What is the Equilibrium Constant?

Core Concepts

In this article about chemical equilibrium, we will learn about the equilibrium constant, also often referred to as the equilibrium coefficient. This article explains what the equilibrium constant is, why it is significant, and how to calculate it. We introduce the equilibrium equation, and you will briefly learn about the reaction quotient. We also define equilibrium, and discuss the difference between dynamic equilibrium and chemical equilibrium.

Topics Covered in Other Articles

Review: What is dynamic equilibrium?

So let’s define equilibrium. Dynamic equilibrium, or chemical equilibrium, refers to the state a chemical reaction is in when the forward and reverse reactions are at equal rates, meaning that the concentrations of products and reactants both remain constant. The rate in which products are being formed from the reactants is the same as the rate where products are being broken back down into reactants.

Keep in mind that this does not mean that the concentrations of reactants and products necessarily must be the same, just that the rates of formation are equal, so there is no overall change in concentration.

A related concept is called homeostasis, which when a living organism maintains their pH, metabolism, temperature, etc. in a narrow range through a self-regulating process. This state may or may not be in actual equilibrium.

Dynamic Equilibrium vs Chemical Equilibrium

Dynamic equilibrium and chemical equilibrium are equivalent in most circumstances. Chemical equilibrium refers to a chemical reaction, where the reactants and products are in equilibrium. Take for example a physical process with no chemical changes, such as gases diffusing between containers. If that reached equilibrium, then that would be a form of dynamic equilibrium, but not a form of chemical equilibrium.

What is the equilibrium constant? What does it represent?

The equilibrium constant, $K$ , represents the extent of a reaction when it is at equilibrium. It uses the concentrations and coefficients of each reactant and product to form a ratio. From the K value, we can understand whether a reaction favors the reactants or products more, and therefore where the position of equilibrium lies.

If $K > 1$ , the position of equilibrium lies to the right, meaning the formation of the products is favored in the reaction.

If $K < 1$ , the position of equilibrium lies to the left, meaning the formation of the reactants are favored.

In a similar vein, if $K = 1$ , this indicates that the position of equilibrium lies directly in the center, so neither the products nor the reactants are favored.

It is worth noting that the value of $K$ , and what it indicates about the reaction, changes with temperature. This is because the solubility of a substance can increase or decrease depending on the temperature; because of this, the concentrations, and therefore the equilibrium of reaction, can also change with temperature.

The Equilibrium Equation

How to calculate the equilibrium constant

Consider the following equilibrium equation:

$\begin{gather*} aA + bB \rightleftarrows cC + dD \end{gather*}$

The equilibrium constant can be found using the following formula

$\begin{gather*} K = \frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}} \end{gather*}$

The concentrations of all the products are in the numerator and the concentrations of all the reactants are in the denominator; each component is then raised to the power of their specific coefficient. The equilibrium constant is equal to the rate constant of the forward reaction divided by the rate constant of the reverse reaction.

This relation, the equilibrium constant, is known as the law of mass action. The law states, firstly, that the rate of a chemical reaction is directly proportional to the concentrations of its reactants. It also states, secondly, and more importantly for us right now, that the ratio of concentration of reactants and products is constant for a reaction at equilibrium. This constant is known as the equilibrium constant, $K$ .

It is also worth noting that the equilibrium constant expression is equal to the ratio of the forward rate constant to the reverse rate constant:

$\begin{gather*} K = \frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}} = \frac{k_{\text{forward}}}{k_{\text{reverse}}} \end{gather*}$

Specifically, we know this by looking at the rate laws for the forward and reverse reactions:

$\begin{align*} {\text{Rate}_{\text{forward}}&=k_{forward}[A]^{a}[B]^{b}} \\ {\text{Rate}_{\text{reverse}}&=k_{reverse}[C]^{c}[D]^{d}} \end{align*}$

When equilibrium is reached, the forward and reverse reaction rates equal one another, which we can use to solve for the rate constant $K$ :

$\begin{align*} {\text{Rate}_{\text{forward}}&=\text{Rate}_{\text{reverse}}} \\ {k_{forward}[A]^{a}[B]^{b}&=k_{reverse}[C]^{c}[D]^{d}} \\ {k_{forward}&=k_{reverse}\frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}=k_{reverse}K} \\ {K&=\frac{k_{\text{forward}}}{k_{\text{reverse}}}} \end{align*}$

This equation also explains why K indicates the extent of reaction. If $K > 1$ then the numerator, represented by the concentration of products, is greater. If $K < 1$ then the denominator, represented by the concentration of reactants, is greater.

Keep in mind when calculating K!

In order to calculate the equilibrium coefficient, write out the balanced chemical equilibrium equation first. Given the law of mass action, the coefficients of each substance matter, so it is important that the equation is properly balanced.

In addition, be sure to keep the expression of concentrations of all solutions consistent. This is very important because K is found using the concentrations of products and reactants. Most commonly, solutions are expressed in $\text{M}$ , or occasionally in moles if all solutions are in the same volume. Solids, liquids, and solvents are assigned a value of 1, so their concentrations do not affect the value of $K$ . In this manner, only aqueous solutions and gases count towards the equilibrium constant expression.

How to find the equilibrium constant from other, known K values

It is also possible to find the equilibrium constant for a reaction using the $K$ values for other, known reactions. There are a few set rules to follow in these cases:

When adding 2 or more reactions together to create a new one, multiply all the existing $K$ values together to obtain the equilibrium constant for the new reaction. $K_{\text{new}} = K_{1} \cdot K_{2} \cdot K_{3} \cdot$ …
When multiplying an entire reaction by a numerical factor $N$ , raise the existing $K$ value to the power of $N$ to obtain the new constant. $K_{\text{new}} = \left(K_{\text{original}}\right)^{N}$
When reversing a reaction, take the reciprocal of the existing $K$ value to obtain the new one. $K_{\text{new}} = \frac{1}{K_{\text{original}}}$ .
When performing a combination of these operations, apply each step as needed. For example, if a reaction X is first reversed, then added to another reaction Y, then $K_{\text{new}} = \left(\frac{1}{K_{\text{X}}}\right) \cdot \left(K_{Y}\right)$

Related: What is the Reaction Quotient?

The reaction quotient, $Q$ , is a very similar concept to the equilibrium constant, with one significant difference. The reaction quotient is a measure of the extent of reaction at any given moment, not solely at equilibrium. This means that it takes into account the relative amounts (concentrations) of reactants and products at that moment in time. In this vein, with the concentrations of reactants and products representing their actual concentrations at the given time, the following represents the reaction quotient expression:

$\begin{gather*} Q = \frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}} \end{gather*}$

Equilibrium Constant Practice Problems

Problem 1

You observe the following acid-base reaction:

$\begin{gather*} CH_{3}COOH + NH_{3} \rightleftarrows CH_{3}COO^{-} + NH_{4}^{+} \end{gather*}$

After reaching equilibrium, you observe the following concentrations:

$\begin{align*} {[CH_{3}COOH] &= 0.012\text{M}} \\ {[NH_{3}] &= 0.0064\text{M}} \\ {[CH_{3}COO^{-}] &= 1.7\text{M}} \\ {[NH_{4}^{+}] &= 1.8\text{M}} \end{align*}$

Problem 2

Reaction Z involves the sum of two step-wise equilibrium-controlled reactions, X and Y. Reaction T also takes place through two step-wise equilibrium-controlled reactions, first involving the reverse of reaction Y followed reaction V.

Consider the following relationship of rate constants: $K_{Y} < K_{X} < 1 < K_{V}$