BIO134: Chemical Kinetics
K AndoSchool of Chemistry, University of Birmingham
http://www.chem.bham.ac.uk/labs/ando/bio134/

Last updated: February 18, 2005

Contents
1

Thermodynamics
1.1 The 1st and 2nd Laws . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2 Gibbs Free Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.3 Standard Gibbs Free Energy and Equilibrium Constant . . . . . . . . .

3
3
4
5

2

Reaction Rate and Rate Law
2.1 Rate of reaction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.1.1 Stoichiometry . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2 Rate Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.3 Temperature Dependence . . . . . . . . . . . . . . . . . . . . . . . . .
2.3.1 Arrhenius law . . . . . . . . . . . . . . . . . . . . . . . . . . .

7
7
8
9
10
10

3

Data analysis
3.1 Unimolecular reaction A → P . . . . . . . . . . . . . . . . . . . . . .
3.1.1 Zero order reaction . . . . . . . . . . . . . . . . . . . . . . . .
3.1.2 First order reaction . . . . . . . . . . . . . . . . . . . . . . . .
3.1.3 Second order reaction . . . . . . . . . . . . . . . . . . . . . .
3.1.4 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.1.5 Half-life Method . . . . . . . . . . . . . . . . . . . . . . . . .
3.2 More than one species involved . . . . . . . . . . . . . . . . . . . . . .
3.2.1 Isolation Method . . . . . . . . . . . . . . . . . . . . . . . . .
3.2.2 Initial Rate Method . . . . . . . . . . . . . . . . . . . . . . . .

12
12
13
13
14
14
15
17
17
17

1

4

Simple Reactions
4.1 Reversible reaction approaching to equilibrium . . . . . . . . . . . . .
4.2 Pre-equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.3 Consecutive reaction . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.4 Rate-determining process . . . . . . . . . . . . . . . . . . . . . . . . .

20
20
21
23
25

5

Steady-state approximation
5.1 Application . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.2 Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.3 Michaelis-Menten equation . . . . . . . . . . . . . . . . . . . . . . . .

25
26
27
27

Prerequisites• Basic knowledge of A-level Kinetics. (e.g., S-Cool AS&A2 Level Chemistry
Revision Guide, http://www.s-cool.co.uk/topic index.asp?subject id=7)

Readings• Housecroft and Constable, Chemistry, 2nd ed., Prentice Hall, Chapter 14.
• Atkins, The Elements of Physical Chemistry, 3rd ed., Oxford, Chapter 10-11.

Acknowledgments
Many of the diagrams in this lecture note have been downloaded from the web pages ofthese textbooks.

2

1

Thermodynamics

Let us first fix some terminologies.
Open systems can exchange both energy (heat) andmatter with its surroundings.
Closed systems can exchange energy (heat) but not matter with its surroundings.
Isolated systems can exchange neither energy (heat) nor
matter with its surroundings.

1.1

The 1st and 2nd Laws

The 1st and 2nd laws of thermodynamics can be stated as follows. It is important tonote that these apply for energy and entropy within an isolated system.
The 1st law :The total energy Etot of an isolated system conserves: ∆Etot = 0.
The 2nd law :The total entropy Stot of an isolated system tends to increase: ∆Stot ≥ 0.
in which ∆Etot and ∆Stot are changes of Etot and Stot associated with processes occuring in the system.
A typical example of the 2nd law is seen in the diagram below as spontaneous mixingof the two kinds of gases.

The entropy of the system is closely related to the randomness of the system. In thisexample of gas mixing, the randomness increases spontaneously, which means that theentropy also increases. In order to define the entropy more precisely in terms of therandomness, we need to learn Statistical Mechanics. However, this is beyond the scopeof this lecture course, and we will content with the qualitative connection between theentropy and the randomness.
An important thing to note here is that the spontaneous mixing occurs even when
the gases are ideal, i.e., no interaction between the molecules. In this case, there is noenergy change (gain or loss) upon mixing of the gases. This means that, the mixing
3

is not driven by the energy gain but by the entropy gain in terms of the 2nd law ofthermodynamics.

1.2

Gibbs Free Energy

Let us now consider a System in contact with
Surroundings (i.e., heat bath). We assume that the System is closed and the total system of ‘System and Surroundings’ is isolated. The System can thus exchange
heat with the Surroundings. The Surroundings is, on theother hand, supposed to be large enough such that its temperature T is kept constant.
From the 1st law of thermodynamics,∗ the enthalpy change of the total system iszero: ∆Htot = 0. Because the total enthalpy change is the sum of the enthalpy changeof the System and the Surroundings, ∆Htot = ∆Hsys + ∆Hsurr , it follows that
∆Hsys = −∆Hsurr(1)

In other words, the heat is exchanged solely between the System and the Surroundings
and goes nowhere else (because the total system is isolated). The enthalpy gain of the
System is precisely equal to the loss of the Surroundings.
On the other hand, the 2nd law predicts that the entropy of the total system tends toincrease;∆Stotal = ∆Ssys + ∆Ssurr ≥ 0(2)However, this does not appear very useful as it only deals with the total system andnot the System in which the processes of our interest are taking place. We are moreinterested in predicting what will happen in the System rather than in the total systemthat include the Surroundings. We therefore wish to separate our discussion on the
System from the Surroundings. To this end, it is useful to define the Gibbs Free Energyof the System by∆Gsys = ∆Hsys − T ∆Ssys(3)This is useful because we can show that the 2nd law ∆Stot ≥ 0 is essentially equivalentto ∆Gsys ≤ 0. Thanks to this, we only need to look at the quantities (enthalpy andentropy) of the System and do not have to be bothered by the Surroundings.
The equivalence of ∆Stot ≥ 0 and ∆Gsys ≤ 0 can be demonstrated as follows. The
essential point is that the heat bath (= the Surroundings) is large enough such that the∗Here we assume to be dealing with solution phase systems, as is usually the case in biochemistry,
such that the volume change of the system may be neglected. In this case, it would be justified to consider
that the enthalpy and the internal energy are identical. In other words, we neglect pV in H = U + pV .

4equilibrium is maintained in the Surroundings in the course of the processes occuringin the System. Therefore, we may put∆Gsurr ∼=0∆Hsurr ∼= T ∆Ssurr⇒(4)

Putting this and Eq (1) into Eq (5), it follows∆Gsys = −T (∆Ssurr + ∆Ssys ) = −T ∆Stot(5)

Because the absolute temperature T (in Kelvin) is always positive, Eq (5) means that the
2nd law ∆Stotal ≥ 0 is equivalent to ∆Gsys ≤ 0. In other words, we can consider that
∆Gsys effectively contains the influences from the Surroundings, as can be understoodfrom the derivation above.
In summary, the direction of the chemical processes isrelated to the Gibbs free energy change such that
• The reaction in the System occurs spontaneously only
if ∆Gsys < 0• The System is in equilibrium when ∆Gsys = 0

1.3

Standard Gibbs Free Energy and Equilibrium Constant

Let us consider a reaction between x moles of X and y moles of Y producing z molesof Z.
xX + yY −→ zZ(6)
The Gibbs free energy change of the reaction is given by(aZ )z∆G = ∆G + RT ln Q = ∆G + RT ln(aX )x (aY )y
o

o




(7)

where aX,Y,Z are the activities of X, Y and Z, Q is the reaction quotient, R (= 8.314 ×
10−3 kJ mol−1 K−1 ) is the gas constant, and ∆Go is the standard free energy of thereaction. At equilibrium, the free energy change of the reaction vanishes, ∆G = 0, andthe reaction quotient Q is equal to the equilibrium constant K. Then, Eq (7) becomes
0 = ∆Go + RT ln K⇒

∆Go = −RT ln K

(8)Out of equilibrium, the reaction quotient Q is not equal to the equilibrium constant K;
in other words, the activities aX,Y,X are different from their equilibrium values. ∆G isthen not equal to zero, and its sign determines the spontaneous direction of the reaction:
forward or backward reaction when ∆G < 0 or > 0, respectively.
5

Question:

The equilibrium constant of a reaction,
C6 H5 OH(aq) −→ C6 H5 O− (aq) + H+ (aq)is K = 1.3 × 10−10 at 298 K.
Q1. Calculate ∆Go .
Q2. A mixture of [C6 H5 OH] = 0.1 mol dm−3 and [C6 H5 O− ] = [H+ ] = 10−5 mol dm−3
was prepared. Will the reaction occur spontaneously?
In this way, ∆G determines the direction of the reaction. However, it is important
to note that the rate of the reaction is not determined by ∆G.

62

Reaction Rate and Rate Law

2.1

Rate of reaction

The graphs above show the change along time of the concentrations of the reactant
and the product. The rate of the loss of the reactant at a particular moment is given by
the slope of the tangent of the graph at that moment. (See also the diagram below.)
Rate of loss of reactant (R) = −

d[R]
dt

(9)

Similarly, the rate of the formation of the product at that moment is given by
Rate of formation of product (P) = +

d[P]
dt

(10)

Note the minus sign in (9): the reactant concentration [R] is decaying along time so thatits derivative d[R]/dt is negative, and therefore Eq (9) overall is positive.
In general, reaction rates change along time. As the graph below shows, the rateis usually the largest at the beginning (t = 0) and gradually decreases along time as
the reaction proceeds and the concentrations approach to their asymptotic (equilibrium)
values.

7

2.1.1

Stoichiometry

Let us consider a reaction A → B, where one mole of A produces one mole of B. The
rates of the loss of A and the formation of B are related as
A −→ B

⇒

−

d[A]
d[B]
=
dt
dt

(11)

Now, how should (or should not) this be modified for a different reaction A → 2B ? In
this case, because two moles of B is produced as one mole of A is lost, the rate of theformer is twice as large as the latter. Therefore, we findA −→ 2Bd[A]
1 d[B]
=
dt
2 dt

(12)

d[A]
d[B]
= (?)
dt
dt

(13)

⇒

−

⇒

−

Similarly, for a reaction 2A → B,
2A −→ B
(Find the numerical coefficient.)8

Question:
Consider the reaction
2 N2 O5 (g) −→ 4 NO2 (g) + O2 (g)Find the numerical coefficients in−d[N2 O5 ]d[NO2 ]
d[O2 ]
= (?)
= (?)
dt
dt
dt

Here is a useful formula for general reactions
n1 R1 + n2 R2 + · · · −→ m1 P1 + m2 P2 + · · ·(14)

We simply put the inverse of the molar coefficients in front of the derivatives d[R]/dt
etc., remembering the minus signs for the reactants;−

2.2

1 d[R1 ]1 d[R2 ]1 d[P1 ]1 d[P2 ]=−= ··· = +
=+
= ···
n1 dt
n2 dt
m1 dt
m2 dt

(15)

Rate Law

As we have seen, the reaction rates changes along time, normally decreases along withthe progress of the reaction. This implies that the rate decreases as the number of reactant molecules decreases. Normally, as the concentrations † of the reactants decreases,
the chance of their reactive encounter will also decrease. (On the other hand, the increase of the product concentrations may inhibit the reaction.)
We often observe that the reaction rates are proportional to the concentrations or
the square of the concentrations of the reactants. We sometimes observe the reaction
rates proportional to the cube or the square-root of the concentrations. These may bedescribed in a form,
Rate = k[A]n [B]m · · ·(16)where the coefficient k is called the rate constant of the reaction. n and m are theorders with respect to the reactants A and B, respectively. They are most often 1 or 2,
and sometimes 3 or 1/2. We also define the overall order by n + m.
†

For gas phase reactions, we usually consider partial pressures of the species instead of concentrations.
The partial pressure is proportional to the concentration for ideal gases at a given temperature and volume,
pV = nRT ⇒ p = RT (n/V ).

9

It is very important to note that the orders cannot be predicted simply from thestoichiometry of the reaction. Let us see the following two examples:
Example 1.
2NO(g) + O2 (g) −→ 2NO2 (g)(17)

Observed rate law: Rate = k [NO]2 [O2 ]Example 2.
2SO2 (g) + O2 (g) −→ 2SO3 (g)(18)

1[SO2 ]
Observed rate law: Rate = k [SO2 ][SO3 ]− 2 = k p[SO3 ]

Note that these two reactions have the same stoichiometry of
2A + O2 −→ 2AO(19)

but the observed rate laws (the orders) are totally different. This means that we cannot predict the rate law just by looking at the equation of the reaction. Generally, therate laws must be determined from experiments. The experimental methods and their
theoretical background will be discussed in the next section.
As we will see in the subsequent discussions, different rate laws come from different
mechanisms of the reaction. Namely, different involvements of the elementary steps
and the associated intermediate species. In other words, by carefully examining therate law, we expect to be able to clarify the mechanism of the reaction. This is indeed
the major motivation to study Chemical Kinetics. From the analysis of the rate laws
derived from experiments, we can discuss and infer the mechanism of the reactions.
Question: State the orders with respect to each specie and the overall order for Examples 1 and 2. Does the rate increase or decrease as the concentration of SO3 increases in
Example 2? What could be the chemical reason for this?

2.3

Temperature Dependence

2.3.1

Arrhenius law

Normally, the reaction rates increase by raising the temperature. This sounds natural
since the reactant molecules will move more actively in higher temperature, and hence
the chance of their reactive encounter will be larger. It was found from a number of10experiments that this situation is described quantitatively by the following formula forthe rate constantk = Ae−Ea /RT(20)
in which Ea and A are called an activation energy and a pre-factor.‡ R is the gas constant(= 8.314 × 10−3 kJ mol−1 K−1 ).

By taking the natural logarithms (loge or ln) of both sides,
ln k = ln A −

Ea
RT

(21)

Therefore, by plotting ln k against 1/T , we will obtain a straight line as shown in thenext diagrams. From the slope and the intercept of the plot we can calculate the activation energy Ea and the pre-factor A.

‡

The pre-factor A is sometimes called a frequency factor.

11Equations (20) and (21) are called the Arrhenius equation and the plot of ln k against
1/T the Arrhenius plot.
The Arrhenius law is closely related to the Boltzmann’s law. In thermal equilibrium at temperature T , the molecules can take various energies.§ The distribution of thenumber of molecules plotted against energy E will look the following diagrams.

As seen, the number of molecules decreases as the energy increases (in the high energy
region). According to the picture of the activation energy barrier lying between the reactant and product states, the reaction rate will be proportional to the number of molecules
having higher energy than Ea , i.e., in the shaded area of the diagram. By raising thetemperature the molecules will receive energy from the surroundings, and then the distribution graph will shift toward the right, i.e., toward the higher energy. This results in
an increase of the number of molecules above Ea and hence the increase of the reaction
rate. This behaviour is described quantitatively by the Arrhenius equation (20) or (21).

3Data analysis

Suppose that we have carried out a series of experiments to observe changes along timeof the concentrations of the reactants and products as in the diagrams in Section 2.1
From these and other supplementary data, we attempt to determine the rate law, i.e., theorders with respect to the species involved and the rate constant.

3.1

Unimolecular reaction A → P

We begin with a very simple reaction in which only one specie A is involved in thereactant.
A −→ Product(s)(22)§

The temperature T is proportional to the average kinetic energy of the molecules. The energy ofeach molecule may deviate from this average.

12

We do not consider the backward reaction. We also assume that there is no intermediate
species or intermediate steps involved in the reaction. That is, (22) is a single step
reaction. This is an example of the elementary reaction.
In this case, the rate law will contain only the concentration of A,
Rate = k[A]n(23)

Because the reaction rate is that of the loss of A, −d[A]/dt, the above equation is a
simple differential equationd[A]−= k[A]n(24)dtwhich can be integrated mathematically.
3.1.1

Zero order reaction

Let us first consider the zero order case, n = 0. Because [A]0 = 1, (24) becomes
d[A]
=k
(25)
dt
This means that the gradient d[A]/dt is a constant −k. Therefore, [A] decays at aconstant rate and the plot of [A] along time will be a straight line.
−[A]t = [A]0 − kt

(26)

Here and hereafter, we denote the concentration of A by [A]t when we wish to emphasize its dependence on time. The initial concentration at t = 0 is [A]0 .
3.1.2

First order reaction

For n = 1, we haved[A]
= −k[A]dt
The solution¶ is an exponential function(27)[A]t = [A]0 e−kt

(28)ln[A]t = ln[A]0 − kt(29)This is equivalent to
Therefore, the plot of ln[A]t against time t is a straight line.
¶

For simplicity we write A(t) instead of [A]t . It follows from (27) that
Z
Z
dA
dA
= −kA ⇒= −k dt ⇒ ln A(t) = −kt + C
dtAPutting t = 0, we find C = ln A(0). We thus get (29) which is equivalent to (28).

133.1.3

Second order reaction

For n = 2,d[A]
= −k[A]2
dt

(30)

[A]0
1 + kt[A]0

(31)

1
1
=
+ kt
[A]t
[A]0

(32)

The solution isk[A]t =
Or equivalently,

Therefore, the plot of 1/[A]t against time t is a straight line.
3.1.4•

Summary
0th order:

Rate = − d[A]=kdt
⇒ [A]t = [A]0 − kt
⇒ Plot of [A]t against time t is linear•

1st order:

Rate = − d[A]
= k[A]dt⇒ [A]t = [A]0 exp(−kt)⇒ ln[A]t = ln[A]0 − kt

⇒ Plot of ln[A]t against time t is linear•

2nd order:

Rate = − d[A]
= k[A]2dt⇒ [A]t =[A]0
1 + kt[A]0

⇒

1
1
=
+ kt
[A]t
[A]0

⇒ Plot of 1/[A]t against time t is linear
In this way, the order n and the rate constant k can be determined from the graphs of[A], ln[A] or 1/[A] against time. Suppose we have obtained from experiments the concentration of A as a function of time. We then plot [A], ln[A] or 1/[A] against time. If
one of these was found to be a straight line, the order n is 0, 1 or 2, respectively.
k

From (30),dA= −kA2
dt

Z
⇒

−

dA
=k
A2

Z
dt

⇒

1
= kt + CA(t)Putting t = 0, we find C = 1/A(0). This immediately gives (32) and after rearranging we get (31).

14For the first and second order reactions, the raw data of [A]t will look like

Both curves show smooth decay in which the gradient (= rate) decreases along time as
the concentration [A] decreases. This is simply because the rate is proportional to [A]or [A]2 . In other words, the chance of the reaction will decrease as we lose the reactant
molecules along with the progress of the reaction.
3.1.5

Half-life Method

The half-life is the time taken for the concentration of the reactant [A] to halve. It
may depend on the starting concentration [A]. For example, for zero order reactions
where [A]t is a linear function of t, the half life decreases along time, i.e., as the starting
concentration decreases. (Confirm this by drawing a graph.) In contrast, for the first15

order reactions the half life is a constant that is independent of the starting concentration.
See the diagrams below.

This is a special characteristic of the fist order reactions. Let us denote the half life byτ . Then Eq (28) at t = τ is written as[A]0= [A]0 exp(−kτ )2
By dividing both sides by [A]0 and taking the natural logarithms, we find

(33)

1
ln 2
= exp(−k τ ) ⇒ exp(k τ ) = 2 ⇒ k τ = ln 2 ⇒ τ =(34)
2
kAs seen in the previous diagram, the decay profile of the concentration [A] for thesecond order reaction is quite different from the first order reaction. It is easy to show
from Eq (32) that the half life for the second order reaction is proportional to the inverseof the starting concentration, τ = 1/k[A]0 . This means that the half-life increases as thereaction proceeds.
In summary,•

0th order:

half-life = [A]0 /2k∝ [A]0•

1st order:

half-life = ln 2/kindependent of [A]0•

2nd order:

half-life = 1/k[A]0∝ 1/[A]0

Question: Confirm these behaviours (qualitatively) on graphs. Derive these mathematical formulas of the half-life from Eqs (26) - (32).
16

3.2

More than one species involved

In the previous section we assumed that only one reactant specie A is involved in thereaction. This is of course too restrictive; in reality we much more often encounter
reactions involving more than one species. This implies that the data analysis will be
more complicated. For example, how can we determine the orders n and m when therate law is given by
Rate = k[A]n [B]m(35)
which means that the differential equation now contains two unknown time-dependent
variables [A]t and [B]t ? Two representative strategies to work around this problem are
the isolation method and the initial rate method.
3.2.1

Isolation Method

Suppose that we have prepared a large excess of one of the reactants, say A, comparedto B. That is, the initial concentrations are such that [A]0  [B]0 . In this case, we may
consider that [A] will stay large in the course of the reaction so that its concentration
change may be neglected approximately.
[A]t ' [A]0(= constant)

Then the rate law will be approximated asRate ' k[A]n0 [B]m ≡ k 0 [B]mwhere k 0 is defined by k 0 = k[A]n0 . In this way, the unknown variable [A] has been
(approximately) eliminated from the rate law. Now we can apply the methods in theprevious section in order to determine the order m and the new rate constant k 0 .
In the next set of experiments, we prepare a large excess of B compared to A, andrepeat similar procedure to determine the order n. Finally, from n and k 0 we determine
the original rate constant k.
In this way, the contributions from each of the reactant species can be isolated one
by one.
3.2.2

Initial Rate Method

Another method to avoid the complication coming from the involvement of multiple
species is to consider only the initial rates (rates at t = 0).

17

The essential point is that the concentrations of all the reactant species are precisely
known at t = 0 as we have prepared them.
(Initial Rate) = k[A]n0 [B]m0
Therefore, we do not have to be concerned with the complicated changes of the concentrations of the multiple species that will occur in t > 0.
For example, to determine the order n, we measure the initial rate of the decay of[A], that is, −d[A]/dt at t = 0. We repeat this for different initial concentrations [A]0 as
shown in the diagram below. It is important to keep [B]0 to the same value in this seriesof experiments. (Otherwise we will mess up the analysis!)18

Example:

The following data have been obtained for a reaction A + B → P.
[A]0
/ mol dm−3
0.2
0.4
0.2

[B]0
/ mol dm−3
0.2
0.2
0.6

Initial Rate
/ mol dm−3 s−1
5 × 10−5
1 × 10−4
1.5 × 10−4

Find the orders with respect to A and B and the rate constant k.
As illustrated by this Example, the analysis is easy when the orders are immediately found to be simple integers. For otherwise and more general cases, the following
procedure is useful.
By taking the logarithms of both sides of(Initial Rate) = k[A]n0 [B]m0
we obtain∗∗log(Initial Rate) = log k + n log[A]0 + m log[B]0
Therefore, the rate constant k and order n can be determined from a plot of log(Initial Rate)against log[A]0 . Note that [B]0 should be kept to the same value in this analysis. By exchanging A and B, we can determine the order m from another set of experiments.

∗∗

Remember that log XY = log X + log Y and log X n = n log X. Here the base can be anything, so
we can use either log10 or ln depending on our particular conveniences.

19

4

Simple Reactions

4.1

Reversible reaction approaching to equilibrium

Let us consider a reversible reaction between A and B
kf ,kb

A *
) B

(36)

If we start with pure A, i.e., [A]0 6= 0 and [B]0 = 0, part of A
will be converted to B until they reach the equilibrium statein which the concentrations are [A]∞ and [B]∞ . (These are[A]t and [B]t at time t = ∞. Although we use the symbolfor infinity, ∞, this practically means a sufficiently long timecompared to the time scale of the reaction.) The ratio between
these two is the equilibrium constant of this reaction.
K=

[B]∞
[A]∞

(37)

Note the difference from the reactions treated in the previous section
A −→ P
with no backward reaction. In this case, [A] will decay tozero, i.e., [A]∞ = 0. This is in contrast with the reversible
reaction (36), in which finite amount of A will be restored dueto the existence of the backward reaction and hence [A]∞ 6= 0
and the equilibrium constant K is finite.
We will assume that both the forward and backward rate constants kf and kb are firstorder. Therefore, the forward reaction contributes to the loss of A and gain of B as
d[A]
d[B]
=+
= kf [A]dt
dt
Similarly, the backward reaction contributes to the gain of A and loss of B as
−

(38)

d[A]
d[B]
=−
= kb [B](39)dt
dt
Overall, the rate of the change of [A] is a combination of these two contributions suchasd[A]= −kf [A] + kb [B](40)dt
In sufficiently long time t = ∞, the system will relax to the equilibrium state inwhich the concentrations do not change any more. That is, d[A]/dt = 0 at t = ∞.
+20

Then, Eq (40) will be††0 = −kf [A]∞ + kb [B]∞

⇒

[B]∞
kf
=
[A]∞
kb

(41)

As seen in Eq (37), the left hand side of the last equation is equal to the equilibrium
constant K. We therefore find a relation
K=

[B]∞
kf
=
[A]∞
kb

(42)

Although this might appear a simple equation, it is conceptually remarkable because it
finds a relation between the equilibrium property K and the kinetic quantities kf and
kb .

4.2

Pre-equilibrium

Now we introduce, in addition to the reversible process A *) B just considered, a reactive process from B to a final product P.
kf ,kbk

A *) B →r P(43)

This last process will disturb, or even destroy, the equilibrium between A and B. However, we will assume that this process is very slow such that the equilibrium between A
and B is well maintained. In other words, we will consider situations where the ratio[B]/[A] is approximately kept constant. Let us denote this constant by K.
kf
[B]
'K=
[A]
kb

(44)

The last equality has been found in the previous section in Eq (42). We call this the‘pre-equilibrium’ assumption: the equilibrium A *) B prior to the reactive process B→ P is approximately well maintained throuout the reaction.
Question: Sketch (roughly) the changes of [A], [B] and [P] along time t. (Hint: Lookat the graph in the previous page for A *) B, and consider what will happen by introducing the reactive step B → P.)††
We can also derive Eq (41) by the following consideration. In equilibrium at t = ∞, the forward Eq(38) and backward Eq (39) processes will balance, such that

kf [A]∞ = kb [B]∞21

We further assume that the reactive step B → P is a 1st order process:d[P]= kr [B]dt
The overall reaction rate is given by this rate of the product formation.
Rate =d[P]dt

With the pre-equilibrium assumption, Eq (44), this is rewritten as
Rate =d[P]kr kf= kr [B] ' kr K[A] =[A] ∝ [A]
dt
kb

(45)

Therefore, the overall rate law turns out to be 1st order with respect to A.
Now, a question arises: How can we distinguish this from the straightforward 1storder reaction A → P when we have no way to detect the intermediate specie B? Sometimes, though not always, we can get some clue to this question by looking at thetemperature dependence of the reaction rate. To this end, it is useful to look at thefollowing specific example.
Example:Oxidation of nitrogen oxide
2NO(g) + O2 (g) → 2NO2 (g)Experimental facts are:
1. Rate law is given by; Rate = k [NO]2 [O2 ](overall 3rd order)

2. Reaction rate decreases as the temperature increases.
These observations cannot be explained by the simple bimolecular collision mechanism(which predicts a 2nd order rate law k [NO][O2 ]). In particular, the second observationis unusual in view of the ordinary temperature dependence described by the Arrhenius
equation (Section 2.3).
Let us assume a pre-equilibrium mechanism for this reaction:
kf ,kbNO + NO *) N2 O2k

N2 O2 + O2 →r 2NO2
From the pre-equilibrium assumption we get[N2 O2 ]
kf
'K=
2
[NO]
kb

⇒ [N2 O2 ] ' K[NO]2
22

(46)
(47)

Putting this into the rate expression derived from the reactive step EQ (47),
Rate =d[NO2 ]
= kr [N2 O2 ][O2 ]dt

we obtain
Rate = kr K[NO]2 [O2 ](48)

This immediately explains the first fact, the overall 3rd order rate law.
Now, let’s see how this explains the second fact, the unusual temperature dependence of the rate. As seen in Eq (48), the overall rate constant is a product of two
factors, k = kr K. We can naturally assume that the rate constant of the reactive step
kr has a normal temperature dependence such that it increases in higher temperatures.
However, the behaviour of the equilibrium constant K can be different. It may increaseor decrease in higher temperatures depending on the sign ‡‡ of the free energy change∆Go . For example, ∆Go < 0 means that the product state has lower G than the reactant. Generally, by raising the temperature we increase the populations of the higher
energy state. This means that, when ∆Go < 0, increase of the temperature will result in
an increase of the reactant concentrations and hence a decrease of K.
In this way, the overall rate constant k = kr K may contain competing positive
and negative factors of the temerature dependence. The obseved fact for this particular
reaction suggests that the decrease of K is the dominant factor to determine the overalltemperature dependence.

4.3

Consecutive reaction

Let us next consider the following consecutive reaction
k

k

a
b
A −→
B −→
C

The behaviour of the intermediate [B] will depend on the difference between ka and kb .
‡‡oThis can be described more quantitatively by K = e−∆G /RT .

23

Question:

State, with explanation, which diagram corresponds to ka  kb .

Assuming 1st order reactions for both of the two steps, it is possible to solve the rateequations analytically,∗ which gives[A]t = [A]0 e−ka t







ka
[B]t =(−e−ka t + e−kb t )[A]0
k
−
k
a
b






1

−ka t
−kb t

 [C]t = 1 +(kb e− ka e) [A]0ka − kb
The mathematical derivation is far beyond the assumed level of this course. The reason
for showing these equations is because the formula for [C]t is useful to demonstrate the∗

The rate equations for this scheme are:dA
= −ka A,dt

dB
= ka A − kb B,
dt

dC= kb Bdt

The first equation for A(t) is the same as what we have seen previously in Sec 3.1.2. We put the solution
A(t) = A(0)e−ka t in the second equation. This yields an ordinary differential equation for B(t), forwhich the solution method is well established (though we have no space to expand it here). C(t) can beobtained from a relation A(t) + B(t) + C(t) = constant = A(0), which describes the conservation ofgross materials.

24concept of rate-determining step, as will be discussed next.

4.4

Rate-determining process

The above formula for [C]t can be simplified in two limiting cases where one of therate constants is much larger than the other. For example, when ka  kb , we finde−ka t  e−kb t and 1/(ka − kb ) ' 1/ka . Therefore, [C]t can be simplified as[C]t ' (1 − e−kb t )[A]0
This means that the behaviour of [C]t is determined by the rate constant of the slower
step kb .
Question:Show that when ka  kb ,[C]t ' (1 − e−ka t )[A]0

In this way, the slower step determines the overall rate of the product formation.
This is called the Rate-determining or Rate-limiting step. The idea is straightforwardly generalized for reactions involving more than two steps, i.e., more than one
intermediates I1 , I2 , · · · , involved between the reactant R and product P:
R → I1 → I2 → · · · → P

5Steady-state approximation

It is seen in the right panel of the previous diagram that the concentration of B does
not change much but stays low and nearly constant, and therefore d[B]/dt ' 0, whenkb  ka , i.e., when this intermediate specie is highly reactive. This suggests a very
useful and powerful method called the steady-state approximation.
Recipe of the Steady-State Approximation
1. Identify reactive intermediates, I1 , I2 , · · · , in the mechanism
2. Write down
3. Setd[Ii ]
from the assumed mechanism.
dt

d[Ii ]= 0 and solve for [Ii ]. (The solution will be denoted by [Ii ]ss .)dt

4. Deduce, and simplify if possible, the rate law by using [Ii ]ss .
25

5.1

Application

Let us see how the steady-state approximation works for the following scheme
kf ,kbk

A *) B →r P(49)

which has been examined previously by invoking the pre-equilibrium approximation.
The critical assumption there was the reaction from B to P is very slow such that it
does not disturb very much the equilibrium between A and B. We will now relax thisassumption. To this end, let us follow the recipe of the steady-state approximation. As
before, we will assume that all the steps kf , kb and kr are 1st order.
1. We identify B as the reactive intermediate.
2. Noting that B is gained by kf and lost by kb and kr steps,d[B]= kf [A] − kb [B] − kr [B]dt
3. Set the above to be equal to zero, and solve for [B].
kf [A] − kb [B] − kr [B] ' 0 ⇒ [B]ss =kf[A]k b + kr

4. The overall reaction rate is that of the product formation described by the kr step.
We write down this equation and then replace [B] by [B]ss .
Rate =

kr kf
d[P]= kr [B] ' kr [B]ss =[A]dt
kb + kr(50)

Compare this with the result from the pre-equilibrium assumption, Eq (45):Rate (pre-Eq) ' kr K[A] =

kr kf
[A]
kb

The difference is in the denominator of the rate factor. Remember that the fundamental
assumption in the pre-equilibrium case was kr  kb . This means that the denominatorin Eq (50) may be approximated as kb + kr ' kb . We therefore find that the preequilibrium approximation Eq (45) is reproduced as a limiting case of the steady-state
approximation Eq (50). In other words, the steady-state approximation generalizes, andtherefore covers broader applicability than, the pre-equilibrium approximation.

26

5.2

Exercise

Consider a nucleophilic substitution reaction:(CH3 )3 CCl + OH− → (CH3 )3 COH + Cl−
Let us first assume the following mechanism
k1 ,k−1(CH3 )3 CCl *) (CH3 )3 C+ + Cl−k2(CH3 )3 C+ + OH− −→(CH3 )3 COH

Q1. Apply the steady-state approximation for the intermediate (CH3 )3 C+ and derive
the rate law.
Q2. Show that the rate law is 1st-order when [OH− ] is in excess or when k−1  k2Let us next consider an alternative mechanism
kf ,kbkr(CH3 )3 CCl + OH− *(CH3 )3 COH + Cl−) [Cl · · · (CH3 )3 C · · · OH]− −→Q3. Apply the steady-state approximation for the intermediate complex and show that
this mechanism gives a 2nd-order rate law.
Answers:
1.
Rate =

k1 k2 [(CH3 )3 CCl][OH− ]k−1 [Cl− ] + k2 [OH− ]2.
Rate ' k1 [(CH3 )3 CCl] when k−1 [Cl− ]  k2 [OH− ]3.
Rate =

5.3

kf kr[(CH3 )3 CCl][OH− ]kb + kr

Michaelis-Menten equation

Let us consider an enzyme catalysis reaction in which the enzyme E and a substrate S
form an intermediate complex ES and then the reaction occurs to release the product P:
k1 ,k−1

k2
E+S *
E+P
) ES −→

27(51)

This looks similar to the scheme in Eq (49), and hence the steady-state approximation
will givek1 k2
Rate =[E][S]
k2 + k−1(It will be a good exercise to derive and confirm this.) However, this looks very different
from the famous Michaelis-Menten equation:
Rate =

Vmax [S][S] + KM

We notice that one of the major difference is that the enzyme concentration [E] is not
involved in the latter. We therefore try to eliminate [E] in the course of the application of
the steady-state approximation. To this end, it is essential to recognize that the enzyme
is a catalyst. Therefore, it is not destroyed by the reaction and the total number ofenzyme molecules is kept constant. This is described by[E] + [ES] = constant = [E]0(52)

in which [E]0 is the initially prepared concentration of the enzyme. Now we apply therecipe of the steady-state approximation. First we identify that the ES complex is theintermediate. Its change in the scheme (51) is written asd[ES]= k1 [E][S] − k−1 [ES] − k2 [ES]dt
The steady-state approximation sets the right hand side to be equal to zero. But before
doing it, we will eliminate [E] by using Eq (52):k1 ([E]0 − [ES])[S] − k−1 [ES] − k2 [ES] ' 0
k1 [E]0 [S](53)
k1 [S] + k−1 + k2
The overall rate is that of the product formation determined by the final step of thescheme (51):d[P]Rate == k2 [ES]dt
We replace [ES] by Eq (53):⇒ [ES]ss =Rate =

k2 k1 [E]0 [S]
k2 [E]0 [S]Vmax [S]
=
=
k
+k
−1
2
k1 [S] + k−1 + k2
[S] + KM
[S] + k1

in which Vmax = k2 [E]0 is the maximum rate and KM = (k−1 + k2 )/k1 is the Michaelis
constant.
28

Further Readings
Now there would be two ways for you to proceed. One is to review the basics again byreading the recommended textbooks. In particular, it would be a good idea to attempt theworked examples and problems in order to see how the theories work in more practicalsituations and to confirm your understanding of the theories. It is also useful to read
explanations by different authors. This is because no one can explain things perfectly
and you will have to build your own way of understanding by repeating reading andthinking.
Another way forward would be to read advanced books and research papers. One ofrecommendable books is:• A Fersht, Structure and Mechanism in Protein Science, Freeman.
(There are 8 long loan and 2 week loan copies in Barnes Library, Medical School,
Shelfmark QP 551/F.)
I trust that this lecture course has provided sufficient background to get ready for thisand other books and papers.

29