Chemistry · Year 12 · Module 5 · Lesson 10
Calculating Keq & ICE Tables
Apply the ICE framework to multi-step problems, interpret quantitative equilibrium data, and predict the direction of shifts from concentration changes.
1. Cause-and-effect chain — NOx formation in NSW car engines
At the high temperatures inside a car engine (above 1500°C), nitrogen and oxygen from air react: N2(g) + O2(g) ⇌ 2NO(g). This is a key source of NOx air pollution in Sydney and other NSW urban areas, contributing to smog and acid rain. Trace the chain of consequences below. 5 marks
Cause: Combustion in a car engine raises local temperature to >1500°C.
Effect 1: What happens to Keq for the N2/O2/NO equilibrium at this temperature?
Effect 2: How does the equilibrium position shift as the engine temperature rises?
Effect 3: What happens to [NO] in the exhaust gases?
Effect 4: What does a large [NO] at engine temperature tell you about the magnitude of Keq for this reaction at that temperature? Is Keq much greater than 1, close to 1, or much less than 1?
Overall: Why does a catalytic converter help, and how does it change the equilibrium situation?
2. Interpreting equilibrium data — N2O4 ⇌ 2NO2
The graph below shows how Keq for the reaction N2O4(g) ⇌ 2NO2(g) changes with temperature. The forward reaction is endothermic (ΔH = +57 kJ/mol). Study the graph and answer the questions. 9 marks
(a) Describe the trend in Keq as temperature increases from 25°C to 100°C. (2 marks)
(b) Using the graph, estimate Keq at 55°C. Show how you used the graph to obtain your estimate. (2 marks)
(c) Explain why Keq increases with temperature for this equilibrium. In your answer, name the type of reaction and the direction favoured by heating. (3 marks)
(d) At 25°C, Keq = 0.14. Is the equilibrium position product-favoured or reactant-favoured? What does this tell you about the brown colour intensity of the gas in a sealed flask at 25°C? (2 marks)
3. Predict and justify — ICE table with Keq given
For the reaction H2(g) + I2(g) ⇌ 2HI(g) at 430°C, Keq = 54.3. A student mixes [H2]0 = 0.200 mol/L and [I2]0 = 0.200 mol/L with no HI initially present. 7 marks
(a) Before doing any algebra, predict: will there be more H2 or more HI at equilibrium? Justify using Keq. (2 marks)
(b) Set up and complete the ICE table using variable x. (2 marks)
| Row | [H2] (mol/L) | [I2] (mol/L) | [HI] (mol/L) |
|---|---|---|---|
| Initial | |||
| Change | |||
| Equilibrium |
(c) Write the Keq expression in terms of x, simplify (use the square root shortcut), and solve for x. Show all steps. (2 marks)
(d) Verify your answer by substituting equilibrium concentrations back into the Keq expression. (1 mark)
4. Compare direct substitution vs ICE table method
Complete the comparison table to contrast the two methods for calculating Keq. 6 marks
| Feature | Direct substitution | ICE table method |
|---|---|---|
| What data is given? | ||
| Is algebra required? | ||
| When is it used? | ||
| Role of stoichiometry | ||
| Is verification possible? | ||
| One example reaction |
5. Data table — colourimetry experiment, NSW school lab
A student uses a spectrophotometer to determine Keq for N2O4(g) ⇌ 2NO2(g) using three different initial concentrations of N2O4 in separate 1.00 L flasks at 25°C. The [NO2] at equilibrium was measured by absorbance. 6 marks
| Trial | [N2O4]initial (mol/L) | [NO2]eq (mol/L) | [N2O4]eq (mol/L) | Keq |
|---|---|---|---|---|
| 1 | 0.200 | 0.136 | ||
| 2 | 0.400 | 0.204 | ||
| 3 | 0.600 | 0.257 |
(a) Calculate the missing [N2O4]eq and Keq values for each trial. Show working for Trial 1 below. (3 marks)
(b) Do the three Keq values agree? Explain what the data demonstrates about the relationship between Keq and initial concentration. (2 marks)
(c) Identify one controlled variable in this investigation and explain its importance. (1 mark)
Q1 — Cause-and-effect chain
Effect 1: Keq increases (N2 + O2 → 2NO is endothermic; raising temperature shifts Keq toward products, increasing its value).
Effect 2: The equilibrium position shifts to the right (toward more NO production).
Effect 3: [NO] in exhaust gases increases significantly.
Effect 4: A large [NO] in exhaust gases indicates that Keq is relatively large (significantly greater than 1) at engine temperatures — the equilibrium position is product-favoured under these conditions, consistent with the large increase in Keq at high temperatures for this endothermic reaction.
Overall: A catalytic converter promotes the reduction of NO back to N2 and O2 at lower temperatures; this shifts the equilibrium back toward reactants, reducing [NO] in emissions.
Q2 — Graph interpretation
(a) Keq increases continuously as temperature rises from 25°C (Keq = 0.14) to 100°C (Keq ≈ 2.5). The relationship is non-linear (concave upward / exponential-like). 1 mark for trend direction, 1 mark for non-linearity or quoting values.
(b) Interpolating between the data at 45°C (0.50) and 65°C (1.3): at 55°C, Keq ≈ 0.9 (accept 0.8–1.0). 1 mark for a value in this range with a statement about interpolation; 1 mark for showing the method (reading mid-point).
(c) The forward reaction (N2O4 → 2NO2) is endothermic. Increasing temperature supplies energy that drives the endothermic forward reaction; Le Chatelier's principle means the equilibrium shifts right, producing more NO2, so the ratio [NO2]²/[N2O4] — and hence Keq — increases. 1 mark each for: endothermic type named; equilibrium shifts right; Keq increases because ratio shifts in favour of products.
(d) Keq = 0.14 < 1 → reactant-favoured; at 25°C more N2O4 (colourless) than NO2 (brown) is present at equilibrium, so the brown colour is relatively pale/light. 1 mark for reactant-favoured; 1 mark for linking to pale/light brown colour.
Q3 — Predict and justify
(a) Keq = 54.3 > 1, so the equilibrium lies strongly to the right — at equilibrium, [HI] > [H2]; there will be significantly more HI than H2. 1 mark for prediction; 1 mark for linking to Keq > 1.
(b) ICE table: I: H2 = 0.200, I2 = 0.200, HI = 0. C: H2 = −x, I2 = −x, HI = +2x. E: H2 = 0.200−x, I2 = 0.200−x, HI = 2x.
(c) Keq = (2x)²/[(0.200−x)(0.200−x)] = 4x²/(0.200−x)² = 54.3. Take √ both sides: 2x/(0.200−x) = 7.369. 2x = 1.474 − 7.369x. 9.369x = 1.474 × 0.1000... Wait — correctly: 7.369 × 0.200 = 1.4738; 2x + 7.369x = 1.4738; 9.369x = 1.4738; x = 0.1573 mol/L. Equilibrium: [H2] = [I2] = 0.200 − 0.157 = 0.043 mol/L; [HI] = 2(0.157) = 0.314 mol/L. (Accept rounding differences.)
(d) Verification: (0.314)²/[(0.043)(0.043)] = 0.0986/0.00185 = 53.3 ≈ 54.3 ✓ (small discrepancy from rounding).
Q4 — Compare direct substitution vs ICE table
| Feature | Direct substitution | ICE table method |
|---|---|---|
| What data is given? | All equilibrium concentrations directly | Initial concentrations (and sometimes one equilibrium concentration or Keq) |
| Is algebra required? | No — arithmetic only | Yes — solve for x (linear, square root, or quadratic) |
| When is it used? | When equilibrium concentrations of all species are already known | When only some concentrations at equilibrium (or Keq) are known |
| Role of stoichiometry | Only in writing powers in Keq expression | Critical in Change row — sets multiples of x for every species |
| Is verification possible? | Yes — recalculate Keq | Yes — substitute equilibrium row back into Keq expression |
| One example reaction | 2SO₂ + O₂ ⇌ 2SO₃ (all [eq] given) | H₂ + I₂ ⇌ 2HI (Keq given, find [eq]) |
Q5 — Data table
(a) For each trial: Δ[N2O4] = [NO2]eq/2 (stoichiometric ratio 1:2). [N2O4]eq = initial − Δ. Keq = [NO2]²/[N2O4].
Trial 1: Δ = 0.136/2 = 0.068; [N2O4]eq = 0.200 − 0.068 = 0.132 mol/L; Keq = (0.136)²/0.132 = 0.01850/0.132 = 0.140.
Trial 2: Δ = 0.204/2 = 0.102; [N2O4]eq = 0.400 − 0.102 = 0.298 mol/L; Keq = (0.204)²/0.298 = 0.04162/0.298 = 0.140.
Trial 3: Δ = 0.257/2 = 0.1285; [N2O4]eq = 0.600 − 0.1285 = 0.4715 mol/L; Keq = (0.257)²/0.4715 = 0.06605/0.4715 = 0.140.
(b) All three Keq values equal 0.140, confirming that Keq is independent of initial concentration — only temperature determines Keq. 1 mark for "equal (0.140 each)"; 1 mark for linking to temperature-dependence only.
(c) Temperature (held constant at 55°C throughout) — if temperature varied, Keq would change, invalidating the comparison between trials. Accept also volume/pressure of flask (1.00 L).