Question:
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 $\left(E_{D}\right)$, is best described by:
Correct Option: , 3
Solution:
Enthalpy of bond dissociation (kJ/mole) at $298.2 \mathrm{~K}$
For, hydrogen $=435.88$
For, Deuterium $=443.35$
$\therefore \mathrm{E}_{\mathrm{H}} \simeq \mathrm{E}_{\mathrm{D}}-7.5$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.