Equilibrium and Kc

You should already know that many reactions are reversible and do not go to completion, but instead end up as an equilibrium mixture of reactant and products. A reversible reaction that can reach equilibrium is denoted by the symbol ⇌.

Many organic reactions are reversible and will reach equilibrium in time. Take the example ethanoic acid and ethanol where the ester ethyl ethanoate is produced. If the ethanol and ethanoic acid are mixed in a flask (stoppered to prevent evaporation and left, the mixture is eventually obtained in which four substances are present – though a strong acid catalyst is required. The equation now being.

C₂H₅OH + CH₃CO₂H ⇌ CH₃CO₂C₂H₅ + H₂O

The mixture can then be analysed by titrating ethanoic acid with standard alkali. This gives the number of moles of ethanoic acid. From this we can work out the number of moles of the other components, and from this their concentrations if the total volume of the mixture is known.

If several experiments are done with different quantities of starting materials, it is always found to be the ratio:

\[\frac{{[CH3COOCH2CH3(aq)]eqm \times [H2O(aq)]eqm}}{{[CH3COOH(aq)]eqm \times [CH3CH2OH(aq)]eqm}}\]

The ratio is always found to be constant if the temperature is constant. This constant is called the equilibrium constant and has the symbol Kc. The subscript _eqm means that the concentrations have been measured when the equilibrium has been reached. The Chemical Equilibrium subscript _c stands for concentration – the equilibrium constant is a ratio of concentrations.

The equilibrium law is generally written as follows:

aA + bB + cC ⇌ xX + yY + zZ

It is usually expressed like this:

\[\frac{{{{\left[ X \right]}_{eqm}}^x{{\left[ Y \right]}_{eqm}}^y{{\left[ Z \right]}_{eqm}}^z\;}}{{\;{{\left[ A \right]}_{eqm}}^a{{\left[ B \right]}_{eqm}}^b{{\left[ C \right]}_{eqm}}^c}}\]

This expression has a constant value, K_c, provided the temperature is constant. K_c is called the equilibrium constant and is different for different reactions. It changes with temperature. The units of K_c vary, and you must work them out for each reaction by cancelling out the units of each term.

If we take this imaginary example:

A + B ⇌ C

This can be expressed like this:

\[Kc = \frac{{{{[C]}^1}}}{{{{[A]}^1}{{[B]}^1}}}\]

I have written to the power of one this time to demonstrate that they are all to the power of their stoichiometric ratios (the balancing numbers). From now on if it is to the power of one then they can be left without a power!

If you replace the concentrations with the units of concentration then you can work out the overall unit of K_c.

\[Kc(units) = \frac{{mold{m^{ - 3}}}}{{mold{m^{ - 3}} \times mold{m^{ - 3}}}}\]

If you cancel out “like” terms, you will be left with:

\[Kc(units) = \frac{1}{{mold{m^{ - 3}}}}\]

This would then be expressed as:

\[Kc(units) = mo{l^{ - 1}}d{m^3}\]

Because you were left with one over the concentration, then you need to invert the powers so mol¹ becomes mol^-1 and dm^-3 becomes dm³.

Let’s look at a real life example:

0.10mol of ethanol is mixed with 0.10mol of a solution of ethanoic acid and allowed to reach equilibrium. The total volume of the system is 20.0cm³ or 0.020dm³. We find by titration that 0.033mol ethanoic acid is present once equilibrium is reached. From this we can work out the number of moles of the other components present at equilibrium.

At the start:

CH₃CH₂OH (l) + CH₃COOH (l) ⇌ CH₃COOCH₂CH₃ (l) + H₂O (l)

or

\[Kc = \frac{{[CH3COOCH2CH3(aq)]eqm \times [H2O(aq)]eqm}}{{[CH3COOH(aq)]eqm \times [CH3CH2OH(aq)]eqm}}\]

The best way to work out the unknown is by using a table like below. Fill in all the numbers you are given:

 

	CH3CH2OH	CH3COOH	CH3COOCH2CH3	H2O
Initial (mol)	0.10	0.10	0	0
Equilibrium (mol)		0.033
Equilibrium (conc)

Now you need to work out the unknowns. If you look at the mole ratios from the equation you will see that everything is 1:1 (there are no balancing numbers). So if you started with 0.1 of ethanoic acid and finished with 0.033, then it MUST be the same for ethanol.

	CH3CH2OH	CH3COOH	CH3COOCH2CH3	H2O
Initial (mol)	0.10	0.10	0	0
Equilibrium (mol)	0.033	0.033
Equilibrium (conc)

Again because of the 1:1 ratio, if we look at how much is used up (0.1 - 0.033) you will find that it equals 0.067 of each.

	CH3CH2OH	CH3COOH	CH3COOCH2CH3	H2O
Initial (mol)	0.10	0.10	0	0
Equilibrium (mol)	0.033	0.033	0.067	0.067
Equilibrium (conc)

We now need the concentrations which means we need to divide the moles by the total volume. For the purpose of this example we will turn each one into concentrations, but because we have equal numbers of moles on each side of the equation the volumes actually cancel each other out.

	CH3CH2OH	CH3COOH	CH3COOCH2CH3	H2O
Initial (mol)	0.10	0.10	0	0
Equilibrium (mol)	0.033	0.033	0.067	0.067
Equilibrium (conc)	0.033/0.02 = 1.65	0.033/0.02 = 1.65	0.067/0.02 = 3.35	0.067/0.02 = 3.35

Now insert the numbers into our K_c expression:

\[Kc = \frac{{3.35 \times 3.35}}{{1.65 \times 1.65}} = 4.1\;no\,units\]

In this example Kc does NOT have a unit. This is for the same reason we technically did not need to divide each of the moles by volume, because there are the same number of moles on each side of the equation.

\[Kc(units) = \frac{{mold{m^{ - 3}} \times mold{m^{ - 3}}}}{{mold{m^{ - 3}} \times mold{m^{ - 3}}}}\]

Each of the units of concentration cancel out

Sometimes you will not even be given a total volume, and have no way of working it out. In this case it is always good practice to show the examiner that you would be doing this step anyways by dividing each of your mole quantities by V:

\[Kc = \frac{{(0.067/V) \times (0.067/V)}}{{(0.033/V) \times (0.033/V)}}\]

Equilibrium Constants and Le Chatelier’s Principle

Changing concentrations: Equilibrium constants aren't changed if you change the concentrations of things present in the equilibrium. The only thing that changes an equilibrium constant is a change of temperature.

However, even though K_c is not changed, the position of equilibrium is changed if you change the concentration of something present in the mixture. According to Le Chatelier's Principle, the position of equilibrium moves in such a way as to tend to undo the change that you have made. Suppose you have an equilibrium established between four substances A, B, C and D.

A + 2B ⇌ C + D

According to Le Chatelier's Principle, if you decrease the concentration of C, for example, the position of equilibrium will move to the right to increase the concentration again.

The equilibrium constant, Kc for this reaction looks like this:

\[Kc = \frac{{C \times D}}{{A \times {B^2}}}\]

If you have moved the position of the equilibrium to the right (and so increased the amount of C and D), why hasn't the equilibrium constant increased?

If you decrease the concentration of C, the top of the K_c expression gets smaller. That would change the value of K_c. In order for that not to happen, the concentrations of C and D will have to increase again, and those of A and B must decrease. That happens until a new balance is reached when the value of the equilibrium constant expression reverts to what it was before. The position of equilibrium moves - not because Le Chatelier says it must - but because of the need to keep a constant value for the equilibrium constant.

Changing Temperature

This is typical of what happens with any equilibrium where the forward reaction is exothermic. Increasing the temperature decreases the value of the equilibrium constant. Where the forward reaction is endothermic, increasing the temperature increases the value of the equilibrium constant.

The position of equilibrium also changes if you change the temperature. According to Le Chatelier's Principle, the position of equilibrium moves in such a way as to tend to undo the change that you have made. If you increase the temperature, the position of equilibrium will move in such a way as to reduce the temperature again. It will do that by favouring the reaction which absorbs heat. Take this equilibrium as an example:

H₂(g) + I₂(g) ⇌ 2HI(g) ΔH = -10.4kjmol^-1

According to Le Chatelier's Principle the position of equilibrium will move to the left. Less hydrogen iodide will be formed, and the equilibrium mixture will contain more unreacted hydrogen and iodine. Changing the pressure of the system will NOT change the equilibrium constant. We actually use a special equilibrium constant for systems involving gases called K_p which is in another section.

Using a Catalysts

Adding a catalyst does NOT change the equilibrium constant. This is because it increases the rate of both the forward and backwards reaction so K_c will therefore remain constant.