4.1 Constant (T, V, n) Experiments: The Helmholtz Free Energy

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1 Chapter 4 Free Energies The second law allows us to determine the spontaneous direction of of a process with constant (E, V, n). Of course, there are many processes for which we cannot control (E, V, n) as independent variables. For these, we will need alternative statements of the second law. This is where free energies come in. 4. Constant (T, V, n) Experiments: The Helmholtz Free Energy By far, the two most common sets of experimental conditions encountered in chemistry and biology are constant (T, V, n) and constant (T, P, n). We will take up (T, V, n) experiments in this section and (T, P, n) in a later one. Consider the experiment in Fig. 4.. We have a system with (E, V, n) embedded inside a very large bath. We imagine this bath is so huge that its temperature is essentially constant throughout the experiment. The system has fixed V and n, but its energy is allowed to change. Because V is fixed, there is no work involved. The only form of energy exchange is heat. The bath has (E b, V b, n b ), but V b and n b are fixed. The composite system is isolated from the outside, so the total energy E t = E + E b is constant. The composite system obeys the entropy maximum principle because it constitutes a constant (E t, V t, n t ) system. This setup allows us to emulate a constant-temperature environment. To determine the spontaneous direction of the system, we apply the second law to the entire composite system requiring that: S t = S + S b. Let s say during the process, the system changes its energy by E = q. The final energy of the bath is then E b = E b q = E b E. We device a reversible path between the initial and final states of the bath to calculate S b. Since the bath has not changed its temperature or volume, we can simply design an isothermal path at constant volume with an amount of heat q b = E b E b = E injected

2 MICROSCOPIC INTERPRETATION OF THE HELMHOLTZ FREE ENERGY: THE BOLTZMANN DISTRIBUTION Figure 4.: A composite system to emulate a constant-temperature environment. into it. Under constant T, V b and n b, S b = E/T. So we have S t = S E T or T S E, (4.) since the temperature T is always positive. To simplify this, the Helmholtz free energy is defined as: A E T S, (4.2) and we can turn the entropy maximum statement into simply A for constant (T, V, n) processes. Second Law for Constant (T, V, n) Processes: In a constant (T, V, n) system, a process with A is spontaneous. The corresponding variational statement of the second law is: Variational Statement of the Second Law for Constant (T, V, n) Processes: In a system with fixed (T, V, n), if an internal constraint partitions a system into two parts such that the total V and n remain fixed and moving it increases the total A of the system, that change is impossible. (Notice that in our setup, we have not explicitly assumed the system to necessarily have a constant temperature, only the bath. So this statement is more general that it seems.) With this, we can just focus our attention on the system and calculate A of the system alone to determine whether the process is spontaneous, without having to calculate the entropy change of the bath. Because E, T and S are state functions, A is also a state function. 4.2 Microscopic Interpretation of the Helmholtz Free Energy: The Boltzmann Distribution In Ch. 3, we saw that at constant (E, V, n), the spontaneous direction of a process is dictated by the total number of microstates Ω. The second law is

3 CHAPTER 4. FREE ENERGIES 3 a simple statement saying that processes which lead to a net increase in Ω at constant (E, V, n) should be spontaneous. In this section, we will take an excursion back to the microscopic world and ask why the statement of the second law in terms of A is the proper one for (T, V, n) processes. In particular, we will show that the second law can be stated in this case in terms of the properties of the system alone, instead of the system and bath. We return to the system in Fig. 4.. The temperature of the bath is kept constant, but the energy of the system is allowed to fluctuate. Because the energy of the system is not fixed, the number of possible microstates of the composite system is actually larger than when the energy is fixed, because instead of one fixed energy there are now a range of possible energies the system can take on. Accounting for this, the total number of microstates of the composite system, which we will call Q, must be taken to be an integral over all the possible energies of the system (for a system with many particles, the energy is essentially a continuous variable): Q(T ) = de Ω(E)Ω b (E b ) (4.3) where E is the energy of the system, Ω b and E b are the number of microstates and the energy of the bath, respectively. Because the total energy E t is fixed, E b = E t E can be considered to be a function of E. The prime on the integral indicates that it is actually constrained by the temperature T of the bath, so not all possible energies have equal importance in the integral. Therefore, Q should be a function of the temperature of the bath T. To determine how the temperature of the bath constrains the integral, we recall that ( S/ E) V,n = /T. Applying this to the bath s temperature using its entropy S b = k B ln Ω b and its energy E b, we get T = k d ln Ω b (E b ) B, (4.4) de b where we have assumed that V b and n b are fixed so we can consider Ω b to be just a function of a single variable E b. Integrating, we obtain Ω b (E b ) = Ce E b/k B T = Ce (Et E)/k BT, (4.5) where C comes from an integration constant. We set E = to try to find out what value C has, and this yields Ω b (E t ) = Ce Et/kBT, which should not depend on T because the number of microstates of the bath when the energy of the system is E = should be a constant. Calling this C, Eq.(4.3) then becomes: Q(T ) = C de Ω(E)e E/kBT. (4.6) In a constant (T, V, n) process, the second law demands that the total number of microstates of the composite system must not decrease, so Q must increase for a spontaneous process. Notice that when the bath is kept at a fixed T, Eq.(4.6)

4 AN EXAMPLE OF THE APPLICATION OF THE BOLTZMANN DISTRIBUTION expresses the total number of possible microstates of the composite system as a function of the energy of the system E alone and not the entire system or the bath, which accomplishes our goal. Since the number of microstates of the bath Ω b is always positive, Eq.(4.6) suggests that the factor C exp( E/k B T ) can be thought of as some kind of a probability density function for E. In fact, at temperature T, the probability ρ(e) that the system is in a state with energy E is proportional to exp( E/k B T ). This is known as the Boltzmann distribution. And since the energy of the system is not fixed, the observed macroscopic energy of the system must be an average over all the E under this distribution: E m = de E Ω(E)ρ(E) de Ω(E)ρ(E) = de E Ω(E)e E/k B T de Ω(E)e E/k B T, (4.7) where we have used E m for the macroscopic energy to avoid confusion with the integration variable E. This formula is valid only for constant (T, V, n). Notice that this formula can be rewritten in a more compact manner: E m = k B T 2 ( ln Q T ), (4.8) V,n where Q is given by Eq. (4.6). From the formula for the macroscopic energy E m we can obtain the macroscopic entropy S m at temperature T via the thermodynamic relationship ( S m / E m ) V,n = /T, so that at constant V and n: S m = T de T m T + C = T ( ) Em dt, (4.9) T V,n where the integration constant C must be since the third law says S = at T =. Using ρ(e) exp( E/k B T ) and after a little math, we arrive at the result: S m = E m T + k B ln Q. (4.) Rearranging, we have k B T ln Q = E m T S m, and dropping the subscript m we see that Q is related to the Helmholtz free energy A E T S by: k B T ln Q = A = E T S, (4.) where all variables on the right now refer to macroscopic quantities. In a later chapter, we will make use of Eq.(4.) to study phase transitions. 4.3 An Example of the Application of the Boltzmann Distribution In Sect. 3., we saw how the entropy S is related to the number of microstates Ω for a system with constant (E, V, n). In the last section, we have seen that

5 CHAPTER 4. FREE ENERGIES 5 under constant (T, V, n) condition, the total number of microstates is given by Q, and Q is related to the Helmholtz free energy A. For constant (T, V, n), the energy of the system is no longer fixed but will fluctuate, and we saw that the probability that the system is in a state with energy E is given by the Boltzmann distribution ρ(e) exp( E/k B T ). In this section, we will apply this to a simple microscopic model. We will consider an ideal gas with a spatially nonuniform potential. The basic model is the same as that in Sect. 3.2 and with no excluded volume. There are M cells and N particles. The particles are allowed to overlap. In the old model, all cells have the same energy (i.e. ), so every microstate has the same total energy E = no matter where the particles are. We now extend this model to include the possibility that some of the cells have a different energy from the rest. An example is shown in Fig This situation can occur, for instance, when a gas is passed through an affinity column. The gas particles tend to stick to the substrate, so they have a lower potential energy inside the column shown as the yellow cells. Let s say there are M of these yellow cells with potential energy ɛ, and M M cells with the usual energy. (If the yellow cells are sticky, ɛ should be negative.) When N particles go into the system, some of them, say N, will end up in the yellow cells with energy ɛ, and the rest, N N, will end up on the white cells with energy. The energy of such a state will be E = N ɛ, since the N N particles on the white cells have energy. We will stay in the dilute limit by assuming that N M and M. Figure 4.2: Microscopic model of an ideal gas in a spatially nonuniform potential. We will now use Eq. (4.8) to compute the macroscopic energy of the system. Again, we need this formula to calculate E m because the system s energy can fluctuate and the observed macroscopic energy will be an average over the whole distribution. In order to use Eq. (4.8), we must first divide up the microstates into sets, each with the same energy E. This can easily be achieved by grouping the microstates into sets such that every state within the set has the same N, since the energy of all such states is E = N ɛ. The probability of all the states in this set is given by the Boltzmann distribution ρ exp( N ɛ /k B T ). The

6 AN EXAMPLE OF THE APPLICATION OF THE BOLTZMANN DISTRIBUTION number of microstates in this set is just Ω(E), and Ω = M N N! (M M ) N N. (4.2) (N N )! The first factor comes from the number of possible ways to put N identical objects in the M yellow cells with no excluded volume, and the second factor comes from the number of possible ways to arrange the remaining number of particles N N in the M M white cells. This equation can be rearranged to give: [ ( ) ] N Ω(N ) = (M M ) N M, (4.3) N!(N N )! M M where we have written Ω as a function of N since all states in this set with N particles on the yellow cells have the same energy, and the factor outside the brackets does not change with N. We now take this expression for Ω and put it into Eq. (4.8) to compute E m. Since the energies are discrete for our model here, instead of integrating we can just sum over the different sets of microstates with different N, getting: Q = N N = Ω(N )e Nɛ/k BT = (M M ) N N N = (4.4) ( ) N M e ɛ/k BT, (4.5) N!(N N )! M M where the sum runs over all possible values of N from no particles on the yellow cells to all N particles on the yellow cells. Using the binomial theorem, ( + α) N = N N = N! N!(N N )! αn, (4.6) and letting α(t ) = e ɛ/kbt M /(M M ), we can carry out the summation above and get: Q = (M M ) N ( + α(t )) N (4.7) N! Using this in Eq. (4.8) finally yields: E m = Nɛ α(t ) + α(t ) ; α(t ) = M M M e ɛ/k BT. (4.8) This is the macroscopic internal energy of the system at temperature T. The macroscopic energy varies with temperature because the N particles prefer to distribute themselves on the yellow and white cells differently at different T. Since each particle on a yellow cell has an energy ɛ, the equilibrium number of particles on the yellow cells should be E m /ɛ, and the equilibrium mole fraction

7 CHAPTER 4. FREE ENERGIES 7 y of the particles that are sitting on the yellow cells is thus (E m /ɛ )/N, which is y = α(t ) + α(t ) ; α(t ) = M e ɛ/kbt. (4.9) M M If the yellow cells are sticky, ɛ <. In the low-temperature limit T, y and all the particles prefer the yellow cells. But as the temperature increases, some of the particles go into the white cells, which are higher in energy. In the high-temperature limit T, y M /M, indicating that the particles occupy the yellow and white cells randomly. At other temperatures, the fraction of particles on the yellow cells changes continuously between these two limits. This is the first example of an equilibrium between two different species in the system. You can consider the particles on the yellow cells as species and those on the white cells as species 2. The equilibrium can be represented by a chemical equation: 2 (4.2) The equilibirum constant for such a process would be defined as K = y 2 /y, where y and y 2 are the mole fractions of the two species. The equilibrium constant for this model can be calculated with Eq. (4.9) and y + y 2 =. We will study chemical equilibria in detail in Ch. 6. Before we leave this model, we will re-examine the equilibrium position of this system via another method by using the variational statement of the second law. We imagine using an internal constraint to partition the system, such that we can force N particles onto the yellow cells, and N N onto the white. The variational statement says that if I repartition the system by moving N, the Helmholtz free energy A must not increase. For the unconstrained system, A = k B T ln Q, and Q is given by Eq.(4.4). However, when the constraint is present, it forces N to be a unique value N. Since the energy of the system is just proportional to the number of particles on the yellow cells, enforcing a certain N forces the energy to also take on one unique value E = N ɛ. Therefore, with the constraint on, only one term in the sum in Eq. (4.4) contributes to Q, so Q = Ω(N ) exp( N ɛ /k B T ). After some math, we get the following for A with the internal constraint applied: A = Nk B T [y ln α(t ) y ln y ( y ) ln( y ) + C], (4.2) where y = N /N and C is a constant which does not depend on N or T. A is plotted as a function of y in Fig. 4.3 for α =.5. The minimum of A is at approximately y =.66. Whether the system starts with the constraint s position to the left or to the right of this, the spontaneous direction is for the system to head towards y =.66. We can also find the position of the constraint where A is minimum by differentiating Eq.(4.2) with respect to y, obtaining y = α(t ) + α(t ), (4.22) which, of course, is consistent with the result in Eq.(4.9) derived from the value of E m above.

8 84.4. CONSTANT (T, P, N) EXPERIMENTS: THE GIBBS FREE ENERGY A*/NkT y* Figure 4.3: A as a function of y at fixed T. 4.4 Constant (T, P, n) Experiments: The Gibbs Free Energy Consider the setup in Fig. 4.4 for a constant (T, P, n) experiment. We emulate a constant-t environment as before, but now we also allow the volume of the system to fluctuate by installing a piston. The bath is large, so when the piston moves the bath s volume hardly changes, and its pressure can be assumed constant. The piston can therefore be considered to be under constant external pressure P. Figure 4.4: Controlling the internal pressure of a gas. To determine the spontaneous direction of the system, we apply the second law to the entire composite system: S t = S + S b. During the process, the system changes its energy by E = q + w. The final energy of the bath is E b = E b q = E b E, and the final volume is V b = V b V, where V is the

9 CHAPTER 4. FREE ENERGIES 9 volume change of the system. We device a reversible path between the initial state (E b, V b, n b ) and final state of the bath (E b, V b, n b) to calculate S b. Since the bath has not changed its temperature or pressure, we can simply design an isothermal path at constant pressure with an amount of heat q b and work w b injected into is such that E b = q b + w b = E. Recall that the differential for S is ds = de/t + P dv/t µdn/t. Integrating this at constant T and P fot the bath gives S b = (q b + w b )/T P V/T. Therefore, S t = S E T P V T or T S ( E + P V ). (4.23) To simplify this, the Gibbs free energy is defined as: G E + P V T S = H T S, (4.24) and we can turn the entropy maximum statement into simply G, for a constant (T, P, n) process. Second Law for Constant (T, P, n) Processes: In a constant (T, P, n) system, a process with G is spontaneous. The corresponding variational statement of the second law is: Variational Statement of the Second Law for Constant (T, P, n) Processes: In a system with fixed (T, P, n), if an internal constraint partitions a system into two parts such that the total n remain fixed and moving it increases the total G of the system, that change is impossible. (Like in the case of A, we have not assumed explicity in our setup that the system must have constant pressure, only the bath.) G is a state function. From this point forward, we will focus mostly on the Gibbs free energy and (T, P, n) processes. To be able to use G to determine whether a process is spontaneous, we need to be able to calculate G as a function of (T, P, n). This is the subject of the next three sections. 4.5 Temperature-Dependence of G The strategy for calculating G at any (T, P, n) is the same as the one we have used for S (in fact, this method works for all thermodynamic variables). We first write down the differential dg. We design a reversible path to connect the initial and final states, and we integrate dg along this path. Since G is a state function, G is the same for any reversible path. We use the definition of G E + P V T S to write down its differential: dg = d(e + P V T S) = de + P dv + V dp T ds SdT = SdT + V dp + µdn, (4.25)

10 4.6. PRESSURE-DEPENDENCE OF G where we have used ds = de/t +P dv/t µdn/t. The differential of G suggests that G is a function of (T, P, n), which is consistent with the statement of the second law that at constant (T, P, n), it is G that determines the spontaneity of a natural process. Having Eq.(4.25), we can easily determine G 2 at a temperature T 2 if we know G at T : G = G 2 G = 2 SdT (constant P, n). (4.26) To use this equation, we need to know S as a function of T for constant P and n. From Ch. 3, we know that ds = δq rev /T, so at constant P and n, ds = C p (T )dt/t. Therefore, the double integral of this will then give us G 2 at T 2. Sometimes, it is easier to integrate d(g/t ) instead of dg itself. To do this, we need to know how G/T changes with T : ( ) (G/T ) T P,n = T ( ) G T P,n G, (4.27) T 2 We know ( G/ T ) = S, so if we use G = H T S to rewrite S = (G H)/T, the equation above becomes: ( ) (G/T ) = H T T 2, (4.28) which is known as the Gibbs-Helmhotlz equation. Integrating this over T at constant P and n also gives G at any T. P,n 4.6 Pressure-Dependence of G Using Eq.(4.25), we see that the proper way to obtain G 2 at a pressure P 2 from G at P, at constant T and n is: G 2 G = 2 V dp. (4.29) If the equation of state of the substance is known, this can be integrated. But even though this integration is not difficult, the pressure-dependence of G is usually obtained differently. The reason is that the majority of the change of G with P actually has nothing much to do with the particular gas itself, but is due to something more generic. To see this, we will first apply Eq.(4.29) to an ideal gas. Using P V = nrt we get: G i (T, P ) = G i (T, P ) + RT ln P P, (4.3)

11 CHAPTER 4. FREE ENERGIES where the reference pressure P is often taken to be either bar or atm and we have use n= mole so G is the molar Gibbs free energy. We see that G predominently depends on P like ln P. This behavior is exact for an ideal gas, and approximately so for real gases, and this logarithmic dependence is generic. If the majority of the P -dependence of G comes from this factor, we would want to include it as much as possible into the mechanism for calculating G, so we define a fugacity f, which is a function of T and P, such that the real G for a real gas is given by G(T, P, n) = G(T, P, n) + RT ln f P. (4.3) The fugacity is something like a non-ideal equivalence to the pressure. Since Eq.(4.3) is the definition of the fugacity, we cannot get the actual G from it without additional sources of information about the gas itself. But if the equation of state of the gas is known, we can then calculate G at any pressure as follows. Using Eq.(4.29) with state at P = and state 2 being the state at (T, P ) for which we want to calculate G: G(T, P ) G(T, ) = This equation obviously also works for an ideal gas Subtracting, we have G(T, P ) G i (T, P ) = G i (T, P ) G i (T, ) = [ G(T, ) + V dp ] [ V dp. (4.32) RT P G i (T, ) + dp. (4.33) ] RT P dp. (4.34) When the pressure is zero, G(T, ) = G i (T, ), so ( G(T, P ) G i (T, P ) = V RT ), (4.35) P and using Eqs.(4.3) and (4.3): RT ln f(t, P ) P = dp ( V RT ). (4.36) P Sometimes we deine the fugacity coefficient γ f/p to get the more compact formula: Z ln γ(t, P ) = dp, (4.37) P where Z P V /RT is the compressibility of the real gas. Eq.(4.37) can be integrated using the virial equation of state. Once the fugacity of the gas is

12 DEPENDENCE OF G ON THE NUMBER OF MOLES determined at a certain T and P, it can then be used to calculate G there. If the pressure is low, we can truncate the virial equation of state to get Z B (T )P, where B (T ) is the second virial coefficient at temperature T. This leads to the simple approximation: ln γ(t, P ) B (T )P. 4.7 Dependence of G on the Number of Moles The dependece of G on the number of moles of a pure substance is obviously due to the term µdn in Eq.(4.25). µ is the chemical potential; like P dv is the differential pressure-volume work, µdn is the work associated with adding dn moles of particles to the system and can be considered as part of the mechanical work done on the system. We expect that to obtain G for a substance with a certain number of moles n 2, we will just follow the same strategy as in the last two sections and integrate µ from n to n 2 at constant T and P if we know G for n. Though this is certainly correct, it is is unnecessary. Unlike the T - and P -dependence of G, there is a very simply relationship between G and n. Since G E + P V T S and E, V and S are extensive functinos, G must be extensive too. According to Eq.(4.25), dg = SdT + V dp + µdn, but of the three independent variables G depends on, only n is extensive, while P and T are both intensive. Therefore, all the extensivity of G must show up only in its dependence on n. Since µ = ( G/ n) P,T but G is proportional to n, we get the simple reslut that: G(T, P, n) µ(t, P ) = = G(T, P ), (4.38) n i.e. for a pure substance, µ must be just the molar Gibbs free energy. You can contrast this with other thermodynamics variables. For example, ds = T (de + P dv µdn), so S is extensive and it depends on three extensive variables. The entensivity of S must come from all three variables E, V and n, and S should be simply equal to T (E + P V µn)! Notice that in the list of thermodynamics functions we have considered so far, G is special in that only one of its dependent variables is extensive. S is also special in that all three of its dependent variables are extensive.

Chemical Equilibria. Chapter Extent of Reaction

Chemical Equilibria. Chapter Extent of Reaction Chapter 6 Chemical Equilibria At this point, we have all the thermodynamics needed to study systems in ulibrium. The first type of uilibria we will examine are those involving chemical reactions. We will