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Chapter 15 Atkins Physical Chemistry (10th Edition) Peter Atkins and Julio de Paula Chapter 15 Atkins Physical Chemistry (10th Edition) Peter Atkins and Julio de Paula Chapter 15 Atkins Physical Chemistry (10th Edition) Peter Atkins and Julio de Paula Chapter 15 Atkins Physical Chemistry (10th Edition) Peter Atkins and Julio de Paula Chapter 15 Atkins Physical Chemistry (10th Edition) Peter Atkins and Julio de Paula
CHAPTER 15 Statistical thermodynamics Statistical thermodynamics provides the link between the microscopic properties of matter and its bulk properties It provides a means of calculating thermodynamic properties from structural and spectroscopic data and gives insight into the molecular origins of chemical properties 15A The Boltzmann distribution The ‘Boltzmann distribution’, which is used to predict the popu lations of states in systems at thermal equilibrium, is among the most important equations in chemistry for it summarizes the populations of states; it also provides insight into the nature of ‘temperature’ The structure of the Topic separates its key implications from its rather heavy derivation interactions into account This Topic shows how that is done in principle by introducing the ‘canonical ensemble’, and hints at how this concept can be used 15E The internal energy and the entropy The main work of the chapter is to show how molecular partition functions are used to calculate (and give insight into) the two basic thermodynamic functions, the internal energy and the entropy The latter is based on another central equation introduced by Boltzmann, his definition of ‘statistical entropy’ 15F Derived functions 15B Partition functions The Boltzmann distribution introduces the concept of a ‘partition function’, which is the central mathematical concept of the rest of the chapter We see how to interpret the partition function and how to calculate it in a number of simple cases 15C Molecular energies A partition function is the thermodynamic version of a wavefunction, and contains all the thermodynamic information about a system As a first step in extracting that information, we see how to use partition functions to calculate the mean values of the basic modes of motion of a collection of independent molecules 15D The canonical ensemble Molecules interact with one another, and statistical thermodynamics would be incomplete without being able to take these With expressions relating internal energy and entropy to partition functions, we are ready to develop expressions for the derived thermodynamic functions, such as the Helmholtz and Gibbs energies Then, with the Gibbs energy available, we can make the final step into the calculations of chemically significant expressions by showing how equilibrium constants can be calculated from structural and spectroscopic data What is the impact of this material? There are numerous applications of statistical arguments in biochemistry We have selected one of the most directly related to partition functions: Impact I15.1 describes the helix–coil equilibrium in a polypeptide and the role of cooperative behaviour To read more about the impact of this material, scan the QR code, or go to bcs.whfreeman.com/webpub/chemistry/ pchem10e/impact/pchem-15-1.html 15A The Boltzmann distribution Contents 15A.1 Configurations and weights Instantaneous configurations Brief illustration 15A.1: The weight of a configuration (b) The most probable distribution Brief illustration 15A.2: The Boltzmann distribution (c) The relative population of states Example 15A.1: Calculating the relative populations of rotational states (a) 15A.2 605 605 606 607 607 608 608 The derivation of the Boltzmann distribution 608 (a) (b) The role of constraints The values of the constants Checklist of concepts Checklist of equations 609 610 611 611 ➤➤ Why you need to know this material? The Boltzmann distribution is the key to understanding a great deal of chemistry All thermodynamic properties can be interpreted in its terms, as can the temperature dependence of equilibrium constants and the rates of chemical reactions It also illuminates the meaning of ‘temperature’ There is, perhaps, no more important unifying concept in chemistry ➤➤ What is the key idea? The most probable distribution of molecules over the available energy levels subject to certain restraints depends on a single parameter, the temperature ➤➤ What you need to know already? You need to be aware that molecules can exist only in certain discrete energy levels (Foundations B and Topic 7A) and that in some cases more than one state has the same energy The principal mathematical tools used in this Topic are simple probability theory and Lagrange multipliers; the latter is explained in The chemist’s toolkit 15A.1 The problem we address in this Topic is the calculation of the populations of states for any type of molecule in any mode of motion at any temperature The only restriction is that the molecules should be independent, in the sense that the total energy of the system is a sum of their individual energies We are discounting (at this stage) the possibility that in a real system a contribution to the total energy may arise from interactions between molecules We also adopt the principle of equal a priori probabilities, the assumption that all possibilities for the distribution of energy are equally probable ‘A priori’ in this context loosely means ‘as far as one knows’ We have no reason to presume otherwise than that for a collection of molecules at thermal equilibrium, a vibrational state of a certain energy, for instance, is as likely to be populated as a rotational state of the same energy One very important conclusion that will emerge from the following analysis is that the overwhelmingly most probable populations of the available states depend on a single parameter, the ‘temperature’ That is, the work we here provides a molecular justification for the concept of temperature and some insight into this crucially important quantity 15A.1 Configurations and weights Any individual molecule may exist in states with energies ε0, ε1, … For reasons that will become clear, we shall always take the lowest available state as the zero of energy (that is, we set ε0 = 0), and measure all other energies relative to that state To obtain the actual energy of the system we may have to add a constant to the energy calculated on this basis For example, if we are considering the vibrational contribution to the energy, then we must add the total zero-point energy of any oscillators in the system (a) Instantaneous configurations At any instant there will be N0 molecules in the state with energy ε0, N1 in the state with ε1, and so on, with N0 + N1 + … = N, the total number of molecules in the system Initially we suppose that all the states have exactly the same energy The specification of the set of populations N0, N1,… in the form {N0,N1,…} is a statement of the instantaneous configuration of the system The instantaneous configuration fluctuates with time because the populations change, perhaps as a result of collisions At this stage the energies of all the configurations are identical so there is no restriction on how many of the N molecules are in each state 606 15 Statistical thermodynamics We can picture a large number of different instantaneous configurations One, for example, might be {N,0,0,…}, corres ponding to every molecule being in state Another might be {N − 2,2,0,0,…}, in which two molecules are in state The latter configuration is intrinsically more likely to be found than the former because it can be achieved in more ways: {N,0,0,…} can be achieved in only one way, but {N − 2,2,0,…} can be achieved in 12 N (N −1) different ways (Fig 15A.1; see the following Justification) If, as a result of collisions, the system were to fluctuate between the configurations {N,0,0,…} and {N − 2,2,0,…}, it would almost always be found in the second, more likely configuration, especially if N were large In other words, a system free to switch between the two configurations would show properties characteristic almost exclusively of the second configuration A general configuration {N0,N1,…} can be achieved in W different ways, where W is called the weight of the configuration The weight of the configuration {N0,N1,…} is given by the expression W= N! N ! N ! N ! Weight of a configuration (15A.1) with x! = x(x − 1)…1 and by definition 0! = 1 Equation 15A.1 is a generalization of the formula W = 12 N (N −1), and reduces to it for the configuration {N − 2,2,0,…} Justification 15A.1 The weight of a configuration First, consider the weight of the configuration {N–2,2,0,0,…}, which is prepared from the configuration {N,0,0,0,…} by the migration of two molecules from state into state One candidate for migration to state can be selected in N ways There are N − 1 candidates for the second choice, so the total number of choices is N(N − 1) However, we should not distinguish the choice (Jack, Jill) from the choice (Jill, Jack) because they lead to the same configurations Therefore, only half the choices lead to distinguishable configurations, and the total number of distinguishable choices is 12 N (N −1) Now we generalize this remark Consider the number of ways of distributing N balls into bins The first ball can be selected in N different ways, the next ball in N − 1 different ways for the balls remaining, and so on Therefore, there are N(N − 1)…1 = N! ways of selecting the balls for distribution over the bins However, if there are N0 balls in the bin labelled ε0, there would be N0! different ways in which the same balls could have been chosen (Fig 15A.2) Similarly, there are N1! ways in which the N1 balls in the bin labelled ε1 can be chosen, and so on Therefore, the total number of distinguishable ways N = 18 3! Figure 15A.1 Whereas a configuration {5,0,0,…} can be achieved in only one way, a configuration {3,2,0,…} can be achieved in the ten different ways shown here, where the tinted blocks represent different molecules Brief illustration 15A.1 The weight of a configuration To calculate the number of ways of distributing 20 identical objects with the arrangement 1, 0, 3, 5, 10, 1, we note that the configuration is {1,0,3,5,10,1} with N = 20; therefore the weight is W= 20 ! = 9.31 × 108 1! ! 3! 5!10 !1! Self-test 15A.1 Calculate the weight of the configuration in which 20 objects are distributed in the arrangement 0, 1, 5, 0, 8, 0, 3, 2, 0, Answer: 4.19 × 1010 6! 5! 4! Figure 15A.2 The 18 molecules shown here can be distributed into four receptacles (distinguished by the three vertical lines) in 18! different ways However, 3! of the selections that put three molecules in the first receptacle are equivalent, 6! that put six molecules into the second receptacle are equivalent, and so on Hence the number of distinguishable arrangements is 18!/3!6!5!4!, or about 515 million of distributing the balls so that there are N0 in bin ε0, N1 in bin ε1, etc regardless of the order in which the balls were chosen is N!/N0!N1!…, which is the content of eqn 15A.1 It will turn out to be more convenient to deal with the natural logarithm of the weight, ln W, rather than with the weight itself We shall therefore need the expression N! ln W = ln N ! N1 ! N ! ln xy = ln x + ln y = ln ( xy ) = ln x − ln y = ln N !− ln N ! N ! N ! ln N ! − ln N ! − ln N1 ! − ln N ! − = ln N ! − ∑ ln N ! i i 15A The Boltzmann distribution One reason for introducing ln W is that it is easier to make approximations In particular, we can simplify the factorials by using Stirling’s approximation ln x ! ≈ ( x + 12 ) ln x − x + 12 ln 2π Stirling’s approxi mation x ≫ 1 (15A.2a) This approximation is in error by less than per cent when x is greater than about 10 We deal with far larger values of x, and the simplified version ln x !≈ x ln x − x x ≫ 1 Stirling’s approximation (15A.2b) is adequate Then the approximate expression for the weight is ∑ {N ln N − N } = N ln N − N − ∑ N ln N + N [because ∑ N = N ] = N ln N − ∑ N ln N (15A.3) ln W = {N ln N − N } − i i i 607 where E is the total energy of the system The second constraint is that, because the total number of molecules present is also fixed (at N), we cannot arbitrarily vary all the populations simultaneously Thus, increasing the population of one state by demands that the population of another state must be reduced by Therefore, the search for the maximum value of W is also subject to the condition ∑N = N Constant total number of molecules i i Number constraint (15A.5) We show in the next section that the populations in the configuration of greatest weight, subject to the two constraints in eqns 15A.4 and 15A.5, depend on the energy of the state according to the Boltzmann distribution: i i i i i i i i i (b) The most probable distribution We have seen that the configuration {N − 2,2,0,…} dominates {N,0,0,…}, and it should be easy to believe that there may be other configurations that have a much greater weight than both We shall see, in fact, that there is a configuration with so great a weight that it overwhelms all the rest in importance to such an extent that the system will almost always be found in it The properties of the system will therefore be characteristic of that particular dominating configuration This dominating configuration can be found by looking for the values of Ni that lead to a maximum value of W Because W is a function of all the Ni, we can this search by varying the Ni and looking for the values that correspond to dW = 0 (just as in the search for the maximum of any function), or equivalently a maximum value of ln W However, there are two difficulties with this procedure At this point we allow the states to have different energies The first difficulty that results from this change is the need to take into account the fact that the only permitted configurations are those corresponding to the specified, constant, total energy of the system This requirement rules out many configurations; {N,0,0,…} and {N − 2,2,0,…}, for instance, have different energies (unless ε0 and ε1 happen to have the same energy), so both cannot occur in the same isolated system It follows that in looking for the configuration with the greatest weight, we must ensure that the configuration also satisfies the condition ∑N ε = E Constant energy i i i Energy constraint (15A.4) Ni e − βε = N ∑e − βε i i i Boltzmann distribution (15A.6a) The denominator on eqn 15A.6a is denoted q and called the partition function: q= ∑e Definition − βε i i Partition function (15A.6b) At this stage the partition function is no more than a convenient abbreviation for the sum; but in Topic 15B we see that it is central to the statistical interpretation of thermodynamic properties Equation 15A.6a is the justification of the remark that a single parameter, here denoted β, determines the most probable populations of the states of the system We confirm in Topic 15D and anticipate throughout this Topic that β= kT (15A.7) where T is the thermodynamic temperature and k is Boltzmann’s constant In other words: The temperature is the unique parameter that governs the most probable populations of states of a system at thermal equilibrium Brief illustration 15A.2 The Boltzmann distribution Suppose that two conformations of a molecule differ in energy by 5.0 kJ mol−1 (corresponding to 8.3 zJ for a single molecule; 1 zJ = 10−21 J), so conformation A lies at energy and conformation B lies at ε = 8 zJ At 20 °C (293 K) the denominator in eqn 15A.6a is ∑e i − βε i = + e − ε /kT = + e −(8.3×10 −21 J)/(1.381×10−23 J K −1 )×(293 K ) =1.13 608 15 Statistical thermodynamics The proportion of molecules in conformation B at this temperature is therefore N B e −(8.3×10 = N −21 J )/(1.381×10−23 J K −1 )×(293 K ) 1.13 = 0.11 or 11 per cent of the molecules Self-Test 15A.2 Suppose that there is a third conformation a further 0.50 kJ mol−1 above B What proportion of molecules will now be in conformation B? Answer: 0.10, 10 per cent (c) The relative population of states If we are interested only in the relative populations of states, the sum in the denominator of the Boltzmann distribution need not be evaluated, because it cancels when the ratio is taken: N i e − βε = = e − β (ε −ε ) Nj e − βε i i Thermal equilibrium j j Boltzmann population ratio (15A.8a) That β ∝ 1/T is plausible is demonstrated by noting from eqn 15A.8a that for a given energy separation the ratio of populations N1/N0 decreases as β increases, which is what is expected as the temperature decreases At T = 0 (β = ∞) all the population is in the ground state and the ratio is zero Equation 15A.8a is enormously important for understanding a wide range of chemical phenomena and is the form in which the Boltzmann distribution is commonly employed (for instance, in the discussion of the intensities of spectral transitions, Topics 12A and 14A) It tells us that the relative population of two states falls off exponentially with their difference in energy A very important point to note is that the Boltzmann distribution gives the relative populations of states, not energy levels Several states might correspond to the same energy, and each state has a population given by eqn 15A.6 If we want to consider the relative populations of energy levels rather than states, then we need to take into account this degeneracy Thus, if the level of energy εi is gi-fold degenerate (in the sense that there are gi states with that energy), and the level of energy εj is gj-fold degenerate, then the relative total populations of the levels are given by g N i g i e − βε = = i e − β (ε −ε ) Nj g j e − βε g j i i j j Thermal equili brium, degene racies Boltzmann population ratio (15A.8b) Example 15A.1 Calculating the relative populations of rotational states Calculate the relative populations of the J = 1 and J = 0 rotational states of HCl at 25 °C Method Although the ground state is non-degenerate, the level with J = 1 is triply degenerate (MJ = 0, ±1); see Topic 12B From Topic 12B, the energy of state with quantum number ( J + 1) Use B = 10.591 cm −1 A useful relation is J is ε J = hcBJ kT/hc = 207.22 cm−1 at 298.15 K Answer The energy separation of states with J = 1 and J = 0 is ε1 − ε = 2hcB The ratio of the population of a state with J = 1 and any one of its three states MJ to the population of the single state with J = 0 is therefore N J ,M J = e −2hcBβ N0 The relative populations of the levels, taking into account the three-fold degeneracy of the upper state, is NJ = 3e −2hcBβ N0 Insertion of hcB β = hcB /kT = (10.591 cm −1)/(207.22 cm −1) = 0.0511… then gives NJ = 3e −2×0.0511… = 2.708 N0 We see that because the J = 1 level is triply degenerate, it has a higher population than the level with J = 0, despite being of higher energy As the example illustrates, it is very important to take note of whether you are asked for the relative populations of individual states or of a (possibly degenerate) energy level Self-test 15A.3 What is the ratio of the populations of the levels with J = 2 and J = 1 at the same temperature? Answer: 1.359 15A.2 The derivation of the Boltzmann distribution We have remarked that ln W is easier to handle than W Therefore, to find the form of the Boltzmann distribution, we look for the condition for ln W being a maximum rather than dealing directly with W Because ln W depends on all the Ni, when a configuration changes and the Ni change to Ni + dNi, the function ln W changes to ln W + d ln W, where d ln W = ∑ i ∂ ln W ∂N i dN i All this expression states is that a change in ln W is the sum of contributions arising from changes in each value of Ni 15A The Boltzmann distribution (a) The role of constraints At a maximum, d ln W = 0 However, when the Ni change, they so subject to the two constraints ∑ε dN = ∑dN = i i Constraints (15A.9) i i i The first constraint recognizes that the total energy must not change, and the second recognizes that the total number of molecules must not change These two constraints prevent us from solving d ln W = 0 simply by setting all (∂ ln W/∂Ni) = 0 because the dNi are not all independent The way to take constraints into account was devised by the French mathematician Lagrange, and is called the method of undetermined multipliers (The chemist’s toolkit 15A.1) All we need of that method is as follows: • Each constraint is multiplied by a constant and then added to the main variation equation • The variables are then treated as though they were all independent • The constants are evaluated at the end of the calculation The chemist’s toolkit 15A.1 The method of undetermined multipliers Suppose we need to find the maximum (or minimum) value of some function f that depends on several variables x1, x 2, …, xn When the variables undergo a small change from xi to xi + δxi the function changes from f to f + δf, where n δf = ∂f ∑ ∂x δx i i =1 i 609 Because δg is zero, we can multiply it by a parameter, λ, and add it to the preceding equation: n ∂f ∂g ∑ ∂x + λ ∂x δx = i i =1 i i This equation can be solved for one of the δx, δxn for instance, in terms of all the other δx i All those other δx i (i = 1, 2, …, n − 1) are independent, because there is only one constraint on the system But λ is arbitrary; therefore we can choose it so that ∂f ∂g ∂x + λ ∂x = n n (A) Then n−1 ∂g ∂f ∑ ∂x + λ ∂x δx = i i =1 i i Now the n − 1 variations δxi are independent, so the solution of this equation is ∂g ∂f ∂x + λ ∂x = i = 1, 2, …, n − i i However, eqn A has exactly the same form as this equation, so the maximum or minimum of f can be found by solving ∂g ∂f ∂x + λ ∂x = i = 1, 2, …, n i i If there is more than one constraint, g1 = 0, g2 = 0, …, and this final result generalizes to ∂g ∂g ∂f ∂x + λ1 ∂x + λ2 ∂x + = 0, i = 1, 2, …, n i i i with a corresponding multiplier, λ1, λ 2, … for each constraint At a minimum or maximum, δf = 0, so then n ∂f ∑ ∂x δx = i =1 Thus, as there are two constraints we introduce the two constants α and − β and write i i If the xi were all independent, all the δx i would be arbitrary, and this equation could be solved by setting each (∂f/∂x i) = 0 individually When the xi are not all independent, the δxi are not all independent, and the simple solution is no longer valid We proceed as follows Let the constraint connecting the variables be an equation of the form g = 0 The constraint g = 0 is always valid, so g remains unchanged when the xi are varied: n δg = i = ∂ ln W dN + α ∂N i i ∑dN − β ∑ε dN i i ∂ ln W + α − βε i dN i = ∂N i ∑ i i i i All the dNi are now treated as independent Hence the only way of satisfying d ln W = 0 is to require that for each i, ∂g ∑ ∂x δx = i =1 ∑ i i ∂ ln W ∂N i + α − βε i = (15A.10) 610 15 Statistical thermodynamics when the Ni have their most probable values We show in the following Justification that ∂ ln W N = − ln i ∂N i N (15A.11) ∂N j = δij ∂N i with δ ij the Kronecker delta (δ ij = 1 if i = j, δ ij = 0 otherwise) Then It follows from eqn 15A.10 that − ln If i ≠ j, Nj is independent of Ni, so ∂Nj/∂Ni = 0 However, if i = j, ∂Ni/∂Ni = 1 Therefore, Ni + α − βε i = N ∑ and therefore that j N i α −βε =e N ∂Nj lnN j = ∂N i (15A.12) i = which is very close to being the Boltzmann distribution Justification 15.A.2 The derivative of the weight Equation 15A.3 for W is ln W = N ln N − ∑N ln N i i There is a small housekeeping step to take before differentiating ln W with respect to Ni: this equation is identical to i ln W = N ln N − ∑ j ∑ j (1/N N j ) ∂N j / ∂Ni δij ∂ ln Nj ∂Nj ln N + N j j ∂N i ∂N i δ ij δ ln N + ∂Nj = j ij ∂N i ∑δ ( ln N +1) ij j j = ln N i + On bringing the two terms together we can write N ∂ ln W = ln N + − (ln N i + 1) = − ln i N ∂N i as in eqn 15A.11 ∑N ln N j j because all we have done is to change the ‘name’ of the states from i to j This step makes sure that we not confuse the i in the differentiation variable (Ni) with the i in the summation Now differentiation of this expression gives j ∂ ln W ∂(N ln N ) = − ∂N i ∂N i ∑ ∑N = ∑Ne α − βε i i i = Neα i ∑e i − βε i Because the N cancels on each side of this equality, it follows that (1/N ) ∂N / ∂Ni ∂ ( N ln N ) ∂N ∂ ln N = ln N + N ∂N i ∂N i ∂N i eα = ∂N = ln N + ∂N i The (blue) ln N in the first term on the right in the second line arises because N = N1 + N2 + … and so the derivative of N with respect to any of the Ni is 1: that is, ∂N/∂Ni = 1 The second term on the right in the second line arises because ∂(ln N)/∂Ni = (1/N)∂N/∂Ni The final is then obtained in the same way as in the preceding remark, by using ∂N/∂Ni = 1 For the derivative of the second term we first note that ∂ ln Nj ∂Nj = ∂N i Nj ∂Ni ∑ e − βε i = ln N + At this stage we note that N= ∂(Nj ln Nj ) ∂N i The derivative of the (blue) first term on the right is obtained as follows: j (b) The values of the constants (15A.13) i and therefore N i α −βε e − βε =e = eα e − βε = N ∑ e − βε Boltzmann distribution i i i i (15A.14) i which is eqn 15A.6a The development of statistical concepts of thermodynamics begins with the Boltzmann distribution, with quantum theory (Chapter 7) providing insight into and ways of calculating the energies εi in eqn 15A.14 15A The Boltzmann distribution 611 Checklist of concepts ☐ 1 The principle of equal a priori probabilities assumes that all possibilities for the distribution of energy are equally probable ☐ 2 The instantaneous configuration of a system of N molecules is the specification of the set of populations N0, N1, … of the energy levels ε0, ε1, … ☐ 3 The Boltzmann distribution gives the numbers of molecules in each state of a system at any temperature ☐ 4 The relative populations of energy levels, as opposed to states, must take into account the degeneracies of the energy levels Checklist of equations Property Equation Comment Equation number Boltzmann distribution N i /N = e − βεi /q β = 1/kT 15A.6a Partition function q= see Topic 15B 15A.6b gj, gj are degeneracies 15A.8b ∑e − βε i i Boltzmann population ratio N i /Nj = ( g i /g j )e − β (εi −ε j ) 15B Molecular partition functions Contents 15B.1 The significance of the partition function 15B.2 Contributions to the partition function Brief illustration 15B.1: A partition function The translational contribution Brief illustration 15B.2: The translational partition function (b) The rotational contribution Example 15B.1: Evaluating the rotational partition function explicitly Brief illustration 15B.3: The rotational contribution Brief illustration 15B.4: The symmetry number (c) The vibrational contribution Brief illustration 15B.5: The vibrational partition function Example 15B.2: Calculating a vibrational partition function (d) The electronic contribution Brief illustration 15B.6: The electronic partition function (a) Checklist of concepts Checklist of equations 612 613 614 615 over the available energy levels (Topic 15A) In that Topic we introduce the concept of partition function, which is developed here You need to be aware of the expressions for the rotational and vibrational levels of molecules (Topics 12B and 12D) and the energy levels of a particle in a box (Topic 8A) 616 616 617 617 619 620 620 621 621 621 622 622 ➤➤ Why you need to know this material? Statistical thermodynamics provides the link between molecular properties that have been calculated or derived from spectroscopy and thermodynamic properties, including equilibrium concepts The connection is the partition function Therefore, this material is an essential foundation for understanding physical and chemical properties of bulk matter in terms of the properties of the constituent molecules ➤➤ What is the key idea? The partition function is calculated by drawing on calculated or spectroscopically derived structural information about molecules ➤➤ What you need to know already? You need to know that the Boltzmann distribution expresses the most probable distribution of molecules The partition function q = ∑ e − βε is introduced in Topic 15A i simply as a symbol to denote the sum over states that occurs in the denominator of the Boltzmann distribution (eqn 15A.6a, pi = e − βε /q , with pi = Ni/N), but it is far more important than that might suggest For instance, it contains all the information needed to calculate the bulk properties of a system of independent particles In this respect q plays a role for bulk matter very similar to that played by the wavefunction in quantum mechanics for individual molecules: q is a kind of thermal wavefunction This Topic shows how the partition function is calculated in a variety of important cases in preparation for seeing how thermodynamic information is extracted (in Topics 15C and 15E) i i 15B.1 The significance of the partition function The molecular partition function is q= ∑e − βε i Definition Molecular partition function (15B.1a) states i where β = 1/kT As emphasized in Topic 15A, the sum is over states, not energy levels If gi states have the same energy εi (so the level is gi-fold degenerate), we write q= ∑g e i levels i − βε i Alternative definition Molecular partition function (15B.1b) where the sum is now over energy levels (sets of states with the same energy), not individual states Also as emphasized in Topic 15A, we always take the lowest available state as the zero of energy and set ε0 = 0 15b Molecular partition functions Brief illustration 15B.1 A partition function Suppose a molecule is confined to the following non-degenerate energy levels: 0, ε, 2ε, … (Fig 15B.1; later we shall see that this array of levels is used when considering molecular vibration) Then the molecular partition function is Partition function, q 1.4 1.5 1.2 q = + e − βε + e −2 βε + = + e − βε + (e − βε )2 + ε 613 0.5 Temperature, kT/ε 1 Temperature, kT/ε 10 Figure 15B.3 The partition function for a two-level system as a function of temperature The two graphs differ in the scale of the temperature axis to show the approach to as T → 0 and the slow approach to as T → ∞ 3ε 2ε ε Figure 15B.1 The equally-spaced infinite array of energy levels used in the calculation of the partition function A harmonic oscillator has the same spectrum of levels We have derived the following important expression for the partition function for a uniform ladder of states of spacing ε: The sum of the geometrical series 1 + x + x 2 + … is 1/(1 − x), so in this case q= 10 pi = 0 Temperature, kT/ε 1 − e − βε Uniform ladder Partition function (15B.2a) We can use this expression to interpret the physical significance of a partition function To so, we first note that the Boltzmann distribution for this arrangement of energy levels gives the fraction, pi = Ni/N, of molecules in the state with energy εi as 1 − e − βε This function is plotted in Fig 15B.2 Partition function, q q= 10 Figure 15B.2 The partition function for the system shown in Fig 15B.1 (a harmonic oscillator) as a function of temperature Self-test 15B.1 Suppose the molecule can exist in only two states, with energies and ε Derive and plot the expression for the partition function Answer: q = 1+ e − βε , Fig.15 B.3 e − βε q i = (1− e − βε )e − βε i Uniform ladder Population (15B.2b) Figure 15B.4 shows how pi varies with temperature At very low temperatures (large β), where q is close to 1, only the lowest state is significantly populated As the temperature is raised, the population breaks out of the lowest state, and the upper states become progressively more highly populated At the same time, the partition function rises from towards 2, so we see that its value gives an indication of the range of states populated at any given temperature The name ‘partition function’ reflects the sense in which q measures how the total number of molecules is distributed—partitioned—over the available states The corresponding expressions for a two-level system derived in Self-test 15B.1 are q = 1+ e − βε pi = e − βε q i = Two-level system e − βε 1+ e − βε Partition function (15B.3a) i Two-level system Population (15B.3b) 644 15 Statistical thermodynamics Checklist of equations Property Equation Comment Equation number Internal energy U(T) = U(0) − (N/q)(∂q/∂β)V = −N(∂ ln q/∂β)V Independent molecules 15E.2b Heat capacity CV = Nk β (∂2 ln q /∂β )V Independent molecules 15E.5 T ≫ θM 15E.6 CV , m = (2 + R* + 2 V* )R Boltzmann formula for the entropy S = k ln W Definition 15E.7 The entropy in terms of the partition function S ={U − U(0)}/T + Nk ln q Distinguishable molecules 15E.8a S = {U − U(0)}/T + Nk ln(q/N) Indistinguishable molecules 15E.8b Sackur–Tetrode equation Sm (T ) = R ln(Vm e5/2 / NA Λ3 ) Entropy of a monatomic perfect gas 15E.10a Residual molar entropy Sm(0) = R ln s s is the number of equivalent sites 15E.14b 15F Derived functions entropy, which in terms of the canonical partition function are given by Contents 15F.1 The derivations Example 15F.1: Deriving an equation of state Example 15F.2: Calculating a standard Gibbs energy of formation from partition functions 15F.2 645 645 646 Equilibrium constants 647 The relation between K and the partition function (b) A dissociation equilibrium Example 15F.3: Evaluating an equilibrium constant (c) Contributions to the equilibrium constant 647 648 648 648 (a) Checklist of concepts Checklist of equations ∂ lnQ U (T ) = U (0) − ∂β V 650 650 ➤➤ Why you need to know this material? The power of chemical thermodynamics stems from its deployment of a variety of derived functions, particularly the enthalpy and Gibbs energy It is therefore important to relate these to structural features through partition functions One hugely important quantity that is illuminated in this way is the equilibrium constant ➤➤ What is the key idea? The partition function provides a link between spectroscopic and structural data and the derived functions of thermodynamics, particularly the equilibrium constant ➤➤ What you need to know already? This Topic develops the discussion of internal energy and entropy (Topic 15E) You need to know the relations between those properties and the enthalpy (Topic 2B) and the Helmholtz and Gibbs energies (Topic 3C) The final section makes use of the relation between standard Gibbs energy and the equilibrium constant (Topic 6A) Classical thermodynamics makes extensive use of various derived functions Thus, in thermochemistry it focuses on the enthalpy and, provided the pressure and temperature are constant, in discussions of spontaneity it focuses on the Gibbs energy In this Topic we see how these properties can be related to and understood in terms of partition functions All these properties are derived from the internal energy and the S(T ) = Internal energy (15F.1a) U (T ) −U (0) + k lnQ T Entropy (15F.1b) These two general expressions can be adapted for collections of independent molecules by writing Q = q N for distinguishable molecules and Q = q N/N! for indistinguishable molecules (as in a gas) 15F.1 The derivations The Helmholtz energy, A, is defined as A = U − TS This relation implies that A(0) = U(0), so substitution of the expressions for U(T) and S(T) leads to the very simple expression A(T ) = A(0) − kT lnQ Helmholtz energy (15F.2) An infinitesimal change in conditions changes the Helmoltz energy by dA = −pdV − SdT (this is the analogue of the expression for dG derived in Topic 3D (eqn 3D.7, dG = Vdp − SdT) Therefore, it follows that on imposing constant temperature (dT = 0), the pressure and the Helmholtz energy are related by p = −(∂A/∂V)T It then follows from eqn 15F.2 that ∂ lnQ p = kT ∂V T Pressure (15F.3) This relation is entirely general, and may be used for any type of substance, including perfect gases, real gases, and liquids Because Q is in general a function of the volume, temperature, and amount of substance, eqn 15F.3 is an equation of state of the kind discussed in Topic 1C Example 15F.1 Deriving an equation of state Derive an expression for the pressure of a gas of independent particles Method We should suspect that the pressure is that given by the perfect gas law To proceed systematically, substitute the 646 15 Statistical thermodynamics explicit formula for Q for a gas of independent, indistinguishable molecules Answer For a gas of independent molecules, Q = q N/N! with q = V/Λ 3: NkT ∂q kT ∂ q N = q ∂V T q N ∂V T NkT ∂(V /Λ3 ) NkT nN A kT nRT = = = = V V V V /Λ3 ∂V T The calculation shows that the equation of state of a gas of independent particles is indeed the perfect gas law, pV = nRT Self-test 15F.1 Derive the equation of state of a sample for which Q = q Nf/N!, with q = V/Λ , where f depends on the volume Answer: p = nRT/V + kT(∂ ln ƒ/∂V)T At this stage we can use the expressions for U and p in the definition H = U + pV to obtain an expression for the enthalpy, H, of any substance: ∂ lnQ ∂ lnQ H (T ) = H (0) − + kTV ∂β V ∂V T Enthalpy (15F.4) The fact that eqn 15F.4 is rather cumbersome is a sign that the enthalpy is not a fundamental property: as shown in Topic 2B, it is more of an accounting convenience For a gas of independent particles U − U(0) = 23 nRT and pV = nRT Therefore, for such a gas, H − H (0) = 25 nRT (15F.5) One of the most important thermodynamic functions for chemistry is the Gibbs energy, G = H − TS = A + pV We can now express this function in terms of the partition function by combining the expressions for A and p: ∂ lnQ G(T ) = G(0) − kT lnQ + kTV ∂V T Gibbs energy (15F.6) This expression takes a simple form for a gas of independent molecules because pV in the expression G = A + pV can be replaced by nRT: G(T ) = G(0) − kT lnQ + nRT (15F.7) Furthermore, because Q = q and therefore ln Q = N ln q − ln N!, it follows by using Stirling’s approximation (ln N! = N ln N − N) that we can write N/N!, (15F.8) = G(0) − nRT ln q + kT (N ln N − N ) + nRT = G(0) −nRT ln q N with N = nNA Now we see another interpretation of the Gibbs energy: because q is the number of thermally accessible states and N is the number of molecules, the difference G(T) − G(0) is proportional to the logarithm of the average number of thermally accessible states per molecule It will turn out to be convenient to define the molar partition function, qm = q/n (with units mol−1), for then kT ∂Q N ! kT ∂(q N /N !) ∂ lnQ = N = p = kT q ∂V T ∂V T Q ∂V T = G(T ) = G(0) − NkT ln q + kT ln N !+ nRT G(T ) = G(0) − nRT ln qm NA Independent Gibbs energy molecules (15F.9) To use this expression, G(0) is identified with the energy of the system when all the molecules are in their ground state, E0 To calculate the standard Gibbs energy, the partition function has its standard value, q m< , which is evaluated by setting the molar volume in the translational contribution equal to the standard molar volume, so q m< =(Vm< /Λ3 )q Rq V with Vm< = RT /p < Example 15F.2 Calculating a standard Gibbs energy of formation from partition functions Calculate the standard Gibbs energy of formation of H 2O(g) at 25 °C Method Write the chemical equation for the formation reaction, and then the expression for the standard Gibbs energy of formation in terms of the Gibbs energy of each molecule; then express those Gibbs energies in terms of the molecular partition function of each species Ignore molecular vibration as it is unlikely to be excited at 25 °C Take numerical values from the Resource section together with the following rotational constants of H 2O: 27.877, 14.512, and 9.285 cm−1 Take for the atomization energy of H 2O the value −237 kJ mol−1 Answer The chemical reaction is H2 (g) + 12 O2 (g) → H2O(g) Therefore, ∆ f G < = Gm< (H2O, g) − Gm< (H2 , g) − 12 Gm< (O2 , g) Now write the standard molar Gibbs energies in terms of the standard molar partition functions of each species J: Gm< (J) = E0,m (J) − RT ln q m< (J) NA V< q m< (J) = q mT< (J)q R (J) = m q R (J) Λ(J) 15F Derived functions 15F.2 Equilibrium Therefore q < (H ) − E0,m (H2 ) − RT ln m NA q < (O ) − 12 E0,m (O2 ) − RT ln m NA q m< (H2 O)/NA (a) The relation between K and the {V < /N Λ(H2O)3 } R (H2O) ln m < A {Vm /N A Λ(H2 )3 } R (H2 ) q q partition function q m< (H2)/NA 1/2 × {Vm< /N A Λ(O2 )3 } R (O2 ) q = ∆Em − RT ln q m< (O2)/NA N A1/2 {Λ(H2 )Λ(O2 )1/2 /Λ(H2O)}3 Vm