Respuesta :
Given data:
Sublimation of K
K(s) ↔ K(g) ΔH(sub) = 89.0 kj/mol
Ionization energy for K
K(s) → K⁺ + e⁻ IE(K) = 419 Kj/mol
Electron affinity for Cl
Cl(g) + e⁻ → Cl⁻ EA(Cl) = -349 kj/mol
Bond energy for Cl₂
1/2Cl₂ (g) → Cl Bond energy = 243/2 = 121.5 kj/mol
Formation of KCl
K(s) + 1/2Cl₂(g) → KCl(s) ΔHf = -436.5 kJ/mol
To determine:
Lattice energy of KCl
K⁺(g) + Cl⁻(g) → KCl (s) U(KCl) = ?
Explanation:
The enthalpy of formation of KCl can be expressed in terms of the sum of all the above processes, i.e.
ΔHf(KCl) = U(KCl) + ΔH(sub) + IE(K) + 1/2 BE(Cl₂) + EA(Cl)
therefore:
U(KCl) = ΔHf(KCl) - [ΔH(sub) + IE(K) + 1/2 BE(Cl₂) + EA(Cl)]
= -436.5 - [89 + 419 + 243/2 -349] = -717 kJ/mol
Ans: the lattice energy of KCl = -717 kj/mol
The Lattice energy of KCl is -717 kJ/mol.
Given Here,
Enthalpy of sublimation of Potassium [tex]\rm \bold{ \Delta H(sub ) }[/tex] = 89.0 kJ/mol
Ionization Energy for Potassium IE(K) = 419 kJ/mol
Electron affinity for Chlorine is EA(Cl) = −349 kJ/mol
Bond energy of Chlorine, BE(Cl) = 243 kJ/mol
Enthalpy of formation for KCl, [tex]\rm \bold{ \Delta H(f) }[/tex] = −436.5 kj/mol .
The Lattice energy of KCl can be calculated from the formula
[tex]\rm \bold {U(KCl) = \Delta Hf(KCl) - [ \Delta H(sub) + IE(K) + \frac{1}{2}BE(Cl_2) + EA(Cl)]}[/tex]
U( KCl) = -436.5 - [89 + 419 + 243/2 -349]
U( KCl) = -717 kJ/mol
Hence we can calculate that the Lattice energy of KCl is -717 kJ/mol.
To know more about Lattice Energy, refer to the link:
https://brainly.com/question/18222315?referrer=searchResults
