Methanol Plant: Positive ΔG° but Still Viable?
An industrial chemistry class is evaluating a proposed methanol plant. The core reaction is CO(g) + 2H2(g) ⇌ CH3OH(g). At 298 K the standard enthalpy of reaction is −90.5 kJ mol−1 and the standard entropy of reaction is −219 J K−1 mol−1. The proposed operating conditions are 520 K and 250 atm with a copper/zinc oxide catalyst. A company engineer claims the reaction is 'not spontaneous at 520 K because the Gibbs energy is positive' and recommends shutting the plant. The class must test that claim using thermodynamic data and equilibrium reasoning, then decide whether the plant could still be viable. The plant needs at least 15% methanol at equilibrium to be economical, and the feed is stoichiometric CO and H2. Assume ideal gas behaviour and that ΔH° and ΔS° are independent of temperature over this range.
1. Calculate ΔG° at 298 K and at 520 K. Use the values to assess the engineer's claim.
Read the worked answer
Use ΔG° = ΔH° − TΔS°, with ΔS° = −0.219 kJ K−1 mol−1.
At 298 K: ΔG° = −90.5 − 298(−0.219) = −90.5 + 65.3 = −25.2 kJ mol−1. Negative, so the reaction is spontaneous under standard conditions at 298 K.
At 520 K: ΔG° = −90.5 − 520(−0.219) = −90.5 + 113.9 = +23.4 kJ mol−1. Positive, so the reaction is non-spontaneous under standard conditions at 520 K.
The engineer is correct only for standard conditions (all gases at 1 bar). The plant does not operate at standard conditions: it uses 250 atm. Because the reaction has 3 mol of gas on the left and 1 mol on the right, high pressure makes the reaction quotient Q much smaller than 1, so ΔG = ΔG° + RT ln Q becomes more negative. The claim does not by itself show the plant is unviable.
2. Determine the temperature at which ΔG° = 0. Explain what this means for the operating temperature and why the plant might still choose 520 K.
Read the worked answer
Set ΔG° = 0: 0 = ΔH° − TΔS°, so T = ΔH°/ΔS° = (−90.5 kJ mol−1)/(−0.219 kJ K−1 mol−1) = 413 K.
Below 413 K, ΔG° is negative and the forward reaction is spontaneous under standard conditions. Above 413 K, ΔG° is positive and the standard reaction is non-spontaneous. At 520 K the standard Gibbs energy is already positive, so high temperature is thermodynamically unfavourable for equilibrium yield.
However, 520 K may be chosen for kinetics: the catalyst needs a high temperature to work at a useful rate. The plant can compensate for the unfavourable equilibrium by using 250 atm, which shifts the equilibrium towards CH3OH (fewer gas moles) and makes ΔG negative under actual conditions. Thus 520 K is a compromise between rate and equilibrium yield, not evidence that the plant cannot work.
3. The catalyst is added, but the engineer says 'a catalyst changes ΔG, so it will fix the positive value'. Evaluate this statement and explain the role of the catalyst in the plant.
Read the worked answer
A catalyst does not change ΔH°, ΔS°, ΔG° or the equilibrium constant. It provides an alternative pathway with lower activation energy, so both forward and reverse rates increase and equilibrium is reached faster. It cannot make a thermodynamically unfavourable reaction favourable.
In the plant, the catalyst shortens the time needed to reach equilibrium at 520 K. Without it, the rate at 520 K might still be too slow for an economical reactor. The yield is improved by high pressure and by removing methanol as it forms, not by the catalyst changing ΔG. The engineer's statement is therefore wrong: the catalyst affects kinetics, not thermodynamics.