Q.

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

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a

290 kJ mol-1

b

380 kJ mol-1

c

245 kJ mol-1

d

425 kJ mol-1

answer is C.

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Detailed Solution

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

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