Worked examples: Calculating equilibrium constants | Equilibrium | AP Chemistry | Khan Academy

An equilibrium constant can be calculated from experimentally measured concentrations or partial pressures of reactants and products at equilibrium. As an example, let's look at the reaction where N2O4 in the gaseous state turns into 2NO2, also in the gaseous state.

Let's say we do an experiment and we allow this reaction to come to equilibrium at a temperature of 100 degrees Celsius. At equilibrium, the concentration of NO2 is 0.0172 molar, and the concentration of N2O4 is 0.00140 molar.

To calculate the equilibrium constant for this reaction at 100 degrees Celsius, we first need to write the equilibrium constant expression. We can write the equilibrium constant expression by using the balanced equation. We start by writing the equilibrium constant, which is symbolized by K. Since we're dealing with concentrations, we're calculating Kc.

Kc is equal to the products over reactants. This would be the concentration of NO2, and since there's a coefficient of 2 in front of NO2, this is the concentration of NO2 raised to the second power divided by the concentration of our reactant N2O4. Since there's an implied 1 in front of N2O4, this is the concentration of N2O4 raised to the first power.

Next, we plug in our equilibrium concentrations. The equilibrium concentration of NO2 is 0.0172, so let's plug that in. This is equal to (0.0172)^2 divided by the equilibrium concentration of N2O4, which was 0.00140. Therefore, we plug that in as well, 0.00140.

When we solve this, we get that Kc is equal to 0.211, and this is at 100 degrees Celsius. It's important to always give the temperature when you're giving a value for an equilibrium constant because an equilibrium constant is only constant for a particular reaction at a particular temperature.

It's also important to note that K, the equilibrium constant, doesn't have any units. So we would just say that Kc is equal to 0.211 at 100 degrees Celsius for this particular reaction.

Let's calculate the equilibrium constant for another reaction. In this reaction, carbon dioxide reacts with hydrogen gas to produce carbon monoxide and H2O. Since everything is in the gaseous state, experimentally it's easier to work with partial pressures than to work with concentrations.

So instead of calculating Kc, we're going to calculate Kp, where the "P" stands for pressure. We're trying to find Kp at 500 Kelvin for this reaction. To help us find Kp, we're going to use an ICE table, where "I" stands for the initial partial pressure in atmospheres, "C" stands for the change in the partial pressure, also in atmospheres, and "E" is the equilibrium partial pressure.

Let's say that a mixture of carbon dioxide, hydrogen gas, and H2O are placed in a previously evacuated flask and allowed to come to equilibrium at 500 Kelvin. Let's say the initial measured partial pressures are 4.10 atmospheres for carbon dioxide, 1.80 atmospheres for hydrogen gas, and 3.20 atmospheres for H2O. Since we didn't add any carbon monoxide in the beginning, the initial partial pressure of that would be zero.

After the reaction comes to equilibrium, we measure the partial pressure of H2O to be 3.40 atmospheres. So that's why we have 3.40 in the equilibrium part on the ICE table under H2O. The initial partial pressure of H2O is 3.20 atmospheres, and the equilibrium partial pressure is 3.40. So H2O has increased in partial pressures.

We can go ahead in here and write plus x for an increase in the partial pressure of H2O, and 3.20 plus x must be equal to 3.40. So x is equal to 0.20. Therefore, the partial pressure of water increased by 0.20. We could either write plus x in here on our ICE table, or we could just write plus 0.20.

Now that we know the change in the partial pressure for H2O, we can use this information to fill out the rest of our ICE table. For example, the mole ratio of carbon monoxide to H2O is 1 to 1. So if we gained plus 0.20 for H2O, we're also going to gain plus 0.20 for carbon monoxide.

If we're gaining for our two products here, the net reaction is moving to the right to increase the amount of products, which means we're losing reactants. We can figure out by how much by looking at the mole ratios again. For both of our reactants, we have ones as coefficients in the balanced equation.

So if it's plus x for both of our products, it must be minus x for both of our reactants. Since x is 0.20, it'd be minus 0.20 for the change in the partial pressure for both of our reactants. Therefore, the equilibrium partial pressure of carbon dioxide would be 4.10 minus 0.20, which is 3.90, and for H2, it would be 1.80 minus 0.20, which is 1.60.

For carbon monoxide, we started off with zero, and we gained positive 0.20. Therefore, the equilibrium partial pressure is 0.20. As the net reaction moved to the right, we lost some of our reactants and we gained some of our products until the reaction reached equilibrium, and we got our equilibrium partial pressures.

At equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction, and therefore these equilibrium partial pressures remain constant. Now that we know our equilibrium partial pressures, we're ready to calculate the equilibrium constant Kp.

So we need to write an equilibrium constant expression. Kp is equal to the products over reactants. For our products, we would have the partial pressure of carbon monoxide, and since the coefficient is 1 in front of carbon monoxide in the balanced equation, it'd be the partial pressure of carbon monoxide raised to the first power times the partial pressure of our other product, which is H2O. Once again, the coefficient is 1, so that's the partial pressure raised to the first power.

All of this is divided by we think about our reactants next, and they both have coefficients of 1 in the balanced equation. So it would be the partial pressure of carbon dioxide times the partial pressure of hydrogen gas. The partial pressures in our equilibrium constant expression are the equilibrium partial pressures, which we can get from the ICE table.

So the equilibrium partial pressure of carbon monoxide is 0.20. The equilibrium partial pressure of H2O is 3.40. We can plug in the equilibrium partial pressures for carbon dioxide and the equilibrium partial pressure for hydrogen gas as well. Here we have the equilibrium partial pressures plugged into our equilibrium constant expression, and when we solve this, we get that Kp is equal to 0.11 at 500 Kelvin.