1. Δ_rH > 0 and Δ_rS > 0 2. Δ_rH > 0 and Δ_rS 0 4. Δ_rH < 0 and Δ_rS < 0

Answer: 4. Δ_rH < 0 and Δ_rS < 0 2Cl(g)→ Cl₂(g) forms a Cl-Cl bond. Bond formation releases energy, so Δ_rH<0 . Two particles become one, so disorder decreases and Δ_rS<0 . Answer: option (4).

Topic: Second law, entropy and spontaneity· Chemical Thermodynamics

For the reaction, 2Cl(g) → Cl_2(g), the correct option is :

1. \(\Delta_rH > 0\) and \(\Delta_rS > 0\)

2. \(\Delta_rH > 0\) and \(\Delta_rS < 0\)

3. \(\Delta_rH < 0\) and \(\Delta_rS > 0\)

4. \(\Delta_rH < 0\) and \(\Delta_rS < 0\)

Doubtly Asked 30 Oct 2026 NEET UG 2020 Practice test neet Save WhatsApp Report

1 answer

Best answer

Answer: 4. \(\Delta_rH < 0\) and \(\Delta_rS < 0\)

\(\mathrm{2Cl(g)\rightarrow Cl_2(g)}\) forms a Cl-Cl bond.

Bond formation releases energy, so \(\Delta_rH<0\).
Two particles become one, so disorder decreases and \(\Delta_rS<0\).

Answer: option (4).

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

Was this answer helpful?

For the reaction, 2Cl(g) → Cl_2(g), the correct option is :

1 answer

Your answer