Question

Born-Fajans-Haber Cycle Suppose a chemist discovers a new metallic element and names it "Xtrinsium" (Xt) Xt exhibits chemical behaviour similar to an alkaline earth Xt(s) + Cl2(g) → XtCl2(s) Lattice energy for XtCl2 First Ionization energy of Xt Second Ionization energy of Xt Electron affinity of Cl Bond energy of Cl2 Enthalpy of sublimation (atomization) of Xt 2260. kJ/mol 430. kJ/mol 731 kJ/mol -348.7 kJ/mol 239 kJ/mol 170. kJ/mol Use the above data to calculate ΔHof for Xtrinsium chloride.

Answer 1

Xt(s) + Cl2(g) -------------------> XtCl2(s)        $\Delta$ Hf =

same reaction takes place six different stages

Xt(s) --------------------> Xt(g)            $\Delta$ H1 = 170KJ    sublimation energy

Xt(g) ---------------> Xt⁺(g) + e^-          $\Delta$ H2 = 430KJ first ionisation energy

Xt^+(g)------------------> Xt^2+ (g) +e^- $\Delta$ H3 = 731KJ/mole   second ionisation energy

Cl2(g)-------------------------> 2Cl(g)            $\Delta$ H4 = 239KJ   bond energy of Cl2

2Cl(g) + 2e^- -----------> 2Cl^-                    $\Delta$ H5 = -348.7*2   = -697.4KJ electron affinity of Cl

Xt^2+ (g) + 2Cl^- (g) -----------> XtCl2(s)    $\Delta$ H6 = -2260KJ/mole

from Hess law

$\Delta$ H⁰f    = $\Delta$ H1 + $\Delta$ H2 + $\Delta$ H3 + $\Delta$ H4 + $\Delta$ H5 + $\Delta$ H6

   $\Delta$ H⁰f          = 170+430+731+239-697.4 -2260

$\Delta$ H⁰f        = -1387.4KJ/mole >>>>>answer