Possible reactions

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

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

DrStupid

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

[itex]
\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}
[/itex]

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

[itex]
\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}
[/itex]

Borek

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

DrStupid

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

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

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:

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

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

[itex]
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]
[/itex]

this leads to

[itex]
\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 }}
[/itex]

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.