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.)
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.
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. (If you would like learn
more about thermodynamics, consider visiting
The Second Law.)
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.
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.
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"!)
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.
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.
G + B