Chemistry · Ch 6 — Equilibrium

Homogeneous Equilibria

6.4

Homogeneous Equilibria

A chemical equilibrium is called homogeneous when all the reactants and products exist in the same physical phase. This is the simplest type of equilibrium to analyse because the entire system is uniform — there is no interface between phases to complicate the concentrations.

The textbook gives three clear examples:

The Haber process in the gas phase:

$\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$

Here, nitrogen, hydrogen, and ammonia are all gases.
Ester hydrolysis in aqueous solution:

$\text{CH}_3\text{COOC}_2\text{H}_5(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{CH}_3\text{COOH}(aq) + \text{C}_2\text{H}_5\text{OH}(aq)$

All species except water are dissolved in water; water itself is the solvent and is present in vast excess.
The blood-red thiocyanate complex formation:

$\text{Fe}^{3+}(aq) + \text{SCN}^-(aq) \rightleftharpoons \text{Fe(SCN)}^{2+}(aq)$

All three species are ions in aqueous solution.

In each case, every reactant and product occupies the same phase — either all gases or all in aqueous solution. This uniformity allows us to write a single expression for the equilibrium constant using concentrations (or partial pressures for gases) without worrying about phase boundaries.

Note

The textbook explicitly notes that water in the ester hydrolysis example is a pure liquid, not a solute. In the equilibrium expression for reactions involving pure liquids or solids, their concentration is taken as constant and incorporated into the equilibrium constant. So for the ester reaction, the equilibrium constant is written as:

$K_c = \frac{[\text{CH}_3\text{COOH}][\text{C}_2\text{H}_5\text{OH}]}{[\text{CH}_3\text{COOC}_2\text{H}_5]}$

Water does not appear because its concentration remains essentially unchanged.

The rest of this section builds the equilibrium constant expressions for several homogeneous reactions, showing how to write them correctly from the balanced chemical equation.

Writing Equilibrium Expressions for Homogeneous Reactions

For any homogeneous reaction at equilibrium, the law of mass action gives the equilibrium constant. The general form for a reaction:

$aA + bB \rightleftharpoons cC + dD$

is:

K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}

where square brackets denote molar concentrations at equilibrium. The exponents are the stoichiometric coefficients from the balanced equation.

The textbook works through several examples to illustrate this.

Example 1: The Haber process

$\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$

The equilibrium constant expression is:

K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}

Notice that the coefficient 2 on ammonia becomes the exponent 2, the coefficient 1 on nitrogen becomes exponent 1 (usually omitted), and the coefficient 3 on hydrogen becomes exponent 3.

Example 2: Ester hydrolysis

$\text{CH}_3\text{COOC}_2\text{H}_5(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{CH}_3\text{COOH}(aq) + \text{C}_2\text{H}_5\text{OH}(aq)$

As noted, water is a pure liquid and does not appear. The expression is:

K_c = \frac{[\text{CH}_3\text{COOH}][\text{C}_2\text{H}_5\text{OH}]}{[\text{CH}_3\text{COOC}_2\text{H}_5]}

Example 3: Thiocyanate complex formation

$\text{Fe}^{3+}(aq) + \text{SCN}^-(aq) \rightleftharpoons \text{Fe(SCN)}^{2+}(aq)$

The expression is:

$K_c = \frac{[\text{Fe(SCN)}^{2+}]}{[\text{Fe}^{3+}][\text{SCN}^-]}$ …

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