Question

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 1

Given,
`H_(2(g)) rarr 2H_((g)), DeltaH=104 ....(1)`
`Cl_(2(g)) rarr 2Cl_((g)), DeltaH=58kcal ...(2)`
`HCl_((g)) rarr H_((g))+Cl_((g)), DeltaH =103 kcal ....(3)`
Heat of formation for `HCl`
`(1)/(2) H_(2(g))+(1)/(2)Cl_(2(g)) rarr HCl_((g)), DeltaH=?`
Divide `eqs. (1)` and `(2)` by 2, and then add
`(1)/(2)H_(2(g))+(1)/(2)Cl_(2(g)) rarr H_((g))+Cl_((g)), DeltaH=81kcal ....(4)`
Subtracting equation `(3)` from `eq. (4)`
`ul ( {:(HCl_((g)),rarr,+H_((g)),Cl_((g)),,DeltaH=103kcal .....(3)),(-,,-,-," -"):})`
`(1)/(2)H_(2(g))+(1)/(2)Cl_(2(g))rarr HCl_((g)),DeltaH=-22.0kcal`
`:. ` Enthalpy of formation of `HCl` gas `=-22.0kcal`