Using the Born- Haber Cycle, the OH_f° of KBr Is

Multiple Choice

Using the Born- Haber cycle, the OH_f° of KBr is equal to _ _.

A) OH_f°[K (g) ] + OH_f°[Br (g) ] - I₁ - E(Br) + OH_lattice
B) OH_f°[K (g) ] + OH_f°[Br (g) ] + I₁(K) + E(Br) + OH_lattice
C) OH_f°[K (g) ] - OH_f°[Br (g) ] + I₁(K) - E(Br) + OH_lattice
D) OH_f°[K (g) ] + OH_f°[Br (g) ] + I₁(K) + E(Br) - OH_lattice
E) OH_f°[K (g) ] - OH_f°[Br (g) ] - I₁(K) - E(Br) - OH_lattice