The bond dissociation enthalpy of the free radical HS is

Name: JEE Advanced Chemistry
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Question

Prashant Kotian · Accepted Answer

Correct answer: C. HS(g) → H(g) + S(g). ΔH° = ΔfH°(H) + ΔfH°(S) - ΔfH°(HS) = 218 + 277 - 138 = 357 kJ/mol.

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