← Unit 3: Equilibrium, acids and redox reactions

QLDChemistrySyllabus dot point

Topic 1: Chemical equilibrium systems

Derive and apply the equilibrium law expression (Kc) for homogeneous reactions, including calculating Kc from equilibrium concentrations and predicting the position of equilibrium from the value of Kc

A focused answer to the QCE Chemistry Unit 3 dot point on the equilibrium constant. Sets out the equilibrium law expression, works through Kc calculation from a stimulus ICE table (the dominant IA1 question type), interprets the value of Kc in terms of extent, and addresses why Kc is temperature-dependent but pressure-independent.

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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+bBβ‡ŒcC+dDaA + bB \rightleftharpoons cC + dD at equilibrium:

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

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 N2O4(g)β‡Œ2NO2(g)N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}:

Kc=[NO2]2[N2O4]K_c = \frac{[NO_2]^2}{[N_2O_4]}

For CaCO3(s)β‡ŒCaO(s)+CO2(g)CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}:

Kc=[CO2]K_c = [CO_2]

(Both solids excluded; only the gas appears.)

For Fe(aq)3++SCN(aq)βˆ’β‡ŒFeSCN(aq)2+Fe^{3+}_{(aq)} + SCN^-_{(aq)} \rightleftharpoons FeSCN^{2+}_{(aq)}:

Kc=[FeSCN2+][Fe3+][SCNβˆ’]K_c = \frac{[FeSCN^{2+}]}{[Fe^{3+}][SCN^-]}

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.

N2O4 NO2
Initial 1.00 0.00
Change -x +2x
Equilibrium 1.00 - x = 0.69 2x = 0.62

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

Kc=(0.62)20.69=0.3840.69=0.56Β mol/LK_c = \frac{(0.62)^2}{0.69} = \frac{0.384}{0.69} = 0.56 \text{ 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

Kc range Extent Interpretation
Kc >> 1 (e.g. > 10^3) Products dominate Reaction goes essentially to completion
Kc approximately 1 Comparable amounts Significant amounts of both sides
Kc << 1 (e.g. < 10^-3) Reactants dominate Reaction barely proceeds

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.

Reaction delta-n Kc units
H2 + I2 leftrightarrow 2HI 0 dimensionless
2SO2 + O2 leftrightarrow 2SO3 -1 mol^-1 L
N2O4 leftrightarrow 2NO2 +1 mol L^-1

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=[CO2][H2][CO][H2O]=(0.20)(0.20)(0.10)(0.10)=4.0Q = \frac{[CO_2][H_2]}{[CO][H_2O]} = \frac{(0.20)(0.20)}{(0.10)(0.10)} = 4.0

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.

Common traps

Including pure solids or pure liquids in the Kc expression. Exclude them. The classic trap is the limestone equilibrium, where the only term in Kc is [CO2].

Forgetting to use equilibrium concentrations rather than initial ones. Substituting initial values gives Q, not Kc.

Wrong stoichiometric exponents. The exponents are the equation coefficients, not the moles consumed. For 2NO2, the exponent is 2.

Reporting Kc with the wrong unit count. Either give the calculated units explicitly or state Kc as dimensionless. Inconsistent unit work loses marks.

Confusing Kc with rate constant k. The rate constant k is a kinetics term linking rate to concentration; Kc is the ratio of equilibrium concentrations. They are unrelated except through specific microscopic relationships beyond Unit 3.

In one sentence

The equilibrium constant Kc is the ratio of product to reactant equilibrium concentrations, each raised to its stoichiometric coefficient, that is constant for a homogeneous reaction at a given temperature, with a large value indicating products are favoured and a small value indicating reactants are favoured, and only temperature changes its value.

Past exam questions, worked

Real questions from past QCAA papers on this dot point, with our answer explainer.

2023 QCAA-style6 marksAt 1100 K, 0.40 mol of H2(g) and 0.40 mol of CO2(g) are placed in a 2.0 L sealed flask. They react according to H2(g) + CO2(g) <-> H2O(g) + CO(g). At equilibrium, [CO] is measured as 0.10 mol/L. (a) Construct an ICE table and determine the equilibrium concentrations of all species. (b) Calculate Kc at 1100 K. (c) Comment on the position of equilibrium based on the calculated value.
Show worked answer β†’

A 6-mark answer needs the ICE table set up correctly, the Kc calculation with units, and a value-based interpretation.

(a) ICE table. Initial concentrations are [H2]0 = 0.40 / 2.0 = 0.20 mol/L, [CO2]0 = 0.20 mol/L, [H2O]0 = [CO]0 = 0.

H2 CO2 H2O CO
Initial (mol/L) 0.20 0.20 0.00 0.00
Change -x -x +x +x
Equilibrium 0.20 - x 0.20 - x x x

Equilibrium [CO] = 0.10, so x = 0.10. Therefore [H2]eq = [CO2]eq = 0.10 mol/L; [H2O]eq = [CO]eq = 0.10 mol/L.

(b) Kc calculation.

Kc=[H2O][CO][H2][CO2]=(0.10)(0.10)(0.10)(0.10)=1.0K_c = \frac{[H_2O][CO]}{[H_2][CO_2]} = \frac{(0.10)(0.10)}{(0.10)(0.10)} = 1.0

Kc is dimensionless because moles of gas are balanced in this reaction (delta-n = 0).

(c) Interpretation. Kc = 1.0 indicates the products and reactants are present in comparable amounts at equilibrium; neither side is strongly favoured. The extent of reaction is moderate (around 50 percent conversion of starting reactants in this case).

Markers reward the ICE table fully completed, correct substitution into the Kc expression, and an interpretation that links the numerical value to extent rather than just restating it.

2022 QCAA-style3 marksThe reaction 2SO2(g) + O2(g) <-> 2SO3(g) has Kc = 280 at 1000 K and Kc = 2.5 x 10^25 at 298 K. (a) Predict whether the forward reaction is exothermic or endothermic. (b) Justify your prediction using the change in Kc with temperature.
Show worked answer β†’

A 3-mark answer needs the prediction with thermodynamic reasoning.

(a) Prediction. The forward reaction is exothermic.

(b) Justification. Kc decreases dramatically as temperature increases (from 2.5 x 10^25 at 298 K to 280 at 1000 K). For an exothermic forward reaction, heat is a product; increasing temperature shifts equilibrium left (Le Chatelier), so the ratio [products]/[reactants] falls, and Kc falls. The data are consistent with that pattern.

An endothermic forward reaction would show Kc increasing with T, so the observed decrease rules that out.

Markers reward the prediction, the temperature-Kc relationship for exothermic processes, and explicit connection to the numbers given.

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