Possible reactions

[greyd927]

I am currently working on an un-named project and have hit a road block. I have a very shaky background in chemistry and, while I can do the research myself, I don't know where to sart.

I am trying to satisfy a chemical equation which (if my terminology isn't as rusty as i believe
) is autocatlytic:

A+B=C+D
C+E=F+A
D+F=E+B

They have to react into certain states as well but I don't want to start my search too refined.
Is there a name for such a reaction? Has such an equation bean considered before? Basically how can I start researching into this problem.

[Borek]

You wrote three equations, yet you ask for one. Can't say I follow.

I only guess that you are looking for a multistage process, in which initial reactants are also between final products?

[greyd927]

Sorry by equation I meant set of equations. And yes, I'm trying to find a set of reactions that is in theory never ending. But how do I start researching?

[Borek]

And yes, I'm trying to find a set of reactions that is in theory never ending.

You can already stop searching - there will be no such example. You overall reaction is

nothing -> F + E

and for obvious reasons is impossible.

That said, it can be possible to find reactions where you need more A and B at the beginning, something like

2A + 2B -> F + E + A + B

where original reactants are also between products, but then your overall reaction is

A + B -> E + F

so I am not sure if it fits what you are looking for.

I have a gut feeling such things are present in the biochemical cycles, but that's not my league.

[greyd927]

It is possible you are right; chemistry isn't my background, but i feel like you might be enterpreting what i asked differently then i meant so i just wanted to double check.

I want to create a series of chemical reactions

After completing the series i will be left with the same two chemicals i started out with

In theroy the reactions should continue cycleing forever

So if A + B = two new compounds (lets call them C + D) then C and D would go through another set of reactions which would result with the products A + B and the cycle would restart.

I origonaly had:
A+B=C+D
C+E=A
D+F=B

But realized while A and B were constantly being used up and then replenished, E and F were only being used and so i changed my idea to:
A+B=C+D
C+E=F+A
D+F=E+B

Im confused where you say nothing = F + E because I dont have an equation that results in F + E, and all my equations contain reactants.

(Everytime someone says something says it cant be done i always try anyways: It worked for einstein so the idea has to have some meit :) )

[Ygggdrasil]

Stuart Kauffman has done some interesting work on autocatalytic reactions although I am not very familiar with it. It may be worth looking up some of his papers. It may also be worth looking up some papers on oscillatory reactions like the BZ reaction (http://en.wikipedia.org/wiki/Belousov%E2%80%93Zhabotinsky_reaction or http://online.redwoods.cc.ca.us/instruct/darnold/deproj/Sp98/Gabe/).

Edit: For an example of a biological system that undergoes oscillation, see the following paper on the KaiC protein, which is involved in setting the circadian rhythm of cyanobacteria.

Rust et al. (2007) Ordered Phosphorylation Governs Oscillation of a Three-Protein Circadian Clock. Science 318:809. doi:10.1126/science.1148596 (http://dx.doi.org/10.1126/science.1148596) PMC:2427396 (http://www.ncbi.nlm.nih.gov/pmc/articles/PMC2427396/).

[Borek]

In theroy the reactions should continue cycleing forever

No. Thermodynamics doesn't allow that. That would be chemical equivalent to perpetual motion. There are systems where oscillations are observed (like Belousov–Zhabotinsky reaction Ygg linked to), but they are powered by some underlying overall process, which runs only till the initial reactants are consumed.

Im confused where you say nothing = F + E because I dont have an equation that results in F + E, and all my equations contain reactants.

I am referring to the overall equation describing your system,

For multistep processes we can "sum" the equations to see the overall reaction. For example if we have two reactions:

A + B -> C + D

and

C + D -> E + F

we can add these equations (simple mathematical trick) and what we get is

A + B + C + D -> C + D + E + F

But in this equation C and D are on both sides (and in the same quantities) - so we can cancel them, and the overall reaction is

A + B -> E + F

(see http://en.wikipedia.org/wiki/Solvay_process for an example of such multistep process described by a simple overall reaction).

When the same approach is applied to the system of reactions you proposed, it turns out E and F are appearing but nothing is consumed - which is a clear violation of the principle of the mass conservation.

(Everytime someone says something says it cant be done i always try anyways: It worked for einstein so the idea has to have some meit :) )

Beware - this is the simplest path to become crackpot. There are places where we don't know what to expect, so someone stating "it is impossible" can be wrong. But there are places where we have rock solid laws and principle, tested and retested - and assuming something can be possible if it goes against these laws is just a waste of time.

[greyd927]

Ygggdrasil those are great references thank you.


Perpetual motion is not possible because engery is lost (transferred), but what am I loseing which prevents me from creating this chemical system?

What if I utilized a biological agent to fix this loss? It should work then yes?

[Borek]

Perpetual motion is not possible because engery is lost (transferred), but what am I loseing which prevents me from creating this chemical system?

You are not losing, you are generating mass out of nothing.

What if I utilized a biological agent to fix this loss? It should work then yes?

I guess you mean you will be adding something to the system so that the reaction can run? Then it can be possible. Mass in, mass out.

[greyd927]

I though abou that once but dismissed the though thinking that in my set
A+B=C+D
C+E=F+A
D+F=E+B
i have mass: there are four chemicals A,B,E, and F, constantly interacting. so instead of creating mass wouldn't i just be changing the existing chemicals over and overagain?

the goal is constant interaction

[Ygggdrasil]

Any mixture of chemicals will have some equilibrium point at which the free energy of the system is minimized. Over time, the system will approach this equilibrium point, at which point, all the concentrations will remain stable. While most systems will approach equilibrium monotonically, under certain conditions you can create systems far from equilibrium that will oscillate (but still slowly approach equilibrium). The only way to keep a set of reactions cycling indefinitely is to couple them to a thermodynamically favorable process. For example, in the biological example I gave, the system is driven by the hydrolysis of ATP (i.e. the net chemical reaction after one cycle is x ATP --> x ADP + x PO43-).

[greyd927]

So using simply chemical reactions a constantly cycling system would stabilize into equilibrium.

lets say then i use some biological component instead of using all chemicals; for example E and f in my previously posted system. Since the biological components would turn C and D back into A and B, and the biological component would likely draw some energy from another source such as sun light,this should prevent the system from reaching equilibrium. Unless I'm missing something else?
A biological component will not act as quickly, but i could work that issue out.

A+B->C+D
C+E->F+A
D+F->E+B

[Ygggdrasil]

Biological components obey the same laws of chemistry as any other molecules. The fact that the components are biological is not going to change anything. In order to get the reactions to cycle indefinitely, you need some input of energy.

If one of the reactions (biological or non-biological) is powered by light, then yes, it would be theoretically possible to get a cycling reaction.

[greyd927]

Is there a relatively simple way to explain why they stop at an equilibrium? As in where does the energy go? I'm kind of hoping it is lost in heat energy because I can over come that :)

[greyd927]

Second thought: Or does it depend on the situation?

[Ygggdrasil]

As the system approaches equilibrium, some of the energy is irreversibly lost as heat. In addition, the entropy of the system irreversibly increases.

[greyd927]

I wont be converting the heat energy into any form of mechanical work so entropy shouldn't be a problem.

lets say the first reaction is exothermic, while the second and third are endothermic
A+B->C+D
C+E->F+A
D+F->E+B

then take the entire system container and magnetically suspend it in a vacuum chamber (Designed and tested such a design for another project, it should work here to.

While the suspended container would obviously draw and absorb the heat energy from the exothermic reaction, what if the design also forwarded this absorbed energy to the endothermic reaction? in theory would this come close to solving the heat energy loss? It is not a perfect solution obviously, as perfect heat transfer is more theory than reality, but it would come close and limit the need for added external energy right?

[Borek]

You can't selectively "forward the energy" from one reaction to another if they all happen in the same container.

You are trying to find a way to cheat on thermodynamics. It won't work, you are wasting time.

[greyd927]

Ok fine last question: if the energy lost as heat was re added into the system with another source, would it work or is it energy lost another was as well?

[Borek]

Ok fine last question: if the energy lost as heat was re added into the system with another source, would it work

It would just keep the container at another constant temperature. A higher one.

or is it energy lost another was as well?

:confused:

[DrStupid]

I am trying to satisfy a chemical equation which (if my terminology isn't as rusty as i believe) is autocatlytic:

A+B=C+D
C+E=F+A
D+F=E+B


As already mentioned by Ygggdrasil this process will result in an equilibrium. The stoichiometry would require


\begin{array}{l}
\left[ A \right] + \left[ C \right] = \left[ A \right]_0 + \left[ C \right]_0 \\
\left[ B \right] + \left[ D \right] = \left[ B \right]_0 + \left[ D \right]_0 \\
\left[ E \right] + \left[ F \right] = \left[ E \right]_0 + \left[ F \right]_0 \\
\end{array}


Thus there are only three concentrations left to be determined. With three equations it should be possible to solve this problem:


\begin{array}{l}
\left[ {\dot A} \right] = - k_1 \cdot \left[ A \right] \cdot \left[ B \right] + k_2 \cdot \left( {\left[ A \right]_0 + \left[ C \right]_0 - \left[ A \right]} \right) \cdot \left[ E \right] = 0 \\
\left[ {\dot B} \right] = - k_1 \cdot \left[ A \right] \cdot \left[ B \right] + k_3 \cdot \left( {\left[ B \right]_0 + \left[ D \right]_0 - \left[ B \right]} \right) \cdot \left( {\left[ E \right]_0 + \left[ F \right]_0 - \left[ E \right]} \right) = 0 \\
\left[ {\dot E} \right] = - k_2 \cdot \left( {\left[ A \right]_0 + \left[ C \right]_0 - \left[ A \right]} \right) \cdot \left[ E \right] + k_3 \cdot \left( {\left[ B \right]_0 + \left[ D \right]_0 - \left[ B \right]} \right) \cdot \left( {\left[ E \right]_0 + \left[ F \right]_0 - \left[ E \right]} \right) = 0 \\
\end{array}

[Borek]

The stoichiometry would require


\begin{array}{l}
\left[ A \right] + \left[ C \right] = \left[ A \right]_0 + \left[ C \right]_0 \\
\left[ B \right] + \left[ D \right] = \left[ B \right]_0 + \left[ D \right]_0 \\
\left[ E \right] + \left[ F \right] = \left[ E \right]_0 + \left[ F \right]_0 \\
\end{array}


Please elaborate. Could be I am missing something painfully obvious, but I don't see how you got there.

[DrStupid]

Please elaborate. Could be I am missing something painfully obvious, but I don't see how you got there.

A is converted into C in the fist reaction and C is converted back to A in the second reaction. Thus the sum of the concentrations of A and C must be constant. The same applies to B and D in the first and third reaction as well as to E and F in the second and third reaction.

Now I also added the resulting differential rate equations to my last posting. I am to lazy to solve this system of non-linear equations but I run a numerical simulation of this system with different sets of rate constants and starting concentrations. It always reaches an equilibrium.

[greyd927]

im sorry i meant to say "way" not "was"
is the energy lost in another way other than heat energy; and also if this energy loss could be compensated would the system then avoid equilibrium

[Borek]

No, it will always reach an equilibrium.

[Ygggdrasil]

One of the consequences of the second law of thermodynamics is that, as a system approaches equilibrium, its free energy decreases (a measure that incorporates both the chemical potential energy of the system and its entropy). The mixture of reactants and products that produces a system with the minimum free energy is the equilibrium state. Chemical systems will sponaneously move toward equilibrium, and this movement toward equilibrium is irreversible -- increasing the free energy of the system to move away from equilibrium requires performing work on the system (not just supplying heat energy).

[greyd927]

entropy is when you convert the energy into mechanical work right? Im not doing that though. Then again I'm not sure what entropy is completely to begin with haha

What kind of work would need to be performed on the system? Are you saying that with this "work" the system can be moved away from equilibrium?

[Ygggdrasil]

Entropy has to do with the disorder of the system. For example, if you start out with a system that contains purely A and B, this system has a much lower entropy that a system that is a mixture of A,B,...,F.

The work is some sort of external energy supplied to the reaction. The process basically just needs to involve a decrease in free energy greater than the amount of free energy you want to add into the system by moving the system away from equilibrium. This energy can be supplied in may ways, for example, coupling your reaction to a thermodynamically favorable reaction. You could also include some reactions that are driven by electrical energy or by light so that these could be possible external sources of energy.

[greyd927]

for example, coupling your reaction to a thermodynamically favorable reaction.

Would this slow the approach to equilibrium or, if done correctly, avoid it?
and couple how?
im assumeing by "thermodynamically favorable" you mean exothermic?

[Ygggdrasil]

Would this slow the approach to equilibrium or, if done correctly, avoid it?
and couple how?
im assumeing by "thermodynamically favorable" you mean exothermic?

By thermodynamically favorable, I mean an exergonic process (ΔG < 0).

You basically want this exergonic process to drive the regeneration of one (or more) of the species, which can prevent the system from reaching equilibrium.

[DrStupid]

Just to finish my kinetic approach:

In addition to my stoichiometric equations above there is also this one:


\left[ A \right] - \left[ B \right] + \left[ E \right] = \left[ A \right]_0 - \left[ B \right]_0 + \left[ E \right]_0


The two required equations for the remaining components A an B result from the condition for the equilibrium:


k_1 \cdot \left[ A \right] \cdot \left[ B \right] = k_2 \cdot \left[ C \right] \cdot \left[ E \right] = k_3 \cdot \left[ D \right] \cdot \left[ F \right]


this leads to


\left[ B \right] = \frac{{\left( {\left[ A \right]_0 + \left[ C \right]_0 - \left[ A \right]} \right) \cdot \left( {\left[ A \right]_0 - \left[ B \right]_0 + \left[ E \right]_0 - \left[ A \right]} \right)}}{{\left[ A \right] \cdot \left( {1 + \frac{{k_1 }}{{k_2 }}} \right) - \left[ A \right]_0 - \left[ C \right]_0 }}


Now there is only one concentration left but unfortunately I can not solve the corresponding equation. Thus a numeric simulation seems to be the easiest way to get the equilibrium composition from kinetic parameters.