Lattice Energies AP Material Lattice Energy l Lattice

Lattice Energies AP Material

Lattice Energy l Lattice energy: The energy required to completely separate a mole of a solid ionic compound into its elements in their standard states l It is a measure of just how much stabilization results from arranging oppositely charged ions in an ionic solid

l The principal reason that ionic compounds are stable is the attraction between oppositely charged ions l This attraction draws the ions together, releasing energy and causing ions to form a solid array, or lattice l http: //www. youtube. com/watch? v=VBRe. Ojo 3 ri 8&feature=related

l The formation of ionic compounds is highly exothermic (releasing heat and light): Na(s) + ½ Cl 2 (g) → Na. Cl(s) l ∆ H = -788 k. J/mol The reverse reaction is highly endothermic: Na. Cl(s) → Na(s) + ½ Cl 2(g) ∆ H = 788 k. J/mol

Calculating Lattice Energy Eel = к. Q 1 Q 2 d where, l К is a constant (8. 99 x 109 J-m/C 2) l Q 1 and Q 2 are the charges on the ions l d is the distance between their centers

l l l Therefore, the magnitude of the lattice energy is dependant upon: the charges on the particles (Q 1 and Q 2) the distance between the nuclei

Summary l For a given arrangement of ions, the lattice energy increases as the charges on the ions increase and as their radii decrease

Example: Arrange in order of increasing lattice energy a) Na. F, Cs. I and Ca. O Cs. I < Na. F < Ca. O b) Zr. O 2, Mg. F 2, Ca. F 2 and Ca. F 2 < Mg. F 2 < Zr. O 2 c) Li. F, Na. F, Ca. F 2, Al. F 3 Na. F < Li. F < Ca. F 2 < Al. F 3 d) Ca. Cl 2, Li. Cl, Al 2 O 3, Na. Cl < Li. Cl < Ca. Cl 2 < Al 2 O 3

The Born-Haber Cycle l Lattice energy cannot be determined by experiment l It is calculated by envisioning the formation of an ionic compound occurring in steps (Hess’s Law) l Called the Born-Haber Cycle after Max Born and Fritz Haber (German scientists)

Steps in the formation of Na. Cl 1: Na(s) + ½ Cl 2(g) → Na. Cl(s) ∆Hf = -411 k. J 2: Na(s) → Na(g) ∆H°f = 108 k. J 3: 1/2 Cl 2(g) → Cl(g) ∆H°f = 122 k. J 4: Na(g) → Na+(g) + e I 1(Na) = 496 k. J 5: Cl(g) + e → Cl-(g) E(Cl) = -396 k. J 6. Overall : ΔH 6= Σ Δ H 1 to Δ H 5