Question 11 · Complements de química URV Exam Bridging Courses in Chemistry

Tema: Sólidos, energía reticular y fuerzas intermoleculares · Del examen Exam Bridging Courses in Chemistry

Enunciado

Question 11 (1.25 points). Lattice energy and covalent contribution in LiF

Lithium fluoride (LiF) crystallizes with the NaCl structure type. Calculate both the experimental lattice energy (U0expU_0^{\text{exp}}) using a Born-Haber cycle and the theoretical lattice energy (U0theorU_0^{\text{theor}}) using the Born-Landé equation. From their difference, evaluate the non-ionic (covalent) percentage contribution in LiF.

Thermodynamic and structural data:

ΔHf∘(LiF)=−617 kJ/mol\Delta H_f^\circ(\text{LiF}) = -617\text{ kJ/mol}

ΔHsub(Li)=161 kJ/mol\Delta H_{\text{sub}}(\text{Li}) = 161\text{ kJ/mol}

ΔHion(Li)=520 kJ/mol\Delta H_{\text{ion}}(\text{Li}) = 520\text{ kJ/mol}

ΔHdiss(F2)=154 kJ/mol\Delta H_{\text{diss}}(\text{F}_2) = 154\text{ kJ/mol}

EA(F)=−328 kJ/mol\text{EA}(\text{F}) = -328\text{ kJ/mol}

r(Li+)=0.68 A˚, r(F−)=1.33 A˚r(\text{Li}^+) = 0.68\text{ \AA},\ r(\text{F}^-) = 1.33\text{ \AA}

Atomic numbers: Z(Li)=3, Z(F)=9Z(\text{Li}) = 3,\ Z(\text{F}) = 9

NA=6.023⋅1023 mol−1N_A = 6.023 \cdot 10^{23}\text{ mol}^{-1}

e=4.8⋅10−10 esu (statC)e = 4.8 \cdot 10^{-10}\text{ esu (statC)}

1 A˚=10−8 cm; 1 erg=10−7 J1\text{ \AA} = 10^{-8}\text{ cm};\ 1\text{ erg} = 10^{-7}\text{ J}

Structure

Madelung constant (AA)

Cesium chloride · 1.76267

Sodium chloride · 1.74756

Blende of zinc · 1.63806

Wurtzite · 1.64132

Fluorite · 2.51939

Antifluorite · 2.51900

Rutile · 2.40800

Configuration ion

Born exponent (nn)

He · 5

Ne · 7

Ar, Fe2+^{2+} · 9

Kr, Ag+^+ · 10

Xe, Au+^+ · 12

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Enunciado: Question 11 · Complements de química URV Exam Bridging Courses in Chemistry

Solución

Según los criterios oficiales de corrección

Calculate experimental lattice energy (U0expU_0^{\text{exp}}) using a Born-Haber cycle

Solución: 1047 kJ/mol

Calculate theoretical lattice energy (U0theorU_0^{\text{theor}}) using the Born-Landé equation

Solución: 1006 kJ/mol / 1005.5 kJ/mol

From their difference, evaluate the non-ionic (covalent) percentage contribution in LiF

Solución: 3.9 / 3.92 / 4.0

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