1.

Bond dissociation enthalpy of `H_(2)` , `Cl_(2)` and `HCl` are `434, 242` and `431KJmol^(-1)` respectively. Enthalpy of formation of `HCl` isA. `-93" kJ mol"^(-1)`B. `245" kJ mol"^(-1)`C. `93" kJ mol"^(-1)`D. `-245" kJ mol"^(-1)`

Answer» Correct Answer - A
`H_(2)+Cl_(2)rarr 2HCl` ltBrgt `DeltaH_("reaction")=Sigma(BE)_("reactant")-Sigma(BE)_("product")`
`=[(BE)_(H-H)+(BE)_(Cl-Cl)]-[2(BE)_(H-Cl)]`
`=434+242-(431)xx2`
`=-186kJ`
As `DeltaH_("reaction")=-186kJ`
So enthalpy of formation of HCl
`=(-186kJ)/(2)`
`=-93" kJ mol"^(-1)`


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