Pentcho wrote:
>So far I have been trying to explain this in terms of rate constants, but
>the proof is complicated so the discussion naturally shifted to more
>interesting psychological matters. In fact there is a much more simple and
>convincing proof - let me present it.
>>At equilibrium the rates of the forward and back reaction of A + B <-> C
>are equal:
>> v+ = k+(A)(B) = k-(C) = v- /1/
>>On adding a catalyst, the rates may increase dramatically but the second
>law requires that the equilibrium do not shift, i.e. the equilibrium
>concentrations remain the same. No kinetic course of the catalysed forward
>and back reactions guarantees this - on the contrary, any possible one
>shifts the equilibrium.
There is no point in bothering with the rest of the posting, because the
misunderstanding is right here. It's not at all easy to discuss science
with someone who takes so little interest in understanding the standard
point of view before trying to knock it down, but we have to try. In terms
of transition-state theory, any rate constant k+ can be expressed as
k+ = (qT/h)K++
where q is Boltzmann's constant (usually written as a bold k, but as I
can't do bold in this message and we already have a k+ we'll use q
instead), T is the absolute temperature, h is Planck's constant and K+++
(where the first + indicates we are talking about the forward reaction and
++ represents a double dagger) is the pseudo-equilibrium constant derived
in the usual thermodynamic way from the different in Gibbs energy between
the reactant(s) and the transition state. (To understand where this
equation comes from look at any textbook on chemical kinetics: I use Gordon
Hammes "Principles of Chemical Kinetics", Academic Press 1978, but I'm sure
more recent books will say essentially the same thing).
OK, now in this equation q and h are universal constants, so we can forget
them, and T is constant for any given set of conditions, so we can forget
it as well. Thus k+ is determined just by K+++.
For the reverse reaction under the same conditions there will be a
different rate constant k- defined by a different pseudo-equilibrium
constant K-++, but for any given transition state the ratio K+++/K-++ is
completely determined by the difference in Gibbs energies between reactants
and products, and is, indeed the equilibrium constant K for the reaction.
It doesn't matter if we are talking about the true transition state or
about some arbitrary state very far from the principal pathway taken by the
reaction; it only matters that it is the same state for both. It follows
that whatever pathway the reaction actually takes, the ratio of k+/k- for
that pathway is equal to K. So all that is required to conclude that the
experimental k+/k- is equal to K is an assumption that the forward and
reverse reactions follow the same mechanism.
Now, you can object that you don't believe in transition-state theory (and
if I know Pentcho he will), and you can object that you don't believe that
forward and reverse reactions follow the same mechanism (and if I know
Pentcho he will), but what you cannot do is assert that
> the rates may increase dramatically but the second law requires that the
>equilibrium do not shift, i.e. the equilibrium concentrations remain the
>same. No kinetic course of the catalysed forward and back reactions
>guarantees this
On the contrary, *any* assumed course of the catalysed forward and reverse
reactions guarantees this.
The experimental reason or believing all this is that no one has ever
managed to show that a catalyst does anything other than accelerate forward
and reverse reactions to exactly the same extent. The theoretical reasons
for believing it lie in the principle of detailed balance (similar to but
not quite the same as the principle of microscopic reversibility) developed
if I remember rightly by the American physicist Richard Tolman in the
1920s, and in transition-state theory, developed in the 1930s by the
Mexican chemist Henry Eyring.
Incidentally, the difficulties of putting a double dagger in an email
message echo the reasons why the double dagger was chosen in the first
place. The authors of the paper wanted to put an asterisk, but they didn't
have one on their typewriter, so they used a + sign and an = sign on top of
one another, intending this to be interpreted by the printer as *, but the
printer thought it was supposed to be a double dagger. Thus one of the most
widely recognized symbols in physical chemistry has its origins in a
printer's error.
Athel
Email: athel at ibsm.cnrs-mrs.fr
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MCA chapter from my book:
http://ir2lcb.cnrs-mrs.fr/lcbpage/athel/mcai.htm
