I saw a formula in an old edition of Encyclop?dia Britannica about reaction velocity. I like to code formula calculators in win32asm and i coded it, seems working. :) But there were no examples nor explanations existing. And in a newer edition of En.Bri. the formula is not mentioned. Anybody knows where and how to use this formula?

exp( -( qA + qB ) / ( R * T ) ) * N * md^2 * sqrt( 8 * pi * R * T * ( 1 / mA + 1 / mB ) ) * moB * moA

dia1,dia2 atom diameters
md mean diameter =( dia1 + dia2 ) / 2
T temperature in Kelvin
mA,mB molecular masses
qA,qB energies of activation
moA,moB molarities
R constant (22,4/273)
N constant of avogadro 6,02*10^23

I hope that was understandable
Posted on 2003-08-28 11:37:34 by inFinie
As always, forgot to attach the code.:rolleyes:
Posted on 2003-08-28 11:39:53 by inFinie
In case not clearly understandable from text math here is a picture:
Posted on 2003-09-02 15:34:21 by inFinie

Sorry for replying late,but I think your mathematical expression seems to be incomplete:
all this statement must be equal to an expression something like this:

- ------=...

This differential,the reaction rate is about the variation of specie A vs time,inFinie
your equation is about a little advanced physical chemistry / topic:collision theory of
bimolecular gas reactions.

Maybe,you should study primarly the basic kinetic theory of chemical reactions:first-order
reactions,consecutive reactions etc...

I am glad to see here someone interested in chemistry.

PS:I sended you a private message
Posted on 2003-09-04 12:59:36 by Vortex
Hi All!

I'll admit that I studied this in the 1970's, but my (ancient) physical chemistry text gives the rate constant k for a bimolecular reaction in the gas phase as k = Zq. Z is number of colliding molecules per cubic centimeter per second in a system containing one mole of reactant per liter, and q is the fraction of that number which is activated (possesses enough energy to react). The formula given is the one that describes Z where the reacting species are different. The formula is listed as Na*Nb*(((Da + Db)/2)^2)*sqrt(((8*pi*R*T*(Mb + Ma))/(Ma*Mb))) where Na is the number of molecules of a (molarity*Avogadro's number), Nb is the number of molecules of b (molarity*Avogadro's Number), Da is the molecular diameter of a, Db is the molecular diameter of b, pi = 3.14159....., R is the gas constant in the appropriate units, T is the temperature on the absolute scale (usually Kelvin), Ma is the molecular weight of a (usually in gram moles), and Mb is the molecular weight of b (usually in gram moles).

Now that we have Z, let us turn to q, which is the fraction of the molecules possessing enough energy to react. This number will be the sum of the fractions for each species, so for a N'a/Na = exp((-Ea/R*T)). For N'b/Nb the fraction is exp((-Eb/R*T)); therefore, the sum is exp((-(Ea + Eb)/R*T)). Multiplying these two terms gives the formula posted, where Ea is the activation energy for a, Eb is the activation energy for b, R is the gas constant in appropriate units, and T is the temperature on the absolute scale. Of course, Avogadro's Number is approximately 6.0229 * 10^23.

In conclusion, the formula given is the one given for calculating the rate constant for a bimolecular reaction in the gas phase where the molecules are different. The final rate is dependent on the type of reaction (first order, second order, chain or consecutive, etc) and the concentrations of the reactants. Naturally, this calculation is approximately exact for ideal gas phase reactions. Where the real systems differ from the ideal system, a probability (fudge) factor is also required to calculate the rate constant.

Posted on 2003-10-20 15:19:07 by cdquarles
Thanks for that reply, really. That is what i looking for. Although i don't need it, but let the curiousity conquer the world ;)
Posted on 2003-10-21 05:50:54 by inFinie
Hi inFinie,

You are most welcome. As one of my professors once said: "A teacher is a student and a student is a teacher."

Posted on 2003-10-21 12:25:32 by cdquarles
Posted on 2003-11-04 13:12:41 by cykerr
Hi cykerr,

I know that the statement is circular, but: to teach something you must know something, and to know something you must study something. To study something you must ask and answer questions; and by asking and answering questions, you show someone else something they didn't know.

Posted on 2003-11-04 18:34:24 by cdquarles