# Relationship between equilibrium constant reaction quotient and gibbs energy

### Free Energy and Equilibrium - Chemistry LibreTexts The difference between delta e of the reaction quotient for the reaction G at this point is equal to the standard-state free energy of reaction, delta. Let us consider a reversible reaction aA + bB l L + m M Let the What is the relation between Gibbs energy and equilibrium constant?. Instead, as reverse reaction would take place; ΔG is 0, reaction has achieved equilibrium; at this point, there is no longer any free energy left to.

Isomerization of butane The standard molar free energy change for this very simple reaction is —2. The green curve adds the free energy of mixing to the above sum; its minimum defines the equilibrium composition. The answer is that no matter how low the free energy of the products, the free energy of the system can be reduced even more by allowing some of the products to be "contaminated" i. An interesting corollary of this is that any reaction for which a balanced chemical equation can be written can in principle take place to some extent, however minute that might be.

If you know the entropies and the enthalpies of formation of a set of substances, you can predict the equilibrium constant of any reaction involving these substances without the need to know anything about the mechanism of the reaction. Calculate the equilibrium constant for the reaction from the following thermodynamic data: What does the value of Go tell us about the following reaction?

## 19.7: Free Energy and the Equilibrium Constant

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.

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. We're gonna plug this into our equation and we're gonna solve for delta G. So delta G is equal to delta G zero, the standard change in free energy. At 25 degrees C delta G zero is negative 33 kilojoules. So we have negative We need to make that joules so I'm gonna say times 10 to the third joules. Plus R is the gas constant, so that's 8. Since the gas constant is in joules per mole we need to make the delta G zero in joules per mole too. Next we have the temperature. The temperature of our reaction is 25 degrees C, that needs to be in Kelvin. So 25 degrees C is Kelvin. So we write in here Kelvin, so Kelvin would cancel out here. Then we're going to multiply that by the natural log of Q. The natural log where Q is equal to. Let's do the math.

## Changes in free energy and the reaction quotient

Let's start with the natural log of. The natural log of. We're gonna add that to delta G zero, so plus negative And that should give us our change in free energy, which is equal to negative So delta G is equal to negative It's just easier to think in kilojoules.

And remember this is moles of our reaction, so for this specific reaction where one mole of nitrogen combines with three moles of hydrogen to give us two moles of ammonia.

• Free Energy and Equilibrium

So how it's written, delta G for the reaction how it's written is equal to negative Delta G is negative so we know our reaction is spontaneous in the forward direction. The reaction is spontaneous in the forward direction which means we're going to make more of our products.

Let's think about how far away we are from equilibrium. Q, our reaction quotient, is.

### Gibbs Free Energy

For this reaction at 25 degrees C the equilibrium constant, which would be KP, is equal to 6. Our value for the reaction quotient was. That's much, much smaller. That's much smaller than the equilibrium constant. So Q is much smaller than K.

So we know we have too many reactants and too few products. We have a driving force, delta G is negative. We have a driving force to make more of our products. So the reaction moves forward to make more products here. So we know that, by just looking at the sign for delta G, which we know is spontaneous. We also know that from comparing the reaction quotient Q to K. So because Q is not equal to K we're not at equilibrium here.

We are away from equilibrium and we have a spontaneous reaction. The reaction proceeds to the right, we make of-- overall, I should say. We make more of our products and we lose some of our reactants.

What happens to Q? As the reaction progresses we make more of our products so the numerator should increase, and we lose some of our reactants so the reactants should decrease. And so you can think about that, an increase in the numerator, a decrease in the denominator, that means that as the reaction progresses, as the reaction moves to the right to make more products, Q should increase. So as Q increases what happens to delta G?

Let's think about that next. If we get an increase in Q what happens to the change in free energy?