Calculate the bond enthalpy of HCl. Given that the bond enthalpies of `H_(2)` and `Cl_(2)` are `430KJmol^(-1) and 242KJmol^(-1)` respectively and `Delta_(f)H^(0)HCl` is `-91KJmol^(-1)`)

Explanation

By using the relation.
`Detla_(f)H^(@) = SigmaDelta_(f)H^(@)` ( Products) ` - Sigma Delta_(f)H^(@)`( Reactants)
Here , we are given `H_(2) (g) rarr2H(g) ,Delta _(b) H^(@) = + 430 kJ mol^(-1)` ....(i)
`Cl_(2)(g) rarr 2Cl(g) , Delta _(b) H^(@) = + 242kJ mol^(-1) ` .....(ii)
We aim at ` HCl(g) rarr H(g() + Cl(g) ,Delta_(b) H^(@) = ?` .....(iii)
Evidently, for reaction (iii)
`DeltaH = Sigma Delta_(f) H^(@) (`Poroduct) `- Sigma Delta_(f) H^(@) ` ( Reactants ) ` =[ Delta _(f) H^(@) (H)+ Delta_(f) H^(@) ( Cl) ] - [ Delta _(f) H^(@) (HCl)] ` ....(iv)
From equation (i) and(ii) ,
`Delta H = Sigma( + 430 kJ ) = 215 kJ mol^(_1) , Delta _(f) H^(@) (Cl) = (1)/(2) (+242kJ ) = +121 kJmol^(-1)`
Also ,we are given `Delta_(f) H^(@) (HCl) = -91 kJ mol^(-1)`
Putting these value in eqn. (iv) , we get `Delta _(H-Cl) H^(@) = [ + 215+121] - [-91] = 427kJ mol^(-1)`
Secondmethod , By using Hess's law
We are given `: (i) H_(2)(g) rarr 2H(g) , Delta_(r)H^(@)= + 430 kJ mol^(-1)`
`(ii) Cl_(2)(g) rarr 2Cl(g), Delta_(r) H^(@) = + 242 kJ mol^(-1)`
`(iii) (1)/(2) H_(2)(g) +(1)/(2) Cl_(2)(g) rarr HCl(g), Delta_(r)H^(@) = -91 kJ mol^(-1)`
We aim at `HCl(g) rarr H(g) + Cl(g), Delta_(r) H^(@)=?`
`(1)/(2) xx` eqn. (i)` + (1)/(2) xx `eqn. (ii) - eqn. (iii) gives the required result.
Third Method, By applying the relation
`Delta_(r)H^(@) = Sigma` Bond Enthalpies of Reactants `- Sigma` Bond Enthalpies of Products
For the formation of HCl,
`(1)/(2) H_(2)(g)+(1)/(2) Cl_(2)(g) rarr HCl(g) , Delta_(r) H^(@) = Delta_(f) H^(@)`
`:. Delta_(r) H^(@) = Sigma ` B.E. ( Reactants) `- Sigma ` B.E. ( Products ) `= (1)/(2) Delta_(H-H) H^(@) +(1)/(2) Delta _(Cl- Cl) H^(@) - Delta _(H-Cl) H^(@)`
`- 91 = (1)/(2) xx 430 + (1)/(2) xx 242 - Delta_(H- Cl) H^(@)`
`:. Delta _(H-Cl) H^(@) = 215+ 121 + 91 =427 kJ mol^(-1)`