enthalpy change...please help

Question

please help

Consider the reaction:
ICl(s)I(g) +Cl(g) △H= +210kJmol-1
Explain why this value of △H is not the same as that of the I---Cl bond enthalpy.


20 marks will be given to the best solution

Answer 1

The energy needed to break one mole of specific covalent bonds in gaseous state is called bond enthalpy.
i.e. ICl(g) → I(g) + Cl(g) ... ΔH = bond enthalpy of I-Cl bond
All reactants and products are in gaseous state. This is because there is no interactions (attractions and repulsions) between the gaseous particles (ideal gases), and thus all energy absorbed is used to break the bond.

Consider the equation you gave :
ICl(s) → I(g) + Cl(g) ... ΔH = +210 kJ mol-1
ICl is in solid state. 210 kJ mol-1 of energy is used to vaporize ICl(s) to ICl(g) and then break the bond. Therefore, the ΔH value is NOT the bond enthalpy of I-Cl bond.