Reminder: Chemical Equilibrium In this system the ground state of B lies 1 quantum of energy below the ground state of A. This means that the overall ground state of the combined set of energy levels is 0B. The most probable configuration of this new system is n0A = 2, n0B = 6, n1A = 1, n1B = 1 with W = 2520 microstates Note that, when adding up the number of quanta, you must count 0A as being at 1 quantum since it is 1 quantum above the overall ground state 0B. That is, 0A has the same energy as 1B. We proceed likewise for the excited states of A. The equilibrium numbers of the two species now reflect the difference in the molecules. We now find more molecules of type B than of type A, in the most probable distribution.

Reminder: Chemical Equilibrium So our statistical approach to chemical equilibrium is determine the equilibrium configuration over the combined reactant and product states, then calculate the relative number of product and reactant molecules. CHEMICAL EQUILIBRIUM IS DETERMINED SIMPLY BY THE SYSTEM ADOPTING THE CONFIGURATION WITH THE GREATEST NUMBER OF MICROSTATES. Our next (and final) task is to put all this into a more general form. The most probable configuration is given by In this expression, n0* is the occupation number of the lowest energy state of A or B (which is e0B in the previous example). Similarly, q is the partition function considering both A and B states, and N = NA + NB is the total number of molecules in the system.

Reminder: Chemical Equilibrium We can now express the equilibrium ratio of products to reactants (equal to the equilibrium constant for ideal gases) as Notice that the same ground state (which is e0B in the previous example) is used in both the numerator and denominator of this expression. Continuing with this convention, the numerator is now simply the molecular partition function for B, qB. The denominator just requires a little rearrangement:-

Reminder: Chemical Equilibrium We substitute this into out equilibrium constant expression to get Chemical equilibrium, and particularly the equilibrium constant, is determined by the ratio of the number of thermally accessible states of reactants and products, i.e. qB/qA times the ratio of occupation numbers of the ground states of the molecules. The final necessary step in getting from molecular states to the equilibrium constant is the calculation of the molecular partition function for molecules with multiple degrees of freedom. To do this we will assume, as we did for heat capacity, that the different degrees of freedom do not affect one another. emolecule = etranslation + erotation + evibration + eelectronic

emolecule = etranslation + erotation + evibration + eelectronic Chemical Equilibrium The equilibrium constant expression is What does this tell us about the equilibrium constant? At low temperatures, the partition functions tend to 1, so the energy difference between reactant and product ground states becomes the determining factor. At high temperatures the exponential factor tends to 1, and the ratio of partition functions – the number of thermally accessible reactant and product states - determines KP. The final necessary step in getting from molecular states to the equilibrium constant is the calculation of the molecular partition function for molecules with multiple degrees of freedom. To do this we will assume, as we did for heat capacity, that the different degrees of freedom do not affect one another. emolecule = etranslation + erotation + evibration + eelectronic

Molecular Partition Functions If we make use of the assumption that the different degrees of freedom don’t couple so that we are allowed to sum over the different sets of energy levels independently of one another, then the partition function in the expanded form   This expression for q can be expressed more compactly as q = qtrans  qrot  qvib  qelec All we now need is to obtain analytic expressions for partition functions for the missing degrees of freedom, qrot and qelec. Here we present these quantities for the case of an ideal diatomic gas. [Note that it is not necessary to learn the following equations. If needed in an exam, they will be provided. You are expected, however, to know what assumptions have been used in their derivation and to be able to use them in a calculation.]

Rotational Partition Function As for translational motion, we cannot carry out the sum over rotational states exactly. Again we shall make use of the assumption that T >> rot. Then we find qrot = T/rot for heteronuclear molecules e.g. NO, CO, HCl, etc. qrot = T/(2rot) for homonuclear molecules e.g. H2, N2, etc.   Why the difference? A proper explanation involves considering the symmetry of the rotational wave functions. A simplistic rationalization is that, in the case of the homonuclear diatomic, we really can’t distinguish between the molecule before and after a rotation of 180 and so have overcounted the number of thermally accessible rotational states by a factor of 2. We divide qrot by 2 to correct for this. To summarise qrot = T/( rot) for a rigid rotor assuming that T >> rot with the symmetry number,  = 1 for a heteronuclear diatomic, and  = 2 for a homonuclear diatomic.

Electronic States qelec  g(0) As elec is typically > 104K, we usually need only worry about the ground state and its degeneracy g(0), so qelec  g(0) i.e. the number of accessible electronic states per molecule is simply equal to the number of states with energy 0. We already know the partition functions for translation and vibration:-

Chemical Equilibrium We can now expand the equilibrium constant expression into For an isomerization reaction, the translational partition functions always cancel because the masses of A and B must be equal. Substituting and cancelling common parameters yields We can frequently simplify the equilibrium expression even further, depending on specific conditions for particular reactions.

Sample Examination Data Page

Flash Quiz! What approximations or conditions are required for the following relationships to hold?

Answer What approximations or conditions are required for the following relationships to hold? N,E sufficiently large that one configuration dominates. If we are in that equilibrium configuration, we have these occupancies. None Definition N,E sufficiently large that one config domniates T>>Q Small, noninteracting particles T << Q Harmonic potential The diff degrees of freedom do not affect each other. Microstates are equally probable

Chemical Equilibrium: Worked Example Problem Write down an expression for KP when the only difference between the two isomers is the vibrational frequency of a single bond with A > B. Which species predominates at equilibrium, and how do you expect KP to vary with temperature? As there is no difference in ground state energies, the exponential term is equal to one. With no change to the translational, rotational or electronic partition functions in going from reactants to products, these terms will cancel out in the ratio. This leaves With A > B we must have KP > 1 - i.e. the smaller vibrational frequency of B allows for a greater number of thermally accessible states and hence isomer B is favoured. As the temperature decreases, the effect of this difference decreases and KP  1 (below the characteristic vibrational temperature of B).

Summary Next Lecture You should now be able to Write down and explain the general definition of the equilibrium constant for an isomerization reaction, make appropriate cancellations, and calculate the equilibrium constant. Next Lecture Entropy