JEE Main 2023ChemistryThermodynamics (C)MediumNumerical

JEE Main 2023 — Thermodynamics (C) Question with Solution

JEE Main 2023 (31 Jan Shift 1)

Question

The enthalpy change for the conversion of 12Cl2 (g) to Cl-(aq) is (-)
kJmol-1 (Nearest integer)
Given : Δdis HCl2( g)0=240kJmol-1.
ΔegHCl(g)o=-350kJmol-1,
ΔhydHCl(g)-o=-380kJmol-1

Enter your answer

Show full solutionCorrect answer: 610
Correct answer
610

Step-by-step explanation

12Cl2( g) → Cl(g) → Cl(g)- → Cl(aq.)-

Apply Hess's law,

∆H = 12 ∆dis HCl2 (g)0 + ∆egHCl(g)o + ∆hydHCl(g)-o

⇒ ∆H=12× 240 + (-350) + (-380)

⇒ ∆H = -610

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About this question

This is a previous-year question from JEE Main 2023, covering the Thermodynamics (C) chapter of Chemistry. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.