Methanol Synthesis: Thermodynamic Feasibility versus Industrial Conditions
An industrial team is evaluating a methanol synthesis loop: CO(g) + 2 H2(g) ⇌ CH3OH(g). The reactor can operate between 300 and 550 K and at 50–100 atm over a Cu/ZnO/Al2O3 catalyst. The team has standard thermodynamic data at 298 K: ΔHf°(CO,g) = -110.5 kJ mol^-1, S°(CO,g) = 197.7 J mol^-1 K^-1; ΔHf°(H2,g) = 0, S°(H2,g) = 130.7 J mol^-1 K^-1; ΔHf°(CH3OH,g) = -201.0 kJ mol^-1, S°(CH3OH,g) = 239.8 J mol^-1 K^-1. The engineer claims that because the reaction is exothermic it should be run at the lowest possible temperature, but the plant manager insists on 500–550 K. Before changing conditions, they need to know whether the reaction is spontaneous at standard conditions, at what temperature it ceases to be spontaneous, and how pressure affects the decision.
1. Calculate ΔH°, ΔS° and ΔG° for the reaction at 298 K. State whether the reaction is spontaneous under standard conditions.
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For CO(g) + 2 H2(g) → CH3OH(g):
ΔH° = ΣΔHf°(products) - ΣΔHf°(reactants) ΔH° = (-201.0) - [(-110.5) + 2(0)] = -90.5 kJ mol^-1.
ΔS° = 239.8 - [197.7 + 2(130.7)] ΔS° = 239.8 - 459.1 = -219.3 J mol^-1 K^-1 = -0.2193 kJ mol^-1 K^-1.
The entropy change is negative because 3 mol of gas are converted into 1 mol of gas.
At 298 K: ΔG° = ΔH° - TΔS° = -90.5 - 298(-0.2193) ΔG° = -90.5 + 65.4 = -25.1 kJ mol^-1.
ΔG° is negative, so the reaction is spontaneous under standard conditions at 298 K.
2. Estimate the temperature at which ΔG° becomes zero under standard conditions. Calculate K at 500 K and interpret the result.
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Set ΔG° = 0: 0 = ΔH° - TΔS° T = ΔH° / ΔS° = (-90.5 kJ mol^-1) / (-0.2193 kJ mol^-1 K^-1) ≈ 413 K.
Above about 413 K, the -TΔS° term dominates and ΔG° becomes positive under standard conditions.
At 500 K: ΔG° = -90.5 - 500(-0.2193) = -90.5 + 109.7 = +19.2 kJ mol^-1.
Using ΔG° = -RT ln K: ln K = -19200 / (8.314 × 500) = -4.62 K ≈ 9.9 × 10^-3.
K is much less than 1 at 500 K, so under standard conditions the equilibrium lies towards the reactants. The reaction is not thermodynamically favourable at 500 K and 1 bar standard conditions.
3. The plant operates at 500–550 K and high pressure rather than at 298 K. Reconcile this with your thermodynamic results.
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At 298 K the standard Gibbs energy change is favourable, but the reaction rate is far too slow for an industrial process. The catalyst also requires a substantial temperature to be active. Operating at 500–550 K is a compromise: it gives a useful rate, although it reduces the equilibrium constant because the reaction is exothermic.
High pressure compensates thermodynamically. The reaction reduces the number of gas moles (3 → 1), so increasing pressure shifts the equilibrium to the right according to Le Chatelier. In terms of ΔG, the reaction quotient Q for gases decreases as total pressure increases, making ΔG = ΔG° + RT ln Q more negative. Thus 50–100 atm allows a practical conversion even at 500–550 K, where ΔG° alone would predict reactants.