Question

9. Use the data provided below to calculate the lattice energy of RbCl. Is this value greater or less than the lattice energy of NaCl? Explain. Electron affinity of Cl = –349 kJ/mol 1st ionization energy of Rb = 403 kJ/mol Bond energy of Cl2 = 242 kJ/mol Sublimation energy of Rb= 86.5 kJ/mol ?Hf [RbCl (s)] = –430.5 kJ/mol Rb (s) + ½ Cl2 (g) ? RbCl (s) This equation can be written as the sum of the following: Rb (s) ? Rb (g) sublimation ?H1= +86.5 kJ/mol Rb (g) ? Rb+ + e– 1st ionization energy ?H2= +403 kJ/mol ½ Cl2 (g) ? Cl bond energy of Cl2 ?H3= +121 kJ/mol Cl + e– ? Cl– electron affinity of Cl ?H4= –349 kJ/mol Rb+ + Cl– ? RbCl lattice energy ?H5= ??? Rb (s) + ½ Cl2 (g) ? RbCl (s) ?Hf = –430.5 kJ/mol ?Hf = ?H1 + ?H2 + ?H3 + ?H4 + ?H5 Lattice energy = ?H5 = ?Hf – (?H1 + ?H2 + ?H3 + ?H4) = –430.5 – (86.5 + 403 + 121 – 349) = –692 kJ

          9. Use the data provided below to calculate the lattice energy of RbCl. Is this value greater or less than the lattice energy of NaCl? Explain.

Electron affinity of Cl = –349 kJ/mol
1st ionization energy of Rb = 403 kJ/mol
Bond energy of Cl2 = 242 kJ/mol
Sublimation energy of Rb= 86.5 kJ/mol
?Hf [RbCl (s)] = –430.5 kJ/mol
Rb (s) + ½ Cl2 (g) ? RbCl (s)

This equation can be written as the sum of the following:

Rb (s) ? Rb (g) sublimation ?H1= +86.5 kJ/mol
Rb (g) ? Rb+ + e– 1st ionization energy ?H2= +403 kJ/mol
½ Cl2 (g) ? Cl bond energy of Cl2 ?H3= +121 kJ/mol
Cl + e– ? Cl– electron affinity of Cl ?H4= –349 kJ/mol
Rb+ + Cl– ? RbCl lattice energy ?H5= ???
Rb (s) + ½ Cl2 (g) ? RbCl (s) ?Hf = –430.5 kJ/mol

?Hf = ?H1 + ?H2 + ?H3 + ?H4 + ?H5

Lattice energy = ?H5 = ?Hf – (?H1 + ?H2 + ?H3 + ?H4)
= –430.5 – (86.5 + 403 + 121 – 349)
= –692 kJ

9. Use the data provided below to calculate the lattice energy of RbCl. Is this value greater or less than the lattice energy of NaCl? Explain.

Electron affinity of Cl = –349 kJ/mol
1st ionization energy of Rb = 403 kJ/mol
Bond energy of Cl2 = 242 kJ/mol
Sublimation energy of Rb= 86.5 kJ/mol
?Hf [RbCl (s)] = –430.5 kJ/mol
Rb (s) + ½ Cl2 (g) ? RbCl (s)

This equation can be written as the sum of the following:

Rb (s) ? Rb (g) sublimation ?H1= +86.5 kJ/mol
Rb (g) ? Rb+ + e– 1st ionization energy ?H2= +403 kJ/mol
½ Cl2 (g) ? Cl bond energy of Cl2 ?H3= +121 kJ/mol
Cl + e– ? Cl– electron affinity of Cl ?H4= –349 kJ/mol
Rb+ + Cl– ? RbCl lattice energy ?H5= ???
Rb (s) + ½ Cl2 (g) ? RbCl (s) ?Hf = –430.5 kJ/mol

?Hf = ?H1 + ?H2 + ?H3 + ?H4 + ?H5

Lattice energy = ?H5 = ?Hf – (?H1 + ?H2 + ?H3 + ?H4)
= –430.5 – (86.5 + 403 + 121 – 349)
= –692 kJ

Added by Lori R.

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