Calculate the lattice energy of LiF(s), given that: (i) Δ H text sub for Lithium is 155.2 kJ mol -1 (ii) Δ H text diss of 1 / 2 mole of F 2( g ) is 75.3 kJ (iii) I.E of Lithium =520 kJ mol -1 (iv) E.A. of Fluorine =333 kJ mol -1 (v) Δ H f for LiF ( s )=-594.1 kJ mol -1

Question

Calculate the lattice energy of LiF(s), given that: (i) Δ H text sub for Lithium is 155.2 kJ mol -1 (ii) Δ H text diss of 1 / 2 mole of F 2( g ) is 75.3 kJ (iii) I.E of Lithium =520 kJ mol -1 (iv) E.A. of Fluorine =333 kJ mol -1 (v) Δ H f for LiF ( s )=-594.1 kJ mol -1 (A) -1011.6 kJ mol -1 (B) +2056.6 kJ mol - (C) -906.6 kJ mol -1 (D) +573.6 kJ mol -1

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<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/2ba30fe095c4c9daa33d958644ec8bd8-.png" /> ∴ Δf=Δ H sub+(1/2) Δ H diss+ I.E. + E.A. +Δ H I.E ∴ Δ H I.E.=-1011.6 kJ / mol

$- 1011.6 k J m o l^{- 1}$

$+ 2056.6 k J m o l^{-}$

$- 906.6 k J m o l^{- 1}$

$+ 573.6 k J m o l^{- 1}$

$∴ Δ_{f} = Δ H_{s u b} + \frac{1}{2} Δ H_{d i ss} + I . E . + E . A . + Δ H_{I . E}$

$∴ Δ H_{I . E .} = - 1011.6 k J / m o l$

Q. Calculate the lattice energy of LiF(s), given that: (i) ΔHsub ​ for Lithium is 155.2kJmol−1 (ii) ΔHdiss ​ of 1/2 mole of F2​(g) is 75.3kJ (iii) I.E of Lithium =520kJmol−1 (iv) E.A. of Fluorine =333kJmol−1 (v) ΔHf​ for LiF(s)=−594.1kJmol−1

−1011.6kJmol−1

+2056.6kJmol−

−906.6kJmol−1

+573.6kJmol−1

∴Δf​=ΔHsub​+21​ΔHdiss​+I.E.+E.A.+ΔHI.E​ ∴ΔHI.E.​=−1011.6kJ/mol

$- 1011.6 k J m o l^{- 1}$

$+ 2056.6 k J m o l^{-}$

$- 906.6 k J m o l^{- 1}$

$+ 573.6 k J m o l^{- 1}$

$∴ Δ_{f} = Δ H_{s u b} + \frac{1}{2} Δ H_{d i ss} + I . E . + E . A . + Δ H_{I . E}$

$∴ Δ H_{I . E .} = - 1011.6 k J / m o l$