The standard heat of formation of CaBr2 is -675 kJ/mol. The first ionization energy of Ca...

The standard heat of formation of CaBr2 is -675 kJ/mol. The first ionization energy of Ca is 590 kJ/mol and its second ionization energy is 1145 kJ/mol. The heat of sublimation of Ca[Ca(s)→Ca(g)] is 178 kJ/mol. The bond energy of Br2 is 193 kJ/mol, the heat of vaporization of Br2(l) is 31 kJ/mol, and the electron affinity of Br is -325 kJ/mol.

Calculate the lattice energy of CaBr2.

Solutions

Expert Solution

heat of romation

Ca(s) + Br2(l) -----------------> CaBr2(s) H = -675Kj/mole

heat of sublimation

Ca(s) ----------------> Ca(g) H1 = 178Kj/mole

The first ionization energy of Ca

Ca(g) ---------------> Ca^+(g) + e^- H2 = 590Kj/mole

The second ionization energy of Ca

Ca^+(g) ---------------> Ca^2+(g) + e^- H3 = 1145Kj/mole

heat of vaporization of Br2(l)

Br2(l) -----------------> Br2(g) H4 = 31Kj/mole

The bond energy of Br2

Br2(g) -------------> 2Br(g) H5 = 193Kj/mole

the electron affinity of Br

2Br + 2e^- -------------> 2Br^- H6 = 2*-325 = 650KJ/mole

lattice energy

Ca^2+ (g) + 2Br^-(g) ----------------> CaBr2(s) H7 = ?

from Hess law

H = H1 + H2 + H3 + H4 + H5 + H6

-675 = 178+590+ 1145+31+193-650 + H7

H7 = -2162KJ/mole

lattice energy of CaBr2 = -2162KJ/mole

queen_honey_blossom answered 1 year ago

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