At 298.2 K the relationship between enthalpy of bond dissociation (in kJ mol$^{-1}$) for hydrogen ($E_H$) and its isotope, deuterium ($E_D$), is best described by :
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Isotope effect: Heavier isotopes form stronger bonds due to lower zero-point vibrational energy.
Step 1: Understanding the Concept:
Bond dissociation enthalpy depends on the zero-point energy (ZPE) of the molecule.
ZPE is lower for heavier isotopes because they have lower vibrational frequencies. Step 2: Detailed Explanation:
Deuterium (\(D\)) is twice as heavy as Hydrogen (\(H\)).
Because \(D\) has a higher mass, the zero-point energy of \(D_2\) is lower than that of \(H_2\).
A lower starting energy level means more energy is required to reach the dissociation limit.
Consequently, the \(D-D\) bond is slightly stronger than the \(H-H\) bond.
Experimental values: \(E_H \approx 435.9 \text{ kJ/mol}\) and \(E_D \approx 443.4 \text{ kJ/mol}\).
The difference is approximately \(7.5 \text{ kJ/mol}\).
Thus, \(E_H \approx E_D - 7.5\). Step 3: Final Answer:
The relationship is \(E_H \approx E_D - 7.5\).
