IF the bond energies of H – H , Br – Br and H – Br are 433 , 192 and 364 kJ mol⁻¹ respectively, the ΔH° for the reaction
H₂(g) + Br₂(g) → 2HBr(g) is

Options

(a) -261 kJ
(b) ⁺103 kJ
(c) ⁺261 kJ
(d) -103 kJ

Correct Answer:

-103 kJ

Explanation:

H – H + Br – Br → 2H – Br,
(433) + (192) (2 x 364)
=625 =728
(Energyabsorbed) (Energy released)
Net energy released = 728 – 625 = 103 KJ. i.e. ΔH = -103 KJ.