### Video Transcript

In this video, we’ll learn about the equilibrium constant for concentration, we’ll learn how to construct the equilibrium constant for any reaction, and we’ll learn how to calculate it.

Let’s take a look at this equilibrium reaction. Our reactant, dinitrogen tetroxide, is a colorless gas, while our product, nitrogen dioxide, is a yellow-brown gas. Let’s say we put some of the dinitrogen tetroxide gas in a flask and then seal the flask so that we have a closed system, which means that it can’t exchange matter with the surroundings. Over time, we would observe the formation of some yellow-brown gas in the flask. This yellow-brown gas is of course the product nitrogen dioxide. Eventually, we’d see that the amount of yellow-brown gas in the flask is constant over time. In other words, the concentration of nitrogen dioxide will be constant.

At this point, we know that equilibrium has been reached because for any reaction occurring in a closed system, equilibrium is reached when the concentration of our products and reactants are no longer changing. Because the concentrations of our reactants and products are constant at equilibrium, we should be able to use these constant concentrations to somehow quantify equilibrium for a reaction. Quantifying equilibrium is exactly the job of the equilibrium constant. For a general reaction, where these small letters represent the stoichiometric coefficients and the capital letters represent the chemical species involved in the reaction, you can create an expression for the equilibrium constant for concentration as follows.

We’ll start off with the equilibrium concentrations of the products in the numerator. We’ll put the equilibrium concentrations of the reactants in the denominator. Concentrations, as indicated by the square brackets, might be in units of molar, moles per liter, or moles per cubic decimeter To complete our equilibrium constant expression, we need to raise all the concentrations to the power of their stoichiometric coefficients.

It’s important that our reaction is occurring in a closed system. Otherwise, the concentrations of our products and reactants might change over time if they are able to escape the container. You’ll often see the temperature specified for the equilibrium constant. This is because the position of equilibrium can shift to the left or right depending on the temperature if the reaction is endo- or exothermic.

When constructing our equilibrium constant expression, we don’t include any chemical species that are solids or liquids. This is because the concentration of solids and liquids don’t change very much over the course of the reaction. We’ll only include aqueous or gaseous chemical species in our equilibrium constant expression. Let’s try constructing the equilibrium constant expression for the reaction between dinitrogen tetroxide and nitrogen dioxide. The concentration of the product, nitrogen dioxide, goes on top, and the concentration of our reactant, dinitrogen tetroxide, will go in the denominator.

Nitrogen dioxide has a stoichiometric coefficient of two, so we’ll raise its concentration to the power of two. Dinitrogen tetroxide has an implied stoichiometric coefficient of one, so we’ll raise the concentration to the power of one. Or we could just leave it off because raising anything to the power of one doesn’t change its value.

So, here is the equilibrium constant expression for this reaction, the concentration of nitrogen dioxide squared divided by the concentration of dinitrogen tetroxide. From here, we could easily calculate the value of the equilibrium constant if we knew the equilibrium concentrations of nitrogen dioxide and dinitrogen tetroxide. If we were to do that, what exactly would the value of the 𝐾 c for this reaction tell us? The value of the equilibrium constant is like a ratio of the amount of products to the amount of reactants at equilibrium. However, it’s not the true ratio because we’re raising the concentrations to the powers of the stoichiometric coefficients. The value of 𝐾 c is always the same for a reaction no matter how much of the reactants or products we start off with. This means that 𝐾 c can tell us information about the extent of the reaction.

To understand what we mean by extent of reaction, let’s say we have a very simple equilibrium reaction between a reactant and a product. The equilibrium constant expression for this reaction will be the concentration of the product divided by the concentration of the reactant. If the value of 𝐾 c for this reaction is greater than one, that means the value in the numerator is larger than the value in the denominator. In other words, the concentration of the products at equilibrium is greater than the concentration of the reactants. So, the products are favored for this reaction, or you might say that the equilibrium lies to the right.

If the value of 𝐾 c is less than one, that means that the value in the denominator is larger than the value in the numerator. In other words, the concentration of reactants is greater than the concentration of products at equilibrium. So, the reactants are favored in this case, or you might say the equilibrium lies to the left. Though the value of 𝐾 c doesn’t increase linearly, we can generally say that a very large 𝐾 c corresponds to there being even more products at equilibrium, and a very small 𝐾 c corresponds to there being even more reactants at equilibrium.

As an extreme example, here we have the reaction of water forming hydrogen gas and oxygen gas. The 𝐾 c for this reaction is very small, eight times 10 to the minus 41. In this case, our reactant, water, is so favored that it generally doesn’t form hydrogen and oxygen gas, and we say that there is no reaction that occurs.

On the other end of the spectrum, we have a very large 𝐾 c, 1.3 times 10 to the sixth. Here, our products are so favored that we generally say the dissociation of hydrochloric acid into hydrogen and chlorine ions goes to completion. Now, let’s see how we can solve problems that involve the equilibrium constant. Perhaps the simplest type of problem we’ll encounter is one where we’re directly given the equilibrium concentrations of our products and reactants. That’s the case in this problem.

Calculate 𝐾 c for the reaction N2 gas plus three H2 gas in equilibrium with two NH3 gas if the equilibrium mixture contains 0.982 moles per cubic decimeter of NH3 gas and 0.193 moles per cubic decimeter each of N2 gas and H2 gas.

The reaction in this problem is part of the Haber process, where ammonia, which is used to create fertilizers, is synthesized from nitrogen and hydrogen gases. This problem gives us the equilibrium concentrations, so we can easily calculate 𝐾 c by plugging these values into the 𝐾 c expression. Let’s start this problem off by creating our 𝐾 c expression. The concentration of our product ammonia goes on top, and the concentration of our reactants go on the bottom. And we’ll raise all of these concentrations to the power of their stoichiometric coefficients. Now that we have this expression, we can solve for 𝐾 c by plugging in our equilibrium concentrations.

Now, we just have a little bit of math to do. Let’s start off by raising everything to the appropriate powers and then combining those terms on the bottom, which will give us this expression. Dividing 0.964324 by 0.001387 gives us 695.01 for the numerical portion of our answer, which we can round to 695. Now, let’s deal with the units. Recall that we can simplify exponents that have the same base by subtracting the exponent in the denominator from the exponent in the numerator.

For the moles, we have an exponent of two in the numerator and four in the denominator. Two minus four leaves us with the power of negative two. For the decimeters, we have the power of negative six in the numerator and negative 12 in the denominator. Negative six minus negative 12 gives us the power of six. So, the 𝐾 c for the reaction in this problem is 695 moles to the negative two decimeters to the sixth.

Though we just calculated the units for 𝐾 c, which many problems will ask us to do, 𝐾 c is actually a unitless quantity. Although the reasons for this are beyond the scope of this video, however, at higher levels, a discussion on dimensionality will explain why this is. Often, problems asking us to solve for 𝐾 c will give us initial concentrations. These types of problems are an extra challenge because we need to solve for the equilibrium concentrations first before we can solve for 𝐾 c.

Let’s work an example.

Hydrogen sulfide can be decomposed by heat according to the following equation: two H2S gas in equilibrium with two H2 gas plus S2 gas. 0.50 mole of hydrogen sulfide was heated at a steady temperature in a container of volume 2.0 cubic decimeters until an equilibrium was reached, where 0.38 mole of H2S still remained. Calculate 𝐾 c for this reaction.

This problem has given us the initial amount of our reactant, the equilibrium amount of our reactant, and the volume of the reaction container. But we don’t have any information about the equilibrium amounts or concentrations of our products. We can use an ICE table to help us solve for the equilibrium amounts of the chemical species in our reaction. The I in ICE stands for initial. These are the initial amounts or concentrations of our reactants and products. The problem told us we started off with 0.50 moles of hydrogen sulfide, and we didn’t start off with any hydrogen or sulfur gas. C in ICE stands for change. This is the amount that each chemical species changes by on the way to equilibrium. The amount of hydrogen sulfide gas will decrease to form the products. So, the change here will be negative.

Hydrogen and sulfur gas will be formed during this reaction, so the change for these will be positive. Each chemical species will change by some amount 𝑥, which is multiplied by the value of the stoichiometric coefficient. E in ICE stands for equilibrium. This is the equilibrium amount or concentration of each chemical species. This will be equal to the initial amount or concentration of the chemical species plus the amount it changes by over the course of the reaction.

The problem also gave us the equilibrium amount of a hydrogen sulfide. If we set this amount equal to the equilibrium value of hydrogen sulfide in the ICE table, we should be able to use that to solve for 𝑥. First, we’ll subtract 0.5 from both sides. Multiplying both sides by negative one gives us two 𝑥 is equal to 0.12, where 𝑥 is equal to 0.06. Now, we have the equilibrium amounts of all the chemical species, but we need equilibrium concentrations to solve for 𝐾 c. The problem gave us the volume of the container, which we can use to find the concentration. Let’s add another row to our ICE table to solve for the concentration.

We just need to take those equilibrium amounts that we just found and divide it by the volume of the container. Solving for the concentration of H2S gas gives us 0.19 moles per cubic decimeter. We can repeat that for the other chemical species to find their concentrations, which gives us 0.06 for the equilibrium concentration of hydrogen gas and 0.03 for the equilibrium concentration of sulfur gas.

Let’s create our 𝐾 c expression. We have the concentration of hydrogen gas squared times the concentration of sulfur gas divided by the concentration of hydrogen sulfide, which is squared. Now, we can plug those equilibrium concentrations in. Next, we can raise everything to the appropriate powers and combine those two terms on the top. This gives us 0.0029916 mole per cubic decimeter. Converting our answer into scientific notation and rounding gives us 3.0 times 10 to the minus three moles per cubic decimeter for the value of 𝐾 c for this reaction.

Let’s wrap up this video by summarizing the important points. The equilibrium for a reaction can be quantified. For a general reaction, we can create an expression for 𝐾 c by dividing the concentrations of the products by the concentrations of the reactants and raising everything to the power of their stoichiometric coefficients. 𝐾 c indicates the extent of reaction. A 𝐾 c for a reaction with a value less than one tells us that the reactants are favored. Equilibrium lies to the left-hand side. A value of 𝐾 c that is greater than one for a reaction tells us that the products are favored. Equilibrium lies to the right-hand side.