Question -
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
Answer: Option D : D ΔH∘=∑B.E(reactants)−∑B.E(products)=[B.E(H−H)+B.E(Br−Br)]−[2B.E(H−Br)]=(433+192)−(2×364)=625−728=−103kJ
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