Chemistry - Thermochemistry Question with Solution | TestHub
ChemistryThermochemistryBond EnthalpyMedium2 minQB
ChemistryMediuminteger
Find bond enthalpy of (in ) using the following information:
Resonance energy of
Answer:
725
Solution:
700 + 500 – 2ε – 150 = –400
1200 – 2ε – 150 = –400
2ε = 1450
ε = 725 kJ/mol
7 + 2 + 5 = 14 = 1 + 4 = 5 Ans.
Stream:JEESubject:ChemistryTopic:ThermochemistrySubtopic:Bond Enthalpy
⏱ 2mℹ️ Source: QB
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