Lesson Explainer: The Equilibrium Constant for Concentration Chemistry

In this explainer, we will learn how to construct and calculate the equilibrium constant for concentration.

We can quantify the equilibrium that exists between reactants and products in terms of their concentrations by calculating and using the equilibrium constant ๐พ๏Œผ. The equilibrium constant for concentration expresses a value related to the ratio between the concentration of the reactants and the concentration of the products at equilibrium. The equilibrium constant, as the name implies, relates to a closed system that does not allow any matter exchange with its surroundings; therefore, the concentrations of the products and reactants do not change once the system is at equilibrium.

Definition: Closed System

It is a system that does not exchange matter but can exchange energy with its surroundings.

In its simplest form, the equilibrium constant for concentration, ๐พ๏Œผ, can be written as ๐พ=[][].๏Œผproductsreactants

If we extend this to the general equation A+BC+D in which an equilibrium is formed between two reactants, A and B, and two products, C and D, then ๐พ๏Œผ would be written as follows: ๐พ=.๏Œผ[C][D][A][B]

Remember that, in chemistry, we use square brackets to indicate concentration, which is most commonly measured in moles (M), moles per cubic decimetre (molโ‹…dmโˆ’3), or moles per litre (mol/L).

When writing equations for ๐พ๏Œผ from actual chemical equations, it is essential to take note of the stoichiometry. Letโ€™s look closely at the equation for ๐พ๏Œผ for the equilibrium between nitrogen dioxide and dinitrogen tetroxide: 2๐พ=NO()NO()[NO][NO]224242gg๏Œผ๏Šจ

We can see that the concentration of nitrogen dioxide is raised to the power of two and that this mirrors the value of the stoichiometric coefficient of nitrogen dioxide in the original chemical equation. In a slightly more complicated example that reinforces this point, the common equilibrium reaction that occurs as part of the Haber process in the generation of ammonia has an equilibrium constant for concentration expressed as N()+H()NH()[NH][H][N]223322ggg32๐พ=๏Œผ๏Šจ๏Šฉ

Definition: Equilibrium Constant for Concentration

For the equation ๐‘Ž๐‘๐‘๐‘‘A+BC+D the equilibrium constant for concentration, ๐พ๏Œผ, is the product of the concentrations of C and D divided by the product of the concentrations of A and B, each raised to the power of its respective stoichiometric coefficient: ๐‘Ž, ๐‘, ๐‘, or ๐‘‘.

Example 1: Determining the Equation for ๐พ๐‘ from a General Reaction

Consider the general reaction ๐‘Ž๐‘๐‘๐‘‘A+BC+D

What is the expression for the equilibrium constant for the reversible reaction of this general reaction? Take [A] and [B] to be the molar concentrations of the reactants, [C] and [D] to be the molar concentrations of the products, and ๐‘Ž, ๐‘, ๐‘, and ๐‘‘ to be the stoichiometric coefficients in the balanced equation.

  1. ๐พ=๐‘๐‘‘๐‘Ž๐‘๏Œผ[C][D][A][B]
  2. ๐พ=๏Œผ๏Œผ๏Œฝ๏Œบ๏Œป[C][D][A][B]
  3. ๐พ=๏Œผ๏Œบ๏Œป[A][B]
  4. ๐พ=๏Œผ๏Œบ๏Œป๏Œผ๏Œฝ[A][B][C][D]
  5. ๐พ=๏Œผ๏Œผ๏Œฝ[C][D]

Answer

At its simplest level, we know that ๐พ๏Œผ is the equilibrium constant for concentration and that, like all equilibrium constants, it is represented in some fashion by the products being divided by the reactants. We know that in this general reaction equation, the products are C and D; therefore, we immediately know that we can discount answers C, D, and E.

We also know that when writing the equilibrium constant of the concentration, each of the concentrations of the reactants and the products is raised to the power of its stoichiometric coefficient. Answer A has the stoichiometric coefficients multiplied by the concentrations, whereas answer B has concentrations raised to the powers of the stoichiometric coefficients, which is what we are looking for. Hence, answer B is the correct answer.

It is important to note that some questions may involve a chemical reaction in which one of the products and/or reactants may be in the solid phase. In this case, the chemical species in the solid phase is not included in the value for ๐พ๏Œผ, nor is the concentration of pure water when used as a solvent (HO()2l).

Example 2: Writing the Equation for ๐พ๐‘ in the Reduction of Iron(II) Ions by Silver Ions

Silver ions can be reduced by iron(II) ions in the net ionic equation Fe()+Ag()Fe()+Ag()2++3+aqaqaqs

What is the correct equation for ๐พ๏Œผ for this reaction?

Answer

When writing the equation for ๐พ๏Œผ, we must initially follow the overall principle of writing ๐พ๏Œผ equations, which is that the product of the concentrations of the products should be divided by the product of the concentrations of the reactants.

However, in this particular chemical reaction, we must be mindful of the state symbols of each of the different chemical species. The silver product is solid silver metal and should not be included in the equation for ๐พ๏Œผ, as the effective concentration of a solid stays constant throughout the reaction.

As such, the only product in this reaction that is included in the expression for ๐พ๏Œผ is the concentration of iron(III) ions, which should be divided by the product of the concentration of iron(II) ions and silver ions. Hence, the correct answer is ๐พ=.๏Œผ[Fe][Fe][Ag]3+2++

At higher levels, the discussion of a topic known as dimensionality explains why equilibrium constants are often being expressed without a unit, but that is beyond the scope of this explainer.

However, in certain cases, units for ๐พ๏Œผ may need to be calculated. As such, it is important to be able to multiply and divide moles per cubic decimetre (molโ‹…dmโˆ’3) or moles per litre (mol/L) as well as understand what happens to these values when they are raised to different powers. Letโ€™s examine again the reaction between nitrogen and hydrogen. The ๐พ๏Œผ equation in question is ๐พ=.๏Œผ[NH][H][N]32232

Given the equation for ๐พ๏Œผ above, the final unit would be resolved as follows: ๏€บโ‹…๏†๏€บโ‹…๏†๏€บโ‹…๏†=โ‹…โ‹…=1โ‹…=โ‹….moldmmoldmmoldmmoldmmoldmmoldmmoldm๏Šฑ๏Šฉ๏Šจ๏Šฑ๏Šฉ๏Šฉ๏Šฑ๏Šฉ๏Šจ๏Šฑ๏Šฌ๏Šช๏Šฑ๏Šง๏Šจ๏Šจ๏Šฑ๏Šฌ๏Šฑ๏Šจ๏Šฌ

Example 3: Calculating the Units of ๐พ๐‘ Equations

Find the unit for the ๐พ๏Œผ equation ๐พ=.๏Œผ[HI][H][I]222

  1. No unit
  2. molโ‹…dmโˆ’3
  3. molโˆ’1โ‹…dm3
  4. mol2โ‹…dmโˆ’6
  5. molโˆ’2โ‹…dm6

Answer

When concentrations are represented using square brackets in chemistry, we can assume that the unit we are discussing is moles per cubic decimetre (molโ‹…dmโˆ’3) or moles per litre (mol/L), unless we are told otherwise.

In this equation, the unit of the products on top is raised to the power of 2, whereas the units at the bottom are multiplied by each other. We can see in the equation below how these units would cancel out, meaning that answer A, โ€œNo unit,โ€ is the correct answer: ๏€บโ‹…๏†๏€บโ‹…๏†๏€บโ‹…๏†=โ‹…โ‹…โŸถ.moldmmoldmmoldmmoldmmoldmnounit๏Šฑ๏Šฉ๏Šจ๏Šฑ๏Šฉ๏Šฑ๏Šฉ๏Šจ๏Šฑ๏Šฌ๏Šจ๏Šฑ๏Šฌ

In terms of actually calculating ๐พ๏Œผ, there are a variety of ways in which the calculation could be attempted. For example, a question may give you all the concentrations required, meaning that all that would be left to do would be to input the data into the ๐พ๏Œผ equation.

Letโ€™s imagine that during the production of ammonia, the equilibrium mixture contains 0.982 molโ‹…dmโˆ’3 of NH()3g and 0.193 molโ‹…dmโˆ’3 of each H()2g and N()2g. The reaction is N()+3H()2NH()223ggg

So, the equation for ๐พ๏Œผ must be ๐พ=.๏Œผ[NH][H][N]32232

If we substitute the correct numbers into the equation, we can simply calculate the value of ๐พ๏Œผ as follows: ๐พ==[0.982][0.193][0.193]=695โ‹….๏Œผ๏Šจ๏Šฉ๏Šฑ๏Šจ๏Šฌ[NH][H][N]moldm32232

When calculating ๐พ๏Œผ, it is important to ensure that the correct information is extracted from the question. The numbers that need to be used in the calculation should be concentrations, so it may be necessary to calculate the concentration from information in the question before calculating ๐พ๏Œผ.

It may also be the case that all the values needed at equilibrium are not given, but initial values of concentrations or moles are given instead. In this situation, a table is often the most sensible way to clearly and carefully calculate the numbers that will need to eventually be put into the ๐พ๏Œผ equation. One popular method of creating a table is known as the ICE method, which we will use here; however, there are others.

Hydrogen sulfide can be decomposed by heat according to the equation 2HS()2H()+S()222ggg0.50 mol of hydrogen sulfide is heated at a steady temperature in a container of volume 2.0 dm3 until an equilibrium in which 0.38 mol of HS2 still remains is reached.

2HS22H2S2
Initial, I (Moles at Start) 0.500.000.00

In the first row of the table above, we put the โ€œInitial,โ€ which gives us the letter I. These are the values from the start of the experiment before the reaction takes place and equilibrium is reached.

Change, Cโˆ’2๐‘ฅ+2๐‘ฅ+๐‘ฅ
Equilibrium, E (Moles at Equilibrium)0.380.120.06

Next, we fill out the next two rows, โ€œChangeโ€ and โ€œEquilibrium.โ€ The change ๐‘ฅ is multiplied by the stoichiometric coefficient from the reaction equation 22HS()H()+S()222ggg

We can consider the change in the reactants to be negative and the change in the products to be positive, hence the positive and negative values in the row โ€œChange.โ€

We are told that 0.38 mol of HS2 remains at equilibrium; therefore, 0.12 mol must have been converted. We now know that โˆ’2๐‘ฅ=โˆ’0.12.mol

From here, we can calculate that 2๐‘ฅ=0.12mol and ๐‘ฅ=0.06mol.

Concentration at Equilibrium0.382=0.190.122=0.060.062=0.03

In the final row of the table, we use the volume, given to us in the question, to calculate the concentration at equilibrium.

With this information, we can calculate ๐พ๏Œผ as follows: ๐พ==0.06ร—0.030.19=2.99ร—10โ‹….๏Œผ๏Šจ๏Šจ๏Šฑ๏Šฉ๏Šฑ๏Šฉ[H][S][HS]moldm22222

It is also possible that a question may provide you with the amount of substance but not the volume. At first, this may cause you to quite rightly think that there is no solution, as concentration cannot be calculated without a volume. However, in these cases, the volumes will often cancel out, making them an unnecessary piece of information. This can be seen in the following example.

In this example, 0.50 mol of hydrogen gas and 0.40 mol of iodine gas are heated and an equilibrium is allowed to establish. The equilibrium mixture contains 0.64 mol of hydrogen iodide: H()+I()2HI()22ggg

Again, using the ICE method, the following table can be completed.

H2I22HI
Initial, I (Moles at Start)0.500.400.00
Change, Cโˆ’๐‘ฅโˆ’๐‘ฅ+2๐‘ฅ
Equilibrium, E (Moles at Equilibrium)0.180.080.64
Concentration at Equilibrium0.18/๐‘‰0.08/๐‘‰0.64/๐‘‰

Now, we have the values to substitute into the ๐พ๏Œผ equation as follows: ๐พ==[0.64/๐‘‰][0.18/๐‘‰][0.08/๐‘‰]=[0.64][0.18][0.08]=28.44.๏Œผ๏Šจ๏Šจ[HI][H][I]222

Using the examples above, we can see how we are able to calculate values of ๐พ๏Œผ or use values of ๐พ๏Œผ to calculate concentrations of products or reactants at equilibrium. Furthermore, it is vitally important that we have some understanding of what the value of ๐พ๏Œผ is telling us, now that we have quantified it.

If the value of ๐พ๏Œผ is greater than one (๐พ>1)๏Œผ, we know that the equilibrium lies to the right-hand side and that the equilibrium favors the products (forward reaction). If the value of ๐พ๏Œผ is less than one (๐พ<1)๏Œผ, we know that the equilibrium lies to the left-hand side and favors the reactants (backward reaction). While the manner in which the value of ๐พ๏Œผ increases or decreases is not always linear, generally, the greater or smaller the value of ๐พ๏Œผ, the further the equilibrium lies in favor of either the products or the reactants.

Example 4: Calculating the Value of ๐พ๐‘ for a Reaction Involving the Oxides of Sulfur

At equilibrium, ๐พ=32โ‹…๏Œผ๏Šฑ๏Šง๏Šฉmoldm at 325 K for the following reaction involving oxides of sulfur: 2SO()+O()2SO()223ggg

Find the value of ๐พ๏Œผ, including units, at 325 K for the reaction: 2SO()2SO()+O()322ggg

  1. 0.03125 molโ‹…dmโˆ’3
  2. 0.32 molโ‹…dmโˆ’3
  3. 0.32 molโˆ’1โ‹…dm3
  4. 3.125 molโˆ’1โ‹…dm3

Answer

At first inspection, it may appear that we are missing a lot of information in this question. How could we possibly start to calculate ๐พ๏Œผ for the second equation with no information about concentrations? However, if we look closely, we will see that the second reaction is, in fact, the reverse reaction of the first equation. If we were to write out the equations for ๐พ๏Œผ for both of these equations, they would be as follows.

For equation 1, ๐พ=.๏Œผ[SO][SO][O]32222

For equation 2, ๐พ=.๏Œผ[SO][O][SO]22232

As we can see, the second value of ๐พ๏Œผ is, in fact, the reciprocal of the first equation. As the temperature is the same, we can find the reciprocal of 32, which will give us the value of ๐พ๏Œผ for the second equation. Remember, however, that we also need to find the reciprocal of the units. So, our final answer would be as follows: ๐พ=132โ‹…=0.03125โ‹….๏Œผ๏Šฑ๏Šฉ๏Šฑ๏Šฉmoldmmoldm

Key Points

  • Equilibria can be quantitatively evaluated.
  • The equilibrium constant for concentration, ๐พ๏Œผ, is the product of the concentrations of the products divided by the product of the concentrations of the reactants, each raised to the power of its respective stoichiometric coefficient.
  • Tables and the ICE method can be used to calculate concentrations at equilibrium for use in ๐พ๏Œผ equations.
  • Large values for ๐พ๏Œผ indicate that an equilibrium lies far to the right-hand side and favors the products, whereas values smaller than 1 indicate that an equilibrium lies to the left-hand side and favors the reactants.

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