NEETChemistryThermodynamicsMCQ+4 / −1
From the following bond energies :
H H bond energy : 431.37 kJ mol1
C C bond energy : 606.10 kJ mol1
C C bond energy : 336.49 kJ mol1
C H bond energy : 410.50 kJ mol1
Enthalpy for the reaction,
will be
H H bond energy : 431.37 kJ mol1
C C bond energy : 606.10 kJ mol1
C C bond energy : 336.49 kJ mol1
C H bond energy : 410.50 kJ mol1
Enthalpy for the reaction,
will be
- A243.6 kJ mol1
- B120.0 kJ mol1
- C553.0 kJ mol1
- D1523.6 kJ mol1
View written solutionFree
Correct answer: B
Hreaction = Σ(Bond enthalpy)reactants
– Σ(Bond enthalpy)products
= [B.E(C-C) + B.E(H-H) + 4B.E(C-H)]
- [B.E(C-C) + 6B.E(C-H)]
= [606.10 + 4(410.50) + 431.37]
– [336.49 + 6(410.50)]
= 2679.47 – 2799.49
= – 120.02 kJ mol–1
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