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)=2435+192+109=368KJ/mol
At standard conditions, if the change in the enthalpy for the following reaction is -109 kJmol^-1 H_2(g)+ Br_2(g)→ 2 HBr_(g) Given that bond energy…
NEET UG 2020 — Chemistry Physical Chemistry
2020mcqhard
At standard conditions, if the change in the enthalpy for the following reaction is −109kJmol−1
H2(g)+Br2(g)→2HBr(g)
Given that bond energy of H2 and Br2 is 435kJmol−1 and 192kJmol−1, respectively, what is the bond energy (in kJmol−1) of HBr?
Official previous-year question
Held on 30 Apr 2020 · Verified 9 Jul 2026.
Options
- A
368
- B
736
- C
518
- D
259
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