Learning Materials
Features
Discover
In Equilibrium Constant, we used the analogy of a sports team to represent an equilibrium. Say you take 18 players to a soccer game. At any one time, you are only allowed to put 11 of them on the pitch; the rest have to sit on the bench. Although you might choose to make substitutions during the game, the number of players on the pitch and the number of players on the bench don't change. This means that the ratio between their relative amounts doesn't change - it is a constant. We can find it using the number of players on the pitch and the number of players on the bench.
Millions of flashcards designed to help you ace your studies
Sign up for freeYou know the starting concentrations of all of the species involved in a reversible reaction but don't know any of the equilibrium concentrations. Is it possible to calculate Kc?
True or false? To calculate the equilibrium constant Kc in a homogenous system, you need to find out the equilibrium concentrations of all of the species involved.
True or false? To calculate Kc, you need to know the concentration of any catalysts involved in the reaction.
You know the starting concentrations of all of the species involved in a reversible reaction but don't know any of the equilibrium concentrations. Is it possible to calculate Kc?
True or false? To calculate the equilibrium constant Kc in a homogenous system, you need to find out the equilibrium concentrations of all of the species involved.
True or false? To calculate Kc, you need to know the concentration of any catalysts involved in the reaction.
Achieve better grades quicker with Premium
Geld-zurück-Garantie, wenn du durch die Prüfung fällst
Did you know that StudySmarter supports you beyond learning?
Review generated flashcards
Start learning or create your own AI flashcards
Team Calculating Equilibrium Constant Teachers
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.
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, Keq.
The equilibrium constant, Keq, is a value that tells us the relative amounts of reactants and products in a system at equilibrium.
As the name suggests, Keq 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:
For the reaction \(aA(g)+bB(g) \rightleftharpoons cC(g)+dD(g)\), Kc and Kp 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 Reversible Reaction and Dynamic Equilibrium for more.
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 Kc, or partial pressure, in which case we'd work out Kp. 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:
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.
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 H2 and 1.5 M Cl2 is left to reach equilibrium. At equilibrium, the system contains 1.2 M H2. Work out Kc 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 | H2 + Cl2 ⇌ 2HCl | |||
| Concentration (M) | Initial | 2.0 | 1.5 | 0.0 |
| Change | ||||
| Equilibrium | 1.2 | |||
We know both the initial and the equilibrium concentrations of H2. This means that the concentration of H2 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 H2:Cl2:HCl in the equation is 1:1:2. For each mole of H2 that reacts, one mole of Cl2 also reacts and two moles of HCl are formed. Therefore, if the concentration of H2 has decreased by 0.8 M, then the concentration of Cl2 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 | H2 + Cl2 ⇌ 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 Cl2 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 | H2 + Cl2 ⇌ 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 Kc. For this reaction, Kc 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 Kp, not Kc.
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.
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 Kp, we ignore any species that aren't gaseous. Here's an example.
$$FeO(s)+CO(g)\rightleftharpoons Fe(s)+CO_2(g)$$
Note that this is a heterogeneous equilibrium. We're calculating Kc, 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 CO2. Here's the table, with the values given in the question filled in for you:
| Species | CO ⇌ CO2 | ||
| Concentration (M) | Initial | 9.6 x 10-2 | 1.3 x 10-2 |
| Change | |||
| Equilibrium | 5.5 x 10-2 | ||
The concentration of CO2 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 CO2 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 ⇌ CO2 | ||
| 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 ⇌ CO2 | ||
| 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 Kc. Remember that in heterogeneous equilibria, Kc ignores any pure solids or liquids. Our expression therefore only features CO and CO2:
$$K_c=\frac{[CO_2]_{eqm}}{[CO]_{eqm}}$$ $$K_c=\frac{(5.5\times 10^{-2})}{(5.4\times 10^{-2})}$$ $$K_c=1.02$$
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.
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:
We can rearrange the equation to solve for Keq:
$$ K_{eq}=e^{\frac{{-\Delta G}^\circ}{RT}}$$
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 Keq 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.
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:
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.
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 Kc for heterogeneous equilibria, we ignore any pure solids or liquids in the system. When calculating Kp for heterogeneous equilibria, we ignore any species that aren't gaseous.
Sign up for free to gain access to all our flashcards.
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 Kc, we use the expression Kc = [C]eqmc [D]eqmd / [A]eqma [B]eqmb. 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 Kc. 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(Keq). 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 Kc using the expression Kc = [C]eqmc [D]eqmd / [A]eqma [B]eqmb. Kp 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 Kc using the expression Kc = [C]eqmc [D]eqmd / [A]eqma [B]eqmb. However, in heterogeneous reactions, we ignore any pure solids or liquids - you can leave them out of the expression altogether. Kp 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.
At StudySmarter, we have created a learning platform that serves millions of students. Meet the people who work hard to deliver fact based content as well as making sure it is verified.
Lily Hulatt is a Digital Content Specialist with over three years of experience in content strategy and curriculum design. She gained her PhD in English Literature from Durham University in 2022, taught in Durham University’s English Studies Department, and has contributed to a number of publications. Lily specialises in English Literature, English Language, History, and Philosophy.
Get to know Lily
Gabriel Freitas is an AI Engineer with a solid experience in software development, machine learning algorithms, and generative AI, including large language models’ (LLMs) applications. Graduated in Electrical Engineering at the University of São Paulo, he is currently pursuing an MSc in Computer Engineering at the University of Campinas, specializing in machine learning topics. Gabriel has a strong background in software engineering and has worked on projects involving computer vision, embedded AI, and LLM applications.
Get to know Gabriel
Vaia 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 moreSave explanations to your personalised space and access them anytime, anywhere!
Sign up with Email Sign up with AppleBy signing up, you agree to the Terms and Conditions and the Privacy Policy of Vaia.
Already have an account? Log in
Sign up to highlight and take notes. It’s 100% free.
Explanations 
Textbooks 
Exams 
GCSE 
IGCSE 
AS 
A Level 
International A Level 
University Admissions Tests 
Flashcards 
Vaia AI 
Notes 
Study Plans 
Study Sets 
Find a job 
Find a degree 
Student Deals 
Magazine 
Mobile App 
