Equilibrium Constants

As its name suggests, “equilibrium” is a balance – a balance between the forward and reverse rates of a chemical reaction. But because of the relationship between rate and concentration, the balance exists also between the concentrations of the reactants and products.

We have learned that, at equilibrium, the concentrations of reactants and products stop changing (because the forward and reverse rates equalize). We can expect that for any given reaction, the concentrations of the reactants and the products at equilibrium will be related to one another, and that relationship will be determined by the intrinsic rates of the forward and reverse reactions. For a reaction with a fast forward rate and a slow reverse rate, for example, we expect that at equilibrium, the concentrations of the products will be high and that of the reactants will be low.

It’s much easier to measure concentrations than rates, and, as it turns out, it’s much easier to characterize the situation at equilibrium using concentrations. In fact, all you need is a balanced equation and measured values for the concentrations of reactants and products at equilibrium. We will now learn how to analyze reactant and product concentrations at equilibrium using a reaction's equilibrium constant.

Without knowing it, you have already encountered equilibrium constants in the form of K_s (from lesson 3) and K_a and K_b (from Lesson 7). These constants (the solubility product constant and the acid/base dissociation constant) were situational examples of a broader concept: for any equilibrium process, an equilibrium constant K can be associated with it, which governs the reactant and product concentrations. It may be useful to review those topics now.

Writing Equilibrium Constants

Consider the reaction below.

N₂ (g) + 3 H₂ (g) ⇌ 2 NH₃ (g)

It is a reversible reaction, as indicated by the double arrows, which means it will come to equilibrium when the forward and reverse rates are equal to each other.

Leaving aside the effects of pressure and temperature, the only thing that will affect the reverse rate of reaction is the concentration of NH₃, since it is the only product. The forward rate of the reaction, however, will be affected by both the concentrations of N₂ and H₂. So, if we had a given concentration of NH₃, we could imagine many combinations of N₂ and H₂ concentrations that could be in equilibrium with it. However, if a given equilibrium system has more of one reactant, say N₂, then it would need to have less of the other reactant in order for the forward rate to be the same. We would also expect, since three and H₂ molecules are involved in the reaction, that the H₂ concentration would affect the forward rate more strongly.

Also, if we had an equilibrium system with less NH₃, then we would need to have correspondingly less N₂ and H₂, otherwise the forward reaction would be faster than the reverse, and we would not be at equilibrium. This means that the concentrations of all three reactants must be related to each other mathematically, with the amounts of the reactants inversely proportional to each other, with more weight given to substances with larger coefficients, and with the overall amount of products inversely proportional to the overall amount of reactants.

This relationship is captured by writing an equilibrium constant for the reaction. The equilibrium constant takes the form of a fraction, where the product concentrations go on top (in the numerator) and the reactant concentrations go on the bottom (in the denominator), multiplied together. Each substance concentration is raised to the power of its coefficient (powers of one are usually assumed by default, like coefficients of one). And the fraction is all set equal to a number (K, or sometimes K_eq) which has a fixed value. You can see this illustrated below (remember that the brackets around a substance represent its concentration).

K_eq = [NH₃]²/[N₂]*[H₂]³

The equilibrium constant, K_eq, has a value that is influenced by the temperature and by the total pressure. In this lesson, we will ignore the variation of K_eq with T and P and assume in the calculations and problems that we do that these variables don’t change.

One important thing to note about writing equilibrium constants is that we only include substances in the gas or aqueous phase in the expression, not solids or liquids. We observed this when writing K_sp and K_a expressions in Lessons 3 and 7.

Na₂S (s) ⇌ 2 Na⁺ (aq) + S^2- (aq) HF (aq) + H₂O (l) ⇌ H₃O⁺ (aq) + F^- (aq)

K = [Na⁺]²*[ S^2-] K = [H₃O⁺]*[ F^-]/[HF]

As you may remember, the solid sodium sulfide is omitted from its K_sp expression, and the liquid water is omitted from HF's K_a expression. In the case of the K_sp, this results in a denominator of just 1, which we simplify away.

The reason for this is that, in the case of pure solids and liquids, they are not distributed evenly throughout the reaction space, so changing the amount of a solid or liquid does not change its effective concentration in a reaction. In the case of solvents (like water above), the solvent is spread throughout the reaction space, but because there is so much of it, its concentration does not appreciably vary, and so we generally ignore it (this introduces a very small amount of error in calculations, but makes them much simpler).

Practice

Write equilibrium constant expressions for the following reactions.

1. Pb(OH)₄^2- (aq) ⇌ Pb²⁺ (aq) + 4 OH^- (aq)

2. P₄ (s) + 6 H₂ (g) ⇌ 4 PH₃ (g)

3. BaCl₂ (s) + 3 O₂ (g) ⇌ Ba(ClO₃)₂ (s)

Answers

1. All substances are aqueous, so all are included. Be sure to get the power right for hydroxide K = [Pb²⁺]*[OH^-]⁴/[Pb(OH)₄^2-]

2. The P₄ is a solid, so it is excluded. K = [PH₃]⁴/[H₂]⁶

3. In this case, the only non-solid substance is a reactant, making the K somewhat odd-looking K = 1/[O₂]³

Calculations With Equilibrium Constants

As we have seen in Lessons 3 and 7 with solubility products and acid dissociation constants, we write equilibrium constants in order to do calculations involving the concentrations of reactants and products. We will now go over and generalize the process of doing these kinds of calculations. This section should largely feel familiar from those lessons, as the same principles we learned in them will apply.

We will start with an example that will hopefully simplify things by avoiding complexities of chemical formulas. Consider the simple reaction below and its equilibrium constant.

2 A (g) + B (g) ⇌ A₂B (g) K = [A₂B]/[A]²*[B]

We mix some A and B together and let this reaction come to equilibrium. Then we measure the reactant and product concentrations and find the following values.

[A] = 0.50 M [B] = 1.0 M [A₂B] = 2.0 M

What is the value of the equilibrium constant?

To solve this, we plug in these values to our expression for K up above, like so. As you can see, we are leaving the units (M) out of the calculation entirely. For perhaps the first time in this class, you have my express permission to ignore units!

K = 2.0/(0.5²*1.0)

K = 8.0

This is one type of problem you will be asked to solve using K expressions: given the balanced equation and all the relevant concentrations at equilibrium, you should be able to write an expression for K and plug in concentrations to find its value.

Terminology Note: The "value" of a constant refers to a number (8.0 in this example). The "expression" for a constant means writing it in terms of the variables used to calculate it, like we did a few lines up.

You may also be asked to calculate the concentration of one reactant/product, given the value of K and concentrations for all the other reactants/products at equilibrium. This type of problem is slightly more complicated than the one above, as it will require you to rearrange your expression a little to solve for a concentration, but this is the only important difference.

For example, given the equation below, if K has a value of 167.5 and the concentration of N₂O₄ is measured to be 0.850 M, then what must the concentration of NO₂ be?

2 NO₂ (g) ⇌ N₂O₄ (g) K = [N₂O₄]/[NO₂]²

We simply plug in the values we know for K and [N₂O₄], then rearrange and solve for [NO₂].

167.5 = 0.850/[NO₂]²

[NO₂]² = 0.850/167.5

[NO₂]² = sqrt(0.850/167.5)

[NO₂]² = 0.0712 M

As long as we have a value for K and are dealing only with concentrations at equilibrium, these types of problems require only some algebra skills to master

ICE Table Problems

There is a more difficult type of problem that we learned to solve in Lessons 3 and 7. Namely, given a K value and the initial concentrations of species (as opposed to equilibrium concentrations), find concentrations at equilibrium. We had to solve problems of this type to determine the solubility of ionic compounds using the value of K_sp, or to find the pH of a solution given its initial acid/base concentration. To do so, we used the ICE table approach of thinking about how changes in concentration in different species were related to each other.

You should review those subjects, and you should be able to solve ICE table problems of these types. However, you will not be asked to deal with ICE table problems outside of these settings. You can expect to encounter problems like this in more advanced chemistry coursework, should you pursue it.

Your lab workbook contains many practice problems to hone your skills in working with equilibrium constants. You should complete some until you feel confident in your ability to write an equilibrium constant expression for a reaction from its balanced equation, and to solve for K, or for a concentration, given a number of other values.