1. Calculate the lattice energy of LiF, given the following: Li(s) + ¼F:(g) → LiF(s) AH? =-617 kJ Li'(g) + e(g) → Li(g) Li(s) →Li(g) F(g) → F(g) + e(g) F:(g) →2F(g) AH = -520 kJ AH = 161 kJ AH = 328 kJ АН - 154 kJ

Chemistry & Chemical Reactivity
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Chapter12: The Solid State
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1. Caleulate the lattice energy of LiF, given the following:
Li(s) + %F:(g) → LiF(s)
AH =-617 kJ
AH = -520 kJ
Li"(g) + e(g) → Li(g)
Li(s) →Li(g)
F(g) → F(g) + e (g)
F:(g) →2F(g)
AH = 161 kJ
AH = 328 kJ
AH = 154 kJ
Transcribed Image Text:1. Caleulate the lattice energy of LiF, given the following: Li(s) + %F:(g) → LiF(s) AH =-617 kJ AH = -520 kJ Li"(g) + e(g) → Li(g) Li(s) →Li(g) F(g) → F(g) + e (g) F:(g) →2F(g) AH = 161 kJ AH = 328 kJ AH = 154 kJ
Expert Solution
Step 1

Given

1.) Li(s) + 1/2F2(g) LiF(s) 

enthalpy of formation of lithium fluoride , Hof = -617 kJ

2.) Li(g) Li+(g) + e-(g) 

ionization energy of lithium atom , HI.E = 520 kJ

3.) Li(s) Li(g) 

sublimation energy of lithium, Hsub = 161 kJ

4.) F(g) + e- (g)F-(g)

the electron affinity of fluorine atom, HEA =- 328 kJ

5.) F2(g)  2F(g)

The dissociation energy of F2, Hdissociation = 154 kJ

 

To calculate the lattice energy of LiF 

Li+(g) + F-(g)  LiF(s)

 

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