## The equilibrium constant K_{c}

Equilibrium reaction do not go to completion but will reach equilibrium with time. The reaction between ethanoic acid CH_{3}COOH and ethanol C_{2}H_{5}OH to produce ethyl ethanoate CH_{3}COOC_{2}H_{5} and water H_{2}O is a typical equilibrium reaction which proceeds as follows:

CH_{3}COOH + C_{2}H_{5}OH ⇌ CH_{3}COOC_{2}H_{5} + H_{2}O

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

present.

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:

$$\frac{[CH_3COOC_2H_5][H_2O]}{[CH_3COOH][C_2H_5OH]}$$

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

For any equilibrium reaction:

aA + bB ⇌ 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, K_{c}.

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

Though this expression can be applied to any reversible reaction,

- K
_{c}is different for different reactions. - K
_{c}changes with temperature. - The units of K
_{c}vary from reaction to reaction.

### Example 1 – writing equilibrium expressions

Write an expression for K_{c} for the following reactions:

- N
_{2}(g) + 3H_{2}(g) ⇌ 2NH_{3}(g) - 2SO
_{2}(g) + O_{2}(g) ⇌ 2SO_{3}(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) + Fe^{2+}(aq) ⇌ Ag(s) + Fe^{3+}(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 K_{c}?

In any equilibrium expression each species within a square bracket represents the concentration of that species in moldm^{–3}. Since the equilibrium expression varies with each reaction, the units of K_{c} therefore depend on the form of the equilibrium expression.

- A + B ⇌ 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, K_{c} has no units.

- 2A + B ⇌ 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 K_{c} are $latex mol^{-1}dm^{3}$.

### Example 2 – determining the units of K_{c}

Write equilibrium expression for the following reaction and state the units of K_{c}.

CO(g) + 2H_{2}(g) ⇌ CH_{3}OH(g)

### Solution

For CO(g) + 2H_{2}(g) ⇌ CH_{3}OH(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 K_{c} 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 dm^{3} 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:

CH_{3}COOH |
+ | C_{2}H_{5}OH |
⇌ | CH_{3}COOC_{2}H_{5} |
+ | H_{2}O |
---|---|---|---|---|---|---|

0.10 mol | 0.10 mol | 0 mol | 0 mol |

### At equilibrium:

Since there are 0.033 mol of CH_{3}C0OH at equilibrium,

- there must also be 0.033 mol of C
_{2}H_{5}0H 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 CH
_{3}COOH used up in this reaction = 0.10 – 0.033 = 0.067 mol. - according to the molar ratio of the equation, when 1 mol of CH
_{3}COOH reacts, 1 mol of CH_{3}COOC_{2}H_{5}and 1 mol of H_{2}O are produced. Therefore, there must be 0.067 mol of each of these at equilibrium.

CH_{3}COOH |
+ | C_{2}H_{5}OH |
⇌ | CH_{3}COOC_{2}H_{5} |
+ | H_{2}O |
---|---|---|---|---|---|---|

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 dm^{3}, we can find the concentrations as follows:

CH_{3}COOH |
+ | C_{2}H_{5}OH |
⇌ | CH_{3}COOC_{2}H_{5} |
+ | H_{2}O |
---|---|---|---|---|---|---|

$latex \frac{0.033}{0.020}$ | $latex \frac{0.033}{0.020}$ | $latex \frac{0.067}{0.020}$ | $latex \frac{0.067}{0.020}$ | |||

1.65 mol/dm^{3} |
1.65 mol/dm^{3} |
3.35 mol/dm^{3} |
3.35 mol/dm^{3} |

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 K_{c} = 4.12.