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### Topic: simultaneous equilibria  (Read 9723 times)

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#### ursus101

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• Mole Snacks: +0/-0 ##### simultaneous equilibria
« on: October 27, 2012, 03:20:12 PM »
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]

 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.

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

Thanks,
Ursus

#### Borek ##### Re: simultaneous equilibria
« Reply #1 on: October 27, 2012, 06:44:05 PM »
 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 3rd degree - if not higher.

Quote
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.

Quote
 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.
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#### curiouscat

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• Mole Snacks: +121/-35 ##### Re: simultaneous equilibria
« Reply #2 on: October 28, 2012, 01:43:04 AM »
$$K_1=\frac{AB}{A.B} \\ K_2=\frac{AA}{A^2} \\ K_3=\frac{BB}{B^2} \\$$

3 eq. 5 unknowns.

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

Edit: Ok, maybe these?

$$A_0=A+AB+2.AA \\ B_0=B+AB+2.BB$$
« Last Edit: October 28, 2012, 01:56:54 AM by curiouscat »

#### Borek ##### Re: simultaneous equilibria
« Reply #3 on: October 28, 2012, 04:50:47 AM »
Writing these equation is not that difficult. Solving them is challenging.

You can use \times in multiplications - $2\times AA$.
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#### curiouscat

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• Mole Snacks: +121/-35 ##### Re: simultaneous equilibria
« Reply #4 on: October 28, 2012, 05:33:46 AM »
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.

Quote
You can use \times in multiplications - $2\times AA$.

Thanks.

The [B] tag doesn't seem to work though. « Last Edit: October 28, 2012, 06:10:20 AM by Borek »

#### Borek ##### Re: simultaneous equilibria
« Reply #5 on: October 28, 2012, 05:44:03 AM »
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.
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#### curiouscat

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• Mole Snacks: +121/-35 ##### Re: simultaneous equilibria
« Reply #6 on: October 28, 2012, 05:59:24 AM »
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 ##### Re: simultaneous equilibria
« Reply #7 on: October 28, 2012, 06:03:24 AM »
...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.

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#### Borek ##### Re: simultaneous equilibria
« Reply #8 on: October 28, 2012, 06:09:59 AM »
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.
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#### curiouscat

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• Mole Snacks: +121/-35 ##### Re: simultaneous equilibria
« Reply #9 on: October 28, 2012, 06:11:21 AM »
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...

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

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

#### ursus101

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• Mole Snacks: +0/-0 ##### Re: simultaneous equilibria
« Reply #10 on: October 28, 2012, 04:16:02 PM »
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:
$[AB] = K_1 x y$
$[AA] = K_2 x^2$
$[BB] = K_3 y^2$

(1) $x + K_1 x y + 2 \times K_2 x^2 = A_0$
(2) $y + K_1 x y + 2 \times K_3 y^2 = B_0$

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,
$[A] = A_0 - x - 2y$
$[B] = B_0 - x - 2z$

(1) $K_1 = \frac {x} {(A_0 - x - 2y)(B_0 - x - 2z)}$

(2) $K_2 = \frac {y} {(A_0 - x - 2y)^2}$

(3) $K_3 = \frac {z} {(B_0 - x - 2z)^2}$

Then solve for x, y, z. Nothing good happens here too...
« Last Edit: October 28, 2012, 04:35:59 PM by ursus101 »

#### Borek ##### Re: simultaneous equilibria
« Reply #11 on: October 28, 2012, 04:19:12 PM »
Numerical approach seems to be the best one.
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#### Borek ##### Re: simultaneous equilibria
« Reply #12 on: October 28, 2012, 04:43:00 PM »
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.
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#### ursus101

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• Mole Snacks: +0/-0 ##### Re: simultaneous equilibria
« Reply #13 on: October 28, 2012, 05:17:05 PM »
Let's see what it may look like (I will be slowly updating this post as I progress).

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

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

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

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

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

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

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

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

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

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

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

(6d) ...
« Last Edit: October 28, 2012, 06:03:51 PM by ursus101 »

#### ursus101

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• Mole Snacks: +0/-0 ##### Re: simultaneous equilibria
« Reply #14 on: October 28, 2012, 06:32:41 PM »
(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.