Calculate the lattice enthalpy of LiF, given that the enthalpy of (i) sublimation of lithium is 155.2 kJ mol –1 (ii) dissociation of 1/2 mol of F 2 is 75.3 kJ (iii) ionization of lithium is 520 kJ mol –1 (iv) electron gain of 1 mol of F(g) is –333 kJ (v) ∆ f H 0 ∆ f H 0 overall is –795 kJ mol –1 - Wired Faculty

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Calculate the lattice enthalpy of LiF, given that the enthalpy of
(i) sublimation of lithium is 155.2 kJ mol^–1(ii) dissociation of 1/2 mol of F₂is 75.3 kJ
(iii) ionization of lithium is 520 kJ mol^–1(iv) electron gain of 1 mol of F(g) is –333 kJ
(v) ∆_f H⁰ ∆_fH⁰ overall is –795 kJ mol^–1

Solution

Using Born-Haber cycle to LiF
$increment subscript straight f straight H to the power of 0 space equals space increment straight H subscript 1 superscript 0 space plus space increment straight H subscript 2 superscript 0 space plus space increment straight H subscript 3 superscript 0 space plus space increment straight H subscript 4 superscript 0 space plus space increment straight H subscript 5 superscript 0 or space space space space increment straight H subscript 5 superscript 0 space equals space increment subscript straight f straight H to the power of 0 space minus space increment straight H subscript 1 superscript 0 space minus space increment straight H subscript 2 superscript 0 space minus space increment straight H subscript 3 superscript 0 space minus space increment straight H subscript 4 superscript 0$
Substituting the values, we have
$increment straight H subscript 5 superscript 0 space equals space minus 795 minus 155.2 minus 75.3 minus 520 plus 333 space space space space space space space space equals negative 1545.5 space plus space 333 space equals space minus 1212.5 increment straight H subscript 5 superscript 0 space equals space Enthalpy space change space for space latitice space formation space from space space space space space space space space space space space space space space space space Li to the power of plus left parenthesis straight g right parenthesis space plus straight F to the power of minus left parenthesis straight g right parenthesis space space rightwards arrow space space LiF left parenthesis straight s right parenthesis semicolon space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space increment straight H subscript 5 superscript 0 space equals space minus 1212.5 space kJ space mol to the power of negative 1 end exponent space space space space space space space space space space$
The reverse of the above equation i.e.
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defines the lattice enthalpy of LiF
Hence lattice enthalpy of LiF = +1212.5 kJ mol^-1

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