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This is a lecture is a series on combustion chemical kinetics for engineers. The course topics are selections from thermodynamics and kinetics especially geared to the interests of engineers involved in combusition
A bit more Physical Chemistry
for Combustion Kineticists
• Arrhenius Equation
• Order of Reactions
• Fitting the parameters
• Steady State
• Pressure Dependence
• Time Scales
Elementary Reaction Rates
Bimolecular Reactions and Collision Theory
How fast do molecules react?
Given a bunch of hard spheres traveling at certain
average velocity, how often to the spheres collide?
With the help of classical collision theory, we can derive the basic Arrhenius rate equation that is used throughout chemical kinetics. This tells
us how fast species react.
The core question is:
How often do a bunch of hard spheres travelling through space collide?
What do we need to know?
- How many there are?
- How large the spheres are?
- How fast they are traveling?
Hard SPhere Collisions
Ball Diameter: !
Balls collide if:
Between the centers less than
How often do two spheres collide?
The key to answering this is to calculate the volume that the moving spheres sweep out. Then the problem becomes how much overlap is there
between these volumes. When the volumes overlap, there is a collision. In a total volume, the probability of overlap is the probability of
Two large spheres equivalent to:
One ball diameter
running into a point
Collisions will be viewed as
A large sphere of diameter 2!
Moving through a density of points
The key to solving this is to do a trick. Two spheres sweeping out a volume is the same as one big sphere (of the combined radiuses) sweeping
out a volume, with the other sphere being single points.
The probability of collision is related to the density of points (of one sphere) and the volume swept out by the large sphere.
So the next question is how fast is this large sphere moving through space?
Rate of Collision
A single ball (radius !), traveling with speed "
sweeps out a cylinder of volume per second
If the concentration of these balls is CA
Then the volume swept out per second per unit volume is:
A large sphere of radius !, traveling with a speed " sweeps out a volume of area times velocity (per second). If we have a concentration of these
balls, then each in this concentration is sweeping out this volume. The volume of each sphere time the total concentration yielding volume ver
Rate of Collision
If the concentration of the points is: CB
Then the frequency of the two hitting is:
points per unit volume
(point per unit volume)(volume per second) =
points hit per second
collisions per second
The other balls, which have been transformed to a set of points, have a concentration (or density) of the concentration of B. The frequency of
them ‘hitting’ is simply the volume swept out by A multiplied by the density of B, yielding points hit per second, meaning collisions per second
relative velocity according to Maxwell-Boltzmann statistics:
ma + mb
Substituting for collision frequency:
So how fast are these molecules (spheres moving?). For this we can use Maxwell-Boltzmann statistics giving the average speed (by integrating
velocity with the distribution) which is essentially a square root dependence on temperature. This expression can be substituted to give the rate
of collision dependent on temperature.
proportional to concentrations
proportional to square root of temperature
So in the ﬁnal expression, we see the expected dependence on the concentrations and also a dependence on temperature.
Only those molecules over a certain energy will react:
The proportion of molecules
at temperature, T, having an
the activation energy:
So now we know how often two molecules collide. This is the ﬁrst prerequisite to a reaction taking place. The molecules have to meet
The second prerequisite is that the bond making and breaking also requires a minimal amount of energy. If the molecule has at least this much
energy, then the molecules react to form the products. Otherwise, no.
The proportion of molecules at temperature, T, that have an energy over a certain activation energy, Ea, is the exponent of the Ea over RT.
Combining with the frequency, the previous derived result, we get a rate of reaction that is proportional to the concentrations and the square
root and the inverse exponential of the temperature.
Combined with the number of collisions:
The general Arrhenius equation is generalized to:
Three constants are needed:
A, n, Ea
Combining the prerequisites of collision and activation energy, we have derived an expression close to the expected Arrhenius equation. The
exact di!erence is the square root dependence on temperature.
But is does at least justify the general Arrhenius equation which has a dependence on three constants, the Arrhenius coe"cient, the
temperature exponent and the activation energy.
Transition State Theory
Reaction occurs only if
Initial energy needed
to be able to win
the ﬁnal exothermic
total energy gain
The transition state represents the state of the reaction at the top of the energy hill. When the molecules have achieved enough energy, the
activation energy, they can react. In the transition state, the reactant bonds are starting to break and new product bonds are starting to from.
When the transition state is reached, the molecules can either continue reacting to form products, or can break apart and go back to reactants.
The transition state is highly unstable and cannot be isolated. It is di!erent from a reactive intermediate which does exist for a length of time,
even though it is very short.
Potential Energy Surface
A + HB −→ AH + B
Trough is reaction path
products and reactants
In reality, the energy diagram is a multidimensional surface of the potential energy of each geometric conﬁguration.
This diagram is a simple potential surface involving three atoms and the reaction (where A is F and B is Cl):
A + HB --> AH + B
The x and y axis are the AH and BH distances.
The single line diagram, as in the previous slide, is drawn through the minimum of the trough.
Some surfaces are more complex and impossible to visualize. But if a potential energy line is drawn, along the minimum of the trough, from
the reactant to activated complex to products, some useful information about complex conﬁgurations can be extracted.
In trying to determine the product distributions of complex reactions, it is useful to look at the energies of the reactants, intermediates and
This is an example of the addition of oxygen to cyclohexane radical. From the diagram, it is possible to justify qualitatively and possibly
quantitatively the product distributions.
The intermediate molecules in the potential energy minima, are reactive intermediates
Role of Quantum Mechanics
But where do the potential energy diagrams, i.e. the energies of the reactants, products and transition states, values come from?
An ever increasing tool for combustion chemists is to use quantum mechanics, QM. Given a particular structure, QM can compute the energy
value of the molecule. One of the di!culties is determining the best molecular structures (usually through some optimization process).
Another more complicated di!culty is that transition states are often very complicated electronically (for example, bonds are not quite
complete and are stretched). This means that higher order, meaning computationally extensive, calculations must be made.
In the study where this diagram was taken, the molecular structures and their corresponding potential energies were computed.
Role of Quantum Mechanics
Interactions are based on
springs and classical potentials.
Another theoretical ﬁeld that is coming into play in combustion kinetics is that of molecular mechanics. This uses the potential energy surfaces
to the next level beyond using the simple trough to examine the transition state. Here even the full vibrational behavior of the species, treating
atoms as connected with springs, are taken into consideration in determining the rate constants.
k = ATn
RTk = Ae
•A Frequency Factor
•n Temperature Coefﬁcient
•Ea Activation Energy
•R Energy Constant
There are basically two common forms of the Arrhenius equation for the rate. One is a simple exponention with three coe!cients, the frequency
factor (or Arrhenius coe!cient) and the activation energy . The temperature dependent version has an additional temperature coe!cient.
logk = logA −
1.Measure rates at various temperatures
2.Plot log k versus 1/RT
3.Intercept is log A
4.Slope is Ea
The constants to the Arrhenius equation are ﬁtted from experimental data.
The rate constant, with or without the temperature coe!cient, is purely a function of temperature. Thus if the rate constant is measured at
di"erent temperatures, the coe!cients can be determined.
If a plot is made of log k versus 1/RT, then the intercept is the log of the A coe!cient and the slope is the activation energy.
With the temperature coefﬁcient
it is not as straightforward,
but basically the same:
logk = logA + nlogT −
Least squares ﬁtting
When the expression is to include the temperature coe!cient, then a simple plot is not possible. However, given enough points, a least squares
ﬁtting is possible.
H2 + OH −→ H2O + H
Temperature rate x 10**16
Plot and read from axis
k = 7.7x10−12 cm3
Atkinson, R.; Baulch, D.L.; Cox, R.A.; Crowley, J.N.; Hampson, R.F, Jr.; Kerr, J.A.; Rossi, M.J.; Troe, J.
Summary of Evaluated Kinetic and Photochemical Data for Atmospheric Chemistry, 2001
(IUPAC Subcommittee on Gas Kinetic Data Evaluation for Atmospheric ChemistryWeb Version December 2001)
Here is a simple example taken from the literature for the reaction of hydrogen and hydroxyl radical forming water and a hydrogen radical.
Experimentally, the temperature dependence of the rate constant was measured. The log of the rate versus 1/RT is plotted.
The ﬁnal expression for the rate constant give the frequency factor, read from the intercept and the activation energy, from the slope.
•Higher Heating Value
•Thermodynamic heat of combustion
•Enthalpy change for the reaction with the same temperature
before and after combustion.
•Lower Heating Value
•Heat of Vaporization subtracted from higher heating value
•water component is in a vapor state after combustion
Heat released when a given amount of fuel is combusted.
A standard measurement used in industry for the heat content of a species, meaning how much heat is released when the species is
combusted. Two values are used.
The ﬁrst is higher heating value. This is the thermodynamic heat of combustion, meaning the enthalpy change of combustion at a constant
The second is the lower heating value. This is where the heat of vaporization is subtracted from the higher heating value.
Heating Value Relationship
Higher Heating Value =
Lower Heating Value +
The lower heating value is useful for boilers
where in the end the water is evaporators
The di!erence between the higher and the lower heating value is basically related to the heat of vaporization of water. It is for this reason that
the lower heating value is useful for boilers.
NIST Kinetic Data
There are published compilations, for example from NIST, of rate constant data extracted from the literature.
From this plot, one can see that there is still a degree of experimental error. This can lead to some ﬂexibility in the determination of the rate
This in turn can lead to some optimization when putting together the reactions in a larger mechanism.
Standards for Combustion
Standard (Accepted) Reference
Atkinson, R.; Baulch, D.L.; Cox, R.A.; Crowley, J.N.;
Hampson, R.F, Jr.; Kerr, J.A.; Rossi, M.J.; Troe, J.
Summary of Evaluated Kinetic and Photochemical Data for
Atmospheric Chemistry, 2001
(IUPAC Subcommittee on Gas Kinetic Data Evaluation for
Atmospheric Chemistry Web Version December 2001)
One of the standard references for rate contants is the paper published by Baulch et. al.
These are used whenever possible. If these constants are not used in a mechanism, then the modeler must give a very good reason.
Subject to Errors
H2 + M --> H + H + M
H2 + O2 --> HO_2 + H
H+O2 --> O + OH
O+H2 --> H + OH
H_2+OH --> H_2O + H
O+H2O --> OH + OH
H+H+M --> H2 + M
O+O+M --> O2 + M
H+O + M --> OH + M
H+OH --> H2O
(more involving HO2 and H2O2)
•Cannot isolate reactions
•Assumptions in associated reactions
The determination of fundamental rate constants can be subject to several sources of errors. Along with systematic errors in
measurements, it may be hard to actually isolate the particular reaction out of a system. Often one has to make assumptions
about the other reactions in the system in order to derived the isolated reaction. It can happen that as new results are obtained,
these assumptions must be modiﬁed and the rate recalculated.
But what happens with more complicated
with sets of reactions?
(as in combustion)
Rate of products production:
A + B −→ products
dt = k[A][B]k = Te
From the discussion of bimolecular collision theory, we have a rate equation for a single reaction of species A and species B
combining to form products.
d[products]/dt = k[A][B]
where the rate constants has a temperature dependence:
k = T**n exp(Ea/RT)
But combustion mechanisms consist of many of such reactions, of di!erent orders (number of reactants). The following
discussion looks at dealing with the more complicated cases and simpliﬁcations that can be made.
A −→ BUnimolecular:
Reversible Unimolecular: A ←→ B
= −kf [A] + kr[B]
= kf [A] − kr[B]
Let us look at the rate equations for unimolecular transformations between species A and species B.
For the ﬁrst case of non-reversible transformation of A to B, the rate of production is simply minus the rate times its concentration, i.e. the
concentration is decreasing. The increase of B is also dependent on how much A there is.
A reversible reaction of one reactant and one product is also ﬁrst order.
Thought as two independent reactions, one in the forward and one in the reverse, we get the expected equations.
But now, for example, the molecule A has two sources and sinks. In the forward direction, A is being converted to B at the rate of -k[A] (note
the negative sign). However, at the same time, it is being created by B being converted to A at the rate of - k[B].
Thus the rate of production of A is the sum of the forward and reverse directions: - k_f[A] + k_r[B]
A + B −→ CRecombination:
Just as in the previous case, the total rate of production is the sum of all the sources and sinks of the molecular species.
For the forward reaction, there is only one source of each rate of production, namely plus or minus k[A][B].
= −kf [A][B] + kr[C]
= −kf [A][B] + kr[C]
= +kf [A][B] − kr[C]
A + B ←→ C
However, for the reversible case, once again, there is the e!ect of both sides of the reaction.
In general, to determine the rate of production of the species
- Collect all the reactions involving that species.
- Add all the rates (with concentrations depending on order) producing the species
- Subtract all the rates that are consuming the species.
For a complex mechanism, one has a set of di!erential equations, one for each species. On the left hand side is the di!erential and on the right
hand side a polynomial expression in terms of rates and species concentrations.
Rate of Production
dt = +2k1[H2] H2 + M −→ H + H + M
+k2[H2][O2] H2 + O2 −→ HO2 + H
−k3[H][O2] H + O2 −→ O + OH
+k4[O][H2] O + H2 −→ H + OH
+k5[H2][OH] H2 + OH −→ H2O + H
H + H + M −→ H2 + M
−k9[H][O] H + O + M −→ OH + M
−k10[H][OH] H + OH −→ H2O
A more typical case, for even a simple example like one part of the hydrogen mechanism, the rate of production of just one species can be
quite complex. One can see that the set of di!erential equations that must be solved, for even a simple molecule like hydrogen, can be quite
complex. Each species di!erential is equal to a complex polynomial.
So n complex di!erential equations must be solve to determine the time evolution of a species within a given mechanism. This should give an
indication why kinetics is a di"cult computational task when dealing with, for example, reactive ﬂows.
Need Simplifying assumptions
not only for solving the system when set up
when determining fundamental rate constants.
but also, for example,
Quasi First Order
Steady State Assumption
Sometimes, with simplier systems, there are alternatives to having to solve numerically a complex series of di!erential equations. These are
often critical assumptions used to estimate fundamental rate constants. Two common methods are
- Quasi First Order
- Steady State Assumption
A + B −→ C
Assume [B] is constant
Often, for example, [B] >> [A]
dt = −k[A][B] = −k [A]
dt = −k[A][B] = −k [A]
dt = +k[A][B] = +k [A]
where: k' = k[B]
In a bimolecular reaction, sometimes one of the reactants, for example B, has a comparatively large concentration. The
consequence of this is that it total concentration does not change much, the pool approximation. This simpliﬁes the second
order equations to a quasi ﬁrst order equations.
For example, suppose that [C] is 1000 moles/liter and that [B] is 1 mole/liter. Even if the reaction, A + B --> C, went to
completion meaning all the B would be used up, [C] would still be fairly close to 1000, i.e. with some assumptions, 999 moles/
This means that $k[B]$ is relatively constant and only [A] is changing signiﬁcantly. Thus the rate equation can be thought of a
being basically ﬁrst order in A.
allows algebraic solving reduces the complexity
(after a rapid initial buildup)
It reaches an equilibrium.
The concentration of a species does not change
If molecule is in steady state, then that means that its concentration is not changing. This means that the consumption and
depletion rates are the same. It is important to note that molecule does indeed have a non-zero steady state concentration.
Mathematically, a molecule is in steady state when d[A]/dt = 0. This fact can be very useful in simplifying the set of di!erential
equations of a reactive system. For each molecule in steady state, a di!erential equation is eliminated and is replaced by a purely
algebraic equation (which is simpler to solve).
Steady State: N
O + N2 −→fast
NO + N
N + O2 −→slow
NO + O
The production of N is slow, but its destruction is fast
Reaches an equilibrium (of low concentration)
Low concentrations of N
A typical two reaction example is one where a fast reaction is followed by a slow reaction. This one is taken from the NOx
mechanism. Here we assume that the N has reached an equilibrium state within these two reactions.
Steady State: N
Note that it is not zero..
= 0 = k1[O][N2] − k2[N][O2]
O + N2 −→fast
NO + N
N + O2 −→slow
NO + O
Setting d[N]/dt to zero, leaves an algebraic equation where [N] can be solved for. The computations show that ﬁrst, the steady
state concentration of N is non-zero and is dependent on [O], [N2] and [O2].
More assumptions have to be made to solve the system completely or to be able to make some computations.
A + B ←→ C + D
kf [A][B] = kr[C][D]
0 = d(products)
dt = d(reactants)
dt = kf [A][B] − kr[C][D]
Under Equilibrium Conditions
Consider the reversible bimolecular reaction involving the reactants A and B and the products C and D.
Under equilibrium conditions, the rate of production of the reactants is the same as the rate of production of the products.
Setting the di!erentials equal, we get an expression relating the
forward and reverse rate constants and concentrations of the reactants and products.
Collecting the rate terms on one side and the concentration
terms on the other side, we have the expression for
a deﬁned quantity, the Equilibrium Constant.
The equilibrium constant gives the ratio of the reactant concentrations to the product concentrations. For example, a large
equilibrium constant means that the reaction goes close to completion and a signiﬁcant portion of the species went to products.
Gibbs Free Energy
The greatest amount of mechanical work which can be
obtained from a given quantity of a certain substance in a
given initial state, without increasing its total volume or
allowing heat to pass to or from external bodies, except
such as at the close of the processes are left in their
Useful work from an isothermal, isobaric system
∆G = ∆H − T∆S
By deﬁnition it is G=H-TS. In thermodynamic deﬁnitions, the term H-TS occurs frequently enough to justify deﬁning a new
The Gibbs Free Energy is a thermodynamic quantity that represents the useful work from an isothermal, isobaric system. The
quote is how Gibbs himself deﬁned it.
Connection to Thermodynamics
For an ideal gas, there is a connection
to equilibrium constant
(we will take this as a deﬁnition)
∆G = −RTlnKp
We can relate the equilibrium constant,
or the forward and reverse rate constants
to an equation involving entropy and enthalpy
The connection between thermodynamics and kinetics can be found with the Gibbs Free Energy. In this lecture this will be taken as a deﬁnition.
The relationship holds for ideal gases. Note that this is the equilibrium with respect to partial pressures. The relationship between the
equilibrium deﬁned with concentrations has some extra constants and is dependent on the number of reactants and products.
The signiﬁcance of its use is to be able to relate the equilibrium constant, and even the forward and reverse rates to the thermodynamic
quantities of entropy and enthalpy. For example, in numeric programs solving kinetic systems, if the thermodynamics of the reaction is known,
i.e. enthalpy and entropy, and if the forward reaction rate is known, then the reverse reaction rate can be derived.
Conditions For Equilibrium
∆G = −RTlnKp
Kp = e
Kp = e
Kp = e
∆G = ∆ − T∆S
From the two expressions for the Gibbs Free Energy, we can derive what conditions favor reaction completion to products.
Solving for K in terms of the Gibbs Free Energy, we get an exponential. Then substituting the expression for G and separating, we get two
exponentials, one with Enthalpy and one with Entropy.
Conditions for Equilibrium
Kp = e
Exothermic Reaction∆H < 0
Greater molecular chaos∆S > 0
If the equilibrium constant is greater than zero, meaning then the reaction favors the products, the forward rate is greater than the reverse
An exponential is greater than one if the term in the exponential is also greater than zero.
So the ﬁrst term says that if the change in enthalpy is less than zero, meaning that the reaction is exothermic, then the products are favored.
The second term says that if the change in entropy is greater than zero, meaning there is greater molecular chaos, then products will also be
Note that this is not a proof, but rather an example of why the Gibbs Free Energy was deﬁned in the ﬁrst place.
Collisions dependent on other molecules represented by M
A hitting molecule can just transfer its translational energy
Reactants + M ←→ Products + M
Energy needed for reaction can come from collisions
Reactions depend on collisions. Some of the translational energy involved in the collisions can be transfered to the internal energy of the
molecule instigating a bond breaking or making. Remember in the discussion of the bimolecular collision theory, for a reaction to occur, a
certain amount of energy must be present.
One type of collision is when the colliding molecules rearrange their bonds among themselves. Another type of collision is when energy is
transfered to the molecule putting it in an excited state, which leads to a change in bonding. The di!erence is that the molecule that the hitting
molecule does not change, it just transfers some of its energy.
Since the molecule transferring energy does not change and since, for example, many molecules can have this e!ect, the molecule is
represented by a generic M.
A + M −→ka
+ M −→k−a
A + M
Unimolecular decomposition is only possible
If enough energy is available to break the bond
Energy provided by collision with M
Unimolecular decomposition is only possible if enough energy is available to break the bond. This energy is provided by collisions with a
generic set of molecules M.
The Lindemann model is based on the production of an intermediate excited state of a molecule, here seen as A* produced by a collision with a
generic (set) of molecules represented by M. This can be represented by a set of three reactions, production of the activated species, its reverse
and the activating species going to products.
Quasi-steady state of activated species
= 0 = ka[A][M] − k−a[A∗
][M] − ku[A∗
k−a[M] + ku
k−a[M] + ku
From the system of three reactions, the rate of production of the activated species can be written. Since this is an intermediate species, and its
decay is very fast, we can assume that the species is in quasi steady state and from that we can derive an algebraic expression for the
concentration of this activated species:
[A*]= k_a[A][M]/(k(-a)[M] + ku)
which can be substituted back in the equation
for the production of the product species:
d[Products]/dt = k_u [A*] = k_u k_a[A][M]/(k(-a)[M] + ku)
Fall Off Limits
At high pressure, the rate go asymptotically to a constant, k(inf)
At low pressure, the rate has a constant slope, ka[A][M].
In the transition zone the line with the slope bends to the asymptotic constant.
The Lindeman model recognizes that some kinetic relationships cannot be represented with a single rate constant. There are
several expressions within combustion kinetics which take this e!ect into account.
Fitting Low Pressure
Low Pressure Range:
2nd Order Kinetics
(concentration of collision partners is small)
k−a[M] << ku
k−a[M] + ku
At low pressure, the concentration of collision partners is small and the conversion to products dominates. So the term with M disappears in
the denominator leaving the expression ka[A][B], meaning second order kinetics.
This produces the slope of the line dependent on M in the graph.
Fitting High Pressure
k−a[M] >> ku
(ﬁrst order kinetics)
k−a[M] + ku
= kinf [A]
+ M −→k−a
A + M
At high pressure, the M term is larger and the reverse dominates. Thus the term with M in the denominator dominates leaving ﬁrst order
kinetics in the ﬁnal expression.
This produces the ﬂat ‘fall o!’ behavior shown in the graph.
Fitting Fall off
The two fall o! coe"cients can be derived experimentally by plotting the log of the species A versus the log of the rate constant rate constant.
Chemical Time Scales
especially in relation to other processes
How long does a chemical process take?
Chemical time scales are important for gaining insight into to total combustion process. For example, if the chemical time scale is fast
compared to di!usion processes, the chemical process goes to completion (equilibrium) before it di!uses very far. This has important
implications for modeling, meaning the chemistry can be simpliﬁed if the chemical processes are very fast with respect to all other processes.
Combustion time Scales
•chemical times can be fast compared to transport
•Reaction goes to completion fast
•Gas is primarily burn or unburnt
•chemical times can be same order of
magnitude as other processes
•Resulting in more complex mixing
On a computational level, varying time scales of di!erent processes allow simplifying assumptions to be made. For example, fast chemical time
scales allow that less detail need be known for the processes reactants and complete combustion products.
At high temperatures, the chemical times can be fast with regard to transport processes. The consequence is that the kinetics can be
considered as either burnt gases or unburnt gases, meaning the intermediate chemistry does not play an important role.
A challenge now for chemical kinetics is the lower temperatures. Not only are these processes important, but since the chemical processes are
on the same order of magnitude as other processes, such as transport, they are di"cult to simplify and the mixing of intermediate states
Relation to Rate Constant
A −→ products
[A] = [A]0e−kt
It takes an inﬁnite amount of time to reach zero.
How is the chemical time scale quantiﬁed?
How is the chemical time scale quantiﬁed?
Of course, the chemical time scale is directly inﬂuenced by the kinetic rate constant. The higher rate constant, the faster the process and thus
the smaller the chemical time scale. Thus there must be an inverse relationship.
However, for example, for a ﬁrst order reaction, the reactant concentration decays exponentially, but it still takes an inﬁnite amount of time to
reach zero. This is not a useful measure.
Chemical Time Scale is how long it takes to reach:
= .37 63% of the initial
Since the time scale is a relative concept, it is not entirely important exactly at what point we decide. By convention (this is also used in
electronics), the time scale is determined to be the point at which the concentration is 1/e of the original. How much time is needed to consume
about 63% of the original substance.
Solving the original ﬁrst order equation we ﬁnd that the ﬁrst order chemical time scale is the inverse of the rate constant.