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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. |
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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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