Section 9 – Part II
9.3.2. First Order Kinetics Continued…….
Rate = - d [A] / dt = k [A]
[A] = [A]0 e
-kt
ln ([A]/[A]0) = - k t
Plot of ln { [A] / [A]0 } vs. t
gives a straight line with
a slope that is given by – k
[A] = [A]0 e
-kt
Exponential decay of [A]
with time
For a 1st order reaction the
concentration of A would
be [A] =
[A]0/2
at t = t 1/2
When t = t 1/2 the equation becomes:
ln { ([A]0/2 )/ [A] 0 } = - k t 1/2
t
1/2
=
t
1/2
(ln 2) / k =
0.693 / k
is independent of [A]
0
For example, assume the elimination of
a drug follows 1st order kinetics: If
we start with an initial plasma
concentration of 100 g/mL:
[Cp]0 = 100 g/ml
if k = 0.35 hr–1 and t1/2 = 0.693/0.35
then t1/2 = 2 hours
how much remains after 6 hours ?
Following:
100 g/ml
2 hrs.
50 g/ml
4hrs.
25 g/ml
6 hrs.
12.5 g/ml
8hrs.
6.25 g/ml
What Happens if Initial Concentration of Drug
in Plasma is Different ?
kel = 0.347 hr
Initial
Conc.
–1
then
t1/2 = 0.693/0.35 = 2 hours
200 g/mL
100 g/mL
50 g/mL
t =0
100 g/mL
50 g/mL
25 g/mL
t =2
50 g/mL
25 g/mL
12.5 g/mL
t =4
25 g/mL
12.5 g/mL
6.25 g/mL
t =6
12.5 g/mL
6.25 g/mL
3.13 g/mL
t=8
How Does Rate Constant For Elimination
Effect Concentration of Drug in Plasma ?
t1/2 = 0.693/ kel = ?
kel
0.17 hr-1
t1/2
4.08 hrs.
[Cp]0 = 100 g/mL
0.347 hr-1
0.7 hr-1
2 hrs.
1 hour
In each case what is the concentration of drug in plasma
following 8 hours ?
“The larger the rate constant
the more rapid the decay”.
Atkins Phys. Chem 7th Edn. Page 871
Radioactive Decays Follow First Order Kinetics:
• we can use this as a way to determine age of something………
as in problem #12.6 in Chang Text.
Radiocarbon Age Determination continued………
• all plants and animals contain carbon
• most carbon is 12C (stable) , but a small fraction is 14C (radioactive)
• 14C is generated by interaction in upper atmosphere of cosmic rays
with gas molecules, resulting in neutrons which bombard 14N
to produce 14C
(14N7 + 1n0  14C6 + 1p1)
• 14C is radioactive and emits a beta particle according to the following
reaction:
14
14
6
7
C 
N
0
+ -1 
• the radioactive half-life for this reaction is 5730 years
…….the natural abundance of 14C is 1.1. X 10-13 mol % in living matter.
…….the mol % of 14C for all living matter is assumed to be the same
……when matter dies, it no longer exchanges material with the
environment and the mol % of 14C will decrease according to 1st order
decay kinetics.
……when matter dies [14C] becomes [14C]0
and decays with t 1/2 = 5730 years
If we measure the [14C] at some time post-death, we can estimate
elapsed t since t0 (its “age”)****………..
****for this determination we are limited by the minimum amount of
material we can measure reliably
…….limit is approximately 10 half lives (57,000 years old)
9.3.3. Second Order Reactions
1) 2nd Order Reaction with One Reactant:
A  Products
Second order in [A]
and
Second order overall
Rate = - d [A] / dt = k [A]2
•
Rate is proportional to [A] raised to second power
•
k units are M-1 s-1

[A] t
[A] 0
d [A]/[A]2 =
-
t
kdt
0
(1 / [A]) – (1 / [A]0) = kt
or
(1/[A]) = kt + (1/[A]0)
( 1/[A] )
= kt + ( 1/[A]0 )
Plot of 1/[A] versus t gives a straight line with slope equal to k
To obtain t1/2 for second order reaction we have to set [A] = [A]0/2
{ 1/ ([A]0/2)} = k t1/2 + ( 1/[A]0 )
t1/2 = 1 / (k [A]0)
• therefore t1/2 for second order reaction is concentration dependent
Second Order Reaction
t1/2 = 1 / (k [A]0)
……..this is unlike first-order reaction where t1/2 = ln 2 / k
………in the 2nd order reaction the half-life of a substance is
dependent on the initial concentration present
……species that decay by 2nd order kinetics may persist at low
concentrations for long periods of time
….since their half-lives are long when their concentration is low….
Example: many environmentally harmful substances decay by 2nd order
kinetics…….so very difficult to get rid of them completely
2) 2nd Order Reaction with Two Reactants:
A + B  Products
Rate = - d [A] /dt = - d [B] / d t = k [A] [B]
• Reaction is first order in A, first order in B and second order overall
Let [A] and [B] be defined as follows:
[A] = [A]0 – x
[B] = [B]0 - x
where x (in mol/L) is the amount of A and B that has been consumed
within time t.
Rate = - d [A] /dt = - d ([A]0 – x) / d t = dx/dt = k [A] [B]
Rate = - d [A] /dt = k ([A]0 – x) ([B]0 - x)
Rearrangement of the above equation and integration with partial
functions yields the following equation*:
1
([B]0 – [A]0)
ln [B][A]0
= kt
[A][B]0
(Chang text page 454)
* Chang says: “Derivation of this equation was performed with
assumption that : [A]0  [B]0
Not true !!
Here is the derivation:
2nd Order Kinetics – Class II
In general, for a ≠ b
A
a–x
+
B
→
b–x
P
x
v = k2 [A] [B]
If x = concentration of each species reacted, then:
[A]0 = a
[A] = a – x
[B]0 = b
[B] = b – x
dx
= k2 (a – x) (b – x)
dt
Thus:
dx
= k2 dt
(a  x )(b  x )
Method Of Partial Fractions:
Let:
1
A
B
=
+
(a  x )(b  x )
(a  x )
(b  x )
=
A(b  x )  B(a  x )
(a  x )(b  x )
Thus:
(bA + aB) – (A + B) x = 1
And:
bA + aB = 1
A + B = 0
2 equations and 2 unknowns
A =
1
(b  a )
B =
1
(b  a )
Therefore:
dx
=
(a  x )(b  x )
 1  dx
 1 
dx
+ 
= k2 dt



 ( b  a )  (a  x )
 (b  a )  (b  x )
Integrate both sides:
dx
 1 
 1 
= 

 
 ln (a – x)
 b  a  (a  x )
ba
dx
 1 
 1 
= 

 
 ln (b – x)

b

a
b
a
(
b

x
)




k2  dt = k2 t + C
Therefore:
 1 

 [ ln(b – x) – ln(a – x) ] = k2 t + C
b

a


When:
t=0
then:
x=0
and hence:
 1  b
C = 
 ln  
ba a
Final Integrated Rate Law is:
 1   (b  x ) 
 1  b
 – 

 ln 
 ln   = k2 t
 b  a   (a  x ) 
ba a
Or:
 1   (b  x ) 
 1  a
 + 

 ln 
 ln   = k2 t
 b  a   (a  x ) 
ba b
Or:
 1   a (b  x ) 
 = k2 t

 ln 
 b  a   b (a  x ) 
Or:

  [B][A ]0 
1

 ln 
 = k2 t
 [B]0  [A]0   [A][B]0 
2nd Order Reaction with One Reactant:
A  Products
( 1/[A] ) = k t + ( 1/[A]0 )
Slope = k
2nd Order Reaction with Two Reactants:
A + B
1
 Products
ln [B][A]0
= kt
([B]10 – [A]0) ln [B][A]
[A][B]
00 = k t
([B]0 – [A]0)
[A][B]0
Slope = k ([B]0 - [A]0)
To this point we have discussed the following types
of reactions:
ORDER
CHEMICAL EQUATION
RATE LAW
0
A  products
rate = k
1
A  products
rate = k [A]
2
A  products
rate = k [A]2
2
A + B  products
rate = k [A] [B]
Similar Table
also in Chang
text page 458
Atkins Text page 875
Special Case for Second Order Reactions:
PSEUDO-First Order Reactions
….…..when one of the reactants is present in GREAT EXCESS
we can consider its concentration to be approximately “constant”
An EXAMPLE: Hydrolysis of Acetyl Chloride
CH3COCl (aq) + H2O (l)  CH3COOH (aq) + HCl (aq)
…..concentration of water is very high – approx. 55 M
……concentration of acetyl chloride is approx. 1M or less
Therefore amount of water consumed is NEGLIGIBLE relative to
the amount of water initially present (i.e. change in conc. of water
is negligible).
Rate = - d [CH3COCl] / dt = k’ [CH3COCl] [H2O]
= k [CH3COCl]
where k = k’ [H2O]
Reaction appears to follow 1st order kinetics…..called pseudo-first
order kinetics (basically we assume conc. of one reactant is cnst.)
How do we determine the rate law ?
….there are several methods for determining the rate law
or order of a reaction……….
(1) Isolation Method: concentration of all reactants except one are
in large excess…..
if reactants are present in excess then their
concentration can be considered to be constant
e.g. Rate = k[A][B] with B in large excess
= k’ [A]
where k’ = k [B]0 (we approximate [B] = [B]0 )
…..this is a pseudo-first order rate law………
……the change in rate is then due to reactant A ….once we determine the
order with respect to A we then determine for B
……for determination of reaction order for B ….we use A in excess….etc.
# 12.22 Chang Text: The following data were collected for the
reaction between hydrogen and nitric oxide at 700°C:
2 H2(g) + 2 NO(g)  2 H20(g) + N2(g)
Experiment
1
2
3
[H2]/M
0.010
0.0050
0.010
[NO]/M
0.025
0.025
0.0125
Initial Rate/M s-1
2.4 x 10-6
1.2 x 10-6
0.6 x 10-6
(a) What is the rate law for the reaction ?
(b) Calculate the rate constant for the reaction.
9.4. Molecularity of a Reaction
• reactions rarely occur as described by a balanced chemical equation
• in most cases the overall reaction is actually the sum of several steps
• sequence of steps by which reaction occurs is referred to as the
reaction mechanism
• the mechanism of the reaction describes how chemical bonds are
broken and formed, how molecules approach one another etc.
(what’s happening at the molecular level)
Consider the decomposition of hydrogen peroxide:
2H2O2 (aq)  2H2O(l) + O2(g)
…..when the above reaction is catalyzed by Iodide ions the rate law
is known to be as follows:
Rate = k [H2O2] [I-]
Reaction is first order wrt. H2O2 and I-.
However, studies suggest that the decomposition of hydrogen peroxide
actually takes place in two steps:
step (1)
step (2)
H2O2 +
I-
k1
 H2O + IO-
H2O2 +
IO-
k2

H2O + O2 + I-
• each of these steps is called an “elementary step” and describes what
is happening at the molecular level
Knowing the rate law for the decomposition is: Rate = k [H2O2][I-]
….this rate law indicates that step 1 is the rate determining step.
….that is the rate for the first step is much slower than that for the
second step ………k1  k2
….therefore overall rate of decomposition is controlled by step 1.
Rate = k1 [H2O2] [I-] = k [H2O2] [I-]
Also, in looking at steps 1 and 2 and overall equation we see that
IO- does not appear in the overall equation. Therefore, IO- is
referred to as an intermediate.
…….an intermediate is formed in an early elementary step and
consumed in a later elementary step………
…….an intermediate is formed in an early elementary step and
consumed in a later elementary step………
(e.g on previous slide was IO- )
By contrast, a catalyst appears as a reactant in the first
elementary step ……forms an intermediate…….and is regenerated
at the end of the reaction.
Molecularity of Reactions…………….
……we are going to cover unimolecular, bimolecular and termolecular
reactions………
Unimolecular Reactions
Examples of unimolecular reactions………..cis-trans isomerization,
thermal decomposition, ring opening, racemization.
Unimolecular means…..a single molecule shakes itself apart or goes
through a rearrangement……..
e.g Isomerization of cyclopropane into propene
CH3CH = CH2
Unimolecular reactions usually follow 1st order kinetics
(i.e. First order rate law)
……how do these reactions occur ?
…..presumably one would guess
that these reactions occur
as a result of BINARY
COLLISION through which
the molecules acquire enough
energy to change forms………
….if this is the case why isn’t
this a second order reaction ?
Mechanism proposed by Frederick Lindemann (1922)
…….a reactant molecule, A, collides with another reactant molecule
and only one becomes energetically excited at the expense of the
other molecule…..
A + A
k1

A +A*
A * is activated molecule.
k2
A *  product
….the activated molecule can then go on to form product.
Another reaction that may occur is deactivation of A*
A
+ A*
k-1

A +A
k2
A *  product
….the rate of product formation is given by the following expression:
d[product] / dt = k2 [A*]
…..A* is an energetically excited species and so has a short lifetime,
limited stability. ….
The rate of change of [A*] is given by steps producing A* and steps
leading to removal of A*
….by the steady-state approximation this rate of change must be 0
d[A*]/dt = 0 = k1[A] 2 - k-1 [A] [A*] – k2 [A*]
formation
deactivation
product formation
Lindemann – Hinshelwood Mechanism
Excited species
may be deactivated
Species is excited by
collision with A
Excited species may
decay to form product
Atkins Text Figure 25.19
d[A*]/dt = 0 = k1[A] 2 - k-1 [A] [A*] – k2 [A*]
Solving for [A*] we obtain……
[A*]
=
k1 [A]2 / {k2 + k-1 [A]}
So the rate of product formation is:
d[product] / dt = k2 [A*]
= k2 k1 [A]2 / {k2 + k-1 [A]}
Two limiting cases………
1- At high pressures ( 1 atm) most A* molecules will be deactivated…
rather than forming product………… k-1 [A] [A*]  k2 [A*]
k-1 [A]  k2
d[product] / dt = k2 [A*]
= k1 k2 [A]2 / {k2 + k-1 [A]}
At high pressures………….. k-1 [A]  k2
….so in this case the rate is given by the following equation:
d [product] /dt
=
(k1 k2 /k-1 ) [A]
…..reaction is first order with respect to A.
2- At low pressures ….. 0.5 atm……most of the activated reactant
forms product rather than being deactivated.
……in this case the following is true:
k-1 [A] [A*]  k2 [A*]
or k-1 [A]  k2
The rate becomes ………………
d [product] /dt
……..second order with respect to A.
=
k1 [A]2
Bimolecular Reactions:
• an elementary step that involves two reactant molecules
• two molecules collide and exchange energy, atoms, groups of atoms
or undergo some type of change
e.g. H + H2  H2 + H
Termolecular Reactions:
• an elementary step that involves three reactant molecules
• three molecules collide and exchange energy, atoms, groups of atoms
or undergo some type of change
• probability of a three-molecule collision is low so only a few reactions
are termolecular.
All known termolecular reactions involve nitric oxide as one of the
reactants……………
2 NO(g) + X2(g)  2 NOX (g)
X = Cl, Br, I.
Elementary steps with molecularity of greater than three have not
been identified to date.
# 12.22 Chang Text: The following data were collected for the
reaction between hydrogen and nitric oxide at 700°C:
2 H2(g) + 2 NO(g)  2 H20(g) + N2(g)
Experiment
1
2
3
[H2]/M
0.010
0.0050
0.010
[NO]/M
0.025
0.025
0.0125
Initial Rate/M s-1
2.4 x 10-6
1.2 x 10-6
0.6 x 10-6
(a) What is the rate law for the reaction ?
(b) Calculate the rate constant for the reaction.
(c) Suggest a plausible mechanism for the reaction that is consistent
with the rate law. (Hint: assume that the oxygen atom is the
intermediate).
Solution:
Earlier we found rate law for this reaction:
The rate law is:
Overall Rxn.
R = k [NO]2[H2]
2H2(g) + 2NO(g)  2H20 (g) + N2(g)
Hint: assume that the oxygen atom is the intermediate.
(remember an intermediate is formed in an early elementary
reaction and consumed in a later elementary reaction).
…..the rate law suggests that the slow “rate determining step”
involves:
2 molecules of NO and 1 of H2 …………
So a possible mechanism could be ………..
H2 + 2NO  N2 + H2O + O
H2 + O  H2O
slow step
fast step
#12.21 Chang Text – An excited ozone molecule, O3* in the
atmosphere can undergo one of the following reactions:
k
(1) fluorescence
k
(2) decomposition
k
(3) deactivation
O3* 1 O3
O3* 2 O + O2
3
O3 + M
O3* + M 
where M is an inert molecule. Calculate the fraction of ozone
molecules undergoing decomposition in terms of the rate constants.
The fraction of molecules undergoing decomposition is:
=
Rate of decomposition
Rate of fluorescence + Rate of decomposition + Rate of deactivation
= k2 [O3* ] / { k1 [O3* ] + k2 [O3* ] + k3 [O3* ] [M]}
= k2 / { k1 + k2 + k3 [M]}
9.5. Parallel Reactions (First-Order)
Parallel reactions are of the following type:
k2
A
↓
C
k1
→
B
The substance A can decompose by either of two
paths, giving rise to different products B and C.
Parallel Reactions
A
k2
k1
→
B
↓
C
–
d[A ]
= k1 [A] + k2 [A] = (k1 + k2) [A]
dt
d[B]
= k1 [A]
dt
(2)
d[C]
= k2 [A]
dt
(3)
(1)
Equation (1) solution – 1st order in A
 [A] 
 = – (k1 + k2) t
ln 
[
A
]

0 
Or:
[A] = [A]0 exp[ – (k1 + k2) t]
[A] decays exponentially
d[B]
= k1 [A] = k1 [A]0 exp[ – (k1 + k2) t]
dt
d[C]
= k2 [A] = k2 [A]0 exp[ – (k1 + k2) t]
dt
If:
[B]0 = [C]0 = 0
Separate variables and integrate:
[ B]
t
[ B ]0
0
 d[B] = k1 [A]0  exp[(k  k ) t ]  dt
1
[B] – [B]0 =
2
k 1[A]0
exp[ – (k1 + k2) t]
 (k 1  k 2 )
(evaluated at t = t and t = 0)
=
k 1[A]0
k 1[A]0
exp[ – (k1 + k2) t] –
(1)
 (k 1  k 2 )
 (k 1  k 2 )
Therefore:
[B] =
k 1[A]0
{ 1 – exp[ – (k1 + k2) t] }
(k1  k 2 )
Similarly:
[C] =
k 2 [A]0
{ 1 – exp[ – (k1 + k2) t] }
(k1  k 2 )
Note that B and C are formed in a constant
ratio throughout the course of the reaction:
k1
[B]
=
k2
[ C]
Extension to 3 or more parallel reactions
is straightforward
Kinetics curves for reactant and products
for two first-order reactions in parallel.
9.6. Series Reactions (First-Order)
As we discussed previously, some reactions proceed through formation
of an intermediate (I)……..
A
ka

I
kb

P
For example consider decay of Uranium (U-239)………….
239U
2.35 min

239Np
2.35 day

239Pu
If we consider reaction above the rate of unimolecular decomposition
of A is
d[A] /dt = - ka [A]
and A is not replenished. The intermediate I is formed from A
at at a rate (ka[A]) but decays to P at a rate of kb[I].
Series reactions (first-order) are of the type:
k1
A
→
k2
B
→
C
Compound A reacts to form B which then goes on
to form C.
Series Reactions (First Order)
k2
k1
A → B → C
v1 = –
d[A]
= k1 [A]
dt
[A] = [A]0 exp(–k1t)
d[B]
= k1 [A] – k2 [B]
dt
More difficult to solve.
Assume that [B]0 = 0
d[B]
= k1 [A]0 exp(–k1t) – k2 [B]
dt
Rearrange to:
d[B]
+ k2 [B] = k1 [A]0 exp(–k1t)
dt
(1)
Now since:
d
d[B]
{ [B] exp(k2t) } =
exp(k2t) + [B] k2 exp(k2t)
dt
dt
= {
d[B]
+ k2 [B] } exp(k2t)
dt
Multiply both sides of (1) by exp(k2t):
{
d[B]
+ k2 [B] } exp(k2t) = k1 [A]0 exp(–k1t) exp(k2t)
dt
= k1 [A]0 exp[(k2 – k1)t]
Therefore:
d
{ [B] exp(k2t) } = k1 [A]0 exp[(k2 – k1)t]
dt
Separate the variables and integrate;
Both sides all terms are time dependent:
t
t
 d{[B] exp(k t )} = k1 [A]0  exp[(k  k ) t ]  dt
t 0
2
t 0
2
1
Left Hand Side becomes:
[B] exp(k2t)
(evaluated at t = t and t = 0)
= [B] exp(k2t) – [B]0
Right Hand Side becomes:
k 1[A ]0
exp[(k2 – k1)t]
(k 2  k 1 )
=
(evaluated at t = t and t = 0)
k 1[A ]0
{ exp[(k2 – k1)t] – 1 }
(k 2  k 1 )
Therefore:
[B] exp(k2t) – [B]0 =
k 1[A]0
{ exp[(k2 – k1)t] – 1 }
(k 2  k 1 )
Since [B]0 = 0 , we have:
k 1[A ]0
{exp(–k2t)} { exp[(k2 – k1)t] – 1 }
(k 2  k 1 )
[B] =
=
k 1[A]0
[ exp(– k1t) – exp(– k2t) ]
(k 2  k 1 )
v2 =
=
d[C]
= k2 [B]
dt
k 1k 2 [A]0
[ exp(– k1t) – exp(– k2t) ]
(k 2  k 1 )
[C] and t can be separated here;
Thus for [C]0 = 0
k 1k 2 [A]0 t
0 d[C] = =
  [ exp(– k1t) – exp(– k2t) ] dt
(k 2  k 1 ) 0
[C]
Then:
t
t
k 1k 2 [A ]0
{  exp(k 1 t )dt –  exp( k 2 t )dt }
[C] =
0
0
(k 2  k 1 )
=
k 1k 2 [A]0
exp( k 1 t )
exp(k 2 t )
{
–
}
(k 2  k 1 )
 k1
k 2
(evaluated at t = t and t = 0)
=
k 1k 2 [A ]0
exp( k 1 t )
1
exp(k 2 t )
1
{
–
–
+
(k 2  k 1 )
 k1
 k1
k 2
 k2
=
k 1k 2 [A]0
1
1
exp(k 1 t )
exp(k 2 t )
{
–
–
+
}
(k 2  k 1 )
k1
k2
k1
k2
}
=
k 1k 2 [A]0
(k 2  k 1 )
(k 2 exp(k 1 t )  k 1 exp(k 2 t ))
{
+
}
(k 2  k 1 )
k 1k 2
k 1k 2
= [A]0 { 1 –
1
[ k2 exp(–k1t) – k1 exp(–k2t) ] }
(k 2  k1 )
Final Result:
[C] = [A]0 { 1 –
1
[ k2 exp(–k1t) – k1 exp(–k2t) ] }
(k 2  k 1 )
Figure 1 illustrates the pattern
Figure 1
Kinetic curves for reactant, A , intermediate, B , and
product, C , for two first-order reactions in series
A → B → C
[A]
[B]
[C]
decreases exponentially
rises, reaches a maximum then decreases to 0 as 2nd step removes B
curve starts with 0 slope (v2 = 0)0 then undergoes an inflection
d 2 [ C]
= 0) when [B]max is reached.
(where
dt 2
[C] approaches a limiting value as reaction nears completion
2 Limiting Cases:
(1)
k1 >> k2
2nd is the Rate Determining Step
(Figure 2)
A converts to B rapidly
B undergoes 1st order conversion to C (controlled by k2)
[A]
[B]
[C]
drops rapidly
builds rapidly, then drops exponentially
formation nearly identical to that of a single step
1st order reaction
Note:
[C] = [A]0 { 1 –
if k1 >> k2
1
[ k2 exp(–k1t) – k1 exp(–k2t) ] }
(k 2  k 1 )
≈ [A]0 { 1 –
1
[ – k1 exp(–k2t) ] }
 k1
= [A]0 { 1 – exp(–k2t) }
Figure 2
Kinetic curves for reactant, A, intermediate, B , and
product, C , for two first-oder reactions in series,
with the 2nd as the rate-determining step
A → B → C
(2)
k1 << k2
1st is the Rate Determining Step
(Figure 3)
Slow conversion of A to B
Then rapid conversion of B to C
[B]
remains low throughout
C essentially appears as A disappears.
Velocity of reaction followed by either
formation of C or disappearance of A
Note:
[C] = [A]0 { 1 –
if k1 << k2
1
[ k2 exp(–k1t) – k1 exp(–k2t) ] }
(k 2  k 1 )
≈ [A]0 { 1 –
1
[ k2 exp(–k1t) ] }
k2
= [A]0 { 1 – exp(–k1t) }
Figure 3
Kinetic curves for reactant, A , intermediate , B, and
product, C , for two first-order reactions in series with
the 1st as the rate-determining step
A → B → C
Steady-State Approximation
The steady-state approximation assumes that, after an initial
induction period, an interval during which the concentrations of
intermediates, I , rise from zero, and during the major part of
the reaction, the rates of change of the concentrations of all
reaction intermediates are negligibly small.
d[I] / dt = 0
This approximation greatly simplifies the discussion of reaction
schemes and can be applied to a large variety of different types
of reactions, as we shall see.
The basis of the steady-state
approximation. It is supposed
that the concentrations of all
intermediates remain small and
hardly change during most of
the course of the reaction.
Atkins: Physical Chemistry, 7th Edition,
Figure 25.16
Steady-State Approximation
For the case of two first-order reactions in series,
with the first as the rate-determining step (k1 << k2)
Apply the Steady-State Approximation to
d[B]
:
dt
d[B]
= k1 [A] – k2 [B] ≈ 0
dt
Then:
k 
[B] ≈  1  [A]
 k2 
And:
k 
d[C]
= k2 [B] ≈ k2  1  [A] = k1 [A]
dt
 k2 
Therefore:
d[C] = k1 [A] = k1 [A]0 exp(– k1 t)
Integrate both sides of the equation:
t
t
0
0
[C] =  d[C] = k1 [A]0  exp(k 1 t )  dt
t
 exp(k 1 t ) 
= k1 [A]0 

  k1  0
 exp( k 1 t )
1 
= k1 [A]0 




k
k

1
1
= [A]0 [ – exp(– k1 t) + 1 ]
= [A]0 [ 1 – exp(– k1 t) ]
Same result as before for: Case 2 (k1 << k2) !!!
Example: Blood Clotting
An injury sets off a cascade of reactions involving a series of
proteolytic (protein cutting) enzymes, each of which activates a target
protein. Overall process involves a cascade of at least 8 such steps
(shown in following figure).
Final step is conversion of fibrinogen to fibrin (the clotting material).
This is catalyzed by the proteolytic enzyme thrombin.
Thrombin is not normally present in blood but is converted from
prothrombin by the action of another proteolytic enzyme and Ca2+ ion.
Thrombin concentration increases dramatically after wounding and then
decreases again once the clot is formed, very much like that of an
intermediate in a series of reactions.
Concentration of active thrombin after wounding. The clotting rate
for the conversion of fibrinogen to fibrin is a direct measure of the
thrombin concentration. [From: R. Biggs, Analyst 78, 84 (1953)]
The cascade serves serves to accelerate the clotting mechanism so
that it can be switched from fully off to fully on in less than 1 minute.
Thrombin concentration in blood must not remain high or it will cause
fibrin to form in circulatory system and block the normal flow of blood.
Antithrombin deactivates thrombin usually within a few minutes after
it has reached its maximum level. Action of thrombin restricted to the
period when it is critically needed !!
A = Prothrombin
B = Thrombin
C = Inactive Thrombin
Appearance of active proaccelerin turns on 1st step (A → B)
Formation of antithrombin turns on 2nd step
Rate of clotting
indicates
[Thrombin]
(B → C)
during the process.