# Derive the relationship between gibbs free energy and equilibrium constant

### Free Energy and the Equilibrium Constant - Chemistry LibreTexts

Learn the concepts of Relationship Between Free Energy And Equilibrium Constant Gibbs energy change ΔG is related to the electrical work done by the cell. Derivation of relationship between equilibrium constant and Gibbs .. It would seem that this violates the definition of the Gibb's free energy. Apr 13, How I derive the relationship between Gibbs free energy function and For a reaction that occurs at constant temperature and pressure, the.

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 K. So, we're trying to synthesize ammonia here, and at Kelvin, or 25 degrees C, the standard change in free energy, delta-G zero, is equal to negative So, let's write down our equation that relates delta-G zero to K. Delta-G zero is negative This is equal to the negative, the gas constant is 8.

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 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 K, so, we write 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 So, we're going to divide that by negative 8. And so, we get So, now we have So, how do we solve for K here?

Well, we would take E to both sides. So, if we take E to the So, let's take E to the 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?

## Connection between \(E_{cell}\), ∆G, and K

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.

• 19.7: Free Energy and the Equilibrium Constant
• Standard change in free energy and the equilibrium constant

So now, our reaction is at Kelvin, so we're still trying to make ammonia here, and our goals is to find the equilibrium constant at this temperature.

At 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. Go therefore describes this reaction only when all three components are present at 1 atm pressure. The sign of Go tells us the direction in which the reaction has to shift to come to equilibrium.

The fact that Go is negative for this reaction at 25oC means that a system under standard-state conditions at this temperature would have to shift to the right, converting some of the reactants into products, before it can reach equilibrium.

### Connection between \(E_{cell}\), ∆G, and K - Chemistry LibreTexts

The magnitude of Go for a reaction tells us how far the standard state is from equilibrium. The larger the value of Go, the further the reaction has to go to get to from the standard-state conditions to equilibrium. Assume, for example, that we start with the following reaction under standard-state conditions, as shown in the figure below. If we could find some way to harness the tendency of this reaction to come to equilibrium, we could get the reaction to do work.

The free energy of a reaction at any moment in time is therefore said to be a measure of the energy available to do work. When a reaction leaves the standard state because of a change in the ratio of the concentrations of the products to the reactants, we have to describe the system in terms of non-standard-state free energies of reaction.

The difference between Go and G for a reaction is important. There is only one value of Go for a reaction at a given temperature, but there are an infinite number of possible values of G.

### Relationship Between Free Energy And Equilibrium Constant - Study Material for IIT JEE | askIITians

The figure below shows the relationship between G for the following reaction and the logarithm to the base e of the reaction quotient for the reaction between N2 and H2 to form NH3. They therefore describe systems in which there is far more reactant than product. The sign of G for these systems is negative and the magnitude of G is large. The system is therefore relatively far from equilibrium and the reaction must shift to the right to reach equilibrium.

Data on the far right side of this figure describe systems in which there is more product than reactant.