1.

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»

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



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