LECTURE 22. STATISTICAL THERMODYNAMICS So far we have discussed thermo in fairly simple terms that allowed us to do two useful things: • Predict reaction spontaneity from ΔG = ΔH – TΔS • Perform simple calculations of ΔG, ΔH, ΔS, B.E., w This was all put together in worksheets and permitted the ability to quickly feel useful with thermodynamics. But--It really played fast and loose with the underlying theory of thermo. It is all fine, it was just sloppy. Kind of like you didn’t earn it. So now you will earn it, with more sophisticated lectures on thermodynamic theory • Statistical thermodynamics • Internal energy • Entropy, equilibria, and free energySTATISTICAL THERMO To this point in thermo we have dealt entirely with bulk properties of a system Provides a bulk V, T, P, G, H, S, E of a system But what about if we deal with the system one molecule at a time and ask questions about is E or S one molecule at a time. This is the study of statistical thermodynamics. And the short answer to the question is this: To find internal energy: To find internal entropy: E = ½ kT S=k ln W Which is the energy per which is the absolute entropy molecule per degree per molecule in a of motion freedom system. in a system The constant, k, in both equations is the Boltzmann constant = 1.38x10-23 J/KThe Origin of Internal Energy Consider an atom It can move 3 ways in a 3-D coordinate Now hook that atom up to other atoms to make a molecule. O Î H H If each atom has three directions in which to move, then N atoms will have 3N ways to move. So for example, water has 9 total different kinds of movement. If you start playing with the vextors, you see that some times all the vectors for the atom are going in the same direction, and the molecule can translate. Other times the vextors oppose each other and you get vibrations. In fact there are three fundamentally different kinds of motion. • Translation- the molecule change location • Rotation – the molecule spins • Vibration- the molecule twists If you are bored, try to visualize these movements Translation: water moving to the left because all 3 atoms move left. There are 3 ways all molecule can move.Rotation: is water spinning counterclockwise. There are 3 ways most molecules can rotate. Linear molecules can only rotate 2 distinct ways. Vibration: is water vibrating. There are 3N -6 ways most molecules vibrate (6= 3 trans. + 3 rotates) 3N-5 ways linear molecules vibrate ( 5=3 trans.+ 2 rotates) How much energy is associated with each movement? The amount is dependent on temperature. From equipartition theory we learn the average energy per motion is, E= ½ kT T= Kelvin K= Boltzman constant So, the Translation of a molecule are 3 x ½ kT = (3/2) kT Rotation of a non-linear molecule are 3 x ½ kT = (3/2) kT Rotation of a linear molecule are 2 x ½ kT = (2/2) kT So a non-linear molecule (most molecules) have 3/2 kT + 3/2kT = 3kT of motional energy at room temperature. A linear molecule (like CO2 or I2-) has 3/2 kT + 2/2 kT= 5/2 kT of motional energy at room temperature. But what about vibration?For our purposes, at room temperature there isn’t enough energy to excite vibration so we ignore them. So what if you have more than one molecule? Just add up the motional energies of the individual atoms. Look at the example below for a mole of molecules. ******Time out for something utterly fascinating: Did you know R (the ideal gas constant) is actually = kN where N is, Avagadros’ number. ***** So one molecule = 3/2 kT + 3/2 kT = 3kT of energy but one mole = 3/2 RT + 3/2 RT = 3RT of energy Therefore, A mole of non-linear gas molecules at room temperature has 3RT= 7.5 kJ of motional energy.Absolute Entropy As temperature, K, decreases, the motional energy of matter, J, decreases. At that point, in the absence of motion, we start to form perfectly ordered crystals. T 0 O O O And we see that S decreases as well. Î O O O We can think, though, about how the complexity of an individual molecule figures into the order. O=O, when flipped, is still O=O O=C=O, when flipped , is still O=C=O but, C= O, when flipped, is now O= C So in terms of the ability to order a compound, we would say that O=O and CO2 have a lower absolute entropy then CO. For example, there is only one way to order molecules of CO. But there are two ways to order molecules of CO. O=C=O O=C=O C=O O=C O=C=O O=C=O C=O O=C One way to orient O=C=O vs. C=O C=O C=O C=O Two ways to orient C=O This means that there are 14 = 1 possible orientations of four molecules of CO2 And there are This means that there are 24 = 16 possible orientations of four molecules of CO Boltzmann generalized this way of describing the extent of the disorder: S (Absolute Entropy) = k ln W where W = the number of possible orientation = # of orientations #of moleculesSo when I have 4 molecule of O2, I have 14 = 1 orientation 4 molecule of CO, I have 24= 16 orientations And calculating the absolute entropy, S, for those four molecules: S= k ln W = 1.38x10-23 J/K ln 1 = 0 J/K for O2 S= k ln W= 1.38x10-23 J/K ln 16 == 3.8 x 10-23 J/K for CO If I have a mole of molecules it works the same way. What is the S for a mole of BH3, which has 1 orientation? S = 1.38x10-23 ln (1 6.02x10^23) = 0 J/K So we now have a way to find the absolute entropy of a system just by knowing • How many particles are involved • How they orient. Unfortunately this latter consideration is complicated by the physical system. Consider: • I can look at a symmetrical molecules and figure there is one way to pack it. • I can look at an asymmetric molecules and figure its orientation. H Cl Br B has 3 C has 4 Cl Cl H Br• But
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