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Topic: why is Rate law like that???  (Read 10207 times)

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Offline 2810713

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why is Rate law like that???
« on: December 30, 2005, 10:56:15 AM »
Why is the expression of mass action law like that???
I mean like->
  for a reaction
   aA + bB ----> cC + dD    
rate of forward reaction [R] = k[A]ab


Specefically, why does that stoichiometric coefficient of reactant goes to the superscript???
Why not [R] = k[A]a*b or something else???

The law of mass action tells that the rate would increase faster than the eqn. i've written below that.
But how can we explain this...
for at given pt of time the gradient of the curve=[R]
[R]= (1/a)*d[A]/dt= 1/b* (dB/dt) .[this i is clear to me.]
& thus
[R]= (1/a)*d[A]/dt=k[A]x y

thus
  (1/a)*(1/[A]x)*d[A]/dt =  k* {integral t ranging from 0 to t of (d/dt)*dt}y

but [R]= (1/a)*d[A]/dt= 1/b* (dB/dt), thus,

(d/dt)=b/a * d[A]/dt

now,

 (1/a)*(1/[A]x)*d[A]/dt =  k* {integral t ranging from 0 to t of (b/a*
d[A]/dt)*dt}y


now substituting d[A]/dt as a function of time f(t), we will get
an eqn, that can be intergated for dA on L.H.S. and dt on R.H.S. we get a function in terms of [A] in LHS and that in terms of t on RHS .
now putting  [A] as a function of time {=integral f(t) for o to t}and getting the relation between a,b and x,y   may be possible.

I don't know the f(t) please help me find out how can the eqn of law of mass action can be derived. Or tell me what intuition makes you accept this law so that it might serve me a good way to be comfortable while learning kinetics...



2810713
« Last Edit: January 03, 2006, 03:35:38 PM by geodome »

Offline Donaldson Tan

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Re:why is mass-action law like that???
« Reply #1 on: January 03, 2006, 02:10:24 AM »
what talking you?

the mass-action law is for chemical equilibrium. the mass-action law states that the equilibrium expression for a chemical system (aA + bB <->cC + dD) at a fixed pressure and temperature is given as K = [C]c[D]d/[A]ab

it is derrived in the follow manner:

aA + bB <->cC + dD
forward rate = kf.[A]ab
backward rate = kb.[C]c[D]d

at equilibrium, forward rate = backward rate

kf.[A]ab = kb.[C]c[D]d

By convention, K = kf/kb
=> K = [C]c[D]d/[A]ab
« Last Edit: January 03, 2006, 02:10:44 AM by geodome »
"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

Offline 2810713

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Re:why is mass-action law like that???
« Reply #2 on: January 03, 2006, 05:15:44 AM »
Sorry, if I've used wrong terminology. I's reffering to the expressions of forword and backword rates.
As u said-
aA + bB <->cC + dD
forward rate = kf.[A]ab
backward rate = kb.[C]c[D]d


So, why are Rf and Rb expressions like that...


hrushikesh

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Re:why is Rate law like that???
« Reply #3 on: January 03, 2006, 08:06:41 PM »
Consider the bimolecular elementary reaction A + B --> C

From an intuitive perspective, we can think of a reaction as a collision between molecules of A and B.  If A and B collide with enough enery and the proper orientation, they will form C.  Now, the probability of a collision between A and B should be proportional to the concentrations of A and B.  Since I live in Los Angeles, my professors use this analogy:  car accidents are more frequent during rush hour simply because there are more cars on the road.

So, if you double the concentration of A, the rate should double and if you double the concentration of B, the rate should also double.  These hypothesis can and have been experimentally verified.  Now, consider the additive model for kinetics, that the rate of formation of C is given by:

d[C]/dt  = k([A] + )

Based on our reasoning, if we double [A], the rate should double also.  However, that is not the case.  Consider, k = 1, [A] =  1, and = 1.  Under these condition, d[C]/dt = 2.  However, if we take [A]=2, then d[C]/dt = 3, which does not agree with our predictions.

However, the model, d[C]/dt = k[A] , does agree with our predictions.  I'll leave this to you to verify.

Now, what about the reaction 2A --> C.  Well, let's re-write the expression to resemble our original reaction: A + A --> C.  Now, we can just write the same rate expression, but replace with [A], giving:

d[C]/dt = k[A][A] = k[A]2

If we extend this principle, we see that for aA + bB --> cC + dD,

(1/c)d[C]/dt = k[A]ab

Offline 2810713

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Re:why is Rate law like that???
« Reply #4 on: January 04, 2006, 01:03:36 AM »
Thanksn
  n=infinity
      !!!


hrushikesh

Offline Juan R.

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Re:why is Rate law like that???
« Reply #5 on: January 15, 2006, 07:57:46 AM »
Why is the expression of mass action law like that???
I mean like->
  for a reaction
   aA + bB ----> cC + dD    
rate of forward reaction [R] = k[A]a[ B ]b

2810713

1)

Nomenclature is not adequate. The "[R]" means concentration of R and above the units are (concentration / time) because the 'k' have units, therefore write just R, where R means 'Rate'.

2)

It is not true that
Quote
the mass-action law is for chemical equilibrium.

This is a common error of modern chemical literature. The mass-action law was introduced in 1864 by Guldberg and Waage to describe their observation about the rate of chemical reactions.

R = k rho_1 rho_2

the rhos often were not equal to real masses (or concentrations) and therefore were named 'active masses'; therein the name of the law. Today, we know why active masses are not equal to real masses; it is due to activity effects in solution and today the above law is best expressed in terms of activities in chemical kinetics textbooks.

R = k a_1 a_2

3)

The usual expressions in terms of concentrations are valid only as approximation and the typical discussion in terms of binary collisions really valid for phase gas kinetics. This is the reason that chemical dynamics is really developed only for gas phase kinetics systems. Next, i cite some of approximations in usual kinetic laws of chemistry:

- Valid only for macroscopic size matter

- Valid for constant volume systems

- Valid for 'slow' chemical processes

- Valid for systems in thermal equilibrium

- Absence of long correlations (e.g. strong gravitatory fields)

4)

Why has then the mass action law of chemistry that generic form?

The explanation is via canonical science. See [1] for a first pionnering version, [2] for a students level discussion of some basic stuff and the non-technical page [3] for some recent advances.

The basic idea is as follow. The state of any system can be represented via a vector state (we use the same Keizer original nomenclature for backward compatibility, but often the conceptual schema is different).

Any process is by definition a change in the vector state. See first equation of [3] or 2nd paragraph of section III of [1]. The rate of the process is a superposition of two contributions: direct and inverse.

Take for example the direct contribution. The rate is equal to the product of its 'dinamical' rate by the posibility that process can happen. For example, imagine that rate for the chemical process A + B --> C is 45 in arbitrary units. If you have chemicals A and B spatially separated then the rate is 45·0 = 0 and there is not reaction!

The posibility for a process is directly related to the entropy of the system, i.e. to the number of favourable atomic-molecular* configurations. The section III of [1] is still good for some details, but modern interpretation of entropy [4] is more general and does not limited to macroscopic matter.

Therefore, the rate for a process is 'dinamical' · 'favourable quantum states' or

V = Omega · exp[···]

See equation (168) of [2]. Second equation of [3] or equation (1) of [1] for the full expression (which is complicated and this board has no mathematical capabilities far from sub and sup)

The chemical reaction law R = k [A]a[ B ]b is a special case of canonical equation. In more general situations the canonical expression continue being valid but the kinetic law is not (for example in many processes in biological membranes).

Moreover, the Arrhenius law (and generalizations) are also derived from the canonical science, whereas in chemical kinetics textbooks simply postulated.

See section IV-B of [1] for a derivation of chemical kinetic laws from canonical equation but remember that is an old approach and some concepts are outdated by new version.

The advantage of the canonical form is that is unified (explain from elementary particles to cosmology), is more general and theoretically sound that chemical kinetic law of mass action**.

[1] Keizer, Joel. J. Chem. Phys. 1976, 64(4), 1679.

[2] http://www.britannica.com/eb/article-52910. See equations (167) and (168).

[3] http://www.canonicalscience.com/en/researchzone/canonical.xml. This is a XML page with MathML technology for mathematics, and some old browsers or nonstandard (e.g. IE) cannot render it.

[4] http://www.canonicalscience.com/en/researchzone/nanothermodynamics.xml This is a XML page with MathML technology for mathematics, and some old browsers or nonstandard (e.g. IE) cannot render it.

* Atomic-molecular here means microscopic structure of matter and is valid in situations where there are not molecules or atoms, for example elementary particles or Dp-branes.

** I show this in Geodome post "rate constants and chemical engineering"
« Last Edit: January 15, 2006, 08:06:48 AM by Juan R. »
The first canonical scientist.

Offline 2810713

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Re:why is Rate law like that???
« Reply #6 on: February 01, 2006, 09:48:29 PM »
Thanks for your reply but my browser isn't letting me see those pages.
I've derived it using probability.
rate=k[probability of collision] = k'[A]nm

This proof is longer. But, before posting it . I wanted to see that canonical science- proof.
What do you think about my route of proving?

hrushikesh

Offline Juan R.

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Re:why is Rate law like that???
« Reply #7 on: February 05, 2006, 11:13:26 AM »
Thanks for your reply but my browser isn't letting me see those pages.
I've derived it using probability.
rate=k[probability of collision] = k'[A]n[ B]m

This proof is longer. But, before posting it . I wanted to see that canonical science- proof.
What do you think about my route of proving?

hrushikesh

I already said that proof (correct for ideal gases) is not valid in more general situations. I expressed some criticism in above point 3.

You equation offers wrong replies in many biological reactions for example.

Moreover, which is the equation when there is chemical noise?

What is more, you are asuming a reaction inside a volume. What is the equation for reactions in surfaces? Note that there is not [A] = moles A / Volume because the volume is zero on the surface

Moreover even corrected for surfaces that equation you are using offers the incorrect result in many situations covered by ART theory and people has developed SRT theory.

Etc.

Again you cannot derive general equations via specific aproaches. What is more, you are deriving an equation valid only for chemical reactions. What is the equation for mass flow? What is the reaction for thermal flow? Etc. Canonical theory equation is valid for more general situations. In fact, it is valid for All situations, neutron scattering, particle physics reactions, Boltzman kinetics, ecology..

No problem if you cannot acess to above webpages. You can see

[1] Keizer, Joel. J. Chem. Phys. 1976, 64(4), 1679.

or

[2] http://www.britannica.com/eb/article-52910. See equations (167) and (168).

for some related information and basic thoughts. The first reference at research level and the other at educative level.
« Last Edit: February 12, 2006, 08:58:59 AM by Juan R. »
The first canonical scientist.

Offline 2810713

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Re:why is Rate law like that???
« Reply #8 on: February 07, 2006, 06:20:28 AM »
Thanks, its nice to discuss always...

i sort of used probability of finding  required reaction/process-unit [ balanced left side of the reaction] in the given volume/ area/ point!

What's SRT theory?

within a couple of days i'll just check out some more books and web and then return...


hrushikesh

Offline Juan R.

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Re:why is Rate law like that???
« Reply #9 on: February 12, 2006, 08:56:59 AM »
What's SRT theory?

SRT is an acronym for Statistical Rate Theory, which is a theory of rates developed by physicists in the last part of 20th century. SRT is better than chemical kinetics in a class of systems, but poor in others.

In a next thread, (one week or so) I will introduce a short resume of different expressions for rates (including SRT) of several processes of interest in chemistry, physics, or biology. The resume will highlight the typical errors produced when we use some rate expression outside of its field of applicability. I believe that this may be of high educative interest, since often textbooks assume that rate theories are perfect (in fact, some decades ago the chemical rate theory was popularly named “Absolute Rate Theory” meaning that was the definitive expression) when are not.

I will also introduce some basic discussion of basic stuff, for example why the computation of the probability for a collision as being equal to [A]^a^b is valid just as an approximation, being in general an incorrect assumption (part of the experimental success of SRT over ART is due to a correct computation of the probability for a collision of reactants).
« Last Edit: February 12, 2006, 09:01:57 AM by Juan R. »
The first canonical scientist.

Offline 2810713

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Re:why is Rate law like that???
« Reply #10 on: February 14, 2006, 03:57:06 AM »
so its statistical rate theory...'llbe wt.ing for those new things your going to introduce...i'll try to get some info about them untill u post it here.








hrushikesh

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