Chemistry - Thermochemistry Question with Solution | TestHub
ChemistryThermochemistryBond EnthalpyMedium2 minQB
ChemistryMediumsingle choice
Passage / Comprehension
The bond dissociation enthalpy of the first H-S bond in hydrogen sulphide is 376.0 kJ/mol. The enthalpies of formation of H2S(g) and S(g) are -20.0 and 277.0 kJ/mol, respectively. The bond dissociation enthalpy of H-H bond is 436.0 kJ/mol.
The enthalpy of formation of the free radical HS is
Options:
Answer:
A
Solution:
The enthalpy of formation for H is calculated as Δf H(H) = 436.0/2 = 218.0 kJ/mol.
For the reaction H₂S → HS + H, we have:
376.0 = Δf H(HS) + 218.0 - (-20.0)
Therefore, Δf H(HS) = 376.0 - 238.0 = 138 kJ/mol. This corresponds to option (a).
Stream:JEESubject:ChemistryTopic:ThermochemistrySubtopic:Bond Enthalpy
⏱ 2mℹ️ Source: QB
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