For the exothermic equilibrium reaction N2 + 3 H2 ⇌ 2 NH3, the total gas moles decrease. To obtain high single-pass equilibrium conversion, the synthesis should be operated at which pressure regime?

Introduction / Context:
Ammonia formation reduces the number of gas molecules (from 4 to 2) and releases heat. Le Chatelier’s principle predicts how pressure and temperature influence equilibrium. Understanding this is central to converter design and loop pressure selection.

Given Data / Assumptions:

• Reaction: N2 + 3 H2 ⇌ 2 NH3, exothermic.
• Total moles decrease across the reaction.
• Goal: maximize equilibrium conversion in a single pass.

Concept / Approach:
Increasing pressure shifts equilibrium toward the side with fewer moles (ammonia). Lowering temperature also favors ammonia but slows kinetics; thus practical plants use elevated pressure and moderate temperature to balance rate and equilibrium. Therefore, high pressure improves single-pass conversion significantly versus low or atmospheric pressure.

Step-by-Step Solution:

Verification / Alternative check:
Equilibrium calculations show rising ammonia mole fraction with pressure at fixed temperature, consistent with industrial practice (converter pressures typically many tens to over one hundred bar).

Why Other Options Are Wrong:

• Low/atmospheric pressure: shift equilibrium toward reactants.
• Very high temperature: lowers equilibrium NH3 despite faster kinetics.
• Vacuum: opposite of desired effect.

Common Pitfalls:
Equating kinetic advantages of high temperature with better equilibrium—these effects oppose each other for exothermic reactions.

Final Answer:
high pressure