At 298.2 K the relationship between enthalpy of bond dissociation (in kJ mol?¹) for hydrogen (E?) and its isotope, deuterium (E?), is best described by:
At 298.2 K the relationship between enthalpy of bond dissociation (in kJ mol?¹) for hydrogen (E?) and its isotope, deuterium (E?), is best described by:
Option 1 -
Eʜ = Eᴅ
Option 2 -
Eʜ = ½ Eᴅ
Option 3 -
Eʜ = 2Eᴅ
Option 4 -
Eʜ ≈ Eᴅ - 7.5
Asked by Shiksha User
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1 Answer
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Correct Option - 4
Detailed Solution:B.D.E H-H < B.D.E D-D
B.D.E H-H = 435.9 kJ / mole
B.D.E D-D = 443.4 kJ / mole
E H ≈ E D - 7.5
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