JEE Main 2021ChemistryThermodynamics (C)MediumNumerical

JEE Main 2021 — Thermodynamics (C) Question with Solution

JEE Main 2021 (26 Aug Shift 1)

Question

The Born-Haber cycle for KCl is evaluated with the following data:
ΔfHΘ for KCl=−436.7kJmol−1;ΔsubHΘ for K=89.2kJmol−1;

Δionization HΘ for K=419.0kJ mol −1;Δelectron gain HΘ for Cl(g)=−348.6kJmol−1

Δbond HΘ for Cl2=243.0kJmol−1
The magnitude of lattice enthalpy of KCl in kJmol-1 is (Nearest integer)

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Show full solutionCorrect answer: 718
Correct answer
718

Step-by-step explanation

So, ΔfHΘ=ΔfHΘKCl=Δsub HΘ+Δ1st ionisation HΘK+1/2ΔbondHΘCl2(g)+Δelectron gain HΘ+Δlattice HΘ-436.7=89.2+419+12243+-348.6+Δlattice HΘ

Δlattice HΘ=-717.8 kJ mole-1≈-718 kJ mole-1

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About this question

This is a previous-year question from JEE Main 2021, 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.