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 (Analyse/Evaluate)
8 marks Analyse–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 (Analyse/Evaluate)
7 marks Analyse–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 markΔG° 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).