Chemistry • Year 12 • Module 5 • Lesson 12
Reaction Quotient Q
Apply the Q vs Keq framework to real concentration data, graphs, disturbance scenarios and Australian industrial contexts.
1. Interpret concentration data, H2 + I2 ⇌ 2HI at 430 °C
For the reaction H2(g) + I2(g) ⇌ 2HI(g), Keq = 54.3 at 430 °C. The table below shows five different reaction mixtures (not at equilibrium) in a sealed vessel. 8 marks
| Mixture | [H2] (mol/L) | [I2] (mol/L) | [HI] (mol/L) | Q (calculate) | Q vs Keq | Direction of shift |
|---|---|---|---|---|---|---|
| A | 0.10 | 0.10 | 0.00 | |||
| B | 0.00 | 0.00 | 0.30 | |||
| C | 0.20 | 0.20 | 0.93 | |||
| D | 0.05 | 0.05 | 0.09 | |||
| E | 0.020 | 0.020 | 0.148 |
1.1 Complete the table by calculating Q for each mixture, comparing it to Keq = 54.3, and stating the direction of shift. Show working for Mixture C below. 5 marks
1.2 Identify which mixture (A, B, C, D or E) is already at equilibrium. Justify your answer with reference to Q and Keq. 2 marks
1.3 For Mixture B, explain in one sentence what Q = ∞ indicates about the starting conditions. 1 mark
2. Graph interpretation, Q approaching Keq over time
The graph below models how the reaction quotient Q changes over time as the system H2(g) + I2(g) ⇌ 2HI(g) approaches equilibrium from two different starting conditions. Keq = 54.3 at 430 °C. 8 marks
Stylised model for H₂ + I₂ ⇌ 2HI at 430 °C (Kₑₓ = 54.3). Time axis is qualitative.
2.1 Describe the trend in Q for the blue (solid) curve between time 0 and equilibrium. State what happens to H2, I2 and HI concentrations to cause this trend. 3 marks
2.2 Both curves converge on the same Q value at equilibrium, even though they started from opposite extremes. Explain why this must be the case. 2 marks
2.3 Estimate (from the graph) the approximate time at which each mixture reaches equilibrium. At this point, what is the value of Q for both mixtures? 2 marks
2.4 A student claims: “The red curve (starting from products) reaches equilibrium faster because it has further to travel.” Identify the error in this reasoning. 1 mark
3. Cause-and-effect chain, adding NH3 to the Haber process
The Haber process at Incitec Pivot’s Gibson Island (Brisbane) plant runs the reaction N2(g) + 3H2(g) ⇌ 2NH3(g). Suppose the system is at equilibrium and an operator adds extra NH3 to the reactor. Use Q reasoning to complete the cause-and-effect chain below. 6 marks
Effect on Q expression (1 mark)
New Q vs Keq (1 mark)
Direction of shift (1 mark)
What happens to [NH3], [N2], [H2] (1 mark)
What happens to Q as the system moves to the new equilibrium? (1 mark)
Overall outcome for the operator (LCP consistent?) (1 mark)
4. Case study, phosphate saturation in the Murray–Darling Basin
Read the passage, then answer the question below. 5 marks
Water quality scientists monitoring phosphate levels in the Murray–Darling river system use Q (known in water chemistry as the ion activity product, IAP) to assess whether calcium phosphate is precipitating or dissolving in river sediments. The relevant equilibrium is:
Ca3(PO4)2(s) ⇌ 3Ca2+(aq) + 2PO43−(aq), Ksp = 2.07 × 10−33
After a heavy-rainfall event washes fertiliser runoff into the Darling River, scientists measure [Ca2+] = 4.0 × 10−3 mol/L and [PO43−] = 6.5 × 10−7 mol/L. They use Q (= IAP = [Ca2+]3[PO43−]2) to decide whether to intervene to reduce phosphate loading. If Q > Ksp, calcium phosphate will precipitate, removing phosphate from solution; if Q < Ksp, no precipitation occurs and dissolved phosphate remains available to fuel algal blooms.
4.1 Calculate Q (IAP) for the river water described above. Show full working. 2 marks
4.2 Compare Q to Ksp and predict whether calcium phosphate will precipitate under these conditions. State what this means for the risk of an algal bloom. 2 marks
4.3 In one sentence, explain why monitoring Q rather than just measuring [PO43−] alone gives scientists more useful information about the river’s phosphate dynamics. 1 mark
5. Predict and justify, diluting an equilibrium mixture
4 marks
A sealed flask contains an equilibrium mixture of H2(g), I2(g), and HI(g) at 430 °C with Keq = 54.3. The volume of the flask is suddenly doubled (all concentrations are halved simultaneously). The reaction is H2(g) + I2(g) ⇌ 2HI(g).
Using the Q expression, predict whether the equilibrium will shift left, shift right, or remain unchanged after the volume doubles. Justify your prediction by calculating the new Q in terms of the original equilibrium concentrations and comparing it to Keq. Show your reasoning algebraically.
Q1, Concentration data table
A: Q = 0²/(0.10×0.10) = 0. Q < Keq. Shift right.
B: Q = 0.30²/(0×0) = ∞ (denominator zero). Q > Keq. Shift left.
C: Q = 0.93²/(0.20×0.20) = 0.8649/0.0400 = 21.6. Q < Keq = 54.3. Shift right. Working sample: numerator = (0.93)² = 0.8649; denominator = (0.20)(0.20) = 0.0400; Q = 0.8649/0.0400 = 21.6.
D: Q = 0.09²/(0.05×0.05) = 0.0081/0.0025 = 3.24. Q < Keq. Shift right.
E: Q = 0.148²/(0.020×0.020) = 0.021904/0.000400 = 54.8 ≈ 54.3. Q = Keq (within rounding). At equilibrium, no net shift.
1.2: Mixture E is at equilibrium; Q ≈ 54.3 = Keq within experimental rounding, so no net shift occurs.
1.3: Q = ∞ means no reactants (H2 and I2) are present, the mixture consists of pure products only (HI), so the denominator of Q is zero and Q is mathematically infinite; the system must shift left to form reactants.
Q2, Graph interpretation
2.1: The blue (solid) curve rises from Q = 0 at time 0, initially steeply then with decreasing gradient, levelling off asymptotically at Keq = 54.3. This occurs because the reaction shifts right: [HI] increases (numerator of Q increases) while [H2] and [I2] decrease (denominator decreases), so Q increases progressively. [3 marks: 1 describe shape; 1 HI increases numerator; 1 H2/I2 decrease denominator]
2.2: Both curves converge on Keq = 54.3 because Keq is a constant at a fixed temperature (430 °C) and represents the single equilibrium composition the system tends toward regardless of starting conditions. [1 Keq is constant at fixed T; 1 equilibrium is the same destination regardless of starting point]
2.3: From the graph, both mixtures reach approximate equilibrium at around time 5–6 (arbitrary units). At that point Q = 54.3 for both mixtures.
2.4: The student is wrong because the rate of approach to equilibrium depends on reaction kinetics (concentrations, temperature, activation energy), not on how far Q is from Keq. “Further to travel” (in Q-units) does not mean faster in real time; both mixtures can reach equilibrium at similar times depending on rate constants.
Q3, Cause-and-effect chain: adding NH3
Effect on Q expression: NH3 is in the numerator of Q (= [NH3]²/([N2][H2]³)). Adding NH3 increases [NH3] → numerator increases → Q increases.
New Q vs Keq: Q > Keq.
Direction of shift: Shift left (reverse direction).
Concentrations: [NH3] decreases; [N2] and [H2] increase as NH3 decomposes.
Q as system shifts: Q decreases back toward Keq as [NH3] falls and [N2]/[H2] rise, until Q = Keq at the new equilibrium.
Overall outcome: The shift is consistent with Le Chatelier’s Principle, the system opposes the increase in NH3 by converting it back to reactants. The final [NH3] is higher than the original but lower than immediately after addition. [N2] and [H2] are slightly higher than their original values.
Q4, Murray–Darling phosphate case study
4.1 Calculation:
Q = [Ca2+]³ × [PO43−]²
= (4.0 × 10−3)³ × (6.5 × 10−7)²
= (6.4 × 10−8) × (4.225 × 10−13)
= 2.7 × 10−20 [2 marks: 1 correct substitution; 1 correct answer with units noted as mol5/L5]
4.2: Q = 2.7 × 10−20 > Ksp = 2.07 × 10−33. Therefore Q > Ksp → the solution is supersaturated with respect to calcium phosphate → precipitation will occur. This is good news for algal bloom risk: as Ca3(PO4)2 precipitates, dissolved [PO43−] is reduced, limiting the nutrient available to fuel algal growth. [1 Q > Ksp stated; 1 precipitation → reduced dissolved phosphate → lower bloom risk]
4.3: Q incorporates both [Ca2+] and [PO43−] together in one expression, so it captures the combined effect of all relevant ions on whether precipitation is thermodynamically favoured, whereas monitoring [PO43−] alone cannot predict whether precipitation will actually occur without knowing [Ca2+] as well.
Q5, Dilution of H2 + I2 ⇌ 2HI
Let the original equilibrium concentrations be [H2] = a, [I2] = b, [HI] = c (where Q = c²/(ab) = 54.3).
After doubling volume, all concentrations halve: [H2] = a/2, [I2] = b/2, [HI] = c/2.
New Q = (c/2)² / ((a/2)(b/2)) = (c²/4) / (ab/4) = c²/ab = 54.3.
Because the stoichiometry is 1:1:2 and the number of moles of gas is equal on both sides (1 + 1 = 2), the (1/2) factors cancel identically in numerator and denominator. Therefore the new Q = original Q = Keq. The equilibrium does not shift. This is unique to reactions where Δn(gas) = 0; for reactions where Δn ≠ 0, dilution would shift the equilibrium toward the side with more moles of gas.