Species A with half the density of states per unit energy than species B will (at temperatures high enough to populate many levels of each) have half B's partition function. So we will expect to find the system occupying B-states twice as often as A-states. Or, the equilibrium constant for the reaction A=B (a unimolecular rearrangement, say) will be K=2, and the mole fractions of equilibrium A and B will be X(A)=1/3 and X(B)=2/3.
This assumes that there is no energy penalty (or reward) for the A=B transition. Why? Because we've calculated z(A) and z(B) assuming that the zero-point energy level of each is 0. If the reaction isn't thermoneutral, we're in trouble.
Maybe not. If, say, the zero-point energy of B lies E' kJ/mol above that of A, we must add E' to each energy level of B. Since that value is common to every B energy level and ex+y = ex ey, we can factor e-E'/kt (which doesn't depend on any summation index) out of z altogether, and correct the equilibrium constant as K = [z(B)/z(A)]e-E'/kt.
Here E' = Eo(B) - Eo(A) =
Eo.
This ruse will work for more complex reactions involving many species since each will contribute it's own zero-point energy difference from some convenient reference, and all we have to watch out for is how many molecules contribute in the reaction under consideration?
Why is that? Because we've lied to you about that original W = N! /
(n!)
by assuming that we could label each molecule in each state.
But identical molecules are indistinguishable unless they're in different states!
So we'll have to make sure that they are. That turns out to be easy.At any reasonable temperature, there'll be many orders of magnitudes more translational energy levels than there are molecules to occupy them.
So we can (conceptually) use the ( nX, nY, nZ) triples to distinguish between different molecules once we know how to count translation W. (Nash's "pink pages" have a good discussion of that.)
Since "Chemistry is the Science of all things whose negative logarithms vary inversely with T," we're moved to look at
- lnK =
G/RT
and ask how this most significant of chemical thermodynamics energies
(Gibb's Free) relates to statistical mechanics, and in the above reaction,
it would beG = E' - RT ln{ z(B)/z(A) }
by a trivial algebraic substitution.Presumably, if we can do that for
G,
we should be able to do it for G or E or S! Since
n=0
(no gas moles created or destroyed) in the above, G = H - TS =
E + (PV) - TS =
E + RTn - TS (T was presumed constant too.)=
E - TS. So playing with -RT lnK =
E - RTln{ z(B)/z(A) } = G = E - TS,
we conclude that, in this instance,S = R ln{ z(B)/z(A) },
positive if B has more states available than A, and negative if vice versa.
Makes sense. But we shouldn't memorize this S
expression when that for S itself is almost as simple!