laboratory 1009178 640 Equilibrium expression and the equilibrium constant (A level Chemistry)

The equilibrium constant Kc

Equilibrium reaction do not go to completion but will reach equilibrium with time. The reaction between ethanoic acid CH3COOH and ethanol C2H5OH to produce ethyl ethanoate CH3COOC2H5 and water H2O is a typical equilibrium reaction which proceeds as follows:


If ethanol and ethanoic acid are mixed in a flask, stoppered to prevent evaporation and left for several days with a catalyst, an
equilibrium mixture will form in which all four substances are

We may then be able to analyse the equilibrium mixture by titrating the ethanoic acid with a standard alkali. This will allow us to work out the number of moles of ethanoic acid in the equilibrium mixture and also the number of moles of the other components. Using the total volume of the mixture we can then calculate the concentrations of the components.

If we repeat the experiment several times, each time starting with different quantities of reactants, we find that the ratio:


will remain constant, whereby $latex [CH_3COOC_2H_5]$ means concentration of $latex CH_3COOC_2H_5$.

For any equilibrium reaction:

aA + bB โ‡Œ yY + zZ

At equilibrium, the ratio $latex \frac{[Y]^y[Z]^z}{[A]^a[B]^b}$ is a constant, provided the temperature is kept constant. We call this constant the equilibrium constant, Kc.

$$K_c\ = \frac{[Y]^y[Z]^z}{[A]^a[B]^b}$$

Though this expression can be applied to any reversible reaction,

  • Kc is different for different reactions.
  • Kc changes with temperature.
  • The units of Kc vary from reaction to reaction.

Example 1 โ€“ writing equilibrium expressions

Write an expression for Kc for the following reactions:

  • N2(g) + 3H2(g) โ‡Œ 2NH3(g)
  • 2SO2(g) + O2(g) โ‡Œ 2SO3(g)

For the first equation:

$$K_c\ = \frac{[NH_3]^2}{[N_2][H_2]^3}$$

For the second equation:

$$K_c\ = \frac{[SO_3]^2}{[SO_2]^2[O_2]}$$

In equilibrium reactions involving a solid, we ignore the concentration of the solid, because its concentration remains constant, however much solid is present. That is because a solid is not part of the solution. For example:

Ag+(aq) + Fe2+(aq) โ‡Œ Ag(s) + Fe3+(aq)

The equilibrium expression for this reaction is:

$$K_c\ = \frac{[Fe^{3+}(aq)]}{[Ag^+(aq)][Fe^{2+}(aq)]}$$

What are the units of the equilibrium constant Kc?

In any equilibrium expression each species within a square bracket represents the concentration of that species in moldmโ€“3. Since the equilibrium expression varies with each reaction, the units of Kc therefore depend on the form of the equilibrium expression.

If the reaction is as follows:
  • A + B โ‡Œ Y + Z

The units are found algebraically as follows:

$$K_c\ = \frac{[Y][Z]}{[A][B]}$$

$$Units\ of\ K_c\ = \frac{(moldm^{-3})(moldm^{-3})}{(moldm^{-3})(moldm^{-3})}$$

They all cancel out. Which means for such a reaction, Kc has no units.

If the reaction is as follows:
  • 2A + B โ‡Œ 2Y

The units are found algebraically as follows:

$$K_c\ = \frac{[Y]^2}{[A]^2[B]}$$

$$Units\ of\ K_c\ = \frac{(moldm^{-3})^2}{(moldm^{-3})^2(moldm^{-3})}$$

Which simplifies to:

$$Units\ of\ K_c\ = \frac{1}{(moldm^{-3})}$$

$$ = mol^{-1}dm^{3}$$

Which means for such a reaction, The units of Kc are $latex mol^{-1}dm^{3}$.

Example 2 โ€“ determining the units of Kc

Write equilibrium expression for the following reaction and state the units of Kc.

CO(g) + 2H2(g) โ‡Œ CH3OH(g)


For CO(g) + 2H2(g) โ‡Œ CH3OH(g)

$$K_c\ = \frac{[CH_3OH]}{[H_2]^2[CO]}$$

$$Units\ of\ K_c\ = \frac{(moldm^{-3})}{(moldm^{-3})^2(moldm^{-3})}$$

$$ = \frac{1}{(moldm^{-3})^2}$$

$$ = mol^{-2}dm^{6}$$

How to calculate the value of Kc for the reaction between ethanol and ethanoic acid

0.10 mol of ethanol is mixed with 0.10 mol of ethanoic acid and
allowed to reach equilibrium. The total volume of the system is
made up to 0.020 dm3 with water. The system is allowed to reach equilibrium and then titrated. It is found that 0.033 mol of ethanoic acid is present once equilibrium is reached.

From this data we can analyse the reaction as follows:

At the start of the reaction:

0.10 mol 0.10 mol 0 mol 0 mol

At equilibrium:

Since there are 0.033 mol of CH3C0OH at equilibrium,

  • there must also be 0.033 mol of C2H50H at equilibrium because the equation tells us that they react 1:1 and we started with the same number of moles of each.
  • the number of moles of CH3COOH used up in this reaction = 0.10 โ€“ 0.033 = 0.067 mol.
  • according to the molar ratio of the equation, when 1 mol of CH3COOH reacts, 1 mol of CH3COOC2H5 and 1 mol of H2O are produced. Therefore, there must be 0.067 mol of each of these at equilibrium.
0.033 mol 0.033 mol 0.067 mol 0.067 mol

Next, we need to find the concentrations of the components at equilibrium. We know that the volume of the system is 0.020 dm3, we can find the concentrations as follows:

$latex \frac{0.033}{0.020}$ $latex \frac{0.033}{0.020}$ $latex \frac{0.067}{0.020}$ $latex \frac{0.067}{0.020}$
1.65 mol/dm3 1.65 mol/dm3 3.35 mol/dm3 3.35 mol/dm3

Using the formula $$K_c\ = \frac{[CH_3COOC_2H_5][H_2O]}{[CH_3COOH][C_2H_5OH]}$$

$$K_c\ = \frac{(3.35 mol/dm^3)(3.35 mol/dm^3)}{(1.65 mol/dm^3)(1.65 mol/dm^3)}$$

The units all cancel out, leaving Kc = 4.12.