1.

For one mole of NaCl(s) the lattice enthalpy is : Na(s) + (1)/(2)Cl_(2)(g) overset(+108.4 kJ//mol)rarr Na(g) + (1)/(2)Cl_(2)(g) overset(+495.6 kJ//mol)rarr Na^(+)(g) + (1)/(2)Cl_(2)(g) overset(121 kJ//mol)rarr Na^(+)(g) + Cl(g) overset(-348.6 kJ//mol)rarr Na^(+)(g) + Cl^(-)(g) overset(Delta H^(theta)" lattice")rarr NaCl("solid") underset(-411.2 kJ/mol)larr Na(s) + (1)/(2)Cl_(2)(g)

Answer»

`- 788 KJ//mol`
`+ 878 kJ//mol`
`+ 788 kJ//mol`
`- 878 kJ//mol`

Solution :`Delta_(f) H = - 411.2 kJ mol^(-1)`
`- 411.2 = 108.4 + 495.6 + 121 - 348.2 + Delta H_("LATTICE")^(@)`
:. `Delta H_("lattice")^(@) = - 411.2 - 108.4 - 495.6 - 121 + 348.2`
`= - 788 kJ mol^(-1)`


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