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Matthew Schwartz Statistical Mechanics, Spring 2019
Lecture 8: Free energy
1 Introduction
Using our various ensembles, we were led to a rather simple dierential relation between energy and other state variables, Eq. (55) of Lecture 7:
dE = TdS ¡ PdV + dN (1)
The terms on the right are dierent ways the energy can change: the rst term is heat, since dS = dQT , the second term is work from expanding volume, the third term is the energy associated with bond formation or particle creation. Holding any two of the four dierentials xed we can derive from this equation lots of relations @S @E
V ;N
T
@S
@N
E ;V
T
@S
@V
E ;N
P
T
@E
@S
V ;N
= T ;
@E
@V
S ;N
= ¡P ;
and so on. Some of these are more useful than others. They are collectively known as Maxwell relations. The way to think about the Maxwell relations is that they relate small changes of E ; S ; V and N , with T ; P and giving the relative size relating the small changes. So the functions depend only on the other dierential variables.
E = E(S ; V ; N ); S = S(E ; V ; N ); V = V (E ; S; N ); N = N (E ; S ; V ) (3)
The variables T ; P ; are derived quantities. They are calculated by taking partial derivatives with respect to the independent variables. That is, if one of the functional forms is known, such as S(E ; V ; N ), we can nd T ; P and by taking derivatives. Since partial derivatives commute, we can derive some additional relations amount derivatives: @T @V
S ;N
@V
@E
@S
V ;N
S ;N
@S
@E
@V
S ;N
V ;N
@P
@S
V ;N
and so on. Keep in mind that the general relations amount E ; S ; V ; N ; T ; P ; hold for any system. The rst 3 equations in Eq. (2) are the de nitions of T ; and P. The others follow mathematically from these using elementary multivariate calculus. Starting with the speci cation of a system in any ensemble (micro, canonical, or grand canonical), we can compute these quantities and the relations will hold. We often use monatomic ideal gases to check these general relations. For example the entropy of a monatomic ideal gas is
S(E ; V ; N ) = NkB
ln
V
N
ln
4 mE 3 Nh^2
Then 1 T
@S
@E
V ;N
NkB E
=) E =
NkBT (6)
which is the statement 32 NkBT , is the average kinetic energy of a gas. Another one is,
P = T
@S
@V
E ;N
= T
NkB V
=) PV = NkBT (7)
which is the ideal gas law, and so on.
When we have a function like S(V ; E ; N ), we can easily see how the system responds when we change V ; E and N. In physical situations, we are often much more interested in knowing how our system responds to changes in temperature (when we heat it) or pressure (when we compress it), or, for chemical reactions in particular, how to characterize equilibrium properties of a system when T and P are held xed. So we would like the system to be described by functions that depend explicitly on T and P rather than E and V. In this lecture, we construct new variables (free energies) that depend on T and P. The four new potentials we introduce are
Helmoltz free energy : F E ¡ TS (8)
Enthalpy : H E + PV (9)
Gibbs free energy : G E + PV ¡ TS (10)
Grand free energy : E ¡ TS ¡ N (11)
These are general de nitions of these new variables, holding for arbitrary systems. We will work through these, constructing them, showing what the depend on, and then exploring their physical signi cance. As a warning, this business of dependent and independent variables is going to be one of the most confusing and infuriating things in the course. Unfortunately, it is essential to understanding thermodynamics so there's no getting around it. We'll try to be as clear about what is going on as possible. As another piece of advice going forward: sequester in your mind all the complicated subtleties with entropy (ergodicity, Boltzmann's H theorem, Lodschmidt's paradox, Landauer's principle, etc.). If you really reach down deep to understand the foundations of the second law, all these things are important. However to use statistical mechanics, and thermodynamics, in physics, chemistry, astronomy and so on, we de ne systems based on macroscopic quantities (P ; V ; N ; T , ...), entropy is extensive (Stot = S 1 + S 2 ), the second law holds without subtlety (Stot > 0), and ; T and P are constant for systems in equilibrium.
2 Euler identity
Before getting started with all the new potentials, there is actually a very nice relation we can derive among all the dierent interdependent variables using only Eq. (1) and the fact that entropy is extensive. From Eq. (1) we see that S = S(E ; V ; N ) (which we already knew since that's how we set up the microcanonical ensemble in the beginning). E ; V ; N and S are all extensive quantities. When we double the size of a system, they all double. This is in contrast to P ; and T which are intensive quantities. When we double the system, they do not change. It's actually very hard to make an extensive function. Any extensive function f(x) must satisfy f (cx) = cf (x) for any c. Let's write f (x) as a Taylor series, f (x) =
P
n anx
n. Then for f(x) to be
extensive, we must have
c
X
n
anxn^ = cf (x) = f(cx) =
X
n
cnanxn^ (12)
The only way this can hold for all x is if an = 0 for n =/ 1. That is, f(x) = ax for some a. We can write this conclusion in a suggestive way ,that any extensive function of one variable must satisfy.
f(x) =
df dx
x (13)
Are thermodynamic functions have logarithms in them, so we can't obviously write them as Taylor expansions. Nevertheless, the constraint still holds. Extensivity requires
S(cE ; cV ; cN ) = cS(E ; V ; N ) (14)
2 Section 2
and so on. It is perhaps also worth pointing out to the mathematically-oriented crowd that the operation of
replacing the dependent variable S in E(S ; V ; N ) with a new dependent variable T = @E @S is known as a Legendre transform. Another example of a Legendre transform is going from a Lagrangian
L(q_; q) that depends on velocity q_ to a Hamiltonian H(p; q) that depends on momentum p = @L@q_. The Hamiltonian does not depend on q_ just as F does not depend on S. Before continuing, let me try to address a common pitfall. You might ask why is there no dE in Eq. (23)? Similarly, you could ask why is there no dT in Eq. (1)? The answer is that the non- trivial content in Eq. (1) is precisely that there is no dT (and no d or dP either). Of course T does vary, so if we have a nonzero dN ; dV and dS then dT is probably nonzero as well. The point is that dT is not an independent variation. Eq. (1) says that we don't need to know what dT is to compute dE, we just need dS, dV and dN. Similarly, the content of Eq. (23) is that we don't need to know dE or dS, it is enough to know dV , dN and dT. E is a dependent variable in Eq. (23), so its variation is determined by the variation of the other, independent variables.
3.1 Free energy for work
The Helmholtz free energy is one of the most useful quantities in thermodynamics. Its usefulness stems from the fact that dV , dN and dT are readily measurable. This in contrast to E(S ; V ; N ) which depends on entropy that is hard to measure and in contrast to S(E ; V ; N ) which depends on energy that is hard to measure. Helmholtz free energy is particularly powerful for systems at constant temperature where d F = d E ¡ TdS. In previous courses you have studied mechanical systems using energy. Mechanical systems all xed degrees of freedom, so S = 0 and F = E. Free energy is a generalization of energy whose importance is revealed by working at nite T. Suppose we have some system in thermal contact with a heat bath, and we want the system to do work by pushing a piston (which in turn can lift a weight, or whatever). We do not demand that the pushing be reversible, so we are considering an arbitrary isothermal expansion. If the system were isolated, as it pushes its energy would go down and it would cool, but since its in contact with a heat bath, it instead draws in energy from the bath and turns that heat into work. If heat Q is drawn in, the change in entropy of the bath is Sbath = ¡QT. By the second law of thermodynamics,
we know also that Ssystem > QT , with the equality holding only for a reversible expansion. If work W is done on the piston by the expansion, then, by conservation of energy, the change in energy of the system is given by Esystem = ¡W + Q. Then the free energy change (at constant T ) is:
Fsystem = Esystem ¡ T Ssystem 6 (¡W + Q) ¡ T
Q
T
= ¡W (26)
That is,
Fsystem 6 ¡W (27)
where the inequality becomes an equality if and only if the expansion is done reversibly. So we see that when work is done at constant temperature, the free energy of the system is depleted to do the work. This is why free energy is called free: it is the energy available to do work. In an insulated system (not heat exchange), the energy of the system is used for work, but in an isothermal system, it is the free energy that is used for work. The minimum that the free energy will go down by is W itself, which happens if and only if the work is done reversibly. For another perspective, consider the case where no work is done by the system on the surround- ings. We can always do this by including whatever the work would be done on as part of the system. But there are also many examples where work is not relevant, such as when gases mix together, or some chemical reactions occur, or a system settles down after some perturbation. When W = 0, we get from Eq. (27) that Fsystem 6 0. Thus, in a system kept at constant temperature, interacting with the surroundings only through an exchange of heat (i.e. no work), the Helmholtz free energy never increases. As the system settles down towards equilibrium, F will decrease until equilibrium is reached when it stops decreasing (if it could decrease more by a uctuation, it would, and then it could never go up again). Therefore, in an isolated system kept at constant temperature, the equilibrium is the state of minimum Helmholtz free energy.
4 Section 3
To be extra clear, let us emphasize that free energy refers to the free energy of the system only, F = Fsystem. So to nd the equilibrium state we minimize the free energy of the system, ignoring the heat bath. Indeed, this is why free energy is powerful: it lets us talk about the system alone. For a concrete example, consider a system of two gases separated by a partition, initially with dierent pressures P 1 and P 2 and dierent volumes V 1 and V 2 with V 1 + V 2 = V , in thermal contact with a heat bath. Then, since F = F (T ; N ; V ) and T and N are xed, the minimization condition is
0 = dF =
@F
@V 1
dV 1 +
@F
@V 2
dV 2 = P 1 dV 1 + P 2 d(V ¡ V 1 ) = (P 1 ¡ P 2 )dV 1 (28)
Thus the pressures are equal at equilibrium if P 1 =/ P 2 then changing V would lower F. Of course, we knew this already; previously, we derived the pressure equality from maximization of the total entropy at constant energy. Here we are deriving it from minimization of free energy of the system at constant temperature. The two are equivalent. Indeed, the second law of thermodynamics is equivalent to the minimization of free energy. In general however, it is much easier to deal with systems at constant temperature than at constant entropy, and to minimize the free energy of the system rather than to maximize the total entropy. In summary, Free energy is to a constant T system what E is to a mechanical system. Free energy is the available energy to do work at constant T. In a system kept at constant T , interacting with the surroundings only through an exchange of heat (i.e. no work), the Helmholtz free energy never increases. In an isolated system at constant T , free energy is minimized in equilibrium. Free energy refers to the free energy of the system only F = Fsystem.
3.2 Free energy and the partition function
Next, consider how to compute free energy from the partition function in the canonical ensemble.
Recall that in the canonical ensemble, S = hE i T +^ kBln^ Z. So for an isolated system where^ hE^ i^ =^ E we immediately get that
F = ¡kBT ln Z (^) (29)
So the free energy is (- kBT times the logarithm of) the partition function. It therefore carries all that useful information about the spectrum that the partition function has the partition function and free energy are both very powerful. But it is the same power, since they are the same function. Another way to write the relation in Eq. (29) is
e¡^ F^ = Z =
X
e¡^ E^ (30)
This has the interpretation that the free energy is the energy a system would have if there were only one microstate. In essence, all the possible degrees of freedom of the system are summed over, and they are all encoded in the free energy. This gives the free energy the interpretation of an eective potential. For example, the Lennard-Jones potential for the energy between atoms is an eective potential it comes from integrating out all the electronic and van der Waals uctuations that generate the interatomic force. If you haven't seen it already, the idea of eective potentials is an enormously powerful concept in physics. We won't use it to much this course, but it guides essentially all of condensed matter physics and much of particle physics. For a monoatomic ideal gas recall that
Z = e¡N
V
N
N
2 m h^2
^3
2 N (31)
So
F = ¡kB T ln Z = ¡NkBT
ln
V
N
ln
2 m h^2
Helmholtz Free Energy 5
What is the equilibrium state of this system? To nd the equilibrium, we want the free energy to be stationary when we vary x, so we need @F @x
@E
@x
¡ T
@S
@x
Now the energy in the gas doesn't depend on x, since we are at constant temperature. So @E @x comes only from the spring, where we get @E @x =^ Fpiston^ =^ ¡kx. We use the curly^ F^ for force to distinguish it from free energy. Fpiston is the force acting on the gas by the piston. The piston doesn't have
any entropy, so @S @x comes entirely from the gas. Now, the volume of the gas is^ V^ =^ V^0 +^ Ax^ where A is the area of the piston. So @S @x
@S
@V
@V
@x
P
T
A =
Fgas T
In the last step we used that pressure times area is force. So Eq. (37) becomes
@F @x
= Fpiston ¡ Fgas (39)
Setting the variation of the free energy to zero, @F@x = 0, implies Fpiston = Fgas. This of course makes complete sense we can compute the equilibrium point of the spring by demanding the the forces are equal, or we could nd the equilibrium by minimizing the free energy. The result is the same.
3.4 Energy (non)-minimization
It is common lore to think of energy being minimized by physical systems: a ball rolls down to the bottom of a hill and stays there. But energy is conserved, so where does this common sense lore come from? It comes from free energy! All those systems in which energy is minimized are really minimizing free energy. You may never have thought about the gas in air surrounding the ball, but it it weren't for the gas, or the molecules in the dirt that can heat up due to friction, the ball would just roll right back up the hill. The gas-spring example hopefully illustrated the point that there is no tendency to minimize energy. Total energy is conserved, in spontaneous motion, adiabatic motion, or whatever. The energy E in the de nition F = E ¡ TS is not the total energy but rather the energy of the system. We assume the system is in thermal equilibrium, so energy in the form of heat can leave the system into the surroundings, or enter the system from the surroundings. Thus E may not be zero. But the energy of system itself does not tend towards a minimum. The tendency to minimize free energy is entirely because of a tendency to maximize entropy. This is clearest if we write
F = ¡T
4 ||{z}S} Ssys
E
T
|||||||||||||||||||||| |||||||||||||||||||||||||||||{z}}}}}}}}}}}}}}}}}}}}}}}}} }}}}}}}}}}}}}}}}}}}}}}}}}} Ssurr
The S term is the entropy change of the system. The second term TE = QT is the entropy change of the surroundings. Their total is maximized, so entropy is maximized, and free energy is minimized. When we think of a spring with friction slowly stopping, we think it is minimizing energy. Indeed, it is minimizing energy, but that is because it is maximizing ¡ET = S. In summary, energy minimization is really free energy minimization, which really is entropy maximization.
4 Enthalpy
We use the symbol H for enthalpy. It is de ned as
H E + PV (41)
So, using Eq. (1),
dH = dE + PdV + VdP = TdS + VdP + dN (42)
Enthalpy 7
Thus, (^) @H @S
P ;N
= T ;
@H
@P
S ;N
= V ;
@H
@N
P ;S
Enthalpy is a concept useful at constant pressure. It is a function H = H(S ; P ; N ). Recall that we had two heat capacities related to how much temperature rises when heat dQ is put in at constant V or constant P
CV =
@Q
@T
V
; CP =
@Q
@T
P
For constant V , no work is done, since W = PdV. So CV =
@E @T
V
. For constant P , the gas has to
expand when heat is absorbed to keep the pressure constant, so work is done and the energy goes down. The total energy change is E = Q ¡ P V. In other words, H = Q. Thus,
CP =
@H
@T
P
Thus enthalpy plays the role at constant pressure that energy does at constant volume.
For a monatomic ideal gas, since PV = NkBT and E = 32 NkBT we get immediately that
H = E + PV =
NkBT (46)
so, from Eq. (45) CP = 5 2 NkB^ in agreement with what we found in Lecture 5.
4.1 Enthalpy of chemical bonds
Enthalpy is especially useful in chemistry, where pressure is nearly always constant but volume is not under control. For example, when you open the door to your room, the volume of the air changes, but the pressure is the same. Chemicals in a solution are also at constant pressure, as are biological reactions in cells. When you mix two solutions, the volume accessible to each changes, but the pressure is the same. Since chemistry (and biology) takes place at constant pressure and temperature, when a chem- ical reaction occurs the heat released is Q = H. Thus when you measure the heat released or absorbed in a reaction, you are measuring the enthalpy. There are dierent ways to compute the enthalpy of a reaction, to dierent degrees of approx-
imation. Consider for example the hydrogenation of ethene= (C 2 H 4 )= into ethane
(C 2 H 6 )= :
H 2 + C 2 H 4! C 2 H 6 (47)
First of all, you can just look up the enthalpy for this reaction. That is called the standard enthalpy of reaction , where standard means under some reference conditions, like 298K and 1 atm. For this reaction, we can look up at the reaction enthalpy is rH° = ¡136.3 0. kJ mol. The enthalpy decreases, to the reaction is exothermic: heat is released. If we don't have the reaction enthalpy handy, we can compute the enthalpy change by taking the dierences of the enthalpies of the products and the reactants. So in this case, we look up that the enthalpy of formation of each. The enthalpy of formation is the enthalpy change in breaking something down into its constituent atoms (C; S ; Si) or diatomic molecules (H 2 ,O 2 ,N 2 ,F 2 ). So the enthalpy of formation of C or H 2 is 0, by de nition. The enthalpy of formation for C 2 H 4 is
fH° = 52.4 0.5 (^) molkJ and for C 2 H 6 is fH° = ¡ 84 0.4 (^) molkJ. Combining these, the enthalpy of the
reaction is rH° = ¡ (^84) molkJ ¡ 52.4 (^) molkJ = ¡136.4 (^) molkJ in agreement with the directly measured enthalpy
of reaction. That we can add and subtract enthalpies to get the reaction enthalpy is known as Hess's law. There are lots of laws in chemistry.
8 Section 4
For a rst example, consider chemical reactions in solids. Clam shells have a layer of calcite and a layer of aragonite. These minerals are both naturally occurring forms of calcium carbonate CaCO 3. They both have the same chemical composition, but dierent lattice structure. Aragonite forms from calcite at high pressure, but at typical pressures on the surface of the earth, aragonite is unstable and turns into calcite on the 10 million year timescale. The two minerals have dierent densities: aragonite is more dense, at 2.93 (^) cmg 3 than calcite at 2.71 (^) cmg 3. Thus when aragonite converts into calcite, its volume expands, doing work and an enthalpy H = 0.21 (^) molkJ is released at P = 1 atm. The volume change is
V =
2.71 (^) cmg 3
2.93 (^) cmg 3
cm^3 g
Now 1 mole of CaCO 3 weights 100g, and using P = 1 bar = 105 Pa = (^105) mJ 3 = 10 ¡^4 cmkJ 3 we get
P V = 10 ¡^4
kJ cm^3
cm^3 g
100 g mol
= 2.8 10 ¡^4
kJ mol
So in this case P V H and therefore the enthalpy and energy changes are nearly identical. The work done is only a small fraction, 0.1%, of the enthalpy change. Volume change is more important when the total number of molecules is not the same on both sides of a reaction. For example, consider the enthalpy change in the formation of ammonia gas, NH 3 , through the reaction of hydrogen and nitrogen gases:
3 H 2 + N 2! 2 NH 3 (50)
This reaction converts 4 molecules into 2, so n = ¡ 2 , so the volume will go down. Using the ideal gas law at room temperature
(PV) = (n )RT = ¡ 2 8.
J
mol K
298 K = ¡4.
kJ mol
Let us compare this to the enthalpy change. Computing the enthalpy change by adding the bond
enthalpies gives H = ¡ 97 kJ mol for this reaction, not far o from the measured reaction enthalpy change of rH° = ¡91.88 (^) molkJ. So we nd that, (PV) = ¡4.5 (^) molkJ is around 5% of the total enthalpy
change in this case. 5% is small, but not so small that it can be neglected. Indeed, a 5% change in the energetics can have important eects on reaction kinetics. In summary, as a rule of thumb, enthalpies and energies are pretty similar for solids and liquids, diering at the less than a percent level, but for gases the dierence can be relatively large. The dierence between the energy and enthalpy change is essentially given by H ¡ E = (n)RT with
n the change in the number of moles of gas. This is equal to 2.48 (^) molkJ n at room temperature and pressure. These enthalpy changes are included in tabulated enthalpies of formation and reaction (which are dierent for liquids and gases) and are also included in bond enthalpies (de ned for gases and computed by averaging the formation enthalpies of various molecules with that bond).
5 Gibbs free energy
Gibbs free energy is de ned as
G H ¡ TS = E + PV ¡ TS (52)
The dierential of G is
dG = dE + d (PV ) ¡ d(TS) = VdP ¡ SdT + dN (53)
so G = G(P ; N ; T ) and @G @T
P ;N
= ¡S ;
@G
@P
T ;N
= V ;
@G
@N
P ;T
10 Section 5
Gibbs free energy is the constant pressure version of Helmholtz free energy; it is E ¡ TS with E replaced by H. Recall that Helmholtz free energy is useful at constant volume and constant temperature. At constant pressure, as in chemistry and biology, enthalpy and Gibbs free energy are used. Gibbs free energy gives the maximum amount of work that can be done at constant pressure and temperature. At constant pressure and constant temperature, Gibbs free energy has its minimum value at equilibrium.
A powerful function of the Gibbs free energy is that it tells which direction a reaction will go. In a chemical reaction at constant T and P , the amount of heat released is given by the enthalpy change rH. To get the sign right, note that if rH = Hprod ¡ Hreact is positive, rH > 0 , then the products have more enthalpy than the reactants, so the surroundings must put in energy and heat is withdrawn. The entropy change in the surroundings (the air or solution or whatever heat
bath is xing T and P in the rst place) is Ssurround = QT = ¡TrH. The entropy change in the
system is rS so the total entropy change is S = Ssys + Ssurround = rS ¡ T rH = ¡TrG. Since total entropy always increases the reaction can only proceed if the Gibbs free energy change is negative : rG < 0. Thus the sign of the Gibbs free energy change indicates which way the reaction will spontaneously proceed.
For an ideal monatomic gas, we want to express G as a function of P ; N and T. We start by using Eq. (32) to write
G = F + PV = ¡NkBT
ln
V
N
ln
2 mT h^2
+ PV (55)
Then we convert V to P using the ideal gas law V = NkPB Tso
G = ¡NkBT
ln kBT P
ln
2 mT h^2
Then we nd
=
@G
@N
P ;T
= kBT ln n^3 (57)
with = (^2) mkh BT p the thermal wavelength, in agreement with our previous result. This explicit
calculation con rms that for a monatomic ideal gas the chemical potential is the Gibbs free energy per molecule.
Note that we found that G = N for the ideal gas. This relation is actually very general. It follows immediately from the de nition G = E + PV ¡ TS and the Euler relation in Eq. (18) (which relied on the extensivity of entropy). When there are multiple species, G =
P
iN. Thus, similar to how the partition function and the Helmholtz free energy were equivalent, the chemical potential and the Gibbs free energy are equivalent.
5.1 Partial pressure
If we have a gas with N 1 molecules of type 1 and N 2 molecules of type 2, then they both exert pressure on the walls of the container. We call the pressure due to molecules of type i the partial pressure for that type and denote it Pi. The ideal gas law PV = NkBT holds for a single gas or for a mixture of gases. So say there are two gases with N 1 + N 2 = N. Then by the ideal gas law, the partial pressure of gas one is
Pi Ni V kBT (ideal gases) (58)
Since the ideal gas law PV = NkBT holds for the whole mixture, we can divide by it giving
Pi P
Ni N
Gibbs free energy 11
That the partition function is multiplicative in this way follows from assuming the gases are non- interacting: the energy of each molecule is independent of the other molecules. We are not making any other assumption about the molecules though, such as how many degrees of freedom there are. The Helmholtz free energy of the two gas mixture is then
F = ¡kBT ln Z ¡kBT
N 1
ln
V 1
N 1
+ N 2
ln
V 2
N 2
where Stirling's relation N! e¡NN N^ was used in the step. The chemical potential for gas 1 is then
1 =
@F
@N 1
T ;V ;N 2
= ¡kBT ln
V 1
N 1
and similarly for gas 2. To make sure we haven't messed anything up, the Gibbs free energy is
G = F + PV = ¡kBT
N 1
ln
V 1
N 1
+ N 2
ln
V 2
N 2
- (N 1 + N 2 )kBT ||||||||||||||||||||||||| ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||{z}}}}}}}}}}}}}}}}}}}}}}} }}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}} from ideal gas law
= 1 N 1 + 2 N 2 (71)
so that G =
P
i iNi^ as expected. All of the N and V dependence is explicit in the chemical potential, so it is helpful to pull these factors into a separate term. We write
i = ¡kBT ln Vi Ni = ¡kBT ln iV N
Thus,
Ni N = exp
i ¡ G 0 i kBT
where, using NV = (^) kP BT G 0 i = ¡kBT ln
kBT P i
G 0 i is the Gibbs free energy for a single molecule in isolation. To see this, for a single molecule
Ni = N = 1 and so G 0 i = i = 1 G. Eq. (73) is like the relation n = (^) ^13 exp
¡ " kBT
we derived last
lecture for a monatomic ideal gas. But here we computing the fraction N Ni rather than ni = N Vi and we have G 0 i written with pressure explicit rather than volume (so we can hold P xed). From here we get a relation between the fractions of reactants. Let us use the notation
xj Nj N
for the molar fraction of a reactant. Then, for something like
2 A + 3B! 7 C (76)
for which 2 A + 3B = 7C we would nd that the i drop out from the combination
xC^7 xA^2 xB^3
= exp
rG 0 kBT
this is the law of mass action. In the general form, the powers on the left hand side are determined by the stoichiometric coe cients in the reaction equation and rG 0 is the change in Gibbs free energy per particle (i.e. rG 0 = 7G 0 A ¡ 2 G 0 B ¡ 3 G 0 C for this example). Recall that we derived the law of mass action in the previous lecture for monatomic ideal gases, where the exponent had the energy change " per particle and the left hand side had the
concentrations [A] = Nj V rather than the molar fractions. That previous formula is a special case of this more general mass action formula, as you can check.
Gibbs free energy 13
To understand the law of mass action, recall that the total Gibbs free energy change in a reaction at equilibrium is zero, G = 0; otherwise, G could be minimized by moving molecules from one side to another. The overall G has a part per molecule , which is the Gibbs reaction energy rG and a part that depends on the concentrations, encoded in the xi fractions. This second part is entirely entropic, driven by the entropy of mixing, Thus the law of mass action says that in equilibrium these two contributions to G exactly cancel. It thereby lets us gure out the equilibrium concentrations from the Gibbs reaction energy per particle. By Dalton's law, we also have xi = P Pi so Eq. (77) can also be thought of as an equation for the equilibrium partial pressures. Chemists also prefer to use moles rather than particles, so we use rG, the Gibbs reaction energy (in (^) molkJ ) and use RT instead of kBT. The ratio of fractions is also given a name, the equilibrium constant and the symbol K:
K
xC^7 xA^2 xB^3
NC N
NA N
NB N
3 =^
P 4
PC^7
PA^2 PB^3
= e
¡ krGBT^0 = e¡^
rG RT (^) (78)
Let's do an example. In the reaction H 2 O(g) H 2 (g) + 12 O 2 (g) at T = 5000 K the Gibbs reaction
energy is rG = (^118) molkJ. If we start with 1 mol of H 2 O it will decompose into moles of H 2 and
2 moles of^ O^2 leaving^1 ¡^ moles of^ H^2 O. The total number of moles is^1 ¡^ +^ +^2 = 1 +^2. Then
xH 2 xO^ 1/2 2 xH 2 O
1 + (^2)
2 ¡ 1 + (^2)
q
1 ¡ 1 + (^2)
= exp
kJ mol 8.3 (^) molJ K 5000 K
Solve numerically for gives = 0.17. Thus at 5000K, 17% of the water molecules will be decomposed into H 2 and O 2. At room temperature rG = (^228) molkJ and = 10 ¡^27.
This example indicates an important qualitative point about using the Gibbs free energy:
generally rG is of order hundreds of (^) molkJ while RT = 2.5 (^) molkJ at room temperature. So the factor
exp
rG RT
is almost always either very very small, if rG > 0 or very very large if rG < 0. Thus
for exergonic reactions (rG < 0 ), the reaction strongly favors the products, while for endergonic reactions (rG > 0), the reactants are favored. Thus to a good approximation, we can use the rule of thumb that
If we mix some chemicals, the sign of rG tells which way the reaction will go.
This rule of thumb works only when we the concentrations are not exponentially small so that
the exp
¡RTrG
dominates. Of course, if the system is in equilibrium, it will not proceed in
any direction. Or if the concentration of products (or reactants) is small enough, the reaction will proceed in the only direction it can. In such a case, when the reaction proceeds against the direction of rG, the total Gibbs free energy of the system is still decreasing: the entropy of mixing associated with the concentration imbalance dominates over the rG eect from the reaction itself (remember, we pulled the concentration-dependence out in Eq.(72)). In summary, the law of mass action always tells us which way the reaction will proceed given some initial concentrations. The rule-of-thumb only determines the reaction direction when the concentrations are not exponentially small.
5.3 Direction of chemical reactions
We saw that the direction a reaction proceeds is determined by the sign of rG. What do we know about this sign? Although G 0 i and rG are in principle computable from the partition function for a single molecule, this is never actually done. One can easily look up rG under standard conditions P = 1 bar and T = 298 K. This standard value of the Gibbs reaction energy is denoted rG. It is more useful however to look up rH and rS and compute rG via:
rG = rH ¡ T rS (80)
14 Section 5
Note that we have assumed that all of the T dependence is in the explicit factor of T in the de - nition rG = rH ¡ T rS. What about the temperature dependence of rH and rS themselves? We can write rH = rE + r(PV). The rE contribution is from breaking bonds, which is inde-
pendent of T. Solids and liquids have the same volume, so r(PV) = (ngas)RT 8 J mol K ngasT^. Although this depends linearly on T , just like the T rS factor in rG, the numerical values for
rS (^161) molJ K in this example are much bigger. Thus we can neglect the temperature dependence of rH when solids and liquids are involved; for gases, it is a subleading eect. The temperature dependence of rS is generally logarithmic, since S NkBln T and generally very small. In the above discussion of CaCO 3 we used only the sign of rG to determine the reaction direction. The law of mass action tells us the relative concentrations in equilibrium
K = xCaO xCO 2 xCaCO 3
= exp
rG kBT
For T = 298 K, K = exp
130 0.0083 293
= 10 ¡^25 and T = 1500 K, K = exp
65 8.3 1.
= 185, so we see
that the equilibrium concentrations are hugely dierent at these two temperatures. Suppose the system is in equilibrium at some temperature. Then if we add more CaO to the system, xCaO will go up. To keep K the same, the system will adjust to remove CaO and increase CaCO 3. This is an example of a phenomenological observation called
Le Chatelier's principle : a system will work to restore equilibrium by counteracting any change.
6 Osmotic pressure
As a nal example, let us return to the topic of osmotic pressure, introduced in the discussion of entropy of mixing. Osmotic pressure is the pressure resulting from a concentration imbalance on the two sides of a semi-permeable barrier. For example, if you put a raisin in water, the higher sugar content of the raisin as compared to the water forces the water to be drawn into the raisin. The result is that the raisin swells up, almost back to the size of a grape. The drying out of a grape is also osmosis: water ows out of the grape into the air over time, and the grape desiccates. Grocers spray water on their fruits and vegetables to increase the local concentration of water so their produce looks more appealing. How about your ngers getting wrinkled when you stay in the bath too long. Is this osmosis? Say we have water in a U-shaped tube with a semi-permeable membrane in the middle. The membrane allows water to pass but not sugar. Now put some sugar on one side. As water ows into the sugar side, it will increase the pressure on that side, lowering the pressure on the other side. Thus the sugar water will move up. This is a physical eect due to the entropy of mixing.
Figure 4. Osmotic pressure arises when concentrations are dierent on two sides of a semi-permeable barrier. Adding glucose to one side causes a pressure imbalance. This can be compensated by applying an external pressure called the osmotic pressure.
16 Section 6
One way to quantify the eect is by the pressure you need to apply on the sugary side to restore the balance. This pressure is called the osmotic pressure and denoted by the symbol . To compute the osmotic pressure, let us start with some de nitions. A solution is a mixture of solvent and solute , with the solvent being the major component and the solute being a small addition. For example, in sugar water, water is the solvent and sugar the solute. We want to compute the chemical potential of the solvent (water) on both sides, which we can derive from the Gibbs free energy accounting for the entropy of mixing. Let us say that initially there is the same number Nw of solvent (water) molecules on both sides of the barrier and we then add Ns solute (sugar) molecules to one side. Recall that entropy of mixing is the extra entropy that a mixed substance has compared to a pure substance with the same properties (N ; T ; P ). A great thing about the entropy of mixing is that it only depends on whether the things mixing are indistinguishable, not any other other properties of those things (internal degrees of freedom, etc.). For N particles in a volume V , the
entropy is S = kBln
N! V^
N ^ ¡kBN ln N V. For our^ Nw^ molecules of solvent (water) and^ Ns^ molecules of solvent, the entropy of mixing is
Smix = ¡kB
Nw ln Nw V
||||||||||||||||||||||||| |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||{z}}}}}}}}}}}}}}}}}}}}}}}} }}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}} mixed system
¡ (Nw + Ns)ln
Nw + Ns V
||||||||||||||||||||||| ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||{z}}}}}}}}}}}}}}}}}}}}} }}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}} pure system with Nw+Nsparticles
=¡kB
Nw ln Nw Nw + Ns
Note that V dropped out. Indeed, the entropy of mixing only depends on the (^) N^1! factor.
In the limit that Ns Nw we can expand in (^) NNws to nd
Smix = kBNs
1 ¡ ln Ns Nw
+ O
Ns^2 Nw
Let's us write Gw(T ; P ) for the Gibbs free energy of pure solvent (water). Since Ns Nw, the enthalpy of the solute gives a negligible contribution to the total Gibbs free energy, and the only contribution that matters is from the entropy of mixing. So the total Gibbs free energy on the mixed side, generalizing Eq. (63), is
Gmixed = Gw ¡ TSmix = Gw ¡ kB TNs
1 ¡ ln Ns Nw
Thus the chemical potential of the solvent on the mixed side is then
w^ mixed(T ; P ) =
@Gmixed @Nw
T ;P ;Ns
= w(T ; P ) ¡ kBT Ns Nw
where w(T ; P ) = Gw Nw^ is the chemical potential of the pure solvent. Equilibrium requires the chemical potential of the solvent to be the same on both sides of the barrier. So
w(T ; Ppure) = w^ mixed(T ; Pmixed) = w(T ; Pmixed) ¡ kBT Ns Nw
The osmotic pressure we are trying to compute is the pressure dierence = Pmixed ¡ Ppure. In the limit Ns Nw we must have Ppure so that we can expand
w(T ; Pmixed) = w(T ; Ppure + ) = w(T ; Ppure) +
@w @P
T ;Nw
Since Gw = wNw, (^) @w @P
T ;Nw
Nw
@Gw @P
T ;Nw
V
Nw
Combining the last 3 equations gives
w(T ; Ppure) = w(T ; Ppure) +
V
Nw
¡ kBT Ns Nw
Osmotic pressure 17
7 Grand free energy
There is one more free energy that is used sometimes, called the grand free energy . We're not going to use it until Lecture 11, because it's harder to interpret physically, but I include the releant formulas here since they are closely related to the other free energies from this lecture. Recall that the grand canonical partition function is
Z(V ; T ; ) =
X
k
e¡^ "k+^ Nk^ (98)
where the sum is over microstates k with Nk particles and energy "k. We also showed that
¡kBT lnZ = hE i ¡ TS ¡ hN i (99)
We then de ne the grand free energy like the free energy F , but computed using the grand canonical ensemble.
= ¡kBT lnZ (100)
Thus,
(T ; V ; ) = hE i ¡TS ¡ hN i (101)
Recall that F (V ; T ; N ) and now we have (V ; T ; ), so the grand free energy has traded N for in the free energy. This is similar to how we used F (V ; T ; N ) = E ¡ TS to trades entropy for temperature using E(S ; V ; N ). Indeed, we can also write
(T ; V ; ) = F ¡ N (102)
Using Eq. (1) again, we nd
d = SdT ¡ PdV + Nd (103)
An important property of is that it is an extensive (like the other energies, internal energy, Helmholtz free energy, Gibbs free energy) function of only a single extensive variable V. Thus it
must be proportional to V. Since @ @V ;T
= ¡P we then have
= ¡PV (104)
This is a useful relation, similar to G = N for the Gibbs free energy. Like G = N , = ¡PV follows from the Euler relation in Eq. (18). We'll use the grand free energy in quantum statistical mechanics. It is not used in chemistry or for pure thermodynamics computations, since we can just use PV. I only include it here since it is a free energy and naturally part of the free energy lecture. We'll only need the de nition Eq. (100) and the relation (103) in future applications.
Grand free energy 19
