Kinetics
Kinetic data gives insight into reaction mechanisms
A kinetic analysis will describe a relationship between concentrations of all chemical species
before the rate determining step in a given reaction and the rate of the reaction
Consider an irreversible unimolecular reaction
A
k
B
The rate equation will describe the change in concentration versus rate
-d[A]/dt = k[A]
Can also describe in terms of appearance of B
d[B]/dt = k[A]
Need to have variable of [B],
not in [A] if solve d[B]/dt
[A] however is dependent on [B]
[B] = [A]o – [A]
d[B]/dt = k([A]o – [B])
[A]o = concentration of A at
time 0
239 Kinetics
What about a reversible unimolecular reaction?
A
k1
k2
B
[A] = [A]o, [B] = 0 at time = 0
Disappearance of [A]
-d[A]/dt = k1[A] – k2[B]
Cannot solve for d[A]/dt
having a variable of [B]
[B] = [A]o – [A]
-d[A]/dt = k1[A] – k2([A]o – [A])
-d[A]/dt = (k1 + k2)[A] – k2[A]o
240 Kinetics
Consider an irreversible bimolecular reaction
A
B
k
C
-d[A]/dt = k[A][B]
In order to make d[A]/dt
solvable, need to have only
[A] as a variable
[A]o – [A] = [B]o – [B]
[B] = [A] – [A]o + [B]o
-d[A]/dt = k[A]([A] – [A]o + [B]o)
Consider a dimerization
A
A
k
B
-d[A]/dt = k[A]2
Overall therefore a solvable rate equation should be relatively straightforward to write, just
remember to understand the relationships present so the only variable is the term being integrated
241 Kinetics
The kinetic order of a rate equation is the sum of all exponents of the concentrations
appearing in the term
-d[A]/dt = k[A]
Therefore first order
-d[A]/dt = k[A][B]
Therefore second order
Can also classify by the order of a given reactant
The overall second order reaction above is thus second order overall, but first order with respect to [A] and first order with respect to [B]
242 Kinetics
An important experimental determination for a reaction is once a rate equation is written,
how to determine the actual value of rate constant (k)
Reconsider the irreversible first order reaction
A
k
B
-d[A]/dt = k[A]
Rearrange equation to place
like terms on same side
d[A]/[A] = -k dt
Need to integrate equation
∫ d[A]/[A] = ∫ -k dt
ln[A] – ln[A]o = -kt
[A] = [A]oe-kt
Therefore to determine the value of k need to measure the concentration of [A] versus time 243 Kinetics
ln[A] - ln[A]o = -kt
ln [A]
Intercept = ln[A]o
Slope = -k
time
A straightforward observation (if not always seen at first) is that if a graph of ln [A] versus
time yields a straight line, then the reaction must have been a first order reaction
The slope of this graph will be the -k value and the intercept will be ln [A]o (which can be used to double check accuracy since this value should be known)
Can also use this data to determine the half-life (T½) for a reaction
At T½ , [A] = ½[A]o
ln([A]0/2) - ln[A]0 = -kt½ T½ = (1/k) ln ([A]0/{[A]o/2}) = (1/k) ln2
244 Kinetics
Consider how determination of rate constant changes if reaction is reversible
k1
A
B
k2
-d[A]/dt = (k1 + k2)[A] - k2[A]o
Need to integrate
equation
- ∫ d[A]/(k1 + k2)[A] - k2[A]o = ∫ dt
ln
(k1 + k2)[A] - k2[A]o
k1[A]o
At equilibrium
= -(k1 + k2)t
k1[A]e = k2{[A]o - [A]e}
Leads to two equations
1)
2)
[A]e =
ln
k2[A]o
k1 + k2
[A]o - [A]e
[A] - [A]e
= (k1 + k2)t
245 Kinetics
1)
2)
[A]e =
ln
k2[A]o
k1 + k2
[A]o - [A]e
[A] - [A]e
= (k1 + k2)t
In order to solve for the k1 and k2 values:
(k1 + k2) can be determined from equation 2 by measuring [A] versus time and knowing the [A]o and [A]e terms
The individual k1 and k2 values are then determined from equation 1
This is for only a relatively simple reversible unimolecular reaction!
* Can easily realize that this kinetic analysis will become very difficult as the complexity of the reactions increase
* Need approximations to solve more complex reactions
246 Kinetics
As an example consider again a second order reaction
A
k
B
C
-d[A]/dt = k[A]([A] – [A]o + [B]o)
Integrate equation
1
[B]o – [A]o
ln
[A]o([B]o - [A]o + [A])
[A][B]o
= kt
What if we start reaction with one reagent in great excess?
[B]o >> [A]o, which also means [B]o >> [A]
If apply to integral
1
[B]o
ln
[A]o
[A]
= kt
247 Kinetics
1
[B]o
ln
ln
[A]o
[A]
[A]o
[A]
= kt
= [B]okt
Let [B]ok = kobs
(often done in practice, all constants are combined into a kobs term which is the actually measured value)
ln
[A]
[A]o
= -kobst
* Identical to irreversible first order reaction
(called pseudo first order kinetics)
For second order reactions, if one reagent is used in great excess (need > 10 fold) then that reagent will not appear in rate equation
248 Kinetics
Excess reagent term, however, will affect the observed rate constant
A
B
k
C
If [B] is in excess
-d[A]/dt = kobs[A]
kobs = k1[B]
As [B] increases, the kobs will also increase
ln[A] – ln[A]o = -kobst
ln [A]
Intercept = ln[A]o
Slope = -kobs
time
With pseudo first order
approximation, a second order
reaction will follow first order
behavior and the observed rate
constant can be obtained from
a graph of ln [A] versus time
249 Kinetics
Another common approximation to use in kinetic analysis is called a steady-state approximation
Consider a reaction of the type shown
A
k1
k2
B
k3
C
rate equations
Step 1
-d[A]/dt = k1[A] - k2[B]
Step 2
d[C]/dt = k3[B]
Considering [B]
-d[B]/dt = k3[B] + k2[B] - k1[A]
Now we assume d[B]/dt = 0
Steady-state approximation
Therefore [B] does not change appreciably over time
Can occur if either k2 >> k1 or k3 >> k1
250 Kinetics
Means that any reactive intermediate B will not accumulate in concentration over time
Change in concentration of B is small compared to change in concentration of either A or C over time
-d[B]/dt = 0 = k3[B] + k2[B] - k1[A]
Solve for [B]
[B] =
k1[A]
k3 + k2
Remember rate equations
-d[A]/dt = k1[A] - k2[B]
d[C]/dt = k3[B]
Can substitute for
solvable integral
d[C]/dt = -d[A]/dt =
k3k1[A]
k3 + k2
Therefore -d[A]/dt = kobs[A]
Obtain first order
equation
251 Kinetics
Can also use steady-state approximation for nonunimolecular reactions
A
Merely pay attention to
properly write rate equations
k1
k2
k3
B
C
D
-d[A]/dt = k1[A] - k2[B]
-d[B]/dt = 0 = k3[B][C] + k2[B] - k1[A]
Steady-state approximation for [B]
d[D]/dt = k3[B][C]
[B] =
k1[A]
k2 + k3[C]
d[C]/dt = -d[A]/dt =
k3[C]k1[A]
k2 + k3[C]
-d[A]/dt = kobs[A]
The kobs term now includes [C] (which means you would need to run in excess), but otherwise is identical to steady-state approximation for unimolecular case
252 Kinetics
Examples of mechanistic questions answered by proper kinetic analysis
1) Does a reaction proceed through an intermediate?
X
X
X
Or does starting
material react directly
to product?
X
X
X
X
X
If the intermediate is reactive and cannot be detected spectroscopically, how can an experimentalist determine whether it occurs or not?
253 Kinetics
Presence of intermediate can be detected by the kinetic analysis
Write the rate equations for the two possible mechanisms:
With intermediate
A
k1
k2
B
k3
D
P
D represent “alkene”
-d[B]/dt = 0 = k3[B][D] + k2[B] - k1[A]
d[P]/dt = -d[A]/dt =
Direct reaction
A
k
D
k3[D]k1[A]
k2 + k3[D]
-d[A]/dt = kobs[A]
With excess of [D]
P
-d[A]/dt = k[D][A]
-d[A]/dt = kobs[A]
How to detect difference between direct reaction and pathway with intermediate? 254 (both simplify to pseudo first order when approximations are applied)
Kinetics
Often an issue when trying to distinguish pathways, how to run experiment to differentiate between possible pathways?
In this case, report experiments with a double reciprocal plot
With intermediate
kobs =
1/kobs =
Direct reaction
k3[D]k1
kobs = k[D]
k2 + k3[D]
k2
k1k3[D]
+ 1/k1
Slope = k2/k1k3
1/kobs
Intercept = 1/k1
1/kobs = 1/(k[D])
Whether the intercept passes
through the origin
(direct reaction) or has a finite
value (with intermediate) in
double reciprocal plot will
distinguish possible mechanisms
Slope = 1/k
Intercept = 0
1/[D]
255 Kinetics
2) Another question that a proper kinetic analysis can answer is how to increase the rate of a reaction
Consider the synthesis of benzhydrol in a sodium chloride solution
H
Cl H
A
Cl
HO H
k1
H2O
k2
k3
B
C
d[C]/dt = k3[H2O][B]
Use pseudo first order approximation with water
d[C]/dt = kobs[B]
Since B is a reactive intermediate, can apply steady-state approximation for the [B] term
256 Kinetics
d[B]/dt = 0 = k1[A] - k2[B][Cl-] - k3[H2O][B]
[B] =
k1[A]
k3[H2O] + k2[Cl-]
d[C]/dt = -d[A]/dt = k3[H2O][B]
-d[A]/dt =
k1k3[H2O][A]
k3[H2O] + k2[Cl-]
With excess of [H2O]
kobs =
k1k3[H2O]
k3[H2O] + k2[Cl-]
Therefore rate becomes slower as concentration of chloride ion increases
* Known as common ion rate depression
257 Kinetics
3) What does kinetics indicate if we change the type of trapping agent in question 1?
A
Assume pathway with
intermediate
k1
B
k2
kobs =
1/kobs =
k3
D
P
k3[D]k1
k2 + k3[D]
k2
k1k3[D]
+ 1/k1
Slope = k2/k1k3
1/kobs
If trapping agent is more
reactive, then the k3 rate will
be faster
Therefore slope in double
reciprocal plot will be less, but
the intercept will be the same
Intercept = 1/k1
1/[D]
258 Kinetics
What occurs if no longer have steady-state conditions when an intermediate is formed?
A
k1
k2
B
C
The kinetics become much more complicated
A = [A]oe-k1t
B =
[A]ok1
k2-k1
Same as a normal
unimolecular reaction
(e-k1t - e-k2t)
C = [A]o{1 + 1/(k1-k2)[k2e-k1t - k1e-k2t]}
The concentration of A, B and C do not follow as simple of relationship
as when a reactive intermediate is present
259 Kinetics
[C]
Concentration
[B]
[A]
Time
If k1 = 0.001/s and k2 = 0.0005/s, then [B]max = 0.46 [A]o and T[B]max = 15 min
If we want to synthesize B, but it can decompose to another product C, then we would need to determine the k1 and k2 rate constants and then find out when [B]max occurs to stop reaction and then isolate B
260