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Use the following to calculate ΔH°latice of NaCl:
CL₂(g) → 2CL(g) ΔH° = 243kJ
Na(g) → Na⁺(g) + e⁻ ΔH° = 496kJ
Cl(g) + e⁻ → Cl⁻(g) ΔH° = -349kJ
Na(s) + (1/2)CL₂(g) → NaCL(s) ΔH° = -411kJ
Na(s) → Na(g) ΔH° = 109KJ
Compared with the lattice energy of LiF (1050 kJ/mol), is the magnitude of the value for NaCI what you expected? Explain.

Respuesta :

The lattice energy of Nacl = -786kJ LiF is more energetic at the lattice than NaF.

By definition, the gaseous cation and anion forming the corresponding ionic compound release energy termed the lattice energy, the energy contained within the lattice structure.

ΔHlattice =−[ΔH∘f,NaCl(s)+ΔHsub,Na+IE1,Na(g)+12ΔHbond,Cl2(g)−EA1,Cl(g)], =  −[411+107+502+12(242)−355]kJ =−786 kJ. LiF is more energetic at the lattice than NaF. Na > Li size. Compared to Li-F, the distance between Na and F is greater in the case of NaF. The force of attraction between Li and F increases as the distance between them decreases. We must therefore expend a lot of energy to separate them. And it will be known as lattice energy

To learn more about lattice energy:

https://brainly.com/question/18222315

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