Question

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:

• A. E$_{H}$ = E$_{D}$
• B. E$_{H}$ > E$_{D}$
• C. E$_{H}$ < E$_{D}$

Answer 1

Answer: (C)

E$_{H}$ < E$_{D}$

Answer 2

Deuterium (D) is heavier than hydrogen (H). This results in a larger reduced mass for D₂ compared to H₂. Consequently, the vibrational frequency of D₂ is lower than that of H₂. The zero-point energy (ZPE), which is directly proportional to the vibrational frequency, is thus lower for D₂. Since bond dissociation enthalpy ($E$) is given by $D_e - ZPE$ (where $D_e$ is the electronic dissociation energy, which is nearly identical for both), a lower ZPE for D₂ means its bond dissociation enthalpy ($E_D$) is greater than that of H₂ ($E_H$).