What's happening?
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If you saw nothing on the previous page, then your browser does not support
Java applets.
Otherwise, you found yourself looking at a computer simulation of a
binary chemical reaction. There are four types of molecules in this simulation,
red, yellow, green and blue. Each wanders around at random in the box you
see. Every time a pair of red and yellow molecules collide they may react
to form a pair of green and blue molecules, and vice versa. This chemical
reaction can be written like so:
R + Y 
G + B
The double arrow in this reaction indicates that both the
forward
(R + Y --> B + G) and the reverse (B + G --> R + Y) reactions are
possible. This is true of all chemical reactions. However, as is also true
of all chemical reactions, the probability that a reaction will
occur during any given collision is not necessarily the same for the forward
and reverse cases. The probabilities of reaction per collision are called
the reaction rate constants and are abbreviated "k_f" for the forward
reaction and "k_r" for the reverse reaction.
Take command!
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The simulation itself appears directly in your browser window. Initially
everything's frozen, and you need to click your mouse anywhere in the box
to get the simulation to start. (If you click again the simulation will
freeze again.) If all goes right then when the simulation starts a control
panel will also pop up somewhere on your screen. Using this control panel
you can change the initial concentrations (expressed as the number
of molecules "per box") and the reaction rate constants (expressed
as a probability between 0.0 and 1.0). To do so, click your mouse over
the appropriate box, delete the old numbers with your DELETE key and type
in new numbers. Click the "Restart simulation" button to restart the simulation
with your new values.
Note: You are not watching an animation of a pre-computed simulation.
This simulation is running in real time, on your computer, as you watch.
Every time you restart the simulation, the molecules are distributed randomly
in the box and given random velocities.
From the control panel you can also turn on and off a "stripchart" that
records the instantaneous concentrations of the four species of molecule
as the simulation proceeds. Just click on the button labeled "stripchart
ON".
Experiment!
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If you set the k_r = 0 the reaction goes "to completion." What are the
final concentrations (number of molecules in the box) in terms of the initial
concentrations? The answer depends on the initial concentrations and on
the ``stoichiometry'' (the proportion in which reactants are used up) but
not the reaction rate constants. Turn on the stripchart and note the shape
of the concentration curves. They will be roughly exponential curves, as
you can show with a little calculus. ("Roughly" because of the fluctuations
noted below.)
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If, on the other hand, neither k_f nor k_r is zero, then after a while
things settle down (the system "comes to equilibrium") and the equilibrium
concentrations will on average be constant. Chemists denote the
value of these concentrations as the symbol for the molecule surrounded
by brackets, as in [ R ] for the concentration of R molecules and
[
Y ] for the concentration of Y molecules. Any ratio of these constants
will of course also be a constant. The particular ratio in which we put
products over reactants, like so:
[ G ] [ B ]
K = -----------
[ R ] [ Y ]
is called (with great originality) the equilibrium constant. Can
you figure out what K will be in terms of the two reaction rate constants
k_f and k_r? The answer does not depend on the absolute size of k_f or
k_r, or on the initial concentrations of the reactants and products. This
is a remarkable and profound statement. Indeed, the experimental fact
that this statement is true is one of the main reasons we believe the collision-of-molecules
model of chemical reactions at which you are looking is correct.
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If you turn on the stripchart you will see that at equilibrium the concentrations
are not changing, on average. But you will see instantaneous (moment by
moment)
fluctuations in the concentrations. You'll note that the
relative size of these fluctuations gets smaller as you increase the total
number of molecules in the box. This is a very general observation, and
is a fact of enormous importance, because it is utterly impractical to
do a simulation like this for the enormous numbers of molecules that take
part in any real chemical reaction. But since the size of the fluctuations
in things you can observe macroscopically (like the concentrations) decreases
as the number of particles increases, you don't have to. It is an empirical
fact that as the number of particles increases to a very large number
the behaviour of anything observable (like the concentrations as a function
of time) becomes very simple, and depends only on other macroscopically
observable variables, like the temperature and volume of the box. This
empirical fact is essentially a statement of the existence of thermodynamics,
the science of the consequences of hidden "degrees of freedom", such as
molecules the motion of which you cannot see.
Remarkably, deducing the existence of thermodynamics by directly considering
the limit of a simulation like this as the number of molecules is increased
to infinity is profoundly difficult, and I am not actually sure it has
ever been done. We must accept thermodynamics at present on the basis of
experience and intuition alone.
The fluctuations have many other important roles and meanings. I will
mention just one more: the fluctuations are (1) properties of the equilibrium
system, and so can be calculated by relatively simple theory from measurements
of the system at equilibrium, but (2) they tell you how the system behaves
dynamically when it is not at equilibrium (but fairly close). Perhaps
by comparing (in this simulation) how fluctuations away from equilibrium
happen, and how the system approaches equilibrium -- watch the strip chart
-- you will be able to see intuitively the truth of this fundamental
and important observation, which is called the Fluctuation-Dissipation
Theorem and is a bit of a trick to prove mathematically.
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Note that at equilibrium the chemical reactions are still going on just
as fast as during the approach to equilibrium. Equilibrium does
not
mean nothing is happening! Rather, the average reaction rates of the forward
and reverse reactions become the same. The reaction rate is the
overall rate at which molecules react, which is not the same as the reaction
rate constant because you have to take into account the frequency
of collisions. Clearly when there are no red molecules the rate of forward
reaction will be zero no matter what the reaction rate constant is. Similarly
the forward reaction rate will be higher than the reverse reaction rate,
even if k_f < k_r, when there are many R and Y molecules but hardly
any G and B molecules. If you think about it, you will probably realize
that the statement that the system eventually stops changing ("comes to
equilibrium") is equivalent to the statement that the forward and reverse
reaction rates must become identical after a while no matter what the reaction
rate constants are. This realization in turn allows you to calculate the
equilibrium constant in terms of the reaction rate constants.
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Reduce both reaction rate constants equally (e.g. divide both by two) and
observe (with the stripchart) what happens to the approach to equilibrium.
Does it go faster or slower? Why? What happens to the size of the fluctuations
at equilibrium? What happens to the equilibrium concentrations? (The answer
to the last question is "nothing"!)
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Along the same lines, see if you can distinguish the kinetics of
this reaction from the thermodynamics. The "thermodynamics" is expressed
by whether at equilibrium there are more products or reactants (i.e. whether
K is large or small). What determines this? The "kinetics" is whether equilibrium
is reached quickly or slowly. What determines this?
How could you set up this simulation so that the forward reaction
would be thermodynamically favorable (K is big, so there'll be a lot of
products at equilibrium) but kinetically nearly impossible (equilibrium
will take a very long time to be reached)? Try it and see!
How could you set up the system so that product was often formed (kinetically
accessible) but the equilibrium concentration of the product was small
(thermodynamically unfavorable)? Would this be useful? Yes! If you had
another
chemical reaction which started with a product of this reaction,
then the overall thermodynamics of the two coupled reactions could be controlled
by the second reaction. That is, the net reaction R + Y --> G + B + X -->
Y + Z could be favorable because of the second reaction G + B + X --> Y
+ Z. Since the intermediate products G and B are formed quickly it does
not actually matter that very few of them are around at any given moment;
quite a lot of the product Y + Z will nevertheless be formed. This illustrates
the concept of a reactive intermediate. Notice that it is critical
that this intermediate be kinetically accessible.
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Suppose you change the density of the reactants. How does the reaction
rate change? You can see why the reaction rate constant is a more useful
quantity, in a sense, then the reaction rate. But the rate is easier to
measure, of course.