JOHN: When we look at heterogeneous chemical equilibrium,
we have to be a little more careful.
We have to take in account material balances
in a way that we don't need to consider
when we're looking at just homogeneous, for example,
just gas phase equilibrium.
And easiest way to understand this is to look at examples.
So let me write down a reaction.
Looking at this equilibrium,
which solid carbon reacts with carbon dioxide gas phase
to make carbon monoxide in the gas phase.
And, to simplify things, look at the case,
we have ideal gases,
and what that means is, that fugacities,
for example, for CO2 in the mixture
is the same as the partial pressure, CO2,
and likewise, for carbon monoxide.
So if I were to write down
the reaction equilibrium expression,
it would be the product,
fugacity is the partial pressure,
and it's divided by the pure component fugacity,
and I'm writing that down just to remind us
that we're gonna have to write pressures
in terms of, you know, it's a bar for these equations, too.
Be correct.
And we have the pressure of CO2,
also divided by one bar,
and then we have the activity of the carbon,
which is the fugacity of carbon,
half the temperature of reaction we're interested in,
divided fugacity at the same temperature,
but it's standard conditions,
meaning one bar pressure.
Well, when we change the pressure on carbon
we're having almost no effect on its fugacity,
so to a pretty good approximation,
this term is one,
that means this equation reduces the pressure CO squared
divided by the pressure of CO2,
where we're remembering that the units are bar.
So important thing to note is,
if I were to change the amount of carbon in my system,
I don't change anything about this equilibrium expression.
In other words, because carbon is in its own phase,
pure component,
the amounts of carbon does not affect its fugacity.
But for this equilibrium expression to apply,
carbon must be present.
So, for example, let's suppose we start out
with 0.1 moles of carbon solid
and, let's say, 50 moles of CO2,
and we let this go to equilibrium.
Well, of course, depends what temperature,
but let's say we're at a temperature
where equilibrium is significantly to the right.
What's going to happen is all the carbon
is gonna react.
And we're still not in equilibrium.
In other words, we react the tenth of all carbon,
and form a 2/10 of a mole of carbon monoxide,
but because of the conditions I picked,
we're still not equilibrium.
In other words, we're at a condition
where the pressure of CO squared
divided by the pressure of CO2
is less than Ka.
So in order for equilibrium, these satisfied,
we have to have this ratio here equal to Ka,
which would mean we need to make more carbon monoxide,
and react more carbon dioxide.
But we have no carbon left to react.
So we cannot satisfy equilibrium.
Important thing to keep in mind is the fugacity of carbon
is equal to pure component fugacity if carbon is present.
If there's no carbon,
the fugacity is zero,
we can't satisfy the equilibrium expression.
So this is why this is fundamentally different
from gas phase reactions.
If I have a gas phase,
two gases reacting to form a third gas,
I can always satisfy the equilibrium expression
because I can make the fugacity,
for example, of the component and denominator.
If I had A gas plus B gas going to C gas,
I can always satisfy equilibrium by making,
for example, this number small, the mole fraction small,
therefore, partial pressure, sufficiently small
that I can reach the equilibrium expression,
the pressure of C,
pressure of A, pressure B,
I can always make it equal to the equilibrium constant.
That's not true when we're dealing with heterogenous.
You know, other heterogeneous systems,
one that's a very common example is calcium carbonate
decomposing to calcium oxide plus CO2 gas,
calcium oxide being a solid.
And we end up with the same issue
where we could, for example,
react all the calcium oxide away
or all the calcium carbonate,
depending what temperature we're at
and not get to equilibrium.
But here's something that's important is different,
these are both solids,
but we're assuming they're gonna be in different phases.
This is important.
In order to say that the activity
of calcium carbonate is one,
and the activity of calcium oxide is one,
means calcium oxide is in its own phase
and calcium carbonate is in its own phase.
And then there's two solids that do not mix.
So, so the takeaway message here
is we have to keep in mind the material balance,
do we have enough of the component
to satisfy the equilibrium expression.