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Explain the Born-Haber Cycle. OR Lattice Enthalpy. |
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Answer» Solution :Lattice Enthalpy : The lattice enthalpy of an ionic compound is the enthalpy CHANGE which occurs when one mole of an ionic compound dissociates into its ions in gaseous state. `Na^(+) CI_((s))^(-) to Na_((g))^(+) + CI_((g))^(-)` `Delta_("lattice") H^( Theta ) = + 788` kj/mol Difference steps of formation of Na Cl and its related enthalpy can be explained by BornHaber cycle as below. (1) `Na_((s)) to Na((g)),` sublimation of sodium `Delta_("sol")H^( Theta ) = 108.4 "kj mol"^(-1)` (2) Ionization enthalpy, `Na_((g)) to Na_((g))^(+) +e_((g))^(-1) , Delta_(i) H^( Theta ) = 496` kj/mol (3) The dissociation of chlorine, the reaction enthalpy is half the bond dissociation enthalpy. `(1)/(2) CI_(2(g)) to CI_((g)) , (1)/(2) Delta_("bond") H^( Theta ) = 121` kj/mol (4) The electron gained enthalpy, `CI_((g)) + e^(-) to CI_((g))^(-) , Delta_("eg") H^(Theta ) = - 348.6` kj/mol (5) `Na_((g))^(+) + CI_((g))^(-) to Na^(+) CI_((s))^(-) , Delta_(U) H^( Theta ) = (?)` (6) Enthalpy of formation of NaCl, `Na_((s)) + (1)/(2) CI_(2(g)) to NaCI_((s)) , Delta_(f) H^( Theta ) = + 411.2` kj/mol Applying Hess.s LAW, we get, `Delta_("lattice") H^( Theta ) = Delta_(f) H^( Theta ) + Delta_(s) H^( Theta ) + (1)/(2) Delta_("bond") H^( Theta )+ Delta_(i) H^( Theta ) + Delta_("eg") H^( Theta )` `= 411.2 + 108.4 + 121 + 496 - 348.6` `= + 788` kj For `NaCI_((s)) to Na_((g))^( + ) + CI_((g))^(-)` internal ENERGY is smaller by 2RT and is equal to `+ 783` kj mol`""^(-)`. Now we use the value of lattice enthalpy to calculate enthalpy of solution from the expression : `Delta_("sol")H^( Theta ) = Delta_("lattice") H^( Theta ) + Delta_("hyd") H^( Theta )` For one mole of `NaCI_((s))`, lattice enthalpy `=+788` kj/mol and `THEREFORE Delta_("hyd") H^( Theta ) = -784` kj/mol `therefore Delta_("sol") H^( Theta ) = + 788-784=+4` kj/mol The dissolution of `NaCl_((s))` is accompanied by very little heat change.  Figure : Enthalpy diagram for lattice enthalpy of NaCl |
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