The standard enthalpies of formation oxides of the fourth period eleemtns of the

Question

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?

Explanation / Answer

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

data

Hf°[TiO(s)] = Hsub[TiO(s)] + 1/2*Hbe[O2]+ Hie1[Ti] + Hie2[Ti] + Hea1[O] + Hea2[O] + Hlatt[TiO(s)]
Hlatt[TiO(s)] = Hf°[TiO(s)] - ( Hsub[TiO(s)] + 1/2*Hbe[O2]+ Hie1[Ti] + Hie2[Ti] + Hea1[O] + Hea2[O] )

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

Enthalpy of sublimation kJ/mol:Hsub
Ti:473
Cu:339
Zn: 126
Inization energies kJ/mol: Hie1,Hie2
Ti:656, 1309
Cu:745, 1948
Zn: 906, 1734

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

Electron affinities of oxygen:
Hea1[O] = 141 kJ/mol
Hea2[O] = +744 kJ/mol

b)Find the lattice energies of TiO, CuO, and ZnO
Hlatt[TiO(s)] = Hf°[TiO(s)] - ( Hsub[TiO(s)] + 1/2*Hbe[O2]+ Hie1[Ti] + Hie2[Ti] + Hea1[O] + Hea2[O] )
-518-(473+498.4/2+656+1309-144+744) = -3805.2 kJ/mol

Hlatt[CuO(s)] = Hf°[CuO(s)] - ( Hsub[CuO(s)] + 1/2*Hbe[O2]+ Hie1[Ti] + Hie2[Ti] + Hea1[O] + Hea2[O] )
-155-(339+498.4/2+745+1948-144+744) = -4036.2 kJ/mol

Hlatt[ZnO(s)] = Hf°[ZnO(s)] - ( Hsub[ZnO(s)] + 1/2*Hbe[O2]+ Hie1[Ti] + Hie2[Ti] + Hea1[O] + Hea2[O] )
-348-(126+498.4/2+906+1734-144+744) = -3963.2kJ/mol

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