If the bond energies of H-H, Br-Br, and H-Br are 433, 192 and 364 kJ mol-1 respectively, the H for the reaction H2(g) + Br2(g) 2HBr(g) is
- -261 kJ
- +103 kJ
- +261 kJ
- -103 kJ
Answer
(4) -103 kJ
Solution
r H = (bond enthalpies of bonds broken) - (bond enthalpies of bonds formed)
Bonds broken: one H–H + one Br–Br ⇒ 433 + 192 = 625 kJ
Bonds formed: two H–Br ⇒ 2 364 = 728 kJ
∴ r H = 625 - 728
= -103 kJ
Answer: (4)