Question

Use the following data to calculate the lattice energy of calcium oxide. You must write

all thermochemical equations for the steps of the cycle. The enthalpy of formation of

calcium oxide (solid) = - 636 kj/mole The enthalpy of sublimation of calcium= + 192

kj/mole First ionization energy of Ca = + 590 kj/mole Second ionization energy of Ca=

+1145 kj/mole The enthalpy of dissociation of O2 (g) = + 494 kj/mole First electron

affinity of O (g) = - 141 kj/mole Second electron affinity of O (g) = + 845 kj/mole​

Answer 1

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Answer 2

Answer:

Lattice energy of CaO is -3322 kJ/mol

Explanation:

Given:

ΔH = -636 kJ/mol

s = +192 kJ/mol

IE₁ = + 590 kJ/mol

IE₂ = +1145 kJ/mol

D = +494 kJ/mol

EA₁ = -141kJ/mol

EA₂ = +845 kJ/mol

To find:

Lattice energy, U = ?

Step-by-step solution:

We know,

ΔH = s + IE + 1/2D + EA + U

∴ U = ΔH - s - (IE₁ + IE₂) - 1/2D - (EA₁ + EA₂)

⇒ U = [- 636 - (590 + 1145) - 1/2(494) + (141 - 845)] kJ/mol

⇒ U = -3322 kJ/mol