Chemistry - Thermochemistry Question with Solution | TestHub
ChemistryThermochemistryBond EnthalpyHard2 minQB
ChemistryHardsingle choice
Passage / Comprehension
The bond dissociation enthalpy of the first H-Se bond in hydrogen selenide is 305.0 kJ/mol. The enthalpies of formation of H2Se(g) and Se(g) are +30.0 and 227.0 kJ/mol, respectively. The bond dissociation enthalpy of H-H bond is 436.0 kJ/mol.
The enthalpy of formation of the free radical HSe is
Options:
Answer:
C
Solution:
Δf H(H) = 436.0/2 = 218.0 kJ/mol. For H2Se → HSe + H: 305.0 = Δf H(HSe) + 218.0 - 30.0, so Δf H(HSe) = 305 - 188 = 117 kJ/mol. This is option (c).
Stream:JEESubject:ChemistryTopic:ThermochemistrySubtopic:Bond Enthalpy
⏱ 2mℹ️ Source: QB
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