BD,
Again it's really hard to follow your train of thought with equations formatted this way. But from what I can tell you are trying to calculate K from abundances of the isotopes. However the equilibrium constant isn't affected by how much "stuff" is available to react. It's determined from the thermodynamics of the reaction. Thus it should be obvious by now that you can't generalize the equilibrium constant just by the stoichiometry or class of reaction.
For instance -
I couldn't find thermodynamic quantities for the title reaction of this thread, but data for this reaction are widely available
H
2 + D
2 --> 2HD
This is another isotopic substitution reaction. Standard heats for formation and standard entropies are available here:
http://cccbdb.nist.gov/We can see the standard heats of formation for H
2, D
2 and HD are 0, 0 and 0.32 kJ mol
-1, respectively, and the standard entropies are 130.600, 144.960 and 143.799 J K
-1 mol
-1, respectively. The entropies are at 295.15 K and I think the enthalpies are at the same.
Anyway from the data it is easy to calculate that ΔH
f0 = 0.64 kJ mol
-1 and ΔS = 12.038 J K
-1 mol
-1. Two comments here: the enthalpy change is obviously not zero. The reaction is endothermic, and probably this is mostly due to the difference in zero point energies between the reactants and products (more on that in a minute). Second, the enthalpy change is slightly positive, again as we'd expect because there are more possible combinations for HD than there are for the "pure" substances. Order increased, more entropy.
Now you know where we're going next. Standard Gibbs energy change for the reaction can be calculated from
ΔG
0 = ΔH
f0 - TΔS
0Easily it is shown that ΔG
0 = -2.95 kJ mol
-1 at 298.15 K, meaning the reaction is spontaneous (despite being endothermic - the reaction is entropically driven at 298.15K; you can see that the entropy term is much larger in scale than the enthalpy term). And finally we can calculate K from
ΔG
0 = - RT ln K
K is found to be 3.29. Which means that if you start with 1 mole/L each of the reactants, at equilibrium you'll have approximately 0.35 mol/L of the starting materials and about 1.29 moles/L of the product. (And of course using the reaction quotient and calculating ΔG, you can determine the direction the reaction will go from any relative starting concentrations of products and reactants).
You see here that K quite obviously isn't 4.
I surmise that the K value will be a little lower for the reaction you inquired about that that calculated for this one with hydrogen isotopes (a little much closer to 1) because the difference in reduced mass between the starting materials and products will be smaller (mass difference between H
2 and D
2 is pretty large, relatively speaking). However as this is an entropically driven reaction, I think the K values may still be fairly close, because the reduced mass differences will probably mostly affect the enthalpy term.
We can also get an idea of the thermodynamics of this reaction based on the zero-point energies (ZPE). ZPE is the energy of the lowest lying vibrational state. To break a bond, you have to supply the energy difference between the ZPE and the transition state. (Actually, you have to supply less than this because not all molecules are in the lowest state, but we can still get an estimate of what's going on by pretending all molecules are in the lowest lying state.) Thus, to take a mole of H
2 and a mole of D
2 to form 2 moles of HD, you have to supply enough energy to bring 1 mole of H
2 from its ZPE to the transition state and 1 mole of D
2 from its ZPE energy to the transition state. Likewise, when we form 2 moles of HD, we gain an energy equivalent to the difference between the transition state energy and the ZPE of HD (times 2). The overall difference in energy is an approximation of the enthalpy change of the reaction, assuming the gas is dilute and behaves ideally.
Thus if the transition state energy is T, we have an approximation for DH:
ΔH = (T - ΣZPE{reactants}) - (T - ΣZPE{products})
Here, ΣZPE{reactants} = ZPE{H
2} + ZPE{D
2and
ΣZPE{products} = 2*ZPE{HD}
Note here that the transition state energy cancels out so we get that ΔH is only dependent on the differences in the ZPEs for the reactants and products.
ΔH = - ΣZPE{reactants} + ΣZPE{products}
The sign on the ZPE for reactants is negative because it costs us energy to break bonds, and the sign on the ZPE for products is positive because we gain energy from forming them (remember, the sign convention here is opposite what you'd expect - a lower ZPE means MORE energy to break the bond because the bond energy is the difference between this number and the transition state energy, which is very large, relatively speaking): the net energy cost/gain (enthalpy) will be positive if the products have a higher combined ZPE than the reactants and vice-versa.
The ZPE for any species is equal (in the harmonic oscillator approximation) to be 0.5*hcν, where h is Planck's constant and ν is the vibrational frequency in wave numbers (inverse meter).
Again we turn to NIST for the vibrational frequencies and we find that ν for H
2, D
2 and HD is 4401 cm
-1, 3116 cm
-1 and 3813 cm
-1 respectively. (Note, I found out later that the NIST pages actually supply ZPE values for you, so deriving them from vibrational frequencies isn't necessary - results are the same though.) You can see here that the frequency of HD is not exactly in between the pure substances, reflecting the fact that we deal with REDUCED mass, not total mass - this is the primary reason we will find that there is a nonzero ΔH in just a minute.) If we do a little math, we can find that the ZPEs for H
2, D
2 and HD are predicted to be 26.31, 18.63 and 22.79 kJ mol
-1, respectively. Which leads to a ΔH of 0.656 kJ mol
-1.
The ΔH value we calculate by this method is very close to that predicted from the heats of formation values. Any deviation would be because we have made a number of approximations (ideal gas behavior and especially at nonzero temperatures, we will have many states populated above the zero-point level; also we've approximated as harmonic oscillator) but you can see the approximations are really good. Point is that here again the ΔH value is predicted to be nonzero AND the equilibrium is not 4 (and it's not 1).
So while this wasn't the reaction you asked about, I think it is instructive in many ways. I don't think the ΔH value for the chlorine isotopic scrambling reaction will be exactly zero. I think it will be close to zero, but close isn't exact! Equilibrium constant I predict will be a little smaller than the one for the hydrogen reaction, but it will depend a lot on the entropy term. I did all of the above pretty quickly, so I hope there aren't too many errors.
Anyway, I think that's a lot to digest, but if you have any questions, you know where to find me.