Chemistry • Year 12 • Module 5 • Lesson 14
Ka, Kb & Gibbs Free Energy
Build HSC Band 5–6 extended-response technique on Ka, Kb, Kw, ΔG°, and Australian chemical contexts — synthesise data, apply multi-step reasoning, and reach evidence-based evaluative judgements.
1. Data + scenario — acid strength, Kw, and Gibbs free energy (Band 5–6)
8 marks Band 5–6
Scenario. A research team at a CSIRO aquaculture facility is investigating the equilibrium chemistry of ammonia (NH&sub3;) and ammonium (NH&sub4;¹+) in prawn pond water. They measure the following equilibrium data at 25 °C.
| Species | Equilibrium | Equilibrium constant at 25 °C |
|---|---|---|
| Ammonium (NH&sub4;¹+) | NH&sub4;¹+(aq) ⇌ NH&sub3;(aq) + H¹+(aq) | Ka = 5.6 × 10−¹&sup0; |
| Ammonia (NH&sub3;) | NH&sub3;(aq) + H&sub2;O(l) ⇌ NH&sub4;¹+(aq) + OH−(aq) | Kb = ? |
| Water autoionisation | H&sub2;O(l) ⇌ H¹+(aq) + OH−(aq) | Kw = 1.0 × 10−¹&sup4; |
Figure 1. ΔG° versus ln(Keq) for the ammonium dissociation at three temperatures.
Figure 1. ΔG° vs ln(Ka) for NH&sub4;¹+ dissociation at three temperatures. R = 8.314 J mol−¹ K−¹.
Q1. Using the data and figure above, evaluate the statement: “The ammonium/ammonia system is not an equilibrium in the chemical sense — the Ka value is so small that the dissociation is negligible and can be ignored in any practical water chemistry calculation.”
In your response you must:
- Write Ka and Kb expressions for NH&sub4;¹+ and NH&sub3; respectively, and calculate Kb for NH&sub3; using Ka × Kb = Kw.
- Calculate ΔG° for the dissociation of NH&sub4;¹+ at 25 °C using the formula ΔG° = −RT ln Ka; interpret the sign in terms of spontaneity.
- Explain, with reference to Figure 1, how ΔG° changes with temperature and what this implies for the equilibrium position of NH&sub3;/NH&sub4;¹+ at higher temperatures.
- Evaluate the original statement — is it correct? Refer to at least one consequence for Australian aquaculture (CSIRO context).
2. Multi-step calculation & interpretation — Haber process and Gibbs free energy (Band 5–6)
7 marks Band 5–6
The Haber process for synthesising ammonia is central to Australian agricultural fertiliser production. The equilibrium is: N&sub2;(g) + 3H&sub2;(g) ⇌ 2NH&sub3;(g). Measured Keq values are: Keq = 977 at 25 °C (298 K) and Keq = 6.55 × 10−³ at 500 °C (773 K).
Q2. Complete all four parts below. Show all working.
(a) Calculate ΔG° at 25 °C (298 K) for the Haber process. State whether the reaction is spontaneous under standard conditions and identify the implication for equilibrium position. 2 marks
(b) Calculate ΔG° at 500 °C (773 K). Compare its sign and magnitude to part (a) and explain why industry operates the Haber process at 400–500 °C despite a less negative ΔG° at that temperature. 2 marks
(c) A student claims: “Because ΔG° is negative at 25 °C, the Haber reaction is fast at room temperature.” Identify the conceptual error and explain the correct relationship between ΔG° and reaction rate. 2 marks
(d) The Ka for NH&sub4;¹+ at 25 °C is 5.6 × 10−¹&sup0;. Calculate ΔG° for the dissociation of NH&sub4;¹+ at 25 °C. Is the magnitude of this ΔG° larger or smaller than the ΔG° you calculated in part (a)? What does the comparison tell you about relative equilibrium positions? 1 mark
Q1 — Marking criteria (8 marks)
1 mark — Ka expression: Ka(NH&sub4;¹+) = [NH&sub3;][H¹+] / [NH&sub4;¹+]; water excluded because it is the solvent.
1 mark — Kb expression: Kb(NH&sub3;) = [NH&sub4;¹+][OH−] / [NH&sub3;]; water excluded.
1 mark — Correct Kb calculation: Kb = Kw / Ka = 1.0 × 10−¹&sup4; / 5.6 × 10−¹&sup0; = 1.8 × 10−&sup5;.
1 mark — ΔG° calculation at 25 °C: ΔG° = −(8.314)(298) ln(5.6 × 10−¹&sup0;) = −(2477.6)(−21.30) = +52,800 J mol−¹ = +52.8 kJ mol−¹.
1 mark — Correct interpretation: ΔG° > 0 confirms NH&sub4;¹+ dissociation is not spontaneous under standard conditions; Keq << 1 means reactants (NH&sub4;¹+) strongly favoured. The equilibrium lies far to the left.
1 mark — Figure 1 trend: ΔG° becomes more positive as T increases (52.8 → 53.8 → 55.1 kJ mol−¹). Ka therefore increases slightly with temperature, shifting the NH&sub4;¹+ / NH&sub3; equilibrium further toward NH&sub3; at higher T. More un-ionised NH&sub3; exists at 50 °C than at 25 °C.
1 mark — Evaluation of the statement: The statement is incorrect. Ka = 5.6 × 10−¹&sup0; is indeed very small, but in an aquaculture pond with total ammonia at, say, 1 mg L−¹, even 0.056% existing as NH&sub3; can be acutely toxic to crustaceans and fish because NH&sub3; crosses biological membranes. The equilibrium is very real and has direct practical consequences.
1 mark — Australian aquaculture link: CSIRO Marine & Atmospheric Research has quantified NH&sub3; toxicity thresholds for Australian barramundi and prawns. At elevated pond temperatures (common in Queensland summer), the shift toward more NH&sub3; is significant enough to require active management (aeration, pH adjustment, water exchange) — demonstrating that even a “negligible” Ka cannot be ignored in practice.
Q2(a) — ΔG° at 25 °C (2 marks)
ΔG° = −RT ln Keq = −(8.314)(298) ln(977) = −(2477.6)(6.884) = −17,060 J mol−¹ = −17.1 kJ mol−¹ [1 mark]. ΔG° < 0 ⇒ forward reaction is spontaneous under standard conditions; Keq = 977 >> 1 confirms products (NH&sub3;) strongly favoured at equilibrium at 25 °C [1 mark].
Q2(b) — ΔG° at 500 °C (2 marks)
ΔG° = −(8.314)(773) ln(6.55 × 10−³) = −(6427)(−5.03) = +32,300 J mol−¹ = +32.3 kJ mol−¹ [1 mark]. At 500 °C, ΔG° is positive and large; Keq << 1 so the reaction disfavours NH&sub3; production under standard conditions. Industry operates at 400–500 °C not for thermodynamic reasons (equilibrium yield is poor at high T) but for kinetic ones: the iron catalyst requires elevated temperature to achieve an acceptable reaction rate. The trade-off between yield and rate is managed by pressure (which increases Keq for this reaction) and continuous removal of NH&sub3; product [1 mark].
Q2(c) — Conceptual error (2 marks)
The student confuses thermodynamic spontaneity with kinetic rate [1 mark]. ΔG° determines whether a reaction is energetically favoured (can proceed without external energy input) but says nothing about how fast it proceeds. At 25 °C, N&sub2; and H&sub2; react negligibly slowly without a catalyst, even though ΔG° is negative, because the activation energy is very high. Thermodynamics and kinetics are independent — a reaction can be highly spontaneous yet extremely slow [1 mark].
Q2(d) — ΔG° for NH&sub4;¹+ dissociation (1 mark)
ΔG° = −(8.314)(298) ln(5.6 × 10−¹&sup0;) = −(2477.6)(−21.30) = +52,800 J mol−¹ = +52.8 kJ mol−¹ [1 mark]. This is much larger in magnitude (and opposite in sign) than ΔG° for the Haber process (−17.1 kJ mol−¹). The larger positive ΔG° for NH&sub4;¹+ dissociation confirms its equilibrium lies much further to the left (Keq = 5.6 × 10−¹&sup0; vs 977 for the Haber process).