Question

The bond dissociation energy of gaseous H₂, Cl₂ and HCI are 104, 58 and 103 kcal/ mol respectively. Calculate the enthalpy of formation of HCI gas.

Answer 1

For the reaction,

H₂ (g) + Cl₂ (g) → 2HCl(g)

For reactants

Bond energy of mole of H-H bond = 104 kcal/mol

Bond energy of mole of Cl-Cl bond = 58 kcal/mol

For products

Bond energy of moles of H-Cl bond = −2 × 103

= −206 kcal/mole

∴ Enthalpy change for the reaction:

∆_rH = 104 + 58 − 206 = −44 kcal/mole

∴ Enthalpy of formation of one mole of HCl gas

= \(\frac{-44}{2}\) = -22 kcal/mole