Chemistry · Year 12 · Module 5 · Lesson 10
HSC Exam Practice
Calculating Keq & ICE Tables
Short answer
1.Short answer — Keq fundamentals
Define Keq for a general reversible reaction, including the role of stoichiometric coefficients in the expression.
Identify the one factor that changes the value of Keq for a given reaction, and explain why concentration changes do not change Keq.
At 430°C, Keq = 54.3 for H2(g) + I2(g) ⇌ 2HI(g). The equilibrium concentrations are: [H2] = 0.0213 mol/L, [I2] = 0.0213 mol/L, [HI] = 0.157 mol/L. Verify that these concentrations give the stated Keq.
Describe the purpose of each row in an ICE table (Initial, Change, Equilibrium) for the reaction N2O4(g) ⇌ 2NO2(g). Include what determines the entries in the Change row.
For the equilibrium 2SO3(g) ⇌ 2SO2(g) + O2(g) at 600°C, the equilibrium concentrations are: [SO3] = 0.780 mol/L, [SO2] = 0.220 mol/L, [O2] = 0.110 mol/L.
(a) Write the Keq expression.
(b) Calculate Keq to three significant figures.
(c) Account for what the magnitude of Keq indicates about the equilibrium position.
Outline the 5% rule used in ICE table calculations. Under what conditions is it applied, and what must a student do if the rule is violated?
Data response
2.Multi-step ICE calculation — N2O4 system (E4)
The equilibrium N2O4(g) ⇌ 2NO2(g) has Keq = 0.500 at 55°C. A chemist seals 0.400 mol of N2O4 in a 1.00 L flask at 55°C with no NO2 initially present.
(a) Set up the ICE table for this reaction. Define x and write the Equilibrium row in terms of x.
(b) Write the Keq expression in terms of x and set it equal to 0.500.
(c) Before solving algebraically, test whether the simplifying assumption (x ≪ 0.400) is valid using the 5% rule threshold. State whether you will use the assumption or the quadratic formula.
(d) Solve for x and state the equilibrium concentrations of N2O4 and NO2. Show full working.
(e) Verify your answer. State one assumption you made about the system.
3.Data-driven short answer — temperature vs Keq for NO formation (E2)
The table below shows Keq values for N2(g) + O2(g) ⇌ 2NO(g) at various temperatures relevant to internal combustion engines. The forward reaction is endothermic (ΔH = +181 kJ/mol).
| Temperature (°C) | Keq |
|---|---|
| 25 | 1.0 × 10−30 |
| 1000 | 7.5 × 10−9 |
| 1500 | 1.0 × 10−5 |
| 2000 | 4.0 × 10−4 |
(a) Describe the trend in Keq as temperature increases, and explain this trend using the sign of ΔH. (2 marks)
(b) Even though Keq at 2000°C is still very small (4.0 × 10−4), significant concentrations of NO form inside car engines in NSW. Explain this apparent contradiction, referring to the concept of equilibrium position. (2 marks)
(c) Identify one atmospheric consequence of elevated [NO] in vehicle exhaust gases in NSW urban areas, and write a balanced equation for the relevant reaction. (2 marks)
Extended response
4.Extended response (E3)
Analyse the role of the ICE table method in enabling chemists to calculate equilibrium concentrations and determine Keq when direct measurement is not possible. In your response, use the equilibrium H2(g) + I2(g) ⇌ 2HI(g) as a detailed worked example, and evaluate the conditions under which the simplifying assumption is acceptable.
Chemistry · Year 12 · Module 5 · Lesson 10
Answer Key & Marking Guidelines
Section 1 · Short answer · 2 marks · Band 3
Sample response. Keq is the equilibrium constant for a reversible reaction, defined as the ratio of the product of equilibrium concentrations of products to the product of equilibrium concentrations of reactants, with each concentration raised to the power of its stoichiometric coefficient in the balanced equation. For aA + bB ⇌ cC + dD: Keq = [C]c[D]d / ([A]a[B]b).
Marking notes. 1 mark for ratio of products over reactants (equilibrium concentrations); 1 mark for stoichiometric powers applied to each concentration.
Section 1 · Short answer · 3 marks · Band 3–4
Sample response. Temperature is the only factor that changes Keq. When the temperature changes, the activation energy barrier for the forward and reverse reactions changes unequally (dependent on ΔH), altering the ratio of forward and reverse rate constants and therefore Keq. Concentration changes do not change Keq because the system responds by shifting its equilibrium position (via Le Chatelier's principle) until the ratio of equilibrium concentrations returns to the same value — Keq is a constant at fixed temperature regardless of initial concentrations.
Marking notes. 1 mark for naming temperature; 1 mark for explaining mechanism (ΔH causes unequal change to forward/reverse rates); 1 mark for explaining why concentration changes do not change Keq (system re-equilibrates to same ratio).
Section 1 · Short answer · 2 marks · Band 3–4
Sample response. Keq = [HI]² / ([H2][I2]) = (0.157)² / (0.0213)(0.0213) = 0.02465 / 4.537 × 10−4 = 54.3 ✓. The calculated value matches the given Keq, confirming the concentrations are correct.
Marking notes. 1 mark for correct substitution into Keq expression with correct powers; 1 mark for correct numerical result ≈ 54.3 with verification statement.
Section 1 · Short answer · 3 marks · Band 3–4
Sample response. The Initial row records the starting molar concentrations of each species before the system shifts toward equilibrium (products = 0 if starting from pure reactants). The Change row shows the stoichiometric change in concentration as equilibrium is approached — entries are expressed as multiples of a variable x with signs determined by the direction of reaction (reactants decrease, products increase) and magnitudes set by the stoichiometric ratio (N2O4 : NO2 = 1 : 2, so Change = −x and +2x respectively). The Equilibrium row is calculated as Initial + Change and gives the actual equilibrium concentrations substituted into the Keq expression.
Marking notes. 1 mark each for: Initial row (starting concentrations, products = 0 explained); Change row (stoichiometric ratios determine multiples of x, signs determined by direction); Equilibrium row (Initial + Change, used in Keq expression).
Section 1 · Short answer · 5 marks · Band 4
Sample response.
(a) Keq = [SO2]²[O2] / [SO3]²
(b) Keq = (0.220)²(0.110) / (0.780)² = (0.04840)(0.110) / 0.6084 = 5.32 × 10−3 / 0.6084 = 8.75 × 10−3
(c) Keq = 8.75 × 10−3 < 1, indicating the equilibrium is reactant-favoured: at 600°C, the mixture contains significantly more SO3 than SO2 and O2 — the forward decomposition of SO3 proceeds only slightly.
Marking notes. (a) 1 mark for correct expression (products over reactants, correct powers). (b) 2 marks: 1 for correct substitution, 1 for correct answer 8.75 × 10−3 (accept 8.74–8.76). (c) 2 marks: 1 for Keq < 1 → reactant-favoured; 1 for explaining concentration implications (more SO3, less SO2 and O2).
Section 1 · Short answer · 2 marks · Band 4
Sample response. When Keq is very small, x (the shift toward equilibrium) may be negligible compared to the initial concentration — the simplifying assumption treats (initial − x) ≈ initial, avoiding a quadratic. The 5% rule checks validity: if (x / initial) × 100% < 5%, the assumption is acceptable. If the rule is violated (≥5%), the student must use the quadratic formula to solve for x exactly.
Marking notes. 1 mark for stating the assumption and its purpose (avoid quadratic when Keq very small); 1 mark for the 5% rule check and consequence of failure (use quadratic).
Section 2 · Data response (E4) · 8 marks · Band 4–5
Sample response.
(a) ICE table: N2O4: I = 0.400, C = −x, E = 0.400 − x. NO2: I = 0, C = +2x, E = 2x.
(b) Keq = (2x)² / (0.400 − x) = 4x² / (0.400 − x) = 0.500.
(c) Check 5% rule: Keq/[initial] = 0.500/0.400 = 125% ≫ 5% — assumption INVALID. Must use quadratic formula.
(d) 4x² = 0.500(0.400 − x) = 0.200 − 0.500x. 4x² + 0.500x − 0.200 = 0. Quadratic formula: x = [−0.500 ± √(0.250 + 3.200)] / 8 = [−0.500 ± √3.450] / 8 = [−0.500 ± 1.857] / 8. Positive root: x = 1.357 / 8 = 0.170 mol/L. [N2O4]eq = 0.400 − 0.170 = 0.230 mol/L. [NO2]eq = 2(0.170) = 0.340 mol/L.
(e) Verification: (0.340)² / 0.230 = 0.1156 / 0.230 = 0.502 ≈ 0.500 ✓. Assumption: constant temperature (55°C), ideal gas behaviour, sealed fixed-volume system.
Marking notes. (a) 1 mark: correct ICE table with 2x for NO2. (b) 1 mark: correct Keq expression in terms of x. (c) 1 mark: 5% rule test, correct conclusion that quadratic is required. (d) 3 marks: 1 for rearranging to quadratic form; 1 for applying quadratic formula correctly; 1 for correct x and equilibrium concentrations (accept ±0.005 mol/L). (e) 1 mark for verification ≈ 0.500; 1 mark for stating a valid assumption.
Section 2 · Data-driven short answer (E2) · 6 marks · Band 4–5
Sample response.
(a) Keq increases dramatically with temperature — from 1.0 × 10−30 at 25°C to 4.0 × 10−4 at 2000°C (a factor of ~1026 over this range). Because ΔH = +181 kJ/mol (endothermic), increasing temperature supplies energy that drives the forward reaction; Le Chatelier's principle predicts that Keq increases as more product (NO) is favoured at higher temperatures. 1 mark for trend (increases with temperature, quoting values); 1 mark for endothermic explanation linking ΔH to Keq direction.
(b) Even though Keq = 4.0 × 10−4 at 2000°C is small, the starting concentrations of N2 and O2 in air are very large (~0.78 mol/L N2 and ~0.21 mol/L O2) — a small Keq applied to large initial concentrations still produces significant [NO] at equilibrium. Furthermore, the extremely high temperature favours a much larger equilibrium position than at room temperature, and the system may not reach equilibrium before exhaust gases are expelled (kinetic trapping). 1 mark for large initial concentrations argument; 1 mark for temperature-favoured equilibrium shift compared to 25°C.
(c) Acid rain formation: NO oxidises to NO2 in the atmosphere; NO2 then reacts with water — e.g. 3NO2(g) + H2O(l) → 2HNO3(aq) + NO(g), producing nitric acid that lowers pH of rainfall, damaging ecosystems and infrastructure in NSW. Accept also: smog formation / photochemical smog (NO + VOCs + UV → ozone + PANs). 1 mark for naming acid rain (or photochemical smog); 1 mark for a correct balanced equation.
Section 3 · Extended response (E3) · 7 marks · Band 5–6
Marking notes (7 marks).
- 1 mark — explains why direct measurement is not always possible: at equilibrium concentrations of all species may not be directly observable (e.g. colourless gases like H2 and I2; only one species measurable by colourimetry); ICE allows calculation of unknown concentrations from a known initial condition and one measured equilibrium value.
- 1 mark — correct ICE table set-up for H2 + I2 ⇌ 2HI: I row states initial concentrations; Change row shows −x, −x, +2x (stoichiometric ratio 1:1:2 explicitly referenced); Equilibrium row = (a−x), (b−x), 2x.
- 1 mark — Keq expression written correctly: Keq = (2x)² / [(a−x)(b−x)], with the square root shortcut when a = b demonstrated (2x/(a−x) = √Keq).
- 1 mark — worked numerical example correctly solved: e.g. a = b = 0.100 mol/L, Keq = 54.3; x = 0.0787 mol/L; [H2] = [I2] = 0.0213 mol/L; [HI] = 0.157 mol/L.
- 1 mark — verification step included and correctly evaluated (substituting equilibrium concentrations gives ≈54.3).
- 1 mark — valid evaluation of the simplifying assumption: applicable only when Keq ≪ [initial] (Keq/[initial] < 0.05 → less than 5% deviation); for the H2/I2/HI system Keq = 54.3 is NOT small relative to 0.100 mol/L (ratio 543%) so the assumption cannot be applied and the square root (or quadratic) method is required.
- 1 mark — cohesive analysis that explicitly links ICE method to the problem of indirect measurement: states that ICE is necessary when equilibrium concentrations cannot all be directly measured and concludes that the method's reliability rests on: (i) correct stoichiometric Change row; (ii) verification step; (iii) valid choice of algebraic method depending on Keq magnitude.