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QJEEM21Cha8f2: JEE Main 2021 Chemistry - Class 11 JEEM21Cha8f2 Solution: At 298.2 K the relationship between enthalpy of bond…

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 :

Multiple Choice Options

  • A EH=12ED{E_H} = {1 \over 2}{E_D}
  • B EH=ED{E_H} = {E_D}
  • C EH≃ED−7.5{E_H} \simeq {E_D} - 7.5
  • D EH=2ED{E_H} = 2{E_D}

Step-by-Step Verified Solution

Enthalpy of bond dissociation (kJ/mole) at 298.2 K

For, Hydrogen = 435.88

For, Deuterium = 443.35

∴\therefore EH≃ED−7.5{E_H} \simeq {E_D} - 7.5

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