Deduce the equilibrium expression (in terms of Keq) for homogeneous reactions, and perform calculations to find the value of Keq and concentrations of substances within an equilibrium system

What this dot point is asking

NESA wants you to write the equilibrium expression for a given reaction, calculate the value of Keq from experimental data, and use Keq to find equilibrium concentrations. This is the highest-yielding calculation dot point in Module 5 and appears as a 4-6 mark question almost every year.

The answer

The equilibrium expression

For the general reaction:

The equilibrium constant in terms of concentration is:

Products on top, reactants on bottom, each raised to its stoichiometric coefficient.

Rules:

• Pure solids and pure liquids are excluded from the expression. Their "concentration" is essentially constant.
• Solvents (water in dilute aqueous reactions) are also excluded.
• Aqueous and gaseous species are included.
• Keq is temperature-dependent only. Concentration and pressure changes do not change Keq, only the position of equilibrium.

Interpreting the value

The magnitude of Keq tells you where the equilibrium lies.

• If $K_{eq} \gg 1$, equilibrium lies to the right (products favoured).
• If $K_{eq} \approx 1$, there are comparable concentrations of reactants and products.
• If $K_{eq} \ll 1$, equilibrium lies to the left (reactants favoured).

Keq has no fixed unit (the unit depends on the change in moles of gas/aqueous species in the equation). In HSC answers, write the numerical value and, if asked, state whether the equilibrium favours reactants or products.

The reaction quotient Q

For a system not yet at equilibrium, the same expression is evaluated using the current concentrations and is called the reaction quotient Q.

• If $Q < K_{eq}$: too many reactants. Reaction proceeds forward to reach equilibrium.
• If $Q > K_{eq}$: too many products. Reaction proceeds reverse.
• If $Q = K_{eq}$: system is already at equilibrium.

The ICE method

Almost every Keq calculation uses an ICE table (Initial, Change, Equilibrium). The change row uses the stoichiometric coefficients with a variable $x$ for the extent of reaction. The same ICE method underpins the Ksp calculations for sparingly soluble salts and the weak-acid pH calculations you meet later in the module.

Worked example

For the reaction $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$ at 700 K, Keq = 49. If 1.00 mol of $H_2$ and 1.00 mol of $I_2$ are placed in a 1.00 L sealed flask, calculate the equilibrium concentrations.

Step 1: Write the expression.

Step 2: ICE table.

Step 3: Substitute and solve.

Take the square root of both sides (both sides are perfect squares):

Step 4: Final concentrations.

$[H_2] = [I_2] = 1.00 - 0.778 = 0.222$ mol/L.
$[HI] = 2 \times 0.778 = 1.556$ mol/L.

Check. $\frac{(1.556)^2}{(0.222)(0.222)} = \frac{2.42}{0.0493} \approx 49$. Correct.

Common traps

Forgetting the stoichiometric exponent. $K_{eq} = \frac{[HI]^2}{[H_2][I_2]}$. The "2" on HI is essential.

Including pure solids or liquids. For $CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}$, $K_{eq} = [CO_2]$ only. Solids are excluded.

Using moles instead of concentration. Always convert moles to mol/L by dividing by volume.

Wrong direction of Keq when reaction is reversed. $K_{reverse} = 1 / K_{forward}$. If a question gives Keq for one direction, you must invert it for the reverse.

Assuming Keq changes with pressure or concentration. It does not. Only temperature changes Keq.

In one sentence

The equilibrium constant Keq is the ratio of equilibrium product concentrations to reactant concentrations, each raised to its stoichiometric coefficient, with solids and pure liquids excluded, and is calculated using an ICE table that tracks initial concentrations, the change set by stoichiometry, and the final equilibrium values.