Question

Calculate the lattice energy for NaF (s) given the following sublimation of energy +109 kj/mol Hf...

Calculate the lattice energy for NaF (s) given the following

sublimation of energy +109 kj/mol

Hf for F (g) + 77 kj/mol

first ionization energy of Na (g) +495

electron affinity of F(g) -328

enthalpy of formation of NaF(s) -570 kj/mol

A. -177

B. 192

C. -804

D. -1047

E. -923

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Answer #1

     Na + 1/2 F2 -------> NaF H = -570Kj/mole

this reaction takes place five different steps

sublimation of energy +109 kj/mol

Na(s) --------> Na(g)     H1 = 109Kj/mole

first ionization energy of Na (g) +495

   Na(g) ---------> Na+ + e-   H2 = 495Kj/mole

bond energy of F2

F2 ------> 2F   = H = 155Kj/mole

1/2F2 -----> F   = H3   = 155/2 = 77.5Kj/mole

electron affinity of F(g) -328

F + e- -------> F-      H4 = -328Kj/mole

latticw energy

Na+ + F- -------> NaF    H5 =

H = H1 + H2 + H3 + H4 + H5

-570   = 109 +495 +77.5 -328+H5

H5 =-923KJ/mole

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