JEE Main 2021ChemistryThermodynamics (C)EasyMCQ

JEE Main 2021 — Thermodynamics (C) Question with Solution

JEE Main 2021 (25 Jul Shift 1)

Question

At 298.2 K the relationship between enthalpy of bond dissociation (in kJ mol-1) for hydrogen EH and its isotope, deuterium ED, is best described by:

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Show full solutionCorrect option: C
Correct answer
CEH=ED-7.5

Step-by-step explanation

Enthalpy of bond dissociation kJ/mole at 298.2 K

For, hydrogen =435.88

For, Deuterium =443.35

∴ EH≃ED-7.5

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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.