Equilibrium constants heading

Calculating equilibrium constants, Kc

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:

expression for the equilibrium constant

Kc 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 Kc.

An equation for the Haber process is shown below:

Balanced symbolic equation: N2(g) + 3H2(g) 2NH3(g)
Now from this balanced symbolic equation we can write the expression for the equilibrium constant, Kc: expression for the equilibrium contant for the Haber process

Calculating Kc

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, Kc 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 Kc, this is shown below:

The equilibrium constant, Kc 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: PCl3(g) + Cl2(g) PCl5(g)

The equilibrium constant, Kc 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)
CH 3COOH(aq) + CH3CH2OH(aq) CH3COOCH2CH3(l) + H2O(l)

Calculate the equilibrium constant, Kc 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:

CH 3COOH(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:

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 Kc 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:

CH 3COOH(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 Kc 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 Kc 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.
CH 3COOH(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:

CH 3COOH(aq)   +   CH3CH2OH(aq) CH3COOCH2CH3(l) + H2O(l)
     (1-x) mol   +       (1-x) mol   ⇌       x mol      +       x mol
Recall that Kc 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, Kc 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: H2(g) + I2(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 Kc 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 Kc 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, Kc 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:

H2(g) + I2(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 Kc gives:


Key Points

Practice questions

Check your understanding - Questions Equilibrium constants


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