Given that bond energies of H-H and Cl-Cl are 430 kJ mol-1 and 240 kJ mol-1 respectively and Hf for HCl is -90 kJ mol-1, bond enthalpy of HCl is
- 380 kJ mol-1
- 425 kJ mol-1
- 245 kJ mol-1
- 290 kJ mol-1
Answer
(2) 425 kJ mol-1
Solution
Formation reaction: 12H2(g) + 12Cl2(g) HCl(g), Hf = -90 kJ mol-1
Hf = B.E.(reactants) - B.E.(products)
-90 = 12(430) + 12(240) - B.E.(H-Cl)
-90 = 215 + 120 - B.E.(H-Cl)
⇒ B.E.(H-Cl) = 335 + 90
= 425 kJ mol-1
Answer: (2)