At 298.2 K the relationship between


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:

  1. $\mathrm{E}_{\mathrm{H}}=\frac{1}{2} \mathrm{E}_{\mathrm{D}}$

  2. $\mathrm{E}_{\mathrm{H}}=\mathrm{E}_{\mathrm{D}}$

  3. $\mathrm{E}_{\mathrm{H}} \simeq \mathrm{E}_{\mathrm{D}}-7.5$

  4. $E_{H}=2 E_{D}$

Correct Option: , 3


Enthalpy of bond dissociation (kJ/mole) at $298.2 \mathrm{~K}$

For, hydrogen $=435.88$

For, Deuterium $=443.35$

$\therefore \mathrm{E}_{\mathrm{H}} \simeq \mathrm{E}_{\mathrm{D}}-7.5$

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