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Sigrid O'Kon

Feb 20, 2025

Calculate the lattice energy of lithium oxide from the following data?

Enthalpy of sublimation of lithium: +159 kJ/molBond energy of oxygen: +498 kJ/molFirst ionization energy of lithium: +520 kJ/molΔHgain for the gain of the first electron by oxygen: -141 kJ/mol ΔHgain for the gain of the second electron by oxygen: +744 kJ/mol Hf lithium oxide: 598 kJ/mol the answer is 2808 kJ/molplease show work, as i need to know this for an exam.
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Use Born-Haber(-Fajans) Cyclehttp://en.wikipedia.org/wiki/Born%E2%80%93Haber_cy... 2Li^+(g) + O^2- (g) → Li2O(g) U = lattice energy Break down into steps (Hess’s Law)2Li(s) + ½O2 → Li2O(s) ΔHf = -598 kJ mol^-1Li(s) →ΔHsub(Li)→ Li(g) ΔHsub(Li)→Li(g) →ΔHIE1(Li)→Li^+ + e⁻ ΔHIE1(Li) = 520 kJ mol^-1 O2 →½ΔHBE(O2)→ O ½ΔHBE(O2) = ½498O + e⁻→ΔHEA1(O)→ O^- ΔHEA1(O) = -141 kJ mol^-1O^- + e⁻→ΔHEA2(O)→ O^2- ΔHEA2(O) +744 kJ mol^-1ΔHf = U + 2ΔHsub(Li) + 2IE1 + ½ΔHdissO2 + EA1(O) + EA2(O)-598 = U + 2×159 + 2×520 + ½×498 - 141+ 744-598 = U + 2210U = -2808 kJ mol^-1 has correct thermodynamic sign but is usually dropped... Show More
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