Chapter 9 Lecture Notes
Ok, after only eight chapters of thermodynamics we finally get to run a few reactions.
The first issue we need to deal with is the definition of the change in the Gibbs free
energy for a reaction.
Chapter 9.1
In order to define the change in Gibbs free energy for a reaction, we must first come up
with some quantitative measure of how far along the reaction is. We call this the extent
of the reaction, . The change in extent of reaction is easy to define for a simple reaction
like AB. It is simply the number of moles of A that has been converted to B. For a
simple reaction like this, d has the same value as dn and so we can write that dG =
AdnA + BdnB = -Ad + Bd. The minus sign comes from the fact that when the
reaction goes in the left to right direction, the amount of A is decreasing, while the
amount of B is increasing. Looking at these equations, it is reasonable to suggest that:
 G 

   B   A



 P,T
This is the slope of the free energy with respect to the extent of the reaction. This
relationship will have one region where the sign is negative, one point where the value is
zero and one region where the value is positive. If we look at a plot of G as a function of
 G 
  0 is the minimum of the curve. This is the
, we can see that the point where 
   P ,T
point where the chemical potentials of A and B
were the same and A and B are in equilibrium with
one another. On one side of the minimum, the slope
is negative and on the other side, the slope is
positive. In any case, the reaction is spontaneous
(dG is negative) as long as we are moving towards
the minimum, equilibrium position. Note that rG
is a molar quantity -- it has units of Joules per mole
just like the chemical potential. All reactions move
towards equilibrium. The question is, how do we
know where the equilibrium position is?
In order to understand how we can find the
minimum and what the Gibbs free energy of a reaction depends on, let's first start with a
reaction that converts one ideal gas into another. OK, you say, what do you mean
converts one ideal gas into another -- ideal gases do not undergo reactions. Well, that is
true. But we can consider the reactants and the products as ideal gases even if the
somewhere in between a reaction occurred. Ok, let's start out by saying that the reaction
enthalpy is zero. At constant pressure, this means that no heat is given off or taken up by
the system. Thus, the only thing that determines what the ratio of product to reactant
should be is the entropy term. We already said that the entropy term (the mixing term) is
the most favorable when the mixture is half and half.
Gmix  nRT  X A ln X A  X B ln X B 
Remember that the Gibbs free energy of mixing is not a molar quantity and depends on n
(unlike the reaction Gibbs free energy). Also, the Gibbs free energy of mixing is defined
relative to pure A and B. Anyway, by playing with the numbers a bit in the above
equation, you can show yourself that the free energy of mixing is most negative when the
amounts of A and B are the same.
Thus, if we have a reaction AB and there is no
enthalpy term, we would expect the system to have the
minimum Gibbs free energy when the mole fraction of
A and B are each 0.5. Think about it -- is A can form B
and B can form A and there is no other forces involved
(no heat/enthalpy) that would favor one product over
the other, probability would just state that eventually
you will have statistically the same amount of both A
and B present -- this is the lowest free energy state. (This, of course, will not be true if
the enthalpy for the reaction is not zero). Ok, but usually there is an enthalpy for the
reaction. What then? Lets work through this for an ideal gas reaction:
 r G   B   A   B  RT ln
PB
P
  A  RT ln A

P
P
PB
PA
What we normally do at this point is to give the first two terms a special name. Since it is
the difference between the chemical potentials at standard conditions, we refer to it as the
Gibbs free energy of reaction at standard conditions or the standard Gibbs free energy of
reaction.
 r G    B   A
  B   A  RT ln
 r G   r G   RT ln
PB
PA
We can see that there are two parts to the equation.
One part is just the Gibbs free energy for converting
one mole of A to B at standard conditions (1 bar of
both A and B). The second term is our mixing term
which goes through zero when the amounts of A and B
are equal to one another. The first term you look up in
a book and is specific for a specific reaction. The
second term you calculate knowing the partial
pressures of A and B in the gas. Now the minimum
absolute Gibbs free energy will occur at the bottom of the curve (see left). At the bottom
of the curve, the slope is zero. Thus, the lowest free energy will occur when the reaction
free energy (which remember is the slope of that curve) is equal to zero. At this point,
the reaction will neither go forwards or backwards and we call this equilibrium. So at
equilibrium:
P
 r G  0   r G  RT ln B
PA
at this point, the ratio of PA and PB has a special value which is characteristic of the
reaction and is referred to as the equilibrium constant for the reaction. At equilibrium:
P
K eq  B
PA
This is it. This is the point of the chapter. We can now relate thermodynamic quantities
to concentrations of molecules and we can see that there will be a characteristic ratio of
concentration of reactants and products that will exist for any reaction called the
equilibrium constant. We can also see that at equilibrium:
0   r G   RT ln K eq
 r G    RT ln K eq
K eq  e

 r G
RT
We still need to generalize the equilibrium constant for reactions more complicated that
A going to B, but what we will find is that it all fundamentally works the same way.
First, however, let me point out one thing that often confuses students and leads to low
scores on exams. Equilibrium constants can be any positive, non-zero number. Many
student make the mistake of thinking that when  r G  0 that Keq = 1. This is not true in
general. As you can see from the equations above, Keq depends on the value of  r G  and
is only 1 in the very unusual case that  r G   0 . For the vast majority of reactions, the
standard free energy of the reaction is not zero and so Keq is different from 1.
OK, lets do this for a more general reaction. This time let's assume that we have a raction
occurring in a liquid. The reaction is A + 2BC + 3D. We have:
dG   C dnC   D dn D   A dn A   B dn B
  C d  3 D d   A d  2 B d
Whoa, where did all the 2's and 3's come from? The extent of reaction is always defined
in terms of the stochiometry of the reaction. In other words, it is on a per mole basis.
Thus, if I am making 3 moles of D or using up 2 moles of B, dnD = 3d and dnB = 2d.
Anyway, we can now see that:
 G 

   C  3 D   A  2 B   r G
   P ,T
Remember that for a liquid
 A   *A  RT ln a A
Where aA is the activity of A (in terms of activity coefficients, the activity is just
 m
a A  A  A , a unitless quantity). Substituting we have
m
 r G   C  3 D   A  2 B
  C  3 D   A  2 B  RT ln aC  3RT ln a D  RT ln a A  2 RT ln a B
  r G   RT ln aC  RT ln a D   RT ln a A  RT ln a B 
3
  r G   RT ln
2
aC a D 3
a A a B 2
For a simple situation in which the activity coefficients are unity, we can substute in
molality ratios (mA/m0, etc.) for activities and get:
3
 mC  m D 
    
m  m 

 r G   r G  RT ln 
2
 m A  m D 
    
 m  m 
 rG

mC m D 3
 RT ln 
m m A mB 2
Notice that there is still a left over m0 term in the equation. This is because of the way
the units worked out. It is really there, but since we have picked our standard state as
one, we usually do not write it (the book dropped this out long since). However, without
it there, the units do not work -- we would end of taking the logarithm of inverse molality
which is nonsense.
We will often talk in terms of molarity, particularly when the solvent is water. In this
case (dropping the standard molarity term that makes the units cancel) we would have:
C D3
 r G   r G   RT ln
AB2
(Be careful, the standard reaction Gibbs free energy is now for 1 M concentrations of
everything instead of 1 molal concentrations.) At equilibrium, the reaction Gibbs free
energy is zero and our equilibrium constant is just:
3

C D 
K eq 
AB 2
Remember, this all assumes that the activity coefficients are one. If not, you must leave
them in the equation and use activities, as above. Hopefully, you have seen these results
before in general chemistry. One can of course write a general equation for the reaction
Gibbs free energy in terms of stochiometry coefficients and sums and products for any
reaction, but looking at these examples, I think you will the the idea (the general
expressions are in the book on page 219 if you want to use them).
Chapter 9.2 and 9.3
These sections consider the effects of pressure and temperature on the equilibrium
constant. The effect of pressure is simple. There is none. Changing the pressure may
well change the amounts of individual components in the reaction, but the ratio at
equilibrium will remain the same.
How about temperature. You can do this several different ways. The book uses the
Gibbs-Helmholtz equation
d  r G  / T 
 H
 r 2
dT
T

 r G / T  R ln K eq
d ln K eq 
dT
or
d ln K eq 
1
d 
T 

1 r H 
R T2

r H 
R
The latter equation is quite useful, because if you plot the log of the equilibrium constant
(which is just a ratio of concentrations that you can often measure) vs. 1/T, the slope of
the resulting line is the reaction enthalpy. Perhaps a simpler way of seeing this is:
 r G    r H   T r S 
RT ln K eq   r H   T r S 
ln K eq
r H  r S


RT
R
Looking at the last equation, it is pretty clear that if you were to plot ln(Keq) vs. 1/T that
the slope of the resulting line would be the standard reaction enthalpy divided by R and
the intercept would be the standard reaction entropy divided by R. This is called a van't
Hoff plot and is a common way of getting enthalpies and entropies of reactions.
If you know the Keq at one temperature and want to calculate it at another (assuming the
the reaction enthalpy and entropy do not change much over the temperature range of
interest), just subtract the last equation at T2 from that equation evaluated at T1:
r H 
ln K eq (T2 )  ln K eq T1  
R
1 1
  
 T2 T1 