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## Chemistry library

### Course: Chemistry libraryÂ >Â Unit 12

Lesson 1: Equilibrium constant

# The equilibrium constant K

Reversible reactions, equilibrium, and the equilibrium constant K. How to calculate K, and how to use K to determine if a reaction strongly favors products or reactants at equilibrium.Â

## Key points

• A reversible reaction can proceed in both the forward and backward directions.
• Equilibrium is when the rate of the forward reaction equals the rate of the reverse reaction. All reactant and product concentrations are constant at equilibrium.
• Given a reaction $\text{aA}+\text{bB}â\text{cC}+\text{dD}$, the equilibrium constant ${K}_{\text{c}}$, also called $K$ or ${K}_{\text{eq}}$, is defined as follows:
${K}_{\text{c}}=\frac{\left[{\text{C]}}^{\text{c}}{\text{[D]}}^{\text{d}}}{\left[\text{A}{\right]}^{\text{a}}\left[\text{B}{\right]}^{\text{b}}}$
• For reactions that are not at equilibrium, we can write a similar expression called the reaction quotient $Q$, which is equal to ${K}_{\text{c}}$ at equilibrium.
• ${K}_{\text{c}}$ and $Q$ can be used to determine if a reaction is at equilibrium, to calculate concentrations at equilibrium, and to estimate whether a reaction favors products or reactants at equilibrium.

## Introduction: reversible reactions and equilibrium

A reversible reaction can proceed in both the forward and backward directions. Most reactions are theoretically reversible in a closed system, though some can be considered to be irreversible if they heavily favor the formation of reactants or products. The double half-arrow sign we use when writing reversible reaction equations, $â$, is a good visual reminder that these reactions can go either forward to create products, or backward to create reactants. One example of a reversible reaction is the formation of nitrogen dioxide, ${\text{NO}}_{2}$, from dinitrogen tetroxide, ${\text{N}}_{2}{\text{O}}_{4}$:
${\text{N}}_{2}{\text{O}}_{4}\left(g\right)â2{\text{NO}}_{2}\left(g\right)$
Imagine we added some colorless ${\text{N}}_{2}{\text{O}}_{4}\left(g\right)$ to an evacuated glass container at room temperature. If we kept our eye on the vial over time, we would observe the gas in the ampoule changing to a yellowish orange color and gradually getting darker until the color stayed constant. We can graph the concentration of ${\text{NO}}_{2}$ and ${\text{N}}_{2}{\text{O}}_{4}$ over time for this process, as you can see in the graph below.
Initially, the vial contains only ${\text{N}}_{2}{\text{O}}_{4}$, and the concentration of ${\text{NO}}_{2}$ is 0 M. As ${\text{N}}_{2}{\text{O}}_{4}$ gets converted to ${\text{NO}}_{2}$, the concentration of ${\text{NO}}_{2}$ increases up to a certain point, indicated by a dotted line in the graph to the left, and then stays constant. Similarly, the concentration of ${\text{N}}_{2}{\text{O}}_{4}$ decreases from the initial concentration until it reaches the equilibrium concentration. When the concentrations of ${\text{NO}}_{2}$ and ${\text{N}}_{2}{\text{O}}_{4}$ remain constant, the reaction has reached equilibrium.
All reactions tend towards a state of chemical equilibrium, the point at which both the forward process and the reverse process are taking place at the same rate. Since the forward and reverse rates are equal, the concentrations of the reactants and products are constant at equilibrium. It is important to remember that even though the concentrations are constant at equilibrium, the reaction is still happening! That is why this state is also sometimes referred to as dynamic equilibrium.
Based on the concentrations of all the different reaction species at equilibrium, we can define a quantity called the equilibrium constant ${K}_{\text{c}}$, which is also sometimes written as ${K}_{\text{eq}}$ or $K$. The $\text{c}$ in the subscript stands for concentration since the equilibrium constant describes the molar concentrations, in $\frac{\text{mol}}{\text{L}}$, at equilibrium for a specific temperature. The equilibrium constant can help us understand whether the reaction tends to have a higher concentration of products or reactants at equilibrium. We can also use ${K}_{\text{c}}$ to determine if the reaction is already at equilibrium.

## How do we calculate ${K}_{\text{c}}$â ?

Consider the balanced reversible reaction below:
$\text{aA}+\text{bB}â\text{cC}+\text{dD}$
If we know the molar concentrations for each reaction species, we can find the value for ${K}_{\text{c}}$ using the relationship
${K}_{\text{c}}=\frac{\left[{\text{C]}}^{\text{c}}{\text{[D]}}^{\text{d}}}{\left[\text{A}{\right]}^{\text{a}}\left[\text{B}{\right]}^{\text{b}}}$
where $\left[\text{C]}$ and $\text{[D]}$ are equilibrium product concentrations; $\left[\text{A}\right]$ and $\left[\text{B}\right]$ are equilibrium reactant concentrations; and $\text{a}$, $\text{b}$, $\text{c}$, and $\text{d}$ are the stoichiometric coefficients from the balanced reaction. The concentrations are usually expressed in molarity, which has units of $\frac{\text{mol}}{\text{L}}$.
There are some important things to remember when calculating ${K}_{\text{c}}$:
• ${K}_{\text{c}}$ is a constant for a specific reaction at a specific temperature. If you change the temperature of a reaction, then ${K}_{\text{c}}$ also changes.
• Pure solids and pure liquids, including solvents, are not included in the equilibrium expression.
• ${K}_{\text{c}}$ is often written without units, depending on the textbook.
• The reaction must be balanced with the coefficients written as the lowest possible integer values in order to get the correct value for ${K}_{\text{c}}$.
Note: If any of the reactants or products are gases, we can also write the equilibrium constant in terms of the partial pressure of the gases. We typically refer to that value as ${K}_{\text{p}}$ to tell it apart from the equilibrium constant using concentrations in molarity, ${K}_{\text{c}}$. In this article, however, we will be focusing on ${K}_{\text{c}}$.

## What does the magnitude of ${K}_{\text{c}}$â  tell us about the reaction at equilibrium?

The magnitude of ${K}_{\text{c}}$ can give us some information about the reactant and product concentrations at equilibrium:
• If ${K}_{\text{c}}$ is very large, ~1000 or more, we will have mostly product species present at equilibrium.
• If ${K}_{\text{c}}$ is very small, ~0.001 or less, we will have mostly reactant species present at equilibrium.
• If ${K}_{\text{c}}$ is in between 0.001 and 1000, we will have a significant concentration of both reactant and product species present at equilibrium.
By using these guidelines, we can quickly estimate whether a reaction will strongly favor the forward direction to make productsâvery large ${K}_{\text{c}}$âstrongly favor the backward direction to make reactantsâvery small ${K}_{\text{c}}$âor somewhere in between.

## Example

### Part 1: Calculating ${K}_{\text{c}}$â  from equilibrium concentrations

Let's take a look at the equilibrium reaction that takes place between sulfur dioxide and oxygen to produce sulfur trioxide:
$2{\text{SO}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)â2{\text{SO}}_{3}\left(g\right)$
The reaction is at equilibrium at some temperature, $\text{T}$, and the following equilibrium concentrations are measured:
$\begin{array}{rl}\left[{\text{SO}}_{2}\right]& =0.90\phantom{\rule{0.167em}{0ex}}\text{M}\\ \\ \left[{\text{O}}_{2}\right]& =0.35\phantom{\rule{0.167em}{0ex}}\text{M}\\ \\ \left[{\text{SO}}_{3}\right]& =1.1\phantom{\rule{0.167em}{0ex}}\text{M}\end{array}$
We can calculate ${K}_{\text{c}}$ for the reaction at temperature $\text{T}$ by solving following expression:
${K}_{\text{c}}=\frac{\left[{\text{SO}}_{3}{\right]}^{2}}{\left[{\text{SO}}_{2}{\right]}^{2}\left[{\text{O}}_{2}\right]}$
If we plug our known equilibrium concentrations into the above equation, we get:
$\begin{array}{rl}{K}_{\text{c}}& =\frac{\left[{\text{SO}}_{3}{\right]}^{2}}{\left[{\text{SO}}_{2}{\right]}^{2}\left[{\text{O}}_{2}\right]}\\ \\ & =\frac{\left[1.1{\right]}^{2}}{\left[0.90{\right]}^{2}\left[0.35\right]}\\ \\ & =4.3\end{array}$
Note that since the calculated ${K}_{\text{c}}$ value is between 0.001 and 1000, we would expect this reaction to have significant concentrations of both reactants and products at equilibrium, as opposed to having mostly reactants or mostly products.

### Part 2: Using the reaction quotient $Q$â  to check if a reaction is at equilibrium

Now we know the equilibrium constant for this temperature: ${K}_{\text{c}}=4.3$. Imagine we have the same reaction at the same temperature $\text{T}$, but this time we measure the following concentrations in a different reaction vessel:
$\begin{array}{rl}\left[{\text{SO}}_{2}\right]& =3.6\phantom{\rule{0.167em}{0ex}}\text{M}\\ \\ \left[{\text{O}}_{2}\right]& =0.087\phantom{\rule{0.167em}{0ex}}\text{M}\\ \\ \left[{\text{SO}}_{3}\right]& =2.2\phantom{\rule{0.167em}{0ex}}\text{M}\end{array}$
We would like to know if this reaction is at equilibrium, but how can we figure that out? When we aren't sure if our reaction is at equilibrium, we can calculate the reaction quotient, $Q$:
$Q=\frac{\left[{\text{SO}}_{3}{\right]}^{2}}{\left[{\text{SO}}_{2}{\right]}^{2}\left[{\text{O}}_{2}\right]}$
At this point, you might be wondering why this equation looks so familiar and how $Q$ is different from ${K}_{\text{c}}$. The main difference is that we can calculate $Q$ for a reaction at any point whether the reaction is at equilibrium or not, but we can only calculate ${K}_{\text{c}}$ at equilibrium. By comparing $Q$ to ${K}_{\text{c}}$, we can tell if the reaction is at equilibrium because $Q={K}_{\text{c}}$ at equilibrium.
If we calculate $Q$ using the concentrations above, we get:
$\begin{array}{rl}Q& =\frac{\left[{\text{SO}}_{3}{\right]}^{2}}{\left[{\text{SO}}_{2}{\right]}^{2}\left[{\text{O}}_{2}\right]}\\ \\ & =\frac{\left[2.2{\right]}^{2}}{\left[3.6{\right]}^{2}\left[0.087\right]}\\ \\ & =4.3\end{array}$
Because our value for $Q$ is equal to ${K}_{\text{c}}$, we know the new reaction is also at equilibrium. Hooray!

## Example 2: Using ${K}_{\text{c}}$â  to find equilibrium compositions

Let's consider an equilibrium mixture of ${\text{N}}_{2}$, ${\text{O}}_{2}$ and $\text{NO}$:
${\text{N}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)â2\text{NO}\left(g\right)$
We can write the equilibrium constant expression as follows:
${K}_{\text{c}}=\frac{\text{[NO}{\right]}^{2}}{\left[{\text{N}}_{2}\right]\left[{\text{O}}_{2}\right]}$
We know the equilibrium constant is $3.4Ă{10}^{â21}$ at a particular temperature, and we also know the following equilibrium concentrations:
$\left[{\text{N}}_{2}\right]=\left[{\text{O}}_{2}\right]=0.1\phantom{\rule{0.167em}{0ex}}\text{M}$
What is the concentration of $\text{NO}\left(g\right)$ at equilibrium?
Since ${K}_{\text{c}}$ is less than 0.001, we would predict that the reactants ${\text{N}}_{2}$ and ${\text{O}}_{2}$ are going to be present in much greater concentrations than the product, $\text{NO}$, at equilibrium. Thus, we would expect our calculated $\text{NO}$ concentration to be very low compared to the reactant concentrations.
If we know that the equilibrium concentrations for ${\text{N}}_{2}$ and ${\text{O}}_{2}$ are 0.1 M, we can rearrange the equation for ${K}_{\text{c}}$ to calculate the concentration of $\text{NO}$:
$\left[\text{NO}\right]=\sqrt{K\left[{\text{N}}_{2}\right]\left[{\text{O}}_{2}\right]}$
If we plug in our equilibrium concentrations and value for ${K}_{\text{c}}$, we get:
$\begin{array}{rl}\left[\text{NO}\right]& =\sqrt{K\left[{\text{N}}_{2}\right]\left[{\text{O}}_{2}\right]}\\ \\ & =\sqrt{K\left[{\text{N}}_{2}\right]\left[{\text{O}}_{2}\right]}\\ \\ & =\sqrt{\left(3.4Ă{10}^{â21}\right)\left(0.1\right)\left(0.1\right)}\\ \\ & =5.8Ă{10}^{â12}\phantom{\rule{0.167em}{0ex}}\text{M}\end{array}$
As predicted, the concentration of $\text{NO}$, $5.8Ă{10}^{â12}\phantom{\rule{0.167em}{0ex}}\text{M}$, is much smaller than the reactant concentrations $\left[{\text{N}}_{2}\right]$ and $\left[{\text{O}}_{2}\right]$.

## Summary

• A reversible reaction can proceed in both the forward and backward directions.
• Equilibrium is when the rate of the forward reaction equals the rate of the reverse reaction. All reactant and product concentrations are constant at equilibrium.
• Given an equation $\text{aA}+\text{bB}â\text{cC}+\text{dD}$, the equilibrium constant ${K}_{\text{c}}$, also called $K$ or ${K}_{\text{eq}}$, is defined using molar concentration as follows:
${K}_{\text{c}}=\frac{\left[{\text{C]}}^{\text{c}}{\text{[D]}}^{\text{d}}}{\left[\text{A}{\right]}^{\text{a}}\left[\text{B}{\right]}^{\text{b}}}$
• For reactions that are not at equilibrium, we can write a similar expression called the reaction quotient $Q$, which is equal to ${K}_{\text{c}}$ at equilibrium.
• ${K}_{\text{c}}$ can be used to determine if a reaction is at equilibrium, to calculate concentrations at equilibrium, and to estimate whether a reaction favors products or reactants at equilibrium.

## Want to join the conversation?

• This article mentions that if Kc is very large, i.e. 1000 or more, then the equilibrium will favour the products.

I thought that if Kc is larger than one (1), then that's when the equilibrium will favour the products. Conversely, if Kc is less than one (1), the equilibrium will favour the reactants.
• If Kc is larger than 1 it would mean that the equilibrium is starting to favour the products however it doesnt necessarily mean that that the molar concentration of reactants is negligible.
for example - is the value of Kc is 2, it would mean that the molar concentration of reactants is 1/2 the concentration of products. In this case though the value of Kc is greater than 1, the reactants are still present in considerable amount.
Khan academy was trying to show us all the extreme cases, so the case in which Kc is 1000 the molar concentration of reactants is so less that practically the equilibrium has shifted almost completely to the product side and vice versa in case of Kc being 0.001.
And if you read carefully, they dont say that when Kc is very large products are favoured but they are saying that when Kc if very large mostly products are present and vice versa.
Hope this helps :-)
• why aren't pure liquids and pure solids included in the equilibrium expression?
• Equilibrium constant are actually defined using activities, not concentrations. The activity of pure liquids and solids is 1 and the activity of a solution can be estimated using its concentration. So, pure liquids and solids actually are involved, but since their activities are equal to 1, they don't change the equilibrium constant and so are often left out.
• "Kc is often written without units, depending on the textbook."

When Kc is given units, what is the unit?
• Depends on the question.
For example, in Haber's process: N2 +3H2<---->2NH3
Kc=[NH3]^2/[N2][H2]^3
Using molarity(M) as unit for concentration:
Kc=M^2/M*M^3=M^-2
i.e Kc will have the unit M^-2 or Molarity raised to the power -2.
Hope you can understand my vague explanation!!
• Any suggestions for where I can do equilibrium practice problems?
• Try googling "equilibrium practise problems" and I'm sure there's a bunch. Some will be PDF formats that you can download and print out to do more. Sorry for the British/Australian spelling of practise.
• ââI get that the equilibrium constant changes with temperature. I don't get how it changes with temperature. The formula for calculating Kc or K or Keq doesn't seem to incorporate the temperature of the environment anywhere in it, nor does this article seem to specify exactly how it changes the equilibrium constant, or whether it's a predicable change. With this in mind, can anyone help me in understanding the relationship between the equilibrium constant and temperature?
• You forgot main thing. Kc depends on Molarity and Molarity depends on volume of the soln, which in turn depends on 'temperature'. Hope you get it!
• If the equilibrium favors the products, does this mean that equation moves in a forward motion? Or would it be backward in order to balance the equation back to an equilibrium state?
• If it favors the products then it will favourite the forward direction to create for products (and fewer reactants). As the reaction proceeds, the reaction will approach the equilibrium, and this will cause the forward reaction to decrease and the backward reaction to increase until they are equal to each other.
• Any videos or areas using this information with the ICE theory?
• Check out 'Buffers, Titrations, and Solubility Equilibria'.
• Say if I had H2O (g) as either the product or reactant. Would I still include water vapor (H2O (g)) in writing the Kc formula?
• YES! Very important to know that with equilibrium calculations we leave out any solids or liquids and keep gases. So with saying that if your reaction had had H2O (l) instead, you would leave it out!
• What happens if Q isn't equal to Kc?