- [Instructor] Understanding the concept
of nonstandard free energy changes
is really important when it comes to a chemical reaction.
For this generic chemical reaction,
the reactants turn into the products.
And nonstandard free energy change is symbolized by delta G.
And notice, this delta G
doesn't have the naught superscript,
and therefore, it's nonstandard change in free energy.
The reason why this concept is so important
for a chemical reaction is by calculating
the nonstandard change in free energy,
we can figure out which direction
the net reaction will proceed.
When delta G is less than zero,
so when delta G is negative,
the forward reaction is thermodynamically favored.
Therefore, the reactants will turn into the products.
And the amount of products will increase,
and the amount of reactants will decrease.
When delta G is greater than zero,
so when delta G is positive,
the forward reaction is not thermodynamically favored,
which means the reverse reaction is favored.
So the net reaction goes to the left to increase
the amount of reactants and decrease the amount of products.
As long as there's a difference in free energy
between the reactants and the products,
the net reaction will proceed either to the left
or to the right.
However, when there's no difference in free energy
between the reactants and the products,
or delta G is equal to zero,
the reaction is at equilibrium.
And when the reaction is at equilibrium,
the concentrations of reactants
and products remain constant.
So it's useful to think about nonstandard change
in free energy as the driving force
for a chemical reaction.
As long as there's a difference in free energy
between reactants and products,
the net reaction will move one direction or the other
to the left or to the right.
However, when there's no difference in free energy
between reactants and products,
there's no more driving force
and the reaction is at equilibrium.
Next, let's look at the equation
that relates nonstandard change in free energy, delta G,
to standard change in free energy, delta G naught.
Remember, the superscript naught refers
to the substances being in their standard states
under a pressure of one atmosphere.
So delta G, or the nonstandard change in free energy,
refers to the instantaneous difference
in free energy between the reactants and the products.
So when that reaction moves to the left or to the right,
this value is always changing.
It's the driving force for the reaction.
Delta G naught is a different situation.
Delta G naught is a constant at a certain temperature.
And that's because it's referring to the difference
in free energy between reactants and products
in their standard states.
And so this value will remain the same
as the net reaction moves to the left or to the right.
To calculate the nonstandard change
in free energy, delta G,
it's equal to the standard change
in free energy, delta G naught, plus RT natural log of Q,
where R is the ideal gas constant,
T is the temperature in Kelvin,
and Q is the reaction quotient.
Next, let's calculate delta G for a chemical reaction.
And this is the reaction for the synthesis of ammonia gas
from nitrogen gas and hydrogen gas.
Our goal is to calculate delta G at 25 degrees Celsius
at the moment in time when the partial pressures
of all three gases are one atmosphere.
So one atmosphere for nitrogen, for hydrogen,
and for ammonia.
At 25 degrees Celsius, delta G naught for this reaction
is equal to negative 33.0 kilojoules per mole of reaction.
So our goal is to calculate delta G.
And we know delta G naught,
but we need to figure out what Q is at this moment in time,
when the partial pressures of all the gases
are one atmosphere.
We can get the expression for the reaction quotient Q
from the balanced equation.
So it would be the partial pressure of ammonia.
And since there's a two, it'd be the partial pressure
of ammonia raised to the second power
divided by the partial pressure of nitrogen.
And since there's a one as a coefficient
in the balanced equation,
it's the partial pressure of nitrogen raised
to the first power,
times the partial pressure of hydrogen.
And since we have a three here,
it'd be the partial pressure of hydrogen raised
to the third power.
Since all three gases have a partial pressure
of one atmosphere, when we plug in one atmosphere,
we find that Q is equal to one.
And the natural log of one is equal to zero.
So this second term here in our equation goes to zero.
And at this moment in time,
delta G is equal to delta G naught.
And since delta G naught is equal
to negative 33.0 kilojoules per mole of reaction,
since delta G is negative at this moment in time,
the reaction is thermodynamically favored
in the forward direction.
So the reactants will react together to form
more of the products.
The reason why delta G is equal to delta G naught
at this moment in time is because our reactants and products
are all in their standard states.
By convention for a gas,
standard state refers to the pure gas at a pressure
of one atmosphere.
And since all of the partial pressures are equal
to one atmosphere, that gave us Q is equal to one,
which made the second term equal to zero.
So only when the substances are in their standard states
is delta G equal to delta G naught.
So if we had a different set of partial pressures,
delta G wouldn't be equal to delta G naught.
Let's calculate delta G for the same reaction
at 25 degrees Celsius.
However, this time, instead of having partial pressures
of one atmosphere, all three gasses at this moment in time
have a partial pressure of two atmospheres.
Let's start by calculating the reaction quotient Q.
So when we plug in two atmospheres
for the partial pressures of our gasses,
we find that Q is equal to .25.
Next, to calculate delta G, we need to plug everything
into our equation.
So delta G naught is still equal to negative 33.0 kilojoules
per mole of reaction.
So we can see, we plugged that in here.
The ideal gas constant R is 8.314 joules
per Kelvin mole of reaction.
But to keep our units the same in kilojoules,
you can see I've converted the ideal gas constant
into kilojoules per Kelvin mole of reaction.
So it's .008314.
We said the temperature was 25 degrees Celsius.
So 25 plus 273 is equal to 298 Kelvin.
And the reaction quotient Q is equal to .25.
So be the it'd natural log of .25.
So Kelvin will cancel out.
And when we do the calculation,
we find that delta G is equal
to negative 36.4 kilojoules per mole of reaction.
Since delta G is negative at this moment in time,
the reaction is thermodynamically favorable
in the forward direction.
So the net reaction will move to the right
to make more products,
and to decrease the amount of reactants.
An increase in the amount of ammonia means an increase
in the partial pressure of ammonia.
And a decrease in the amount of reactants
means a decrease in the partial pressures
of the reactants.
Therefore, when the net reaction goes to the right,
the reaction quotient Q increases.
So we started with a relatively small value for Q of .25.
And we know that Q is going to increase.
Let's just jump ahead to a point,
and let's say that Q is equal to 1.0 times 10 to the third,
and let's plug that into our equation
and see what happens to delta G.
So delta G naught stays the same
at negative 33.0 kilojoules per mole of reaction.
The ideal gas constant's the same.
The temperature's the same.
So all we've done is increase Q.
Once again, Kelvin cancels out,
and we find that delta G,
the instantaneous difference in free energy,
is equal to negative 15.9 kilojoules per mole of reaction
at this moment in time.
Since delta G is negative, once again,
the forward reaction is favored.
So the reaction moves to the right to increase products
and to decrease reactants.
So again, Q will continue to increase.
And think about what's happening to delta G.
It was negative 36.4, and now it's negative 15.9.
And as Q keeps increasing, delta G keeps getting closer
and closer to zero.
Q will keep increasing until delta G is equal to zero
and the reaction is at equilibrium.
So as long as Q is not equal to the equilibrium constant K
for this reaction, the reaction is not at equilibrium.
But eventually, Q will be equal
to the equilibrium constant K.
And if we were to plug that value into our equation,
we would find that delta G is equal to zero.
And therefore, the reaction is at equilibrium.
Let's summarize the difference between
the nonstandard change in free energy, delta G,
and the standard change in free energy, delta G naught.
Delta G naught is talking about the difference
in free energy between reactants and products
when reactants and products are in their standard states.
And since that value is a constant,
if the temperature is constant,
notice for all of our calculations,
delta G naught was equal
to negative 33.0 kilojoules per mole of reaction.
Delta G is talking about the instantaneous difference
in free energy between reactants and products.
And as long as there's a difference in free energy,
there's a driving force for the net reaction
to go to the left or to the right.
And eventually, when delta G is equal to zero,
the reaction is at equilibrium.
Delta G is only equal to delta G naught
when the reactants and products
are in their standard states.