MCQ
At $298.2\, \mathrm{~K}$ the relationship between enthalpy of bond dissociation (in $\mathrm{kJ} \,\mathrm{mol}^{-1}$ ) for hydrogen $\left(\mathrm{E}_{\mathrm{H}}\right)$ and its isotope, deuterium $\left(\mathrm{E}_{\mathrm{D}}\right)$, is best described by:
  • A
    $E_{H}=\frac{1}{2} E_{D}$
  • B
    $E_{H}=E_{D}$
  • C
    $E_{H}=2 E_{D}$
  • $E_{H} \simeq E_{D}-7.5$

Answer

Correct option: D.
$E_{H} \simeq E_{D}-7.5$
d
Enthalpy of bond dissociation $(\mathrm{kJ} / \mathrm{mole})$ at $298.2\, \mathrm{~K}$

For, hydrogen $=4.35 .88$

For, Deuterium $=443.35$

$\therefore E_{H}=E_{D}-7.5$

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