Easy MCQ +4 / -1 PYQ · JEE Mains 2021

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 :

  1. A ${E_H} = {1 \over 2}{E_D}$
  2. B ${E_H} = {E_D}$
  3. C ${E_H} \simeq {E_D} - 7.5$ Correct answer
  4. D ${E_H} = 2{E_D}$

Solution

Enthalpy of bond dissociation (kJ/mole) at 298.2 K <br><br>For, Hydrogen = 435.88<br><br>For, Deuterium = 443.35<br><br>$\therefore$ ${E_H} \simeq {E_D} - 7.5$

About this question

Subject: Chemistry · Chapter: p-Block Elements · Topic: Group 13: Boron Family

This question is part of PrepWiser's free JEE Main question bank. 300 more solved questions on p-Block Elements are available — start with the harder ones if your accuracy is >70%.

Drill 25 more like these. Every day. Free.

PrepWiser turns these solved questions into a daily practice loop. Chapter-wise drills, full mocks, AI doubt chat. No auto-renew.

Start free →