In this screen cast we are going to look at a simple gas phase reaction, namely A in the
In this screen cast we are going to look at a simple gas phase reaction, namely A in the

gas phase reacts to 2B molecules in the gas phase.
gas phase reacts to 2B molecules in the gas phase.

And what we are going to do is relate the equilibrium constant to the mole fractions.
And what we are going to do is relate the equilibrium constant to the mole fractions.

And to get the equilibrium constant we have to calculate the Gibbs free energy change
And to get the equilibrium constant we have to calculate the Gibbs free energy change

and tables are available at 25 C, 298.
and tables are available at 25 C, 298.

so for our reaction delta G is the sum of the stoichiometric coefficient times the Gibbs
so for our reaction delta G is the sum of the stoichiometric coefficient times the Gibbs

free energy change of formation for each of the components at standard conditions.
free energy change of formation for each of the components at standard conditions.

This is important, when we say standard conditions for a gas this would be 1 bar pressure, ideal
This is important, when we say standard conditions for a gas this would be 1 bar pressure, ideal

gas, and at the temperature where we are doing the calculation which in this case is 298
gas, and at the temperature where we are doing the calculation which in this case is 298

kelvin.
kelvin.

And for our particular reaction this just becomes stoichiometric coefficient product
And for our particular reaction this just becomes stoichiometric coefficient product

is 2, Gibbs free energy of formation of B minus standard conditions.
is 2, Gibbs free energy of formation of B minus standard conditions.

Gibbs free energy of formation of A, and both of these are formation at 298.
Gibbs free energy of formation of A, and both of these are formation at 298.

Now this Gibbs free energy formation at 298, divided by the gas constant and the temperature
Now this Gibbs free energy formation at 298, divided by the gas constant and the temperature

is the log of the equilibrium constant.
is the log of the equilibrium constant.

And we want token in mind that the equilibrium constant is dimensionless.
And we want token in mind that the equilibrium constant is dimensionless.

That is why we can take a log in this expression.
That is why we can take a log in this expression.

We can write this equilibrium constant and this delta G as minus the sum, again stoichiometric
We can write this equilibrium constant and this delta G as minus the sum, again stoichiometric

coefficients of log of the fugacity of each component in the mixture, divided by the fugacity
coefficients of log of the fugacity of each component in the mixture, divided by the fugacity

of the pure component, again at standard conditions, the zero indicating standard conditions.
of the pure component, again at standard conditions, the zero indicating standard conditions.

So important points, this is mixture.
So important points, this is mixture.

And this is pure component at standard conditions.
And this is pure component at standard conditions.

And then we can write this expression, I am going to take this stoichiometric coefficient,
And then we can write this expression, I am going to take this stoichiometric coefficient,

make it an exponential.
make it an exponential.

So I have a minus sign.
So I have a minus sign.

Sum of the log of the fugacity ratio, raised to the stoichiometric coefficient.
Sum of the log of the fugacity ratio, raised to the stoichiometric coefficient.

So let me re-write this and then we will do some more simplification.
So let me re-write this and then we will do some more simplification.

So I have the sum of the logs which I can re-write as the log of the products.
So I have the sum of the logs which I can re-write as the log of the products.

And this fugacity ratio, raised to the stoichiometric coefficient.
And this fugacity ratio, raised to the stoichiometric coefficient.

And I am able to do this because essentially what I am using as the mathematical relationship
And I am able to do this because essentially what I am using as the mathematical relationship

is the log of a plus the log of a is the log of a times b.
is the log of a plus the log of a is the log of a times b.

So if i compare these terms inside the log I am saying the equilibrium constants equal
So if i compare these terms inside the log I am saying the equilibrium constants equal

the product of the fugacity ratio, raised to the stoichiometric coefficient.
the product of the fugacity ratio, raised to the stoichiometric coefficient.

We could also write this as the product of the activities raised to their stoichiometric
We could also write this as the product of the activities raised to their stoichiometric

coefficients.
coefficients.

So for the reaction A goes to 2B, this fugacity ration becomes the fugacity of B in the mixture.
So for the reaction A goes to 2B, this fugacity ration becomes the fugacity of B in the mixture.

Divided by fugacity of B, standard state, squared and the fugacity of A in the mixture,
Divided by fugacity of B, standard state, squared and the fugacity of A in the mixture,

fugacity of pure A at standard state.
fugacity of pure A at standard state.

If we have an ideal gas, this fugacity in the mixture is just the mole fraction times
If we have an ideal gas, this fugacity in the mixture is just the mole fraction times

the pressure.
the pressure.

Now the standard state fugacity, which we already said that is an ideal gas at 1 bar,
Now the standard state fugacity, which we already said that is an ideal gas at 1 bar,

so that means it's value is 1 bar.
so that means it's value is 1 bar.

This is squared.
This is squared.

Same thing in the denominator.
Same thing in the denominator.

Divided by 1 bar.
Divided by 1 bar.

So the convention is we don't continue to carry through this one bar but we have to
So the convention is we don't continue to carry through this one bar but we have to

remember that pressure must be now expressed in bar for this to make sense.
remember that pressure must be now expressed in bar for this to make sense.

So we have yB^2 over yA times pressure.
So we have yB^2 over yA times pressure.

And we remember that this is dimensionless.
And we remember that this is dimensionless.

So we are understanding that this is pressure divided by 1 bar.
So we are understanding that this is pressure divided by 1 bar.

So we have to write this in terms of bar.
So we have to write this in terms of bar.

Well we could rewrite this to help visualize how things change when we change pressure.
Well we could rewrite this to help visualize how things change when we change pressure.

Gas phase reaction.
Gas phase reaction.

This says if we increase the pressure, So left side smaller, then we must decrease the
This says if we increase the pressure, So left side smaller, then we must decrease the

mole fraction of B so this means the reaction goes to the left when the pressure increases.
mole fraction of B so this means the reaction goes to the left when the pressure increases.

Of course when I write this I should write this as a reversible reaction.
Of course when I write this I should write this as a reversible reaction.

And so, important things to keep in mind, this equilbrium constant is dimensionless
And so, important things to keep in mind, this equilbrium constant is dimensionless

an pressure can have a significance effect on the equilibrium conversion when there is
an pressure can have a significance effect on the equilibrium conversion when there is

a phase change for a gas phase reaction.
a phase change for a gas phase reaction.