JEE Mains · Chemistry · STD 11 - 5. Thermodynamics and thermochemistry
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 \(E_{H}=2 E_{D}\)
- D \(E_{H} \simeq E_{D}-7.5\)
Answer & Solution
Correct Answer
(D) \(E_{H} \simeq E_{D}-7.5\)
Step-by-step Solution
Detailed explanation
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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