Question

At $298.2\, K$ the relationship between enthalpy of bond dissociation (in $kJmol ^{-1}$ ) for hydrogen $\left(E_{H}\right)$ and its isotope, deuterium $\left(E_{D}\right)$, is best described by:

• A. $E_{H}=\frac{1}{2} E_{D}$
• B. $E_{H}=E_{D}$
• C. $E_{H} \simeq E_{D}-7.5$
• D. $E_{H}=2 E_{D}$

Answer 1

Answer: (C)

Enthalpy of bond dissociation $( kJ / mole )$ at $298.2\, K$
For, hydrogen $=435.88$
For, Deuterium $=443.35$
$\therefore E_{H} \simeq E_{D}-7.5$