Foundations of Chemical Kinetics
Lecture 30:
Transition-state theory in the solution phase
Marc R. Roussel
Department of Chemistry and Biochemistry
Transition-state theory in solution
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We revisit our original derivation of the thermodynamic
formalism of transition-state theory.
k
We decompose the reaction A + B −
→ product(s) into
K‡
k‡
−−
*
A+B)
−
− TS −→ product(s)
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For this representation, v = k ‡ [TS].
The equilibrium constant K ‡ is related to the activities of the
transition state and reactants by
aTS
K‡ =
aA aB
γTS [TS]c ◦
=
γA γB [A][B]
Transition-state theory in solution (continued)
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The derivation is completed as follows:
K ‡ γA γB
[A][B]
c ◦ γTS
K ‡ γA γB
∴ v = k ‡ [TS] = k ‡ ◦
[A][B]
c γTS
K ‡ γA γB
∴ k = k‡ ◦
c γTS
For an ideal solution, all the activity coefficients are equal to
unity, and we get the limiting value
[TS] =
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k0 = k ‡
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Therefore
k = k0
K‡
c◦
γA γB
γTS
Transition-state theory in solution
Constancy of first-order rate constants
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If we repeat the above derivation for a first-order reaction
R → product(s), we get
k = k0
γR
γTS
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In a unimolecular reaction, the reactant and transition state
are typically very similar, the difference between the two
involving small changes in charge distribution and geometry.
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Accordingly, γR ≈ γTS , so various factors that affect the
activity coefficients (e.g. pressure, ionic strength) should have
little effect on the rate constant.
Transition-state theory in solution
Constancy of first-order rate constants (continued)
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Now imagine slowly decreasing the density of the solvent.
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In the limit, this operation converts a reaction in solution to
one in the gas phase.
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By the argument above, the ratio of activity coefficients
shouldn’t change much, so the rate constant should be
more-or-less constant throughout this operation.
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This explains the experimental observation that, for first-order
reactions, k is roughly the same whether the reaction occurs
in the gas phase or in solution.
Variation of rate constant with pressure
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We return to the ideal solution case.
In the thermodynamic formulation of transition-state theory,
we have, for second-order reactions,
∆‡ G ◦
kB T
k0 = ◦ exp −
c h
RT
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Recall that
dG = V dp − S dT
or, for a reaction,
d∆G = ∆V dp − ∆S dT
so that
∂∆G = ∆V
∂p T
Variation of rate constant with pressure (continued)
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For k0 , we have
kB T
∆‡ G ◦
= ln
−
c ◦h
RT
‡
◦
1 ∂∆ G =−
RT
∂p T
ln k0
∂
∴
ln k0 ∂p
T
=−
∆‡ V ◦
RT
where ∆‡ V ◦ is the change in the molar volume (change in
volume of solution per mole of reaction) on accessing the
transition state.
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If ∆‡ V ◦ > 0, the rate constant decreases as p increases.
The reverse is true if ∆‡ V ◦ < 0.
Gas-phase vs solution-phase rate constants
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We now consider the thermodynamic formulation of
transition-state theory, which gives, for second-order reactions,
kB T
∆‡ G ◦
k0 = ◦ exp −
c h
RT
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The above equation should apply both in solution
(k0,s ↔ ∆‡ Gs◦ ) and in the gas phase (k0,g ↔ ∆‡ Gg◦ ).
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We want to figure out how the two rate constants are related.
Gas-phase vs solution-phase rate constants (continued)
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Now consider the following thermodynamic cycle:
∆‡ G ◦
s
A(sol) + B(sol) −−−−→
TS(sol)
x

∆ G ◦ (TS)

−∆solv G ◦ (A)−∆solv G ◦ (B)y
 solv
A(g) + B(g)
∆‡ Gg◦
−−−−→ TS(g)
where ∆solv G ◦ is the standard free energy of solvation.
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Reading off the cycle, we have
∆‡ Gs◦ = ∆‡ Gg◦ + ∆solv G ◦ (TS) − [∆solv G ◦ (A) + ∆solv G ◦ (B)]
= ∆‡ Gg◦ + ∆‡ ∆solv G ◦
Gas-phase vs solution-phase rate constants (continued)
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From
k0,s =
∆‡ Gs◦
kB T
exp
−
c ◦h
RT
and
∆‡ Gs◦ = ∆‡ Gg◦ + ∆‡ ∆solv G ◦
we get
k0,s
!
∆‡ Gg◦
kB T
∆‡ ∆solv G ◦
= ◦ exp −
exp −
c h
RT
RT
∆‡ ∆solv G ◦
= k0,g exp −
RT
Gas-phase vs solution-phase rate constants (continued)
k0,s
∆‡ ∆solv G ◦
= k0,g exp −
RT
TS better solvated than reactants: ∆‡ ∆solv G ◦ < 0
In this case, the argument of the exponential
involving this solvation term is positive and
k0,s > k0,g .
Reactants better solvated than TS: ∆‡ ∆solv G ◦ > 0
We then have the exponential of a negative value, so
k0,s < k0,g .
Gas-phase vs solution-phase rate constants (continued)
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Under what conditions might we expect ∆‡ ∆solv G ◦ < 0?
One example might involve hydrophobic reactants and TS.
If the TS is more compact (exposes less surface area to the
solvent) than the two reactants, which will often be the case,
then each of the individual ∆solv G ◦ terms will be positive, but
the solvation free energy of the transition state would be
much smaller than that of either reactant.
Differences in solvation free energies of ∼ −10 kJ mol−1 are
not unreasonable under these conditions, which would give
∆‡ ∆solv G ◦
exp −
= 56
RT
at 25 ◦ C.
Rate enhancements of this order of magnitude have been seen
in some reactions.