Using the Born-Haber cycle, the ΔH_f° of KBr is equal to __________. A)ΔH_f°[K (g)] + ΔH_f°[Br (g)] + I_l(K)+ E(Br)+ ΔH_lattice B)ΔH_f°[K (g)] - ΔH_f°[Br (g)] - I_l(K)- E(Br)- ΔH_lattice C)ΔH_f°[K (g)] - ΔH_f°[Br (g)] + I_l(K)- E(Br)+ ΔH_lattice D)ΔH_f°[K (g)] + ΔH_f°[Br (g)] - I_l - E(Br)+ ΔH_lattice E)ΔH_f°[K (g)] + ΔH_f°[Br (g)] + I_l(K)+ E(Br)- ΔH_lattice

Question

Quizplus · Accepted Answer

The Born-Haber cycle for the formation of an ionic compound like KBr involves the enthalpy of formation of gaseous ions, ionization energy, electron affinity, and lattice energy. The correct expression accounts for the energy required to form gaseous ions and the energy released when the lattice forms.

Using the Born-Haber Cycle, the ΔHf° of KBr Is Equal

Using the Born-Haber Cycle, the ΔH_f° of KBr Is Equal