# Inorganic Reaction Mechanisms

```Inorganic Reaction Mechanisms
Chapter 1 – The Fundamentals
2011
Inorganic Reaction Mechanisms
Chapter 1
A Review of the Fundamentals
Scope of this course:
1.1

Reactions in solution (and particularly aqueous solution).

“Classical inorganic” reactions which occur at metal centres in coordination complexes.
The Arrhenius equation
For the generalized 2nd order reaction
k2
A + B 
P
the experimental rate law is
d [P]
 k2[A][B]
dt
The second order rate constant, k2, is a function of temperature. Its dependence on
temperature is given by the Arrhenius equation, which is based on the empirical observation
that rates of reactions increase with temperature. This then led to the concept of an activation
energy barrier in the reaction pathway. The height of barrier is given by the activation energy,
Ea .
k2 = A.exp[-Ea/RT]
ln k2 = -(Ea/RT) + constant
where A is the rate constant when Ea is zero, often called the pre-exponential factor. In
1
Inorganic Reaction Mechanisms
Chapter 1 – The Fundamentals
2011
reactions that require collision between two reactants p and q, A is the diffusion limited rate
constant and depends on a number of factors, including the size of p and q, and their diffusion
coefficients D.
A = 4π( rp + rq) (Dp + Dq) NA/1000
where rp and rq are the radii of reacting species, Dp and Dq are their diffusion coefficients, NA is
the Avogadro number, and the division by 1000 gives units of cm2 s-1.
1.2
Activated Complex Theory (Transition State Theory)
A more recent treatment of the theory of reaction kinetics is Activated Complex or Transition
State Theory. In this approach, developed by Eyring, Randall, Wilkins and others, a typical
reaction is considered to proceed through an activation barrier populated by the species C‡.
The assumption made is that C‡ can be treated as if it were a stable molecular species; hence all
the laws of thermodynamics will apply. Consider the encounter between two reactive species,
A and B, to form an encounter complex A∙∙∙B that may then proceed forward to the activated
complex C‡ or dissociate back to A and B (Ke is the encounter equilibrium constant).
k1
k

 C 
A + B 
P

k1

What is the relationship between the microscopic rate constants, k1, k-1 and k‡ and the
macroscopic rate constant, k2 that we saw in the previous section? In other words, how are the
2
Inorganic Reaction Mechanisms
Chapter 1 – The Fundamentals
2011
theoretical rate constants that are based upon a postulated mechanism related to the
experimentally determined rate constant? The answer to questions such as this is, arguably, the
single most important function of reaction kinetics because the answer will tell us something
about the mechanism of the reaction.
The probability of forming C‡, the activated complex at the top of the activation barrier, is
determined by the equilibrium constant for its formation (K‡), and hence the Gibbs energy:
K‡ = k1/k-1 = [C‡]/[A][B]
Gact = G‡ = -RT ln K‡ = -RT ln (k1/k-1)
The rate of formation of products is given by:
d[P]/dt = k‡ [C‡] = k‡ K‡[A][B]
Hence, equating this and the experimental rate law we end up with
k2[A][B] = k‡ K‡[A][B]
and so
k2= k‡ K‡
The reaction is considered to occur in the first vibrational transition of the activated complex.
This is given by

kBT
h
where kB and h are the Boltzmann and Planck constants, respectively, and T is the Kelvin
temperature. kBT/h has a value of ~1013 s-1. The decay of C‡ is assumed to go with equal
probability forward or back, so that:
3
Inorganic Reaction Mechanisms
Chapter 1 – The Fundamentals
2011
k‡ = k-1 = kBT/h
Substituting:
k2= k‡ K‡
k2 = kBT/h . K‡
We introduce the transmission coefficient, κ, to take into account the probability that collision
of A and B lead to the activated complex C‡.
k2 = κ kBT/h . K‡
In the absence of information to the contrary, κ is taken to be about 1.
Since
K‡ = exp[-G‡/RT]
it follows that
k2 = κ . kBT/h . exp[-G‡/RT]
or, if κ~ 1,
k2 = kBT/h . exp[-G‡/RT]
which is known as the Eyring equation.
Since C‡ is treated as if it were a stable molecule and the laws of thermodynamics apply, then
clearly
G‡ = H‡ - TS‡
and substituting (assuming κ~1) gives
4
Inorganic Reaction Mechanisms
Chapter 1 – The Fundamentals
2011
k2 = kBT/h . exp[-H‡/RT] . exp[S‡/R]
It follows immediately that
kh
 H    1   S 
ln  2   
   

 R  T   R 
 kB T 
and a plot of ln(k2h/kBT) against 1/T should be a straight line with slope -ΔH‡/R and intercept
ΔS‡/R.
The Eyring equation shows that an increase in reaction rate can be achieved by:
 ΔS‡ values which are more positive (i.e., an increase in entropy in going from the
products to the transition state);
 ΔH‡ values which are less positive (i.e., reactions in which the enthalpy change in going
from reactants to the transition state is as small as possible).
It can be shown (Atkins, Physical Chemistry, 8th ed., p. 883) that Ea and H‡ are related by the
expression
Ea = H‡ + 2RT
Example 1.
In ethanol, an unstable cis-azoalkane decomposed at a rate which was followed by
observing the rate of evolution of N2. The following values were obtained:
T/0C
k / &times; 10-4 s-1
-24.82
-20.73
-17.02
-13.00
-8.95
1.22
2.31
4.39
8.50
14.3
Calculate the enthalpy, entropy, activation energy and Gibbs energy of activation at -200C.
5
Inorganic Reaction Mechanisms
Chapter 1 – The Fundamentals
1.3
2011
The Kinetic Salt Effect
In solution for the reaction
k1

 C
A + B 

k1
K
aC
aA aB

γ C
γA γB
&times;
[C ]
 KγK
[A][B]
But, as we have seen,
k2= k‡ K‡
so
k2 = k‡K/Kγ
In dilute solutions, all γi ~ 1 so Kγ = 1. We will call the rate constant where this condition holds
k2o and hence
k2o = k‡K
from which it immediately follows that
k2 = k2o/Kγ
From the Debye-H&uuml;ckle theory of electrolytes, we know that
log  j   Az 2j I 1/ 2
where A is a constant which depends on the dielectric constant of the medium and the
temperature (A = 0.509 M-2 in aqueous solution at 25 oC) and I 
1
mj z j2 is the ionic strength

2
6
Inorganic Reaction Mechanisms
Chapter 1 – The Fundamentals
2011
of the medium (mj is the molality – in units of mol kg-1 – and zj is the charge of the jth species). If
we take the density of water to be 1.00 g cm-3 we can equate molality and molarity.
Example 2. Calculate the ionic strength of a buffer solution that is made up of 50.0 mM Na3PO4
and 100.0 mM Na2HPO4.
Since
k2 = k2o/Kγ
and
Kγ 
γ C
γA γB
then
log k2 = log k2o + log γA + log γB – log γC
log k2 = log k2o - AI&frac12;(zA2 + zB2 – zC2)
Since zC = zA + zB,
log k2 = log k2o - AI&frac12;(zA2 + zB2 – (zA + zB)2)
log k2 = log k2o +2A(zA zB) I&frac12;
log (k2/k2o) = 2A(zA zB) I&frac12;
This is the kinetic salt effect: the dependence of the rate constant for a reaction between ions
on the ionic strength of the medium. A plot of log (k2/k2o) against √I should be a straight line
7
Inorganic Reaction Mechanisms
Chapter 1 – The Fundamentals
2011
with slope 2A or 1.018 M-&frac12;. The value of the slope can give information about the species
log(k2/k20)
involved in the formation of the transition state.
0
I1/2
8
Inorganic Reaction Mechanisms
Chapter 1 – The Fundamentals
Example 3.
follows:
2011
The rate constant for the base hydrolysis of [Co(Br)(NH3)5]2+ varies with ionic strength as
I /mol kg-1
0.005
0.010
0.015
0.020
0.025
0.030
k/ko
0.718
0.631
0.562
0.515
0.475
0.447
What can be deduced about the charge of the activated complex in the rate-determining step?
Example 4. In an experimental study of a bimolecular reaction in aqueous solution, the second-order rate
constant was measured at 250C and a variety of ionic strengths. It is known that a singly charged anion is
involved in the rate determining step. What is the charge on the other ion?
1.4
I /mol kg-1
0.0025
0.0037
0.0045
0.0065
0.0085
k /M-1 s-1
1.05
1.12
1.16
1.18
1.26
The Effect of Pressure.
The pressure of a system can influence the rate of a reaction. We have from Activated Complex
Theory that
k2 = kBT/h . exp[-G‡/RT]
and this can be rewritten as
ln k2 + ln(kBT/h) = -G‡/RT
We carry out a partial differentiation with respect to P at constant T. Starting from the
fundamental thermodynamic relationship dG = VdP – SdT, we rewrite this as dG/RT = VdP/RT –
SdT/RT. We do a partial differentiation with respect to P at constant T, and can show (see
9
Inorganic Reaction Mechanisms
Chapter 1 – The Fundamentals
2011
lecture notes) that
  ln k2 


 P T
=
-V 
RT
A graph of ln k2 against P will be a straight line graph with slope -V‡/RT.
10
```