Chemical Reaction Rate Formula

[Squark]

Hello dear group.

I have recently encountered the following expression for the rate of a
chemical
reaction with activation energy E:

k_B T / h exp(-E/RT)

Here k_B is the Boltzman coefficient, R is the gas constant and h is
Planck's
constant.

1) What is R doing there? Isn't that supposed to be k_B? It doens't
make
sense since R has "moles" in it's dimensions and k_B doesn't.

2) Except the above mischief, the dimensions work out right. Besides
that,
though, I can't make sense of the formula. The exp(-E/RT) part seems
reasonable (modulo issue 1) but what is k_B T / h? In particular, what
is h
doing in the denominator?? This expression becomes infinite in the
classical
limit so it can't be the tunnel effect that's doing this (which would
yield an
opposite tendecy: it is probably neglected here).

Thx for any help!

Best regards,
Squark

[Richard Saam]

Squark wrote:
> Hello dear group.
>
> I have recently encountered the following expression for the rate of a
> chemical
> reaction with activation energy E:
>
> k_B T / h exp(-E/RT)
>
> Here k_B is the Boltzman coefficient, R is the gas constant and h is
> Planck's
> constant.
>
> 1) What is R doing there? Isn't that supposed to be k_B? It doens't
> make
> sense since R has "moles" in it's dimensions and k_B doesn't.
>
> 2) Except the above mischief, the dimensions work out right. Besides
> that,
> though, I can't make sense of the formula. The exp(-E/RT) part seems
> reasonable (modulo issue 1) but what is k_B T / h? In particular, what
> is h
> doing in the denominator?? This expression becomes infinite in the
> classical
> limit so it can't be the tunnel effect that's doing this (which would
> yield an
> opposite tendecy: it is probably neglected here).
>
> Thx for any help!
>
> Best regards,
> Squark
>

Just analyzing the equation dimensionally
All units cancel except for Planck's constant
A frequency 'nu' must be introduced
such that h*nu is energy
as k_B*T is energy
and k_B*T / (h*nu) is unitless

_____________

h joule sec

K_B erg / degree Kelvin (Boltzmann Constant)

R = L*k_B (Gas Constant)

T Kelvin (absolute temperature)

L /mole (Avogadro's Number)

E joule/mole (Arrhenius Activation Energy)

nu /sec (frequency)

(k_B*T /(h*nu))*exp(-E/(R*T))

or

(k_B*T /(h*nu))*exp(-E/(L*k_B*T))

as you wrote the formula, it equals frequency (nu)


nu = (k_B*T /h)*exp(-E/(L*k_B*T))


Arrhenius Activation Energy represents an energy barrier
to a process represented by (k_B*T /(h*nu)) in this case
which is ordinarily indicated thermodynamically (-delta free energy) to proceed.

Richard

[Calvin D. Ritchie]

Squark wrote:
>have recently encountered the following expression for the rate of a
>chemical
>reaction with activation energy E:
>
>k_B T / h exp(-E/RT)
>
>Here k_B is the Boltzman coefficient, R is the gas constant and h is
>Planck's constant.

The E in that equation is a Gibbs Free Energy, G. The activation
energy is related to an Enthalpy, H: E_a=H* + RT, and G=H-TS. I'm
being a little sloppy in that the G, H, and S in these equations are
Delta(G) , etc., but I'll continue to use the simple letters with the
understanding that you'll put in the deltas.

Let's go one thing at a time:
(1) The factor of (k_B T / h ) arises as a part of transition state
theory, which postulates thermodynamic equilibrium between reactants
and "transition state", and decomposition of the transition state to
products with a universal rate constant (k_B T / h ). That universal
rate constant is calculated from the sum over states for translational
motion along the "reaction coordinate" within some small, one
dimensional, region. Essentially, it is the quantum statistical
mechanical motion of a particle in a 1-D box. There's a bit of
hand-waving that goes into the derivation; Chemists are a gullible
bunch :-), but, essentially, one of the vibrational modes of a
"normal" molecule becomes the translation along the reaction
coordinate for a transition state.

(2) Since the transition state is assumed to be at equilibrium with
reactants, there is an equilibrium constant, call it K*, governed by
the free energy difference between reactants and transition state
(That * is almost universally written as a double-dagger in the
Chemical literature.).
The rate of reaction is then equal to the concentration of transition
states multiplied by the universal rate, (k_B T / h ). The rate
constant, then, becomes k_rate=(k_B T / h )K*. Now, all you need is
the relationship between (Gibbs) free energy difference, G*, between
reactants and transition state: G*=-RTLn(K*). Note that G* is in units
for per mole; that's why the R is there rather than Boltzmann's
constant k_B.

The final expression is: k_rate=(k_B T / h )Exp(-G*/RT).

That's probably more than you really wanted to know, already, but if
you're interested in the gory details, see: "Physical Organic
Chemistry, The Fundamental Concepts", 2nd Ed., by C. D. Ritchie, pgs.
49-52, and 289-293, Marcel Dekker, Inc., 1990. and/or "Mechanism and
Theory in Organic Chemistry", 3rd Ed., by T. H. Lowry and K. S.
Richardson, pg. 203 ff, Harper and Row, (1987).

Don Ritchie

[Souvik]

> I have recently encountered the following expression for the rate of a
> chemical
> reaction with activation energy E:
>
> k_B T / h exp(-E/RT)
>
> Here k_B is the Boltzman coefficient, R is the gas constant and h is
> Planck's
> constant.
>
> 1) What is R doing there? Isn't that supposed to be k_B? It doens't
> make
> sense since R has "moles" in it's dimensions and k_B doesn't.

The E then must be "Activation Energy per mole".

-Souvik

[Uncle Al]

Squark wrote:
>
> Hello dear group.
>
> I have recently encountered the following expression for the rate of a
> chemical
> reaction with activation energy E:
>
> k_B T / h exp(-E/RT)
>
> Here k_B is the Boltzman coefficient, R is the gas constant and h is
> Planck's
> constant.
>
> 1) What is R doing there? Isn't that supposed to be k_B? It doens't
> make
> sense since R has "moles" in it's dimensions and k_B doesn't.
>
> 2) Except the above mischief, the dimensions work out right. Besides
> that,
> though, I can't make sense of the formula. The exp(-E/RT) part seems
> reasonable (modulo issue 1) but what is k_B T / h? In particular, what
> is h
> doing in the denominator?? This expression becomes infinite in the
> classical
> limit so it can't be the tunnel effect that's doing this (which would
> yield an
> opposite tendecy: it is probably neglected here).
>
> Thx for any help!
>
> Best regards,
> Squark

Boltzmann's constant is per molecule, the Universal Gas Constant is
per mole of molecules. Looks like an attempt to quantify quantum
noise in the pre-exponential factor by statistical thermodynamics. Is
somebody playing with molecular beams?

http://www.shodor.org/UNChem/advanced/kin/arrhenius.html
Arrhenius calculators

--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/qz3.pdf

[Squark]

Calvin D. Ritchie wrote:
> The E in that equation is a Gibbs Free Energy, G.

Right. That already confuses me, since I'm having hard time making
precise mathematical sense of the free energy _as a function of the
state of the system_.
I have no problem with free energy at thermodynamic equilbrium,
that's completely well defined. I'm also OK with free-energy of a
certain _mixed state_, that is, a phase-space probability distribution
in the classical case or a density operator in the quantum case (the
entropy of a mixed state is defined via S = - k_B Tr (rho ln rho) in
the
quantum case and S = - k_B integral (rho ln rho) in the classical
case; the internal energy U is just the expectation value of the
Hamiltonian; pressure P is... hmm... I guess it can be defined but I
don't want to think about it right now; G = U + PV - TS)
However, in this context, what are the mixed states we are
considering?

> The activation
> energy is related to an Enthalpy, H: E_a=H* + RT

Hmm... What is the * doing there? Is it a typo? What's the physical
meaning of that anyway? At zero enthalpy difference (that's Delta H
there, right?) you get positive activation energy? Maybe it should be
H - RT (so you substract the energy supplied by thermal fluctuations or
something like that)?

> (1) The factor of (k_B T / h ) arises as a part of transition state
> theory, which postulates thermodynamic equilibrium between reactants
> and "transition state", and decomposition of the transition state to
> products with a universal rate constant (k_B T / h ). That universal
> rate constant is calculated from the sum over states for translational
> motion along the "reaction coordinate" within some small, one
> dimensional, region.

There is one way I can recover that constant. Namely, we have
Delta E * Delta t >= hbar / 2 where Delta t is the time for a quantum
state to become a distinguisable state (i.e. an orthogonal one) and
Delta E is the indeterminancy of energy (the standard deviation). So I
can think of this as if the molecule, having energy of order E = k_B T
"tries to decompose" every Delta t time where E * Delta t = hbar
(Delta E is bounded by something of order E since E >= 0). However I
don't quite understand which sum over states are you considering here.

> Note that G* is in units
> for per mole; that's why the R is there rather than Boltzmann's
> constant k_B.

Yep, I guessed that after writing the post.

Thx a lot for your help!

Best regards,
Squark

[Calvin D. Ritchie]

Squark wrote:
>That already confuses me, since I'm having hard time making
>precise mathematical sense of the free energy _as a function of the
>state of the system_.

If you're looking for precise mathematical sense, transition state
theory is poor hunting ground ;-). The theory was cooked up by Eyring
(J. Chem. Phys., V3, 107 (1935)) and Polanyi (sorry, don't have that
ref. handy) with considerable mathematical laxity.

Perhaps the hardest thing to accept about the theory is the basic
postulate that the species/state, called the transition state, is at
thermodynamic equilibrium with the reactant (and product) states. The
transition state is defined as those 'configurations' of the system
lying within some small length along the coordinate of negative
curvature at the lowest energy saddle point connecting reactants and
products on the potential energy surface of the system. Since the
'lifetime' of the transition state is then calculated to be on the
order of E-12 secs., that equilibrium postulate seems strange.
IMO, the main reason that the theory gained wide (never
universal among Physical Chemists) acceptance is that its use in
calculating isotope effects
on reaction rates (the "Bigeleisen equation"), leads to very good
agreement with experiment There has been a large amount of work in the
past 25 years to try to come up with something better (see, for
example the work of D. G. Truhlar (Ann. Rev. Phys. Chem., V35, 189
(1984) is the latest I have handy, but there is a lot of more recent
work by Truhlar and others), but SFAIK, nothing really contradicts at
least near equilibrium for non-pathologic systems.

CDR wrote
>> The activation
>> energy is related to an Enthalpy, H: E_a=H* + RT

and Squark asked
>Hmm... What is the * doing there? Is it a typo? What's the physical
>meaning of that anyway? At zero enthalpy difference (that's Delta H
>there, right?) you get positive activation energy? Maybe it should be
>H - RT (so you substract the energy supplied by thermal fluctuations
>or
>something like that)?

The activation energy, E_a, is an experimentally measured quantity in
the empirical Arhennius equation:
k_rate=ZExp(-E_a/RT),
where Z is also a measured quantity. All this is, really, is an
empirical equation stating the experimental facts that the log of the
rate constants for reactions are almost always observed to be
inversely proportional to temperature ..... at least, over fairly
small temperature ranges.

The H* (not a typo) is the enthalpy difference between reactants and
transition state. If you go through the usual business of
dLn(K*)/d(1/T)=-H*/R and the relation of k_rate to K*, and compare
that with the empirical equation, you get the relationship between H*
and E_a that I wrote.

Squark continued
>There is one way I can recover that constant. Namely, we have
>Delta E * Delta t >= hbar / 2 where Delta t is the time for a quantum
>.................................... However I
>don't quite understand which sum over states are you considering
>here.

There's a bit of a stretch in getting that constant! The "system"
called the transition state is a normal system with a potential energy
surface, but with 3N-1 ordinary degrees of freedom. That one missing
ordinary degree of freedom is the motion along the reaction coordinate
(the coordinate of negative curvature at the saddle point ..... etc.,
etc.). Motion along that coordinate is taken as translational motion
within a 1-D box. Now, of course, that box has no real walls.
Nevertheless, transition state theory proceeds as if the walls, which
really define the extent within which the transition state exists,
were real. After all the smoke and dust settle, it turns out that the
length of the box cancels out of the final expression, anyway.

To put a trite remark to new use: Transition State Theory isn't
hard to understand ...... it's hard to believe!!

Don Ritchie

[Squark]

Calvin D. Ritchie wrote:
> >> The activation
> >> energy is related to an Enthalpy, H: E_a=H* + RT
> The activation energy, E_a, is an experimentally measured quantity in
> the empirical Arhennius equation:
> k_rate=ZExp(-E_a/RT),
> where Z is also a measured quantity.

If we match that with the transition theory equation we get

Z exp(-E_a / RT) = (k_B/hbar) T exp(-Delta G / RT)

Choosing T_0 an arbitary temperature:

Z exp(-E_a / RT) = (k_B T_0 / hbar) (T / T_0) exp(-Delta G / RT)

Z exp(-E_a / RT) = (k_B T_0 / hbar) exp(-Delta G / RT + T / T_0)

Z exp(-E_a / RT) = (k_B T_0 / hbar) exp[-(Delta G - RT ln(T / T_0)) /
RT]

exp[(-E_a + Delta G - RT ln(T / T0)) / RT] = (k_B T_0 / hbar Z) =
const(T)

(-E_a + Delta G - RT ln(T / T0)) / RT = const(T)

Now E_a = Delta H + RT, Delta G = Delta H + (Delta S) T hence

[(Delta S) T - RT - RT ln(T / T0)] / RT = const(T)

(Delta S) / R - 1 - ln(T / T0) = const(T)

(Delta S) / R - ln(T / T0) = const(T)

Delta S = R ln(T / T0) + const(T)

S = R ln T + const(T)

Where "const(T)" is something that doesn't depend on T.
That equation works for an ideal gas, I think. Is it always a good
approximation, for some reason?
Also, the RT summand in the definition of E_a seems to amount
only to a redefinition of Z. Why do we need it, then? Merely a
convention?

Another question: one would expect the rate of a reaction of type
A + B -> ... to depend on the frequency of collisions between a
molecule of type A and a molecule of type B. However, there seems
to be no expression for this fact in the transition state theory
formula.
How does it work out?

Best regards,
Squark

[Squark]

I wrote:
> Delta S = R ln(T / T0) + const(T)
>
> S = R ln T + const(T)
>
> Where "const(T)" is something that doesn't depend on T.
> That equation works for an ideal gas, I think.

Sorry, that should be

Delta S = R ln T + const(T)

and I'm not sure how to relate it to the S(T) ~ ln T property
of an ideal gas since here it is the _difference_ in entropies
between a (meta)equilibrium and a transition state.
So, what _is_ the justification for that?

Best regard,
Squark

[Calvin D. Ritchie]

Squark wrote:
>Z exp(-E_a / RT) = (k_B/hbar) T exp(-Delta G / RT)

o.k.

>Choosing T_0 an arbitary temperature

o.k., but I don't see why you want to introduce that. Are you getting
ready for a Taylor expansion, or something? It's not necessary.

>Z exp(-E_a / RT) = (k_B T_0 / hbar) (T / T_0) exp(-Delta G / RT)
>Z exp(-E_a / RT) = (k_B T_0 / hbar) exp(-Delta G / RT + T / T_0)

??????? (T / T_0) ?=? exp(T / T_0) ???????

Try just going with derivatives w.r.t. T on both sides of
Z exp(-E_a / RT) = (k_B/hbar) T exp(-Delta G / RT).

It's a little less messy to do as I suggested in the previous post and
use
RT Ln(K*)=-G*, d(Ln(K*))/d(1/T)=-H*/R,
and work from there.
Once you accept that K* and G* are "good" thermodynamic
quantities, everything else is exact thermodynamics. See the
following, however.


>Another question: one would expect the rate of a reaction of type
>A + B -> ... to depend on the frequency of collisions between a
>molecule of type A and a molecule of type B. However, there seems
>to be no expression for this fact in the transition state theory
>formula.
>How does it work out?

There's a huge difference between rate of a reaction and rate constant
for a reaction. For a "second order" reaction, of the sort you're
proposing, the rate is the product k_rate[A][B] where [A] and [B] are
concentrations, or, more accurately "activities" of the reactants A
and B.
In transition state theory, the rate of reaction is equal to the
concentration (NOT activity) of the transition state times that (k_B
T/h) rate of conversion of transition state to products. The
concentration of transition state is governed by K*=[TS]/[A][B], where
all bracketed quantities are activities. If activities are not equal
to concentrations, as they generally are not except for "infinitely
dilute" solution, the concentration of transition state is
K*[A][B]/(activity coefficient of ts).
For second order reactions, the fastest reactions possible are
those called "diffusion controlled", which is just the rate at which
collisions can occur. In solution or in gas, this generally works out
to give rate constants on the order of E+10 to E+12, same order of
magnitude as k_B T/h for most gas phase cases.

Don Ritchie

[Squark]

Calvin D. Ritchie wrote:
> Squark wrote:
> >Z exp(-E_a / RT) = (k_B T_0 / hbar) (T / T_0) exp(-Delta G / RT)
> >Z exp(-E_a / RT) = (k_B T_0 / hbar) exp(-Delta G / RT + T / T_0)
>
> ??????? (T / T_0) ?=? exp(T / T_0) ???????

I got it wrong here but I corrected myself later (missed that
particular
equation when correcting) so the conclusion Delta S = R ln T + const
still holds.

> There's a huge difference between rate of a reaction and rate constant
> for a reaction.

Well, (k_B T / hbar) exp(Delta G / RT) has dimension suitable for the
rate constant of a first order reaction (that is 1 / time) but not of a
second
order reaction (1 / time x mole).

> For a "second order" reaction, of the sort you're
> proposing, the rate is the product k_rate[A][B] where [A] and [B] are
> concentrations, or, more accurately "activities" of the reactants A
> and B.

What is "activity"?

> In transition state theory, the rate of reaction is equal to the
> concentration (NOT activity) of the transition state times that (k_B
> T/h) rate of conversion of transition state to products. The
> concentration of transition state is governed by K*=[TS]/[A][B], where
> all bracketed quantities are activities. If activities are not equal
> to concentrations, as they generally are not except for "infinitely
> dilute" solution, the concentration of transition state is
> K*[A][B]/(activity coefficient of ts).

So K* has to have units of 1 / mole, so it can't be just exp(-Delta G /
RT).
What is it, then?

Another question:
Some sources use purely classical consideration to estimate the
reaction
rate. For instance, given a reaction of type A <-> B, they split the
phase
space into parts corresponding to A and B (the attraction basins of the
corresponding local energy minima, say) and compute the "flux" across
the boundary. More sophistical approaches involve following sample
boundary crossing trajectories for some time, to make sure they don't
"cross back" (and following them backwards in time as well, to make
sure
they are originated in the right part of the phase space).
One example is an article by D. Chandler appearing in the Journal of
Chemical Physics 68(6) (1978) "Statistical mechanics of isomerization
dynamics in liquids and the transition state approximation".
This seems to be in extreme contrast to the (k_B T / hbar) exp(-Delta G
/ RT)
since the later expression becomes infinite (!) in the classical limit,
suggesting that a purely classical analysis of reaction rates is
meaningless.
How can the discrepancy be settled?

Best regards,
Squark

[Calvin D. Ritchie]

Squark wrote:
> so the conclusion Delta S = R ln T + const
>still holds.

O.K., you defined (k_B T_0 / hbar Z) =const(T).
The correct thermodynamic expression is
Delta(S*)=RLn(Z)-RLn(k_BT/h)-R
There are no ideal gas assumptions in that.

Sloppy (I'll blame it on ASCII limitations) notation, however, is
getting
in the way and is to blame for your confusion on the next point.

>Well, (k_B T / hbar) exp(Delta G / RT) has dimension suitable for the
>rate constant of a first order reaction (that is 1 / time) but not of
>a
>second
>order reaction (1 / time x mole).

Perhaps it would be less confusing, at the start, to use the equation:

k_rate=(k_B T/h)K*

which shows immediately that the rate constant has concentration units
if
the equilibrium constant does, as in the conversion, say, of two
reactants into a transition state.

>What is "activity"?

Even though I've been in this business a few more than a few years, I
always have to stop and think, then go back to basics to find, why the
equation: Delta(G^o)=-RTLn(K_eq) makes sense even when K_eq has
concentration units in it.
The "basics" needed here are for classical (not even statistical)
chemical thermodynamics as expounded by G. N. Lewis. Let's use
chemical
potential of species j, mu_j, and chemical potential of species j in
the
chosen standard state, Muo_j. Then, the activity of species j, a_j, is
defined by the equation:

mu_j=Muo_j + RTLn(a_j).

Now, for any Chemical reaction at equilibrium, the chemical potentials
of
the reactants must equal those of the products. Take a reaction

A + B = C;

at equilibrium,

mu_C = mu_A + mu_B;
Muo_C + RTLn(a_C)=Muo_A + Muo_B + RTLn(a_A a_B).

Just rearrange, and

Muo_C - Muo_A - Muo_B = RTLn((a_A a_B)/a_C),

and define

Muo_C - Muo_A - Muo_B = Delta(G^o);
(a_C/(a_A a_B))= K,

and there you have it: Delta(G^o)=-RTLn(K); Chew on that one for a
while!!!

Usually, not always, the standard states are defined as something
close
to the experimental conditions where K is measured, and activity is
defined as concentration times activity coefficient:

a_j = c_j gamma_j.

If that is done, then activity coefficients are near unity corrections
for "non-ideality".
One of my favorite examples of how activity coefficients need not be
near
unity is the following:
Take a separatory funnel in which you have equal volumes of water and
hexane. Now, toss in a small amount of salt (NaCl) and shake to
dissolve.
At equilibrium, the chemical potential of the salt in the water must
equal the chemical potential of the salt in the hexane. Choose
solution
of salt in water at low concentration as the standard state of the
salt.
What do you think the activity coefficient is for the salt in hexane??

>Another question:
>Some sources use purely classical consideration to estimate the
>reaction
>rate. .......................
>This seems to be in extreme contrast to the (k_B T / hbar) exp(-Delta
>G
>/ RT)
>since the later expression becomes infinite (!) in the classical
>limit,
>suggesting that a purely classical analysis of reaction rates is
>meaningless.
>How can the discrepancy be settled?

I haven't read such treatments, so can't know exactly what they're
doing.
The easiest conceptual treatment of reaction rate is exactly what you
seem to be getting at. Consider a general system, having the requisite
number of atoms, not yet distinguished as reactants or products. Now,
populate all configurations of those atoms (some arranged into
reactants,
some into products, etc., with various kinetic energies) by "simple"
Boltzmann statistics. Now, all you have to do is decide which you want
to call reactants and which you wish to call products, and the
Boltzmann count gives you the equilibrium constant. For rates, you
have to decide on where the dividing boundary is placed, and how
rapidly Boltzmann
statistics are established.
I don't see how you can get away from quantum mechanical
calculation
somewhere in all that. For example, even if someone handed you the
potential energy surface for such a system and claimed to have
obtained
it without QM, when you go to use it, you're almost certainly going to
have to calculate zero-point energies in the reactant and product
"basins". If you want rate, you're going to have to do something about
how fast the system moves from one point to another of phase space
.....
often through some vibrational mode. Even rotations, particularly when
"hindered", must be considered as quantized. Only the translations in
3-D
space might be handled as a classical continuum.
In fact, the above is just about what transition state theory
does. The messy question of dividing boundary is settled by observing
that the largest number of transitions from reactant to product will
occur through the lowest energy saddle point. The question of rate of
population of states is done by the 1-D box treatment. That, not
coincidentally, turns out to be close to an average sort of
vibrational frequency.

Don Ritchie

[Squark]

Calvin D. Ritchie wrote:

> O.K., you defined (k_B T_0 / hbar Z) =const(T).
> The correct thermodynamic expression is
> Delta(S*)=RLn(Z)-RLn(k_BT/h)-R
> There are no ideal gas assumptions in that.

Can you show that expression holds for any choise of phase-space and
Hamiltonian? Unfortunatelly, I'm not completely sure how to interpret
it
since it's not clear what is S*.
I understand how the entropy of a meta-stable equilibrium can be
defined.
A meta-stable equilibrium can be represented by the phase-space
distribution rho = chi_A exp(-beta H) / Z_A where chi_A is the
characteristic function of some phase-space volume surrounding a local
minimum of H (say, the attractor basin of that local minimum) and Z_A
is
a the normalization constant

Z_A = integral (chi_A exp(-beta H))

Such a distribution has an associated entropy

S_A = -k_B integral (rho ln rho)

However, S* is the entropy of a "transition state" i.e. it refers to
something like the critical point lying between the two local minima
(more precisely, the maximum of H on the path connecting the two
minima which has the minimal value of this maximum). I can't imagine
what phase-space distribution can be associated with _that_.

> >Well, (k_B T / hbar) exp(Delta G / RT) has dimension suitable for the
> >rate constant of a first order reaction (that is 1 / time) but not of
> >a
> >second
> >order reaction (1 / time x mole).
>
> Perhaps it would be less confusing, at the start, to use the equation:
>
> k_rate=(k_B T/h)K*
>
> which shows immediately that the rate constant has concentration units
> if
> the equilibrium constant does, as in the conversion, say, of two
> reactants into a transition state.

The rate constant would have dimensions of concentration / time since
(k_B T / hbar) has dimensions of 1 / time. Ok, so for A -> B we get
K* = [A] / [B] = exp(-Delta G / RT). For A + B -> C we get
K* = [A][B] / [C] = N_Avogadro exp[-(G_C - G_A - G_B)/RT] (given
that [X] is measured in 1 / mole i.e. molecule per mole). Is that
right?

> >What is "activity"?

> Let's use
> chemical
> potential of species j, mu_j, and chemical potential of species j in
> the
> chosen standard state, Muo_j.

What is the "chosen standard state"? Just an arbitrary choice of
offset?

> Then, the activity of species j, a_j, is
> defined by the equation:
>
> mu_j=Muo_j + RTLn(a_j).

i.e. a_j = exp((mu_j - Muo_j) / RT). For me that looks better :-) Aha,
now
I see why we need the activities, it's because we've been ignoring the
chemical potential in our exp(Delta G / RT) expression, i.e. it's "as
if"
we've been using the canonical ensemble rather than the grand
canonical ensemble.

> Now, for any Chemical reaction at equilibrium, the chemical potentials
> of
> the reactants must equal those of the products.

Of course, since the chemical potential of the reactants / products
separately is only an approximate notion: the corresponding molecule
numbers are not truly conserved (because of the reaction).

> ...
> (a_C/(a_A a_B))= K,
>
> and there you have it: Delta(G^o)=-RTLn(K); Chew on that one for a
> while!!!

Hold it. The activity is dimensionless. The equilibrium constant has
dimensions. Is N_Avogradro lurking somewhere here?

> ...
> Take a separatory funnel in which you have equal volumes of water and
> hexane. Now, toss in a small amount of salt (NaCl) and shake to
> dissolve.
> At equilibrium, the chemical potential of the salt in the water must
> equal the chemical potential of the salt in the hexane. Choose
> solution
> of salt in water at low concentration as the standard state of the
> salt.
> What do you think the activity coefficient is for the salt in hexane??

Umm, doesn't it depend on the "small amount" and the
"low concentration"?

> >Another question:
> >Some sources use purely classical consideration to estimate the
> >reaction
> >rate. .......................
> >This seems to be in extreme contrast to the (k_B T / hbar) exp(-Delta
> >G
> >/ RT)
> >since the later expression becomes infinite (!) in the classical
> >limit,
> >suggesting that a purely classical analysis of reaction rates is
> >meaningless.
> >How can the discrepancy be settled?
>
> ...
> I don't see how you can get away from quantum mechanical
> calculation
> somewhere in all that. For example, even if someone handed you the
> potential energy surface for such a system and claimed to have
> obtained
> it without QM,

This is _clearly_ impossible! However, the final expression

(*) (k_B T / hbar) exp(-Delta G / RT)

contains energies (through Delta G),
hence we can consider the limit in which the potential energy surface
is
held fixed and hbar is scaled to 0. In this limit, (*) becomes
infinite, however, there seems to exist a perfectly good classical
limit
through the phase-space approach. In other words, given a potential
energy surface, we _can_ do all computations classically and get a
meaningful result (whether or not it's a good approximation and the
result is useful). However, (*) would not yield any meaningful result
in the same limit (Delta G fixed, hbar goes to 0)!

Best regards,
Squark

[Calvin D. Ritchie]

Squark wrote
>Can you show that expression holds for any choice of phase-space and
>Hamiltonian? Unfortunately, I'm not completely sure how to interpret
>it
>since it's not clear what is S*.

S* is the entropy difference between reactants and transition state.
Again, ASCII is getting in the way. For all of the following, put a
delta before the thermo quantity and a superscript 0 to indicate that
we're talking about "standard state" quantities; i.e. related to some
(arbitrary!!, but consistency is essential) choice of standard states.
I (and Eyring and Polanyi) take S* as an 'ordinary' thermodynamic
state function.

Forget about all that fancy-schmancy quantum statistical thermo; let's
stick with good old old-fashioned, bookkeeping, thermodynamics (steam
engines, cannon barrels, etc.). I'll assume that we're working at
constant pressure, and, therefore, the pertinent thermo quantities are
G, H, and S. G=H-TS. dG/dT=-S (that's a total, not partial,
derivative!! dH/dT=TdS/dT.). If you want other independent variables,
use Maxwell's relations (Yeah, that's the same J. C. Maxwell; the
relations are just Legendre transformations.) O.K.? Now simple
calculus and empirical dZ/dT=0:

Ln(k_rate)=Ln(Z)-E_a/RT
=Ln(k_B T/h)+Ln(K*)=Ln(k_B T/h)-G*/RT;

d[Ln(k_rate)]/d(1/T)=-E_a/R
=-T-G*/R-TS*/R=-T-G*/R-(H*-G*)/R=-T-H*/R.

That's the first result, which I think you already got.
Now, go back to the first equation and substitute E_a=H*+RT:

Ln(Z)-H*/RT-1=Ln(k_B T/h)-G*/RT,

and rearrange:

Ln(Z)-Ln(k_B T/h)-1=H*/RT-G*/RT=S*/R.

>However, S* is the entropy of a "transition state" i.e. it refers to
>something like the critical point lying between the two local minima
>(more precisely, the maximum of H on the path connecting the two
>minima which has the minimal value of this maximum). I can't imagine
>what phase-space distribution can be associated with _that_.

I keep repeating: transition state theory postulates that the
transition state is an ordinary "species" with all of the usual phase
space EXCEPT that one 'special' coordinate, the "reaction coordinate".
That's the coordinate orthogonal to the hypersurface separating
'reactants' and 'products'. What Eyring and Polanyi do is argue that
you don't really have to consider any regions other than those in the
vicinity of what you're calling a "critical point".

You're just going to have to look at a good book on this. I gave you
two references, but almost any undergraduate text on Physical
Chemistry will have a section on transition state theory.
Unfortunately, most modern texts don't do more than a cursory
treatment of plain vanilla thermodynamics, and almost all of them make
the mistake of saying that activity coefficients are "near unity" or
are "small corrections for deviations from ideality".

>The rate constant would have dimensions of concentration / time since
>(k_B T / hbar) has dimensions of 1 / time. Ok, so for A -> B we get
>K* = [A] / [B] = exp(-Delta G / RT). For A + B -> C we get
>K* = [A][B] / [C] = N_Avogadro exp[-(G_C - G_A - G_B)/RT] (given
>that [X] is measured in 1 / mole i.e. molecule per mole). Is that
>right?

No. First, you have your equilibrium expressions upside-down. Second,
I don't know what you mean by "(G_C - G_A - G_B)". The thing called
"chemical potential" is a partial molar (or molal, or ....) free
energy. It is the CHANGE in free energy of a SYSTEM with increase in
molarity (or molality, or...) of a specific thing. It will almost
always depend on particular species, concentration, temperature, etc.,
etc. For example, you'd find that a curve showing free energy of a
system as a function of concentration of some one species is almost
always curved. The chemical potential is a slope.
The business of standard states, activities, and activity
coefficients merits careful study. I wrote the basic relationship for
the specific case of A+B=C in terms of those quantities in my last
post. You should be able to generalize that to any stoichiometry.


>Of course, since the chemical potential of the reactants / products
>separately is only an approximate notion: the corresponding molecule
>numbers are not truly conserved (because of the reaction).

Nothing is forever :-))


>> Take a separatory funnel in which you have equal volumes of water
>> and
>> hexane. Now, toss in a small amount of salt (NaCl) and shake to
>> dissolve.
>> At equilibrium, the chemical potential of the salt in the water
>> must
>> equal the chemical potential of the salt in the hexane. Choose
>> solution
>> of salt in water at low concentration as the standard state of the
>> salt.
>> What do you think the activity coefficient is for the salt in
>> hexane??

>Umm, doesn't it depend on the "small amount" and the
>"low concentration"?

Not much. I said to choose standard state of "low" concentration in
water, which should be close to whatever you get by throwing in a
"small amount". The point of the example is that
activity=concentration x activity coefficient. For water, the activity
coefficient will be close to unity. Now, the concentration of salt in
water is going to be many orders of magnitude greater than the
concentration in hexane. (The ratio of concentrations will be roughly
constant over a moderate range.) See the point?


>In other words, given a potential
>energy surface, we _can_ do all computations classically and get a
>meaningful result (whether or not it's a good approximation and the
>result is useful). However, (*) would not yield any meaningful result
>in the same limit (Delta G fixed, hbar goes to 0)!

The h comes into transition state theory from the partition function
for motion of the TS along the reaction coordinate. That 1-D partition
function is "calculated" as f_RC=(d/h)(2 Pi m_ts k_B T)^(1/2), where d
is the length in which the TS exists, and m_ts is some kind of
"effective mass" for the motion. I guess that the classical limit
would be just (1/2)k_B T for that, but I'll leave it to you to go back
and find what that does to transition state theory and where you can
drop h without hitting infinity. There have been many studies of
"balls rolling on surfaces". I just haven't paid much attention to
them.

Don Ritchie

[Richard Saam]

Calvin D. Ritchie wrote:
> Squark wrote
>
>>Can you show that expression holds for any choice of phase-space and
>>Hamiltonian? Unfortunately, I'm not completely sure how to interpret
>>it
>>since it's not clear what is S*.
>
>
> S* is the entropy difference between reactants and transition state.
> Again, ASCII is getting in the way. For all of the following, put a
> delta before the thermo quantity and a superscript 0 to indicate that
> we're talking about "standard state" quantities; i.e. related to some
> (arbitrary!!, but consistency is essential) choice of standard states.
> I (and Eyring and Polanyi) take S* as an 'ordinary' thermodynamic
> state function.
>
> Forget about all that fancy-schmancy quantum statistical thermo; let's
> stick with good old old-fashioned, bookkeeping, thermodynamics (steam
> engines, cannon barrels, etc.). I'll assume that we're working at
> constant pressure, and, therefore, the pertinent thermo quantities are
> G, H, and S. G=H-TS. dG/dT=-S (that's a total, not partial,
> derivative!! dH/dT=TdS/dT.). If you want other independent variables,
> use Maxwell's relations (Yeah, that's the same J. C. Maxwell; the
> relations are just Legendre transformations.) O.K.? Now simple
> calculus and empirical dZ/dT=0:
>
> Ln(k_rate)=Ln(Z)-E_a/RT
> =Ln(k_B T/h)+Ln(K*)=Ln(k_B T/h)-G*/RT;
>
> d[Ln(k_rate)]/d(1/T)=-E_a/R
> =-T-G*/R-TS*/R=-T-G*/R-(H*-G*)/R=-T-H*/R.
>
> That's the first result, which I think you already got.
> Now, go back to the first equation and substitute E_a=H*+RT:
>
> Ln(Z)-H*/RT-1=Ln(k_B T/h)-G*/RT,
>
> and rearrange:
>
> Ln(Z)-Ln(k_B T/h)-1=H*/RT-G*/RT=S*/R.
>
>
>>However, S* is the entropy of a "transition state" i.e. it refers to
>>something like the critical point lying between the two local minima
>>(more precisely, the maximum of H on the path connecting the two
>>minima which has the minimal value of this maximum). I can't imagine
>>what phase-space distribution can be associated with _that_.
>
>
> I keep repeating: transition state theory postulates that the
> transition state is an ordinary "species" with all of the usual phase
> space EXCEPT that one 'special' coordinate, the "reaction coordinate".
> That's the coordinate orthogonal to the hypersurface separating
> 'reactants' and 'products'. What Eyring and Polanyi do is argue that
> you don't really have to consider any regions other than those in the
> vicinity of what you're calling a "critical point".
>
> You're just going to have to look at a good book on this. I gave you
> two references, but almost any undergraduate text on Physical
> Chemistry will have a section on transition state theory.
> Unfortunately, most modern texts don't do more than a cursory
> treatment of plain vanilla thermodynamics, and almost all of them make
> the mistake of saying that activity coefficients are "near unity" or
> are "small corrections for deviations from ideality".
>
>
>>The rate constant would have dimensions of concentration / time since
>>(k_B T / hbar) has dimensions of 1 / time. Ok, so for A -> B we get
>>K* = [A] / [B] = exp(-Delta G / RT). For A + B -> C we get
>>K* = [A][B] / [C] = N_Avogadro exp[-(G_C - G_A - G_B)/RT] (given
>>that [X] is measured in 1 / mole i.e. molecule per mole). Is that
>>right?
>
>
> No. First, you have your equilibrium expressions upside-down. Second,
> I don't know what you mean by "(G_C - G_A - G_B)". The thing called
> "chemical potential" is a partial molar (or molal, or ....) free
> energy. It is the CHANGE in free energy of a SYSTEM with increase in
> molarity (or molality, or...) of a specific thing. It will almost
> always depend on particular species, concentration, temperature, etc.,
> etc. For example, you'd find that a curve showing free energy of a
> system as a function of concentration of some one species is almost
> always curved. The chemical potential is a slope.
> The business of standard states, activities, and activity
> coefficients merits careful study. I wrote the basic relationship for
> the specific case of A+B=C in terms of those quantities in my last
> post. You should be able to generalize that to any stoichiometry.
>
>
>
>>Of course, since the chemical potential of the reactants / products
>>separately is only an approximate notion: the corresponding molecule
>>numbers are not truly conserved (because of the reaction).
>
>
> Nothing is forever :-))
>
>
>
>>>Take a separatory funnel in which you have equal volumes of water
>>>and
>>>hexane. Now, toss in a small amount of salt (NaCl) and shake to
>>>dissolve.
>>>At equilibrium, the chemical potential of the salt in the water
>>>must
>>>equal the chemical potential of the salt in the hexane. Choose
>>>solution
>>>of salt in water at low concentration as the standard state of the
>>>salt.
>>>What do you think the activity coefficient is for the salt in
>>>hexane??
>
>
>>Umm, doesn't it depend on the "small amount" and the
>>"low concentration"?
>
>
> Not much. I said to choose standard state of "low" concentration in
> water, which should be close to whatever you get by throwing in a
> "small amount". The point of the example is that
> activity=concentration x activity coefficient. For water, the activity
> coefficient will be close to unity. Now, the concentration of salt in
> water is going to be many orders of magnitude greater than the
> concentration in hexane. (The ratio of concentrations will be roughly
> constant over a moderate range.) See the point?
>
>
>
>>In other words, given a potential
>>energy surface, we _can_ do all computations classically and get a
>>meaningful result (whether or not it's a good approximation and the
>>result is useful). However, (*) would not yield any meaningful result
>>in the same limit (Delta G fixed, hbar goes to 0)!
>
>
> The h comes into transition state theory from the partition function
> for motion of the TS along the reaction coordinate. That 1-D partition
> function is "calculated" as f_RC=(d/h)(2 Pi m_ts k_B T)^(1/2), where d
> is the length in which the TS exists, and m_ts is some kind of
> "effective mass" for the motion. I guess that the classical limit
> would be just (1/2)k_B T for that, but I'll leave it to you to go back
> and find what that does to transition state theory and where you can
> drop h without hitting infinity. There have been many studies of
> "balls rolling on surfaces". I just haven't paid much attention to
> them.
>
> Don Ritchie
>
>
>

Fellows:

ref: Aquatic Chemistry, Stumm and Morgan
An interesting variation on the theme
is nucleation and crystal growth
from solution

J = A exp (- delta G / kT)

A is collisions per volume per time

J is successful nucleating collisions per volume per time

delta G = (4 pi r^3 / V) kT ln (a/ao) + 4 pi r^2 gamma

V is molecular Volume ( ~ a few Angstrom cubed)

r is nucleated crystal radius

a/ao = saturation ratio

gamma is nucleus surface free energy / area
(on the order of 100 mJ/m^2 for inorganic material

inspection indicates that a crystal nucleus must pass a critical radius
before energetics will allow the crystal to increase in size spontaneously.

Practically speaking, seed crystals are added to supersaturated solutions
to initiate nucleation.
Sometimes mechanical agitation does it like shaking a can of soda.

This is all in the realm of molecules in solvent (typically water).
I wonder if the principle applies in other dimensional regimes.

Richard

[Calvin D. Ritchie]

Richard Saam wrote:
>ref: Aquatic Chemistry, Stumm and Morgan
>An interesting variation on the theme
>is nucleation and crystal growth
>from solution

These kinds of phenomena are known as "first order phase transitions"
(having a discontinuity in entropy) and have been treated in all kinds
of ways; none very successful SFAIK. Having done more than my fair
share of crystallizations/re-crystallizations, I wouldn't trust any
theoretical treatments that don't include the phase of the moon,
number of clouds in the sky, and a few other things. There's a story
told around the light of the Bunsen burners about old sugar chemists
shaking their beards over the flasks in which they were trying to
crystallize some new sugar. All visiting sugar chemists were also
asked to shake their beards in the hope that they might have a
different set of "seed" crystals.

On the other hand, there has been amazing progress in treating
(Understanding may be a different question; see Weinberg, TQTF Vol 2,
pg 148.) second-order phase transitions and I am just beginning to get
into that. There's a good book available on such many body problems,
free, at www.physics.rutgers.edu/~coleman. Zee, QFT in a Nutshell,
also has a little on such things.

Don Ritchie

[Richard Saam]

Calvin D. Ritchie wrote:
> Richard Saam wrote:
>
>>ref: Aquatic Chemistry, Stumm and Morgan
>>An interesting variation on the theme
>>is nucleation and crystal growth
>
>>from solution
>
> These kinds of phenomena are known as "first order phase transitions"
> (having a discontinuity in entropy) and have been treated in all kinds
> of ways; none very successful SFAIK. Having done more than my fair
> share of crystallizations/re-crystallizations, I wouldn't trust any
> theoretical treatments that don't include the phase of the moon,
> number of clouds in the sky, and a few other things. There's a story
> told around the light of the Bunsen burners about old sugar chemists
> shaking their beards over the flasks in which they were trying to
> crystallize some new sugar. All visiting sugar chemists were also
> asked to shake their beards in the hope that they might have a
> different set of "seed" crystals.
>
> On the other hand, there has been amazing progress in treating
> (Understanding may be a different question; see Weinberg, TQTF Vol 2,
> pg 148.) second-order phase transitions and I am just beginning to get
> into that. There's a good book available on such many body problems,
> free, at www.physics.rutgers.edu/~coleman. Zee, QFT in a Nutshell,
> also has a little on such things.
>
> Don Ritchie
>

Model presented again with proper sign
and addition of Go energy variable.

**************
J = A exp (- delta G / kT)

A is collisions per volume per time

J is successful nucleating collisions per volume per time

delta G = -(4 pi r3 / V) kT ln (a/ao) + 4 pi r2 gamma + Go

V is molecular Volume ( ~ a few Angstrom cubed)

r is nucleated crystal radius

a/ao = saturation ratio

gamma is nucleus surface free energy / area
(on the order of 100 mJ/m2 for inorganic material
*********************

I would agree that
"phase of the moon,
number of clouds in the sky"
may have something to do with it

The energies are extremely low
and any known or unknown Go could influence the nucleation reaction.

I have modeled oil water coalescence phenomena
with ortho kinetic principle after Miron Smoluchowski
to arrive at drop collisions per volume per time (A)
where gamma ranges over many orders of magnitude (.0001 to 73 mJ/m^2)
A approach with probabilistic input
and analysis of probabilistic output
achieves some conformance with experiment or practice.

In final synergy between nucleation theory and practice
Sugar does pour
paper is a controlled white
oil is separated from water and
floc settles in the clarifier, etc.

I have heard anecdotally that NASA is somewhat disappointed in its
crystallization experiments conducted in the orbiting space station
where all the Go's were to be minimized.

Looking at the model,
I cannot resist the thought of applicability to the Big Bang.
Was there an initial gamma "surface tension" that had to be overcome
before the Universe "inflated" spontaneously?

Richard