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## AP®︎/College Chemistry

### Course: AP®︎/College Chemistry > Unit 9

Lesson 5: Free energy and equilibrium# Free energy and equilibrium

The standard change in free energy, Δ

*G*°, for a reaction is related to its equilibrium constant,*K*, by the equation Δ*G*° = -*RT*ln*K*. When Δ*G*° < 0,*K*> 1, and the reaction is product-favored at equilibrium. When Δ*G*° > 0,*K*< 1, and the reaction is reactant-favored at equilibrium. When Δ*G*° = 0,*K*= 1, and the reaction is at equilibrium under standard state conditions. Created by Jay.## Want to join the conversation?

- Isn't ΔG° constant? How can it be different for different temperatures, if it's defined to be for 25°C(4 votes)
- The standard state simply refers to a particular special state of the reactants and products, and this has nothing to do with temperature.

These special states are:

1) liquid/solid -> must be pure and at 1 atm.

2) gas -> must be at 1 atm.

3) solution -> must be 1 molar in concentration.

When a chemical reaction occurs and both the products and reactants have these states, we say it occurs at STP.

What about the temperature? It can be anything, but by convention, we use 25 degrees Celsius (298K).(1 vote)

- Is this equation somehow related to the Arrhenius equation?(1 vote)

## Video transcript

- [Instructor] Let's say
we have a generic reaction where reactants turn into products. And our goal is to think
about the relationship between free energy and this reaction when it comes to equilibrium. First, we need to consider the equation that allows us to calculate
non-standard changes in free energy delta G. We can think about delta G as the instantaneous
difference in free energy between reactants and products. And from the equation we see that delta G is equal to delta naught plus RT ln of Q. Delta G naught is the
standard change in free energy between reactants and products. This value refers to the
difference in free energy between reactants and products
in their standard states at a specified temperature. R is the ideal gas constant, T is the absolute temperature in Kelvin, and Q is the reaction quotient. At equilibrium, the instantaneous
difference in free energy between reactants and products is zero, which means there's no more
driving force for the reaction. And at equilibrium,
the reaction quotient Q is equal to the equilibrium constant K. So we can plug in zero for
delta G in our equation. And we can plug in K for Q, that gives us zero is
equal to delta G naught plus RT ln of K. If we subtract RT ln of K from both sides, we get delta G naught is equal to negative RT ln of K. This equation is extremely useful because now we have a relationship between free energy and equilibrium. If we know the standard
change in free energy for a particular reaction
at a particular temperature, we can calculate the equilibrium
constant for that reaction. Next, let's go ahead
and go through the math to solve for the equilibrium constant K. To solve for K, we need to first divide
both sides by negative RT. So on the right side, negative RT would cancel out. We could then rewrite it as ln of K is equal to negative delta
G naught divided by RT. And to get rid of the ln, we need to take e to both sides, which gives us the following equation. The equilibrium constant K is equal to e to the negative
delta G naught divided by RT. So if we know the value for delta G naught for a particular reaction, and we know that temperature, we can calculate the equilibrium
constant for that reaction at that particular temperature. Let's calculate the equilibrium constant K for the synthesis of ammonia
from nitrogen and hydrogen. And let's do this for three
different temperatures. Well, let's start with
a temperature of 298K. delta G naught at this temperature is equal to - 33.0 kilojoules
per mole of reaction. So to solve for the
equilibrium constant K, we need to plug in the
value for delta G naught. We also need to plug in
the temperature in Kelvin and the ideal gas constant R. The ideal gas constant is equal to 8.314 joules
per mole of reaction Kelvin. And since we're using joules in the units, we need to make sure to convert
the units for delta G naught into joules per mole of reaction. So - 33.0 kilojoules per mole of reaction is equal to - 33,000
joules per mole of reaction still with three significant digits. And once we plug in our
temperature of 298 Kelvin, notice how Kelvin cancels out and also joules per mole
of reaction cancels out. When don't we do the math, we find that the equilibrium constant K is equal to e to the 13.3 power, which is equal to 6 times 10 to the fifth. Notice how equilibrium
constants don't have units. When e is raised to a power, the result has the same
number of significant figures as there are decimal places in the power. Since we have only one
decimal place in this power, we have one significant
figure for our final answer. Since the value for K for
this reaction at 298 Kelvin is much greater than one, that tells us at equilibrium
there are more products than there are reactants. So there's a lot more ammonia than there is nitrogen or hydrogen, when this reaction reaches
equilibrium at 298 Kelvin. So whenever delta G naught is negative, or you could say delta G
naught is less than zero. The equilibrium constant
is greater than one, and products are favored over
reactants at equilibrium. For the same reaction for
the synthesis of ammonia, let's calculate the equilibrium constant K at a different temperature, this time it's 1000 Kelvin. And that 1000 Kelvin delta
G naught for this reaction is equal to + 106.5 kilojoules
per mole of reaction. Just like the previous example, we're gonna plug everything
into our equation to calculate K. So we need to plug in delta G naught, the temperature which
is 1000 Kelvin this time and the ideal gas constant. So we plug in the ideal gas constant, the temperature and delta G naught. Once again, we had to convert from kilojoules per mole reaction into joules per mole of reaction. So joules per mole of reaction cancels out and Kelvin cancels out. That gives us K is equal
to e to the - 12.81 power, which is equal to 2.7 times 10 to the - 6. Because we have two decimal
places in the power. The final answer is to
two significant figures. For this example, K is much less than one which tells us at equilibrium, there are far more reactants
than there are products. So at 1000 Kelvin, if this reaction at equilibrium we'd have a lot more of our reactants, nitrogen and hydrogen, and only a small amount
of our product ammonia. So whenever delta G naught is positive, or you could say delta G
naught is greater than zero. The equilibrium constant
K is less than one, which means that equilibrium, there are a lot more reactants
than there are products. And for our last example, let's calculate the equilibrium constant K for the same reaction. This time at 464 Kelvin, at that temperature delta G naught for this reaction is equal to zero. Calculating K is a lot
easier for this example, because if we plug in
zero for delta G naught, K is equal to each of the zero
power which is equal to one. And when the equilibrium
constant is equal to one, that means that equilibrium
there's significant amounts of both reactants and products. So if this reaction were to
reach equilibrium at 464 Kelvin, there'll be a significant
amount of both our reactants, which are nitrogen and hydrogen, and our product which is ammonia. Let's summarize what we've
learned about the relationship between free energy and equilibrium. The standard at change in free energy for a reaction delta G naught is related to the equilibrium
constant K by this equation. And if we are using this equation, we know the reaction is at equilibrium because we found that equation by setting the instantaneous
change in free energy delta G equal to zero. It's very easy to think that delta G without the superscript naught, and delta G naught are the same thing. However, they are very
different quantities. For all three of the situations
that we talked about, delta G without the
superscript was equal to zero, because the reaction was at equilibrium. However, because delta G naught is the difference in free energy between reactants and products
and their standard states, delta G naught does not
have to be equal to zero at equilibrium. delta G naught is a constant
at a particular temperature, just like the equilibrium constant K. And if delta G naught is less than zero, K is greater than one, which means at equilibrium, there are more products than reactants. If delta G naught at a
particular temperature is greater than zero. That means the equilibrium
constant is less than one, which means that equilibrium there are more reactants than products. And when delta G naught is
equal to zero K is equal to one, which means significant
amounts of both reactants and products at equilibrium.