Reversible reactions reach an equilibrium when the rate of the forward reaction equals
that of the reverse one. Doesn't mean that the reactions stop, but it does mean that
the concentrations of the reactants and products are constant unless the conditions are changed.
So let's say we have a gas reaction that we'll write in terms of our coefficients as well
as our products and reactants. If we want, for a gas reaction, to write the equilibrium
constant, we write it in terms of the partial pressure. So this would be the partial pressure of C
big C, raised to the little c. The same with D. And notice it's products over reactants.
So one of the things that we have to be careful of are those coefficients that our partial
pressures are raised too. So we are going to look at an example of a gas reaction, and
this is the formation of ammonia. We would normally write this as the partial pressure
of ammonia squared, because there are two of them. Divided by the partial pressure of nitrogen
times the partial pressure of hydrogen cubed. In material and energy balances, however we
often write the K in terms of mole fractions, but you have to be really careful in using
mole fractions, because you need to remember that this is really a partial pressure.
You might try and write it like this, just using the mole fractions however this is incorrect.
So let's write out our partial pressures in terms of our y, and I am going to use y sub
i for each one of these mole fractions. So this is squared. This is not squared, because
it is just N2, and this one is cubed. Therefore if we want to write it only in terms of mole
fractions, we have to make sure that those extra pressure terms stay in there. So instead
we are going to write it like this, and when we cancel out the pressures we find that this
has to be multiplied by 1 over P squared. Let's take a look at an example. In this problem
we're going to use the extent of reaction to find the number of moles at equilibrium.
The number of moles of nitrogen is going to equal the 4 that came in minus the extent of
reaction. The number of moles of H2 at equilibrium is going to equal 8 minus 3 times the extent
of reaction. And the reason we use that 3, again if we go up and look at the reaction
the stoichiometric coefficient is 3. The number of moles at equilibrium of ammonia, and that
because there is nothing that comes in, is just 2 times the extent of reaction. And we are going to
put an e right next of these, since we are in equilibrium. Now let's set up the expression.
Since we need the mole fractions, we are going to need the total moles. So if we add these
up, we're going to end up with 12 minus 2 times the extend of reaction. Let's write all
of the mole fractions in terms of the extent of reaction. So here are our mole fractions
at equilibrium. We can put it into our equilibrium expression. This is what it looks like, however
you have to be careful, because remember we have that pressure term. So this whole
thing has to be multiplied by 1 over P squared. My suggestion is that at this point, you put it in some kind
of computer program to solve for this extent of reaction. You might suspect that
there's more than one possible answer, but if you look at your possibilities, what you
will find is that the extent of reaction at equilibrium is 2.07. In order to find our
compositions, which is what we were looking for, we go back to how we described our mole
fractions, and we put in the extent of reaction that we calculated. And one way to check that
you have used the right extent of reaction is that as happens here, the sum of your
mole fractions equals 1.