Ask A Scientist Chemistry Archive

name         Abir A.
status       student
age          17

Question -   If you raise the temperature of a system, how does the
disassociation constant change? Is it at all affected?
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     If you are referring to dissociation constants in water, say weak
acids and bases, the answer is they sometimes increase, they sometimes
decrease, and sometimes they change very little, depending upon the
compound. From chemical thermodynamics we know that
ln(K2) - ln(K1) = -H/R*[1/T1 - 1/T2] where K1 and K2 are the dissociation constants of the
dissociation at temperatures T2 and T1 kelvins, respectively, H is the heat
of dissociation, and R is the gas constant R=1.987 cal/K*mol.
    The complexity arises because water is weird! It turns out that H ~ 0
kcal/mol +/- 2 kcal and it too changes significantly with changes in
temperature-- it is sometimes position and sometimes negative.
    From chemical thermodynamics we also know:
H2 - H1 = D[Cp]*[T2 - T1] where H2 and H1 are the heats of dissociation at
temperatures T2 and T1, respectively, and D[Cp] is the change in the heat
capacity of the products minus the reactants for the dissociation reaction.
That is a directly measurable quantity. It turns out that the change in the
heat of dissociation per degree change that is D[Cp] about the same size as
H itself and sometimes larger!!
    The degree to which a weak acid or base dissociates is governed, not by
energetics (despite what you may find in text books), but by solvation
effects that are reflected in the entropy, not the energy.
    As you may have guessed by now you touched upon a topic that is near and
dear to my heart. If you want the whole story look up"Actual Effects
Controlling the Acidity of Carboxylic Acids" J. of Chemical Education,
Vol. 48 (5) pg. 338 (1971).

Vince Calder
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