August 05, 2024, 07:17:14 PM
Forum Rules: Read This Before Posting

### Topic: Rate expressions: Short introduction to standard chemical rate expressions  (Read 23270 times)

0 Members and 1 Guest are viewing this topic.

#### Juan R.

• Chemist
• Full Member
• Posts: 148
• Mole Snacks: +24/-3
• Gender:
##### Rate expressions: Short introduction to standard chemical rate expressions
« on: February 21, 2006, 01:16:56 PM »
Next, i list some useful expressions for the rate of processes of interest in chemistry, physics, or biology. This review is not aimed to be complete. It assumes basic knowledge of statistical mechanics, and kinetic theory. Details can be found in standard literature. Anyone needing more information in some topic or having doubts can write here or send me a personal message.

What is the rate for the next process?

Br2 + I2 => 2 BrI

Textbooks assert it is

R = k [Br2 ] [I2 ]

What is then the rate for the next process?

Br2 + H2 => 2 BrH

One could write an expression similar to that of above, but will not work. The actual rate (valid for standard empirical conditions) is

R = {k [H2 ] [Br2 ] 1/2} /  {1 + k´ [HBr] / [Br2 ]}

The rate expression for the gas-phase reaction of Br2 with H2 can be explained by a complicated series of steps, each one following simple expressions as that for the I2. The series of steps is called a mechanism and the above reaction called complex. The lesson here is that rate expression is very dependent of the mechanism of the process you are studying.

The great failure of the education on chemical kinetics is the dogmatic belief of most kineticians on that any simple (often called elementary) process (i.e. mechanism with a simple step)

aA + bB => products

follows a rate expression

R = k [Ab ] [Bb ]

With R = –1/ a d[A] / dt = –1/ b d / dt

which is not correct as we will next.
« Last Edit: February 21, 2006, 01:34:58 PM by Juan R. »
The first canonical scientist.

#### Juan R.

• Chemist
• Full Member
• Posts: 148
• Mole Snacks: +24/-3
• Gender:
##### Rate expressions: Volume changing
« Reply #1 on: February 21, 2006, 01:39:00 PM »

The rate in function of rate of concentrations may be corrected as follow

R = –1/ a d[A] / dt + [A]/{aV} dV / dt

Where V is the total volume.

Many chemical reactions are studied in solution and since liquids are “incompressible” to a good approximation, it is acceptable to do dV / dt = 0, recovering the standard chemical expression.

Volume changing is usual for liquid-phase reactions and for gas reactions in open systems –for example for flames–. In those cases, one cannot do dV = 0.
The first canonical scientist.

#### Juan R.

• Chemist
• Full Member
• Posts: 148
• Mole Snacks: +24/-3
• Gender:
##### Rate expressions: Relativistic corrections
« Reply #2 on: February 21, 2006, 01:41:19 PM »

Usual rate is valid for relatively slow processes, in rapid processes there is relativistic corrections may be taken in account. In  we discussed a bit about corrections to rates from the perspective of relativistic rational thermodynamics, extended thermodynamics, and kinetic theory of far from equilibrium gases.
The first canonical scientist.

#### Juan R.

• Chemist
• Full Member
• Posts: 148
• Mole Snacks: +24/-3
• Gender:
##### Rate expressions: Chemical noise
« Reply #3 on: February 21, 2006, 01:45:52 PM »

Take the process

A => products

The standard rate expression is

R = k [A]

With R = – d[A] / dt

However, this is ignoring the role of chemical noise (Brownian motion). A more accurate expression is

R = k [A] + Frandom

There are several theoretical expressions for computing Frandom that I will not write here due to his complexity (this board has not great mathematical capacities). Anyone interested can search expressions computed, for example, via standard Mori theory (sometimes called Zwanzig-Mory theory due to its complementation of related Zwanzig theory).

The relationship between Frandom and k is stated by standard fluctuation-dissipation theorem. For example for Brownian-like noise

<Frandom(t) Frandom(t’)> 2 k <([A] – [A] eq) 2>eq delta(t – t’)

where “< >” denotes a statistical average and where delta is the Dirac function.

Usual rate expression is only valid when one neglects the fluctuations.
The first canonical scientist.

#### Juan R.

• Chemist
• Full Member
• Posts: 148
• Mole Snacks: +24/-3
• Gender:
##### Rate expressions: Memory effects
« Reply #4 on: February 21, 2006, 01:47:08 PM »
Above expression for the chemical reaction with noise is a special case of so-called Langevin equations:

da(t) / dt = – Tetha a(t) + Frandom(t)

for some dynamical variable a (e.g. concentration of a chemical, the position of a dust particle, reorientation of a dipole in a fluid, etc.)

In general, Langevin equations (and chemical rate equations) are valid when memory effects are neglected. A more general expression is

da(t) / dt = – Int0 t K(s) a(t – s) ds + Frandom(t)

where K is called the memory kernel, which accounts for the effects on the variable a of the past history of the system. Those kernels are very useful in the study of materials with memory. Note that rate for variable a depends of the value of a in the past (t – s). Above equations are called generalized Langevin equations and are broadly used in physics.

In macroscopic systems –without anomalous memory, or histeresis phenomena, or similar– we can apply the standard Markovian regime, then

a(t – s) -> a(t)

and

Int0 t K(s) ds-> Int0 infinity K(s) = Tetha

And one recovers the usual expression without memory (e.g. the standard chemical reaction rate)

One can understand what is the limit of applicability of chemical rate theory with a simple analysis of the limit taken. The absence of correlations in time is usual in chemical systems due to thermal randomness and, therefore, one could imagine that substitution a(t – s) -> a(t) may be good enough.

However, a half of the above limit is based in the substitution of the upper limit of the integral from t to infinity. This means that you are assuming an average description of phenomena to long times when compared with characteristic decay time of function K(s). Therefore, the computed Tetha (or the ks of chemical kinetics) will be valid after of some characteristic “microscopic” time. This is very easy to see.

Take the chemical expression

d[A](t) / dt = – k [A](t)

and integrate it. Imagine that reaction began to t = t0. After t0 the expression is

[A](t) = exp{– k (t – t0)} [A](t0)

but before is

[A](t) = [A](t0)

and in t = t0 there exists a discontinuity in the change of concentration. In fact, there is not derivative at t = t0. This is unphysical –note that the discontinuity implies an infinite “acceleration”–! The problem is that the exponential expression of above is only valid after some microscopic time. Near the very beginning of the reaction, the rate expression is of the non-Markovian type and the discontinuity is eliminated.
« Last Edit: February 21, 2006, 01:49:08 PM by Juan R. »
The first canonical scientist.

#### Juan R.

• Chemist
• Full Member
• Posts: 148
• Mole Snacks: +24/-3
• Gender:
##### Rate expressions: Very small systems
« Reply #5 on: February 21, 2006, 01:51:15 PM »

Usual chemical rates are valid for macroscopic amounts of matter. In small systems –for example at nanoscale– there exist a series of exotic phenomena I do not discuss here. Simply to state that in the nanoscale, laws as equality of chemical potentials at equilibrium are not longer valid. This is not so strange in the nanoscale, zero law of thermodynamics is not valid. I and other researchers computed size corrections to usual macroscopic laws. For example, in gas-phase systems, two small systems at equilibrium have not equal pressures. This is an active research hot topic and I do not will discuss here since it is needed the introduction of very advanced topics.

However, we can understand some size effects in a easy manner. Take the simple bimolecular reaction

A + A => products

The standard rate expression would be

R = k [A] [A]

Now, rewrite it as

R = k [A] [A] = k’ nA nA

Where nA denotes the number of molecules of specie A. Volume, atomic-molecular masses, etc. were introduced into the new k’. Imagine a very, very small system composed by one single molecule of A! The reaction for obtaining products is, according to above expression,

R = k’ nA nA = k’

But above reaction involves two molecules of A! For a single molecule there is not possibility of reaction and R may vanish. The problem is that a more realistic expression is

R = k’ nA (n – 1)A

Then when there is a single molecule R = 0, which is correct.

Of course, for very, very large amounts of matter (n – 1) -> n and this is the reason that one can write

R =aprox= k’ nA nA = k [A] [A]

In usual chemical conditions the standard kinetic rate R = k [A] [A] is good enough, but for small systems such as chemical phenomena in bio-membranes, the equation does not work. Note also that in a macroscopic system, the standard rate expression is not valid for large times just when there is a small amount of reactant and difference between (n – 1)  and n begins to be noticeable.
The first canonical scientist.

#### Juan R.

• Chemist
• Full Member
• Posts: 148
• Mole Snacks: +24/-3
• Gender:
##### Rate expressions: Diffusion
« Reply #6 on: February 21, 2006, 01:54:05 PM »

Now take again above bimolecular process

A + A => products

Imagine that I say you that reaction is a macroscopic dimerization in solution in a closed vessel (therefore, volume is constant), that there exists not memory effects, that chemical noise is to be neglected and that reaction is sufficiently slow that there is not measurable relativistic corrections. Would I write

partial rhoA / partial t = – k rhoA,

with the rho the density, as the correct rate expression?

Well, maybe not! If reaction is slow, really slow, then the expression for the rate is not that of above but

partial rhoA / partial t = DA grad2 rhoA

because the process is controlled by chemical diffusion.
The first canonical scientist.

#### Juan R.

• Chemist
• Full Member
• Posts: 148
• Mole Snacks: +24/-3
• Gender:
##### Rate expressions: Computing collisions
« Reply #7 on: February 21, 2006, 01:55:05 PM »

It is usually admitted in many textbooks that probability for a bimolecular (or biatomic) collision is computed as

Prob = [molecule1] [molecule2]

This is valid for gas-phase systems and for high-dilution situations only.

The best way to see this is from classical statistical mechanics. The basic classical equation for the colliding evolution of a molecule 1 is

partial rho1 / partial t = Tr2 LV rho12

rho1 is the classical distribution function for the molecule. Tr denotes the integration in the phase space (Q, P) of the dynamics. The integration is done over the classical degrees of freedom of the molecule 2 (i.e. coordinates Q and impulses P). LV is the interaction Liouvillian, which is simply the Poisson bracket for the intermolecular potential. The most interesting entity is rho12, which denotes s the classical distribution function for the collision pair.

For gaseous systems at very high dilution,

rho12 =aprox= rho1 rho2.

This is the famous Boltzmann’s assumption of molecular chaos (called by him Stosszahlansatz). It is an approximation valid for diluted gases only and the equation derived from the introduction of the assumption in above classical equation named Boltzmann equation. Of course, the Boltzmann equation is also valid for gases in high dilution. For most condensed phase systems the Boltzmann equation is just wrong.

In general the probability for a collision is not rho1 rho2 because that approximation is assuming statistical independence, that is, absence of correlations between the molecules; this is only valid for high-diluted systems with short range potentials.

A lot of stuff could be said about computing probabilities for collisions. I have no time for a serious review of all has been published in the topic since is a very sophisticated topic. However, I will introduce some basic ideas.

Boltzmann like equations (e.g. chemical bimolecular reaction rates) are local in time and space and assume absence of correlations before the collision.

In ionised gases (i.e. plasmas), a basic equation is Balescu one. Therein the distribution function for the colliding pair appears in the denominator instead of in the numerator. This feature is characteristic of systems with collective effects, that is, systems where molecules (in this case ions) are not randomly distributed but acting as a whole. In collective phenomena. Usual methods based in two-body or three-body processes are not valid in collective phenomena. Precisely this is one of technical difficulties are founding in the developing of tokamak reactors.

The more general expression for a collision may include non-local effects in space and time. The introduction of temporal non-localities is done via non-Markovian equations (Resibois-Prigogine kinetic equation is a characteristic example). Prigogine was Nobel Prize for Chemistry in 1977. A problem of his non-Markovian kinetic equation is that is very difficult to be solved even with most advanced mathematical techniques.

Boltzmann equation and hidrodynamics are valid in certain length regimes. Hydrodynamics correspond to situations when mean free path for molecules (or atoms) is much more larger than range of interactions, whereas the range of inhomogeneities (gradients in concentration for instance) is much more larger than mean free path.

However, the mean free path is much more larger than the range of inhomogeneities in rarified gases. In that case, Boltzmann-like equations (or hydrodynamics) are again not valid. It is really interesting to compare the large mean free path collision term with the Boltzmann collision term. The former (assuming molecular chaos) is non-local in space

Collision= rhoalpha Int {rhoj(r)} dr.

That is, molecule alpha can collide with the molecule j sited at some arbitrary distance. In the own Prigogine words [1]:

Quote
Molecule alpha interacts, so to speak, with an “average molecule” coming from an arbitrary distance

Some of those collisions at-a-distance are valid even with short-range forces due to large correlations. Those phenomena are often called “dynamical bubbles”.
The first canonical scientist.

#### Juan R.

• Chemist
• Full Member
• Posts: 148
• Mole Snacks: +24/-3
• Gender:
##### Rate expressions: Ionic strength effects
« Reply #8 on: February 21, 2006, 01:57:50 PM »

In solutions of non-zero ionic strength, the rate for

A + B => products

may be written as

R = k aA aB

with a the chemical activity. Some textbooks write

R = k’ [A]

being k’ = k gammaA gammaB

where gamma are the activity coefficients. That is correct, of course; that I find perplexing is that many textbooks do not state that k’ is not a constant and continue calling to k’ a “rate constant” (see for example [2]).
The first canonical scientist.

#### Juan R.

• Chemist
• Full Member
• Posts: 148
• Mole Snacks: +24/-3
• Gender:
##### Rate expressions: SRT
« Reply #9 on: February 21, 2006, 02:00:10 PM »

During decades chemists have assumed the dogma that the rate for the macroscopic reaction without diffusion

A + B = C + D

is always

R = k [A] – k’ [A]                          (2)

However, there were many examples where the equation did not fit to experimental data. Ward said [3]:

Quote
When this predicted concentration was compared with measurements strong disagreement was found during the initial period of the reaction […] However by this time the rate expression in Eq. (2) occupied such a hallowed position in chemical kinetics that the experimentalists chose to decide that then was something wrong with the experimental technique, rather than to decide that there was something wrong with the rate expression given in Eq. (2); […]

Then Ward developed the SRT. The SRT rate for above chemical reaction is

R = Kx {(Ke aA aB / aC aD) – (aC aD / Ke aA aB)}

As usual the as denote activities, Kx is the exchange rate constant, and Ke the equilibrium constant.

The SRT expression fits experimental data, whereas chemical kinetic theory (ART) offers wrong results. In [3] are plotted experimental data and theoretical ART and SRT predictions compared for electron exchange

*V2+ + V3+ = *V3+ + V2+

*MnO4 2– + MnO4 = * MnO4 + MnO4 2–

*Ag+ + Ag2+ = *Ag2+ + Ag+
Ward also detail different “hints” used by chemists for maintaining the (a prior assumed) validity of the ART. Particularly amazing is the case of Gordon and coworkers. Since usual rate expression does not fit experimental data on silver in HClO4, they decided change the mechanism and proposed a disproportionating mechanism involving Ag3+. However, no measurement of presence of Ag3+ was reported in their work...

I know cases that are more recent where for maintaining the validity of the ART, investigators fitted data using NEGATIVE activation energies, which is not just an absurdity but contradicts the basic assumption of ART that the activation energies are positives.

Unfortunately, many chemists rejected SRT during decades and heavily attacked Ward and coworkers. Just recently [4], they begin to admit that SRT is better than the traditional ART of chemical kinetics in a broad range of systems. In a recent comparison [4] in adsorption kinetic theories, we can read:

Quote
we would say that the obtained results strongly favour using the new SRT approach rather than the classical ART approach.

Part of the relative success of SRT approach is based in a correct computation of the probability for the collisions, taking in account molecular correlations.
The first canonical scientist.

#### Juan R.

• Chemist
• Full Member
• Posts: 148
• Mole Snacks: +24/-3
• Gender:
##### Rate expressions: More improvements
« Reply #10 on: February 21, 2006, 02:00:46 PM »

One of my previous professors in the University is working in the study of molecular processes (e.g. synthesis of sulphur chains) in interstellar clouds. He uses gas-phase chemical kinetics, equilibrium statistical thermodynamics and classical mechanics, quantum chemistry, specific molecular dynamics methods, and the GAUSSIAN in his research. He uses not general relativity, for example, since interstellar clouds are in regions of weak gravitational fields.

Now imagine a physical chemist, an astrophysicist, or an exobiologist interested in understanding certain chemical processes near a “strong” gravitational field. According to general relativity, spacetime is appreciably curved near strong gravitational fields, and there we cannot use the typical flat-spacetime mathematics.

For example, the temporal partial derivatives  (partial rho / partial t) used in standard chemical kinetics have not physical meaning in a curved spacetime, and they may be substituted by their covariant versions.
The first canonical scientist.

#### Juan R.

• Chemist
• Full Member
• Posts: 148
• Mole Snacks: +24/-3
• Gender:
##### Rate expressions: Canonical Rate Theory and References
« Reply #11 on: February 21, 2006, 02:01:22 PM »

SRT as all others rate theories discussed above are special cases of the rate expression predicted by canonical science.

[1] Prigogine, Ilya. Non-equilibrium statistical mechanics. John Wiley & Sons, Inc.: New York, 1966. P. 216.

[2] Moore, John W.; Pearson, Ralph G. Kinetics and Mechanism (3rd Ed.) John Wiley & Sons, Inc. 1981. P. 245 or 272.

[3] Ward, C. A. J. Chem. Phys. 1983, 79(11) 5605.

[4] Rudzinski, Wladyslaw; Panczyk, Tomasz. Langmuir 2002, 18, 439.
The first canonical scientist.

#### Donaldson Tan

• Editor, New Asia Republic
• Retired Staff
• Sr. Member
• Posts: 3177
• Mole Snacks: +261/-13
• Gender:
##### Re:Rate expressions: Volume changing
« Reply #12 on: February 22, 2006, 03:07:38 PM »

The rate in function of rate of concentrations may be corrected as follow

R = –1/ a d[A] / dt + [A]/{aV} dV / dt

Where V is the total volume.

Many chemical reactions are studied in solution and since liquids are “incompressible” to a good approximation, it is acceptable to do dV / dt = 0, recovering the standard chemical expression.

Volume changing is usual for liquid-phase reactions and for gas reactions in open systems –for example for flames–. In those cases, one cannot do dV = 0.
Isn't dV negligible, unless during a chemical reaction, the liquid vapourises?

Above expression for the chemical reaction with noise is a special case of so-called Langevin equations:

da(t) / dt = – Tetha a(t) + Frandom(t)

for some dynamical variable a (e.g. concentration of a chemical, the position of a dust particle, reorientation of a dipole in a fluid, etc.)

In general, Langevin equations (and chemical rate equations) are valid when memory effects are neglected. A more general expression is

da(t) / dt = – Int0 t K(s) a(t – s) ds + Frandom(t)

where K is called the memory kernel, which accounts for the effects on the variable a of the past history of the system. Those kernels are very useful in the study of materials with memory. Note that rate for variable a depends of the value of a in the past (t – s). Above equations are called generalized Langevin equations and are broadly used in physics

What is Frandom? It seems to be a function that quantifies chemical noise? What is the s-variable in the da(t)/dt expression? what does a(t) represent?
"Say you're in a [chemical] plant and there's a snake on the floor. What are you going to do? Call a consultant? Get a meeting together to talk about which color is the snake? Employees should do one thing: walk over there and you step on the friggin� snake." - Jean-Pierre Garnier, CEO of Glaxosmithkline, June 2006

#### 2810713

• Regular Member
• Posts: 63
• Mole Snacks: +2/-1
• I'm a llama!
##### Re:Rate expressions: Short introduction to standard chemical rate expressions
« Reply #13 on: February 23, 2006, 09:03:15 PM »
its bonus i've saved it on my PC i'll read it and ask if i find any difficulty understanding it...thanks alot...
hope others too enjoy it...

#### Juan R.

• Chemist
• Full Member
• Posts: 148
• Mole Snacks: +24/-3
• Gender:
##### Re:Rate expressions: Volume changing
« Reply #14 on: February 26, 2006, 09:53:46 AM »
Isn't dV negligible, unless during a chemical reaction, the liquid vapourises?

I copied the quote from the Laidler. However, i fail to understand you, could you be more explicit please.

What is Frandom? It seems to be a function that quantifies chemical noise? What is the s-variable in the da(t)/dt expression? what does a(t) represent?

In Langevin theory Frandom represents a random force quantifying variations from deterministic evolution.

the s is simply a variable of integration. Take expresion

da/dt = ka

or

da/a = kdt

The most general integration is

a(t) = Texp (Int_0^t k ds) a(0)

the s is nothing except math, you cannot write dt because then you confound with upper limit of integral. If k is independent of t then

a(t) = exp (k (t-0)) a(0)

I already said a(t) represents -but perhaps you did not read- and offered two or three examples on that. If a is chemical concentration then Frandom quantifies randommnes on the concentration

If a is the position of a dust particle then Frandom quantifies Brownian motion.

If have more doubts please return
The first canonical scientist.