Statistical Thermodynamics

Introduction

to

Statistical

Thermodynamics

1

1. The Boltzmann’s distribution

Given three distinguishable particles, A, B and C.1 Each one of these particles is allowed

to have the energy 0, x, 2x, 3x, … etc. Suppose that the total energy (internal energy) of the

system consisting of these three particles is constant and equals 3x. The question now is how this

energy can be distributed over the three particles; or, rephrased, what the various possibilities of

distributing the particles over the energy levels are so that the total energy is 3x. As the following

diagram shows, there are 10 possible ways to achieve the required state. These possible

arrangements are also called complexions. The above required state is thus represented by ten

complexions. It is important to notice that the energy of any given particle is, as a result of the

continuous collisions between these particles and the accompanying energy exchange, not

constant with time; the total energy, however, is (isolated system).

The fundamental assumption of statistical thermodynamics is that all possible

complexions are equally probable. In other words, there is no preference of one arrangement over

the other. Any particle is likely to have one energy value as the other.

1 For example, atoms of a solid are distinguishable because the atoms are localized and each atom is characterized by its position, unlike the atoms of a gas.

{{C

CC

C

C

C

C

CC

C

B

BB

B

B

B

B

B

B

B AA

AA

AA

AA

A

A

{

X Y Zn3=0n2=0n1=3n0=0

n3=1n2=0n1=0n0=2

n3=0n2=1n1=1n0=1

2

3x

2x

1x

0

E

The 10 complexions shown above represent three micro-states X, Y and Z. Each micro-

state is defined by its own population distribution (i.e. the ni profile, where ni is the number of

particles occupying the ith energy level). Consequently, one complexion corresponds to the micro-

state X, three complexions correspond to Y and six complexions correspond to Z. We conclude

therefore that it is more probable to find the system in the micro-state Z than in the micro-states

X or Y. In general, the higher the number of complexions corresponding to one state is, the more

probable is this state.

Our next task is to derive a general formula to calculate the number of complexions

corresponding to a specific state, i.e. that with n0 particles in the energy level 0, n1 particles in the

energy level 1, n2 particles in the energy level 2, and so on. The above problem is identical with

having a sac containing N distinguishable balls (e.g. of different colors). The balls are to be

distributed in a certain number of boxes so that n balls come in the first box, n’ come in the

second box, n’’ in the third box, and so on. The number of different possibilities achieving that

state can be shown to be given by

(1.1)

Notice that the number of possibilities has been divided by the number of permutations of

the particles in a certain energy level among themselves because the order of pulling out the balls

is not important. To elucidate this point further consider the state X in the system described

above. The first energy level contains three particles A, B and C. It is not important whether A

then B then C were pulled out of the sac or A then C then B or B then C then A … etc.1 The six

possibilities (see footnote 1) reduce thus to a single complexion! Applying the above equation

for the states X, Y and Z gives2

Equation (1.1) has very important consequences as the number of particles N grows very

large. Suppose the system contains one mole of particles. The number of possible arrangements

(complexions) of having these Nav particles distributed over Nav energy levels (i.e. a single

particle per energy level, as in the Z state) is Nav!. The number of possible arrangements

1 These are six possibilities ABC, ACB, BAC, BCA, CAB and CBA.2 0!=1.

3

(complexions) of having all these Nav particles in the same energy level (as in the X state) is 1.

The chance of finding the system in the X state is now practically zero (compared to 10% when

the total number of particles was just 3). In other words, the system is expected to be found

entirely in the state with maximum number of complexions (the Z state in this case). This state is

not just the most probable; it is the stable state that can only be formed. It is the state of

equilibrium. Small fluctuation from this state may occur, however with less probability. Suppose

that one of the Nav drops from its energy level to the one below. As a result another particle must

jump from its energy level to the one above in order to keep the energy constant. The probability

of finding the system in this state is W=N!/4. If two particles drop to a lower energy level and

two jump to a higher one, then W=N!/8.

The question now arises whether it is possible to determine the most probable state (the ni

profile over the energy levels). Mathematics delivers the answer. We are looking for the

maximum value of W which is again the maximum value of lnW.1 The first derivative must thus

be zero. We make also use of the fact that W is a state function having an exact differential.

Two constraints now apply:

According to Lagrange, we introduce the undetermined multipliers and . Multiplying

the two constraints with and , respectively, and adding dlnW leads to equation 1.2.

(1.2)

Equation 1.2 is actually an equation system:

1 It is easier to manage the problem in the logarithmic form.

4

which can be summarized as

(1.3)

In the above equation system, there are i variables with two conditions (constant N and

constant E). That means there are i-2 independent variables. Now let us choose and so that

The equation system 1.3 reduces thus to

Since these i-2 variables are independent, then changing one of them doesn’t affect the

others. This means that the sum of all terms that involve that specific variable i must,

independently of the others, be zero, i.e.:

(1.4)

We’ll try now to evaluate (dlnW/dni) and will thereby make use of the Stirling’s theorem:

If N is very large, the above formula reduces to

Applying the Stirling’s theorem in equation 1.1 yields

5

and

(1.5)

Substituting equation 1.5 in equation 1.4 gives

Since a is constant, ea is constant and is set equal to A:

(1.6)

Equation 1.6 is the so-called Boltzmann’s distribution. This is the distribution of the most

probable state which represents the equilibrium distribution. Fluctuations from this state are

limitingly small.

If the ith energy level were degenerate with the degeneracy gi, equation 1.6 can be shown

to become

(1.7)

Equation 1.7 can be applied also even if the energy states are not exactly degenerate but

are very close to each other and fall within the energy range of d. In such case, one speaks of

energy bundles of d-width and gi is the number of energy states within this bundle. This is very

useful when treating continuous energies as will be shown in the case of translational energy.

With help of equation 1.6 or equation 1.7, the fraction of particles with the energy i can

be determined (this is the probability of finding the particle in the ith energy level):

(1.8)

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and the ratio of particles in the ith energy level to those in the jth energy level is given by

(1.9)

The summation ( ) in the above expressions is called the partition

function and is given the symbol q. This function is very important in statistical thermodynamics.

When known, all thermodynamic properties of the system can be calculated. More to this issue

and to the physical meaning of the partition function is found in section ??.

The constant A in the Boltzmann’s distribution (eq.1.6 or 1.7) is determined as follows:

7

2. Entropy and disorder

As explained in section 1, the most stable state is that with maximum number of

complexions. On the other hand and according to the laws of thermodynamics, the equilibrium

state is characterized by maximum entropy. It seems therefore plausible to assume that entropy is

associated with the number of complexions. In the language of mathematics,

Let us now consider two subsystems A and B:

Differentiating with respect to WB gives:

Differentiating with respect to WA gives:

For conveniency, the constant W0 is set to be zero and we are left with the famous

equation for statistical entropy engraved on Boltzmann’s tombstone in Vienna:

(2.1)

Above equation relates entropy with disorder. The larger the number of complexions

corresponding to a given state, the larger is the disorder in this state. Disordered states are more

stable because they can be achieved in more possible ways than ordered ones. The constant k in

equation 2.1 is the so-called Boltzmann’s constant. Interestingly, Boltzmann himself didn’t

determine its value. It was Planck who first made in 1900 an estimation of this value from his

solution of the black body radiation problem.

A B

8

By correlating statistical mechanics with thermodynamics, the value of can be

determined:

Suppose now that the system is supplied slowly with heat. The number of particles will

not thereby change (i.e. dni=0) but their distribution on the various i energy levels will. The

heat supplied to the system (dQ) is then equal to the change in the system energy (idni).

Substituting in the above expression yields thus:

The definition of entropy in thermodynamics is given by

Then, by applying equation 2.1

Comparing the two expressions for dlnW/dQ gives

and the Boltzmann’s distribution now reads

(2.2)

We’ll try now to evaluate the Boltzmann’s constant k. Consider a particle in three-

dimensional box with the dimensions a, b and c. The energy of such particle is given by quantum

mechanics to be

9

p, q and r are thereby positive integers (quantum numbers). Applying the above

expression for energy in the Boltzmann’s distribution gives

All constants in the exponent can be reduced to a single constant . Thus,

Because the energy levels are very close to each other, the summation can be replaced by

integration. The resulting integration is a standard one.

The same procedure is applied over the other two summations over q and r. The total

number of particles is then given by

From Boltzmann’s distribution

From the kinetic gas theory

Now comparing with the ideal gas law (pV=nRT) yields

10

The Bolzmann’s constant is thus the general gas constant per particle.

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3. The partition function

3.1 The molecular partition function and its interpretation

It is a sum of Boltzmann factors, , that specify how the particles are partitioned

throughout the accessible states. The numerical value of the partition function represents the

effective number of energy levels thermally accessible for the particle. To elucidate this point,

consider a two-level system1; the lower energy level has the energy zero and the second energy

level has the energy 1. The partition function reads

At very low temperatures (T→0), the second term approaches zero (e-∞) and q equals 1.

This means that the particle only exists in the first energy level. As the temperature is increased,

the second term increases and the value of q increases. A q-value higher than 1 means that the

particle has now also access to the second energy level but with lower probability than for

existing in the first energy level. As T approaches ∞, goes to unity and q =2. This means

that the particle can occupy the two levels with equal probability.

Another important example is a system with infinite number of equidistant energy levels

at 0, , 2, 3, …etc, where is the energy difference between two adjacent energy levels.

This resembles the vibrational energy levels of harmonic oscillator. The corresponding partition

function q reads

1 An example of two level system is the electronic spin energy in presence of external magnetic field that leads to lifting of degeneration.

12

The number of thermally accessible levels depends clearly on the ratio kT/ where kT

represents the thermal energy supplied by the surroundings. The higher the thermal energy (the

higher the temperature), the more are the levels that the molecule can exist in. Note that as the

energy level increases, the probability to exist at that energy level decreases (Boltzmann’s

distribution).

The vibrational energy of harmonic oscillator differs slightly from the ladder system

described above since the energy of the zeroth level is not zero. According to the laws of

quantum mechanics the energy of the zeroth vibrational level is the zero point energy which is

equal to 1/2ho, where o is the fundamental frequency of oscillation. The Boltzmann’s

distribution requires however that the energy of the zeroth level is zero. In order to apply the

Boltzmann’s distribution to the vibrational energy of harmonic oscillator, the vibrational energies

must be shifted donwards so that o becomes zero. This is achieved by subtracting the zero point

13

energy from the actual vibrational energies. Taking in consideration that the energy difference

between any two adjacent energy levels of the harmonic oscillator is ho

(3.1)

Exercise: Calculate the fraction of 1H35Cl (o=2886 cm-1) molecules present in the zeroth

vibrtional energy level at room temperature and at 1000ºC. Do the same calculation for the 127I-35Cl molecules (search for the fundamental frequency!). Compare the results and explain the

observed variation.

The partition function of rotation is more difficult to compute. For a linear molecule, the

energy of rotation is given by

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where I is the moment of inertia, B the rotational constant and J the rotational quantum number

with values 0, 1, 2, …etc. The rotational levels show 2J+1-fold degeneracy. The partition

function thus reads

(3.2)

Given the value of B, the partition function can be evaluated numerically (i.e. for each

value of J, the term is calculated, the terms are then summed up; you will see that the series

above converges to a certain value).

Exercise: Use Excel or Origin software to evaluate the rotational partition function of 1H35Cl at

25ºC. B=10.591 cm-1.

When the thermal energy kT is much larger than the energy difference between two

neighboring rotational levels, the sum in equation 3.2 can be approximated by an integral.

Equation 3.2 becomes

(3.3)

Exercise: Evaluate the rotational partition function of 1H35Cl at 25ºC using equation 3.3.

To derive an expression for the partition function of translation per degree of freedom, we

consider the particle to behave as a particle in a box. Its energy is thus given by

where a is the box length and p is a positive integer. The partition function can be approximated

by an integer because the energy levels are very close to each other (continuum).

with ,

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(3.4)

The quantity is given the symbol and called the thermal wavelength. For

three dimensions;

Notice that the partition function of translation depends on the size of the container.

Exercise: Calculate the translational partition function of an H2 molecule confined to 100 cm3 at

25ºC.

In the following it is shown that the total partition function of a molecule is the product of

individual partition functions. Assuming that the various energy contributions of a molecule are

independent (Born-Oppenheimer Approximation):

(3.5)

The same procedure can be used to show that the total partition function of any energy

contribution is the product of the individual partition functions per degree of freedom:

(3.6)

(3.7)

3.2 Significance of the partition function

The importance of the molecular partition function lies in the fact that it contains all

information needed to calculate all macroscopic thermodynamic properties of systems of

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independent particles1,2. Since the molecular partition function is related to a single molecule, so-

called system partition functions Q are defined for systems containing N particles (as the case is

in real systems). It can be shown that for a system containing N distinguishable independent

particles Q=qN. For N indistinguishable independent particles, Q=qN/N! because permutations of

the particles among themselves have to be counted.

a) Internal Energy

Taking in consideration that the internal energy at zeroth level is not necessarily

zero:

b) Entropy

1 For systems of interacting molecules, the so-called canonical partition functions are used but this is out of the scope of this introduction.2 In this sense, it resembles the wave function in quantum mechanics that contains all information about the quantum system. Partition functions are thus some sort of “thermal wave functions”

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c) Helmholtz free energy

d) Pressure

e) Free Gibbs energy

The free Gibbs energy of indistinguishable ideal gas particles requires special attention.

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The last equation gives a new interpretation for the free Gibbs energy. It is proportional to

the logarithm of the number of thermally accessible states per molecule. Introducing the molar

partition function qm when N=Nav yields

(3.8)

Exercise: Calculate the entropy of a collection of N independent I2 molecules (o=214.6 cm-1) at

25ºC assuming harmonic oscillator behavior.

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For I2 at 25ºC:

From graph, S = 8.4 Jmol-1K-1.

Exercise: Evaluate the molar entropy of N two level systems and plot the resulting expression.

Exercise: Calculate the vibrational contribution to entropy of Br2 at 600 K given that the

wavenumber of the vibration is 321 cm-1. Calculate the vibrational energy of the system.

20

Collision

Theory

21

4.1 Translational Energy Distribution

In this section, the Maxwell-Boltzmann distribution is to be derived. But first of all, some

thoughts about the Planck's constant h have to be considered. The unit of h is J s. In basic SI units

which is equal to the unit of linear moment p=m.v (kgm/s) multiplied by the unit of position x. h,

thus, has the dimension of moment × distance (p×x).

Let's consider now the so-called phase plane for one dimensional motion1. This is the

plane constructed by plotting the linear moment of motion versus the position.

The smallest unit in such a plane is h (the cyan rectangle). Each discrete state (defined by

the values of p and x) is represented by such a rectangle in the phase plane, corresponding to an

energy i=px2/2m.

Next step is to determine the degeneracy gi of gaseous molecules. These are the energy

states that are so close to each others and lie within the energy interval d. Obviously, this is the

number of states (rectangles) belonging to that bundle represented by the rectangle dp×dx. Thus,

1 For higher dimensions we speak of phase space which is, as an example, a six dimensional space in the case of three dimensional motion (x,y,z,px,py,pz).

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For three dimensional motion:

Applying equations 1.8 and 3.4, taking thereby in consideration that the separation

between the energy levels of translational motion are so small that the summation can be replaced

by integration

The last equation above gives the fraction of particles with kinetic energy i located

between 0-a irrespective of position. It is usually written in the form

(4.1)

`The velocity distribution according to equation (4.1) is shown below for N2 molecules at

298 K and 1500 K. The y-axis represents thereby the fraction of particles with velocity vx. Note

that the curve is symmetrical and extends to infinity in both directions (the direction of the one

dimensional motion is thereby taken in consideration). The average speed is zero because motion

in opposite directions cancels each other. As the temperature is increased, the curve broadens

since higher temperatures means higher kinetic energies which in turn means higher velocities.

The curve center remains at zero because no direction is preferred over the other. Notice the

absolute value of the speed with which the molecules move.

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For three dimensions,

(4.2)

Transfering equation 4.2 into spherical coordinate system ( ) gives

(4.3)

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Equation 4.3 represents now the fraction of particles with speed v irrespective of

direction. The volume element over which integration is carried out (4v2dv) is a very thin shell

around the origin with thickness dv and radius v.

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The average speed of gas particles can be evaluated according to

Substituting equation 4.3 gives

with

(4.4)

The Maxwell-Boltzmann distribution (equation 4.3) can also be written in terms of the

kinetic energy of gas particles:

Equation 4.3 becomes

(4.5)

The average energy can be shown to be

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4.2 Rate of reaction

The basic idea in collision theory is simple and intuitional. For a chemical reaction to

occur, the reactant molecules must "meet" or "collide". They must encounter at a distance short

enough to enable energy exchange. Let us consider the following bimolecular reaction to

elucidate these points.

As long as the A2 and B2 molecules are kept in separate containers (e.g. beakers), no

chemical change will take place. Mixing the two reactants together is a prerequisite for the

reaction to occur because it gives the reactant molecules the chance to "meet". However mere

colliding is not enough. As can be concluded from the reaction equation above, chemical

reactions are processes in which bonds are redistributed, i.e. present bonds are broken and new

are formed. But breaking bonds requires energy and the collision theory states that the required

energy can be obtained from the energy exchange in the collision. The colliding pair must thus

have energy sufficient to overcome such a barrier. If not, the collision is said to be non-reactive.

According to the above concept, the rate of reaction must be proportional to the rate of

collision. We'll show below how this is done for bimolecular collisions of gas particles A and B.

Imagine a gas phase system containing the particles A and B as shown schematically in the figure

below. The A particles are assumed to be stationary whereas the B particles move horizontally

along the drawn cylinder. The question to be answered now is: with which A particles is this

specific B particle going to collide. A simple geometric consideration reveals that collisions occur

only if the separation d between the centers of A and B particles is not larger than the sum of the

radii of A and B (d ≤ rA+rB). The cross-sectional area of the hypothetical cylinder confining all A

particles with which the B particle is going to collide is called the cross-section of collision and

is given by

The next question is about the number of collisions that this B particle is going to make in

one second if it moves with an average speed . This is the question about the number of A

particles the center of which is confined in the above described hypothetical cylinder with length

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x, which is the distance moved by particle B in one second. Obviously, this depends on the

concentration of A. We define NA, the number of A molecules per unit volume. The number of A

particles confined in the cylinder with length x is given by

The expression above gives the number of collisions of one B particle with A particles in

one second. Considering that there are NB particles per unit volume, the following equation for

the number of collisions of A and B, ZAB, in one second and per unit volume is given:

In the above derivation, the A particles were assumed to be stationary. This, however, is

not the case. This is why in the above expression the average velocity of B relative to A must be

substituted instead of the actual velocity of B.

(4.6)

x

A

A

A

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For a 90º-collision between A and B particles, the average velocity is given by

Equation 4.6 becomes

(4.7)

For a 90º-collisions between like molecules, the average velocity is

Equation 4.6 becomes

(4.8)

However, not all these collisions (given by equations 4.7 or 4.8) are reactive. A certain

energy barrier, b, must be overcome, i.e. only collisions with ≥ b are reactive:

vrel

29

In the simplest form of the collision theory where only two dimensions are considered

(collision takes place in a plane), the fraction of particles fulfilling the above condition can be

determined from the Maxwell-Boltzmann distribution (equation 4.5 but for two dimensions)1:

The number of reactive collisions is thus:

(4.9)

Now let's consider the elementary reaction

A + B → product

(4.10)

The constants in the pre-exponential term in equation 4.10 are summarized as BAB:

(4.11)

Equation 4.11 explains the temperature dependence of the rate constant. As temperature is

increased, the fraction of particles with ≥ b increases as dictated by the Maxwell-Boltzmann

distribution.

1 The general equation is

where 2s equals the number of dimensions.

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Equation 4.11 predicts also a temperature dependence of the pre-exponential factor. In the

following a comparison with Arrhenius equation is conducted:

If Eb >>Ea, then Eb ≈ Ea. Under such conditions, Aobs=T1/2BAB. The latter is often used to

check the agreement of the kinetic values predicted by collision theory with those experimentally

observed. The table below summarizes such a comparison for a selected set of gas phase

reactions. As can be seen, the agreement is good only in simple cases (reactions #1 and #5). One

suggestion to explain the deviation of experimental values from those calculated by the collision

theory was based on the idea that in addition to energy requirements there are some steric

requirements for a collision to be reactive. The steric requirements are clearly shown in the

figure below.

H2+I2 → 2HI 6.78 7.38

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NO + O3 → NO2 + O2 4.80 6.90

D + H2 → DH + H 5.99 7.32

CH3 + H2 → CH4 + H 4.25 7.27

CH3 + CH3 → C2H6 6.32 6.62

C2H4 + butadiene → cyclohexene 2.80 7.20

To take these steric requirements into account, the so-called steric factor p must be

inserted in equation 4.11. It takes values between 0-1. p=1 means that no such requirements exist

and any collision with ≥ b is reactive. p=0 means that all collisions regardless of their energies

are non-reactive; the steric requirements are so high that they are never fulfilled.

(4.12)

Experimental rate constants compared to the ones predicted by collision

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theory for gas phase reactions

Reaction A ZAB Steric factor

2ClNO → 2Cl + 2NO 9.4 109 5.9 1010 0.16

2ClO → Cl2 + O2 6.3 107 2.5 1010 2.3 10-3

H2 + C2H4 → C2H6 1.24 106 7.3 1011 1.7 10-6

Br2 + K → KBr + Br 1012 2.1 1011 4.3

Special attention should be paid to the last reaction in the above table. The steric factor is

much larger than 1. The above reaction is an example of the so-called harpoon reactions. When

the K atom is close enough to the bromine molecule, an electron (the harpoon) flips from the K

atom to the bromine molecules forming ions. The ions attract each other producing KBr. The

collisional cross section is thus largely increased in comparison to the geometric ones.This means

that the collisional cross section is much larg

However, with the introduction of the steric factor, the collision theory has lost its

prediction power since the steric factor can not be computed. However, the simple pictorial

model with intuitive concepts keeps reserving an important place for the collision theory in

chemical kinetics.

Reaction probability: Energy exchange between particles depends on the way collisions

occur. In the derivation of equation 4.10, it was assumed that the collisions are head-on or central

collisions. This is not always true. Consequently, collisions are reactive only if the energy

component in the direction of collision (and not the total energy) is larger than the barrier. The

magnitude of this component is determined by the cosine of the angle between the direction of

this component of the relative velocity, vrel,A-B, and the direction of relative velocity, vrel.

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The existence of an energy threshold, a, implies that there is a maximum value of a,

above which no reaction occurs.

The collisional cross section becomes now

It is smaller than the pure geometric one applied in equation 4.10.

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4.3 Molecularity of reactions

Elementary reactions are those reactions that proceed in a single step, i.e. reactions that

result from a simple collision process (A particle collides with B particle producing a product

molecule). Elementary reactions can be monomolecular, bimolecular or termolecular. Higher

molecularities are not possible because the probability of four or higher number of particles to

meet by chance at one point is infinitely small that it is practically zero.

Bimolecular reactions are easy to understand: Two particles collide with each other. But

what does mean to have a monomolecular reaction. A single particle doesn't react. It needs

another particle to gain energy from in order to be able to overcome the energy barrier. In

photochemical reactions, this second particle is a photon. In normal gas phase reactions, the

second particle can be another reactant particle or an inert particle M.

This issue was addressed by Lindemann who proposed the following mechanism for so-

called monomolecular reactions:

The first step represents the activation of the reactant molecule by collision with an inert

particle M (or another reactant molecule A). The A particle gains energy and becomes excited

(A*). Applying the steady state approximation yields:

(4.13)

1

-1A + M A* + M

2A* product

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Normal pressure is considered to be high. Thus, [M] is high and it follows that

k-1[M]>>k2, and a first order kinetics is obtained.

However, at low pressures, k-1[M]<<k2 and a second order kinetics is observed.

The result is experimentally verified in accordance with Lindemann predictions. An

example is N2O5 → NO2 + NO3.

If equation 4.13 is to be written in the form of first-order kinetics, then

kobs is thus a function of the concentration of the inert substance M (or A itself in case the

required energy to overcome the barrier has been provided by the collision A-A).

From the figure above, ko, k1/2 and [M]1/2 can be determined. [M]1/2 is the concentration of

M where kobs equals ko/2 (k1/2).

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Thus, k1 can be experimentally determined. The steric factor for the process can be

calculated applying equation (4.12).

Termolecular reactions need also some attention. According to the hard sphere model

where particles are treated as if they were rigid billiard balls (section 4.2), the probability of

termolecular collisions is practically zero. This is because the two particles depart as soon as they

collide, and the chance of meeting a third particle at the same instant of their collision is

negligibly small. But in real molecules, the two colliding particles stay attached for a short time

which is in the magnitude of a single vibration (10-13s), raising thus the chance of meeting a third

particle.1 In this short time, gas particles move in average a distance of 10-8cm (let's call this

distance ). The cross section for effective collisions is thus a circle with the radius . The

number of collisions is given by:

At room temperature and 1 atm, ZABC is in the range 1026-1027 cm-3s-1. This is two orders of

magnitude smaller than ZAB. This is the reason why elementary termolecular reactions proceed

much slower than the bimolecular ones.

1 Hence, termolecular collisions can be considered to consist of two consecutive bimolecular collisions.

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A last comment concerning termolecular reactions: Recombination reactions, as the one

presented below, are not really bimolecular. Since bond energy releases energy, a third particle

must be present in order to dissipate this energy and stabilize the product molecule. Otherwise the

"formed" bond will break down!

H + H → H2 H=-BE

H + H + M → H2 + M

A

B

C

A

B

C

effective uneffective

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