Free energy and equilibrium | Applications of thermodynamics | AP Chemistry | Khan Academy

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, ΔG.

We can think about ΔG as the instantaneous difference in free energy between reactants and products. From the equation, we see that ΔG is equal to ΔGº plus RT natural log of Q. ΔGº 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. At equilibrium, the reaction quotient Q is equal to the equilibrium constant K. So we can plug in zero for ΔG in our equation and we can plug in K for Q. That gives us 0 = ΔGº + RT natural log of K.

If we subtract RT natural log of K from both sides, we get ΔGº = -RT natural log 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 -RT. On the right side, -RT would cancel out. We can then rewrite it as natural log of K = -ΔGº / RT. To get rid of the natural log, we need to take e to both sides, which gives us the following equation: K = e^(-ΔGº / RT).

If we know the value for ΔGº for a particular reaction and we know the 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.

Let's start with a temperature of 298 K. ΔGº at this temperature is equal to -33.0 kJ/mol of reaction. So to solve for the equilibrium constant K, we need to plug in the value for ΔGº. 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 J/(mol·K).

Since we're using joules in the units, we need to make sure to convert the units for ΔGº into joules per mole of reaction. So -33.0 kJ/mol of reaction is equal to -33,000 J/mol of reaction, still with three significant digits. Once we plug in our temperature of 298 K, notice how K cancels out, and also J/(mol·reaction) cancels out.

When we do the math, we find that the equilibrium constant K is equal to e^13.3 power, which is equal to 6 × 10^5. 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 K is much greater than one, that tells us that 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 K.

So, whenever ΔGº is negative, or you could say ΔGº is less than zero, the equilibrium constant is greater than one, and products are favored over reactants. 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 K. At 1000 K, ΔGº for this reaction is equal to +106.5 kJ/mol of reaction.

Just like the previous example, we're going to plug everything into our equation to calculate K. So we need to plug in ΔGº, the temperature (which is 1000 K this time), and the ideal gas constant. We plug in the ideal gas constant, the temperature, and ΔGº. Once again, we had to convert from kJ/(mol·reaction) into J/(mol·reaction), so J/(mol·reaction) cancels out and K cancels out.

That gives us K = e^(-12.81 power), which is equal to 2.7 × 10^-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 that at equilibrium, there are far more reactants than there are products. So, at 1000 K, if this reaction were at equilibrium, we'd have a lot more of our reactants (nitrogen and hydrogen) and only a small amount of our product (ammonia).

Whenever ΔGº is positive, or you could say ΔGº is greater than zero, the equilibrium constant K is less than one, which means at equilibrium, there are a lot more reactants than there are products. For our last example, let's calculate the equilibrium constant K for this same reaction, this time at 464 K. At that temperature, ΔGº for this reaction is equal to zero.

Calculating K is a lot easier for this example because, if we plug in zero for ΔGº, K = e^0 power, which is equal to one. When the equilibrium constant is equal to one, that means at equilibrium there's significant amounts of both reactants and products. So, if this reaction were to reach equilibrium at 464 K, there'd be a significant amount of both our reactants (nitrogen and hydrogen) and our product (ammonia).

Let's summarize what we've learned about the relationship between free energy and equilibrium. The standard change in free energy for a reaction, ΔGº, is related to the equilibrium constant K by this equation. 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, ΔG, equal to zero.

It's very easy to think that ΔG without the superscript “º” and ΔGº are the same thing; however, they are very different quantities. For all three of the situations that we talked about, ΔG without the superscript was equal to zero because the reaction was at equilibrium. However, because ΔGº is the difference in free energy between reactants and products in their standard states, ΔGº does not have to be equal to zero at equilibrium.

ΔGº is a constant at a particular temperature, just like the equilibrium constant K. If ΔGº is less than zero, K is greater than one, which means at equilibrium there are more products than reactants. If ΔGº at a particular temperature is greater than zero, that means the equilibrium constant is less than one, which means at equilibrium there are more reactants than products. When ΔGº is equal to zero, K is equal to one, which means significant amounts of both reactants and products at equilibrium.