Chemistry - Thermodynamics - II Question with Solution | TestHub
ChemistryThermodynamics - IIDS in chemical reactionHard2 minQB
ChemistryHardmatching list
Match the process of Column I with entropy or enthalpy change in Column II
Column I | Column II |
|---|---|
P) N2(g) + O2(g) → 2NO(g) | i) Δ S ≈ 0 |
Q) 2KI(aq) + HgI2(aq) → K2[HgI4](aq) | ii) Δ S < 0 |
R) PCl5(g) → PCl3(g) + Cl2(g) | iii) Δ H > 0 |
S) NH3(g) + HCl(g) → NH4Cl(s) | iv) Δ H < 0 |
Options:
Answer:
D
Solution:
(P) Δ ng = 0, so Δ S ≈ 0; breaking N≡N and O=O to form the weaker N–O bond is endothermic, Δ H > 0 — i,iii. (Q) Four free ions combine into one complex ion, ordering the solution: Δ S < 0; complex formation is exothermic, Δ H < 0 — ii,iv. (R) One mole of gas gives two moles, so Δ S > 0 (not listed), and PCl5 decomposition is endothermic, Δ H > 0 — iii only. (S) Two gases form a solid, Δ S < 0; the combination is strongly exothermic, Δ H < 0 — ii,iv.
Stream:JEESubject:ChemistryTopic:Thermodynamics - IISubtopic:DS in chemical reaction
⏱ 2mℹ️ Source: QB
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