Chemistry • Year 12 • Module 5 • Lesson 11
Consolidation — ICE Table Mastery
Apply ICE table reasoning to real data, Australian contexts, and multi-step scenarios including Q-first problems.
1. Interpret ICE table data — acid dissociation in carbonated beverages
Carbonic acid (H2CO3) forms when CO2 dissolves in water, and it dissociates according to:
H2CO3(aq) ⇌ H+(aq) + HCO3−(aq) Keq = 4.3 × 10−7 at 25°C
A food scientist measured the initial H2CO3 concentration in four carbonated beverage samples and recorded the equilibrium [H+] found by pH measurement. The table below shows their data. 8 marks
| Sample | [H2CO3]initial (mol L−1) | [H+]eq (mol L−1) | Keq/[initial] (%) | Assumption valid? |
|---|---|---|---|---|
| A (cola) | 0.0340 | 1.21 × 10−4 | ||
| B (sparkling water) | 0.0120 | 7.17 × 10−5 | ||
| C (fruit soda) | 0.0680 | 1.71 × 10−4 | ||
| D (tonic water) | 0.0200 | 9.27 × 10−5 |
1.1 Calculate the Keq/[initial] percentage for each sample and determine whether the simplifying assumption is valid. Fill in the last two columns of the table. Show one sample calculation below. 4 marks
1.2 For Sample A, verify the scientist’s measured [H+] by substituting the ICE table E-row values into the Keq expression. Does the result agree with Keq = 4.3 × 10−7? Show your working. 2 marks
1.3 Explain why the CO2 gas dissolved in the drink is not written as a species in the Keq expression for the H2CO3 dissociation shown above. 2 marks
2. Interpret graph — equilibrium yield of NH3 vs temperature
Incitec Pivot operates ammonia synthesis plants in Australia using the Haber process: N2(g) + 3H2(g) ⇌ 2NH3(g). The graph below shows calculated equilibrium percent yield of NH3 at a total pressure of 200 atm as a function of temperature, based on known Keq values. 9 marks
2.1 Describe the trend shown in the graph. 2 marks
2.2 Using Le Chatelier’s Principle and your knowledge of Keq, explain why the equilibrium yield of NH3 decreases as temperature increases. 3 marks
2.3 Incitec Pivot operates at approximately 450°C even though this gives only ~22% equilibrium yield. Using both the graph data and ICE table reasoning, explain the industrial compromise between yield and rate. In your answer, identify what would happen to Keq if temperature were reduced to 300°C. 4 marks
3. Cause-and-effect chain — ICE table with non-zero initial products
Trace the reasoning chain for the following scenario. Fill in the empty effect boxes. 5 marks
Scenario: A flask contains [H2] = 0.10, [I2] = 0.10, [HI] = 0.50 mol L−1 at 430°C. Keq = 54.3 for H2(g) + I2(g) ⇌ 2HI(g).
Non-zero concentrations of all three species: [H2] = [I2] = 0.10, [HI] = 0.50 mol L−1
Q = [HI]² / ([H2][I2]) =
Q = ____ ; Keq = 54.3 ; Q __ Keq
H2: ______ ; I2: ______ ; HI: ______
3.1 (Overall outcome) Explain in one sentence what happens to [HI] as the system reaches equilibrium, and how this is consistent with Q vs Keq. 2 marks
4. Case study — Applying ICE reasoning to a real problem
The following is a description of an error commonly reported in HSC examiner reports. 6 marks
“Many candidates correctly set up the ICE table and solved for x = 0.0920 mol L−1, but then wrote [NH3]eq = 0.0920 mol L−1 when the equilibrium should have given [NH3]eq = 0.184 mol L−1. Candidates lost 1 mark on the final answer and often 1 further mark because the verification step was skipped.” — HSC Chemistry Examiner’s Report (composite, adapted)
4.1 Identify the two specific errors described in the examiner’s report and explain the chemical reason for each. 4 marks
4.2 For N2 + 3H2 ⇌ 2NH3, if x = 0.0920 mol L−1, write the correct expression for [NH3]eq in terms of x and calculate the correct value. 2 marks
Q1.1 — 5% check for carbonated beverage samples
Keq/[initial] × 100 for each: A: (4.3×10−7/0.0340) × 100 = 1.26×10−3% — valid. B: (4.3×10−7/0.0120) × 100 = 3.58×10−3% — valid. C: (4.3×10−7/0.0680) × 100 = 6.32×10−4% — valid. D: (4.3×10−7/0.0200) × 100 = 2.15×10−3% — valid. All samples: assumption valid (all far below 5%).
Q1.2 — Verification for Sample A
ICE E-row: [H2CO3]eq = 0.0340 − 1.21×10−4 ≈ 0.03388; [H+]eq = [HCO3−]eq = 1.21×10−4. Keq = (1.21×10−4)² / 0.03388 = 1.464×10−8 / 0.03388 = 4.32×10−7 ≈ 4.3×10−7. Yes, consistent. (1 mark for correct substitution; 1 mark for conclusion.)
Q1.3 — CO2 exclusion
The equilibrium shown is specifically for H2CO3(aq) dissociation. CO2(g) is a separate species whose concentration in solution is fixed by Henry’s Law at a given pressure; in a closed system it is treated as a separate pure-gas phase and is not included in the Keq expression for the aqueous dissociation. (Accept: CO2 is a distinct equilibrium — its dissolution is a separate process.)
Q2.1 — Graph trend
As temperature increases from 300°C to 550°C, the equilibrium yield of NH3 decreases sharply (from ~78% to ~10%), with the steepest decline occurring between 300°C and 450°C.
Q2.2 — LCP and Keq explanation
The forward Haber reaction is exothermic (ΔH = −92 kJ mol−1). By Le Chatelier’s Principle, increasing temperature favours the endothermic (reverse) reaction, shifting equilibrium to the left and decreasing [NH3]eq. At the molecular level, higher temperature increases the value of the reverse rate constant more than the forward, so Keq decreases with increasing temperature for this exothermic reaction.
Q2.3 — Industrial compromise
At 300°C, Keq is large (yield ~78%) but reaction rate is negligible — the activation energy for NH3 synthesis is high and the reaction would take impractically long without a catalyst. At 450°C with an iron catalyst, sufficient rate is achieved to make the process economically viable even though equilibrium yield is only ~22%. The product is continuously removed (reducing Q below Keq) so the system continually shifts forward, compensating for the lower equilibrium yield. Reducing T to 300°C would increase Keq and equilibrium yield, but rate would be prohibitively slow.
Q3 — Cause-and-effect chain
Step 2: Q = (0.50)² / (0.10 × 0.10) = 0.25/0.010 = 25. Step 3: Q = 25 < Keq = 54.3. Step 4: System shifts forward (to the right) to increase [HI]. Step 6: H2: −x; I2: −x; HI: +2x. 3.1: [HI] increases as the system shifts forward, which is consistent with Q < Keq — the system must produce more HI to bring Q up to Keq.
Q4.1 — Two errors
Error 1 (stoichiometric conversion): The student solved for x = 0.0920, but for N2 + 3H2 ⇌ 2NH3, the Change row entry for NH3 is +2x (coefficient = 2). Therefore [NH3]eq = 2x = 2(0.0920) = 0.184 mol L−1, not x. The student confused x with the equilibrium concentration. (2 marks)
Error 2 (missing verification): Without substituting back into the Keq expression, arithmetic errors in solving the quadratic or stoichiometric errors go undetected. Verification is the mandatory final step of any ICE table calculation. (2 marks)
Q4.2 — Correct [NH3]eq
[NH3]eq = 2x = 2(0.0920) = 0.184 mol L−1.