At 298.2 K the relationship between enthalpy of bond dissociation (in kJ mol−1) for hydrogen (EH) and its isotope, deuterium (ED), is best described by
A
EH≃ED−7.5
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
EH=2ED
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
EH=12ED
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
EH=ED
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is AEH≃ED−7.5 Enthalpy of bond dissociation of hydrogen = 435.88kJmol−1 Enthalpy of bond dissociation of deuterium = 443.35kJmol−1 EH=ED−7.47 EH≃ED−7.5