[Hovgaard]

Hello everyone.

Please forgive me for possibly noob question, but I didn't find a suitable (for my level of understanding) answer in open search.

Let's say we have a tripartite mixture of A, B and C, miscible in one another.
A and C slowly react with B, yielding semi-stable (AB) and (CB) complexes and water (all products, including water, are well-soluble in the broth) but not with themselves and not with reaction products.
Molar ratios of reactants are:
A - 0.99
B - 0.0099
C - 0.0001, so roughly n(A) = 100n(B) = 10,000n(C), or n(A)/n(B) = 100 = n(B)/n(C)
Let's say we add catalyst, or just observe (measure) the yields for some, long enough, time.
The questions are:
1. If reaction speed in binary mixtures of A+B = (AB) + H20 is 100 times slower than that for (C+B) + H20, given the conditions (temperature, pressure, catalyst, if applicable) are all the same in both bipartite and tripartite mixtures, how will the yields of AB and CB grow in comparison to one another?
2. Will such reactants compete for the B from the start, altering their binary reactions rates, or not?
3. Finally, could you receive a reasonably good (significant) conversion of C to CB in any given moment of time?

Thank you in advance.

[Borek]

Broadly speaking the only way to find out is to create a mathematical model and see what it yields.

There can be not enough data though, kinetics may tell what happens at the beginning, but to find out outcome you may need equilibrium data.

[Hovgaard]

Hello Borek,

thank you for reply.
Of course, in case if I needed some solid quntitative aata, i'd need to do (order, rather) some solid maths.
The thing is, however, that i need only the qualitative approach estimation, the insight in broad perspective.
The way I see it, the equilibria at process start will strongly favor the products, as we have large excess of one component over the other in each pair. Secondly, at the beginning, when most of B is available, the reaction C+B = CB *seems to* flow much faster, as if in the case of separate tanks, because the equilibrium shift due to large excess of reagents are the same in slow and rapid branches.
That is my speculation, of course, based on intuitive logic...

[mjc123]

Are the reactions reversible? You didn't say so in your initial post.

[Hovgaard]

Are the reactions reversible? You didn't say so in your initial post.

Thank you for the reply. Yes, they are. Let's say the're esterification vs hydrolisys (no water is present at initial mixture).

[Orcio_87]

Reaction of C with B is 100 times faster than A with B, but concentration of A is 10 000 times greater, so one way or another AB will be major product.. in the beginning.

Let's say we add catalyst, or just observe (measure) the yields for some, long enough, time.

Reactions are reversible:

AB + H₂O  A + B

CB + H₂O  C + B

So after appropriate time one of them (AB or CB) will undergo hydrolisis releasing B.

If CB is more stable (have lower enthalpy of formation) than AB, then it can be normally obtained.

[Hovgaard]

Reaction of C with B is 100 times faster than A with B, but concentration of A is 10 000 times greater, so one way or another AB will be major product.. in the beginning.

Let's say we add catalyst, or just observe (measure) the yields for some, long enough, time.

Reactions are reversible:

AB + H₂O  A + B

CB + H₂O  C + B

So after appropriate time one of them (AB or CB) will undergo hydrolisis releasing B.

If CB is more stable (have lower enthalpy of formation) than AB, then it can be normally obtained.

Thank you very much!