Question

Born-Fajans-Haber Cycle

Suppose a chemist discovers a new metallic element and names it "Xhaustium" (Xh).
Xh exhibits chemical behaviour similar to an alkaline earth.

Xh(s) + F₂(g) → XhF₂(s)

Lattice energy for XhF2	-2120. kJ/mol
First Ionization energy of Xh	390. kJ/mol
Second Ionization energy of Xh	702 kJ/mol
Electron affinity of F	-327.8 kJ/mol
Bond energy of F2	154 kJ/mol
Enthalpy of sublimation (atomization) of Xh	200. kJ/mol

Use the above data to calculate ΔH°_f for Xhaustium fluoride.

Answer 1

Xh(s) + F₂(g) → XhF₂(s)      ΔH°_{f   =}

Enthalpy of sublimation (atomization) of Xh :

Xh(s) -----------> Xh(g)         ΔH₁     = 200Kj/mole

First Ionization energy of Xh

Xh(g) ------------> Xh^+ (g) + e^-   ΔH₂   = 390KJ/mole

Second Ionization energy of Xh

Xh^+ (g) -------------> Xh^2+ (g)    + e^-       ΔH₃   = 702KJ/mole

Bond energy of F₂

F2(g) -------------> 2F(g)                               ΔH₄    = 154Kj/mole

Electron affinity of F

2F(g) + 2e^- -----------> 2F^-(g)               ΔH₅         = 2*-327.8 kJ/mol    = -655.6KJ/mole

Lattice energy for XhF₂

Xh^2+ (g) + 2F^- (g) -----------> XhF2(s)      ΔH₆        = -2120. kJ/mol

According to Hess law

ΔH⁰f      = ΔH₁ + ΔH₂ + ΔH₃ + ΔH₄ + ΔH₅ + ΔH₆  

             = 200+390+702 + 154-655.6-2120

             = -1329.6KJ/mole >>>>answer

Xh(s) + F₂(g) → XhF₂(s)      ΔH°_{f   =}   -1329.6KJ/mole >>>>answer