This is an example of heterogeneous equilibrium, namely when we have different phases reacting,
in this case a liquid reacting with a solid to form two other solids, and the critical
thing about this is that, in this calculation, we're assuming none of the phases mix.
In other words, these are separate individual crystals, and so we have calcium oxide crystal,
which is different from thorium crystal.
And the question is, then what's the maximum temperature at which calcium liquid is going
to reduce thorium oxide, and we're given delta G's for the reaction at two temperatures.
So what we're going to do is, to calculate the temperature, we are going to first use
the equilibrium expression, and the equilibrium expression, the general form of the equilibrium
expression we're saying the activity of thorium, and in general we would put this symbol to
indicate that it's in a mixture, but since thorium is its own phase, we don't need the
symbol here, and likewise the activity of calcium oxide, and the activity of calcium
liquid squared, and the activity of thorium oxide.
Now, activity is defined as fugacity of thorium at the condition of temperature and pressure
divided by the fugacity of pure thorium at standard conditions, say 1 bar pressure, and
likewise for the other components.
So this is the equilibrium expression.
Now, the fugacity at whatever pressure, and it's not specified here, divided by the fugacity
of that same pure material, same crystal, standard conditions, is essentially the same.
In other words, pressure affects the fugacities of solids and liquids very little, so basically
this equilibrium constant becomes one.
Each of these terms is one, and so our equilibrium constant is one at the conditions where we
have these phases in equilibrium.
Either side of this temperature where the equilibrium constant is one, we're either
going to have reaction going in one direction, so all of one of the components reacts, or
reaction going in the other direction until all of one component reacts.
Now we have K.
We know that the equilibrium constant is equal to the exponential of the change in Gibbs
free energy over RT.
So what we're going to do then is calculate, at the conditions where K equals one corresponds
to the Gibbs free energy change being zero, because the exponential of zero is equal to
one.
Now, the question becomes, at what temperature does delta G equal zero?
So let's use the information we're given, and use the, if you like, the definition of
delta G in terms of delta H at a given temperature, delta S. So we can substitute in for our two
temperatures our two values of Gibbs free energy.
Let's do that.
So I made the substitution.
I've written this in calories instead of kilocalories.
I'm also assuming that delta H did not change with temperature.
That's usually a pretty good assumption, assuming that delta S also is not changed significantly
with temperature.
And therefore I can take these two equations and solve for delta H and delta S to get a
general expression for delta G as a function of temperature.
And if I just subtract these two equations, I can eliminate delta H, so let's do that,
and so I have a value for delta S. I took the first equation, subtracted it from the
second equation, delta H disappeared.
I can now substitute into either one of these equations this delta S value, and if I do
that I can calculate delta H. So a couple things.
Exothermic reaction.
That means, as I raise the temperature, the equilibrium constant becomes smaller, and
that's why the problem statement says estimate the maximum temperature, because as I go above
that temperature where delta G at standard conditions equals zero, as I go above that
temperature, equilibrium is going to shift back to the left.
It's the maximum temperature because it's exothermic.
And so now we just need to take, starting again, delta G is delta H minus T delta S.
We know delta G is zero, we calculated the values for delta H and delta S, so we can
solve for T. So temperature is 1760 Kelvin.
We're below that temperature, and the reaction is going to the right, we're reducing the
thorium oxide.
Above that temperature, reaction is going to the left, and reaction would continue until
we used up all of one of the products, be it either the calcium oxide or thorium, above
that temperature.
In terms of significant figures, this is probably 1800 K based on the information that we were
given.