---
title: "A student studies the gas-phase equilibrium used in sulfuric acid production.  \\(2\\text{SO}_2\\text{(g)}+\\text{O}_2\\text{(g)}\\rightleftharpoons 2\\text{SO}_3\\text{(g)}\\)  After the system reaches equilibrium in a rigid vessel at constant temperature, the student injects inert \\(\\text{Ar(g)}\\). The total pressure increases, but the temperature and volume remain constant. Which statement correctly predicts and explains the effect of adding the \\(\\text{Ar(g)}\\) on the equilibrium position?"
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url: "https://nerd-notes.com/ubq/119407/"
date_modified: "2026-08-19T12:40:10+00:00"
---

# A student studies the gas-phase equilibrium used in sulfuric acid production.

\(2\text{SO}_2\text{(g)}+\text{O}_2\text{(g)}\rightleftharpoons 2\text{SO}_3\text{(g)}\)

After the system reaches equilibrium in a rigid vessel at constant temperature, the student injects inert \(\text{Ar(g)}\). The total pressure increases, but the temperature and volume remain constant. Which statement correctly predicts and explains the effect of adding the \(\text{Ar(g)}\) on the equilibrium position?

A student studies the gas-phase equilibrium used in sulfuric acid production.

\(2\text{SO}_2\text{(g)}+\text{O}_2\text{(g)}\rightleftharpoons 2\text{SO}_3\text{(g)}\)

After the system reaches equilibrium in a rigid vessel at constant temperature, the student injects inert \(\text{Ar(g)}\). The total pressure increases, but the temperature and volume remain constant. Which statement correctly predicts and explains the effect of adding the \(\text{Ar(g)}\) on the equilibrium position?

- **A.** The equilibrium position does not shift because adding \(\text{Ar(g)}\) increases the forward and reverse reaction rates by the same factor.
- **B.** The equilibrium position does not shift because the partial pressure of each reacting gas remains unchanged, so \(Q_p\) remains equal to \(K_p\).
- **C.** The equilibrium shifts toward \(\text{SO}_3\text{(g)}\) because the increase in total pressure favors the side with fewer moles of gas.
- **D.** The equilibrium shifts toward \(\text{SO}_2\text{(g)}\) and \(\text{O}_2\text{(g)}\) because adding \(\text{Ar(g)}\) lowers the mole fractions and therefore the partial pressures of the reacting gases.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/119407/*