Chemistry · Year 12 · Module 5 · Lesson 10
Calculating Keq & ICE Tables
Synthesise quantitative equilibrium reasoning, evaluate experimental data and claims, and reach evidence-based judgements at Band 5–6 level.
1. Data + scenario — Haber process optimisation, extended response
Scenario. The Haber process manufactures ammonia: N2(g) + 3H2(g) ⇌ 2NH3(g), ΔH = −92 kJ/mol. An industrial chemist is comparing two operating conditions (Table 1) to advise her company on which produces more ammonia per batch while keeping Keq economically useful.
Table 1. ICE data for two operating conditions (1.00 L reaction vessel)
| Condition | T (°C) | [N2]0 (mol/L) | [H2]0 (mol/L) | [NH3]0 (mol/L) | [N2]eq (mol/L) |
|---|---|---|---|---|---|
| A | 400 | 1.00 | 3.00 | 0 | 0.752 |
| B | 550 | 1.00 | 3.00 | 0 | 0.931 |
In your extended response, you must:
- Complete the ICE table for each condition and calculate [NH3]eq using stoichiometric ratios.
- Calculate Keq for Condition A and Condition B. Show full working.
- Explain the trend in Keq between A and B, linking it to the exothermic nature of the reaction.
- State which condition produces more ammonia per batch and justify using both the Keq values and equilibrium concentrations from your ICE tables.
- Evaluate the chemist's trade-off: why might industry choose 400°C rather than a higher temperature, despite the lower [NH3]eq at 400°C?
Working space — ICE tables, Keq calculations:
Extended response — write your evaluated answer below: 8 marks
2. Source critique — student misconception
"After I added more N2O4 to the equilibrium flask, the brown colour got darker, which proves that Keq increased. More reactant pushed more product formation, so the equilibrium constant went up. This shows Keq is sensitive to concentration — temperature doesn't really matter for Keq, only for the rate."
— Student blog post, NSW chemistry revision site, 2025.
In your response, you must:
- Identify the two specific scientific claims in the passage that are incorrect or misleading.
- For each error, explain the correct chemistry with reference to the definition and properties of Keq.
- Use an ICE table argument to show what actually happens to [N2O4]eq and [NO2]eq after adding more N2O4, and hence why Keq is unchanged.
- Describe one experimental procedure that would allow a student to correctly determine whether Keq has actually changed.
Extended response: 7 marks
Q1 — Haber process extended response
Marking notes (8 marks):
- Mark 1 — Correct ICE for Condition A: Δ[N2] = 1.00 − 0.752 = −0.248 mol/L; Δ[H2] = −3(0.248) = −0.744 mol/L; Δ[NH3] = +2(0.248) = +0.496 mol/L. [N2]eq = 0.752; [H2]eq = 3.00 − 0.744 = 2.256 mol/L; [NH3]eq = 0.496 mol/L.
- Mark 2 — Correct ICE for Condition B: Δ[N2] = 1.00 − 0.931 = −0.069 mol/L; Δ[H2] = −0.207 mol/L; Δ[NH3] = +0.138 mol/L. [H2]eq = 2.793 mol/L; [NH3]eq = 0.138 mol/L.
- Mark 3 — Correct Keq for A: Keq(A) = [NH3]² / ([N2][H2]³) = (0.496)² / (0.752 × (2.256)³) = 0.2460 / (0.752 × 11.483) = 0.2460 / 8.635 = 0.0285 (accept 0.028–0.029).
- Mark 4 — Correct Keq for B: Keq(B) = (0.138)² / (0.931 × (2.793)³) = 0.01904 / (0.931 × 21.80) = 0.01904 / 20.30 = 9.37 × 10⁻⁴ (accept 9.0–9.5 × 10⁻⁴).
- Mark 5 — Trend explained: Keq(A) > Keq(B); the reaction is exothermic (ΔH = −92 kJ/mol). Increasing temperature from 400°C to 550°C shifts Keq toward reactants (Le Chatelier's principle — the system counters the heat by favouring the endothermic reverse reaction). So Keq decreases with increasing temperature for an exothermic reaction.
- Mark 6 — More NH3 produced per batch at Condition A: [NH3]eq(A) = 0.496 mol/L vs [NH3]eq(B) = 0.138 mol/L. Justified by higher Keq(A): a larger Keq means a product-favoured equilibrium position, so more NH3 at equilibrium.
- Mark 7 — Trade-off evaluation: at higher temperature (Condition B), reaction rate is faster (more energy, more frequent effective collisions), which means equilibrium is reached more quickly even though Keq is smaller. Industry uses moderate temperature (400–500°C) with an iron catalyst to get acceptable yield (Keq reasonably large) at an acceptable rate — choosing neither the highest yield nor the fastest rate, but the most economical balance.
- Mark 8 — Explicit evaluative judgement: states a recommendation (Condition A for higher yield) but acknowledges the rate trade-off and the role of the catalyst, reaching a nuanced conclusion about industrial optimisation.
Q2 — Source critique
Marking notes (7 marks):
- Mark 1 — Error 1 correctly identified: "Keq increased" after adding N2O4 is incorrect. Keq is defined as the ratio of equilibrium concentrations of products to reactants at a given temperature; it does not change when concentration changes.
- Mark 2 — Correct explanation for Error 1: adding N2O4 initially creates a reaction quotient Q < Keq — the system shifts right, producing more NO2, until a new equilibrium is established where the ratio [NO2]²/[N2O4] again equals the original Keq. Keq is unchanged.
- Mark 3 — ICE argument: if initial [N2O4] before addition = A, after adding Δ: new Initial [N2O4] = A + Δ. Change row: −x for N2O4, +2x for NO2. At the new equilibrium, x adjusts until (original [NO2] + 2x)²/(A + Δ − x) = Keq. The equilibrium concentrations change, but Keq is the same value.
- Mark 4 — Error 2 correctly identified: "temperature doesn't really matter for Keq, only for the rate" is incorrect. Temperature is the only factor that changes Keq.
- Mark 5 — Correct explanation for Error 2: temperature changes Keq because it alters the thermodynamic favourability of the reaction — for an endothermic reaction (N2O4 → 2NO2, ΔH = +57 kJ/mol), increasing temperature increases Keq. The student conflates two separate effects: temperature changes both rate (via activation energy / Arrhenius) and Keq (via ΔH).
- Mark 6 — Experimental procedure: measure [NO2]eq by spectrophotometry (absorbance) and [N2O4]eq via ICE table before and after adding N2O4 (at constant temperature). Calculate Keq = [NO2]²/[N2O4] both times — if Keq is unchanged, the student's claim is refuted. Any valid, described procedure at constant temperature is accepted.
- Mark 7 — Quality of overall argument: clear structure identifying two distinct errors, each with correct chemistry and appropriate use of Keq definition and/or ICE framework; reaches an unambiguous conclusion that the student's claim is wrong.
Sample response summary: The student makes two errors. First, Keq does not increase when N2O4 is added — Keq is temperature-dependent, not concentration-dependent. Adding N2O4 shifts the position right (Q < Keq), producing more NO2, but the final ratio of equilibrium concentrations equals the original Keq. Second, temperature is the primary determinant of Keq; the student's claim that "temperature doesn't really matter for Keq" directly contradicts the definition of the equilibrium constant. To test this, a student should calculate Keq before and after adding N2O4 at constant temperature using spectrophotometry — the values will be equal, refuting the claim.