Calculating equilibrium constants

Calculating equilibrium constants, K_c

For any reversible reaction at equilibrium, such as:

aA_(l) + bB_(l) ⇌ cC_(l) + dD_(l)

Where A,B,C and D are the reactants and products, while a, b, c and d are the coefficients in the balanced equation for the reaction, then we can write the equilibrium constant for this reaction as:

K_c is the equilibrium constant and the subscript c refers to concentration of the reactants and products in mol dm^-3. The [] are used to indicate that we are using concentrations in mol dm^-3 for all of the reactants and products.

Example 1 - The Haber process- making ammonia

As an example consider the Haber process for making ammonia. The best place to start is by writing a balanced equation for the Haber process and then from this we can write an expression for the equilibrium contant K_c.

An equation for the Haber process is shown below:

Balanced symbolic equation: N_2(g) + 3H_2(g) ⇌ 2NH_3(g)

Now from this balanced symbolic equation we can write the expression for the equilibrium constant, K_c:

expression for the equilibrium contant for the Haber process

Calculating K_c

Suppose we have a flask filled with an equilbrium mixture of nitrogen, hydrogen and ammonia. The equilibrium mixture contains gases at the following concentrations: [NH₃]= 0.003M, [N₂]= 0.04M and [H₂]= 0.01M. Calculate the equilibrium constant, K_c for ammonia production. Since we know the equilibrium concentrations for all three gases it is simply a case of substituting the equilibrium concentrations into the expression above for K_c, this is shown below:

The equilibrium constant, K_c value of 0.225 is a small number indicating that the equilibrium mixture contains little ammonia and the position of equilibrium lies to the left.

Example 2 - Preparation of phosphorus pentachloride (PCl₅)

Phosphorus pentahloride (PCl₅) can be prepared by reacting phosphorus trichloride (PCl₃) with chlorine gas (Cl₂) according to the equation below:

Balanced symbolic equation: PCl_3(g) + Cl_2(g) ⇌ PCl_5(g)

The equilibrium constant, K_c for this reaction at the given temperature is 35. The equilibrium concentrations are: [PCl₃] = 0.01M and [Cl₂] = 0.015. Calculate the [PCl₅] at equilibrium. The equilibrium expression for the reaction is given below, all that is needed to solve this problem is to rearrange the equilibrium expression and make [PCl₅] the subject of the formula, this is outlined below:

Example 3 - Preparation of the ester ethyl ethanoate

You may remember from your gcse lessons that esters are made by reacting a carboxylic acid with an alcohol. The ester ethyl ethanoate is made by reacting the alcohol ethanol with the ethanoic acid. An equation for this esterification reaction is given below:

ethanoic acid_(aq) + ethanol_(aq) ⇌ ethyl ethanoate_(l) + water_(l)

CH₃COOH_(aq) + CH₃CH₂OH_(aq) ⇌ CH₃COOCH₂CH₃_(l) + H₂O_(l)

Calculate the equilibrium constant, K_c when 0.25mol of ethanol and ethanoic acid are mixed with water and an acid catalyst in a sealed flask, the reaction is heated to a constant temperature and allowed to reach equilibrium . The total volume of the system is 50ml and at equilibrium it was found by titration that there was 0.0825 mol of ethanoic acid present.

The best way to tackle this type of product is by drawing up a table similar to the one below, it is often called an ICE table (Initial, change, equilibrium). You should use this table or one similar to it to record and work out all the concentration of the reactants and products before the reaction starts and after it has reached equilibrium. You can also use it to workout how the concentration of the reactants and products change during the reaction.

reactants and products	ethanol	ethanoic acid	ethyl ethanoate	water
initial concentration (mol dm^-3)
change in concentration (mol dm^-3)
equilibrium concentration (mol dm^-3)

Initially we have the following:

CH₃COOH_(aq) + CH₃CH₂OH_(aq) ⇌ CH₃COOCH₂CH₃_(l) + H₂O_(l)

0.25mol + 0.25mol 0 mol + 0 mol

At equilibrium we know that there is 0.0825 mol of ethanoic acid present. Initially there was 0.25 mol of ethanoic acid present so:

This means that 0.25mol -0.0825mol = 0.1675mol of acid reacted.

Since mole of acid reacts with 1 mole of ethanol, then we can say that: 0.25mol -0.0825mol = 0.1675mol of ethanol reacted

From the stoichiometry of the equation we know that 1 mole of acid reacts with 1 mole of ethanol to produce 1 mole of the ester and 1 mole of water. Since 0.1675mol of acid reacted we know that 0.1675mol of ethanol also reacted to produce 0.1675mol of ester and 0.1675 mol of water.

So at equilibrium we have:

CH₃COOH_(aq) + CH₃CH₂OH_(aq) ⇌ CH₃COOCH₂CH₃_(l) + H₂O_(l)

0.0825mol + 0.0825mol 0.1675mol + 0.1675mol

All that is needed to calculate K_c is to convert the number of moles for each reactant and product at equilibrium into concentrations. This is easily done using the formula below, since we know the total volume of the system was 50ml or 0.05dm^-3.

This gives equilibrium concentrations of:

CH₃COOH_(aq) + CH₃CH₂OH_(aq) ⇌ CH₃COOCH₂CH₃_(l)+ H₂O_(l)

0.0825mol/0.05 + 0.0825mol/0.05 0.1675mol/0.05 + 0.1675mol/0.05

We can now fill in our ICE table using the calculate concentrations from above:

reactants and products	ethanol	ethanoic acid	ethyl ethanoate	water
initial concentration (mol dm^-3)	0.25/0.05 =5	0.25/0.05 =5	0	0
change in concentration (mol dm^-3)	-3.35	-3.35	+3.35	+3.35
equilibrium concentration (mol dm^-3)	0.0825/0.05=1.65	0.0825/0.05=1.65	0.1675/0.05=3.35	0.1675/0.05=3.35

To calculate K_c it is simply a matter of substituting the values for the concentrations of the reactants and products into the expression for the equilibrium constant:

Example 4- Calculating equilibrium amounts

As a varaition on the ester example above, suppose you mixed together 1 mol of ethanoic acid and 1 mol of ethanol and you wanted to know how much ester you would have at equilibrium. We already know K_c for the reaction and we also know the stoichiometry of the reaction.

Start by writing out a shortened version of our ICE table, simply stating the number of moles for each substance that we know:
To begin with we have 1 mol of ethanoic acid and 1 mol of ethanol.

CH₃COOH_(aq) + CH₃CH₂OH_(aq) ⇌ CH₃COOCH₂CH₃_(l) + H₂O_(l)

1mol + 1mol 0 mol + 0 mol

We do not know how many moles of the ester will be produced at equilibrium, so lets just call this x mol. From the stoichiometry of the equation if we produce x mol of ester and water at equilibrium, then this will consume x mol of ethanoic acid and ethanol. This means that at equilibrium we will be left with 1-x mol of ethanoic acid and ethanol:

CH₃COOH_(aq) + CH₃CH₂OH_(aq) ⇌ CH₃COOCH₂CH₃_(l) + H₂O_(l)

(1-x) mol + (1-x) mol ⇌ x mol + x mol

Recall that K_c requires concentrations, however we have only mols. Suppose the volume of the equilibrium mixture was V dm^-3, then we have:

so at equilibrium we have 2/3 mol of ester and water and 1/3 mol of ethanoic acid and ethanol in the equilibrium mixture.

Example 5 - Preparation of hydrogen iodide

You could be asked to calculate an equilibrium constant for a reaction when not all the equilibrium concentrations are known.
However as long as we know the concentration of the reactants and the equilibium concentration of at least one of the reactants or products then it is possible to calculate the equilibrium constant, K_c for the reaction.
As an example consider the reaction of hydrogen and iodine to form hydrogen iodide. The equation of the reaction is shown below:

Balanced symbolic equation: H_2(g) + I_2(g) ⇌ 2HI_(g)

Example:

A mixture of 5 x 10^-3 mol of hydrogen gas and 1.8 x 10^-3 mol of iodine were placed in a flask with a volume of 5 dm^-3. The flask was heated to a known temperature and allowed to reach equilibrium. At equilibrium the [HI] was found to be 2.88 x 10^-4 mol dm^-3. Calculate K_c for this reaction.

reactants and products	H₂	I₂	HI
initial concentration (mol dm^-3)
change in concentration (mol dm^-3)
equilibrium concentration (mol dm^-3)

Remember that when using K_c the "c" refers to concentration in mol dm^-3, however in the problem above we are only given moles and not concentrations. So the first step is to calculate the concentrations from the values given. The concentration is found using the formula:

So simply substitute the values for the number of moles of iodine and hydrogen into the equation to obtain their concentrations in mol dm^-3:

Now fill in the table with all known values of concentration:

reactants and products	H₂	I₂	HI
initial concentration (mol dm^-3)	1.0 x 10^-3	3.6 x 10^-4
change in concentration (mol dm^-3)
equilibrium concentration (mol dm^-3)			2.88 x 10^-4

The next step in finding the equilibrium constant, K_c is to calculate the equilibrium concentrations for hydrogen and iodine. This is easily done.
We need to consider the stoichiometry of the eqution for the reaction:

H_2(g) + I_2(g) ⇌ 2HI_(g)

From this we can see that for every 2 moles of HI produced 1 mole of H₂ and 1 mole of I₂ is consumed. Since the [HI] in the equilibrium mixture is 2.18 x 10^-4M, so to produce this 2.88 x 10^-4/2 mol dm^-3 or 1.44 x 10^-4 mol dm^-3 of each of the reactants will have to be consumed. We can fill this in new data in the table below:

reactants and products	H₂	I₂	HI
initial concentration (mol dm^-3)	1.0 x 10^-3	3.6 x 10^-4	0
change in concentration (mol dm^-3)	-1.44 x 10^-4	-1.44 x 10^-4	+2.88 x 10^-4
equilibrium concentration (mol dm^-3)	8.56 x10^-4	2.16 x 10^-4	2.88 x 10^-4

Substituting the equilibrium concentrations into the expression for K_c gives:

Key Points

We can calculate equilibrium constants if we know the equilibrium concentration of the reactants and products or if we know we are able to calculate the concentrations from other data.