The equilibrium constant Kc and reaction extent (QCE Chemistry Unit 3)

What this dot point is asking

QCAA wants you to write the equilibrium law expression for a homogeneous reaction, calculate Kc from concentration data (often through an ICE-style table in IA1 stimulus), and interpret what the numerical value of Kc says about the position and extent of equilibrium. This is the highest-yielding calculation type in QCE Chemistry Unit 3 and dominates IA1.

The answer

For a homogeneous reaction at equilibrium, the equilibrium law expresses the ratio of product concentrations to reactant concentrations, each raised to the power of its stoichiometric coefficient. This ratio is constant at a given temperature and is called the equilibrium constant, Kc.

The general expression

For a reaction $aA + bB \rightleftharpoons cC + dD$ at equilibrium:

All concentrations are equilibrium values, in mol/L.

Conventions:

• Pure solids and pure liquids are excluded. Their effective concentrations are constants absorbed into Kc.
• Solvents in dilute solution (water) are excluded. [H2O] is treated as constant in aqueous reactions.
• Gases use partial pressures (Kp) or concentrations (Kc). QCE Chemistry uses concentrations.
• Coefficients become exponents. The 2 in 2NO2 becomes [NO2]^2 in the expression.

Worked Kc set-up

For $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$:

For $CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}$:

(Both solids excluded; only the gas appears.)

For $Fe^{3+}_{(aq)} + SCN^-_{(aq)} \rightleftharpoons FeSCN^{2+}_{(aq)}$:

Using an ICE table

ICE (Initial, Change, Equilibrium) is the standard tool for IA1 calculation stimulus.

Method:

1. Convert all initial amounts to concentrations (mol/L) using the flask volume.
2. Write the change as +x or -x according to the stoichiometric ratios.
3. Express equilibrium values as initial plus/minus the change.
4. Substitute equilibrium values into the Kc expression and solve.

Worked example. 1.00 mol N2O4 placed in a 1.0 L flask at 100 degrees C. At equilibrium, [N2O4] = 0.69 mol/L.

So x = 0.31. [NO2]eq = 0.62 mol/L.

The units are mol/L because the equation has 1 reactant mol to 2 product mol on gas counting; (mol/L)^2 / (mol/L) = mol/L.

Interpreting the value of Kc

The bigger the Kc, the further the equilibrium lies to the right. Kc does not tell you anything about reaction rate, only about the position once equilibrium is reached.

What changes Kc

Only temperature changes Kc.

• Adding or removing reactant or product: position shifts, but Kc unchanged.
• Changing volume or pressure: position shifts (for gas systems with delta-n not zero), but Kc unchanged.
• Adding a catalyst: position unchanged, Kc unchanged.
• Changing temperature: position shifts AND Kc changes.

For an exothermic forward reaction, Kc decreases as T increases. For an endothermic forward reaction, Kc increases as T increases. This is the basis of QCAA stimulus questions where you must infer the thermicity of the forward reaction from a Kc vs T table.

Units of Kc

Many QCAA answers omit Kc units and treat the constant as dimensionless. This is correct in strict thermodynamic treatment (concentrations divided by 1 mol/L standard state). QCE Chemistry accepts either approach, but if you assign units they must reflect delta-n = (mol products gas) - (mol reactants gas), with the unit being (mol/L)^delta-n.

Kc vs Q (the reaction quotient)

Q has the same expression as Kc but is calculated for any set of concentrations, not just equilibrium ones. Comparing Q to Kc predicts the direction of the net change:

• Q < Kc: too few products; net forward reaction.
• Q = Kc: at equilibrium; no net change.
• Q > Kc: too many products; net reverse reaction.

This is one of QCAA's data-test favourites: given non-equilibrium concentrations, predict which way the reaction will go.

Worked example: an IA1 stimulus

Stimulus: at 800 K, the reaction CO(g) + H2O(g) leftrightarrow CO2(g) + H2(g) has Kc = 4.0. A reaction mixture contains [CO] = 0.10, [H2O] = 0.10, [CO2] = 0.20, [H2] = 0.20 mol/L.

Question: is the system at equilibrium? If not, in which direction will it proceed?

Answer:

Q = Kc, so the system is already at equilibrium. No net change. Add to the response: forward and reverse rates are equal, but reactions continue at the particle level.

If instead [CO2] = 0.30 and [H2] = 0.30 with the other concentrations unchanged, Q = 9.0 > Kc, so net reverse reaction would proceed until Q = Kc.

Examples in context

Example 1. Haber synthesis at Incitec Pivot Gibson Island. The Gibson Island plant operates the equilibrium $\text{N}_2(g) + 3 \text{H}_2(g) \rightleftharpoons 2 \text{NH}_3(g)$ with $K_c = 6.0 \times 10^{-2}$ at $500^{\circ}\text{C}$. With $[\text{N}_2] = 0.50 \, \text{mol L}^{-1}$ and $[\text{H}_2] = 1.50 \, \text{mol L}^{-1}$, the equilibrium ammonia concentration is $[\text{NH}_3] = \sqrt{K_c \times [\text{N}_2] \times [\text{H}_2]^3} = \sqrt{0.060 \times 0.50 \times 3.375} = 0.318 \, \text{mol L}^{-1}$. Operators run high pressure to push equilibrium forward (compatible with $K_p$ pressure dependence) but accept the moderate $K_c$ as a thermodynamic limit.

Example 2. Iron(III)-thiocyanate ICE-table in QCAA IA2 spectroscopy. A common Year 12 task tracks $\text{Fe}^{3+}(aq) + \text{SCN}^-(aq) \rightleftharpoons \text{FeSCN}^{2+}(aq)$ ($K_c = 1.4 \times 10^2$) by colorimetry. Starting with $[\text{Fe}^{3+}]_0 = [\text{SCN}^-]_0 = 0.0100 \, \text{mol L}^{-1}$, set $x = [\text{FeSCN}^{2+}]_{eq}$. ICE table: equilibrium $[\text{Fe}^{3+}] = 0.0100 - x$, $[\text{SCN}^-] = 0.0100 - x$, $[\text{FeSCN}^{2+}] = x$. So $K_c = x / (0.0100 - x)^2 = 140$. Solving the quadratic gives $x = 0.00692 \, \text{mol L}^{-1}$; equilibrium $[\text{Fe}^{3+}] = 0.00308 \, \text{mol L}^{-1}$. Measured absorbance confirms the value.

Try this

Q1. Write the equilibrium expression $K_c$ for $2 \text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2 \text{SO}_3(g)$. State its units. [2 marks]

• Cue. $K_c = [\text{SO}_3]^2 / ([\text{SO}_2]^2 [\text{O}_2])$; units $\text{L mol}^{-1}$ (or $\text{mol}^{-1} \text{L}$).

Q2. For $\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2 \text{HI}(g)$, $K_c = 54$ at $700 \, \text{K}$. Starting with $1.00 \, \text{mol L}^{-1}$ of each $\text{H}_2$ and $\text{I}_2$, calculate equilibrium $[\text{HI}]$. [4 marks]

• Cue. ICE: $[\text{HI}] = 2x$; $[\text{H}_2] = [\text{I}_2] = 1 - x$. $K_c = (2x)^2 / (1-x)^2 = 54$. $2x/(1-x) = 7.35$, $x = 0.787$; $[\text{HI}] = 1.57 \, \text{mol L}^{-1}$.

Q3. $K_c$ for the reaction $\text{N}_2 + 3 \text{H}_2 \rightleftharpoons 2 \text{NH}_3$ is $0.060$ at $500^{\circ}\text{C}$ and $0.50$ at $400^{\circ}\text{C}$. (a) Predict whether forward is exothermic or endothermic. (b) Justify. (c) Explain why industry runs at higher T despite lower $K_c$. [2+2+3 marks]

• Cue. (a) Exothermic. (b) $K_c$ falls with $T$. (c) Rate vs yield trade-off; higher T faster despite lower equilibrium. Judgement criterion.