CHE-20028: PHYSICAL & INORGANIC CHEMISTRY STATISTICAL THERMODYNAMICS: LECTURE 1 Dr Rob Jackson Office: LJ 1.16 r.a.jackson@keele.ac.uk http://www.facebook.com/robjteaching

Statistical Thermodynamics: topics for lecture 1 A link between quantum mechanics and thermodynamics and kinetics The Boltzmann Distribution Partition Functions Vibrational Translational Rotational che-20028: Statistical Thermodynamics Lecture 1

che-20028: Statistical Thermodynamics Lecture 1 Why is it useful? There are 2 strands of Physical Chemistry: Macroscopic: thermodynamics and kinetics (do reactions occur, and if so, at what rate?) Microscopic: atomic structure-spectroscopy and their explanation by quantum mechanics. These (apparently) unrelated strands can be linked by statistical thermodynamics. E.g. we can calculate the enthalpy of a reaction instead of measuring it ! che-20028: Statistical Thermodynamics Lecture 1

Additional useful reference http://pubs.rsc.org/en/content/ebook/978-0-85404-632-4 ‘Thermodynamics & Statistical Mechanics’ by J M Seddon & J D Gale. (An RSC Primer) Some page references will be given to this book, but alternatives will be given on the Teaching Pages. che-20028: Statistical Thermodynamics Lecture 1

The Boltzmann Distribution - 1 This is the formula which tells us how to calculate the number of molecules in each state of a system at a temperature T: Ni= no. of molecules in a state with energy Ei, N= total number of molecules, k= Boltzmann’s constant, q= partition function. che-20028: Statistical Thermodynamics Lecture 1

The Boltzmann Distribution - 2 We have come across all these terms before except for q, which is defined as: The sum of the exponential of the negative of each energy level divided by kT (This is a general expression for q, but we will look at how to calculate it and what it means later. che-20028: Statistical Thermodynamics Lecture 1

The Boltzmann Distribution - 3 Molecules in a substance are arranged over a range of molecular energy levels, and there are many possible arrangements of molecules that will give a particular energy. The Boltzmann distribution gives the statistically most probable distribution of molecules in each level for a given overall energy E and temperature T. The total number of molecules, N = Ni, and the total energy, E = Ei, is constant. che-20028: Statistical Thermodynamics Lecture 1

che-20028: Statistical Thermodynamics Lecture 1 Boltzmann Distribution example (see Seddon & Gale, p 72-3, ex. 9.2 & 9.3) A diagram will be drawn of a system containing 5 energy levels, spacing 1 x 10-20 J (n=0 to n=4), and we will see how many ways 6 particles can be arranged such that the total energy is 4 x 10-20 J. Once all the combinations are drawn, we have to decide if they are all equally probable. che-20028: Statistical Thermodynamics Lecture 1

How do we decide which distribution has the highest probability? All distributions have equal energy What about Free Energy, G? G = H – TS, where S is entropy Entropy can be thought of as a measure of disorder Which state has the highest entropy? Hint: consider distribution over energy levels. che-20028: Statistical Thermodynamics Lecture 1

che-20028: Statistical Thermodynamics Lecture 1 The Boltzmann Distribution: Occupation of energy levels as a function of temperature As the temperature increases, more energy levels are occupied. q is the partition function, which we will meet later. This is a measure of how many levels are occupied. You can see that q goes from 1.05 to 3.86 as the temperature increases. che-20028: Statistical Thermodynamics Lecture 1

An example of the use of the Boltzmann Distribution The simplest application of the Boltzmann distribution is to calculate the relative number of molecules in two states separated by an energy difference E = E2 – E1 From the equation on slide 4, we consider 2 states E1 and E2, and obtain: che-20028: Statistical Thermodynamics Lecture 1

The boat and chair conformations of cyclohexane The boat conformation of cyclohexane lies 22 kJ mol-1 above the chair conformation (why?) The relative population of the two conformations at a given temperature (e.g. 293 K) can be found from the expression on slide 11: Nboat/Nchair = exp(-22x103/8.314x293) = 1.2 x 10-4 This shows the very low population of the boat conformation at 293K – try a higher temperature! che-20028: Statistical Thermodynamics Lecture 1

che-20028: Statistical Thermodynamics Lecture 1 Note on use of k and R In the previous calculation we used R instead of k. Which to use depends on if we are dealing with single atoms/molecules (k) or moles (R). Make sure you use the right one! Examples will be given. che-20028: Statistical Thermodynamics Lecture 1

Notes on energy levels and degeneracies An important convention is that all energies are expressed relative to the ground state, even if there is a zero point energy. For the harmonic oscillator, energies are: ½ h, 3/2 h, … but written as 0, h, … Energy levels that are degenerate have several states. The number of molecules in a particular state is multiplied by the degeneracy, g, to get the overall population of the energy level (illustrated in examples later). che-20028: Statistical Thermodynamics Lecture 1

Where do partition functions come in? - 1 As long as we only want to calculate the relative populations of levels and states, we do not need the partition function, q because it cancels out in the expression on slide 11. However, if we need to know the actual population of a state, we need to know q. Going back to the expression on slide 6, and remembering that the ground state energy is zero, we can write a new expression for q: che-20028: Statistical Thermodynamics Lecture 1

Where do partition functions come in? - 2 The new expression is: The first term is 1 since E0=0, and exp(0)=1 If T=0 K it follows that q=1 because all terms apart from the first are equal to 0. At 0 K only the ground state would be occupied, so q tells us the number of occupied states at a particular value of T. che-20028: Statistical Thermodynamics Lecture 1

Where do partition functions come in? - 3 If T is very high, however, all the values of Ei/kT ~ 0, so in this case: q= 1+1+1+1+1 … (= the number of states of the molecule). In this case all available states are occupied. Consider an intermediate case, where kT is large compared to E1 and E2 but small compared to E3: q = 1+1+1+ 0 = 3 (so 3 states are occupied). che-20028: Statistical Thermodynamics Lecture 1

An example of the calculation of q In the case of cyclohexane, there are 2 energy levels, at 0 and 22 kJ mol-1 How are the levels populated at different temperatures? q = 1 + exp(-22 x 103/8.314 x T) When T=293 K, q=1.0001, which suggests that only one energy level will be occupied at this temperature, but what about at higher temperatures? note use of R since E is in kJ mol-1 che-20028: Statistical Thermodynamics Lecture 1

Wavelength, wavenumber and frequency units You need to be able to convert between wavelength, wavenumber and frequency, as all these are used! Basic formulas Wavelength and frequency:  x  = c Wavelength and wavenumber: = 1/ Wavenumber and frequency:  = c x Examples for you to try will be given on the problem sheet. che-20028: Statistical Thermodynamics Lecture 1

Examples of partition functions - 1 We can obtain expressions for the partition function which describe particular types of energy level – e.g. vibrational, translational or rotational: 1. The vibrational partition function From quantum mechanics we know that the permitted energy levels, relative to a ground state energy of zero, are: E0=0, E1=h, E2=2h, E3=3h, etc. che-20028: Statistical Thermodynamics Lecture 1

Examples of partition functions - 2 So the expression for q is: q= 1 + exp(-h/kT) + exp(-2h/kT) + exp(-3h/kT) + ... This can be simplified to give: che-20028: Statistical Thermodynamics Lecture 1

Calculation of a vibrational partition function The vibrational partition function of 35Cl2 at 298K For 35Cl2, = 560 cm-1 (wavenumber units*) So  = 3 x 1010 x 560 Hz = 1680 x 1010 Hz q = 1 / [1 - exp (-6.626 x 10-34 x 1.68 x 1013 / 1.381 x 10-23 x 298)] q = 1.07, which tells us that at 298K, one vibrational energy level is occupied. che-20028: Statistical Thermodynamics Lecture 1

Examples of partition functions - 3 2. The translational partition function The partition function for a molecule of mass m in a flask with volume V is given by: An example calculation will be included on the problem sheet. che-20028: Statistical Thermodynamics Lecture 1

Examples of partition functions - 4 3. The rotational partition function For a linear molecule at high temperature, the rotational partition function is written as: Where B is the rotational constant, and  is the symmetry number (1 for an unsymmetrical linear molecule, e.g. HCl, and 2 for a symmetrical linear molecule, e.g. H2). che-20028: Statistical Thermodynamics Lecture 1

Calculation of a rotational partition function For H35Cl, B= 318 GHz* and =1 (unsymmetrical linear molecule) T= 298 K q = (1.381 x 10-23 x 298)/(1 x 6.626 x 10-34 x 318 x 109) q = 19.5, so about 20 rotational states are occupied at 298 K. *Note that B is sometimes given in wavenumber units, so be prepared to convert! che-20028: Statistical Thermodynamics Lecture 1

che-20028: Statistical Thermodynamics Lecture 1 Summary Statistical Thermodynamics has been introduced. The Boltzmann distribution has been discussed and used in calculation of the separation of energy levels. Partition functions have been introduced, and expressions given for vibrational, translational and rotational forms, and some* example calculations carried out. * But see also the problem sheet. che-20028: Statistical Thermodynamics Lecture 1