Reversible, position-specific fractionation
Many chemical reactions are reversible, and run in both directions. When the forward and reverse rates of a reaction are finite,
non-zero, and adjust to compensate for a samll perturbation to the system (
e.g., addition of extra product or substrate), a reaction is said to be in chemical equilibrium.
If the above conditions are met and the distribution of isotopologues
in the product and substrate pools are unchanging, a reaction would be in isotopic equilibrium.
How are position-specific isotope effects, which arise from different reaction rates in different isotopologues,
expressed if forward and reverse reaction rates are equal? Since heavy isotopes slow reaction rates,
isotopologues with heavy substitutions will have longer mean residence times in stable phases compared
to unsubstituted counterparts. At equilibrium, heavy isotopologues will tend to concentrate
in the phase in which they are most stable (i.e. lowest free energy) compared to light isotopologues.
This tends to be the phase with the stiffest bonds, where isotopic substitution gets you the most
"bang for your buck." Often this is the phase where the atom has the highest oxidation state or the lowest coordination number.
These are equilibrium isotope effects.
Check out Schauble 2004
for more qualitative rules of thumb for predicting the direction and size of equilibrium isotope effects.
At higher temperatures, bonds vibrate faster and break more frequently,
and so the small differences in bond strength and reaction rate from heavy isotopic
substitution become less pronounced. Entropy wins out over enthalpy.
The size of equilibirum isotopic fractionations decrease, roughly in proportion to 1/T2.
Since equilibrium isotope effects scale with temperature, they can be used to constrain the temperature
of a system when a equilibrated state is preserved.