The playing card idea then can be used to show the probabilities and equilibrium
constants of the isotope exchange NH3+ND3NH2D+NHD2. The work that follows shows how
the equilibrium constants for the isotope exchange can be determined using the ideas derived
from the playing card illustration.
There are three possible hydrogen locations in the molecule that can have hydrogen (1) or
deteurium (2) in it. so the equilibrium can be viewed as
NH3 <-> NH2D <-> NHD2 <-> ND3
So the equilibrium constant can be calculated using the following equation
NH3 + ND3 <-> NH2D + NHD2
In order to look at how possible distributions effect the equilibrium constants, some outcomes of
a 24 red cards and 24 black cards over these possible 4 molecule species can be calculated as
follows:
The probability of each outcome can be found as it was above: by taking the multiplicity of each
outcome and dividing it by the maximum multiplicity. It is important to note that in the
multiplicity determination, the factorial term is multiplied by (3^n)(3^n) where n is the
maximum number of card piles RR or BB (molecules in the distribution of NH2D or NHD2) and
3 is the total possible places that can be occupied by an H and/ or a D.
The reason why there is a (3^n)(3^n) instead of a singular 3^n (or a 2^n in the above section
about card dealings) is because two species are possible in the equilibrium (NH2D and NHD2).
In the above problem RR and BB combine to make RB, so there is only one exponential term
(2^n). In this problem with NH3 and ND3, the two species can combine to make either an NH2D
or an NHD2, therefore, two 3^n terms are incorporated for the determination of outcome
probability.
It is important to note that the probability determination has a 16! in the numerator of the
multiplicity component because the example outcomes add up to 16 piles.
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Next the Equilibrium constants are determined as they have been through out the maple
document.
In this exercise they are found using the equation form:
The equilibrium constant for this problem is that of the outcome with the largest multiplicity.
The largest multiplicity came from the distribution of 2NH3, 2ND3, 6NH2D, and 6NHD2. The
equilibrium constant for that distribution is 9 and the probability of that distribution is 0.16628.
To verify that the distribution 2,2,6,6 has the largest multiplicity:
It can be seen that the multiplicity of largest value is from the distribution of 2 NH3, 2 ND3,
6 NH2D, and 6 NHD2. It is important to notice that the card distribution that has the largest
multiplicity does not have the largest equilibrium constant. Nor does that distribution have the
smallest constant or a constant equal to 0. This is because of the formula: