
B
 

RT ln
 
B
We can now consider the effects of activity on the dilute solutions by writing:

B
   
1
B
  
RT ln
 
B
Note that the activity in this case approaches the mole fraction as the concentration of the solute approaches zero (the Henry's law dilute-ideal condition).
Most of the time, we would rather deal with either molarity or molality than mole fraction. Molarity is moles per liter of solution. Molality is moles per kg of solution.
For precise work, we deal with molality, since volume depends on temperature (but mass does not). We can put the equation above in terms of activities based on molality. If you pick a standard state (usually 1 mole per kg solvent) and you consider the ratio m
B
/m
0
, where m
B
is the molality of B and m
0
is the standard molality, then we can rewrite our ideal chemical potential for the solute A as:

B
   
1
B
  
RT ln
 m
B m


B
      
B

RT ln m m
B
 where now our new

A
0 is just the chemical potential of A in the standard state (usually 1 molal). We can then define an activity for this as well: a
B
 
B m
B m
 giving:

B
 
  
B
 

RT ln
 
B

One of the important things in this chapter has been the definition of standard states .
Remember that the standard state of a gas was 1 bar, the standard state of a solvent was the pure solvent and the standard state of a solute is 1 molal (1 mole solute per kg solvent). While these definitions are arbitrary, they effect everything we do from here on.