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.

Examples in context

Example 1. Orica Kooragang Island ammonia plant. The Haber synthesis $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$ runs continuously at the Orica Kooragang Island ammonia plant near Newcastle, feeding the explosives and fertiliser markets. Plant engineers monitor $K_{eq}$ at the reactor temperature (around 720 K) using gas chromatography sampling at the outlet. Because $K_{eq}$ at that temperature is roughly $1.6 \times 10^{-4}$, only about 15 percent of the feed converts on a single pass. The unconverted $N_2$ and $H_2$ are separated from $NH_3$ by refrigeration and recycled. The ICE-table arithmetic that students do by hand on the HSC is exactly the calculation the plant control system performs every minute.

Example 2. Macquarie River dissolved carbonate equilibrium. Limestone gorges along the Macquarie River downstream of Wellington release calcium carbonate that dissolves in the slightly acidic stream water. The relevant equilibrium CaCO_{3(s)} + CO_{2(aq)} + H_2O_{(l)} \rightleftharpoons Ca^{2+}_{(aq)} + 2HCO_3^{-}_{(aq)} has $K_{eq}$ that depends only on temperature. WaterNSW field officers measuring $[Ca^{2+}]$ and $[HCO_3^-]$ during summer high-temperature periods compare $Q$ against $K_{eq}$ to predict whether calcite will continue dissolving (when $Q < K_{eq}$) or precipitate out as scale in irrigation infrastructure (when $Q > K_{eq}$). The HSC ICE-table treatment maps directly onto this real monitoring workflow.

Try this

Q1. Define the equilibrium constant $K_{eq}$ and state two factors that do not change its value at constant temperature. [2 marks]

• Cue. Ratio of product to reactant concentrations raised to stoichiometric coefficients at equilibrium; concentration changes and pressure or volume changes do not alter $K_{eq}$.

Q2. At 500 K, $2.00 \text{ mol}$ of $PCl_5$ is placed in a $2.00 \text{ L}$ vessel and decomposes via $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$. At equilibrium $[Cl_2] = 0.30 \text{ mol L}^{-1}$. Calculate $K_{eq}$. [3 marks]

• Cue. Set up the ICE table with initial $[PCl_5] = 1.00$, change $-x$ and $+x$ on each product; $x = 0.30$, then substitute into $K_{eq} = \frac{[PCl_3][Cl_2]}{[PCl_5]}$.

Q3. For $2NO_{2(g)} \rightleftharpoons N_2O_{4(g)}$, $K_{eq} = 4.0$ at 298 K. A mixture has $[NO_2] = 0.50 \text{ mol L}^{-1}$ and $[N_2O_4] = 0.50 \text{ mol L}^{-1}$. (a) Calculate $Q$. (b) State the direction of net reaction. (c) Explain what would happen to $K_{eq}$ if the volume were halved. [2+1+2 marks]

• Cue. (a) $Q = 0.50 / 0.25 = 2.0$. (b) $Q < K_{eq}$, so net forward reaction. (c) Volume changes shift position but leave $K_{eq}$ unchanged.