Note:  For simplicity, this discussion will be confined to the special case in which the two boxes have the same area.  This corresponds to a chemical reaction in a homogeneous system.  The mathematical relationships become considerably more complex if the boxes are of different sizes in the conceptual experiment, or if the reactants and products of a chemical reaction are in different physical phases.

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The jumping of beans back and forth across the barrier from one box to the other is a bit confusing when we consider the rate of transfer.  To make this simpler, we will put all of the beans initially on the high side (left) of the barrier and lower the other side (right) so that jumps from right to left are practically impossible.  We could get the same effect if the boxes were at the same height, and we put some sticky-stuff on the right side.

The transfer of beans would seem to go completely from left-to-right, but there is always some possibility of a jump from right-to-left, so we will say that it has an extremely large equilibrium constant (perhaps approaching infinity).  We can describe the rate of this reaction in terms of the loss of beans from the left side or the gain of beans on the right side

We observed earlier that the rate of beans jumping from one side to the other depends on the number of beans on the side they are leaving.  In the example above, we expect to see a fairly rapid transfer at first, but slowing down as the number of beans remaining on the left becomes smaller and smaller.  The amount of time that it takes to go from 6 beans on the left to 5 is expected to be about 1/6 of the time it takes to go from 1 to 0.  We describe this with a rate equation.

This is actually the equation which is used to describe Nuclear Decay.  It may be integrated to obtain:

For Chemical Reactions, we prefer to use concentrations (instead of the number of beans or molecules), and we use a differential instead of a delta to write the rate equation:

If we reverse the drawing above,

and start with all of the beans on the right side, we can expect a similar relationship:

The rate constants for these transfers are expected to depend on the height of the barrier and on the relative areas on the left and right sides of the box (and we are assuming that these areas are equal).  The basic relationship here is that raising the barrier slows down the transfer, and lowering the barrier speeds up the transfer.  The rate constants also depend on the rate that the beans are jumping around - how many times a second that any one bean jumps - which we will assume is a property of the beans.  Another factor involves the average energy associated with the jumps, which is expected to depend on the temperature. It is possible that the temperature also affects the rate of jumping, but the effect on the energy of the jumps is more important.

We also recognize that the dependence of the rate on the concentration may be a more complex mathematical function.  This mathematical function is often the concentration raised to some power, which is called the order of the reaction.  The equations developed above are for a first-order reaction.

The simplest model for the rate constant of a chemical reaction is the Arrhenius Equation:

T is the temperature in Kelvin, and R is the gas law constant, 8.314 Joules/mole-K.  Rate constants are usually observed to increase rapidly with increasing temperature in good agreement with a positive value for the activation energy in this equation.  The frequency factor then represents the limiting value (at infinite temperature) for the rate constant.

The beans draw their energy for jumping from the thermal energy around them, and this energy is closely related to the temperature.  To consider them in terms of the Arrhenius Equation, we would say that the frequency factor A is the rate at which the beans jump in any upward direction, irrespective of the energy of the jump.  The jumping rate is independent of temperature, and the distribution of the jump velocities (energy) increases with the temperature.  The height of the barrier above the bottom of the box represents the activation energy, Ea.

Considering situations where the difference in the heights of the boxes is not so great as pictured above, we can see that raising or lowering the barrier will change the rates of both the forward and reverse reactions without affecting the eventual equilibrium condition.  This is the effect of a catalyst (lowering the barrier and accelerating the reaction) or an inhibitor (raising the barrier and slowing the reaction).

 

In all of the discussion above, we have only considered one-way transfers.  Now we look at the more common situation in which beans are simultaneously jumping from left-to-right and from right-to-left.  We combine the transfers in a single equation:

If we start with all the beans on the left and none on the right, the number on the left decreases rapidly at first.  The rate slows down as the number on the left decreases and the number on the right increases.  Eventually the rate of change will reach zero, and no further change in the numbers will occur even though the beans continue to jump back and forth.  Actually, there will be some slight fluctuations in the number if we are dealing with a small number of beans.

If we start with all the beans on the right and none on the left, the number on the left increases rapidly at first.  Again, the rate slows down as the number on the right decreases and the number on the left increases, and again the rate eventually becomes zero.

Whether we start with all the beans on the right, all on the left, or any other distribution, the final situation will be described by:

This describes the equilibrium condition, and the terms may be rearranged to:

The equilibrium constant, K, is seen to be related to the ratio of the rate constants for the individual processes.  The equilibrium constant is normally written in terms of concentrations, and that also applies here since we started off assuming that the areas of the two boxes were the same.  We can make this look more like a chemical reaction with an equilibrium constant:

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These relationships become more complicated if the areas of the two compartments are different.  We have defined the rate constants above in terms of the rate of jumping and the probability that a jump will be high enough to clear the barrier.  However, some of those jumps will cross the barrier and some will fall back on the same side that the jump originated.  Half of the jumps that have sufficient energy will result in a transfer if the areas of the two compartments are the same.  If the areas are different, there will be a bias toward the side with the greater area.

When the rate becomes equal to zero, the equilibrium condition is again defined by the ratio of the concentrations on the right and left sides equal to the ratio of rate constants for the forward and reverse transfers.

However, the rate constants discussed earlier without considering the areas of the boxes differ from those considered here by a factor of two.

The Monte Carlo Animation allows a more detailed study of the relationships between the rate constants, the relative areas of the boxes, and the equilibrium distribution.

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