CHE-20028: PHYSICAL & INORGANIC CHEMISTRY STATISTICAL THERMODYNAMICS: LECTURE 2 Dr Rob Jackson Office: LJ 1.16 r.a.jackson@keele.ac.uk http://www.facebook.com/robjteaching

Statistical Thermodynamics: topics for lecture 2 Summary from lecture 1 The molecular partition function Calculation of thermodynamic properties from Statistical Thermodynamics Internal energy Heat capacity Residual entropy Entropy che-20028: Statistical Thermodynamics Lecture 2

The molecular partition function- 1 What contributes to the energy of a molecule? Vibrational (V), translational (T) and rotational (R) modes of motion Electron distribution (E) Electronic and nuclear spin (S) che-20028: Statistical Thermodynamics Lecture 2

The molecular partition function- 2 Remembering from lecture 1 that the partition function q is given by: We can substitute for Ei in this expression: che-20028: Statistical Thermodynamics Lecture 2

The molecular partition function- 3 We then use some mathematical trickery to simplify the expression (ea+b = eaeb): The total energy is the sum of contributions from the different terms, and the total partition function is the product of these contributions: che-20028: Statistical Thermodynamics Lecture 2

Note on electron and spin partition functions While it is not possible to obtain expressions for these, as we have done for the other terms, we note the following: For closed shell molecules, excited states are so high in energy that only the ground state is occupied, and qE= 1 Electron spin makes an important contribution when there are unpaired electrons, since the electron can occupy either spin state, and then qS=2 che-20028: Statistical Thermodynamics Lecture 2

Calculation of internal energy from the partition function - 1 It can be shown that the internal energy is obtained from the derivative of the partition function with respect to temperature: Note that E is the energy with respect to the lowest energy state of the molecule che-20028: Statistical Thermodynamics Lecture 2

Calculation of internal energy from the partition function - 2 So, in general terms we should write the internal energy U as: U = E + U(0), where U(0) is the zero point energy e.g. for a harmonic oscillator U(0) = ½ h We can use these expressions to calculate the internal energy for some example systems: che-20028: Statistical Thermodynamics Lecture 2

The internal energy of a monatomic gas - 1 Only need to consider qT (and assume qE=1) Remember that qT = aT3/2 (where a= (2mk)3/2 V/h3) (lecture 1 slide 23) So (from slide 6) And so che-20028: Statistical Thermodynamics Lecture 2

The internal energy of a monatomic gas - 2 So if N=NA, ET = (3/2)RT U = U(0) + (3/2)RT Example: calculate the internal energy of Argon gas at 300 K U = U(0) + (3/2) x 8.314 x 300 J mol-1 = 3741.3 J mol-1 What would be the result for another monatomic gas? che-20028: Statistical Thermodynamics Lecture 2

The internal energy of a diatomic gas - 1 Neglecting vibrational motion, we just need to calculate the energy contribution of rotational motion, ER: From lecture 1 slide 20, qR = bT (where b=k/(hB)) dqR/dT = b If N=NA, ER = RT che-20028: Statistical Thermodynamics Lecture 2

The internal energy of a diatomic gas - 2 So to summarise: U = U(0) + ET + ER = U(0) + (3/2)RT + RT U = U(0) + 5/2RT Note: both expressions, for the monatomic gas and the diatomic gas, assume the gas to be perfect. The vibrational contribution has been neglected for the diatomic gas. che-20028: Statistical Thermodynamics Lecture 2

How can we make comparisons with experimental values? Heat capacities can be calculated, and compared with experimental values. Remember that CV= dU/dT at constant volume. For a monatomic gas: U = U(0) + (3/2)RT, dU/dT= (3/2)R= 12.47 J mol-1K-1 For a diatomic gas: U = U(0) + (5/2)RT, dU/dT= (5/2)R= 20.79 J mol-1K-1 che-20028: Statistical Thermodynamics Lecture 2

che-20028: Statistical Thermodynamics Lecture 2 Heat capacities Experimental heat capacities Good agreement for inert gases and for nitrogen; agreement gets worse for halogens with increasing departure from perfect gas behaviour. Why does this happen? che-20028: Statistical Thermodynamics Lecture 2

Entropy and Statistical Thermodynamics If we can calculate entropy, this is an important step to being able to calculate further thermodynamic properties We are already familiar with the connection between entropy and the disorder of a system Boltzmann suggested that the entropy, S, of a system should be given by the expression S = k ln W (W is the number of different ways the molecules in a system can be arranged to give the same energy) che-20028: Statistical Thermodynamics Lecture 2

che-20028: Statistical Thermodynamics Lecture 2 Boltzmann's grave in the Zentralfriedhof, Vienna, with bust and entropy formula. Boltzmann's grave in the Zentralfriedhof, Vienna, with bust and entropy formula. che-20028: Statistical Thermodynamics Lecture 2

che-20028: Statistical Thermodynamics Lecture 2 Calculating entropy - 1 The formula makes sense because S = 0 if there is only one way of achieving a given energy (if W = 1, S = ln (1) = 0). Similarly the formula predicts a high entropy if there are many arrangements with the same energy (i.e. if W is large). In most cases W = 1 when T= 0 because there is only one way to achieve zero energy: put all the molecules into the lowest energy level. che-20028: Statistical Thermodynamics Lecture 2

che-20028: Statistical Thermodynamics Lecture 2 Calculating entropy - 2 In this case S = 0 when T = 0, which agrees with the third law of thermodynamics (the entropy of all perfectly crystalline materials is zero at 0 K). There are, however, cases where this is not observed. This occurs when there is more than one energetically equivalent arrangement of molecules when T = 0. che-20028: Statistical Thermodynamics Lecture 2

che-20028: Statistical Thermodynamics Lecture 2 Residual entropy - 1 There are several examples of molecules that can have more than one energetically equivalent arrangement at 0 K Examples include linear molecules where they may be no energetic difference between the arrangement AB AB AB AB and the arrangement AB BA AB AB e.g. solid carbon monoxide, where there is no energetic difference between the arrangements CO CO CO CO and CO CO OC CO che-20028: Statistical Thermodynamics Lecture 2

che-20028: Statistical Thermodynamics Lecture 2 Residual entropy - 2 In this case we say that the substance has a residual entropy. For CO, there are 2 orientations of equal energy, CO or OC, and if there are N molecules, the number of ways of getting the same overall energy is 2N So S = k ln W = k ln 2N = N k ln 2 = R ln 2 Where we assume that we have one mole of molecules. che-20028: Statistical Thermodynamics Lecture 2

Residual entropy - 3 In this case, S = 8.314 x ln 2= 5.76 J K-1 mol-1 The experimental value is 5.0 JK-1 mol-1 Other examples of molecules having a residual entropy include N2O and H2O. The summarised expression is: S = R ln (no. of orientations of equal energy) Example questions are included in the problem sheet. che-20028: Statistical Thermodynamics Lecture 2

The entropy of a monatomic gas To calculate the entropy of a monatomic gas we use an equation called the Sackur-Tetrode* equation: All the symbols have their usual meanings, and p = 105 Pa, (noting that kB = k). * Derived in 1912, anniversary in 2012. che-20028: Statistical Thermodynamics Lecture 2

Sackur-Tetrode equation example Calculate the molar entropy of argon gas (M = 39.948 g mol-1) at 298 K and 105 Pa Divide the calculation into sections and rewrite the equation as: Sm = R ln (A B3/2) A = e5/2 kT/p = 12.182 x 1.381x10-23 x 293 /105 = 5.014 x 10-25 B = 2mkT/h2 = 2 x (39.948 x 1.661 x 10-27) x 1.381 x 10-23 x 298)/(6.626 x 10-34)2 = 3.908 x 1021, so B3/2 = 2.443 x 1032 S = R ln (5.014 x 10-25 x 2.443 x 1032) = 154.84 J mol-1 K-1 (Experimental value is 154.6 J mol-1 K-1) che-20028: Statistical Thermodynamics Lecture 2

Advice on calculations Don’t try to do the whole calculation in one go! This is because most calculators have a maximum index value of 10+/- 99. Don’t try to calculate h3 (or h4) in one go! If your calculation comes out as zero, it’s probably because the calculator can’t handle the numbers. Break down the calculation. che-20028: Statistical Thermodynamics Lecture 2

che-20028: Statistical Thermodynamics Lecture 2 Summary The molecular partition function has been defined. Internal energy and heat capacity have been obtained from the partition function. Residual entropy has been introduced and calculated. The Sackur-Tetrode equation, for calculating entropy, has been introduced and used. che-20028: Statistical Thermodynamics Lecture 2