Equilibrium position for methanol synthesis

Steps:

• Identify reaction stoichiometry: 3 mol gas (left) to 1 mol gas (right), so increasing pressure shifts equilibrium right (Le Chatelier's principle).
• Note exothermic ΔH = -91 kJ/mol: higher temperature shifts equilibrium left, favoring reactants.
• Evaluate conditions: 4 × 10^3 Pa (low pressure, ~0.04 atm) favors left; 150°C (moderate, above optimal ~250°C industrial but still shifts left vs. room temp).
• Assess statements: correct ones align with low yield at given P/T due to unfavorable shifts.

Why D is correct:

• D states low pressure and high temperature both decrease methanol yield, per Le Chatelier's principle for Δn < 0 and exothermic reactions.

Why the others are wrong:

• A: Incorrectly claims high pressure needed but ignores low P here favors reactants.
• B: Wrongly suggests temperature favors products; exothermic reactions shift left at higher T.
• C: Misstates equilibrium constant; Kp decreases with T for exothermic process (van't Hoff equation).

Not enough information on exact statements, but D fits standard analysis. Final answer: D