Essential Thermodynamic Concepts for Biochemists
Neal Woodbury
Arizona State University
Thermodynamics is a very powerful formalism. It is powerful because it applies to
everything. The thing to remember about thermodynamic formalism is that it has nothing
to do with chemistry. As you will see, all the real chemistry is contained in the empirical
constants and fudge factors (if you want chemistry, you need to learn about quantum
mechanics and electrostatics). Thermodynamic formalism is about populations. It really
does not matter whether you are talking about populations of molecules or chickens, the
principles apply equally well (assuming you have something like a mole of chickens so
that the statistical fluctuations average out). Thermodynamic formalism is derived for
ideal gases (collections of particles that do not interact in any way except that they run
into each other elastically – that’s not chemistry). The application of thermodynamics to
everything else simply involves adding the aforementioned empirical constants and fudge
factors.
So lets start at the beginning. In thermodynamics we make two fundamental
assumptions:
1) The total amount of energy in the Universe does not change (OK, mass/energy if
you want to get technical, but biochemistry does not involve any thermonuclear
reactions).
2) The amount of disorder in the Universe is constantly increasing and can never
decrease. Another way of saying this is that things spread themselves around in a
random fashion if you shake them up a bit, and if you have a large number of
things, they are never going to become any more organized. The more you shake,
the more disorganized (random in space and time) the things will become.
Sometimes people refer to the definition of temperature as another “Law”. Basically
temperature is random motion of molecules and is proportional to kinetic energy. The
proportionality constant is kB, the Boltzman constant. At some very low temperature,
motion entirely stops (this is absolute zero in degrees Kelvin).
Now we need to convert these fairly obvious ideas into some kind of mathematical
formalism. This is all thermodynamics is, actually, a mathematical formalism for these
two concepts that can be applied to the world to make quantitative predictions.
For the first Law we have to decide what the physical manifestations of energy are and
how we will measure them. Actually, we never measure absolute energy (except maybe
temperature). We only measure changes in energy. The three kinds of energy changes
that are important to biology are heat (called q), mechanical work (called w) and
electrical work (another kind of work, w). When a system gives off heat, its total energy
decreases (but the total energy of the surroundings increases so the total energy of
everything is constant). When it takes up heat, its energy increases. When a system does
mechanical work on the surroundings (pushes something against a force), its energy
decreases. When the surroundings do work on the system (compression, of the system
for example), the system’s energy increases. Electrical work is like mechanical work
except that we are moving a charge either against or along an electrical potential (for
example, it takes work to move two positive charges closer together).
We normally measure heat through changes in temperature (though the system can give
off or pick up heat without changing temperature if other things happen at the same time).
We normally measure mechanical or electrical work by measuring the force that we are
pushing against and the distance we push against it. We can now start writing these ideas
down in mathematical form:
dq  C p dT
dw  fdx   PdV
dw  zFdE
These are the differential forms of the equations defining the infinitesimally small
changes. An incremental amount of heat (dq) is given in terms of the heat capacity (Cp,
the heat energy something can absorb for every degree in temperature that it increases)
and an infinitesimally small change in temperature (dT). An increment of mechanical
work (dw) is just the force (f) times the distance moved (dx) which if put in terms of
three dimensions is the pressure (P) times the change in volume (dV). You can think of
this in terms of the work required to blow up a balloon. Finally electrical work is just the
energy required to move a particle of charge z through a tiny voltage dE (F is the Faraday
constant and just makes the units come out right). So, we are now ready to state the first
law in terms of the energy of our system:
dUsys = dq + dw
Usys is the energy of the system, often called the internal energy. Any change in U is
either because heat went in or out or because work was done on or by the system. The
change in Utot for the universe is zero (that is actually the first law).
The first law is a statement setting some limitations about what can happen – you cannot
do things which change the total energy of the universe. However, it is far from the
whole story. I can drop an egg on the floor. It breaks into a thousand pieces and makes a
big mess. However, the energy of the Universe did not change. The energy of the falling
egg was converted into random molecular motion when it hit the floor (heat).
Furthermore, the energy of the Universe would not change if exactly the reverse reaction
occurred. If the energy lost as heat when the egg hit the floor jumped back into the egg,
causing it to come back together and fly up into my hand. However, we all know that the
first process (dropping the egg and smashing it) happens, but the second process (having
it spontaneously reassemble and fly back up into my hand) does not. Neither process
violates the first law of thermodynamics. What are we missing here?
Remember that the second law says that if something is going to happen, it must result in
the Universe moving towards an overall state of greater disorder. Well that pretty much
solves the egg problem. The egg got pretty disordered when it hit the ground and it is not
going to reorder itself on its own. Further, if the universe gets more ordered in one place,
it must get even more disordered in another. I could, in principle, put the egg back
together, but I would have to expend considerable energy doing it. That means I would
have to burn up food which will make the atoms in the food much more disordered. In
the end, I will have made the Universe less ordered by putting the egg back together (you
can try this argument out on your Mother when she tells you to clean up your room).
OK, so how are we going to state all this in quantitative terms. We need some measure
of order or disorder. From the egg example, we have some idea that the problem with the
egg putting itself back together is that the energy that resulted in it becoming disordered
(the energy of its motion when it hit the flow) was turned into the random motion of heat.
So heat lost has something to do with it. But it also has something to do with
temperature. Basically, there would not be as much of a change in the amount of random
energy in the system if there was already a lot of random energy to begin with. It other
words, the higher the temperature, the smaller the change in disorder because at high
temperatures there is already a lot of random motion. We therefore measure the change
in disorder as something called entropy (S), which is a ratio of the heat lost (actually, the
heat lost reversibly, but let’s not worry about that just now) to the temperature:
dS = dqrev/T
The second law states that for any process that will actually happen:
S tot  0 
In other words, the disorder of the Universe must increase or the process will not take
place.
Now in chemistry (or biochemistry), it is going to be a lot easier to think of things in
terms of energy than in terms of entropy. Notice that if you multiply entropy by
temperature, it has units of energy (heat). So it might be easier to think about the second
law of thermodynamics in terms of the quantity TdS, or if we consider a macroscopic
change, TS. T is always a positive number if we deal with degrees Kelvin, so we can
say that
TS tot  0
or
 TS tot  0
We give the energy -TStot a special name. We call this energy the free energy change of
the process and usually designate it G. Remember that a free energy change associated
with a process is always a free energy change for the entire universe. It is not the free
energy change of just the molecules involved in the process. It includes the effect of heat
or work done on the surroundings as well. This means that another expression of the
second law of thermodynamics is:
G  0
OK so based on the above discussion, we can see that
G = -TStot
We can now split this up in terms of the entropy change for the system and that for the
surroundings:
G = -TSsur -TSsys
What constitutes the system and the surroundings? Basically, the system is the molecules
involved in our reaction or process and the surroundings is the rest of the universe.
Now we have to come back to the definition of entropy change in terms of heat and
temperature. As long as we agree to keep the temperature of the surroundings constant,
we can arrange things so that heat is transferred to it or from it in a reversible fashion.
This concept of reversible is kind of tricky, but for heat transfer it basically means that
the temperature of the thing under consideration does not change (we have not changed,
in some sense, the ability of the surroundings to transfer heat one way or the other). For
the surroundings, this is certainly true (the surrounding are infinitely large compared to
the system). Thus:
G = -Tqsur/T -TSsys = -qsur -TS
The heat transferred TO the surroundings is just the negative of the heat transferred
FROM the system. Thus we can write:
G = qsys -TS
Biochemistry, thank goodness, occurs at constant temperature and pressure (this just
saved us many pages of grief). If the pressure is held constant, we have another name for
the heat generated or consumed by the system. We call it the enthalpy change and give it
the symbol “H”. Thus we can write an equation which should start to look pretty
familiar to you:
G = H -TS
OK, this is conceptually useful, but most of the practical applications of thermodynamics
in biochemistry rely on understanding its relationship to concentrations of chemicals in
reactions. How are we going to generate a mathematical formalism for this?
This is where what I said in the very first paragraph becomes important. Basically, the
free energy associated with a chemical reaction under some particular set of conditions is
an empirical constant. You cannot calculate it using thermodynamics. You can maybe
calculate it using quantum mechanics and electrostatics, but only for very small
molecules. In biochemistry, essentially everything of importance in this regard is
empirical.
Great. We just went through four pages of this, and NOW I tell you that you just have to
look the value of the free energy up in a book anyway. What’s the point? The point is
that you only have to look the value up in a book for one set of conditions. In principle
(with a few occasional empirical fudge factors that we will discuss later thrown in), you
can use thermodynamics to calculate the free energy under all other sets of conditions.
This brings us to the concept of the standard state. Basically, solution chemists decided a
long time ago to tabulate reactions for the state where everything was at about room
temperature and all the concentrations of the reactants and products were at 1 M, except
for reactions were the solvent is one of the reactants, in which case the standard
concentration of the solvent molecule in the reaction is the concentration of the pure
solvent. Thus, in essentially all biochemical reactions, water is always at its standard
state (which will mean we can ignore it in our thermodynamic calculations for the most
part). You should also note that in biochemistry, unlike chemistry, the standard state of
H+ is 10-7 M (pH 7.0) rather than 1 M. In addition, free energies are reported on a per
mole basis. That is, the free energies we will deal with will actually have units of
energy/mole (the amount of free energy change when one mole of reactant goes to one
mole of product).
OK, now you are saying, hold on a second Dr. W., you are not going to tell me that you
are going to run a reaction on a protein in a solution with 1M protein!! After all that
could be hundreds of kilograms of protein per liter!! Not possible. The answer is that
you do not have to actually ever run the reaction at standard state. The free energies are
measured at some reasonable concentration and then adjusted as described below until
the free energy is what would be expected if you could run the reaction at standard state.
That value is what is reported in the table.
So here is the way to think about the free energy change at standard state. This is
important if you want to understand what is going on. This is the energy of one mole of
reactants being consumed to form products under conditions where NOTHING changes
including the concentrations of the reactants and products. This is possible, after all, if
there is a very large amount of reactant and product to begin with. For example, a
thousand liters of 1M everything will not change concentration appreciable if you convert
one mole of reactant to product (there was a thousand moles of reactant in the one
thousand liters at 1M after all).
We call this standard value of the free energy change G0. You look this up for the
reaction of choice in a table somewhere.
I have not really shown you this, but free energy is a state function. This means that the
energy change between the beginning of a process and the end of a process is
independent of how you get there. So, if we want to know the free energy of a reaction
that is not at the standard state, you can do this in three steps:
1) Determine the free energy change involved in changing the concentration of the
one mole of reactant that is going to react from the real concentration to the
standard state concentration.
2) Run the reaction at standard state giving a molar free energy change of G0
(molar means per mole).
3) Determine the free energy change involved in changing the concentration of the
one mole of product from the standard state concentration back to the real
concentration.
So what is the free energy associated with changing concentrations of a compound? This
is entirely an entropy term. The more spread out something is (lower concentration) the
greater the entropy. This has nothing to do with the reaction itself, which is all tied up in
the G0 term. Yes, OK, we will have to ultimately consider the attraction between
molecules in solutions which is not entirely entropy, but hold off on that – we will invent
another fudge factor to take care of it in a minute. Actually, in biology, most things are
present at low concentrations so that this is not as large an effect as it might otherwise be.
First lets consider particles that do not interact with one another. This is an ideal gas.
Forget for the moment that we do not run enzymatic reactions in the gas phase. From the
point of view of thermodynamics, it really does not matter. We can concentrate
molecules closer together and increase their concentration or let them be farther apart and
decrease their concentration. In biology, molecules are often at pretty low concentrations
and surrounded by a homogenous medium (water) making them act in a sense as though
they were in the gas phase (separated and not interacting with each other).
What does it take to push ideal gas molecules closer together? Now we need that first
law stuff we wrote out above. The only kind of energy that an ideal gas has is kinetic
energy. Movement. Temperature. That’s it, there is nothing else these inert particles are
allowed to do but move. This means that if we hold the temperature constant while we
change the concentrations of the molecules in the reaction, there will be no change in
internal energy. That means that w = -q. Also, because we are holding the temperature
constant during the process of changing concentration, the heat is reversible. So
remember we said that this was entirely entropy thus:
Gcon. change = -TSconc. change = -Tqrev/T = -qrev = w
What is the work associated with compressing a gas at constant temperature? Remember
that:
dw = -PdV
To get w we just integrate between the uncompressed and the compressed state (this is
the only real calculus in this whole document, promise!). But we have a small problem.
As we change V, P is going to change as well. Fortunately, we know that for an ideal
gas, pressure and volume are inversely proportional at constant temperature (ideal gas
law, remember?):
PV=nRT
Or
P = nRT/V
Substituting we see that:
f
Vf
nRT
dV
G  w    PdV  
dV  nRT 
nRT ln 
V
V
 Vi
Vi
Vi
Vi
Vf
Vf
V



Here, Vf and Vi are the final and initial volumes, respectively. Since we are doing all this
on a molar basis (the free energy change is for one mole), n = 1. Also, we are not losing
or gaining material, we are just making it more or less concentrated. Thus the volume
that the material is held in is just inversely proportional to the concentration. In other
words:
Vf/Vi = Ci/Cf
So we can see that the free energy for changing the concentration of a molecule is given
by:
C
G   RT ln  i
C
 f

C
  RT ln  f
C

 i




Remember ln(a/b) = -ln(b/a). We are almost there. Now we have to actually do the three
steps we talked about above.
This is most easily done using an example. Let’s consider the reaction:
A + B  C
The initial concentrations of A, B and C are just CA, CB and CC. Now lets do the three
steps:
1) Change the concentration of one mole of A and one mole of B from their current
concentrations to 1M.
So the free energy to change A from CA to 1M is just:
 1M 
C 
   RT ln  A    RT ln A
G A  RT ln 
 1M 
 CA 
So what’s with [A] all of a sudden? Well we use that nomenclature all the time in
chemistry as though it was the concentration of a chemical. Indeed it has the same value
as the concentration in moles per liter does, but it is actually unitless. It is a ratio of the
concentration in moles per liter divided by 1 mole per liter. It has to be unitless or all
those equations where we take logs of concentration would never work. You can’t take
the log of a unit. So [A] means the concentration of A relative to a solution of one molar
A.
It is now pretty easy to calculate the free energy change for changing the concentration of
B:
GB  RT ln B
2) Next we need the free energy of the chemical reaction itself under standard
conditions. This of course is just G0.
3) Finally we need to take the products formed (in this case C) and change them
back from 1M to the true concentration, CC. This is the same as above but the
opposite sign:
C 
GC  RT ln  C   RT ln CC 
 1M 
The signs are reversed because in this case the initial concentration was 1M and the final
concentration was Cc (the ratio was inverted).
Now, finally, we just add up the free energies for the three steps:
Grxn = GA + GB + G0 + GC
or
 C  
Grxn   RT ln [ A]  RT ln B  G 0  RT ln C   G 0  RT ln 

 AB 
Now this should look familiar. If you are confused about the manipulation of logs,
remember that ln(a) + ln(b) = ln(ab) and ln(1/a) = -ln(a). Anyway, we can generalize this
equation as:
Grxn  G 0  RT ln Q 
Where Q is just the quotent of all the relative concentrations of products multiplied
together divided by all the relative concentrations of the reactants multiplied together.
This is the master equation we use all the time in biochemistry.
Note that if we consider equilibrium, which the situation where the free energy change in
the reaction is zero (it does not spontaneously go in either direction), we have:
0 = G0 + RTln(Keq)
or
G0 = -RTln(Keq)
Where Keq is just the ratio Q when the system has come to equilibrium. Note that this
means that there is a fixed ratio of concentrations for a reaction at equilibrium that is
independent of how much of any reactants or products you started with. This ratio is
called the equilibrium constant. Note that it does depend on temperature.
Fine, fine, but hemoglobin is not an ideal gas! Well that’s true, but first, as long as it
does not see too many other hemoglobin molecules and just floats around by itself in
water, it is not that different from an ideal gas from the point of view of changing
concentrations. Second, to the extent that attractions between molecules do occur, we
realize that this makes it easier to push molecules together than you would otherwise
predict. In other words, the molecules act like they are at a lower concentration than they
really are. No problem, we will just make up a fudge factor (normally less than one, but
not always), called an activity coefficient, and multiply our hemoglobin concentration by
that. Now we have an effective concentration that does, more or less, obey the rules. We
can come up (empirically) with fudge factors like this for all of the molecules we work
with. We refer to this effective concentration as an activity.
OK what can we do with this? The above equations are themselves useful for general
biochemical reactions. Let’s consider a few useful special cases. We deal a lot with pH
equations, membrane potentials and redox equations in biochemistry. They are all just
this equation in disguise. Lets work through them. First redox equations:
Redox reactions are just moving electrons around. This is electrical work and can be
thought of in terms of the energy for moving a charge with or against a voltage. We
stated earlier that the energy for doing this was just equal to -zFE, where z is the charge
(the number of electrons transfer during the reaction), F is the faraday constant for getting
the units right and E is the voltage you are moving the charge through. Thus to convert
from a free energy to a voltage you just divide the free energy by -zF:
G/zF = -G0/zF + -(RT/zF)ln(Q)
or
E = E0 - (RT/zF)lnQ
For a reaction between a reduced material going to an oxidized material, Q is just
([Red]/[Ox]). This gives the familiar redox equation:
E = E0 - (RT/zF)ln([Red]/[Ox])
Now this is actually the voltage for a half reaction. If something is going to be oxidized,
after all, something else must get reduced in the process. But you can just add the
potentials of half reactions together to get complete reactions.
How about membrane transport? Here we have one type of molecule moving across a
membrane. The reaction is
Ain  Aout
We can write as usual:
G = G0 + RTln([Aout]/[Ain])
If there is no voltage across the membrane (usually there is) or if A is not a charged
molecule (and thus does not feel the membrane voltage), then at standard state when
there is the same amount of A on the inside and the outside, the system is at equilibrium.
Equilibrium means that there is not tendency for the reaction to go one way or the other.
In other words the free energy is zero. If G is zero and if Ain = Aout (which results in the
log being zero) then G0 for this reaction must be zero. This makes sense; we would
expect that at the standard state when the inside and outside concentrations were the same
that the free energy of the reaction should be zero. Plugging this in gives the following
for transport of molecules across membranes if the molecules are uncharged of the
membrane has not potential:
G = RTln([Aout]/[Ain])
This “reaction” is ALL entropy. There is no real chemical reaction taking place, you are
just moving molecules from one place to another. If you make the inside concentration
high, the free energy change is negative and flow outward is spontaneous. If you make
the outside concentration high, the free energy change is positive and flow inward is
spontaneous.
But what if we are dealing with a charged molecule and there is a voltage across the
membrane? Then the free energy for transfer of a mole of charge across the membrane is
not zero any more (G0 is not zero). It will depend on the membrane voltage as we
discussed before. People usually refer to membrane voltages as So the energy of
transferring a charge z across a membrane when the concentrations on the two sides are
the same is just zF. The complete equation including both the effect of different
concentrations on the two sides of the membrane (chemical gradients) and a membrane
potential across the membrane is:
G = zF + RTln([Aout]/[Ain])
Note that now at equilibrium, if the membrane potential is not zero, the concentration
gradient of a charged molecule will not be zero. At equilibrium G = 0 so:
RTln([Aout]/[Ain]) = zF

This equation is important for calculating the ion gradient at equilibrium from a
membrane potential or the other way around.
For a more complete description of membrane potential calculations and conventions
concerning the assignment of the sign of the membrane potential and the direction of the
ion transfer see the Membrane Potential lecture notes.
Another equilibrium equation of great importance comes from considering the acid base
reaction:
A + H+  AH+
G = G0 + RTln([AH+]/[A][H+])
If we let this system come to equilibrium (G = 0) we have
 = G0 + RTln([AH+]/[A][H+])
Now lets divide by RT
 = G0/RT + ln([AH+]/[A][H+])
Convert the ln into log10
0 = G0/RT + 2.303log([AH+]/[A][H+])
divide by 2.303
0 = G0/2.303RT + log([AH+]/[A][H+])
split the log up
0 = G0/2.303RT - log([H+]) + log([AH+]/[A])
We call –log([H+]) pH. Rearranging we have (careful about minus signs)
pH = -G0/2.303RT + log([A]/[AH+])
Remember that G0 = -RTln(Keq). The equilibrium constant for this reaction is usually
called KA and the negative log of this equilibrium constant is called pKA. I think you can
see that pKA = -G0/2.303RT if you rearrange things. This gives the familiar equation:
pH = pKA + log([A]/[AH+])
These are the fundamental thermodynamic equations that we use in Biochemistry and a
description of their origins. None of this involves high level math. The concepts
associated with entropy and free energy take some thinking about, but if you consider
them in the terms presented here, I hope it will all start to come together.