Ammonia Reactor at 700 K: When a Positive ΔG° Still Gives Product
In a fertiliser plant the shift engineer is reviewing the ammonia synthesis section. For N2(g) + 3H2(g) ⇌ 2NH3(g) the plant manual quotes ΔH° = −92.4 kJ mol⁻¹ and ΔS° = −198 J K⁻¹ mol⁻¹, effectively constant between 300 K and 900 K. The reactor runs at about 700 K and 200 atm, with ammonia continuously condensed out of the product stream. A trainee, having computed that the standard Gibbs energy change is no longer favourable at the operating temperature, argues that the reactor cannot possibly make ammonia and that the temperature must be lowered to 300 K to make the reaction 'spontaneous'. The shift engineer disagrees, pointing out that at 300 K the catalyst would be inactive and the rate negligible, and that the trainee has confused two different quantities.
1. Calculate ΔG° at 298 K and at 700 K, and find the temperature at which ΔG° = 0. Show every step.
Read the worked answer
Use ΔG° = ΔH° − TΔS°, with ΔS° = −198 J K⁻¹ mol⁻¹ = −0.198 kJ K⁻¹ mol⁻¹.
At 298 K: ΔG° = −92.4 − 298(−0.198) = −92.4 + 59.0 = −33.4 kJ mol⁻¹. At 700 K: ΔG° = −92.4 − 700(−0.198) = −92.4 + 138.6 = +46.2 kJ mol⁻¹. Crossover: 0 = −92.4 + T(0.198), so T = 92.4/0.198 ≈ 467 K.
So under standard conditions the reaction is spontaneous below about 467 K and non-spontaneous above it. The trainee's arithmetic is right: ΔG° at 700 K is indeed positive. The mistake lies in treating ΔG° as if it were the actual ΔG under plant conditions.
2. Using ΔG = ΔG° + RT ln Q, show quantitatively why ammonia is still formed at 700 K and 200 atm.
Read the worked answer
For N2 + 3H2 ⇌ 2NH3, Q = p(NH3)²/[p(N2)·p(H2)³]. At 700 K, RT = 8.314 × 700 = 5.82 kJ mol⁻¹.
Take a realistic reactor composition at 200 atm: p(NH3) = 2 atm, p(N2) = 49.5 atm, p(H2) = 148.5 atm. Then Q = 4/(49.5 × 148.5³) ≈ 4/(1.62 × 10⁸) ≈ 2.5 × 10⁻⁸, so ln Q ≈ −17.5.
Therefore ΔG = 46.2 + 5.82(−17.5) = 46.2 − 101.9 ≈ −55.7 kJ mol⁻¹ - clearly negative, so the forward reaction is spontaneous under these conditions. The high total pressure makes Q very small, and continuous removal of ammonia keeps p(NH3) low and drives Q down further. The positive ΔG° at 700 K only says that a standard-state mixture would not react; the plant never operates at standard state.
3. The manager proposes 900 K to speed things up at the same 200 atm. Estimate ΔG° at 900 K and explain why a compromise temperature near 700 K is chosen instead.
Read the worked answer
At 900 K: ΔG° = −92.4 − 900(−0.198) = −92.4 + 178.2 = +85.8 kJ mol⁻¹. The standard driving force has become even more unfavourable, so the equilibrium constant falls sharply (ln K = −ΔG°/RT = −85 800/(8.314 × 900) ≈ −11.5, K ≈ 1 × 10⁻⁵, compared with ln K ≈ −7.9, K ≈ 3.7 × 10⁻⁴ at 700 K). Physically, the reaction is exothermic with a decrease in gas moles, so raising temperature shifts the equilibrium back towards N2 and H2 by Le Chatelier's principle, and the extra pressure work cannot fully compensate.
A temperature near 700 K is the compromise: high enough for the iron catalyst to give a workable rate, low enough that K is still appreciable, with 200 atm and ammonia removal supplying the rest of the driving force. Running at 900 K would raise the rate but lower the per-pass conversion and waste hydrogen, so the judgement is kinetic versus thermodynamic, not simply 'spontaneous or not'.