Chapter05.pdf

Geochemistry

W. M. White

Chapter 5: Kinetics

Chapter 5: Kinetics: The Pace of Things

5.1 Introduction

hermodynamics concerns itself with the distribution of components among the various phases

and species of a system at equilibrium. Kinetics concerns itself with the path the system takes in

achieving equilibrium. Thermodynamics allows us to predict the equilibrium state of a system.

Kinetics, on the other hand, tells us how and how fast equilibrium will be attained. Although thermo-

dynamics is a macroscopic science, we found it often useful to consider the microscopic viewpoint in

developing thermodynamics models. Because kinetics concerns itself with the path a system takes,

what we will call reaction mechanisms , the microscopic perspective becomes essential, and we will very

often make use of it.

Our everyday experience tells one very important thing about reaction kinetics: they are generally

slow at low temperature and become faster at higher temperature. For example, sugar dissolves much

more rapidly in hot tea than it does in ice tea. Good instructions for making ice tea might then incor-

porate this knowledge of kinetics and include the instruction to be sure to dissolve the sugar in the hot

tea before pouring it over ice. Because of this temperature dependence of reaction rates, low tempera-

ture geochemical systems are often not in equilibrium. A good example might be clastic sediments,

which consist of a variety of phases. Some of these phases are in equilibrium with each other and with

porewater, but most are not. Another example of this disequilibrium is the oceans. The surface waters

of the oceans are everywhere oversaturated with respect to calcite, yet calcite precipitates from sea-

water only through biological activity. At a depth of 2500 m, the ocean is undersaturated with calcite,

yet calcite shells of micro-organisms persist in sediments deposited at these depths (though they do

dissolve at greater depths). Thus, great care must be used in applying thermodynamics to such sys-

tems. Even in the best of circumstances, thermodynamics will provide only a limited understanding of

low temperature geochemical systems. A more complete understanding requires the application of ki-

netic theory. Indeed for such systems, kinetics is the deciding factor controlling their state and evolu-

tion. Even in metamorphic systems, with temperatures in the range of 300-700° C, kinetics factors are

crucially important in determining their final states.

High temperature geochemical systems, such as magmas, are more likely to be in equilibrium, and

thermodynamics provides a reasonable understanding of these systems. However, even at high tem-

peratures, kinetic factors remain important and can inhibit equilibrium. One obvious example of dis-

equilibrium at high temperature is the formation of volcanic glasses. Thermodynamics predicts that

magmas should crystallize as they cool. But where cooling is rapid enough, this does not occur.

Glasses, which in many ways are simply extremely viscous liquids, form instead.

It is perhaps ironic that it is kinetic factors, and a failure to achieve equilibrium, that in the end allow

us to use thermodynamics to make statements about the Earth's interior. As we pointed out in the pre-

ceding chapter, if equilibrium were always achieved, the only rocks we could collect at the surface of

the Earth (which is, after all, the only place we can collect them) would consist of quartz, clays, serpen-

tine, etc.; their petrology would tell us nothing about their igneous or metamorphic histories. Fortu-

nately, kinetic factors allow the original minerals and textures of gneisses, peridotites, lavas, etc. to be

preserved for our study.

The foregoing might suggest that kinetics and thermodynamics are entirely unrelated subjects, and

further, that what we have learned about thermodynamics is of little use in many instances. This is cer-

tainly not the case. As we shall see, transition state theory provides a very strong link between kinetics

and thermodynamics. What we have learned about thermodynamics will prove very useful in our brief

study of kinetics. Furthermore, chemical systems are always governed by a combination of thermody-

namics and kinetics, so a full understanding of the Earth requires the use of both thermodynamic and

kinetics tools. The goal of this chapter is to add the latter to our geochemical toolbox.

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Chapter 5: Kinetics

5.2 Reaction Kinetics

5.2.1 Elementary and Overall Reactions

In thermodynamics, we found that the equilibrium state of a system is entirely independent of the

path taken to reach that state. The goal of kinetics is a description of the manner in which the equilib-

rium state is achieved. This description is inherently path-dependent. Consider for example, the

weathering of anorthite. We can write an overall reaction for this process as:

CaAl 2 Si 2 O 8 + 3H 2 O +CO 2(g) → CaCO 3 + 2Al(OH) 3 + 2SiO 2(qz) 5.1

In nature, however, this process will involve several intermediate steps. These intermediate steps

can include:

H 2 O +CO 2(g) → H 2 CO 3(aq)

5.2

5.3

H 2 CO 3 → HCO 3( aq)

+ H +

–

CaAl 2 Si 2 O 8 + H 2 O +2H + → Si 2 Al 2 O 5 (OH) 4 + Ca (aq)

5.4

2 +

H 2 O + Si 2 Al 2 O 5 (OH) 4 → 2SiO 2(qz) + 2Al(OH) 3

5.5

5.6

HCO 3( aq)

→ CO 3( aq)

+ H +

–

2−

2+ → CaCO 3 5.7

In thermodynamics, equation 5.1 is a perfectly adequate description of the reaction. In kinetics, a de-

scription of an overall reaction such as 5.1, requires a knowledge of the path taken, that is a knowledge

of the steps involved. Reactions 5.2 through 5.7 thus describe the overall reaction 5.1. Reactions 5.2,

5.3, and 5.6 are elementary reactions in that they involve only one step and the reaction as written de-

scribes what occurs on the microscopic level. The remaining reactions are not elementary in that they

each consist of a number of more elementary steps.

5.2.2 Reaction Mechanisms

Reaction 5.4 describes the breakdown of anorthite to form kaolinite plus a free calcium ion. This re-

action involves profound structural changes in the solid phase that are not described by equation 5.4.

A full kinetic description of 5.4 will require some knowledge of the steps involved in these structural

changes. One possibility is that all components are in solution at an intermediate state:

CO 3( aq)

+ Ca (aq)

2−

2+ + 2OH – 5.4a

2H 4 SiO 4(aq) + 2Al(OH) 2( aq + + 2OH – → Si 2 Al 2 O 5 (OH) 4 + 5H 2 O 5.4b

Reaction 5.5, the breakdown of kaolinite to quartz and gibbsite, could involve SiO 2 dissolving, sub-

sequently precipitating as opaline silica, and later transforming to quartz:

Si 2 Al 2 O 5 (OH) 4 + 5H 2 O→ 2H 4 SiO 4(aq) + 2Al(OH) 3 5.5a

H 4 SiO 4(aq) → SiO 2(opal) + 2H 2 O 5.5b

SiO 2(opal) → SiO 2(qz) 5.5c

The description of an overall reaction in terms of elementary reactions is called the reaction mecha-

nism. The rates of truly elementary reactions are path-independent because there is only one possible

path. In this sense, elementary reactions are somewhat analogous to state functions in thermody-

namics. Clearly then, an important step in any kinetic study is determination of the reaction mecha-

nism, that is, to describe the process in terms of elementary reactions. As we shall see, there may be

more than one possible path for an overall reaction, and that several paths may be simultaneously in-

volved. Kinetics can only provide an accurate description of a process if all these paths are known.

CaAl 2 Si 2 O 8 + 6H 2 O + 2H + → 2H 4 SiO 4(aq) + 2Al(OH) 2( aq)

+ Ca (aq)

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Chapter 5: Kinetics

5.2.3 Reaction Rates

Consider a reaction such as the precipitation of dolomite from a solution. We can describe this as:

Ca 2+ + Mg 2+ + 2CO 2 − ® CaMg(CO 3 ) 2

We define the rate of this reaction, ℜ , as the rate at which dolomite is produced:

ℜ ≡ d [CaMg(CO 3 ) 2 ]

Clearly, if dolomite is to be formed, CO 2 − must be consumed in this reaction twice as fast as Ca or Mg.

For every mole of Ca or Mg consumed, exactly two moles of CO 2 − will also be consumed and one mole

of dolomite produced. This being the case, we could equally well express the reaction rate as:

d [CO 2− ]

ℜ = − d [Ca 2+ ]

= − d [Mg 2+ ]

ℜ = − 1

We can now formulate the general rule. For any reaction such as:

a A + b B → c C + d D

5.8

The reaction rate , ℜ , is defined as the change in composition of the reaction mixture with time:

ℜ ≡ − 1

d [A]

= − 1

d [B]

= 1

d [C]

= 1

d [D]

5.9

The brackets denote the concentrations of the species and the negative sign indicates that reactants are

consumed as the reaction proceeds. Thus the rate of a reaction is simply the rate at which a reactant is con-

sumed or product produced divided by its stoichiometric coefficient.

5.2.3.1 The Reaction Rate for an Elementary Reaction: Composition Dependence

Reaction rates will, in general, depend on the concentration of the reactant. To understand this, con-

sider the reaction:

N° + O 2 ® NO + O

5.10

This reaction between free nitrogen at-

oms and oxygen molecules occurs in

the stratosphere (where N° is produced

by high energy collisions involving N 2 )

and contributes to the production of ni-

trous oxide. Let’s assume that reaction

5.10 is an adequate description of this

reaction. In other words, we are as-

suming that 5.10 is an elementary reac-

tion and the reaction mechanism for

the production of NO from nitrogen

and oxygen gas is collision of a N°

molecule and O 2 molecule. For the re-

action to occur, the nitrogen and oxy-

gen molecules must collide with

enough kinetic energy that the mutual

repulsion of the electron clouds is

overcome and the electrons can be re-

distributed into new covalent orbits.

The repulsive force represents an en-

ergy barrier, E B , which will prevent

low energy nitrogen and oxygen atoms

from reacting. Figure 5.1 illustrates

Figure 5.1. A nitrogen atom approaching an oxygen molecule

must have enough kinetic energy to pass through the region

where it is repelled by electrostatic repulsion of the electron

cloud of the oxygen. Otherwise, it will not approach closely

enough so that its electrons can combine with those of oxygen.

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Chapter 5: Kinetics

this point. The reaction rate will

therefore depend on 1) the number

of collisions per unit time, and 2)

the fraction of N and O molecules

having energy greater than the

barrier energy.

Let’s first consider the number

of collisions per unit time. In order

for a ‘collision’ to occur, the elec-

tron clouds must overlap, that is,

they must approach within (r N +

r O 2 ), where r N and r O 2 are the radii

of the nitrogen and oxygen mole-

cules. To make things simple,

imagine the oxygen to be fixed and

the nitrogen in motion. In other

words, our reference frame will be

that of the oxygen molecules. We

can imagine the nitrogen sweeping

out a cross-section with radius (r N

+ r O 2 ) as it travels. If the nitrogen

is travelling at velocity v , in time t , it will sweep out a cylindrical volume (Figure 5.2):

V = v π( r N + r O 2 ) 2 t 5.11

Whether a collision occurs will depend on whether the center of an oxygen molecule falls within this

volume (Figure 5.2). The number of collisions that will occur in this time will be:

C = n O v π( r N + r O 2 ) 2 t 5.12

where n O is the number of oxygen molecules per unit volume. The number of collisions per unit time is

then simply:

Figure 5.2. A nitrogen atom will sweep out a volume V = v π (r N +

r O 2 ) 2 t in time t . Whether a collision occurs will depend on whether

the center (indicated by black dot) of an oxygen atom falls within

this volume.

t = n O v π( r N + r O 2 ) 2 5.13

If there are n N nitrogen atoms and the average velocity between nitrogen and oxygen molecules is – ,

then the number of collisions per unit time is:

(

) 2

c = n N n O v π r N + r O 2

5.14

(

) 2

If we let

k = v π r N + r O 2

then the rate at which collisions occur is:

c = kn N n O 2 5.15

Thus we see that the reaction rate in this case will depend on the concentration of nitrogen, oxygen and a con-

stant that depends on the nature of the reactants. This is a general result.

5.2.3.2 The Reaction Rate for an Elementary Reaction: Temperature Dependence

We now need to estimate the fraction of nitrogen and oxygen atoms having at least the barrier en-

ergy, E B . For simplicity, we will assume that oxygen and nitrogen molecules have an identical energy

distribution. The Boltzmann Distribution Law, which we encountered in Section 2.6.4.1 (equ. 2.84), can

be written to express the average number of molecules having energy level ε i as:

n i = Ae −ε i / k T

5.16

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Chapter 5: Kinetics

where k is Boltzmann's constant and A is a constant (comparing with equation 2.84, we see that A =

n / Q where n is the total number of molecules in the system and Q is the partition function ). In plain

English, this equation tells us that the number of molecules in some energy level i decreases exponen-

tially as the energy of that level increases (Figure 2.9). We want to know the number of molecules with

energy greater than E B . In this case we are dealing with translational energy. The quantum spacings

between translational energy levels are so small that they essentially form a continuum, allowing us to

integrate equation 5.16. Fortunately for us, the integration of 5.16 from ε = E B to infinity has a simple

solution:

∞

∫

e −ε i / k T d ε

= A k Te − E B / k T

5.17

E B

The fraction of molecules with energy greater than E B is just:

∞

∫

e −ε i / k T d ε

= A k Te − E B / k T

A k T

E B

= e − E B / k T

5.18

∞

∫

e −ε i / k T d ε

The rate of reaction will be the rate of collision times the fraction of molecules having energy greater

than E B :

(

) 2 e − E B / k T

ℜ = n N n O v π r N + r O 2

5.19

Now we just need to find a value for velocity. The average velocity can be calculated from the Max-

well-Boltzmann Law*, which gives the distribution of velocities of molecules in a gas. Doing so, we

find that the average velocity is:

8k T

πµ

v =

5.20

where µ is the reduced mass, µ = m N m O 2 /(m N + m O 2 ). Substituting 5.20 into 5.19, our equation for the

reaction rate is:

8k T

(

) 2

πµ e − E B / k T

ℜ = n N n O π r N + r O 2

5.21

8k T

(

) 2

πµ e − E B / k T

Redefining k as:

k = π r N + r O 2

5.22

our reaction rate equation is:

ℜ = kn N n O 2 5.23

Thus the reaction rate in this case depends on the concentration of nitrogen and oxygen and a constant

k , called the rate constant † , which depends on temperature, properties of the reactants, and the barrier

energy.

In a more rigorous analysis we would have to take into consideration atoms and molecules not be-

ing spherically symmetric and that, as a result, some orientations of the molecules are more likely to re-

sult in reaction than others. In addition, a head-on collision is more likely to result in reaction than a

glancing blow, so the collision cross section will be less than π (r N + r O 2 ) 2 . These factors can, however, be

accounted for by multiplying by a constant, called a stearic factor , so the form of our equation, and the

temperature dependence, would not be affected. Values of stearic factors for various reactions range

over many orders of magnitude and can be quite small. In rare circumstances, they can be greater than

1 (implying an effective collision cross section greater than the combined atomic radii).

* So-called because Maxwell proposed it and Boltzmann proved it rigorously.

† To distinguish the rate constant, k , from Boltzmann’s constant, k, we will always write the former in lower case ital-

ics and the latter in roman typeface.

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