What means " Solve it in description? " Do you mean that A reversibly reacts with B in two reactions producing either C or D

A + B = C

A + B = D ?

and [A] is constant?

C = K1*A*B, D = K2*A*B, mass balance B = Bo -D -C , then B = Bo/(1 + K1*A + K2*A)

For me, steady-state assumption contradicts time-dependency behavior. What do you really mean? In addition, start by correctly specifying the stoichiometric equations. Presumably you mean A + B C and A + 2B D, but that's just a guess. I would like to add that at equilibrium, the term "production rate" make no sense.

@ Yuri Geletii , K1 and K2 are equilibrium constants for parallel reactions 1 and 2 right

Kamal Asghar, Try to clarify the question, not only from the stoichiometry point of view, but also explain what two-phase system you mean. There is a difference between the liquid-vapor system and the liquid-liquid system.

Mirosław Grzesik Thanks for your reply, I modified the question and made possible editing because it was little hard to show the equation verbally. Here phase means I want to find degree of freedom F=C-P+2.

Yurii V Geletii Thanks for your reply, I modified the question and made possible editing because it was little hard to show the equation verbally. You can now understand the real equation.

Well, In reaction A and B one can get two different products, but it will be one reaction and not two. It may also be the case that the product of reaction A with B, ie C, will react with B to give D. However, there will be two series reactions. The situation you write about is impossible, except through intermediate products that can be neglected. Furthermore, the Gibbs phase rule does not apply to systems with chemical reactions.

If we assume the stoichiometric scheme in the form: A + B C and A + B D and additionally assume that: (1) both reactions are elementary, (2) the volume of the system is constant, (3) cA=cAo, (4) system is ideal and well mixed, then:

Ad(a) dα/dt=k1cAocBo(1-α-δ)-k-1cBoα, α(0)=0, dδ/dt=k2cAocBo(1-α-δ)-k-2cBoδ, δ(0)=0, where α=cC/cBo, δ=cD/cBo. I believe that an analytical solution is possible here.

Ad (b) As Yurii Geleti wrote. Instead of concentrations, the equilibrium yields of C and D can be used.

Ad (c) Apart from two equations for the chemical equilibrium constants, one should add four equations describing the phase equilibrium. Of course, the description of a non-ideal system will be more complicated than an ideal one. In addition, the Gibbs phase rule does not apply here.

How do we know the rate determining reaction step, of the two elementary reactions and taking the concentration of reactant A as constant, then cant it be combined with the rate constants to be pseudo rate constants-K1Cao = K3 and K2Cao = K4

To add, the critical concentration of B, C and D is CBo ignoring A and considering that no C and D is present initially, then Bt + Ct + Dt = CBo at all times.

Finally, integration of the two expressions for time concentration changes of C and D can be done analytically using the Matlab symbolic toolbox.

Karinate Okiy, In fact, little is known, because the author did not formulate the stoichometric equations. The scheme: A + B C and A + B D only makes sense in the case of intermediate steps, which, for example, due to the high rate, can be omitted. It is not enough to state that there are two reversible chemical reactions, but to state kinetic equations. The mass balance equations that I wrote are just a guess. It is also known that for any time the following stoichiometric relationship must be met: cBo=cB+cC+cC.

If B has two reactive sites then two reactions are possible

A + B C and A + B D

you can add CD, but K of this reaction would be a combination Of Ks of the first two reactions.

There are two different questions, the thermodynamic ratio of final B, C and D, as well the time profiles of B, C, and D (which is a kinetic problem). The derivation of the kinetic equations is possible, but it's not interesting for me (sorry)

A can be constant if A is in large excess or the system is able somehow to keep it constant (like pH in buffered solutions).

The hurricane Zeta passed by Atlanta and killed the internet access for two days.

Yurii V Geletii, Ensuring that the concentration of substrate A is nearly constant is not a problem. One can consider a semi-batch system or, as you write, a large excess of A in relation to B, e.g. 30: 1, which will additionally provide a strong shift of the equlibrium to the right. However, I am interested in a specific example of such two reactions, of course excluding the situation that C and D are isomers.

Dear Mirosław Grzesik ,

"I am interested in a specific example of such two reactions,"

I'm as well, but it's not my thread. I would not like to spend time looking for other example.

Dear Yurii V Geletii,

It wasn't my intention you to look for something for me. You wrote about chemical compounds with "two reactive sites", so I thought you knew such chemicals. Best regards,

Consider this a parallel reaction, at first the reaction of A+B -> C is kinetically favored due to lower activation energy Ea, but is thermodynamically limited at the reaction conditions. Then for longer reaction times, the reaction A+B D is favored due to energetically overcoming the higher energy barrier (activation energy) to progress to form product-D. In all the final reaction mixture will composed of more of product D (higher yield) and less of C (lower yield). Because producing C is thermodynamically limited.

The rate law for the bimolecular reaction A+B=C is

dC/dt =k(CBo-Cc)

Where k is the pseudo kinetic constant- k = k1CAo

For A+B D

Kc = K2/K3 = |CD|/|CAo||CBo-Cc-Cd|

Where Kc is the equilibrium constant

Karinate Okiy , I would avoid to discuss your post for the evident reason.