Lesson Explainer: The Equilibrium Constant for Partial Pressures Chemistry

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

When an equilibrium is established in a closed system, equilibrium constants can be calculated to quantify the position of the equilibrium. When an equilibrium involves gases, we can use the pressure exerted by the different gases in the mixture to calculate the equilibrium constant for partial pressures.

In a similar fashion to other equilibrium constants, the equilibrium constant for partial pressures can most simply be expressed as the partial pressures of the products divided by the partial pressures of the reactants: 𝐾=𝑃𝑃.productsreactants

The 𝑃 in the above equation represents the partial pressure of the respective chemical species. To use a specific chemical equation, 𝐾 would be written as 𝐾=ο€Ίπ‘ƒο†ο€Ήπ‘ƒο…π‘ƒοŒοŠ¨οŠ©NHHN322 for the reaction between hydrogen and nitrogen gas to form ammonia: N()+3H()2NH(g)223gg

Notice that, in the equation for 𝐾 shown above, the individual partial pressures of the products and reactants are each raised to the power of the stoichiometric coefficients of the balanced reaction equation that can give us the following definition for the equilibrium constant for partial pressures.

Definition: Equilibrium Constant for Partial Pressures

The equilibrium constant for partial pressures, 𝐾, is the ratio between the concentrations of products and reactants expressed as partial pressures at equilibrium.

For the equation abcdA+BC+D,𝐾 is calculated as the product of the partial pressures C and D divided by the product of the partial pressures of A and B, where each individual partial pressure is raised to the power of their respective stoichiometric coefficients c, d, a, and b.

Example 1: Writing Equations of 𝐾𝑝 from Chemical Equations

The equilibrium of a mixture of gases can be evaluated by 𝐾, the equilibrium constant for partial pressures.

  1. Consider the equation 2NO()NO()224gg What is the correct equation of 𝐾 for the equilibrium between the two different oxides of nitrogen shown?
  2. Consider the equation 4NH()+7O()4NO()+6HO()3222gggg What is the correct equation of 𝐾 for the equilibrium between ammonia and oxygen shown?

Answer

Part 1

The equilibrium constant for partial pressures is formulated in a similar way to other equilibrium constants in that numerical values relating to the products are divided by the same numerical values of the reactants. As such, we know that the partial pressure for NO24 should be divided by the partial pressure for NO2.

Additionally, we also know that when writing the equation for the equilibrium constant for partial pressures each partial pressure is raised to the power of the stoichiometric coefficient of that chemical from the reaction equation. In this case, the partial pressure of nitrogen dioxide should be raised to the power of 2 as per the reaction equation.

The correct answer is: 𝐾=𝑃𝑃.NONO242

Part 2

Although this example may look more complicated than the previous one, it is very similar, and immediately, like we did above, we can deduce that the partial pressures of the products will be divided by the partial pressures of the reactants.

We also need to raise each of the individual partial pressures to the power of the stoichiometric coefficient from the reaction equation to give the final correct answer.

So, the correct answer is: 𝐾=𝑃𝑃𝑃𝑃.οŠͺοŠͺNOHONHO2232

In order for us to calculate the final value for 𝐾, we need to be able to calculate the individual partial pressures for each of the products and reactants. Some questions may give you the partial pressures, in which case they can be placed directly into the equation and the equation can be solved. However, if you do not have the partial pressures, you will need to calculate them and you can do this by calculating the mole fraction, X.

Definition: Mole Fraction, 𝑋

The mole fraction is the amount of a substance measured in moles, divided by the total amount of all of the substances in the equilibria mixture, also expressed in moles.

If we take a general equation such as A+BC+D, then the mole fraction of A can be expressed as follows, where 𝑛 represents the amount of substance measured in moles: 𝑋=𝑛𝑛,𝑛=𝑛+𝑛+𝑛+𝑛.AAtotaltotalABCDwhere

Example 2: Calculating the Mole Fraction for a Mixture of Gases at Equilibrium

A mixture of gases at equilibrium contains 20 moles of SO2 gas, 40 moles of O2 gas, and 40 moles of SO3 gas. What is the mole fraction of SO2 at equilibrium?

Answer

We define the mole fraction as the amount of a substance, measured in moles, divided by the total amount of all of the substances in the equilibria mixture, also expressed in moles.

For a general equation A+BC+D, the mole fraction for A, 𝑋A, can be written as 𝑋=𝑛𝑛+𝑛+𝑛+𝑛.AAABCD

At equilibrium, there are 20 moles of SO2 gas and the total number of moles of all the chemicals at equilibrium is 100 moles: 40+40+20=100.

If we divide 20 by 100, we get the answer of 0.2, which is the mole fraction for this sample of sulfur dioxide at equilibrium at this temperature: 20100=0.2.

An assumption is made here that all of the different molecules within the mixture are all of the same size and all behave identically. Once we know the mole fraction of one of the products or reactants, we are able to work out its partial pressure as a proportion of the total pressure exerted by all of the gases in the equilibria mixture. This is done by multiplying the mole fraction by the total pressure of the system: 𝑃=𝑋×𝑃.AAtotal

Example 3: Calculating the Partial Pressure of a Gas at Equilibrium

A mixture of gases in a closed vessel at equilibrium contains three different gases: nitrogen (20 moles), hydrogen (45 moles), and ammonia (25 moles). Calculate the partial pressure of hydrogen at 40 atm.

Answer

The question is asking us to calculate the partial pressure of hydrogen; however, in order for us to work out the partial pressure of hydrogen, we first need to know the mole fraction. The mole fraction for hydrogen is calculated as the amount of hydrogen, measured in moles, divided by the total amount of all of the chemicals, also measured in moles, at equilibrium: 𝑋=4520+45+25=0.5.hydrogen

At equilibrium, we have 90 moles in total with hydrogen making up 45 moles; hence, the mole fraction of hydrogen would be 0.5. As the numbers here are relatively simple, we can easily see that multiplying the mole fraction 0.5 by the total pressure of 40 atm gives us the final correct answer of 20 atm for the partial pressure of hydrogen: 0.5Γ—40=20.atm

Once all of the partial pressures have been calculated for each of the reactants and products, these numbers can be put into the equation for 𝐾. Let’s put all of these steps together and imagine we had to calculate the equilibrium constant for partial pressures, 𝐾, for an equilibrium mixture of 30 moles of nitrogen, 25 moles of hydrogen, and 50 moles of ammonia at a pressure of 50 atm: N()+3H()2NH()223ggg

It is good practice to store and process the given information in a table, where we will calculate the mole fraction and the partial pressure, as we have done before.

NitrogenHydrogenAmmonia
Moles at Equilibrium, 𝑛302050
Mole Fraction, 𝑋30100=0.320100=0.250100=0.5
Partial Pressure, 𝑃0.3Γ—50=150.2Γ—50=100.5Γ—50=25

We now need to write our equation for 𝐾 from the reaction equation as follows: 𝐾=𝑃𝑃𝑃.NHHN322

We then substitute in the values we calculated earlier from our table and solve for the correct answer, taking care to include or discount the units depending on the requirements of the question: 𝐾=(25)(10)Γ—15𝐾=0.0416.atm

In terms of the units, the final answer had the units of atmβˆ’2. Pressures will often be given with units of atmospheres, atm, or kilopascals, kPa. We will now break down exactly what happened with the units in the previous example so we are clear on exactly how we arrived at that result: =Γ—==.atmatmatmatmatmatmοŠͺ

Some questions will request that the units be given; others will not. This relates to whether or not it has been assumed that a thermodynamic factor known as the quotient of activities is constant under the conditions at which the equilibrium constant has been determined. However, these quotients are part of a concept known as dimensionality, which is a topic often discussed after high school and is beyond the scope of this explainer. It is best to look at the question you are answering and see whether or not units are requested in the answer.

It may be that a question does not give you all the information that you require for the products and reactants at equilibrium. It may be the case that values are given for the start of the reaction, in which case, another stage of working is necessary to work out how much of each of the gases is present at equilibrium before beginning to calculate the mole fractions. Here, again, using a table is a very useful way of accurately keeping track of the numbers and avoiding mistakes. One popular method of presenting these tables is known as the ICE method, which we will use here.

A typical question may involve a mixture of gases such as sulfur dioxide and oxygen. Let’s imagine that sulfur dioxide and oxygen were introduced into a heated vessel at 500 K and under 45 atm of pressure. Furthermore, the initial reaction mixture contained 18.0 mol of SO2 and 8.00 mol of O2, and, once an equilibrium had been established, 12.00 mol of the SO2 was converted into SO3.

For the reversible reaction between sulfur dioxide and oxygen 2SO()+O()2SO()223ggg how would we calculate the value of 𝐾?

SO2O2SO3
Initial, I
(𝑛, Moles at Start)
18.08.00.00
Change, Cβˆ’2π‘₯βˆ’π‘₯+2π‘₯
Equilibrium, E
(𝑛eq, Moles at Equilibrium)
TotalMoles=6+2+12=20
6.02.012.0
Mole Fraction, 𝑋620=0.3220=0.11220=0.6
Partial Pressure, 𝑃0.3Γ—45=13.50.1Γ—45=4.50.6Γ—45=27.0

The second row in the table is the β€œI” from ICE and relates to the initial concentrations of the reactants in the chemical reaction. Although it is not given to us, we know before the reaction starts that no sulfur trioxide could have been formed, so we know that the number of moles at the start should be zero for sulfur trioxide.

The third row in the table is the β€œC” from ICE and is the change in moles between the start of the reaction and equilibrium. We know that 12 moles of sulfur dioxide are converted into sulfur trioxide, and we use the value β€œ2” here as this relates to the stoichiometric value from the reaction equation. The same stoichiometric values help us fill in the rest of the row.

The next row is the equilibrium row, β€œE” from ICE, and we know that the number of moles of sulfur dioxide at equilibrium must be 6 moles as 12 moles have been converted. If 2π‘₯ is equivalent to 12 moles then β€œπ‘₯” must equal 6, and as such we can calculate the number of moles at equilibrium for all of the remaining reactants and products.

In the final two rows of the table, we calculate the mole fraction and partial pressures as we have done before: 𝐾=𝑃𝑃𝑃.SOSOO322

Writing our equation for 𝐾 and substituting in the values from our table, we can solve the equation and arrive at our answer: 𝐾=(27.0)(13.5)Γ—4.5𝐾=0.889.atm

Example 4: Calculating 𝐾𝑝 at Equilibrium for a Mixture of Nitrogen, Hydrogen, and Ammonia

6.00 mol of N2 gas and 20.00 mol of H2 were allowed to react at 650 K and 50 atm of pressure. At equilibrium, 4 mol of N2 gas had been converted into ammonia according to the following reaction equation: N()+3H()2NH()223ggg

Calculate 𝐾 for this equilibrium, giving your answer to two decimal places in scientific notation.

Answer

We do not have the quantities of all the gases at equilibrium, and so we will need to construct a table and use the ICE method to deduce this information.

N2H2NH3
Initial, I
(𝑛, Moles at Start)
6.0020.000.00
Change, Cβˆ’π‘₯βˆ’3π‘₯+2π‘₯
Equilibrium, E
(𝑛eq, Moles at Equilibrium)
TotalMoles=2+8+8=18
2.08.08.0
Mole Fraction, 𝑋218=0.11818=0.44818=0.44
Partial Pressure, 𝑃0.11Γ—50=5.560.44Γ—50=22.220.44Γ—50=22.22

We can now put these values into the equation for 𝐾: 𝐾=𝑃𝑃𝑃𝐾=22.2222.22Γ—5.56,NHHN322 giving us the correct answer of 8.10Γ—10 atmβˆ’2.

To conclude our discussion of this topic, it is important to think about what 𝐾 is telling us. A value of 𝐾 greater than 1 𝐾>1ο…οŒ corresponds to an equilibrium that lies toward the right-hand side and favors the products (forward reaction), whereas values of 𝐾 less than one 𝐾<1ο…οŒ correspond to equilibria in which the position of the equilibrium favors the reactants (backward reaction).

Key Points

  • Equilibria involving mixtures of gases can be evaluated quantitatively by using partial pressures.
  • The equilibrium constant for partial pressures, 𝐾, is the ratio between the concentrations of products and reactants expressed as partial pressures at equilibrium. It is calculated as the product of the partial pressures of the reaction products divided by the product of the partial pressures of the reactants, where each individual partial pressure is raised to the power of their respective stoichiometric coefficients.
  • In order to calculate partial pressures for individual gases in the mixture, it may be necessary to first work out the mole fraction of the individual gases.
  • Tables and the ICE method are useful ways to track and calculate the amounts of different gases present at equilibrium when the information is not provided by the question.
  • Large values of 𝐾 indicate an equilibrium that favors the right-hand side and production of the products, whereas a value of 𝐾 less than one indicates an equilibrium that favors the left-hand side and a great proportion of reactants.

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