# Relationship of delta and keq

### Equilibrium Constant from Delta G The change in Gibbs Free Energy (ΔG) for any reaction is related to the equilibrium constant Keq by the simple equation: ΔG=−RTlnKeq. The Relationship Between Enthalpy (H), Free Energy (G) and Entropy (S) . When delta G standard equals Zero, the reaction is at Equilibrium. H - delta (TS). If the reaction is run at constant temperature, this equation can be . This equation allows us to calculate the equilibrium constant for any reaction.

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?

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 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. 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 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 K, and our standard change in free energy, delta-G zero, is equal to positive So, delta-G zero is equal to negative RT, natural log of K. This time, we're putting in positive So, let's do the math there.

We'll start with our delta-G zero, which is So, we're going to take that value and divide it by negative 8. So, we have negative 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 So, what is that equal to?

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

E to the negative 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.

Relationship B/w Equilibrium Constant K, Reaction Quotient Q and Gibbs Energy G- Equilibrium (15)

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. 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.

• How does delta G affect keq?
• 19.7: Free Energy and the Equilibrium Constant

Data on the far right side of this figure describe systems in which there is more product than reactant. The sign of G is now positive and the magnitude of G is moderately large. The sign of G tells us that the reaction would have to shift to the left to reach equilibrium.

## Standard change in free energy and the equilibrium constant

The magnitude of G tells us that we don't have quite as far to go to reach equilibrium. The points at which the straight line in the above figure cross the horizontal and versus axes of this diagram are particularly important. The straight line crosses the vertical axis when the reaction quotient for the system is equal to 1. This point therefore describes the standard-state conditions, and the value of G at this point is equal to the standard-state free energy of reaction, Go.