Answer to Question #208210 in Physical Chemistry for Yakubu

Question

Calculate the lattice energy of rubidium flouride from the following data:

Heat of sublimation=+86kjmol

Standard heat of formation of RbF =-548kjmol

Dissociation energy of F²(g)=+155kjmol

Ionization energy of Rb(g)=+402kjmol

Electron affinity of F(g)= -346kjmol

2. Calculate the lattice energy of calcium chloride from the following data: 

Heat of sublimation Ca(g)=+193kjmol

Standard heat of formation of CaCl2(g) =-794kjmol

Dissociation energy of Cl2(g)=+242kjmol

Ionization energy of Ca(g) to Ca²+(g)=+1725kjmol

Electron affinity of Cl(g)= -364kjmol

Answer 1

a) Lattice energy of rubidium flouride=$-251kJ/mol$

b) Lattice energy of calcium chloride=$+1,002kJ/mol$