Main content

### Course: MCAT > Unit 9

Lesson 19: Equilibrium- Equilibrium questions
- Dynamic equilibrium
- Writing equilibrium constant and reaction quotient expressions
- Le Chȃtelier’s principle: Changing concentration
- Le Chȃtelier’s principle: Changing volume
- Le Chȃtelier’s principle: Changing temperature
- Changes in free energy and the reaction quotient
- Standard change in free energy and the equilibrium constant
- Galvanic cells and changes in free energy

© 2024 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# Standard change in free energy and the equilibrium constant

The relationship between standard Gibbs free energy change and the equilibrium constant K. Calculating K when you know the standard free energy of reaction.

## Want to join the conversation?

- I'm confused by the word "equilibrium". From the previous video it was said that at (delta)G = 0, the reaction is in equilibrium, But in this video, eventhought it question is asking for equilibrium constance, (delta)G = -33 kJ.

Well this is what I understand,

The "equilibrium" indicated by (delta)G = 0 is the equilibrium of spontaneity. It means by the energy and entropy of that environment, the reaction rate will be constant both forward and backward.

The "equilibrium" indicated by equilibrium constant K however, is the equilibrium of the concentration. It determines whether the reaction goes forward or backward depending on the qoutient (concentration of the products and reactants)

For instance,

ex1. the reaction has (delta)G = 0, the reaction still goes forward, if Q<K.

ex2. if Q < Kp (the reaction favours products) but (delta)G > 0, the reaction still doesn't occurs, since the reaction is not spontaneous. (unless, energy is added into the system)

ex3 if Q> Kp (the reaction favours reactants, but (delta)g<0 , the reaction occurs (spontaneous), but there are more reactants than products,

Am i understanding it right? please correct me if i'm wrong.(8 votes)- Equilibrium means that there is no observed change in the concentrations of reactants or products. Think about it as when the reaction has completed, even though it isn't exactly like that. For Example, for the equation:

N2(g) + 3H2(g) --> 2NH3(g)

at Equilibrium, the Reaction might not stop occurring, but the concentrations of N2, H2, and NH3 stay relatively the same.

The K value tells how much the concentrations of products and reactants there are.

If K >1, there are more Products than Reactants in Equilibrium.

If K=1, there are the same amount of Products as Reactants in Equilibrium.

If K <1, there are more Reactants than Products in Equilibrium.

Q is basically what the value of the the ratio of Products to Reactants before Equilibrium is established. so if

Q<K, the Reaction shifts to the Right,

Q>K, the Reaction shifts to the Left,

Q=K, the Reaction is at Equilibrium.

Delta G comes into Play when figuring out if the Reaction is Spontaneous.

when delta G>0, the Reaction is non- Spontaneous, but if

delta G <0, the reaction is spontaneous.

When K<1, the reaction favors the Reactants, so the Reaction is not Spontaneous, making delta G >0. but when K >1, the Reaction favors the Products, so it is Spontaneous, making delta G< 0.

delta G = -RTlnK

R=8.31 Joules/(mole*Kelvin)

Hope that clears up some of your confusion!(23 votes)

- Why the ΔG0 has different values if the temperature is always T=298 K ?(3 votes)
- Thats because, ΔG = ΔH - TΔS

Even if T of environment is constant, but the change in the entropy is dependent on change in disorder. Hence the term TΔS is different for every reaction.(3 votes)

- What is Standard change in free energy?(4 votes)
- I thought dG0 was a constant for the reaction at 25C. To then find the new equilibrium at a new temperature you take e^(-dg0/(RT))... Why does dg0 change in the different examples for the same reaction?(4 votes)
- Is ΔG only applicable to reversible chemical reactions?(2 votes)
- ΔG applies to every reaction, but ΔG = 0 only for a reaction at equilibrium.(3 votes)

- I’m a bit confused about Kp and Kc, how can they be equal since isn’t the formula the Kp is Kp=Kc(RT)^delta N(N is the coefficient of the elements) and N is clearly not 0 here(3 votes)
- At5:39he said "since we're dealing with gases, if you wanted to put in a Kp here you could." But because Kp and Kc aren't interchangeable (related by a factor of (RT)^∆n), how do you know if the value you got is Kp or Kc? I know the example given is with gases, but if I had denoted the K as Kc instead, it would supposedly be wrong, right? How is the K in the larger expression -RTlnK defined?(3 votes)
- Over here, the teacher uses the value of Qp instead of Qc to find the value of change in gibbs free energy, isnt that incorrect because Qp and Qc have different values here?(2 votes)
- You can use either Qp or Qc to find free energy. But you would only use Qp for reactions involving gases and Qc for reactions involving solutes dissolved in solution. It would only be incorrect if they used Qp to find free energy in a reaction where all the states were aqueous because, as you pointed out, Qp and Qc can have different values.

Hope that helps.(2 votes)

- Why would increased T cause delta G naught to increase? I figured it would be more - at greater T(2 votes)
- It will depend on the reaction. Higher T will increase deltaG IF deltaS(change in entropy) is negative, and we can see from the example of Haber process equation that this is indeed the case (less gas molecules)(2 votes)

- Me too confused about Kp and Kc, how can they be equal since isn’t the formula the Kp is Kp=Kc(RT)^delta N(N is the coefficient of the elements) and N is clearly not 0 here(2 votes)
- So you are correct that Δn would not be 0 and therefore the Kc an the Kp for this reaction would be different. However, Jay isn't listing the Kc for any of the answers, only the Kp. He explicitly lists it at5:38explaining that since all the chemical species are gases, only the Kp will be used. He drops using the 'p' in Kp for the later problems explicitly but this is the Haber process, a fairly well known chemical reaction, where we can look up the equilibrium values Kp to determine if they are actually Kp values. The literature values for Kp for the Haber process match the values given in the problem so we can conclude that Jay is exclusively using Kp, not Kc.

Hope that helps.(2 votes)

## Video transcript

- [Voiceover] In the previous video, we looked at the relationship between the change in free energy, delta-G, and the reaction quotient, Q. And we plugged in different values for Q and we saw how that affected
our answer for delta-G. The sign of delta-G told us if a reaction was spontaneous or not. We also said that at equilibrium, Q, the reaction quotient is equal to the equilibrium constant, K. And we plugged K into the equation and solved for delta-G. Delta-G was equal to zero. So, we know, at equilibrium, the change in free energy is equal to zero. So, there's no difference in free energy between the reactants and the products. Let's plug in delta-G is equal to zero into our top equation
here, so, we have zero is, zero is equal to delta-G zero, the standard change in free energy, plus R times T, and since
we're at equilibrium, delta-G is equal to zero, this would be the natural log of the
equilibrium constant, K. So, we solve for delta-G zero. Delta-G zero is equal to negative RT, natural log of K. So, we have another very
important equation to think about. Delta-G zero is the standard
change in free energy, or the change in free energy
under standard conditions. R is the gas constant, T is
the temperature in Kelvin, and K is our equilibrium constant. So, if you're using this equation, you're at equilibrium,
delta-G is equal to zero. And we know at equilibrium,
our equilibrium constant tells us something about
the equilibrium mixture. Alright, do we have more products or do we have more reactants at equilibrium. And this equation relates the equilibrium constant
K to delta-G zero, the standard change in free energy. So, delta-G zero becomes a guide to the ratio of the amount of products to reactants at equilibrium,
because it's related to the equilibrium constant
K in this equation. If you're trying to find the
spontaneity of a reaction, you have to use this equation up here, and look at the sign for delta-G. So, if you're trying to find if a reaction is spontaneous or not, use this equation. If you're trying to find or think about the ratio of the amount
of products to reactants at equilibrium, then you wanna
use this equation down here, and that ratio is related to the standard change in
free energy, delta-G zero. Now we're ready to find
some equilibrium constants. Remember, for a specific temperature, you have one equilibrium constant. So, we're going to find
the equilibrium constant for this reaction at 298 K. So, we're trying to
synthesize ammonia here, and at 298 Kelvin, or 25 degrees C, the standard change in free energy, delta-G zero, is equal to
negative 33.0 kilojoules for this balanced equation. So, let's write down our equation that relates delta-G zero to K. Delta-G zero is equal to negative RT, natural log of K. Delta-G zero is negative 33.0 kilojoules, so, let's write in here, negative 33.0, and let's turn that into joules, so times ten to the third, joules. This is equal to the negative, the gas constant is 8.314 joules over moles times K. So, we need to write over here, joules over moles of reaction. So, for this balanced equation, for this reaction,
delta-G zero is equal to negative 33.0 kilojoules. So, we say kilojoules, or
joules, over moles of reaction just to make our units work out, here. Temperature is in
Kelvin, so we have 298 K, so, we write 298 K in here,
Kelvin would cancel out, and then we have the natural log of K, our equilibrium constant, which is what we are trying to find. So, let's get out the calculator and we'll start with the
value for delta-G zero which is negative 33.0 times 10 to the third. So, we're going to divide that by negative 8.314, and we'd also need to divide by 298. And so, we get 13.32. So, now we have 13.32, right, so our units cancel out here, and this is equal to the natural log of the equilibrium constant, K. So, how do we solve for K here? Well, we would take E to both sides. So, if we take E to the 13.32 on the left, and E to the natural
log of K on the right, this would cancel out
and K would be equal to E to the 13.32, so let's do that. So, let's take E to the 13.32, and that's equal to, this would be 6.1, 6.1 times ten to the one, two, three, four, five. So, 6.1, 6.1 times ten to the fifth. And since we're dealing with gases, if you wanted to put in
a KP here, you could. So, now we have an
equilibrium constant, K, which is much greater than one. And we got this value from a negative value for delta-G zero. So, let's go back up to here, and we see that delta-G
zero, right, is negative. So, when delta-G zero is less than zero, so when delta-G zero is negative, what do we get for our
equilibrium constant? We get that our equilibrium constant, K, is much greater than one. So, what does this tell us
about our equilibrium mixture? This tells us that at equilibrium, the products are favored
over the reactants, so the equilibrium mixture contains more products than reactants. And we figured that out by using
our value for delta-G zero. Let's do the same problem again, but let's say our reaction is
at a different temperature. So now, our reaction is at 464 Kelvin, so we're still trying
to make ammonia here, and our goals is to find the equilibrium constant
at this temperature. At 464 Kelvin, the standard
change in free energy, delta-G zero, is equal to zero. So, we write down our equation, delta-G zero is equal to negative RT, a natural log of the
equilibrium constant, K. And this time, for delta-G
zero, we're plugging in zero. So, zero is equal to, we know that R is the gas constant, and we know that the temperature
here would be 464 Kelvin. So, for everything on the
right to be equal to zero, the natural log of K
must be equal to zero. So, we have zero is equal
to the natural log of K. And now, we're solving for K, we're finding the equilibrium constant. So, we take E to both sides. So, E to the zero is equal
to E to the natural log of K. E to the natural log of
K is just equal to K. So, K is equal to E to the zero, and E to the zero is equal to one. So, when delta-G zero is equal to zero, so let's write this down on here, so, when your standard
change in free energy, delta-G zero, is equal to zero, K is equal to one. And that means that at equilibrium, your products and your
reactants are equally favored. Let's do one more example. So, let's find the
equilibrium constant again at another temperature. So, now we're at 1000 K, and our standard change in free energy, delta-G zero, is equal to
positive 106.5 kilojoules. So, delta-G zero is equal to negative RT, natural log of K. This time, we're putting in
positive 106.5 kilojoules, which is positive 106.5 times ten to the third joules is equal to negative, R is our gas constant, 8.314, I'll leave units out of this just to make it a little bit clearer, times the temperature, which is 1000 K, so this would be 1000 Kelvin times the natural log of the
equilibrium constant, K. So, let's do the math there. We'll start with our delta-G zero, which is 106.5 times ten to the third. So, we're going to take
that value and divide it by negative 8.314, and then, we need to divide by 1000, and that gives us negative 12.81. So, we have negative 12.81 is equal to the natural log of the
equilibrium constant. So, to solve for the equilibrium constant, we take E to both sides, and we get that K is equal to E to the negative 12.81. So, what is that equal to? E to the negative 12.81 is equal to 2.7 times
10 to the negative six. So, K, the equilibrium constant, is equal to 2.7 times ten to the negative six. So, when delta-G zero is positive, when the standard change
in free energy is positive, let's write this one down. So, when delta-G zero
is greater than zero, so, when it's positive, your
equilibrium constant, K, is less than one. Alright, so K is less than one. And we know what that
means at equilibrium. The reactants are favored at equilibrium. So, your equilibrium mixture contains more reactants than products.