Use the data given below to construct a Born-Haber cycle to determine the heat of formation of KCl. DH°(kJ)
K(s) → K(g) 89
K(g) → K⁺(g) + e⁻ 418
Cl2(g) → 2 Cl(g) 244
Cl(g) + e⁻ → Cl⁻(g) -349
KCl(s) → K⁺(g) + Cl⁻(g) 717
A) -1119 kJ
B) -997 kJ
C) -437 kJ
D) +631 kJ
E) +158 kJ
Correct Answer:
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