Bond dissociation enthalpies of H2, Cl2 and HCl are 434, 242 and 431 kJ mol-1 respectively. Enthalpy of formation of HCl is
- -93 kJ mol-1
- 245 kJ mol-1
- 93 kJ mol-1
- -245 kJ mol-1
Answer
(1) -93 kJ mol-1
Solution
Formation reaction: 12H2(g) + 12Cl2(g) HCl(g)
fH = Σ B.E.(reactants) − Σ B.E.(products)
= 12(434) + 12(242) - 431
= 217 + 121 - 431
= -93 kJ mol-1
Answer: (1)