Given that bond energies of H - H and Cl - Cl are 430 KJ mol- and 240 KJ mol- respectively and ΔfH for HCI is 90kJmol-1. Bond enthalpy of HCI is

  1. A

    290 kJ mol-1

  2. B

    380 kJ mol-1

  3. C

    425 kJ mol-1

  4. D

    245 kJ mol-1

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    Solution:

    12H2+12Cl2HClΔHf=12ΔHH2+12ΔHCl2ΔHHCl90=12×430+12×240ΔHHCl90=215+120ΔHHClΔHHCl=335+90=425kJ/mol

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