Download Thermodynamics of Ionic Crystal Formation: Born-Haber Cycle and Lattice Energy and more Lecture notes Thermodynamics in PDF only on Docsity!
Thermodynamics of Crystal Formation
! All stable ionic crystals have negative standard enthalpies of formation, Δ H o f , and negative standard free energies of formation, Δ G o f.
Na( s ) + ½ Cl 2 ( g ) ÷ NaCl( s ) Δ H o f = –410.9 kJ Δ G o f = –384.0 kJ
Cs( s ) + ½ Cl 2 (g) ÷ CsCl( s ) Δ H o f = –442.8 kJ Δ G o f = –414.4 kJ
Mg( s ) + ½O 2 ( g ) ÷ MgO( s ) Δ H o f = –385.2 kJ Δ G o f = –362.9 kJ
Ca( s ) + C( s ) + 3 / 2 O 2 ( g ) ÷ CaCO 3 ( s ) Δ H o f = –1216.3 kJ Δ G o f = –1137.6 kJ
! The exothermic and spontaneous formation of ionic solids can be understood in terms of a Hess's Law cycle, called the Born-Haber cycle.
! The lattice energy is the most important factor in making the formation of ionic crystals exothermic and spontaneous.
! Lattice energy, U , is defined as the enthalpy required to dissociate one mole of crystalline solid in its standard state into the gaseous ions of which it is composed; e.g.,
NaCl( s ) ÷ Na+^ ( g ) + Cl–( g ) U = +786.8 kJ
U Defined in this way, lattice energy is a positive (endothermic) quantity.
U Sometimes lattice energy is defined by the reverse reaction, in which case the values are negative (exothermic).
Na+( g ) + Cl-( g )
NaCl( s )
H
o
f = -410.9 kJ
H
o
sub = +107.7 kJ^2 D^ = +121.7 kJ
Na( s ) + 2 Cl 2 ( g )
Na( g ) + Cl( g )
I = +496 kJ A = -349 kJ
- U =?
Born-Haber Cycle for NaCl( s )
Na( s ) 6 Na( g ) Δ H osub = 107.7 kJ Na( g ) 6 Na+^ ( g ) + e –^ I = 496 kJ ½Cl 2 ( g ) 6 Cl( g ) ½ D = 121.7 kJ Cl( g ) + e –^6 Cl–( g ) A = –349 kJ Na+^ ( g ) + Cl–( g ) 6 NaCl( s ) – U =? Na( s ) + ½Cl 2 ( g ) 6 NaCl( s ) Δ H o f = –410.9 kJ
Y Δ H o f = Δ H osub + I + ½ D + A – U
ˆ U = Δ H osub + I + ½ D + A – Δ H o f = 107.7 kJ + 496 kJ + 121.7 kJ + (–349 kJ) – (–410.9 kJ) = 787 kJ
Calculating Lattice Energy
U In principle, the lattice energy for a crystal of known structure can be calculated by summing all the attractive and repulsive contributions to the potential energy.
! For a pair of gaseous ions
where Z +^ , Z –^ = ionic charges r o = distance between ions e = electronic charge = 1.602 × 10–19^ C 4 πεo = vacuum permittivity = 1.11 × 10–10^ C^2 @J –1@m–
! Potential energy is negative for the attraction of oppositely charged ions and positive for repulsion of like-charged ions.
! The potential energy arising from repulsions and attractions acting on one reference ion can be calculated.
! Scaled up to a mole of ion pairs (and with a change of sign) this should equal the lattice energy of the crystal.
Calculating U for NaCl
Consider the potential energy arising from attractions and repulsions acting on a central Na+^ ion of NaCl.
Neighbors Distanc e 6 Cl–^ r o 12 Na+
8 Cl–
6 Na+
24 Cl–
@@@ @@@
L The series in parentheses converges at a value that defines the Madelung constant, M.
Born-Landé Equation
U At r = r o the potential energy must be a minimum, so
U Solving for B gives
U Substituting for B in the equation for the Coulombic and Born contributions to potential energy gives the Born-Landé equation ,
! The value of n can be calculated from measurements of compressibility or estimated from theory.
L For NaCl, n = 9.1 from experiment, and the Born-Landé equation gives U o = -771 kJ/mol.
! In the absence of experimental data, Pauling's approximate values of n can be used.
Ion configuration He Ne Ar, Cu+^ Kr, Ag +^ Xe, Au+ n 5 7 9 10 12
Born-Mayer Equation
! Born-Landé values are approximate.
! Mayer showed that e– r / ρ , where ρ is a constant dependant on the compressibility of the crystal, gives a better repulsion term than 1/ r n.
! Using this improved repulsion term leads to the Born-Mayer equation :
- ρ = 30 pm works well for all alkali metal halides and other simple cases when r o values are in pm.
! Further refinements involve terms for van der Waals (dispersion) energy and evaluation of the zero point energy.
