This constant represents the **equilibrium constant**. Like in our example of a sports team above, we can calculate the equilibrium constant for all sorts of chemical reactions using information about the relative amounts of species in the system at equilibrium.

- This article is about
**calculating the equilibrium constant**in chemistry. - We'll
**define the equilibrium constant**before looking at how you**calculate its value for homogeneous and heterogeneous equilibria**. - You'll be able to practice your skills with the help of our
**worked examples**. - We'll then take a deep dive into
**calculating the equilibrium constant****using Gibbs free energy and standard electrode potential**.

## What is the equilibrium constant?

A reversible reaction in a state of dynamic equilibrium is characterized by two key features: **The rate of the forward reaction equals the rate of the backward reaction,** or equivalently **the concentrations of reactants and products don't change**. Like in our example of a soccer team, this means that **the ratio between the relative amounts of reactants and products doesn't change either**. This ratio is known as the **equilibrium constant, K**_{eq}.

The **equilibrium constant**,** K _{eq}**, is a value that tells us the

**relative amounts of reactants and products**in a system at equilibrium.

As the name suggests, **K _{eq} is a constant**.

**For a**

**certain reaction at a certain temperature**,

**the equilibrium constant is always the same**, no matter how much of the products or reactants you start with.

**But change the temperature and you change the value of the equilibrium constant.**

We measure the equilibrium constant using either **concentration** or **partial pressure**. The important thing to remember is that the equilibrium constant is always worked out using **measurements taken *** at equilibrium*:

- K
_{c}measures the**concentrations****of aqueous or gaseous species**in a reversible reaction at equilibrium. - K
_{p }measures the**partial pressures of gaseous species**in a reversible reaction at equilibrium.

For the reaction \(aA(g)+bB(g) \rightleftharpoons cC(g)+dD(g)\), K_{c} and K_{p} have the following expressions:

$$K_c=\frac{{[C]_{eqm}}^c\space {[D]_{eqm}}^d}{{[A]_{eqm}}^a\space {[B]_{eqm}}^b}\qquad K_p=\frac{{{(P_C)}_{eqm}}^c\space {{(P_D)}_{eqm}}^d}{{{(P_A)}_{eqm}}^a\space {{(P_B)}_{eqm}}^b}$$

We won't go into those expressions in any more detail here, but if you are not sure what all the different symbols mean, check out **Equilibrium**** Constant** for more information. There, you'll also learn about some of the other types of equilibrium constants. On top of that, you'll see how we deal with pure solids and liquids when it comes to the equilibrium constant. That particular knowledge will come in handy later in this article.

If you need a reminder about reversible reactions and dynamic equilibria, don't worry - we have articles on those subjects too. Head over to **R****eversible Reaction **and **Dynamic Equilibrium **for more.** **

## Calculating equilibrium composition from an equilibrium constant

Hopefully, that was a useful reminder about what exactly the equilibrium constant is and how to write equilibrium constant expressions for different reactions. We can now move on to **calculating the equilibrium constant**.

Remember how the equilibrium constant is based on **the relative amounts of reactants and products in a reversible reaction at equilibrium**? To calculate the equilibrium constant, we, therefore, need to take **experimental measurements of all of these species at equilibrium**. We can either measure **concentration**, in which case we'd work out **K**_{c}, or **partial pressure**, in which case we'd work out **K**_{p}. We then substitute the equilibrium measurements into the expression for the equilibrium constant in order to get your final answer.

However, sometimes you don't know the equilibrium concentrations (or partial pressures) of each of the species in a reversible reaction. This doesn't matter. Provided you know **the starting concentrations of all of the species**, and **the equilibrium concentration of *** at least one species, *you can use the balanced chemical equation to work out the unknown equilibrium concentrations of the other species. Here are the steps you should follow:

- Create a table with rows for the
**initial**,**change in**, and**equilibrium concentrations**of all of the reactants and products. Fill in the initial concentrations and any equilibrium concentrations that you know. - Subtract the equilibrium concentration from the initial concentration of a species that you
*do*know to calculate its change in concentration. - Use the balanced chemical equation to calculate the change in concentration of the remaining species involved.
- Use the initial concentration and change in concentration to calculate the equilibrium concentration of each of these species.
- Substitute all of the equilibrium concentration values into the expression for the equilibrium constant to get your final answer.

That sounds a little complicated, but it is much simpler than it sounds. In the next section, we'll go through some worked examples for both homogeneous and heterogeneous equilibria so you can see exactly how the process works.

## Calculating the equilibrium constant for the reaction

First up, let's look at **calculating the equilibrium constant for a reactionb**. In this case, the reaction has a **homogenous equilibria**.

**A closed gaseous system containing 2.0 M H _{2} and 1.5 M Cl_{2} is left to reach equilibrium. At equilibrium, the system contains 1.2 M H_{2}. Work out K_{c} for this reaction, using the following equation to help you:**

$$H_2(g)+Cl_2(g)\rightleftharpoons 2HCl(g)$$

The first step is to create a table with rows for initial, change in, and equilibrium concentrations of each species. It can be helpful to use the chemical equation as headings for the columns. We can then fill in the values that were given to us in the question:

Species | H_{2} + Cl_{2} ⇌ 2HCl | |||

Concentration (M) | Initial | 2.0 | 1.5 | 0.0 |

Change | ||||

Equilibrium | 1.2 |

We know both the initial and the equilibrium concentrations of H_{2}. This means that the concentration of H_{2} has changed by 1.2 - 2.0 = -0.8 M. Note that the concentration has *decreased*, and so this is a negative change.

Now take a look at the balanced chemical equation. The ratio of H_{2}:Cl_{2}:HCl in the equation is 1:1:2. For each mole of H_{2} that reacts, one mole of Cl_{2} also reacts and two moles of HCl are formed. Therefore, if the concentration of H_{2} has decreased by 0.8 M, then the concentration of Cl_{2} has also decreased by 0.8 M. On the other hand, the concentration of HCl has *increased *by 2 x 0.8 = 1.6 M. Let's add these values to our table:

Species | H_{2} + Cl_{2} ⇌ 2HCl | |||

Concentration (M) | Initial | 2.0 | 1.5 | 0.0 |

Change | -0.8 | -0.8 | +1.6 | |

Equilibrium | 1.2 |

We can now use the initial concentration and the change in concentration to work out the equilibrium concentration of each species. The equilibrium concentration of Cl_{2} is 1.5 - 0.8 = 0.7 M, whilst the equilibrium concentration of HCl is 0.0 + 1.6 = 1.6 M. Once again, let's add these to our table:

Species | H_{2} + Cl_{2} ⇌ 2HCl | |||

Concentration (M) | Initial | 2.0 | 1.5 | 0.0 |

Change | -0.8 | -0.8 | +1.6 | |

Equilibrium | 1.2 | 0.7 | 1.6 |

Finally, we can substitute the equilibrium concentrations into the expression for K_{c}. For this reaction, K_{c} takes the following expression:

$$K_c=\frac{{[HCl]_{eqm}}^2}{[H_2]_{eqm}\space [Cl_2]_{eqm}}$$

Substituting in the equilibrium concentrations that we've calculated, we reach our final answer:

$$K_c=\frac{{(1.6)}^2}{(1.2)\space (0.7)}=\frac{64}{21}$$ $$K_c=3.05$$

Note that although we've used concentration in this example, and in our description of the method above, you can carry out the same process using partial pressure instead. In this case, you'd calculate a value for K_{p}, not K_{c}.

Hopefully, that example has helped make the process a little clearer for you. Let's now look at calculating the equilibrium constant for **heterogeneous equilibria**.

## Calculating the equilibrium constant for heterogeneous equilibria

To **calculate the equilibrium constant for heterogeneous equilibria**, we use the same method as you used for homogenous equilibria. However, there's just one slight difference: when writing the expression for the equilibrium constant, we **ignore any pure solids or liquids in the system**. This is because they aren't dissolved in anything and so their concentration is always 1. Therefore, we don't need to include them in our table. Likewise, if we want to find K_{p}, we **ignore any species that aren't gaseous**. Here's an example.

**A system consists of FeO, CO, Fe, and CO**

_{2}. Initially, the concentration of CO is 9.6 x 10^{-2}M and the concentration of CO_{2}is 1.3 x 10^{-2}M. The equilibrium concentration of CO_{2}is 5.5 x 10^{-2}M. Calculate K_{c}for this reaction, using the following equation to help you:$$FeO(s)+CO(g)\rightleftharpoons Fe(s)+CO_2(g)$$

Note that this is a heterogeneous equilibrium. We're calculating K_{c}, and this means that the expression ignores any pure solids or liquids. We can therefore leave them out of our table; we only need to include CO and CO_{2}. Here's the table, with the values given in the question filled in for you:

Species | CO ⇌ CO_{2} | ||

Concentration (M) | Initial | 9.6 x 10^{-2} | 1.3 x 10^{-2} |

Change | |||

Equilibrium | 5.5 x 10 ^{-2} |

The concentration of CO_{2} has increased by (5.5 x 10^{-2}) - (1.3 x 10^{-2}) = 4.2 x 10^{-2} M. Looking at the balanced chemical equation, we can see that CO and CO_{2} are found in a 1:1 ratio, and so this means that the concentration of CO has *decreased* by 4.2 x 10^{-2} M:

Species | CO ⇌ CO_{2} | ||

Concentration (M) | Initial | 9.6 x 10^{-2} | 1.3 x 10^{-2} |

Change | -4.2 x 10^{-2} | +4.2 x 10^{-2} | |

Equilibrium | 5.5 x 10^{-2} |

This gives us an equilibrium concentration for CO of (9.6 x 10^{-2}) - (4.2 x 10^{-2}) = 5.4 x 10^{-2} M:

Species | CO ⇌ CO_{2} | ||

Concentration (M) | Initial | 9.6 x 10^{-2} | 1.3 x 10^{-2} |

Change | -4.2 x 10^{-2} | +4.2 x 10^{-2} | |

Equilibrium | 5.4 x 10^{-2} | 5.5 x 10^{-2} |

Let's now substitute the equilibrium concentrations into the expression for K_{c}. Remember that in heterogeneous equilibria, K_{c} ignores any pure solids or liquids. Our expression therefore only features CO and CO_{2}:

$$K_c=\frac{[CO_2]_{eqm}}{[CO]_{eqm}}$$ $$K_c=\frac{(5.5\times 10^{-2})}{(5.4\times 10^{-2})}$$ $$K_c=1.02$$

Not too tricky, huh? Before we finish this article, let's take a deep dive into two further ways of calculating the equilibrium constant, using **Gibbs free energy** and **standard electrode potential**.

## Calculating equilibrium constant using Gibbs free energy

In the article **Gibbs Free Energy**, you'll learn that Gibbs free energy is a measure of how thermodynamically favorable a reaction is - or in other words, whether it will proceed without you inputting any extra energy. And in the article **Free Energy and ****Equilibrium**, you'll see how free energy relates to equilibria. In fact, you can use Gibbs free energy to find out about the relative amounts of reactants and products in a system at equilibrium.

You can also **calculate the equilibrium constant using Gibbs free energy**. They're related using the following equation:

$$\Delta G^\circ =-RT\space ln(K_{eq})$$

Note the following:

- ΔG° represents the
**standard Gibbs free energy change**^{-1}but for this equation, we must convert it into J mol^{-1}. - R is the
**gas constant**, 8.314 J mol^{-1}K^{-1}. - T is the
**temperature**, in K. - K
_{eq}is the**equilibrium constant**.

We can rearrange the equation to solve for K_{eq}:

$$ K_{eq}=e^{\frac{{-\Delta G}^\circ}{RT}}$$

We'll go through some more problems in the two articles mentioned above, but here's a quick worked example to get you thinking.

**A reaction is carried out at 350 K and has a standard free-energy change of 16 kJ mol ^{-1}. Work out the equilibrium constant K_{eq} for this reaction.**

Well, we know that ΔG° = 16 kJ mol^{-1}, which equals 16,000 J mol^{-1}. We know that T = 350 K, and we also know that the gas constant R = 8.314 J mol^{-1} K^{-1}. Let's substitute these into the equation we derived above:

$$ K_{eq}=e^{\frac{-16000}{(8.314)(350)}}=e^{\frac{-16000}{2910}}$$ $$K_{eq}=4.09\times 10^{-3}$$

That is our final answer.

## Calculating equilibrium constant from electrode potential

One additional way of calculating the equilibrium constant involves **standard electrode potential**. The formula is derived from an equation linking Gibbs free energy and the **standard electrode potential (E°)** of a chemical reaction:

$$\Delta G^\circ =-nFE^\circ$$

Note the following:

- ΔG° represents the
**standard change in Gibbs free energy**, measured in J mol^{-1}. - n represents the
**number of moles of electrons transferred**in the reaction according to the balanced chemical equation. - F is the
**Faraday constant**, 96,485 C (mol e^{-})^{-1}. - E° is the
**standard electrode potential**value of the reaction, measured in V.

If we substitute this into the equation linking Gibbs free energy and the equilibrium constant that we looked above, we end up with the following:

$$E^\circ =\frac{RT}{nF}ln(K_{eq})$$

If you want to see this equation in action, check out the article **Cell Potential and Free Energy**.

That's it for this article. You should now understand how you can **use equilibrium measurements to calculate the equilibrium constant for homogeneous and heterogeneous equilibria**. You should feel comfortable **calculating the equilibrium constant using Gibbs free energy**, and **understand the relationship between standard electrode potential and the equilibrium constant**.

## Calculating Equilibrium Constant - Key takeaways

- The
**equilibrium constant**,**K**, is a value that tells us the_{eq}**relative amounts of reactants and products**in a system at equilibrium. - We
**calculate the equilibrium constant**using**experimental measurements taken at equilibrium**. We use**concentration**to calculate**K**_{c}and**partial pressure**to calculate**K**_{p}. - To calculate the equilibrium constant:
Use the initial and equilibrium concentration of a species that you

*do*know to calculate its change in concentration.Use the balanced chemical equation to calculate the change in concentration and equilibrium concentration of the remaining species involved in the reaction.

Substitute the equilibrium concentration values into the expression for the equilibrium constant to get your final answer.

When calculating

*K*_{c}for**heterogeneous**equilibria, we**ignore any pure solids or liquids**in the system. When calculating*K*_{p}for**heterogeneous**equilibria, we**ignore any species that aren't gaseous**.

###### Learn with 4 Calculating Equilibrium Constant flashcards in the free StudySmarter app

We have **14,000 flashcards** about Dynamic Landscapes.

Already have an account? Log in

##### Frequently Asked Questions about Calculating Equilibrium Constant

How do you calculate the equilibrium constant?

You calculate the equilibrium constant using an expression linking the relative amounts of reactants and products in a system at equilibrium. For example, take the reaction aA + bB ⇌ cC + dD. To find the equilibrium constant K_{c}, we use the expression K_{c} = [C]_{eqm}^{c }[D]_{eqm}^{d} / [A]_{eqm}^{a }[B]_{eqm}^{b}. Simply substitute your equilibrium concentrations of all of the species involved in the reaction into the expression, and you should arrive at a value for K_{c}. Check out the rest of this article for some worked examples.

How do you calculate the equilibrium constant with temperature?

The equilibrium constant varies depending on the temperature of the reaction. If you change the temperature, you'll change the value of the equilibrium constant. This means that when working out the equilibrium constant, you must make sure you specify the temperature of the system.

You can also use the van't Hoff equation to calculate the change in K with temperature. Here you are working under the assumption that the enthalpy and entropy of the system are not changing (significantly) with the change of temperature (usually acceptable approximation).

How do you calculate the equilibrium constant from cell potential?

To calculate the equilibrium constant from cell potential, you use the equation E° = (RT/nF) ln(K_{eq}). Here, E° represents the standard electrode potential (emf) value of the reaction, R represents the gas constant, T is the temperature, n is the number of electrons transferred according to the balanced chemical equation, and F is the Faraday constant.

How do you calculate the thermodynamic equilibrium constant?

For the reaction aA + bB ⇌ cC + dD, we find the equilibrium constant K_{c} using the expression K_{c} = [C]_{eqm}^{c }[D]_{eqm}^{d} / [A]_{eqm}^{a }[B]_{eqm}^{b}. K_{p} is calculated in a similar way but uses equilibrium partial pressure instead of equilibrium concentration.

How do you calculate the equilibrium constant for a heterogeneous reaction?

For the reaction aA + bB ⇌ cC + dD, we find the equilibrium constant K_{c} using the expression K_{c} = [C]_{eqm}^{c }[D]_{eqm}^{d} / [A]_{eqm}^{a }[B]_{eqm}^{b}. However, in heterogeneous reactions, we ignore any pure solids or liquids - you can leave them out of the expression altogether. K_{p} is calculated in a similar way but uses equilibrium partial pressure instead of equilibrium concentration. In addition, we ignore any species that aren't gaseous.

##### About StudySmarter

StudySmarter is a globally recognized educational technology company, offering a holistic learning platform designed for students of all ages and educational levels. Our platform provides learning support for a wide range of subjects, including STEM, Social Sciences, and Languages and also helps students to successfully master various tests and exams worldwide, such as GCSE, A Level, SAT, ACT, Abitur, and more. We offer an extensive library of learning materials, including interactive flashcards, comprehensive textbook solutions, and detailed explanations. The cutting-edge technology and tools we provide help students create their own learning materials. StudySmarter’s content is not only expert-verified but also regularly updated to ensure accuracy and relevance.

Learn more