I have this equation (k*CA)*t/(rohB*r)=F(X) for following reaction:
A(g) + b B(s) --->c C() + d D(s)
k is reaction rate constant (m/s)
CA is concentration of component A (mole A/m3)
rohB is molar density of component B (mole B/m3)
r is radius and is constant (m)
t is time (s)
X is solid conversion of component B (-)
F(X) is a function of X and is without unit (-)
There is still information missing here. Assuming this was an actual elemental reaction, the rate would simply by r=kCACBb.
Since you have solid conversion factors here, there must be another process involved which should be specified somewhere.
Pure mathematical formalism
t =F(X)*(rohB*r)/(k*CA)
X = {Bo - (CAo - CA)/b}/Bo (from mass balance)
Insert X into F(x) and calculate the derivative
dt = {F(X)*(rohB*r)/(k*CA)/d(CA)
You will have the dependence of t versus CA. Reorganize this equation to the dependence of dCA/dt on CA and you are done.
I'm not sure that you need to do all these calculation. It might be simpler to use t =F(X)*(rohB*r)/(k*CA) to find the dependence of CA as a function of time. And then calculate the derivative.
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Ensiyeh Karimi
Pasargad Steel Complex
Hi Mr.Geletii,
Your solution method is completely correct but, X is solid conversion not mole fraction. I think the equation for X must be like this :
X = {(CAo - CA)*b}/Bo
As you see, b coefficient must be multiplied instead of divided. How do you think?
OK , I agree that X = {(CAo - CA)*b}/Bo. But the original equation remains very strange. This reaction seems to be heterogeneous. it should be dependent on stirring rate, size and surface area of B. How was it obtained. What is the theory behind this equation?
Yes, the reaction is heterogeneous, dependent on stirring rate, size and surface area of B. My main reference is (Chemical Engineering Science, 1974, Vol. 29, pp. 630-634). This part of article attached here. By combining equation 6 and 25, you get form (k*CA)*t/(rohB*r)=F(X).
A(g) + b B(s) --->c C(g) + d D(s)
Thanks, Ensiyeh Karimi . I'm not an expert in kinetics of heterogeneous reaction. I looked through the enclosed file and several papers. I noticed that they apply the "mass action law:"
"The rate of a reaction is proportional to the product of the active masses of the reagents involved" (Wiki). At the same time: "For species in solution active mass is equal to concentration. For solids, active mass is taken as a constant." But B is consumed in the reaction and I'm completely confused.
Now, I'm not sure that your question can be answered. You wrote your reaction as irreversible, but if the reaction is reversible, the reaction rate is the difference between forward and reversed reaction rates. In this case you need to know the reaction rate law for the reversed reaction.
I guess you need to create your own kinetic model and derive the reaction rate law.
I don't know if you've seen it, Hong Yong Sohn is still active, even here on RG:
https://www.researchgate.net/profile/Hong-Yong-Sohn
So, if you need help understanding his mechanism, you can ask him directly. Eventually, he might even tell you that he has something better in storage for the scenario.
For the derivation you otherwise require access to the reference 8 from your cited paper.
There is something wrong with your reaction rate term: it is not a rate as there is no time in the expression, so it is purely an equilibrium value. The constant r implies a pore controlled reaction, more common in catalysis than reactions.
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