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At 298.2 K the relationship between enth...

At 298.2 K the relationship between enthalpy of bond dissociation (in kJ `mol^(-1)`) for hydrogen `(E_(H))`and its isotope, deuterium (ED), is best described by :

A

`E_(H)=E_(D)-7.5`

B

`E_(H)=1/2E_(D)`

C

`E_(H)=2E_(D)`

D

`E_(H)=E_(D)`

Text Solution

Verified by Experts

The correct Answer is:
A
Enthalpy of bond dissociation (kJ/mole) at 298.2K
For , hydrogen = 435.88
For , Deuterium = 443.35
`:.E_(H)~~ E_D - 7.5`.
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