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JEE Main 202125 JulyMorning Shift
Exam
JEE Main
Year
2021
Date
25 July
Shift
Morning Shift
Subject
Chemistry
Chapter
Thermodynamics (C)

JEE Main 2021 · 25 July · Morning Shift · Chemistry · Thermodynamics (C)

At 298 . 2K the relationship between enthalpy of bond dissociation...

At 298.2K298 . 2K the relationship between enthalpy of bond dissociation (in kJmol−1kJ{mol}^{- 1}) for hydrogen \left({E}_{H}\right)anditsisotope,deuterium(ED)and its isotope, deuterium \left({E}_{D}\right), is best described by:

Options

  1. AEH=12ED{E}_{H} = \dfrac{1}{2}{E}_{D}
  2. BEH=ED{E}_{H} = {E}_{D}
  3. CEH=ED−7.5{E}_{H} = {E}_{D} - 7 . 5
  4. DEH=2ED{E}_{H} = 2{E}_{D}

Correct answer

C. EH=ED−7.5{E}_{H} = {E}_{D} - 7 . 5

Step-by-step solution

Enthalpy of bond dissociation (kJ/mole)\left(kJ / mole\right) at 298.2K298 . 2K

For, hydrogen =435.88= 435 . 88

For, Deuterium =443.35= 443 . 35

∴EH≃ED−7.5∴ {E}_{H} ≃ {E}_{D} - 7 . 5

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