I think a shovel is a wedge simple machine.
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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
<u>To determine:</u>
Lattice energy of KCl
K⁺(g) + Cl⁻(g) → KCl (s) U(KCl) = ?
<u>Explanation:</u>
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
Answer:
The answer is OXIDATION REACTION.
