Mathematical Models for Equilibrium Simulation

1. Feb 13, 2009

jacksonpeeble

In my high school Honors Chemistry course today, we performed a simulation (along with several real labs) of chemical (dynamic) equilibriums. I'll briefly outline the simulation:

1. Place 42 pennies on the reactants side and 0 on the products side.
2. In the first round, move 1/3 of the pennies in the reactants pile to the products pile. At the same time, move 1/4 of the pennies from the products pile to the reactants pile. Note: Always round down to the nearest whole number, never up.
3. Record data.
4. Repeat steps 2 and 3 until there are no further changes in the numbers of products and reactants.

Obviously, this is a fairly basic example. I got to thinking, however, that this should be relatively simple to model in a mathematical formula. So first (as I always try to do), I attempted to make a written table and reason it out.

P = {(0, 42), (1, 28), (2, 21), ...}
R = {(0, 0), (1,14), (2, 21), ...}

I just made it in a T-chart, but whatever. After I failed to reason it out, I plugged it into a table on my calculator (a TI-Nspire), and tried to regress it. Unfortunately, I don't know what type of regression this even requires.

This isn't something that I'm required to do, I'm just curious and think that I should be able to do it. Can anyone be of assistance? ;-)

2. Feb 14, 2009

GCT

You may be able to write the results of each round as a function time steps.

On the left side

At t=1 x - (1/3)x + (1/4)(1/3)x = (3/4)x

At t=2 Simplifies to (9/16)x = ((3/4)^2)x

Investigate the next step to understand whether the left side fullfills this trend of (3/4)^t)x at t=3

etc... Same for the right side.

A more simple way would be to read your text and derive the equation from explanations of the rate constant and rate laws.

Last edited: Feb 14, 2009
3. Feb 15, 2009

jacksonpeeble

Thanks for taking the time to reply to my question. I took some time to review what you said, and while 3/4 is clearly the ratio, I still do not understand how to write the type of equation that I need for this. I learn best from example; if I were to type this into a calculator to graph it, what exactly would I write?

Thanks again to anyone that can help!

4. Feb 15, 2009

Ygggdrasil

Modeling the return of chemical reactions to equilibrium is a very important tool in chemistry and it is often employed by chemists to study the kinetics of chemical reactions. Since you are in high school, you may not yet have the necessary math to derive the actual equations, so I'll just tell you the end result. The concentration of reactants will decay exponentially to the equilibrium concentration. The products will also rise in a way that depends on an exponential also.

Formally, the equations are written:
$$R(t) = R_o \left(A + B e^{-kt}\right)$$
$$P(t) = R_o B \left(1 - e^{-kt}\right)$$

Where $$A = \frac{k_{-1}}{k_1 + k_{-1}}$$, $$B = \frac{k_1}{k_1 + k_{-1}}$$, and k = k1 + k-1. Ro is the initial amount of reactant, k1 is the rate constant for the forward reaction, and k-1 is the rate constant for the reverse reaction.

These equations are derived by finding the solution to the following system of coupled linear differential equations:
$$\frac{dR}{dt} = -k_1 R + k_{-1}P$$
$$\frac{dP}{dt} = k_1 R - k_{-1}P$$
With the inital conditions R(0) = Ro and P(0) = 0.

Last edited: Feb 15, 2009
5. Feb 16, 2009

Gokul43201

Staff Emeritus
Clearly, it will not, as it is supposed to equilibrate at a non-zero concentration, while (3/4)^t -> 0 for large t. So, in addition to this exponential decay, there will turn out to be a constant term which represents the equilibrium concentration. The derivation follows from solving the coupled linear DEs spelled out by yggdrasil, but that would likely be beyond the high school level.

To JP: At the high school level, one might look into building the evolution out of recursion relations (later in life you will find that this will turn out to be no different than converting the differential equations into difference equations). But for now, if you have the interest and patience (it takes time and effort), you can probably "derive" (by expanding through a number of iterations and noticing a pattern) the above equation pretty nicely by slogging away with long algebraic expressions. You decide if you are up to it...

If you are:

Let $f_P$ and $f_R$ be the fractions of the product and reactant that you move over to the other side in each iteration, then you can write down how much each side gains or loses every turn, in terms of these fractions and how much of each there was at the end of the previous iteration. Add or substract the change in each side to get the new value of each side.

For example, after the n+1'th iteration, you would have, for the reactant (R) side:

$$R_{n+1} = R_{n} - f_RR_{n} + f_PP_{n}$$

Does that make sense? Can you figure out what the equation for the product (P) side would be?

Last edited: Feb 16, 2009
6. Feb 16, 2009

GCT

You are not understanding my approach - you are forgetting that I haven't illustrated the right side ( the product side ) using the same method as I did for the reactant side , once this equation is obtained , the equilibrium time could be solved for . I agree , however , that the standard way to approach this is as Yggradsil demonstrated and I indicated this in my original post.

7. Feb 16, 2009

GCT

I haven't used a TI since I was in high school. I'm certain that such methods exist however it is probably more of a skill than something that could be resolved perfunctorily with a program.

What you need to do is to find the equation that defines the trend of products with respect to time , similar with the equation that was suggested for the reactant side - ((3/4)^t)x where x is the original amount on the reactant side - it should be a lot simpler. Make sense of what is happening on the product side with each step in time , write the algebraic equations out and then simplify them next observe and find common factors with respect to time e.g. observe the reactant side equation.

8. Feb 16, 2009

GCT

The product side is actually not that simple to mess with - I am going to write the results here when I have the time .

9. Feb 16, 2009

Gokul43201

Staff Emeritus
I guess I am not. Are you or are you not saying that the amount of reactant at time t is given by $C(t)=C_0(3/4)^t$, where $C_0$ is the reactant concentration at t=0?

10. Feb 16, 2009

jacksonpeeble

This is exactly what I'm looking for; you're right, it's a little bit heavy for what I've taken so far, but I understand most of it because of the advanced courses I'm in.

I really appreciate the help, everyone! Gokul43201, your post definitely simplified things, and GCT, I appreciate the continual feedback.