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  • At , the relationship between enthalpy of bond dissociation <img alt="\mathrm{(in\: \: \mathrm{kJ} \mathrm{mol}^{-1} )}" src="https://entrancecorner.codecogs.com/gif.latex?%5Cm

At 298.2 \mathrm{~K}, the relationship between enthalpy of bond dissociation \mathrm{(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:
Option: 1 \mathrm{E}_{\mathrm{H}}=\frac{1}{2} \mathrm{E}_{\mathrm{D}}
Option: 2 \mathrm{E}_{\mathrm{H}}=\mathrm{E}_{\mathrm{D}}
Option: 3 \mathrm{E}_{\mathrm{H}} \simeq \mathrm{E}_{\mathrm{D}}-7.5
Option: 4 \mathrm{E}_{\mathrm{H}}=2 \mathrm{E}_{\mathrm{D}}

Answers (1)

best_answer

As we have learnt,

Heavier isotopes form stronger bonds.

Hence, the bond energy for deuterium will be greater than that of hydrogen .
E_{D}> E_{H}

Numerically, these values are given as

E_{D}= 443\cdot 35KJ\, mol^{-1}
E_{H}= 435\cdot 88KJ\, mol^{-1}
\therefore E_{D}-E_{H}\simeq 7\cdot 5KJ\, mol^{-1}

Hence, the correct answer is option (3)

Posted by

sudhir.kumar

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