[ursus101]

Hi all,

I am trying to understand how to find an equilibrium concentration [AB] in the following system. All reactions are in solution. All substrates are soluble (i.e. no heterogeneous phases). Initial concentrations of A and B are Ao and Bo, respectively.

A + B AB, dissociation constant K1
A + A AA, dissociation constant K2
B + B BB, dissociation constant K3

unknowns: [AB], [AA], [BB]

[1] Is it possible to find a general analytical expression for equilibrium concentrations [AB] as a function of (Ao,Bo,K1,K2,K3)? I was trying to denote equilibrium concentrations [AB] and [AA] as unknown variables x and y, respectively, and to build a system of equations using mass balance laws. However, I got stuck instantly.

[2] Assuming I am given numerical values for Ao, Bo, K1, K2, K3, how would I solve this problem?

Thanks,
Ursus

[Borek]

[1] Is it possible to find a general analytical expression for equilibrium concentrations [AB] as a function of (Ao,Bo,K1,K2,K3)?

Not sure if there will exist a nice analytical solution, as what you have here is a set of simultaneous nonlinear equations. At best I would expect to get a polynomial of at least 3^rd degree - if not higher.

I was trying to denote equilibrium concentrations [AB] and [AA] as unknown variables x and y, respectively, and to build a system of equations using mass balance laws. However, I got stuck instantly.

That's the correct approach, but it is not guaranteed to yield a nice solution.

[2] Assuming I am given numerical values for Ao, Bo, K1, K2, K3, how would I solve this problem?

Numerically, for example using some variant of a Newton-Raphson method.

[curiouscat]

[tex]
K_1=\frac{AB}{A.B} \\

K_2=\frac{AA}{A^2} \\
 
K_3=\frac{BB}{B^2} \\
[/tex]

3 eq. 5 unknowns.

What would be the other 2 eq. The Mass balance, yes, but how to formulate?

Edit: Ok, maybe these?

[tex]
A_0=A+AB+2.AA \\

B_0=B+AB+2.BB
[/tex]

[Borek]

Writing these equation is not that difficult. Solving them is challenging.

You can use \times in multiplications - [itex]2\times AA[/itex].

[curiouscat]

Writing these equation is not that difficult. Solving them is challenging.

Analytically, yes. Numerically, should be easy. Something like Matlab should spit out the answer in a jiffy.

You can use \times in multiplications - [itex]2\times AA[/itex].

Thanks.

The [B] tag doesn't seem to work though. 

[Borek]

The [nоbbc][/nоbbc] tag doesn't seem to work though. 

Thanks for the report, I will see what is going on.

As I expected, solving these equations is not easy. Maxima already works for about half an hour and have not yet found the solution.

[curiouscat]

The [nоbbc][/nоbbc] tag doesn't seem to work though. 

Thanks for the report, I will see what is going on.

As I expected, solving these equations is not easy. Maxima already works for about half an hour and have not yet found the solution.

...an analytical closed-form solution may not even exist.

[Borek]

...an analytical closed-form solution may not even exist.

Which is what I suggested in the very first post in the thread.

Time to kill Maxima. Another 15 minutes and nothing has changed.

[Borek]

OK, everything is OK about [B] tag - that is, it works as usual. I guess you copy pasted text from the post formatting explanation - there are some tricks used there to make the post look like I wanted it to look.

I am editing your post to make the tag look correctly. Unfortunately [B] is difficult to use, and there is not much that can be done about it.

[curiouscat]

OK, everything is OK about [B] tag - that is, it works as usual. I guess you copy pasted text from the post formatting explanation - there are some tricks used there to make the post look like I wanted it to look.

I am editing your post to make the tag look correctly. Unfortunately [B] is difficult to use, and there is not much that can be done about it.

Trying again...


[tex]

K_1=\frac{[AB]}{[A][B]} \\  K_2=\frac{[AA]}{[A]^2} \\   K_3=\frac{[BB]}{[B]^2} \\

[/tex]


Yup; works. Thanks. This is why I'd wanted it..

[ursus101]

Hi everyone,

Thanks a lot for your feedback. It seems that there does not exist an analytical solution. I also tried to solve the system of equations but without success.

For my particular task, I tried to denote x = [A], y = [B]. Then, in equilibrium state:
[itex][AB] = K_1 x y[/itex]
[itex][AA] = K_2 x^2[/itex]
[itex][BB] = K_3 y^2[/itex]

(1) [itex]x + K_1 x y + 2 \times K_2 x^2 = A_0[/itex]
(2) [itex]y + K_1 x y + 2 \times K_3 y^2 = B_0[/itex]

Mathematica spits out three solutions for this system -- each two-pages long. None of them seems right to me. Trying to substitute Ao, Bo, K1, K2, K3 with some real numbers results in gibberish -- complex numbers or numerical errors (like division by zero, etc.)

I tried to formulate problem differently, by denoting x = [AB], y = [AA], z = [BB]. Then,
[itex][A] = A_0 - x - 2y[/itex]
[itex][B] = B_0 - x - 2z[/itex]

(1) [itex]K_1 = \frac {x} {(A_0 - x - 2y)(B_0 - x - 2z)}[/itex]

(2) [itex]K_2 = \frac {y} {(A_0 - x - 2y)^2}[/itex]

(3) [itex]K_3 = \frac {z} {(B_0 - x - 2z)^2}[/itex]

Then solve for x, y, z. Nothing good happens here too...

[Borek]

Numerical approach seems to be the best one.

[Borek]

If you solve K₁ for B] K₂ for AA and K₃ for BB, and you put these into the B mass balance, you will get a 2nd degree equation in A and AB only, this can be solved for A. The result can be in turn put into the A mass balance, yielding an equation in AB only. Still not easy to solve, but at least you have one unknown only.

[ursus101]

Let's see what it may look like (I will be slowly updating this post as I progress).

Solving K1 for [B]:
(1) [itex][B] = \frac {1}{K_1} \frac {[AB]} {[A]}[/itex]

Solving K2 for [AA]:
(2) [itex][AA] = K_2 [A]^2[/itex]

Solving K3 for for [BB]:
(3) [itex][BB] = K_3 [B]^2 = K_3 (\frac {1}{K_1} \frac {[AB]} {[A]})^2[/itex]

Put these into the B mass balance:
(4a) [itex][B] = B_0 - [AB] - 2[BB][/itex]

(4b) [itex]\frac {1}{K_1} \frac {[AB]} {[A]} = B_0 - [AB] - 2[BB][/itex]

(4c) [itex]\frac {1}{K_1} \frac {[AB]} {[A]} = B_0 - [AB] - 2K_3 (\frac {1}{K_1} \frac {[AB]} {[A]})^2[/itex]

Solve (4c) for [A] (with Mathematica's help):

(5) [itex][A] = \frac {[AB] \pm [AB] \sqrt {8 B_0 K_3 - 8 K_3 [AB] + 1}} {2 B_0 K_1 - 2 K_1 [AB]}[/itex]

Put the result into the A balance:
(6a) [itex][A] = A_0 - [AB] - 2[AA][/itex]

(6b) [itex]\frac {[AB] \pm [AB] \sqrt {8 B_0 K_3 - 8 K_3 [AB] + 1}} {2 B_0 K_1 - 2 K_1 [AB]} = A_0 - [AB] - 2 K_2 [A]^2[/itex]

(6c) [itex]\frac {[AB] \pm [AB] \sqrt {8 B_0 K_3 - 8 K_3 [AB] + 1}} {2 B_0 K_1 - 2 K_1 [AB]} = A_0 - [AB] - 2 K_2 ( \frac {[AB] \pm [AB] \sqrt {8 B_0 K_3 - 8 K_3 [AB] + 1}} {2 B_0 K_1 - 2 K_1 [AB]} ) ^2[/itex]

(6d) ...

[ursus101]

(6d) I get a solution but it is too long to paste it here. Anyway, the solution does not seem to be the right one -- I guess, there are numerical problems in such situations.