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

Question

Calculate the lattice energy of lithium oxide using the following data:

Enthalpy of sublimation of lithium: +159 kJ/mol
Bond energy of oxygen: +498 kJ/mol
First ionization energy of lithium: +520 kJ/mol
ΔH for the gain of the first electron by oxygen: -141 kJ/mol
ΔH for the gain of the second electron by oxygen: +744 kJ/mol
Standard enthalpy of formation (ΔHf) of lithium oxide: -598 kJ/mol

The answer is 2808 kJ/mol. Please show your work, as I need to understand this for an upcoming exam.

Anonymous · Accepted Answer

Use Born-Haber(-Fajans) Cyclehttp://en.wikipedia.org/wiki/Born%E2%80%93Haber_cy...2Li^+(g)  + O^2- (g)  &rarr; Li2O(g)  U = lattice energy Break down into steps (Hess&rsquo;s Law)2Li(s)  +  &frac12;O2 &rarr; Li2O(s) &Delta;Hf = -598 kJ mol^-1Li(s) &rarr;&Delta;Hsub(Li)&rarr; Li(g) &Delta;Hsub(Li)&rarr;Li(g) &rarr;&Delta;HIE1(Li)&rarr;Li^+  + e&#8315; &Delta;HIE1(Li) = 520 kJ mol^-1 O2 &rarr;&frac12;&Delta;HBE(O2)&rarr; O  &frac12;&Delta;HBE(O2) = &frac12;498O + e&#8315;&rarr;&Delta;HEA1(O)&rarr; O^-  &Delta;HEA1(O) =  -141 kJ mol^-1O^- + e&#8315;&rarr;&Delta;HEA2(O)&rarr; O^2- &Delta;HEA2(O) +744 kJ mol^-1&Delta;Hf  =  U  + 2&Delta;Hsub(Li)  + 2IE1  +  &frac12;&Delta;HdissO2  +  EA1(O) +  EA2(O)-598  = U  + 2&times;159 + 2&times;520 + &frac12;&times;498 - 141+ 744-598  = U  + 2210U = -2808 kJ mol^-1 has correct thermodynamic sign but is usually dropped

Anonymous

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

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