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Suggest and explain an indirect method to calculate lattice enthalpy of sodium chloride crystal. |
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Answer» Solution :Let us use the Born- Haber cycle for determining the LATTICE enthalpy of Nacl as follows : Since the reaction is carried out with REACTANTS in elemental forms and products in their standard states, at I bar, the overcall enthalpy change of the reaction is also the enthalpy of formation for NaCl. Also , the formation of NaCl can be considered in 5 steps . Thesum of the enthalpy changes of these steps is EQUAL to the enthalpy change for the overall reaction from which the lattice enthalpy of NaCl is calculated. Let us calculate the lattice energy of sodium CHLORIDE using Born- Haber cycle  `Delta^(@)H_(f)`=heat of formation of sodium chloride `= -411.3kJmol^(-1)` `Delta^(@)H_(1)`= Heat of sublimation of `Na_((S))=108.7kJ mol^(-1)` `Delta^(@)H_(2)` = ionisation energy of `Na_((g))=495.0 kJ mol^(-1)` `Delta^(@)H_(3)` = dissociation energy of `Cl_(2_((g))) =244kJ mol^(-1)` `Delta^(@)H_(4)` = Electron AFFINITY of `Cl_((g))=-349.0 kJ mol ^(-1)` U = Lattice energy of NaCl `Delta^(@)H_(f) = Delta^(@)H_(1) +Delta^(@)H_(2)+1//2Delta^(@)H_(3)+Delta^(@)H_(4)+Delta^(@)H_(5)` `:. Delta^(@)H_(5)=(Delta^(@)H_(f))-(Delta^(@)H_(1)+Delta^(@)H_(2)+1//2Delta^(@)H_(3)+Delta^(@)H_(4))` `implies Delta^(@)H_(5)=(-441.3)-(108.7+495.0+122-349)` `Delta^(@)H_(5)=(-441.3)-(376.7)` `Delta^(@)H_(5)=-788kj mol^(-1)` The negative p sign in lattice energy indicates that the energy is released when sodium is formed from its constituent gaseous ions `Na^(+) and Cl^(-)` |
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