JEE Main 2023ChemistryThermodynamics (C)EasyNumerical

JEE Main 2023 — Thermodynamics (C) Question with Solution

JEE Main 2023 (13 Apr Shift 1)

Question

A2+B2→2AB.ΔHf=-200 kJ mol-1

AB, A2 and B2 are diatomic molecules. If the bond enthalpies of A2, B2 and AB are in the ratio1:0.5:1, then the bond enthalpy of A2 is_________kJ mol–1 (Nearest integer)

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Show full solutionCorrect answer: 800
Correct answer
800

Step-by-step explanation

First write balanced reaction for the formation of one mole of AB with the help of diatomic molecule:

12A2+12B2→AB     ΔHf=-200 kJ/mol

Now, 

∆Hreaction=12∆HA-A+12∆HB-B-∆HAB     let, ∆HAB=x 

12x+12(0.5x)-x=-200  ⇒  x2+0.25x-x=-200

⇒  -0.25x=-200  ⇒  x=800 kJ/mol

Bond enthalpy =800 kJ/mol.

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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.