The bond dissociation energy of gaseous `H_(2),Cl_(2)` and `HCl` are `104,58` and `103 kcal mol^(-1)` respecitvely. Calculate the enthalpy of formation for `HCl` gas.

The bond dissociation energy of gaseous `H_(2),Cl_(2)` and `HCl` are `

1.	The bond dissociation energy of gaseous `H_(2),Cl_(2)` and `HCl` are `104,58` and `103 kcal mol^(-1)` respecitvely. Calculate the enthalpy of formation for `HCl` gas.
Answer» The required equation is `(1)/(2)H_(2)(g)+(1)/(2)Cl_(2)(g)toHCl(g), " "DeltaH=?` `DeltaH=[(1)/(2)DeltaH_(H-H)+(1)/(2)DeltaH_(Cl-Cl)]-[DeltaH_(H-Cl)]` `=(1)/(2)xx104+(1)/(2)xx58-103=-22` kcal `mol^(-1)`

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The bond dissociation energy of gaseous `H_(2),Cl_(2)` and `HCl` are `104,58` and `103 kcal mol^(-1)` respecitvely. Calculate the enthalpy of formation for `HCl` gas.

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