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Enthalpy

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Question

Calculate the lattice enthalpy(nearest integer value) in $k J {m o l}^{- 1}$ of $L i F$ , given that enthalpy of,
(i) sublimation of lithium is $155.3 k J$ ${m o l}^{- 1}$ ;
(ii) dissociation of $1 / 2$ mole of $F_{2}$ is $75.3 k J$ ;
(iii) ionization enthalpy of lithium is $520 k J$ ${m o l}^{- 1}$ ;
(iv) electron gain enthalpy of 1 mole of $F (g)$ is $- 333 k J$ ;
(v) $Δ_{f} H$ overall is $- 594 k J$ ${m o l}^{- 1}$ .

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Solution

Given:
$Δ_{s u b} H_{L i} = 155.2 k J {m o l}^{- 1}$
$\frac{1}{2} Δ H_{F - F} = 75.3 k J {m o l}^{- 1}$
$Δ_{f} H_{L i} = 520 k J {m o l}^{- 1}$
$Δ_{e g} H_{F} = - 333 k J {m o l}^{- 1}$
$Δ_{f} H_{L i F} = - 594.1 k J {m o l}^{- 1}$
For $L i F$ : $Δ_{f} H_{L i F} = Δ_{s u b} H_{L i} + Δ \frac{1}{2} H_{F - F} + Δ_{i} H_{L i} + Δ_{e g} H_{F} + Δ_{l} H_{L i F}$
or $- 594.1 = 155.2 + 75.3 + 520 - 333 + Δ_{l} H_{L i F}$
$∴$ $Δ_{l} H_{L i F} = - 1011.6 k J$ ${m o l}^{- 1}$

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