Calculate the bond enthalpy of HCl. Given that the bond enthalpies of H2 and Cl2 are 430 kJ mol–1 and 242 kJ mol–1 respectively and ∆fH° for HCl is –91 kJ mol–1.

Question

Calculate the bond enthalpy of HCl. Given that the bond enthalpies of H2 and Cl2 are 430 kJ mol–1 and 242 kJ mol–1  respectively and ∆fH° for HCl is –91 kJ mol–1.

Shiksha Nation · Accepted Answer

By applying the relationΔfH° = ∑ Bond enthalpies of Reactants − ∑ Bond enthalpies of productsFor the formation of HCl,1/2H2(g) + 1/2Cl2(g) → HCl(g), ΔfH° = ΔfH°∴  ΔrH° = ∑ B.E. (Reactants) − ∑ B.E. (Products) =1/2ΔH−HH° +1/2ΔCl−ClH° − ΔH−ClH°−91 =1/2× 430 +1/2× 242 − ΔH−ClH°∴  ΔH−ClH° = 215 + 121 + 91 = 427 kJ mol−1

Calculate the bond enthalpy of HCl. Given that the bond enthalpies of H₂ and Cl₂ are 430 kJ mol^–1 and 242 kJ mol^–1 respectively and ∆_fH° for HCl is –91 kJ mol^–1.

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