In the contact process (SO2 → SO3 over V2O5), which condition thermodynamically favours the oxidation of SO2 to SO3?

Difficulty: Medium

Correct Answer: Low temperature and high pressure

Explanation:

Introduction:Optimizing SO2 oxidation to SO3 is central to sulfuric acid manufacture. Understanding how temperature and pressure shift equilibrium is a direct application of Le Chatelier’s principle and reaction stoichiometry, key thermodynamics for process engineers.

Given Data / Assumptions:

  • Reaction: 2 SO2 + O2 → 2 SO3.
  • Exothermic reaction (heat released).
  • Moles decrease (3 gas moles → 2 gas moles).

Concept / Approach:For exothermic reactions, lower temperature shifts equilibrium toward products. When total gas moles decrease, higher pressure also drives equilibrium to the product side. Practical operation uses a compromise temperature (about 400–450°C) to maintain good kinetics while not sacrificing equilibrium too much.

Step-by-Step Solution:Apply Le Chatelier to heat: exothermic → lower T favours SO3.Apply to moles: fewer moles in products → higher pressure favours SO3.Combine: low T and high P best favour oxidation thermodynamically.

Verification / Alternative check:Equilibrium constants decrease with increasing temperature for exothermic reactions. Pressure dependence follows stoichiometric reduction in total gas moles, pushing equilibrium to SO3 at higher pressure.

Why Other Options Are Wrong:

  • High temperature options shift equilibrium backward for an exothermic reaction.
  • Low pressure disfavors the side with fewer moles (SO3).

Common Pitfalls:Confusing kinetic requirements (need moderate T for rate) with pure equilibrium preference; plant conditions balance both, but the question asks about what “favours” the conversion thermodynamically.

Final Answer:Low temperature and high pressure

More Questions from Chemical Process

Discussion & Comments

No comments yet. Be the first to comment!
Join Discussion