**Chapter 4 Lecture Notes**

Ok, after six 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.

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, x.
The change in extent of reaction is easy to define for a simple reaction like Aß à
B. It is simply the number of moles of A that has been converted to B. For a
simple reaction like this, dx has the
same value as dn and so we can write that dG = m_{A}dn_{A}
+ m_{B}dn_{B} = -m_{A}dx
+ m_{B}dx. 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:

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 x, we can
see that the point where is the minimum of the curve.
This is the 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 D_{r}G
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.

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 Aß à B and there is no enthalpy term (and no change in the inherent entropy of A vs. B), 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 or if A and B have different inherent entropies). Ok, but usually there are additional terms for the reaction. What then? Lets work through this for an ideal gas reaction:

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.

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. The chemical potentials of A and B are equal. At this point, the reaction will neither go forwards or backwards and we call this equilibrium. So at equilibrium:

at this point, the ratio of P_{A} and P_{B} has a special
value which is characteristic of the reaction and is referred to as the
equilibrium constant for the reaction. At equilibrium:

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:

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 that K_{eq} = 1. This
is not true in general. As you can see from the equations above, K_{eq}
depends on the value of and is only 1 in the very
unusual case that . For the vast majority of reactions, the standard
free energy of the reaction is not zero and so K_{eq} 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 + 2Bß à C + 3D. We have:

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, dn_{D} = 3dx and dn_{B}
= 2dx. Anyway, we can now see that:

Remember that for a liquid

Where a_{A} is the activity of A (in terms of activity coefficients,
the activity is just , a unitless quantity).
Substituting we have

For a simple situation in which the activity coefficients are unity, we can
substute in molality ratios (m_{A}/m^{0}, etc.) for activities
and get:

Notice that there is still a left over m^{0} 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:

(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:

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 get the idea.

How do we get the standard Gibbs energies of reaction? Generally they can either be found in a table somewhere or we can calculate them from the standard Gibbs energies of formation. These in turn can be had by combining the standard Gibbs entropy and enthalpy of formation:

D_{F}G^{0} = D_{F}H^{0} - TD_{F}S^{0}

Or you could calculate the stardard enthalpy and entropy of the reaction separately and combine them in a similar way.

The book next goes into the concept of coupled reactions. All this means is that we are going to run two reactions together and make it so that one reaction cannot run without the other. Enzymes do this all the time. They often couple the hydrolysis of ATP (adenosine triphosphate) to some reaction that would otherwise be unfavorable. When the two reactions are coupled, the Gibbs energy changes add. So even if one of the reactions has a positive Gibbs energy change, as long as the other has a sufficiently negative one, it works out ok.

For example, the enzyme citrate synthase catalyzes the conversion of a compound called oxaloacetate along with a molecule of acetyl CoA to make citrate:

Oxaloacetate + Acetyl CoA à Citrate D_{R}G^{0} = -31.4kJ/mole

The equilibrium constant for this reaction is huge, about 3 x 10^{5},
so there is really no way to start with Citrate and get any appreciable amount
of Oxaloacetate. However, there is
another enzyme, called ATP Citrate lyase which couples the hydrolysis of ATP to
the reverse of this reaction. That is to
say that the enzyme only lowers the activation energy between the reactants and
products if ATP is hydrolyzed to ADP and phosphate along with the reaction. So we have:

Citrate à
Oxaloacetate + Acetyl CoA D_{R}G^{0}
= 31.4kJ/mole

ATP + H_{2}O à ADP + Phosphate D_{R}G^{0} = -30.5kJ/mole

_____________________________________________________

Citrate + ATP + H_{2}O à Oxaloacetate + Acetyl
CoA + ADP + Phosphate D_{R}G^{0}
= 0.9 kJ/mole

The Gibbs reaction energy is still slightly positive, but not very. The equilibrium constant for this reaction is about 0.72 which means that if you start with Citrate and ATP you will end up having around half of it convert to Oxaloacetate and the other products (see below). This is good enough if there are other reactions that remove the oxaloacetate.

This is all about how to calculate the composition of a reaction given the standard Gibbs reaction energy. So, lets use the last reaction as an example. Lets say I start with 1 mole of each of the reactants. What is the composition at equilibrium? The equilibrium constant in terms of molarity is:

K_{eq} =
[Oxaloacetate][Acetyl-CoA][ADP][Phosphate]/([Citrate][ATP][H_{2}O]) =
.72

Now, the first thing is we have to visit the concept again of standard
states. Remember that the standard state
of the solvent is the pure solvent (not 1 M).
The molarity of H_{2}O is essentially unchanged from the pure
solvent, so its value is 1. This is
really important and confuses a lot of people.
The standard state for the rest of the compounds is 1 M and therefore
they have the value in this expression equivalent to their molarity. Knowing that we start with 1 M of each the
reactants, we can see that

[Citrate]=[ATP]=X at all times and

[Oxaloacetate]=[Acetyl-CoA]=[ADP]=[Phosphate]=Y

So plugging X and Y into the equilibrium expression we have:

Y^{4}/X^{2} = .72

We also know from mass action that

X + Y = 1

Since we know both X and Y are positive, real values, we can take the square root of the first equation:

Y^{2}/X = .85

Substituting X = 1-Y

Y^{2}/(1-Y) = .85

Solving for Y I get 0.59

So the final concentration of the product oxaloacetate will be 0.59 M at equilibrium.

This is not the same when you use other concentrations. For example, a more realistic concentration would be 0.01 M for all species. In this case, I get that the final concentration of oxaloacetate is about 0.0099 M which means that almost all of the 0.01 M citrate is converted to oxaloacetate under these conditions, even though the Gibbs energy is greater than zero. Think about this. It is important how concentration affects these reactions. Remember, what the Gibbs reaction energy is under standard conditions (everything 1 M) is not what it is under other conditions (like everything 0.01 M). The difference is entropy. Look at the chemical equation. We are taking two things (ignoring water whose concentration does not appreciably change) and converting them into four things. This corresponds to a big entropy increase. As you lower the concentration, this effect of entropy becomes greater and greater (there is more and more “room” for the molecules to explore).

This section basically defines a catalyst as something that accelerates the reaction without changing the reaction Gibbs energy. We really can’t say much more until we learn about kinetics.

Consider the affects 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. One is the Gibbs-Helmholtz equation

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:

Looking at the last equation, it is pretty clear that if you were to plot
ln(K_{eq}) 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 K_{eq} 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 T_{2}
from that equation evaluated at T_{1}:

ln K_{eq}(T_{2})
= ln K_{eq} (T_{1}) + (D_{r}H^{0}/R)(1/T_{1} – 1/T_{2})

This also tells us something that seems reasonable intuitively. For endothermic reactions (positive enthalpy change, takes up heat), increasing the temperature favors the products (the equilibrium constant gets larger). For exothermic reactions (negative enthalpy change, releases heat), increasing the temperature favors the reactants (the equilibrium constant decreases).

**Acid Base Equilibrium**

The application that I want you to understand in detail is the acid/base equilibrium. In this chapter (and the next chapter as well), we will deal entirely with aqueous (water) solutions. For our purposes, an acid is something that donates protons and a base is something that accepts protons. Thus, many molecules can be both donors or acceptors of protons, depending on the situation. Obviously a molecule that initial gives up a proton, acting as an acid, can then take back the proton later, acting as a base.

here HA is an acid on the left side of the equation and H_{2}O is
the base. However, for the reverse reaction H_{3}O^{+} is the
acid and A^{-} is the base. Note that people often simply leave the
water out of chemical equations like this and write:

This is a simple short hand, but it ignores the important point that one never has a bare proton in solution. It is always solvated by water. In fact, all ions in water are heavily solvated with more than one water molecule in close association often. However, it is enough to remember that in a deprotonation/protonation reaction such as this, water can serve as the base and the resulting protonated water molecule as the acid. Water can also serve directly as an acid in the following reaction:

In these equilibria, we would call AH an acid and A^{-} its
conjugate base. We would also refer to B as a base and BH^{+} as its
conjugate acid. Finally, water can act as both the acid and the base in the
same reaction:

OK, now let's write down the equilibrium constants for these reactions in terms of concentrations of reactants and products. Often this is done in terms of molarities of the different components (this is water, so at normal temperatures and low solute concentrations, molarity and molality are essentially the same). For the time being, we will stick to activities, which is more precise, but later we will lapse into molarity. So, for each of these equations:

OK, what happened to all the H_{2}O terms? Why did they all
dissappear? This has to do with standard states. Remember that for solvents,
our standard state is the pure solvent and we define the activity such that in its
standard state, a substance has an activity of one.

Note that in this equation, when the activity of water is one, the chemical potential of water is just its chemical potential in the pure state. Thus, an activity of one is just pure water. The types of solutions we are talking about here have about 55.6 molar water and perhaps 0.1 or 0.2 molar acid or base. Thus, the water is essentially pure and the activity of water is essentially one. Thus, we drop it from our equilibrium equations.

Note that the activity of the ions is relative to their standard states. All of the ions have standard states of 1 molal (essentially 1 molar for water). This can be a confusing point, but it is an important one. Typically the standard state of a solvent is the pure solvent and this is the condition for which the activity is one. However the standard state of the solute is typically 1 molal and activity is always relative to one molal. Remember that activities are always relative to some standard state. They are unitless quantities.

Back to equilibrium constants… K_{W} is a very special equilibrium
constant and at 25 C has a value of. Note
that the product of the concentrations of H_{3}O^{+} and ^{-}_{W} at 25C, regardless of what else might be going on. K_{A}
and K_{B} are numbers that depend on which acid or base we are
considering and you must look these up in books.

From our work with free energies of reactions, you recall that we often had equations with terms of the form:

Because the log of the equilibrium constant of a reaction is such a
fundamental property, we often speak of equilibrium constants in log form.
Because we normally count in base 10, we often use base 10 logs to speak of
equilibrium constants ().
Because most of the reactions that we are speaking of have equilibrium
constants much less than one, we often refer to the -log_{10}(K_{eq})
which we call the pK. Thus, we can speak of pK_{A} = -log_{10}(K_{A})
or of pK_{B} = -log_{10}(K_{B}) or of pK_{W} =
-log_{10}(K_{W}). Of course we can expand this last one:

Thus, we have special names for the negative log of the H_{3}O^{+}
activity and the negative log of the ^{-}_{3}O^{+}
concentration is 1x10^{-5} molar. Instead they will say that the pH is
5.

Now, notice that I can write two different chemical equations for the acid form and the conjugate base form of a molecule:

If I add these two equations, I see that I just get my equation for water:

Also notice that if I multiply the equilibrium constants of the two
equations (K_{A} and K_{B}), I get K_{W}:

This is general, the product of the acid constant and base constant for a
substance and its conjugate base are equal to K_{W}.

One final point. In pure water, the equation

dictates that the concentrations of H_{3}O^{+} and ^{-}

So the pH of pure water is 7.0. (Try measuring the pH of

The kinds of acids and bases we are talking about above are called weak
acids and weak bases because their equilibrium constants are much less than
one. Strong acids and bases have equilbrium constants much greater than one.
Therefore, we usually do not think of these as being in equilibrium. Instead
they stochiometrically convert weak acids and weak bases (such as the ones
described above or water itself) by either donating protons or extracting
protons. A typical example of a strong acid is HCl. HCl added to water
stochiometrically converts H_{2}O to H_{3}O^{+}. A
strong base is something like a metal hydroxide such as NaOH. This
stochiometrically dissociates forming ^{-}

We often talk of titrating a weak acid or weak base. This means adding a strong acid or base and converting the weak acid to a weak base or the weak base to a weak acid.

Remember that AH is also in equilbrium with water

so as the strong base is added, the equilibrium concentrations in this
equation will be perturbed (though the equilibrium constant, K_{A}, is
of course constant). This is called a titration. An example of a titration
curve of a weak acid with a strong base is shown below:

What we need to do now is to understand how this works. We will consider four different regions of the titration. The starting point before any strong base is added, the point near where the amount of base added has converted half of the weak acid to its conjugate base, the point where the amount of strong base added is just equal to the amount of weak acid and the region where there is significantly more strong base than weak acid.

*The starting point.*

What is the pH of a solution with a certain amount of weak acid added to it? The first thing to do is to write down the equilibrium constant:

Now, we need to realize that since we started with all HA, then as long as
enough dissociation occurs so that the H_{3}O^{+} concentration
is much greater than the 10^{-7} present in distilled water, then the
stochiometry of the reaction dictates that [H_{3}O^{+}] = [A^{-}]
and therefore

Taking the log of both sides then gives

Thus, we can easily calculate the pH for adding a certain amount of a weak
acid (as long as it generates protons at a significantly higher concentration
than 10^{-7}M) using this equation.

*The region near the half stochiometric point*.

Now we start to titrate our weak acid with a strong base like NaOH. This will stochiometrically convert the weak acid to its conjugate base. Once a significant amount of the conjugate base has been formed, we can write that:

That last equation you have probably seen before. It is called the
Henderson-Hasselbalch equation. It allows you to calculate the pH from the
ratio of the base to acid forms of the buffer (the weak acid) and its pK_{A}.
As we shall see below, it is in this region of the titration curve that weak
acids or weak bases make good pH buffers, stabilizing the pH in the presents of
other sources or sinks of protons.

*The stochiometric point.*

This is the point were exactly enough strong base has been added to convert
all of the weak acid to its conjugate weak base. At this point, we are really
working now with a weak base (A^{-}), rather than a weak acid so

Thus, the only thing you need to remember about the stochiometric point is that all the weak acid has been converted to the weak base, so now it is the same as having added the pure weak base to the solution.

*More than a stochiometric amount of base.*

If you continue to add base, the extra base will simply determine the pH.
The first thing to do is to determine how much excess base is present (this is
just the amount of strong base added minus the amount of weak acid added in
moles per liter). Next, we assume that the base completely reacts forming a
stochiometric amount of ^{-}^{-}_{3}O^{+}]
= 10^{-14} to determine the pH:

Ok, let's run through a quick example of a titration of a weak acid with a
strong base. Consider actetic acid, CH_{3}COOH as our weak acid and
NaOH as our strong base. We start by adding 0.1 M actetic acid to distilled
water. The pH is now given by:

If we now add 0.05M of a strong base like NaOH, we will convert exactly half of the weak acid to its conjugate weak base. We now use the H-H equation:

so the pH is equal to the pK_{A} when the amount of the acid form is
equal to the amount of the base form of the weak acid.

Now what if we add another 0.05 M NaOH. Now the acid has been stochiometrically converted to base and we use the stiochiometric equation:

If we now add another 0.05 M NaOH, we have gone beyond the end of the tritration and the excess base determines the pH:

This is an important example. You should go through it until you understand it. In addition, try starting all over again using a weak base (such as ammonia) and titrating it with a strong acid (such as HCl). Derive all the equations and the pHs at each of the four major steps starting with 0.1M ammonia and adding HCl.

Now that we have finished with an introduction to thermodynamics, I would like to work through two applications of thermodynamics to biochemical systems that involve linked equilibrium:

Hemoglobin and binding cooperativity