Chemistry • Year 12 • Module 5 • Lesson 9
Writing Keq Expressions
Apply Keq rules to real data, interpret magnitude, perform reciprocal and multiplier calculations, and analyse a Great Barrier Reef case study.
1. Interpret graph — Keq for the Haber process vs temperature
The figure below shows published values of Keq (log scale) for the reaction N2(g) + 3H2(g) ⇌ 2NH3(g) over a range of temperatures. 9 marks
Adapted from: Lide (ed.), CRC Handbook of Chemistry and Physics (2023). Reaction: N2(g) + 3H2(g) ⇌ 2NH3(g), ΔH = −92 kJ mol−1.
1.1 Write the Keq expression for N2(g) + 3H2(g) ⇌ 2NH3(g) and identify the state of all species included. 2 marks
1.2 Describe the trend in Keq as temperature increases from 200°C to 600°C. Estimate the approximate Keq at 400°C from the graph (state the value as a power of 10). 2 marks
1.3 The industrial Haber process uses ~450°C even though a lower temperature gives a much larger Keq. Using only Keq concepts (not reaction rate), explain why a lower temperature is not necessarily ‘better’ for industrial production. 2 marks
1.4 The Haber process uses a high-pressure vessel. Explain why increasing pressure does not change Keq but does affect the equilibrium position for this reaction. 3 marks
2. Trace the effect of reversing and scaling equilibrium equations
Complete each step of the cause-and-effect chain. The original reaction is given; you must supply the expression, calculate the new Keq, and state the implication. Show all working. 8 marks
Starting data: 2SO2(g) + O2(g) ⇌ 2SO3(g), Keq = 1.8 × 106 at 500°C
Step A. Write the Keq expression for the original equation. 1 mark
Step B. Reverse the equation to give: 2SO3(g) ⇌ 2SO2(g) + O2(g). Calculate Keq for the reversed reaction and write the new expression. 2 marks
Step C. Halve all coefficients: SO3(g) ⇌ SO2(g) + ½O2(g). Calculate Keq for this half-coefficient form (working from the reversed equation in Step B). 2 marks
Step D. Overall outcome. A student claims that Keq for SO3(g) ⇌ SO2(g) + ½O2(g) is the same as for 2SO2(g) + O2(g) ⇌ 2SO3(g). Evaluate this claim using your Step C result. 3 marks
3. Case study — limestone dissolution and the Great Barrier Reef
Read the stimulus, then answer the questions using Keq concepts. 7 marks
Stimulus. Coral skeletons are made of calcium carbonate (CaCO3). As ocean water absorbs more atmospheric CO2, the following equilibrium shifts: CaCO3(s) ⇌ Ca2+(aq) + CO32−(aq). This is the dissolution equilibrium for calcium carbonate, and its equilibrium constant is called Ksp (solubility product). As ocean acidification adds H+ to seawater, CO32− is consumed (forming H2CO3), which drives the dissolution equilibrium to the right. Scientists monitoring the Great Barrier Reef have observed a measurable decline in coral calcification rates since pre-industrial times. At Boral Marulan in NSW, limestone quarrying extracts CaCO3 from rock strata that were once marine reef deposits. The same dissolution chemistry that threatens modern reefs once dissolved those ancient limestones, then reprecipitated CaCO3 as geological conditions changed.
3.1 Write the Ksp expression for CaCO3(s) ⇌ Ca2+(aq) + CO32−(aq). Identify any excluded species and the reason for exclusion. 2 marks
3.2 The Ksp of CaCO3 is 3.4 × 10−9 at 25°C. Interpret this value — does it indicate that CaCO3 is highly soluble or sparingly soluble in water? Justify using Keq magnitude reasoning. 2 marks
3.3 Using the stimulus, explain how the removal of CO32−(aq) by ocean acidification shifts the Ksp equilibrium. Does Ksp itself change? Justify your answer. 3 marks
4. Predict and justify
Answer in 3–4 sentences using precise Keq language. 4 marks
Q4. The equilibrium N2(g) + O2(g) ⇌ 2NO(g) has Keq = 1 × 10−30 at 25°C, but Keq = 0.04 at 2000°C (inside a car engine cylinder during combustion). A student says: “Because Keq is still less than 1 at 2000°C, the engine cannot produce significant NO.” Predict whether significant NO is actually produced in car engines and justify your prediction using Keq reasoning.
Q1.1 — Keq expression
Keq = [NH3]2 / ([N2][H2]3). All species (N2, H2, NH3) are gases (g) — all are included. 1 mark for correct expression with exponents; 1 mark for stating all are gases and none are excluded.
Q1.2 — Graph trend + value
Keq decreases strongly as temperature increases — the line falls steeply on the log scale, so each 100°C rise reduces Keq by approximately one to two orders of magnitude [1]. At 400°C, log Keq ≈ 1.7, so Keq ≈ 101.7 ≈ 50 (accept 101.5 to 102.0) [1].
Q1.3 — Industrial temperature reasoning
A lower temperature does give a much larger Keq (more NH3 favoured at equilibrium) [1], but Keq is a thermodynamic statement about the equilibrium position — it does not tell us how fast equilibrium is reached. In practice, the reaction is too slow at low temperatures without a catalyst, so commercially useful production rates require a compromise temperature. From a Keq-only perspective, lower T is better for yield; other factors (reaction rate) determine the practical operating temperature [1].
Q1.4 — Pressure and Keq
Keq depends only on temperature — pressure is a concentration/partial pressure change and does not affect the thermodynamic constant Keq [1]. However, for this reaction, the left side has 1 + 3 = 4 moles of gas and the right side has 2 moles of gas. Increasing pressure reduces volume and increases concentrations of all species [1]. The system responds by shifting equilibrium to the side with fewer moles of gas (right → more NH3) to reduce total moles and relieve the pressure increase — equilibrium position shifts right, but Keq value is unchanged [1].
Q2 — Cause-and-effect chain
Step A. Keq = [SO3]2 / ([SO2]2[O2]) = 1.8 × 106.
Step B. Reverse: Keq(reversed) = 1 / (1.8 × 106) = 5.6 × 10−7. Expression: Keq = ([SO2]2[O2]) / [SO3]2.
Step C. Halving coefficients → raise Keq(reversed) to the power ½: Keq(half) = (5.6 × 10−7)0.5 = 7.5 × 10−4. Expression: Keq = [SO2][O2]0.5 / [SO3].
Step D. The student's claim is incorrect [1]. The original forward reaction Keq = 1.8 × 106; the half-coefficient reversed form has Keq = 7.5 × 10−4 — these differ by ~109 [1]. The two equations describe different directions and different stoichiometries; the Keq expression and its numerical value are completely different. They are not the same because reversing and halving the equation both change Keq [1].
Q3.1 — Ksp expression
Ksp = [Ca2+][CO32−]. Excluded: CaCO3(s) — pure solid, activity = 1 by convention.
Q3.2 — Magnitude interpretation
Ksp = 3.4 × 10−9 is much less than 1 (much less than 10−3) [1]. This means the equilibrium strongly favours reactants — at equilibrium, very little CaCO3 dissolves. CaCO3 is therefore sparingly soluble (very slightly soluble) in water [1].
Q3.3 — Ocean acidification and Ksp
When H+ is added to seawater, it reacts with CO32−: CO32− + H+ → HCO3−. This removes CO32− from the dissolution equilibrium [1]. Because CO32− is a product of CaCO3 dissolution, its removal causes the equilibrium to shift right (forward) to restore the equilibrium ratio, dissolving more CaCO3 — coral skeletons dissolve faster [1]. Ksp itself does not change, because ocean acidification is a concentration change (at approximately constant ocean temperature) and Ksp depends only on temperature [1].
Q4 — Predict and justify
Significant NO is indeed produced in car engines. At 2000°C, Keq = 0.04 — this is much greater than 10−30 (the room-temperature value), meaning the equilibrium position has shifted dramatically toward products compared with room temperature [1]. Although Keq = 0.04 is still less than 1, it is close enough to 1 (within two orders of magnitude) that significant concentrations of both NO and reactants are present at equilibrium at engine temperatures [1]. The student's reasoning misapplies the magnitude guide: ‘Keq < 1’ means reactants are favoured, not that product concentration is negligible — at Keq = 0.04 and the high temperatures and pressures in a cylinder, meaningful amounts of NO form and are released in exhaust gases [2].