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Question: Answered & Verified by Expert

At standard conditions, if the change in the enthalpy for the following reaction is -109 kJmol-1

H2(g)+Br2(g)2HBr(g)

Given that bond energy of H2 and Br2 is 435 kJmol-1 and 192 kJmol-1, respectively, what is the bond energy (in kJ mol-1) of HBr?

ChemistryThermodynamics (C)NEETNEET 2020 (Phase 2)
Options:
  • A 368
  • B 736
  • C 518
  • D 259
Solution:
1209 Upvotes Verified Answer
The correct answer is: 368
ΔH=Σ(B·E)Reactants -Σ(B.E)Products 
-109=B·E(H-H)+B·E(Br-Br)-2×B·E(H-Br)
-109=435+192-2×B·E(H-Br)
B.E(H-Br)=435+192+1092=368 KJ/mol

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