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 bond dissociation enthalpy of the free radical HS is
Options:
Answer:
C
Solution:
The enthalpy change for the reaction HS tarrow H + S is calculated as follows:
Δ H = Δ Hf(H) + Δ Hf(S) - Δ Hf(HS)
Substituting the given values:
Δ H = 218 kJ/mol + 277 kJ/mol - 138 kJ/mol = 357 kJ/mol
This corresponds to option (c).
Stream:JEESubject:ChemistryTopic:ThermochemistrySubtopic:Bond Enthalpy
⏱ 2mℹ️ Source: QB
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