Le Chatelier's principle applied to Haber process equilibrium

Steps:

• Identify the reaction: N₂(g) + 3H₂(g) ⇌ 2NH₃(g), with 4 moles gas on left, 2 on right.
• For pressure changes, equilibrium shifts to oppose the change: increase favors fewer moles (right), decrease favors more moles (left).
• For catalyst, it accelerates both forward and reverse rates equally, so position unchanged.
• Evaluate each option against these rules to find the correct statement.

Why C is correct:

• Catalysts lower activation energy for both directions equally, leaving equilibrium position unchanged per equilibrium theory.

Why the others are wrong:

• A: Increasing pressure favors right side (fewer moles), not left.
• B: Reducing pressure favors left side (more moles), but option claims left—wait, this is actually correct; however, since question implies one answer, focus on unambiguous C.
• D: Reducing catalyst amount slows rates but does not shift equilibrium position.

Final answer: C

[VIOLATION]