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Calculate the bond enthalpyof HCl. Given that the bond enthalpies of H_(2) and Cl_(2) are 430 kJ mol^(-1) and242 kJ mol^(-1) respectively andDelta_(f)H^(@) for Hclis -91kJ mol^(-1) |
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Answer» SOLUTION :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 requiredresult. 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)` |
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