Question

In: Chemistry

The standard enthalpies of formation oxides of the fourth period eleemtns of the d block are...

The standard enthalpies of formation oxides of the fourth period eleemtns of the d block are as follows:

TiO VO MnO FeO CoO NiO CuO ZnO

-518 -431 -385 -264 -239 -241 -155 -348

a)write a general equation suitable for calculating the lattice energy of any of these oxides, based on theoritical data

b)Find the lattice energies of TiO, CuO, and ZnO

c) Overall the most dissimlar enthalpy values are those of TiO and CuO; the most dissimilar enthaly values of neighvoring oxides are those of CuO and ZnO. Which thermodynamic factor is apparently repsonsible for each of these dissimilarities? d)What factor of theoritcial lattice energy calculations is most repsonsible for substantial change in lattice eneregies between TiO and CuO?

Solutions

Expert Solution

a)write a general equation suitable for calculating the lattice energy of any of these oxides, based on theoritical

data

ΔHf°[TiO(s)] = ΔHsub[TiO(s)] + 1/2*ΔHbe[O2]+ ΔHie1[Ti] + ΔHie2[Ti] + ΔHea1[O] + ΔHea2[O] + ΔHlatt[TiO(s)]
ΔHlatt[TiO(s)] = ΔHf°[TiO(s)] - ( ΔHsub[TiO(s)] + 1/2*ΔHbe[O2]+ ΔHie1[Ti] + ΔHie2[Ti] + ΔHea1[O] + ΔHea2[O] )

Enthalpy of formation:ΔHf°
TiO:-518
CuO:-155
ZnO:-348

Enthalpy of sublimation kJ/mol:ΔHsub
Ti:473
Cu:339
Zn: 126
Inization energies kJ/mol: ΔHie1,ΔHie2
Ti:656, 1309
Cu:745, 1948
Zn: 906, 1734

Bond dissociation Enthalpy:
ΔHbe[O2] = 498.4 kJ/mol

Electron affinities of oxygen:
ΔHea1[O] = −141 kJ/mol
ΔHea2[O] = +744 kJ/mol


b)Find the lattice energies of TiO, CuO, and ZnO
ΔHlatt[TiO(s)] = ΔHf°[TiO(s)] - ( ΔHsub[TiO(s)] + 1/2*ΔHbe[O2]+ ΔHie1[Ti] + ΔHie2[Ti] + ΔHea1[O] + ΔHea2[O] )
-518-(473+498.4/2+656+1309-144+744) = -3805.2 kJ/mol

ΔHlatt[CuO(s)] = ΔHf°[CuO(s)] - ( ΔHsub[CuO(s)] + 1/2*ΔHbe[O2]+ ΔHie1[Ti] + ΔHie2[Ti] + ΔHea1[O] + ΔHea2[O] )
-155-(339+498.4/2+745+1948-144+744) = -4036.2 kJ/mol


ΔHlatt[ZnO(s)] = ΔHf°[ZnO(s)] - ( ΔHsub[ZnO(s)] + 1/2*ΔHbe[O2]+ ΔHie1[Ti] + ΔHie2[Ti] + ΔHea1[O] + ΔHea2[O] )
-348-(126+498.4/2+906+1734-144+744) = -3963.2kJ/mol


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