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1) Use the Born-Haber cycle and given data ΔH∘fKCl = -436.5 kJ/mol, IE1(K) = 419 kJ/mol, ΔHsub(K)=89.0 kJ/mol, Cl2(g)bondenergy = 243 kJ/mol, EA(Cl) = -349 kJ/mol to calculate the lattice energy of KCl.
2) Use the Born-Haber cycle and data from Appendix IIB and Table 9.3 in the textbook to calculate the lattice energy of CaO. You may need the following data: electron affinity of oxygen EA1=−141kJ/mol, EA2=744kJ/mol; ionization energy of calcium IE1=590kJ/mol, IE2=1145kJ/mol, the enthalpy of sublimation for calcium ΔHsub=178kJ/mol.

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Answer:

Explanation:

1 ) Let the lattice energy per mole be x

2 K + Cl₂ =  2KCl

Applying Born- Haber cycle

2 x 89 + 2 x 419 + 243 - 2 x 349  + Δx = - 2 x 436.5

178 + 838+ 243 - 698 + Δx = - 873

Δx = 1434 kJ

Lattice energy per mole = 1434 / 2 = 717 kJ / mol

2 )

2Ca + O₂ = 2 CaO

Heat of formation of CaO = - 635 kJ

Let lattice energy be x / mol

Bond energy of oxygen = 498 kJ / mol

178 x 2 + 2 ( 590 + 1145 ) + 498 + 2 ( 744 - 141 ) + x = - 2 x 635

356 + 3470 + 1206 + 498 + 2 x = - 1270

5530 + 2 x = - 1270

x = - 3400  kJ

pstnonsonjoku

The lattice energy for each specie can be calculated form the Born - Haber cycle and Hess law.

a) Given that;

Heat of formation (ΔH∘f) = -436.5 kJ/mol

Ionization energy (IE) = 419 kJ/mol

Heat of sublimation(ΔHsub) = 89.0 kJ/mol

Bond energy (BE) = 243 kJ/mol

Electron affinity (EA) =  -349 kJ/mol

Lattice energy (U) = ?

From Hess law;

ΔH∘f = ΔHsub + BE + IE + EA + U

U = ΔH∘f - [ΔHsub + BE + IE + EA]

U = ( -436.5 kJ/mol) -  [89.0 kJ/mol + 243 kJ/mol +  419 kJ/mol + ( -349 kJ/mol)]

U = -838.5  kJ/mol

b)

Heat of formation (ΔH∘f) =-635 kJ/mol

Ionization energy (IE) =IE1 + IE2 = 1735  kJ/mol

Heat of sublimation(ΔHsub) = 178kJ/mol

Bond energy (BE) = 498 kJ/mol

Electron affinity (EA) =  EA1 + EA2 = 603 kJ/mol

Lattice energy (U) = ?

From Hess law;

ΔH∘f = ΔHsub + BE + IE + EA + U

U = ΔH∘f - [ΔHsub + BE + IE + EA]

U = (-635 kJ/mol) - [ 178kJ/mol + 498 kJ/mol + 1735  kJ/mol + 603 kJ/mol]

U = -3649 kJ/mol

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