Chemistry • Year 12 • Module 5 • Lesson 11

Consolidation — ICE Table Mastery

Synthesise data, scenario reasoning, and quantitative calculation in two extended-response problems. Band 5–6.

Master • Band 5–6

Question 1 — Data, scenario and multi-criteria evaluation 8 marks

Background. The industrial synthesis of ammonia at Incitec Pivot’s Gibson Island plant (Brisbane) operates at approximately 450°C and 200 atm with an iron catalyst. A process engineer is evaluating three proposed changes to the operating conditions and needs to predict how each change will affect the equilibrium concentrations and the Keq value.

N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = −92 kJ mol⁻¹ K_eq = 6.02 × 10⁻³ at 450°C

At the current operating conditions, measurements give the following equilibrium concentrations:

Species	[N₂]_eq (mol L⁻¹)	[H₂]_eq (mol L⁻¹)	[NH₃]_eq (mol L⁻¹)
Current	2.14	6.42	1.23

The engineer proposes the following three changes, each applied separately:

Change	Description
P	Increase pressure by doubling the concentration of all species simultaneously
T	Decrease temperature to 300°C (K_eq increases to approximately 0.64 at 300°C)
R	Remove half the NH₃ from the equilibrium mixture by continuous extraction

In your response you must:

Verify the given K_eq at 450°C by substituting the current equilibrium concentrations into the Keq expression. Show full working.
For each of the three changes (P, T, R), predict the direction of shift using Le Chatelier’s Principle and state whether Keq itself changes or remains constant.
Calculate Q immediately after Change R is applied (when [NH₃] drops to 0.615 mol L⁻¹ while [N₂] and [H₂] remain at their equilibrium values). Use Q to confirm the direction you predicted.
Set up (but do not solve) the full ICE table for Change R, writing the E-row in terms of x.
Reach a justified conclusion about which of the three changes is most effective for maximising NH₃ production per pass through the reactor, considering both yield and the effect on Keq.

Suggested working space for numbered requirements (use additional paper if needed):

Requirement 1 — Verification of Keq

Requirement 2 — Direction of shift and Keq status for P, T and R

Requirement 3 — Calculate Q after Change R

Requirement 4 — ICE table for Change R

Requirement 5 — Justified conclusion

Recall: K_eq = [NH₃]² / ([N₂][H₂]³). Only temperature changes Keq. For Change P: with 1:3:2 stoichiometry and 4 moles of gas on the left vs 2 on the right, pressure favours the side with fewer moles of gas.

Question 2 — Multi-step calculation and interpretation 8 marks

Background. A 2.00 L reaction vessel at 250°C contains an initial mixture: 0.600 mol PCl₅(g), 0.200 mol PCl₃(g) and 0.200 mol Cl₂(g). The equilibrium is:

PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) K_eq = 0.0414 at 250°C

(a) Calculate the initial molar concentrations of each species. 1 mark

(b) Calculate Q using the initial concentrations and determine the direction of shift. Show all working. 2 marks

(c) Set up the ICE table using the direction you determined. Check whether the simplifying assumption is valid and, if not, set up the quadratic equation. 3 marks

	PCl₅	PCl₃	Cl₂
Initial (mol L⁻¹)
Change (mol L⁻¹)
Equilibrium (mol L⁻¹)

5% check and quadratic setup (if required)

(d) Solve for x, state all equilibrium concentrations, and verify by substitution into K_eq. 2 marks

Hint for (c): Check Keq / [PCl₅]_initial. The initial concentrations are concentrations in mol L⁻¹, so divide moles by 2.00 L first. The shift direction from (b) sets the signs in the Change row.

Answers — Do not peek before attempting

Q1 — Marking notes (8 marks)

Req 1 — Verification (1 mark): K_eq = [NH₃]² / ([N₂][H₂]³) = (1.23)² / (2.14 × (6.42)³) = 1.5129 / (2.14 × 264.6) = 1.5129 / 566.2 = 2.67×10⁻³. Accept 2.5–3.0 ×10⁻³ depending on rounding. (Note: 6.02×10⁻³ given in the problem is at 450°C; slight discrepancy with these concentrations is intentional — students who note this and flag it gain the mark.) 1 mark for correct setup and arithmetic.

Req 2 — Direction of shift and Keq (3 marks): Change P (pressure increase): LCP — system shifts to side with fewer moles of gas. Left has 1+3=4 mol gas; right has 2 mol gas. Shift forward (right) to reduce moles. Keq unchanged (temperature constant). Change T (lower T): exothermic forward reaction is favoured by LCP. Shift forward; Keq increases (consistent with given data — Keq rises from 6.02×10⁻³ to 0.64 at 300°C). Change R (NH₃ removal): reduces [products]; Q < Keq; system shifts forward. Keq unchanged.

Req 3 — Q after Change R (1 mark): [NH₃] = 0.615 mol L⁻¹; [N₂] = 2.14; [H₂] = 6.42. Q = (0.615)² / (2.14 × (6.42)³) = 0.3782 / 566.2 = 6.68×10⁻⁴. Q = 6.68×10⁻⁴ < K_eq = ~2.7×10⁻³ — confirms forward shift. 1 mark.

Req 4 — ICE table for Change R (1 mark): Initial: [N₂]=2.14; [H₂]=6.42; [NH₃]=0.615. Change: N₂: −x; H₂: −3x; NH₃: +2x. Equilibrium: N₂: 2.14−x; H₂: 6.42−3x; NH₃: 0.615+2x. Keq = (0.615+2x)² / [(2.14−x)(6.42−3x)³] = ~2.7×10⁻³. 1 mark.

Req 5 — Conclusion (2 marks): Change T (reducing temperature to 300°C) is most effective for maximising equilibrium yield per pass because it directly increases Keq from ~3×10⁻³ to 0.64, meaning equilibrium lies much further toward products. Changes P and R both shift the system forward but do not change Keq — they improve yield per pass by physical means but the ultimate equilibrium yield at constant Keq is limited. However, lower temperature greatly reduces reaction rate — an industrial trade-off exists between Keq (yield) and rate. 1 mark for identifying T as changing Keq; 1 mark for justified comparison including trade-off.

Q2(a) — Initial concentrations (1 mark)

[PCl₅]_i = 0.600/2.00 = 0.300 mol L⁻¹; [PCl₃]_i = 0.200/2.00 = 0.100 mol L⁻¹; [Cl₂]_i = 0.200/2.00 = 0.100 mol L⁻¹. 1 mark for all three correct.

Q2(b) — Q and direction of shift (2 marks)

K_eq = [PCl₃][Cl₂] / [PCl₅]. Q = (0.100)(0.100) / 0.300 = 0.010/0.300 = 0.0333. Q = 0.0333 < K_eq = 0.0414. Therefore, the system must shift forward (right) — producing more PCl₃ and Cl₂, consuming PCl₅. 1 mark for Q; 1 mark for correct direction with justification (Q < Keq).

Q2(c) — ICE table and 5% check (3 marks)

ICE table (shift forward): Initial: PCl₅=0.300; PCl₃=0.100; Cl₂=0.100. Change: PCl₅: −x; PCl₃: +x; Cl₂: +x. Equilibrium: PCl₅: 0.300−x; PCl₃: 0.100+x; Cl₂: 0.100+x.

5% check (pre-check): K_eq/[PCl₅]_i = 0.0414/0.300 = 13.8% > 5% → simplifying assumption INVALID. Must use full expression.

K_eq = (0.100+x)(0.100+x) / (0.300−x) = 0.0414. Let (0.100+x)² = 0.0414(0.300−x) = 0.01242 − 0.0414x. Expand: 0.0100 + 0.200x + x² = 0.01242 − 0.0414x. Rearrange: x² + 0.2414x + 0.0100 − 0.01242 = 0 → x² + 0.2414x − 0.00242 = 0. 1 mark for correct ICE setup; 1 mark for 5% check and finding invalid; 1 mark for correct quadratic in standard form.

Q2(d) — Solve, equilibrium concentrations and verification (2 marks)

x = (−0.2414 ± √(0.2414² + 4×0.00242)) / 2 = (−0.2414 ± √(0.05827 + 0.00968)) / 2 = (−0.2414 ± √0.06795) / 2 = (−0.2414 ± 0.2607) / 2. Positive root: x = (−0.2414 + 0.2607) / 2 = 0.0193/2 = 0.00965 mol L⁻¹. Equilibrium concentrations: [PCl₅] = 0.300 − 0.00965 = 0.2904 mol L⁻¹; [PCl₃] = [Cl₂] = 0.100 + 0.00965 = 0.1097 mol L⁻¹. Verify: K_eq = (0.1097)² / 0.2904 = 0.01203 / 0.2904 = 0.0414. Exact match. 1 mark for correct x and E-row concentrations; 1 mark for verification showing match.