Cristal iónicodepa.fquim.unam.mx/amyd/archivero/Diapo2_35658.pdf · 2019-02-20 · Cristal...

Cristal ioacutenico

sustancia soacutelida cuya estructura puede representarse como

un arreglo infinito y regular de iones cargados

opuestamente La red se mantiene principalmente por

fuerzas electrostaacuteticas no direccionales entre los iones

cargados

El empaquetamiento debe 1) maximizar las atracciones

emtre las cargas opuestas y 2) minimizar las repulsiones

manteniendo a los iones de la misma carga lo maacutes alejado

posible

Cristal ioacutenico

Inorganic energetics Dasent W E New York Cambridge University Press 1982

Na+(g) + Cl-(g) rarr NaCl(s) Ured

Energiacutea red

____________________________________

T = 0 K

Cristal ioacutenico

U0 experimental = Hordmred

Na(s) rarr Na(g) Hatm 108 kJmol

Na(g)- e rarr Na+(g) EI 502

frac12 Cl2(g) rarr Cl(g) Hatm 121

Cl(g) + e rarr Cl-(g) AE -354

Na+(g) + Cl-(g) rarr NaCl(s) Hordmred

Na(s)+Na(g)+frac12Cl2(g) rarr Na(g)+ Na+(g) +Cl(g)+Cl-(g)+NaCl(s) Hatm

Na + EI + HatmCl + AE + Hordmred

+Cl(g) +Na+(g)+Cl-(g)

Na(s) + frac12 Cl2(g) rarr NaCl(s) Hf -411


Na(s) rarr Na(g) Hatm 108kJmol

Na(g) rarr Na+(g) + e EI 502




________________________________________

Na(s) + frac12 Cl2(g) rarr NaCl(s) Hf= -411 kJmol

Hordmred = Hf ndash(HatmNa + EI + Hatm

Cl + AE)

Hordmred = -411 ndash (108 + 502 + 121 -354) = -788 kJmol

U 0 = Ucalc

U0 = Uc + UB + UL + UZ + correccioacuten x T +

U c

Uc = - NA z+z- e2 [6 - 12 + 8 - 6 + 24 ]

4πε0r 2 3 2 5

Uc = - NA z+z- e2 A

4πε0r

UC

Uc = - 6 NA z+z- e2 + 12 NA z+z+ e2 - + -

4πε0r 2 r4πε0

r = d

NaCl(s)

A o

Blenda cuacutebico compacto Ruacutetilo hexagonal compacto

Wurzita Hexagonal compacto Fluorita cuacutebico compacto

Estructura cristalina Ejemplo

Antifluorita K2O K2SLi2O Na2O Na2Se Na2S

Cloruro de cesio CsCl CaS TlSb CsCN CuZn

Fluorita CaF2 UO2 BaCl2 HgF2 PbO2

Arseniuro de niacutequel NiAs NiS FeS PtSn CoS

Perovskita CaTiO3 BaTiO3 SrTiO3

Sal de roca NaCl LiCl KBr RbI AgCl AgBr MgO CaO TiO FeO NiO SnAs UC ScN

Rutilo TiO2 MnO2 SnO2 WO2 MgF2 NiF2

Blenda ZnS CuCl CdS HgS GaP InAsWurtzita ZnS ZnO BeO MnS AgI AlN SiC NH4F

Uc = - NA z+z- e2 A

4πε0rkJmol

A constante de Madelung = 1748

r+ + r- (distancia interioacutenica) = 2814 x10ndash10 m

NA = 6023 x 10 23

e carga del e = 160218 x 10 ndash19 C

0 = 8854188 x 10 ndash12 C2Jm

Uo = - 6023 x 10 23mol-1(1) (1) (160218 x 10 ndash19 C)2(1748)

4π (8 854188 x 10 ndash12 C2Jm) (2814 x10ndash10 m)

C2mol Jmol (1kJ1000J) = kJmol

(C2Jm) m

NaCl(s)

Uc = ndash86319 kJmol

UB

UB infin 1 r n para un par ioacutenico

UB = NAB r n

n coef de Born

n = depende de la densidad de la nube electroacutenica

experimental de la compresibilidad del cristal resistencia de un ioacuten a

acercarse lo maacuteximo a otro

B cte de proporcionalidad que se discierne en el equilibrio cuando la fza

atractiva y repulsiva se igualan r= r0

dU Calc = 0 = d (Uc +UB)

dr dr

U Calc = U0 + UB

B = z+z-e2A (r0n-1)

4πε0n

dU0= d - NA z+z- e2 A + NAB

dr (r= r0) dr (r= r0) 4πε0r rn

dU0= 0

dr (r= r0)

dU0= 0 = NAz+z-e2A - nNAB

dr (r= r0) 4πε0 r02 r0

n+1

De donde se despeja B

U Calc = U0 + UB

d Xm = m Xm-1

dX

UB = NAB

rn

r=r0

Sustituyendo B

UB = NA z+z-e2A

4πε0nr

Para NaCl

A = constante de Madelung (1748)

n = exponente de Born (8)

r0 = r+ + r- = (distancia interioacutenica) = 2814 x10ndash10 m

z+ y z- = cargas catioacuten y anioacuten

NA = 6023 x 10 23


0 = 8854188 x 10 ndash12 C2Jm

Sustituyendo B

B = z+z-e2A (r0n-1)

4πε0n

UB = 10789 kJmol

Ioacuten tipo Ej n

He-He LiH 5

Ne-Ne NaF MgO 7

NaCl 8

Ar-Ar KCl CaS CuCl 9

Kr-Kr RbBr AgBr 10

Xe-Xe CsI 12

promedio

Uo = Uc + UB

Uo = - NA z+z- e2 A + NAB

4πε0r 0 r0n

Uo = - NA z+z- e2 A + NA z+z-e2A

4πε0r 0 4πε0nr0

factorizando

Uo = - NA z+z-e2A (1 ndash 1 )

4πε0r0 n

A = constante de Madelung

n = exponente de Born

r0 = r+ + r- = (distancia interioacutenica)


NA = 6023 x 10 23


0 = 8854188 x 10 ndash12 C2Jm

Ec de Born- Landeacute

2 2

0 0 0

14

AN Az eU

r r

= minus minus

Born-Mayer

UB = NAB e (-r)


n exponente de Born = 8


z+ y z- = 1

NA = 6023 x 10 23


0 = 8854188 x 10 ndash12 C2Jm

Uo = ndash 6023 x 10 23mol-1(1) (1) (160218 x 10 ndash19 C)2(1748) (1 ndash 1 )

4π (8 854188 x 10 ndash12 C2Jm) (2814 x10ndash10 m) 8

Uo = ndash 270257x10-14 ( 7 ) C2mol Jmol

31309x10-20 8 (C2Jm) m

U0 para el NaCl seguacuten Born-landeacute

Uo = ndash 863192692 (78)

Uo = ndash 75529360 Jmol x 1kJ = ndash 75529 kJmol

1000J

Uc = ndash86319 kJmolUB = 86319 = 10789 kJmol

8

Ec Born-Lande

(Uc +UB) -7553 kJmol

U0 para el NaCl

Uc =

ndash86319Hordmred

-788 kJmol 1001095

9585

Ag(s) rarr Ag(g) Hatm 286 KJmol

Ag(g) rarr Ag+(g) + e EI 7308


Cl(g) + e rarr Cl-(g) -AE -354

Ag+(g) + Cl-(g) rarr AgCl(s) Hordmred

_____________________________________________________________________________________________

Ag(s) + frac12 Cl2(g) rarr AgCl(s) Hf -127 kJmol

Hordmred= Hf ndash(HatmAg + EI + Hatm

Cl + AE) = -9108 kJmol

AgCl


Cl + AE)

Uo = - NA z+z-e2A (1 ndash 1n)4πε0r0

el AgCl es un cristal con estructura de NaCl

Uo = Uc + UB

Uo = - 6023 x 10 23mol-1(1) (1) (160218 x 10 ndash19 C)2(1748) (1 ndash 19)


UC= - 820623 kJmol UB= 102578 kJmol

Uo = - 270257x10-14 (1 ndash 19) [Jmol] divide 1000 [kJmol]

3 2933x10-20

Uo = -71804 [kJmol]

Ec Born-Lande

UL Energiacutea de London

2 2

6 8exp

4 r

AA A A

N Az e r C DU N B N N

r r

= minus + minus minus minus

Energiacutea de

London (UL)

6

C

rDonde Componente de las fuerzas de dispersioacuten

dipolo-dipolo inducido

Componente de las fuerzas de dispersioacuten

dipolo- cuadrupolo inducido8

D

r

- NA z+z-e2A (1 ndash 1n)4πε0r0

bullLas expresiones de Born-Landeacute y Born- Mayer no toman en cuenta las fuerzas atractivas de London

Caacutelculo del valor de la constante C

Para un par de iones M+X-

J E Mayer (1933)

1( )

2cristalC kC k C C+minus ++ minusminus= + +

1 2

1 2

3

2 ( )

E EC

E E +minus =

+

23

4C E ++ + += 23

4C E minusminus minus minus=

1 2

1 2

3

2 ( )

E EC

E E +minus =

+

23

4C E ++ + += 23


Donde

α1 y α2 es la polarizabilidad de cada ioacuten

E1 y E2 son las energiacuteas caracteriacutesticas de los dos iones

La energiacutea caracteriacutestica teoacutericamente esta dada por hn donde n es

la frecuencia de absorcioacuten maacutexima en el ultravioleta del ioacuten en el

cristal

Sin embargo en la praacutectica se utilizan solo aproximaciones de eacutestas

energiacuteas

Ccristal = 795 x 10-78 J m6 par

r0 = 2772 x 10-10 m

Entonces

6

0

C

rminus = = -109 eV NA = -1055 kJmol= -0175 x 10-18 Jmol

Para el AgClDatos

Polarizabilidad

αAg+ = 172 x 10-30 m3 (Calculados por Pauling)

αCl- = 345 x 10-30 m3

Energiacutea caracteriacutestica

E+ = 310 x 10-18 J par (El 90 del 2deg potencial de ionizacioacuten = 2148 eV)

E- = 156+ x 10-18 J par (Calculado del espectro de absorcioacuten UV)

NA NA

C+- = 92 x 10-78 J m6 moleacutecula

C++ = 69 x 10-78 J m6 moleacutecula

C-- = 139 x 10-78 J m6 moleacutecula

dipolo-cuadrupolo D

De manera general la contribucioacuten de la interacccioacuten dipolo-

cuadrupolo corresponde al 10-20 del valor de la interaccioacuten

dipolo-dipolo

Interaccioacuten dipolo-cuadrupolo ndash NA = ndash 19 kJ mol

Entonces la energiacutea total de dispersioacuten de London para el

AgCl es

UL= -106 kJ mol - 19 kJ mol = -125 kJmol

UL=

8

D

r

2 2

6 8exp

4 r

AA A A


r r


2 2

6 8exp

4 r

AA A A


r r


Uo = - NA z+z-e2A (1 ndash 1n) mdash

4πε0r0


Uo = Uc + UB + UL

Uo = - 6023 x 10 23mol-1(1) (1) (160218 x 10 ndash19 C)2(1748) (1 ndash 18) ‒125 [kJmol]


Uo = - 270257x10-14 (1 ndash 18) [Jmol] -125 [KJmol]

3 2933x10-20


Uo = -71804 [KJmol] - 125 [kJmol]= - 84304 kJmol

Ec Born-Lande





Ag+(g) + Cl-(g) rarr Ag|Cl(s) Hordmred= Hf ndash(Hatm

Ag + EI + HatmCl + AE)

_____________________________________________________________________________________________


Hordmred = -9108

Ec Born-Lande Ec Born-Lande + UL Hordmred

-71804 KJmol -84304 KJmol -9108

788 956 100

Factores que provocan el aumento en la energiacutea de London

bullUn alto nuacutemero de coordinacioacuten

bullUna distancia internuclear pequentildea

bullIones con altos valores de polarizabilidad

α+

Li+ Be

2+ Cu

+ Zn

2+ Al

3+

0029 0008 043 029 0054

Na+ Mg

2+ Ag

+ Cd

2+ Ga

3+

018 0094 172 109 02

k+ Ca

2+ Au

+ Hg

2+ In

3+

084 047 188 125 073

Rb+ Sr

2+ Tl

3+

142 086 087

Cs+ Ba

2+

244 156

α-

H- F

- O

2-

102 105 392

Cl- S

2-

369 103

Br- Se

2-

481 106

I- Te

2-

716 141

Polarizabilidadx 10-30 m3


Energiacutea de punto cero (UZ)

La U a T = OK donde persisten la vibraciones de la red (frecuencia de Debye)

max

1

2Z AU N hv=

Componentes de la energiacutea de red (kJmol)

Compuesto U0 (calc) UC UB UL UZ ΔH298

(calc)

ΔH 298

(ciclo Born)

NaCl -766 -863 114 -25 8 -776 -787 CsI -580 -619 90 -54 3 -592 -602

AgI -795 -808 138 -128 3 -807 -891

CuBr -870 -925 114 -64 4 -882 -977

108772

NaCl -772 es del 979

CsI -580 es del 964

AgI -795 es del 892

CuBr -870 es del 89

AgCl -840 es del 922

788

AgCl -840 -820 1026 -125 3 -9108

Simplificando la Ec de Born-Lande

U0 = - NAA z+z-e2 (1-1n)

40 (r0)

U0 = - 139 x 10-4 A z+z- (1-1n) (J mmol) r+ + r- (distancia interioacutenica m)

(r+ + r-)

1pm = 1x10-12 m = 10-2 Ả

U0 = - 139 x 105 A z+z- (1-1n) (kJ pm mol) r++ r-(distancia interioacutenicapm)

(r+ + r-)

Ecuacioacuten de Kapustinsky

bull Usada en casos donde los datos experimentales son escasos para el calculode la energiacutea de red

Principalmente cuando no se conoce el tipo de estructura del cristal

Comuacuten para iones complejos


2

0

11

4 2

c aA

c a

Z ZN Ae vU

r r n

= minus

+

2

0

14 2

c aA

c a c a

Z ZN Ae vU

r r r r

= minus

+ +

Donde v es el nuacutemero de iones en la formula simple

Zc y Za son las cargas del catioacuten y el anioacuten

rc y ra son los radios ioacutenicos Pauling o de Goldschmidt

2 2

0 0

11

4

AN AZ eU

r n

= minus minus

2 2

0 0 0

14

AN Az eU

r r

= minus minus

Born-Landeacute

Born-Mayer

ndash

ndash

Generalizacioacuten de los datos del NaCl a otras estructuras

bull AasympANaCl = 174756

bull r0 asymp rc+ra nm

bull n asymp 8

n2 =22 =1

bull2

04

AN e

108 c a

c a

Z ZU v KJ mol

r r=

+-

2

0

11

4 2

c aA

c a

Z ZN Ae vU

r r n

= minus

+

ndash

Del ciclo de Born-Haber

Ejemplo MgF2

kJmol

Mg(s) rarr Hatm 149

rarr EI1 +EI2 7375 y 1450

F2 (g) rarr 2Hatm 79 X2

rarr -2AE -339 X2

rarr Hordmred= _____________________________________________________________________________________________

Mg(s) + F2(g) rarr MgF2(s) Hf -1123 kJmol

-29395

Ecuacioacuten de Born-Landeacute

Ecuacioacuten de Kapustinsky2 2

0 0

11

4 r

AN Az eU

n

= minus

MgF2 tiene estructura de rutilo (hex

comp tetragonal) la constante de

Madelung A= 2408

Distancia internuclear o r0 = r+ + r-

Mg2+(6 86 pm) F- (3116pm) = 202 pm

0202 nm y 202 x 10-10 m

z = 1 2

n=8

U = -28996 kJmol

Zc = +2 Za = -1

v = 3

bullrMg2+ 0078 nm rF- = 0133 nm

bull0211nm

U =- 307109 kJmol

Ejemplo MgF2

U0 = - 139 x 10-4 A z+z- (1-1n) Jmol

(r+ + r-) (m)

108 c a

c a

Z ZU v KJ mol

r r=

+-

-

Hordmred= -29395 100986 1045

Uc = -33139 kJmol y UB = 4142

Radio termodinaacutemico

iquestCoacutemo saber cual es el radio de un ioacuten poliatoacutemico

O iquestcuanta energiacutea requiere la formacioacuten de un ioacuten poliatoacutemico

Sean dos compuestos A= Na2SO4 y B = K2SO4 para los cuales seconocen sus ΔHf y los radios ioacutenicos de los cationes son conocidos

UA ndashUB = [ΔHdegf (NaSO4)- ΔHdegf (K2SO4)] ndash [2 ΔHdegf Na+(g) - 2 ΔHdegf K+

(g)]

Al evaluar la diferencia en la energiacutea de red es posible sustituir

en la ecuacioacuten de Kapustinsky y tener como uacutenica incoacutegnita el

radio del anioacuten Los radios obtenidos mediante este

tratamiento se conocen como radios termodinaacutemicos

kJmol

2ΔH(Na+) 1220

iquestProceso real

AE 1+2 (SO4-2

) = ΔH(SO4-2)

Hordmred

____________________________________________________________________________________

rarr Na2SO4(s) -1658 kJmol Hf

H red Na2SO4 = Hf - 2ΔH (Na+) - ΔH (SO4-2) = -13858 - ΔH (SO4

-2)

H red K2SO4 = Hf - 2ΔH (K+) - ΔH (SO4-2) = -11998 - ΔH (SO4

-2)

kJmol

2ΔH(K+) 1028

iquestProceso real

AE 1+2 (SO4-2

) = ΔH(SO4-2)

Hordmred

____________________________________________________________________________________

rarr K2SO4(s) -1718 kJmol Hf


-2)

H redK2SO4 = Hf - 2ΔH (K+) - ΔH (SO4-2) = -11998 - ΔH (SO4

-2)

H K2SO4 ndash H Na2SO4=

r(Na+) = 116 pm = 0116 nm r(K+) = 152 pm = 0152 nm

Regla de los radios

Aproximacioacuten para predecir el de coordinacioacuten de los iones

La razoacuten de R = r+ r- da una guiacutea de cuantos cationes pueden acomodarse

alrededor de un anioacuten

coordinacioacuten 4 td debajo de 0414

6 oct entre 0414 y 0732

8 cuacutebico entre 0732 y 1

12 dodecaedro 1

Se asumen esferas duras por lo que pueden desviaciones importantes

Ej ZnS 052 pero es 4

Reglas de FAJANrsquoS ndash grado de ionicidad y poder polarizante

Grado de ionicidad depende del poder polarizante (PP) del catioacuten y de la

polarizabilidad (P) de anioacuten

Para un catioacuten An+ PP seraacute ne (43 πrA3)

Para un anioacuten Bm- P seraacute me bull (43 πrB3)

Reglas de Fajans

bull Cuanto maacutes polarizante es el catioacuten mayor seraacute el

caraacutecter covalente del enlace El caraacutecter covalente se incrementa al

decrecer el tamantildeo del Catioacuten yo incrementarse la carga del mismo

bull Cuanto maacutes polarizable es el anioacuten mayor el caraacutecter covalente que

se incrementa al incrementarse el tamantildeo yo la carga del anioacuten

bull La polarizacioacuten del enlace (y por lo tanto el caraacutecter covalente) aumenta

cuando el catioacuten no tiene configuracioacuten de gas noble Esto se da en los

metales de transicioacuten (grupos 3 - 12) y en el bloque f debido a que los

cationes son maacutes polarizantes que los que tienen conf de gas noble

bullEj El Hg2+ es maacutes polarizante que el Ca2+ siendo que tienen praacutecticamente el mismo radio (102 y 100 pm respectivamente)

EjemploNaX

F Cl Br I kJmol

ΔHatm Na

EI

ΔHatm X

-AE

ΔHred

-569 -410 -360 -288 ΔHf

Ejemplo MI

Li Na K Rb Cs kJmol

161

526

108

502

90

424

82

408

78

381

ΔHatm M

EI

106

-301

106

-301

106

-301

106

-301

106

-301

ΔHatm I

-AE

ΔHred

-271 -288 -328 -329 -337 ΔHf

M(s) + frac12 X2(ee) rarr MX(s)

Cs(s) rarr Cs(g) Hatm 78 kJmol

Cs(g) rarr Cs +(g) + e EI 376

frac12 F2(g) rarr F (g) Hatm 79

F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-

(g) rarr CsF (s) Hordmred= U red

________________________________________

Cs(s) + frac12 F2(g) rarr CsF (s) Hf -Hf

Hf = HatmNa + EI + Hatm

Cl + AE + Ured = -531 kJmol




F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf

Sf = Sdeg productos ndash Sdeg reactivos

T = 298 K

Sf = Sdeg298 CsF(s) - Sdeg298 Cs(s) - frac12 Sdeg298 F2

Entropiacutea de los elemento a partir de sus estados

estaacutendar (por elemento) a 298 K (JKmol)






F(g) + e rarr F-(g) AE -333

Cs +(g) + F-


________________________________________

Cs(s) + frac12 F2(g) rarr CsF (s) Hf = -531 kJmol

Gf = Hf ndash T Sf



Sf = 80 -84 ndash 101 JKmol

Sf = - 105 JKmol x 1 kJ1000J o bien

Gf = -531- 298(-105) kJmol

1000

Gf = -531 + 313 = -4997 kJmol

iquestPor que el oacutexido de magnesio no es Mg+Ominus

si podriacutean ahorrarse el gasto de la 2ordf EI del Mg y la 2ordf AE del O

Mg(s) rarr Mg(g) Hatm

Mg(g) rarr Mg +(g) + e EI1

Mg +(g) rarr Mg 2+(g) + e EI2

frac12 O2(g) rarr O (g) Hatm

O(g) + e rarr Ominus(g) AE1

Ominus(g) + e rarr O2minus

(g) AE2

Mg 2+(g) + O2minus

(g) rarr MgO (s) Hordmred=

___________________________________________

Mg(s) + frac12 O2(g) rarr MgO (s) Hf

149 kJmol

738

1450

249

-142

+844

-610

Extra

GASTO

De ENERGIacuteA

iquestPor que el oxido de magnesio no es Mg+Ominus



Mg(g) rarr Mg +(g) + e EI


O(g) + e rarr Ominus(g) AE

Mg +(g) + Ominus(g) rarr MgO (s) Hordmred= U red

____________________________________________


149 kJmol

738

249

-142

108 c a

c a

Z ZU v KJ mol

r r=

+-

Para estimar los radios Mg+

Debe estar entre el del Li+ (90pm) y el Mg+2 (86pm)

y para el O-

debe estar entre el del Cl- (167pm) y O-2 (126pm)

Por lo tanto rc + ra = 0257 a 0212 nm

iquestPor que los halogenuros de Na no son Na2+X2minus

siacute como en los alcalino-terreos una carga maacutes alta

podriacutean generar una Hordmred maacutes negativa

Hatm

EI1

EI2= 4567

Hatm

AE1

Hordmred=

___________________________________________

Na(s) + X2(g) rarr NaX2 (s) Hf

Para estimar radios catioacuten estaacute entre el Ca2+(114pm) y Na+ 116 pm

rcatioacuten = 0115 0001 nm

Para el anioacuten estaacute entre Cl-(0167nm)

Na+(g) + Cl-(g) rarr NaCl(s) Ured

Energiacutea red

____________________________________

T = 0 K

Cristal ioacutenico

















________________________________________



Cl + AE)


U 0 = Ucalc


U c

Uc = - NA z+z- e2 [6 - 12 + 8 - 6 + 24 ]

4πε0r 2 3 2 5

Uc = - NA z+z- e2 A

4πε0r

UC

Uc = - 6 NA z+z- e2 + 12 NA z+z+ e2 - + -

4πε0r 2 r4πε0

r = d

NaCl(s)

A o












Uc = - NA z+z- e2 A

4πε0rkJmol



NA = 6023 x 10 23


0 = 8854188 x 10 ndash12 C2Jm

Uo = - 6023 x 10 23mol-1(1) (1) (160218 x 10 ndash19 C)2(1748)



(C2Jm) m

NaCl(s)


UB


UB = NAB r n

n coef de Born







dr dr

U Calc = U0 + UB

B = z+z-e2A (r0n-1)

4πε0n



dU0= 0

dr (r= r0)


dr (r= r0) 4πε0 r02 r0

n+1


U Calc = U0 + UB

d Xm = m Xm-1

dX

UB = NAB

rn

r=r0

Sustituyendo B

UB = NA z+z-e2A

4πε0nr

Para NaCl





NA = 6023 x 10 23


0 = 8854188 x 10 ndash12 C2Jm

Sustituyendo B

B = z+z-e2A (r0n-1)

4πε0n

UB = 10789 kJmol

Ioacuten tipo Ej n

He-He LiH 5

Ne-Ne NaF MgO 7

NaCl 8


Kr-Kr RbBr AgBr 10

Xe-Xe CsI 12

promedio

Uo = Uc + UB


4πε0r 0 r0n


4πε0r 0 4πε0nr0

factorizando


4πε0r0 n





NA = 6023 x 10 23


0 = 8854188 x 10 ndash12 C2Jm


2 2

0 0 0

14

AN Az eU

r r

= minus minus

Born-Mayer

UB = NAB e (-r)




z+ y z- = 1

NA = 6023 x 10 23


0 = 8854188 x 10 ndash12 C2Jm




31309x10-20 8 (C2Jm) m


Uo = ndash 863192692 (78)


1000J


8

Ec Born-Lande


U0 para el NaCl

Uc =

ndash86319Hordmred

-788 kJmol 1001095

9585






_____________________________________________________________________________________________




AgCl


Cl + AE)



Uo = Uc + UB





3 2933x10-20

Uo = -71804 [kJmol]

Ec Born-Lande


2 2

6 8exp

4 r

AA A A


r r


Energiacutea de

London (UL)

6

C





D

r





J E Mayer (1933)

1( )


1 2

1 2

3

2 ( )

E EC

E E +minus =

+

23

4C E ++ + += 23


1 2

1 2

3

2 ( )

E EC

E E +minus =

+

23

4C E ++ + += 23


Donde





cristal


energiacuteas


r0 = 2772 x 10-10 m

Entonces

6

0

C


Para el AgClDatos

Polarizabilidad


αCl- = 345 x 10-30 m3




NA NA




dipolo-cuadrupolo D



dipolo-dipolo



AgCl es


UL=

8

D

r

2 2

6 8exp

4 r

AA A A


r r


2 2

6 8exp

4 r

AA A A


r r



4πε0r0


Uo = Uc + UB + UL




3 2933x10-20



Ec Born-Lande







_____________________________________________________________________________________________


Hordmred = -9108


-71804 KJmol -84304 KJmol -9108

788 956 100





α+

Li+ Be

2+ Cu

+ Zn

2+ Al

3+

0029 0008 043 029 0054

Na+ Mg

2+ Ag

+ Cd

2+ Ga

3+

018 0094 172 109 02

k+ Ca

2+ Au

+ Hg

2+ In

3+

084 047 188 125 073

Rb+ Sr

2+ Tl

3+

142 086 087

Cs+ Ba

2+

244 156

α-

H- F

- O

2-

102 105 392

Cl- S

2-

369 103

Br- Se

2-

481 106

I- Te

2-

716 141





max

1

2Z AU N hv=



(calc)

ΔH 298

(ciclo Born)

NaCl -766 -863 114 -25 8 -776 -787 CsI -580 -619 90 -54 3 -592 -602

AgI -795 -808 138 -128 3 -807 -891

CuBr -870 -925 114 -64 4 -882 -977

108772


CsI -580 es del 964

AgI -795 es del 892

CuBr -870 es del 89


788

AgCl -840 -820 1026 -125 3 -9108


U0 = - NAA z+z-e2 (1-1n)

40 (r0)


(r+ + r-)

1pm = 1x10-12 m = 10-2 Ả


(r+ + r-)






2

0

11

4 2

c aA

c a

Z ZN Ae vU

r r n

= minus

+

2

0

14 2

c aA

c a c a

Z ZN Ae vU

r r r r

= minus

+ +




2 2

0 0

11

4

AN AZ eU

r n

= minus minus

2 2

0 0 0

14

AN Az eU

r r

= minus minus

Born-Landeacute

Born-Mayer

ndash

ndash




bull n asymp 8

n2 =22 =1

bull2

04

AN e

108 c a

c a

Z ZU v KJ mol

r r=

+-

2

0

11

4 2

c aA

c a

Z ZN Ae vU

r r n

= minus

+

ndash


Ejemplo MgF2

kJmol

Mg(s) rarr Hatm 149

rarr EI1 +EI2 7375 y 1450


rarr -2AE -339 X2

rarr Hordmred= _____________________________________________________________________________________________


-29395



0 0

11

4 r

AN Az eU

n

= minus



Madelung A= 2408


Mg2+(6 86 pm) F- (3116pm) = 202 pm

0202 nm y 202 x 10-10 m

z = 1 2

n=8

U = -28996 kJmol

Zc = +2 Za = -1

v = 3


bull0211nm

U =- 307109 kJmol

Ejemplo MgF2

U0 = - 139 x 10-4 A z+z- (1-1n) Jmol

(r+ + r-) (m)

108 c a

c a

Z ZU v KJ mol

r r=

+-

-

Hordmred= -29395 100986 1045

Uc = -33139 kJmol y UB = 4142






(g)]





kJmol

2ΔH(Na+) 1220

iquestProceso real

AE 1+2 (SO4-2

) = ΔH(SO4-2)

Hordmred

____________________________________________________________________________________



-2)


-2)

kJmol

2ΔH(K+) 1028

iquestProceso real

AE 1+2 (SO4-2

) = ΔH(SO4-2)

Hordmred

____________________________________________________________________________________



-2)


-2)


r(Na+) = 116 pm = 0116 nm r(K+) = 152 pm = 0152 nm

Regla de los radios





6 oct entre 0414 y 0732


12 dodecaedro 1








Reglas de Fajans











EjemploNaX

F Cl Br I kJmol

ΔHatm Na

EI

ΔHatm X

-AE

ΔHred

-569 -410 -360 -288 ΔHf

Ejemplo MI

Li Na K Rb Cs kJmol

161

526

108

502

90

424

82

408

78

381

ΔHatm M

EI

106

-301

106

-301

106

-301

106

-301

106

-301

ΔHatm I

-AE

ΔHred

-271 -288 -328 -329 -337 ΔHf





F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________







F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf


T = 298 K










Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf





Gf = -531- 298(-105) kJmol

1000

Gf = -531 + 313 = -4997 kJmol









(g) AE2

Mg 2+(g) + O2minus


___________________________________________


149 kJmol

738

1450

249

-142

+844

-610

Extra

GASTO

De ENERGIacuteA








____________________________________________


149 kJmol

738

249

-142

108 c a

c a

Z ZU v KJ mol

r r=

+-



y para el O-






Hatm

EI1

EI2= 4567

Hatm

AE1

Hordmred=

___________________________________________





















________________________________________



Cl + AE)


U 0 = Ucalc


U c

Uc = - NA z+z- e2 [6 - 12 + 8 - 6 + 24 ]

4πε0r 2 3 2 5

Uc = - NA z+z- e2 A

4πε0r

UC

Uc = - 6 NA z+z- e2 + 12 NA z+z+ e2 - + -

4πε0r 2 r4πε0

r = d

NaCl(s)

A o












Uc = - NA z+z- e2 A

4πε0rkJmol



NA = 6023 x 10 23


0 = 8854188 x 10 ndash12 C2Jm

Uo = - 6023 x 10 23mol-1(1) (1) (160218 x 10 ndash19 C)2(1748)



(C2Jm) m

NaCl(s)


UB


UB = NAB r n

n coef de Born







dr dr

U Calc = U0 + UB

B = z+z-e2A (r0n-1)

4πε0n



dU0= 0

dr (r= r0)


dr (r= r0) 4πε0 r02 r0

n+1


U Calc = U0 + UB

d Xm = m Xm-1

dX

UB = NAB

rn

r=r0

Sustituyendo B

UB = NA z+z-e2A

4πε0nr

Para NaCl





NA = 6023 x 10 23


0 = 8854188 x 10 ndash12 C2Jm

Sustituyendo B

B = z+z-e2A (r0n-1)

4πε0n

UB = 10789 kJmol

Ioacuten tipo Ej n

He-He LiH 5

Ne-Ne NaF MgO 7

NaCl 8


Kr-Kr RbBr AgBr 10

Xe-Xe CsI 12

promedio

Uo = Uc + UB


4πε0r 0 r0n


4πε0r 0 4πε0nr0

factorizando


4πε0r0 n





NA = 6023 x 10 23


0 = 8854188 x 10 ndash12 C2Jm


2 2

0 0 0

14

AN Az eU

r r

= minus minus

Born-Mayer

UB = NAB e (-r)




z+ y z- = 1

NA = 6023 x 10 23


0 = 8854188 x 10 ndash12 C2Jm




31309x10-20 8 (C2Jm) m


Uo = ndash 863192692 (78)


1000J


8

Ec Born-Lande


U0 para el NaCl

Uc =

ndash86319Hordmred

-788 kJmol 1001095

9585






_____________________________________________________________________________________________




AgCl


Cl + AE)



Uo = Uc + UB





3 2933x10-20

Uo = -71804 [kJmol]

Ec Born-Lande


2 2

6 8exp

4 r

AA A A


r r


Energiacutea de

London (UL)

6

C





D

r





J E Mayer (1933)

1( )


1 2

1 2

3

2 ( )

E EC

E E +minus =

+

23

4C E ++ + += 23


1 2

1 2

3

2 ( )

E EC

E E +minus =

+

23

4C E ++ + += 23


Donde





cristal


energiacuteas


r0 = 2772 x 10-10 m

Entonces

6

0

C


Para el AgClDatos

Polarizabilidad


αCl- = 345 x 10-30 m3




NA NA




dipolo-cuadrupolo D



dipolo-dipolo



AgCl es


UL=

8

D

r

2 2

6 8exp

4 r

AA A A


r r


2 2

6 8exp

4 r

AA A A


r r



4πε0r0


Uo = Uc + UB + UL




3 2933x10-20



Ec Born-Lande







_____________________________________________________________________________________________


Hordmred = -9108


-71804 KJmol -84304 KJmol -9108

788 956 100





α+

Li+ Be

2+ Cu

+ Zn

2+ Al

3+

0029 0008 043 029 0054

Na+ Mg

2+ Ag

+ Cd

2+ Ga

3+

018 0094 172 109 02

k+ Ca

2+ Au

+ Hg

2+ In

3+

084 047 188 125 073

Rb+ Sr

2+ Tl

3+

142 086 087

Cs+ Ba

2+

244 156

α-

H- F

- O

2-

102 105 392

Cl- S

2-

369 103

Br- Se

2-

481 106

I- Te

2-

716 141





max

1

2Z AU N hv=



(calc)

ΔH 298

(ciclo Born)

NaCl -766 -863 114 -25 8 -776 -787 CsI -580 -619 90 -54 3 -592 -602

AgI -795 -808 138 -128 3 -807 -891

CuBr -870 -925 114 -64 4 -882 -977

108772


CsI -580 es del 964

AgI -795 es del 892

CuBr -870 es del 89


788

AgCl -840 -820 1026 -125 3 -9108


U0 = - NAA z+z-e2 (1-1n)

40 (r0)


(r+ + r-)

1pm = 1x10-12 m = 10-2 Ả


(r+ + r-)






2

0

11

4 2

c aA

c a

Z ZN Ae vU

r r n

= minus

+

2

0

14 2

c aA

c a c a

Z ZN Ae vU

r r r r

= minus

+ +




2 2

0 0

11

4

AN AZ eU

r n

= minus minus

2 2

0 0 0

14

AN Az eU

r r

= minus minus

Born-Landeacute

Born-Mayer

ndash

ndash




bull n asymp 8

n2 =22 =1

bull2

04

AN e

108 c a

c a

Z ZU v KJ mol

r r=

+-

2

0

11

4 2

c aA

c a

Z ZN Ae vU

r r n

= minus

+

ndash


Ejemplo MgF2

kJmol

Mg(s) rarr Hatm 149

rarr EI1 +EI2 7375 y 1450


rarr -2AE -339 X2

rarr Hordmred= _____________________________________________________________________________________________


-29395



0 0

11

4 r

AN Az eU

n

= minus



Madelung A= 2408


Mg2+(6 86 pm) F- (3116pm) = 202 pm

0202 nm y 202 x 10-10 m

z = 1 2

n=8

U = -28996 kJmol

Zc = +2 Za = -1

v = 3


bull0211nm

U =- 307109 kJmol

Ejemplo MgF2

U0 = - 139 x 10-4 A z+z- (1-1n) Jmol

(r+ + r-) (m)

108 c a

c a

Z ZU v KJ mol

r r=

+-

-

Hordmred= -29395 100986 1045

Uc = -33139 kJmol y UB = 4142






(g)]





kJmol

2ΔH(Na+) 1220

iquestProceso real

AE 1+2 (SO4-2

) = ΔH(SO4-2)

Hordmred

____________________________________________________________________________________



-2)


-2)

kJmol

2ΔH(K+) 1028

iquestProceso real

AE 1+2 (SO4-2

) = ΔH(SO4-2)

Hordmred

____________________________________________________________________________________



-2)


-2)


r(Na+) = 116 pm = 0116 nm r(K+) = 152 pm = 0152 nm

Regla de los radios





6 oct entre 0414 y 0732


12 dodecaedro 1








Reglas de Fajans











EjemploNaX

F Cl Br I kJmol

ΔHatm Na

EI

ΔHatm X

-AE

ΔHred

-569 -410 -360 -288 ΔHf

Ejemplo MI

Li Na K Rb Cs kJmol

161

526

108

502

90

424

82

408

78

381

ΔHatm M

EI

106

-301

106

-301

106

-301

106

-301

106

-301

ΔHatm I

-AE

ΔHred

-271 -288 -328 -329 -337 ΔHf





F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________







F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf


T = 298 K










Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf





Gf = -531- 298(-105) kJmol

1000

Gf = -531 + 313 = -4997 kJmol









(g) AE2

Mg 2+(g) + O2minus


___________________________________________


149 kJmol

738

1450

249

-142

+844

-610

Extra

GASTO

De ENERGIacuteA








____________________________________________


149 kJmol

738

249

-142

108 c a

c a

Z ZU v KJ mol

r r=

+-



y para el O-






Hatm

EI1

EI2= 4567

Hatm

AE1

Hordmred=

___________________________________________





U 0 = Ucalc


U c

Uc = - NA z+z- e2 [6 - 12 + 8 - 6 + 24 ]

4πε0r 2 3 2 5

Uc = - NA z+z- e2 A

4πε0r

UC

Uc = - 6 NA z+z- e2 + 12 NA z+z+ e2 - + -

4πε0r 2 r4πε0

r = d

NaCl(s)

A o












Uc = - NA z+z- e2 A

4πε0rkJmol



NA = 6023 x 10 23


0 = 8854188 x 10 ndash12 C2Jm

Uo = - 6023 x 10 23mol-1(1) (1) (160218 x 10 ndash19 C)2(1748)



(C2Jm) m

NaCl(s)


UB


UB = NAB r n

n coef de Born







dr dr

U Calc = U0 + UB

B = z+z-e2A (r0n-1)

4πε0n



dU0= 0

dr (r= r0)


dr (r= r0) 4πε0 r02 r0

n+1


U Calc = U0 + UB

d Xm = m Xm-1

dX

UB = NAB

rn

r=r0

Sustituyendo B

UB = NA z+z-e2A

4πε0nr

Para NaCl





NA = 6023 x 10 23


0 = 8854188 x 10 ndash12 C2Jm

Sustituyendo B

B = z+z-e2A (r0n-1)

4πε0n

UB = 10789 kJmol

Ioacuten tipo Ej n

He-He LiH 5

Ne-Ne NaF MgO 7

NaCl 8


Kr-Kr RbBr AgBr 10

Xe-Xe CsI 12

promedio

Uo = Uc + UB


4πε0r 0 r0n


4πε0r 0 4πε0nr0

factorizando


4πε0r0 n





NA = 6023 x 10 23


0 = 8854188 x 10 ndash12 C2Jm


2 2

0 0 0

14

AN Az eU

r r

= minus minus

Born-Mayer

UB = NAB e (-r)




z+ y z- = 1

NA = 6023 x 10 23


0 = 8854188 x 10 ndash12 C2Jm




31309x10-20 8 (C2Jm) m


Uo = ndash 863192692 (78)


1000J


8

Ec Born-Lande


U0 para el NaCl

Uc =

ndash86319Hordmred

-788 kJmol 1001095

9585






_____________________________________________________________________________________________




AgCl


Cl + AE)



Uo = Uc + UB





3 2933x10-20

Uo = -71804 [kJmol]

Ec Born-Lande


2 2

6 8exp

4 r

AA A A


r r


Energiacutea de

London (UL)

6

C





D

r





J E Mayer (1933)

1( )


1 2

1 2

3

2 ( )

E EC

E E +minus =

+

23

4C E ++ + += 23


1 2

1 2

3

2 ( )

E EC

E E +minus =

+

23

4C E ++ + += 23


Donde





cristal


energiacuteas


r0 = 2772 x 10-10 m

Entonces

6

0

C


Para el AgClDatos

Polarizabilidad


αCl- = 345 x 10-30 m3




NA NA




dipolo-cuadrupolo D



dipolo-dipolo



AgCl es


UL=

8

D

r

2 2

6 8exp

4 r

AA A A


r r


2 2

6 8exp

4 r

AA A A


r r



4πε0r0


Uo = Uc + UB + UL




3 2933x10-20



Ec Born-Lande







_____________________________________________________________________________________________


Hordmred = -9108


-71804 KJmol -84304 KJmol -9108

788 956 100





α+

Li+ Be

2+ Cu

+ Zn

2+ Al

3+

0029 0008 043 029 0054

Na+ Mg

2+ Ag

+ Cd

2+ Ga

3+

018 0094 172 109 02

k+ Ca

2+ Au

+ Hg

2+ In

3+

084 047 188 125 073

Rb+ Sr

2+ Tl

3+

142 086 087

Cs+ Ba

2+

244 156

α-

H- F

- O

2-

102 105 392

Cl- S

2-

369 103

Br- Se

2-

481 106

I- Te

2-

716 141





max

1

2Z AU N hv=



(calc)

ΔH 298

(ciclo Born)

NaCl -766 -863 114 -25 8 -776 -787 CsI -580 -619 90 -54 3 -592 -602

AgI -795 -808 138 -128 3 -807 -891

CuBr -870 -925 114 -64 4 -882 -977

108772


CsI -580 es del 964

AgI -795 es del 892

CuBr -870 es del 89


788

AgCl -840 -820 1026 -125 3 -9108


U0 = - NAA z+z-e2 (1-1n)

40 (r0)


(r+ + r-)

1pm = 1x10-12 m = 10-2 Ả


(r+ + r-)






2

0

11

4 2

c aA

c a

Z ZN Ae vU

r r n

= minus

+

2

0

14 2

c aA

c a c a

Z ZN Ae vU

r r r r

= minus

+ +




2 2

0 0

11

4

AN AZ eU

r n

= minus minus

2 2

0 0 0

14

AN Az eU

r r

= minus minus

Born-Landeacute

Born-Mayer

ndash

ndash




bull n asymp 8

n2 =22 =1

bull2

04

AN e

108 c a

c a

Z ZU v KJ mol

r r=

+-

2

0

11

4 2

c aA

c a

Z ZN Ae vU

r r n

= minus

+

ndash


Ejemplo MgF2

kJmol

Mg(s) rarr Hatm 149

rarr EI1 +EI2 7375 y 1450


rarr -2AE -339 X2

rarr Hordmred= _____________________________________________________________________________________________


-29395



0 0

11

4 r

AN Az eU

n

= minus



Madelung A= 2408


Mg2+(6 86 pm) F- (3116pm) = 202 pm

0202 nm y 202 x 10-10 m

z = 1 2

n=8

U = -28996 kJmol

Zc = +2 Za = -1

v = 3


bull0211nm

U =- 307109 kJmol

Ejemplo MgF2

U0 = - 139 x 10-4 A z+z- (1-1n) Jmol

(r+ + r-) (m)

108 c a

c a

Z ZU v KJ mol

r r=

+-

-

Hordmred= -29395 100986 1045

Uc = -33139 kJmol y UB = 4142






(g)]





kJmol

2ΔH(Na+) 1220

iquestProceso real

AE 1+2 (SO4-2

) = ΔH(SO4-2)

Hordmred

____________________________________________________________________________________



-2)


-2)

kJmol

2ΔH(K+) 1028

iquestProceso real

AE 1+2 (SO4-2

) = ΔH(SO4-2)

Hordmred

____________________________________________________________________________________



-2)


-2)


r(Na+) = 116 pm = 0116 nm r(K+) = 152 pm = 0152 nm

Regla de los radios





6 oct entre 0414 y 0732


12 dodecaedro 1








Reglas de Fajans











EjemploNaX

F Cl Br I kJmol

ΔHatm Na

EI

ΔHatm X

-AE

ΔHred

-569 -410 -360 -288 ΔHf

Ejemplo MI

Li Na K Rb Cs kJmol

161

526

108

502

90

424

82

408

78

381

ΔHatm M

EI

106

-301

106

-301

106

-301

106

-301

106

-301

ΔHatm I

-AE

ΔHred

-271 -288 -328 -329 -337 ΔHf





F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________







F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf


T = 298 K










Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf





Gf = -531- 298(-105) kJmol

1000

Gf = -531 + 313 = -4997 kJmol









(g) AE2

Mg 2+(g) + O2minus


___________________________________________


149 kJmol

738

1450

249

-142

+844

-610

Extra

GASTO

De ENERGIacuteA








____________________________________________


149 kJmol

738

249

-142

108 c a

c a

Z ZU v KJ mol

r r=

+-



y para el O-






Hatm

EI1

EI2= 4567

Hatm

AE1

Hordmred=

___________________________________________





U c

Uc = - NA z+z- e2 [6 - 12 + 8 - 6 + 24 ]

4πε0r 2 3 2 5

Uc = - NA z+z- e2 A

4πε0r

UC

Uc = - 6 NA z+z- e2 + 12 NA z+z+ e2 - + -

4πε0r 2 r4πε0

r = d

NaCl(s)

A o












Uc = - NA z+z- e2 A

4πε0rkJmol



NA = 6023 x 10 23


0 = 8854188 x 10 ndash12 C2Jm

Uo = - 6023 x 10 23mol-1(1) (1) (160218 x 10 ndash19 C)2(1748)



(C2Jm) m

NaCl(s)


UB


UB = NAB r n

n coef de Born







dr dr

U Calc = U0 + UB

B = z+z-e2A (r0n-1)

4πε0n



dU0= 0

dr (r= r0)


dr (r= r0) 4πε0 r02 r0

n+1


U Calc = U0 + UB

d Xm = m Xm-1

dX

UB = NAB

rn

r=r0

Sustituyendo B

UB = NA z+z-e2A

4πε0nr

Para NaCl





NA = 6023 x 10 23


0 = 8854188 x 10 ndash12 C2Jm

Sustituyendo B

B = z+z-e2A (r0n-1)

4πε0n

UB = 10789 kJmol

Ioacuten tipo Ej n

He-He LiH 5

Ne-Ne NaF MgO 7

NaCl 8


Kr-Kr RbBr AgBr 10

Xe-Xe CsI 12

promedio

Uo = Uc + UB


4πε0r 0 r0n


4πε0r 0 4πε0nr0

factorizando


4πε0r0 n





NA = 6023 x 10 23


0 = 8854188 x 10 ndash12 C2Jm


2 2

0 0 0

14

AN Az eU

r r

= minus minus

Born-Mayer

UB = NAB e (-r)




z+ y z- = 1

NA = 6023 x 10 23


0 = 8854188 x 10 ndash12 C2Jm




31309x10-20 8 (C2Jm) m


Uo = ndash 863192692 (78)


1000J


8

Ec Born-Lande


U0 para el NaCl

Uc =

ndash86319Hordmred

-788 kJmol 1001095

9585






_____________________________________________________________________________________________




AgCl


Cl + AE)



Uo = Uc + UB





3 2933x10-20

Uo = -71804 [kJmol]

Ec Born-Lande


2 2

6 8exp

4 r

AA A A


r r


Energiacutea de

London (UL)

6

C





D

r





J E Mayer (1933)

1( )


1 2

1 2

3

2 ( )

E EC

E E +minus =

+

23

4C E ++ + += 23


1 2

1 2

3

2 ( )

E EC

E E +minus =

+

23

4C E ++ + += 23


Donde





cristal


energiacuteas


r0 = 2772 x 10-10 m

Entonces

6

0

C


Para el AgClDatos

Polarizabilidad


αCl- = 345 x 10-30 m3




NA NA




dipolo-cuadrupolo D



dipolo-dipolo



AgCl es


UL=

8

D

r

2 2

6 8exp

4 r

AA A A


r r


2 2

6 8exp

4 r

AA A A


r r



4πε0r0


Uo = Uc + UB + UL




3 2933x10-20



Ec Born-Lande







_____________________________________________________________________________________________


Hordmred = -9108


-71804 KJmol -84304 KJmol -9108

788 956 100





α+

Li+ Be

2+ Cu

+ Zn

2+ Al

3+

0029 0008 043 029 0054

Na+ Mg

2+ Ag

+ Cd

2+ Ga

3+

018 0094 172 109 02

k+ Ca

2+ Au

+ Hg

2+ In

3+

084 047 188 125 073

Rb+ Sr

2+ Tl

3+

142 086 087

Cs+ Ba

2+

244 156

α-

H- F

- O

2-

102 105 392

Cl- S

2-

369 103

Br- Se

2-

481 106

I- Te

2-

716 141





max

1

2Z AU N hv=



(calc)

ΔH 298

(ciclo Born)

NaCl -766 -863 114 -25 8 -776 -787 CsI -580 -619 90 -54 3 -592 -602

AgI -795 -808 138 -128 3 -807 -891

CuBr -870 -925 114 -64 4 -882 -977

108772


CsI -580 es del 964

AgI -795 es del 892

CuBr -870 es del 89


788

AgCl -840 -820 1026 -125 3 -9108


U0 = - NAA z+z-e2 (1-1n)

40 (r0)


(r+ + r-)

1pm = 1x10-12 m = 10-2 Ả


(r+ + r-)






2

0

11

4 2

c aA

c a

Z ZN Ae vU

r r n

= minus

+

2

0

14 2

c aA

c a c a

Z ZN Ae vU

r r r r

= minus

+ +




2 2

0 0

11

4

AN AZ eU

r n

= minus minus

2 2

0 0 0

14

AN Az eU

r r

= minus minus

Born-Landeacute

Born-Mayer

ndash

ndash




bull n asymp 8

n2 =22 =1

bull2

04

AN e

108 c a

c a

Z ZU v KJ mol

r r=

+-

2

0

11

4 2

c aA

c a

Z ZN Ae vU

r r n

= minus

+

ndash


Ejemplo MgF2

kJmol

Mg(s) rarr Hatm 149

rarr EI1 +EI2 7375 y 1450


rarr -2AE -339 X2

rarr Hordmred= _____________________________________________________________________________________________


-29395



0 0

11

4 r

AN Az eU

n

= minus



Madelung A= 2408


Mg2+(6 86 pm) F- (3116pm) = 202 pm

0202 nm y 202 x 10-10 m

z = 1 2

n=8

U = -28996 kJmol

Zc = +2 Za = -1

v = 3


bull0211nm

U =- 307109 kJmol

Ejemplo MgF2

U0 = - 139 x 10-4 A z+z- (1-1n) Jmol

(r+ + r-) (m)

108 c a

c a

Z ZU v KJ mol

r r=

+-

-

Hordmred= -29395 100986 1045

Uc = -33139 kJmol y UB = 4142






(g)]





kJmol

2ΔH(Na+) 1220

iquestProceso real

AE 1+2 (SO4-2

) = ΔH(SO4-2)

Hordmred

____________________________________________________________________________________



-2)


-2)

kJmol

2ΔH(K+) 1028

iquestProceso real

AE 1+2 (SO4-2

) = ΔH(SO4-2)

Hordmred

____________________________________________________________________________________



-2)


-2)


r(Na+) = 116 pm = 0116 nm r(K+) = 152 pm = 0152 nm

Regla de los radios





6 oct entre 0414 y 0732


12 dodecaedro 1








Reglas de Fajans











EjemploNaX

F Cl Br I kJmol

ΔHatm Na

EI

ΔHatm X

-AE

ΔHred

-569 -410 -360 -288 ΔHf

Ejemplo MI

Li Na K Rb Cs kJmol

161

526

108

502

90

424

82

408

78

381

ΔHatm M

EI

106

-301

106

-301

106

-301

106

-301

106

-301

ΔHatm I

-AE

ΔHred

-271 -288 -328 -329 -337 ΔHf





F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________







F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf


T = 298 K










Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf





Gf = -531- 298(-105) kJmol

1000

Gf = -531 + 313 = -4997 kJmol









(g) AE2

Mg 2+(g) + O2minus


___________________________________________


149 kJmol

738

1450

249

-142

+844

-610

Extra

GASTO

De ENERGIacuteA








____________________________________________


149 kJmol

738

249

-142

108 c a

c a

Z ZU v KJ mol

r r=

+-



y para el O-






Hatm

EI1

EI2= 4567

Hatm

AE1

Hordmred=

___________________________________________





A o












Uc = - NA z+z- e2 A

4πε0rkJmol



NA = 6023 x 10 23


0 = 8854188 x 10 ndash12 C2Jm

Uo = - 6023 x 10 23mol-1(1) (1) (160218 x 10 ndash19 C)2(1748)



(C2Jm) m

NaCl(s)


UB


UB = NAB r n

n coef de Born







dr dr

U Calc = U0 + UB

B = z+z-e2A (r0n-1)

4πε0n



dU0= 0

dr (r= r0)


dr (r= r0) 4πε0 r02 r0

n+1


U Calc = U0 + UB

d Xm = m Xm-1

dX

UB = NAB

rn

r=r0

Sustituyendo B

UB = NA z+z-e2A

4πε0nr

Para NaCl





NA = 6023 x 10 23


0 = 8854188 x 10 ndash12 C2Jm

Sustituyendo B

B = z+z-e2A (r0n-1)

4πε0n

UB = 10789 kJmol

Ioacuten tipo Ej n

He-He LiH 5

Ne-Ne NaF MgO 7

NaCl 8


Kr-Kr RbBr AgBr 10

Xe-Xe CsI 12

promedio

Uo = Uc + UB


4πε0r 0 r0n


4πε0r 0 4πε0nr0

factorizando


4πε0r0 n





NA = 6023 x 10 23


0 = 8854188 x 10 ndash12 C2Jm


2 2

0 0 0

14

AN Az eU

r r

= minus minus

Born-Mayer

UB = NAB e (-r)




z+ y z- = 1

NA = 6023 x 10 23


0 = 8854188 x 10 ndash12 C2Jm




31309x10-20 8 (C2Jm) m


Uo = ndash 863192692 (78)


1000J


8

Ec Born-Lande


U0 para el NaCl

Uc =

ndash86319Hordmred

-788 kJmol 1001095

9585






_____________________________________________________________________________________________




AgCl


Cl + AE)



Uo = Uc + UB





3 2933x10-20

Uo = -71804 [kJmol]

Ec Born-Lande


2 2

6 8exp

4 r

AA A A


r r


Energiacutea de

London (UL)

6

C





D

r





J E Mayer (1933)

1( )


1 2

1 2

3

2 ( )

E EC

E E +minus =

+

23

4C E ++ + += 23


1 2

1 2

3

2 ( )

E EC

E E +minus =

+

23

4C E ++ + += 23


Donde





cristal


energiacuteas


r0 = 2772 x 10-10 m

Entonces

6

0

C


Para el AgClDatos

Polarizabilidad


αCl- = 345 x 10-30 m3




NA NA




dipolo-cuadrupolo D



dipolo-dipolo



AgCl es


UL=

8

D

r

2 2

6 8exp

4 r

AA A A


r r


2 2

6 8exp

4 r

AA A A


r r



4πε0r0


Uo = Uc + UB + UL




3 2933x10-20



Ec Born-Lande







_____________________________________________________________________________________________


Hordmred = -9108


-71804 KJmol -84304 KJmol -9108

788 956 100





α+

Li+ Be

2+ Cu

+ Zn

2+ Al

3+

0029 0008 043 029 0054

Na+ Mg

2+ Ag

+ Cd

2+ Ga

3+

018 0094 172 109 02

k+ Ca

2+ Au

+ Hg

2+ In

3+

084 047 188 125 073

Rb+ Sr

2+ Tl

3+

142 086 087

Cs+ Ba

2+

244 156

α-

H- F

- O

2-

102 105 392

Cl- S

2-

369 103

Br- Se

2-

481 106

I- Te

2-

716 141





max

1

2Z AU N hv=



(calc)

ΔH 298

(ciclo Born)

NaCl -766 -863 114 -25 8 -776 -787 CsI -580 -619 90 -54 3 -592 -602

AgI -795 -808 138 -128 3 -807 -891

CuBr -870 -925 114 -64 4 -882 -977

108772


CsI -580 es del 964

AgI -795 es del 892

CuBr -870 es del 89


788

AgCl -840 -820 1026 -125 3 -9108


U0 = - NAA z+z-e2 (1-1n)

40 (r0)


(r+ + r-)

1pm = 1x10-12 m = 10-2 Ả


(r+ + r-)






2

0

11

4 2

c aA

c a

Z ZN Ae vU

r r n

= minus

+

2

0

14 2

c aA

c a c a

Z ZN Ae vU

r r r r

= minus

+ +




2 2

0 0

11

4

AN AZ eU

r n

= minus minus

2 2

0 0 0

14

AN Az eU

r r

= minus minus

Born-Landeacute

Born-Mayer

ndash

ndash




bull n asymp 8

n2 =22 =1

bull2

04

AN e

108 c a

c a

Z ZU v KJ mol

r r=

+-

2

0

11

4 2

c aA

c a

Z ZN Ae vU

r r n

= minus

+

ndash


Ejemplo MgF2

kJmol

Mg(s) rarr Hatm 149

rarr EI1 +EI2 7375 y 1450


rarr -2AE -339 X2

rarr Hordmred= _____________________________________________________________________________________________


-29395



0 0

11

4 r

AN Az eU

n

= minus



Madelung A= 2408


Mg2+(6 86 pm) F- (3116pm) = 202 pm

0202 nm y 202 x 10-10 m

z = 1 2

n=8

U = -28996 kJmol

Zc = +2 Za = -1

v = 3


bull0211nm

U =- 307109 kJmol

Ejemplo MgF2

U0 = - 139 x 10-4 A z+z- (1-1n) Jmol

(r+ + r-) (m)

108 c a

c a

Z ZU v KJ mol

r r=

+-

-

Hordmred= -29395 100986 1045

Uc = -33139 kJmol y UB = 4142






(g)]





kJmol

2ΔH(Na+) 1220

iquestProceso real

AE 1+2 (SO4-2

) = ΔH(SO4-2)

Hordmred

____________________________________________________________________________________



-2)


-2)

kJmol

2ΔH(K+) 1028

iquestProceso real

AE 1+2 (SO4-2

) = ΔH(SO4-2)

Hordmred

____________________________________________________________________________________



-2)


-2)


r(Na+) = 116 pm = 0116 nm r(K+) = 152 pm = 0152 nm

Regla de los radios





6 oct entre 0414 y 0732


12 dodecaedro 1








Reglas de Fajans











EjemploNaX

F Cl Br I kJmol

ΔHatm Na

EI

ΔHatm X

-AE

ΔHred

-569 -410 -360 -288 ΔHf

Ejemplo MI

Li Na K Rb Cs kJmol

161

526

108

502

90

424

82

408

78

381

ΔHatm M

EI

106

-301

106

-301

106

-301

106

-301

106

-301

ΔHatm I

-AE

ΔHred

-271 -288 -328 -329 -337 ΔHf





F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________







F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf


T = 298 K










Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf





Gf = -531- 298(-105) kJmol

1000

Gf = -531 + 313 = -4997 kJmol









(g) AE2

Mg 2+(g) + O2minus


___________________________________________


149 kJmol

738

1450

249

-142

+844

-610

Extra

GASTO

De ENERGIacuteA








____________________________________________


149 kJmol

738

249

-142

108 c a

c a

Z ZU v KJ mol

r r=

+-



y para el O-






Hatm

EI1

EI2= 4567

Hatm

AE1

Hordmred=

___________________________________________














Uc = - NA z+z- e2 A

4πε0rkJmol



NA = 6023 x 10 23


0 = 8854188 x 10 ndash12 C2Jm

Uo = - 6023 x 10 23mol-1(1) (1) (160218 x 10 ndash19 C)2(1748)



(C2Jm) m

NaCl(s)


UB


UB = NAB r n

n coef de Born







dr dr

U Calc = U0 + UB

B = z+z-e2A (r0n-1)

4πε0n



dU0= 0

dr (r= r0)


dr (r= r0) 4πε0 r02 r0

n+1


U Calc = U0 + UB

d Xm = m Xm-1

dX

UB = NAB

rn

r=r0

Sustituyendo B

UB = NA z+z-e2A

4πε0nr

Para NaCl





NA = 6023 x 10 23


0 = 8854188 x 10 ndash12 C2Jm

Sustituyendo B

B = z+z-e2A (r0n-1)

4πε0n

UB = 10789 kJmol

Ioacuten tipo Ej n

He-He LiH 5

Ne-Ne NaF MgO 7

NaCl 8


Kr-Kr RbBr AgBr 10

Xe-Xe CsI 12

promedio

Uo = Uc + UB


4πε0r 0 r0n


4πε0r 0 4πε0nr0

factorizando


4πε0r0 n





NA = 6023 x 10 23


0 = 8854188 x 10 ndash12 C2Jm


2 2

0 0 0

14

AN Az eU

r r

= minus minus

Born-Mayer

UB = NAB e (-r)




z+ y z- = 1

NA = 6023 x 10 23


0 = 8854188 x 10 ndash12 C2Jm




31309x10-20 8 (C2Jm) m


Uo = ndash 863192692 (78)


1000J


8

Ec Born-Lande


U0 para el NaCl

Uc =

ndash86319Hordmred

-788 kJmol 1001095

9585






_____________________________________________________________________________________________




AgCl


Cl + AE)



Uo = Uc + UB





3 2933x10-20

Uo = -71804 [kJmol]

Ec Born-Lande


2 2

6 8exp

4 r

AA A A


r r


Energiacutea de

London (UL)

6

C





D

r





J E Mayer (1933)

1( )


1 2

1 2

3

2 ( )

E EC

E E +minus =

+

23

4C E ++ + += 23


1 2

1 2

3

2 ( )

E EC

E E +minus =

+

23

4C E ++ + += 23


Donde





cristal


energiacuteas


r0 = 2772 x 10-10 m

Entonces

6

0

C


Para el AgClDatos

Polarizabilidad


αCl- = 345 x 10-30 m3




NA NA




dipolo-cuadrupolo D



dipolo-dipolo



AgCl es


UL=

8

D

r

2 2

6 8exp

4 r

AA A A


r r


2 2

6 8exp

4 r

AA A A


r r



4πε0r0


Uo = Uc + UB + UL




3 2933x10-20



Ec Born-Lande







_____________________________________________________________________________________________


Hordmred = -9108


-71804 KJmol -84304 KJmol -9108

788 956 100





α+

Li+ Be

2+ Cu

+ Zn

2+ Al

3+

0029 0008 043 029 0054

Na+ Mg

2+ Ag

+ Cd

2+ Ga

3+

018 0094 172 109 02

k+ Ca

2+ Au

+ Hg

2+ In

3+

084 047 188 125 073

Rb+ Sr

2+ Tl

3+

142 086 087

Cs+ Ba

2+

244 156

α-

H- F

- O

2-

102 105 392

Cl- S

2-

369 103

Br- Se

2-

481 106

I- Te

2-

716 141





max

1

2Z AU N hv=



(calc)

ΔH 298

(ciclo Born)

NaCl -766 -863 114 -25 8 -776 -787 CsI -580 -619 90 -54 3 -592 -602

AgI -795 -808 138 -128 3 -807 -891

CuBr -870 -925 114 -64 4 -882 -977

108772


CsI -580 es del 964

AgI -795 es del 892

CuBr -870 es del 89


788

AgCl -840 -820 1026 -125 3 -9108


U0 = - NAA z+z-e2 (1-1n)

40 (r0)


(r+ + r-)

1pm = 1x10-12 m = 10-2 Ả


(r+ + r-)






2

0

11

4 2

c aA

c a

Z ZN Ae vU

r r n

= minus

+

2

0

14 2

c aA

c a c a

Z ZN Ae vU

r r r r

= minus

+ +




2 2

0 0

11

4

AN AZ eU

r n

= minus minus

2 2

0 0 0

14

AN Az eU

r r

= minus minus

Born-Landeacute

Born-Mayer

ndash

ndash




bull n asymp 8

n2 =22 =1

bull2

04

AN e

108 c a

c a

Z ZU v KJ mol

r r=

+-

2

0

11

4 2

c aA

c a

Z ZN Ae vU

r r n

= minus

+

ndash


Ejemplo MgF2

kJmol

Mg(s) rarr Hatm 149

rarr EI1 +EI2 7375 y 1450


rarr -2AE -339 X2

rarr Hordmred= _____________________________________________________________________________________________


-29395



0 0

11

4 r

AN Az eU

n

= minus



Madelung A= 2408


Mg2+(6 86 pm) F- (3116pm) = 202 pm

0202 nm y 202 x 10-10 m

z = 1 2

n=8

U = -28996 kJmol

Zc = +2 Za = -1

v = 3


bull0211nm

U =- 307109 kJmol

Ejemplo MgF2

U0 = - 139 x 10-4 A z+z- (1-1n) Jmol

(r+ + r-) (m)

108 c a

c a

Z ZU v KJ mol

r r=

+-

-

Hordmred= -29395 100986 1045

Uc = -33139 kJmol y UB = 4142






(g)]





kJmol

2ΔH(Na+) 1220

iquestProceso real

AE 1+2 (SO4-2

) = ΔH(SO4-2)

Hordmred

____________________________________________________________________________________



-2)


-2)

kJmol

2ΔH(K+) 1028

iquestProceso real

AE 1+2 (SO4-2

) = ΔH(SO4-2)

Hordmred

____________________________________________________________________________________



-2)


-2)


r(Na+) = 116 pm = 0116 nm r(K+) = 152 pm = 0152 nm

Regla de los radios





6 oct entre 0414 y 0732


12 dodecaedro 1








Reglas de Fajans











EjemploNaX

F Cl Br I kJmol

ΔHatm Na

EI

ΔHatm X

-AE

ΔHred

-569 -410 -360 -288 ΔHf

Ejemplo MI

Li Na K Rb Cs kJmol

161

526

108

502

90

424

82

408

78

381

ΔHatm M

EI

106

-301

106

-301

106

-301

106

-301

106

-301

ΔHatm I

-AE

ΔHred

-271 -288 -328 -329 -337 ΔHf





F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________







F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf


T = 298 K










Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf





Gf = -531- 298(-105) kJmol

1000

Gf = -531 + 313 = -4997 kJmol









(g) AE2

Mg 2+(g) + O2minus


___________________________________________


149 kJmol

738

1450

249

-142

+844

-610

Extra

GASTO

De ENERGIacuteA








____________________________________________


149 kJmol

738

249

-142

108 c a

c a

Z ZU v KJ mol

r r=

+-



y para el O-






Hatm

EI1

EI2= 4567

Hatm

AE1

Hordmred=

___________________________________________





UB


UB = NAB r n

n coef de Born







dr dr

U Calc = U0 + UB

B = z+z-e2A (r0n-1)

4πε0n



dU0= 0

dr (r= r0)


dr (r= r0) 4πε0 r02 r0

n+1


U Calc = U0 + UB

d Xm = m Xm-1

dX

UB = NAB

rn

r=r0

Sustituyendo B

UB = NA z+z-e2A

4πε0nr

Para NaCl





NA = 6023 x 10 23


0 = 8854188 x 10 ndash12 C2Jm

Sustituyendo B

B = z+z-e2A (r0n-1)

4πε0n

UB = 10789 kJmol

Ioacuten tipo Ej n

He-He LiH 5

Ne-Ne NaF MgO 7

NaCl 8


Kr-Kr RbBr AgBr 10

Xe-Xe CsI 12

promedio

Uo = Uc + UB


4πε0r 0 r0n


4πε0r 0 4πε0nr0

factorizando


4πε0r0 n





NA = 6023 x 10 23


0 = 8854188 x 10 ndash12 C2Jm


2 2

0 0 0

14

AN Az eU

r r

= minus minus

Born-Mayer

UB = NAB e (-r)




z+ y z- = 1

NA = 6023 x 10 23


0 = 8854188 x 10 ndash12 C2Jm




31309x10-20 8 (C2Jm) m


Uo = ndash 863192692 (78)


1000J


8

Ec Born-Lande


U0 para el NaCl

Uc =

ndash86319Hordmred

-788 kJmol 1001095

9585






_____________________________________________________________________________________________




AgCl


Cl + AE)



Uo = Uc + UB





3 2933x10-20

Uo = -71804 [kJmol]

Ec Born-Lande


2 2

6 8exp

4 r

AA A A


r r


Energiacutea de

London (UL)

6

C





D

r





J E Mayer (1933)

1( )


1 2

1 2

3

2 ( )

E EC

E E +minus =

+

23

4C E ++ + += 23


1 2

1 2

3

2 ( )

E EC

E E +minus =

+

23

4C E ++ + += 23


Donde





cristal


energiacuteas


r0 = 2772 x 10-10 m

Entonces

6

0

C


Para el AgClDatos

Polarizabilidad


αCl- = 345 x 10-30 m3




NA NA




dipolo-cuadrupolo D



dipolo-dipolo



AgCl es


UL=

8

D

r

2 2

6 8exp

4 r

AA A A


r r


2 2

6 8exp

4 r

AA A A


r r



4πε0r0


Uo = Uc + UB + UL




3 2933x10-20



Ec Born-Lande







_____________________________________________________________________________________________


Hordmred = -9108


-71804 KJmol -84304 KJmol -9108

788 956 100





α+

Li+ Be

2+ Cu

+ Zn

2+ Al

3+

0029 0008 043 029 0054

Na+ Mg

2+ Ag

+ Cd

2+ Ga

3+

018 0094 172 109 02

k+ Ca

2+ Au

+ Hg

2+ In

3+

084 047 188 125 073

Rb+ Sr

2+ Tl

3+

142 086 087

Cs+ Ba

2+

244 156

α-

H- F

- O

2-

102 105 392

Cl- S

2-

369 103

Br- Se

2-

481 106

I- Te

2-

716 141





max

1

2Z AU N hv=



(calc)

ΔH 298

(ciclo Born)

NaCl -766 -863 114 -25 8 -776 -787 CsI -580 -619 90 -54 3 -592 -602

AgI -795 -808 138 -128 3 -807 -891

CuBr -870 -925 114 -64 4 -882 -977

108772


CsI -580 es del 964

AgI -795 es del 892

CuBr -870 es del 89


788

AgCl -840 -820 1026 -125 3 -9108


U0 = - NAA z+z-e2 (1-1n)

40 (r0)


(r+ + r-)

1pm = 1x10-12 m = 10-2 Ả


(r+ + r-)






2

0

11

4 2

c aA

c a

Z ZN Ae vU

r r n

= minus

+

2

0

14 2

c aA

c a c a

Z ZN Ae vU

r r r r

= minus

+ +




2 2

0 0

11

4

AN AZ eU

r n

= minus minus

2 2

0 0 0

14

AN Az eU

r r

= minus minus

Born-Landeacute

Born-Mayer

ndash

ndash




bull n asymp 8

n2 =22 =1

bull2

04

AN e

108 c a

c a

Z ZU v KJ mol

r r=

+-

2

0

11

4 2

c aA

c a

Z ZN Ae vU

r r n

= minus

+

ndash


Ejemplo MgF2

kJmol

Mg(s) rarr Hatm 149

rarr EI1 +EI2 7375 y 1450


rarr -2AE -339 X2

rarr Hordmred= _____________________________________________________________________________________________


-29395



0 0

11

4 r

AN Az eU

n

= minus



Madelung A= 2408


Mg2+(6 86 pm) F- (3116pm) = 202 pm

0202 nm y 202 x 10-10 m

z = 1 2

n=8

U = -28996 kJmol

Zc = +2 Za = -1

v = 3


bull0211nm

U =- 307109 kJmol

Ejemplo MgF2

U0 = - 139 x 10-4 A z+z- (1-1n) Jmol

(r+ + r-) (m)

108 c a

c a

Z ZU v KJ mol

r r=

+-

-

Hordmred= -29395 100986 1045

Uc = -33139 kJmol y UB = 4142






(g)]





kJmol

2ΔH(Na+) 1220

iquestProceso real

AE 1+2 (SO4-2

) = ΔH(SO4-2)

Hordmred

____________________________________________________________________________________



-2)


-2)

kJmol

2ΔH(K+) 1028

iquestProceso real

AE 1+2 (SO4-2

) = ΔH(SO4-2)

Hordmred

____________________________________________________________________________________



-2)


-2)


r(Na+) = 116 pm = 0116 nm r(K+) = 152 pm = 0152 nm

Regla de los radios





6 oct entre 0414 y 0732


12 dodecaedro 1








Reglas de Fajans











EjemploNaX

F Cl Br I kJmol

ΔHatm Na

EI

ΔHatm X

-AE

ΔHred

-569 -410 -360 -288 ΔHf

Ejemplo MI

Li Na K Rb Cs kJmol

161

526

108

502

90

424

82

408

78

381

ΔHatm M

EI

106

-301

106

-301

106

-301

106

-301

106

-301

ΔHatm I

-AE

ΔHred

-271 -288 -328 -329 -337 ΔHf





F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________







F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf


T = 298 K










Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf





Gf = -531- 298(-105) kJmol

1000

Gf = -531 + 313 = -4997 kJmol









(g) AE2

Mg 2+(g) + O2minus


___________________________________________


149 kJmol

738

1450

249

-142

+844

-610

Extra

GASTO

De ENERGIacuteA








____________________________________________


149 kJmol

738

249

-142

108 c a

c a

Z ZU v KJ mol

r r=

+-



y para el O-






Hatm

EI1

EI2= 4567

Hatm

AE1

Hordmred=

___________________________________________





B = z+z-e2A (r0n-1)

4πε0n



dU0= 0

dr (r= r0)


dr (r= r0) 4πε0 r02 r0

n+1


U Calc = U0 + UB

d Xm = m Xm-1

dX

UB = NAB

rn

r=r0

Sustituyendo B

UB = NA z+z-e2A

4πε0nr

Para NaCl





NA = 6023 x 10 23


0 = 8854188 x 10 ndash12 C2Jm

Sustituyendo B

B = z+z-e2A (r0n-1)

4πε0n

UB = 10789 kJmol

Ioacuten tipo Ej n

He-He LiH 5

Ne-Ne NaF MgO 7

NaCl 8


Kr-Kr RbBr AgBr 10

Xe-Xe CsI 12

promedio

Uo = Uc + UB


4πε0r 0 r0n


4πε0r 0 4πε0nr0

factorizando


4πε0r0 n





NA = 6023 x 10 23


0 = 8854188 x 10 ndash12 C2Jm


2 2

0 0 0

14

AN Az eU

r r

= minus minus

Born-Mayer

UB = NAB e (-r)




z+ y z- = 1

NA = 6023 x 10 23


0 = 8854188 x 10 ndash12 C2Jm




31309x10-20 8 (C2Jm) m


Uo = ndash 863192692 (78)


1000J


8

Ec Born-Lande


U0 para el NaCl

Uc =

ndash86319Hordmred

-788 kJmol 1001095

9585






_____________________________________________________________________________________________




AgCl


Cl + AE)



Uo = Uc + UB





3 2933x10-20

Uo = -71804 [kJmol]

Ec Born-Lande


2 2

6 8exp

4 r

AA A A


r r


Energiacutea de

London (UL)

6

C





D

r





J E Mayer (1933)

1( )


1 2

1 2

3

2 ( )

E EC

E E +minus =

+

23

4C E ++ + += 23


1 2

1 2

3

2 ( )

E EC

E E +minus =

+

23

4C E ++ + += 23


Donde





cristal


energiacuteas


r0 = 2772 x 10-10 m

Entonces

6

0

C


Para el AgClDatos

Polarizabilidad


αCl- = 345 x 10-30 m3




NA NA




dipolo-cuadrupolo D



dipolo-dipolo



AgCl es


UL=

8

D

r

2 2

6 8exp

4 r

AA A A


r r


2 2

6 8exp

4 r

AA A A


r r



4πε0r0


Uo = Uc + UB + UL




3 2933x10-20



Ec Born-Lande







_____________________________________________________________________________________________


Hordmred = -9108


-71804 KJmol -84304 KJmol -9108

788 956 100





α+

Li+ Be

2+ Cu

+ Zn

2+ Al

3+

0029 0008 043 029 0054

Na+ Mg

2+ Ag

+ Cd

2+ Ga

3+

018 0094 172 109 02

k+ Ca

2+ Au

+ Hg

2+ In

3+

084 047 188 125 073

Rb+ Sr

2+ Tl

3+

142 086 087

Cs+ Ba

2+

244 156

α-

H- F

- O

2-

102 105 392

Cl- S

2-

369 103

Br- Se

2-

481 106

I- Te

2-

716 141





max

1

2Z AU N hv=



(calc)

ΔH 298

(ciclo Born)

NaCl -766 -863 114 -25 8 -776 -787 CsI -580 -619 90 -54 3 -592 -602

AgI -795 -808 138 -128 3 -807 -891

CuBr -870 -925 114 -64 4 -882 -977

108772


CsI -580 es del 964

AgI -795 es del 892

CuBr -870 es del 89


788

AgCl -840 -820 1026 -125 3 -9108


U0 = - NAA z+z-e2 (1-1n)

40 (r0)


(r+ + r-)

1pm = 1x10-12 m = 10-2 Ả


(r+ + r-)






2

0

11

4 2

c aA

c a

Z ZN Ae vU

r r n

= minus

+

2

0

14 2

c aA

c a c a

Z ZN Ae vU

r r r r

= minus

+ +




2 2

0 0

11

4

AN AZ eU

r n

= minus minus

2 2

0 0 0

14

AN Az eU

r r

= minus minus

Born-Landeacute

Born-Mayer

ndash

ndash




bull n asymp 8

n2 =22 =1

bull2

04

AN e

108 c a

c a

Z ZU v KJ mol

r r=

+-

2

0

11

4 2

c aA

c a

Z ZN Ae vU

r r n

= minus

+

ndash


Ejemplo MgF2

kJmol

Mg(s) rarr Hatm 149

rarr EI1 +EI2 7375 y 1450


rarr -2AE -339 X2

rarr Hordmred= _____________________________________________________________________________________________


-29395



0 0

11

4 r

AN Az eU

n

= minus



Madelung A= 2408


Mg2+(6 86 pm) F- (3116pm) = 202 pm

0202 nm y 202 x 10-10 m

z = 1 2

n=8

U = -28996 kJmol

Zc = +2 Za = -1

v = 3


bull0211nm

U =- 307109 kJmol

Ejemplo MgF2

U0 = - 139 x 10-4 A z+z- (1-1n) Jmol

(r+ + r-) (m)

108 c a

c a

Z ZU v KJ mol

r r=

+-

-

Hordmred= -29395 100986 1045

Uc = -33139 kJmol y UB = 4142






(g)]





kJmol

2ΔH(Na+) 1220

iquestProceso real

AE 1+2 (SO4-2

) = ΔH(SO4-2)

Hordmred

____________________________________________________________________________________



-2)


-2)

kJmol

2ΔH(K+) 1028

iquestProceso real

AE 1+2 (SO4-2

) = ΔH(SO4-2)

Hordmred

____________________________________________________________________________________



-2)


-2)


r(Na+) = 116 pm = 0116 nm r(K+) = 152 pm = 0152 nm

Regla de los radios





6 oct entre 0414 y 0732


12 dodecaedro 1








Reglas de Fajans











EjemploNaX

F Cl Br I kJmol

ΔHatm Na

EI

ΔHatm X

-AE

ΔHred

-569 -410 -360 -288 ΔHf

Ejemplo MI

Li Na K Rb Cs kJmol

161

526

108

502

90

424

82

408

78

381

ΔHatm M

EI

106

-301

106

-301

106

-301

106

-301

106

-301

ΔHatm I

-AE

ΔHred

-271 -288 -328 -329 -337 ΔHf





F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________







F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf


T = 298 K










Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf





Gf = -531- 298(-105) kJmol

1000

Gf = -531 + 313 = -4997 kJmol









(g) AE2

Mg 2+(g) + O2minus


___________________________________________


149 kJmol

738

1450

249

-142

+844

-610

Extra

GASTO

De ENERGIacuteA








____________________________________________


149 kJmol

738

249

-142

108 c a

c a

Z ZU v KJ mol

r r=

+-



y para el O-






Hatm

EI1

EI2= 4567

Hatm

AE1

Hordmred=

___________________________________________





UB = NAB

rn

r=r0

Sustituyendo B

UB = NA z+z-e2A

4πε0nr

Para NaCl





NA = 6023 x 10 23


0 = 8854188 x 10 ndash12 C2Jm

Sustituyendo B

B = z+z-e2A (r0n-1)

4πε0n

UB = 10789 kJmol

Ioacuten tipo Ej n

He-He LiH 5

Ne-Ne NaF MgO 7

NaCl 8


Kr-Kr RbBr AgBr 10

Xe-Xe CsI 12

promedio

Uo = Uc + UB


4πε0r 0 r0n


4πε0r 0 4πε0nr0

factorizando


4πε0r0 n





NA = 6023 x 10 23


0 = 8854188 x 10 ndash12 C2Jm


2 2

0 0 0

14

AN Az eU

r r

= minus minus

Born-Mayer

UB = NAB e (-r)




z+ y z- = 1

NA = 6023 x 10 23


0 = 8854188 x 10 ndash12 C2Jm




31309x10-20 8 (C2Jm) m


Uo = ndash 863192692 (78)


1000J


8

Ec Born-Lande


U0 para el NaCl

Uc =

ndash86319Hordmred

-788 kJmol 1001095

9585






_____________________________________________________________________________________________




AgCl


Cl + AE)



Uo = Uc + UB





3 2933x10-20

Uo = -71804 [kJmol]

Ec Born-Lande


2 2

6 8exp

4 r

AA A A


r r


Energiacutea de

London (UL)

6

C





D

r





J E Mayer (1933)

1( )


1 2

1 2

3

2 ( )

E EC

E E +minus =

+

23

4C E ++ + += 23


1 2

1 2

3

2 ( )

E EC

E E +minus =

+

23

4C E ++ + += 23


Donde





cristal


energiacuteas


r0 = 2772 x 10-10 m

Entonces

6

0

C


Para el AgClDatos

Polarizabilidad


αCl- = 345 x 10-30 m3




NA NA




dipolo-cuadrupolo D



dipolo-dipolo



AgCl es


UL=

8

D

r

2 2

6 8exp

4 r

AA A A


r r


2 2

6 8exp

4 r

AA A A


r r



4πε0r0


Uo = Uc + UB + UL




3 2933x10-20



Ec Born-Lande







_____________________________________________________________________________________________


Hordmred = -9108


-71804 KJmol -84304 KJmol -9108

788 956 100





α+

Li+ Be

2+ Cu

+ Zn

2+ Al

3+

0029 0008 043 029 0054

Na+ Mg

2+ Ag

+ Cd

2+ Ga

3+

018 0094 172 109 02

k+ Ca

2+ Au

+ Hg

2+ In

3+

084 047 188 125 073

Rb+ Sr

2+ Tl

3+

142 086 087

Cs+ Ba

2+

244 156

α-

H- F

- O

2-

102 105 392

Cl- S

2-

369 103

Br- Se

2-

481 106

I- Te

2-

716 141





max

1

2Z AU N hv=



(calc)

ΔH 298

(ciclo Born)

NaCl -766 -863 114 -25 8 -776 -787 CsI -580 -619 90 -54 3 -592 -602

AgI -795 -808 138 -128 3 -807 -891

CuBr -870 -925 114 -64 4 -882 -977

108772


CsI -580 es del 964

AgI -795 es del 892

CuBr -870 es del 89


788

AgCl -840 -820 1026 -125 3 -9108


U0 = - NAA z+z-e2 (1-1n)

40 (r0)


(r+ + r-)

1pm = 1x10-12 m = 10-2 Ả


(r+ + r-)






2

0

11

4 2

c aA

c a

Z ZN Ae vU

r r n

= minus

+

2

0

14 2

c aA

c a c a

Z ZN Ae vU

r r r r

= minus

+ +




2 2

0 0

11

4

AN AZ eU

r n

= minus minus

2 2

0 0 0

14

AN Az eU

r r

= minus minus

Born-Landeacute

Born-Mayer

ndash

ndash




bull n asymp 8

n2 =22 =1

bull2

04

AN e

108 c a

c a

Z ZU v KJ mol

r r=

+-

2

0

11

4 2

c aA

c a

Z ZN Ae vU

r r n

= minus

+

ndash


Ejemplo MgF2

kJmol

Mg(s) rarr Hatm 149

rarr EI1 +EI2 7375 y 1450


rarr -2AE -339 X2

rarr Hordmred= _____________________________________________________________________________________________


-29395



0 0

11

4 r

AN Az eU

n

= minus



Madelung A= 2408


Mg2+(6 86 pm) F- (3116pm) = 202 pm

0202 nm y 202 x 10-10 m

z = 1 2

n=8

U = -28996 kJmol

Zc = +2 Za = -1

v = 3


bull0211nm

U =- 307109 kJmol

Ejemplo MgF2

U0 = - 139 x 10-4 A z+z- (1-1n) Jmol

(r+ + r-) (m)

108 c a

c a

Z ZU v KJ mol

r r=

+-

-

Hordmred= -29395 100986 1045

Uc = -33139 kJmol y UB = 4142






(g)]





kJmol

2ΔH(Na+) 1220

iquestProceso real

AE 1+2 (SO4-2

) = ΔH(SO4-2)

Hordmred

____________________________________________________________________________________



-2)


-2)

kJmol

2ΔH(K+) 1028

iquestProceso real

AE 1+2 (SO4-2

) = ΔH(SO4-2)

Hordmred

____________________________________________________________________________________



-2)


-2)


r(Na+) = 116 pm = 0116 nm r(K+) = 152 pm = 0152 nm

Regla de los radios





6 oct entre 0414 y 0732


12 dodecaedro 1








Reglas de Fajans











EjemploNaX

F Cl Br I kJmol

ΔHatm Na

EI

ΔHatm X

-AE

ΔHred

-569 -410 -360 -288 ΔHf

Ejemplo MI

Li Na K Rb Cs kJmol

161

526

108

502

90

424

82

408

78

381

ΔHatm M

EI

106

-301

106

-301

106

-301

106

-301

106

-301

ΔHatm I

-AE

ΔHred

-271 -288 -328 -329 -337 ΔHf





F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________







F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf


T = 298 K










Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf





Gf = -531- 298(-105) kJmol

1000

Gf = -531 + 313 = -4997 kJmol









(g) AE2

Mg 2+(g) + O2minus


___________________________________________


149 kJmol

738

1450

249

-142

+844

-610

Extra

GASTO

De ENERGIacuteA








____________________________________________


149 kJmol

738

249

-142

108 c a

c a

Z ZU v KJ mol

r r=

+-



y para el O-






Hatm

EI1

EI2= 4567

Hatm

AE1

Hordmred=

___________________________________________





Ioacuten tipo Ej n

He-He LiH 5

Ne-Ne NaF MgO 7

NaCl 8


Kr-Kr RbBr AgBr 10

Xe-Xe CsI 12

promedio

Uo = Uc + UB


4πε0r 0 r0n


4πε0r 0 4πε0nr0

factorizando


4πε0r0 n





NA = 6023 x 10 23


0 = 8854188 x 10 ndash12 C2Jm


2 2

0 0 0

14

AN Az eU

r r

= minus minus

Born-Mayer

UB = NAB e (-r)




z+ y z- = 1

NA = 6023 x 10 23


0 = 8854188 x 10 ndash12 C2Jm




31309x10-20 8 (C2Jm) m


Uo = ndash 863192692 (78)


1000J


8

Ec Born-Lande


U0 para el NaCl

Uc =

ndash86319Hordmred

-788 kJmol 1001095

9585






_____________________________________________________________________________________________




AgCl


Cl + AE)



Uo = Uc + UB





3 2933x10-20

Uo = -71804 [kJmol]

Ec Born-Lande


2 2

6 8exp

4 r

AA A A


r r


Energiacutea de

London (UL)

6

C





D

r





J E Mayer (1933)

1( )


1 2

1 2

3

2 ( )

E EC

E E +minus =

+

23

4C E ++ + += 23


1 2

1 2

3

2 ( )

E EC

E E +minus =

+

23

4C E ++ + += 23


Donde





cristal


energiacuteas


r0 = 2772 x 10-10 m

Entonces

6

0

C


Para el AgClDatos

Polarizabilidad


αCl- = 345 x 10-30 m3




NA NA




dipolo-cuadrupolo D



dipolo-dipolo



AgCl es


UL=

8

D

r

2 2

6 8exp

4 r

AA A A


r r


2 2

6 8exp

4 r

AA A A


r r



4πε0r0


Uo = Uc + UB + UL




3 2933x10-20



Ec Born-Lande







_____________________________________________________________________________________________


Hordmred = -9108


-71804 KJmol -84304 KJmol -9108

788 956 100





α+

Li+ Be

2+ Cu

+ Zn

2+ Al

3+

0029 0008 043 029 0054

Na+ Mg

2+ Ag

+ Cd

2+ Ga

3+

018 0094 172 109 02

k+ Ca

2+ Au

+ Hg

2+ In

3+

084 047 188 125 073

Rb+ Sr

2+ Tl

3+

142 086 087

Cs+ Ba

2+

244 156

α-

H- F

- O

2-

102 105 392

Cl- S

2-

369 103

Br- Se

2-

481 106

I- Te

2-

716 141





max

1

2Z AU N hv=



(calc)

ΔH 298

(ciclo Born)

NaCl -766 -863 114 -25 8 -776 -787 CsI -580 -619 90 -54 3 -592 -602

AgI -795 -808 138 -128 3 -807 -891

CuBr -870 -925 114 -64 4 -882 -977

108772


CsI -580 es del 964

AgI -795 es del 892

CuBr -870 es del 89


788

AgCl -840 -820 1026 -125 3 -9108


U0 = - NAA z+z-e2 (1-1n)

40 (r0)


(r+ + r-)

1pm = 1x10-12 m = 10-2 Ả


(r+ + r-)






2

0

11

4 2

c aA

c a

Z ZN Ae vU

r r n

= minus

+

2

0

14 2

c aA

c a c a

Z ZN Ae vU

r r r r

= minus

+ +




2 2

0 0

11

4

AN AZ eU

r n

= minus minus

2 2

0 0 0

14

AN Az eU

r r

= minus minus

Born-Landeacute

Born-Mayer

ndash

ndash




bull n asymp 8

n2 =22 =1

bull2

04

AN e

108 c a

c a

Z ZU v KJ mol

r r=

+-

2

0

11

4 2

c aA

c a

Z ZN Ae vU

r r n

= minus

+

ndash


Ejemplo MgF2

kJmol

Mg(s) rarr Hatm 149

rarr EI1 +EI2 7375 y 1450


rarr -2AE -339 X2

rarr Hordmred= _____________________________________________________________________________________________


-29395



0 0

11

4 r

AN Az eU

n

= minus



Madelung A= 2408


Mg2+(6 86 pm) F- (3116pm) = 202 pm

0202 nm y 202 x 10-10 m

z = 1 2

n=8

U = -28996 kJmol

Zc = +2 Za = -1

v = 3


bull0211nm

U =- 307109 kJmol

Ejemplo MgF2

U0 = - 139 x 10-4 A z+z- (1-1n) Jmol

(r+ + r-) (m)

108 c a

c a

Z ZU v KJ mol

r r=

+-

-

Hordmred= -29395 100986 1045

Uc = -33139 kJmol y UB = 4142






(g)]





kJmol

2ΔH(Na+) 1220

iquestProceso real

AE 1+2 (SO4-2

) = ΔH(SO4-2)

Hordmred

____________________________________________________________________________________



-2)


-2)

kJmol

2ΔH(K+) 1028

iquestProceso real

AE 1+2 (SO4-2

) = ΔH(SO4-2)

Hordmred

____________________________________________________________________________________



-2)


-2)


r(Na+) = 116 pm = 0116 nm r(K+) = 152 pm = 0152 nm

Regla de los radios





6 oct entre 0414 y 0732


12 dodecaedro 1








Reglas de Fajans











EjemploNaX

F Cl Br I kJmol

ΔHatm Na

EI

ΔHatm X

-AE

ΔHred

-569 -410 -360 -288 ΔHf

Ejemplo MI

Li Na K Rb Cs kJmol

161

526

108

502

90

424

82

408

78

381

ΔHatm M

EI

106

-301

106

-301

106

-301

106

-301

106

-301

ΔHatm I

-AE

ΔHred

-271 -288 -328 -329 -337 ΔHf





F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________







F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf


T = 298 K










Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf





Gf = -531- 298(-105) kJmol

1000

Gf = -531 + 313 = -4997 kJmol









(g) AE2

Mg 2+(g) + O2minus


___________________________________________


149 kJmol

738

1450

249

-142

+844

-610

Extra

GASTO

De ENERGIacuteA








____________________________________________


149 kJmol

738

249

-142

108 c a

c a

Z ZU v KJ mol

r r=

+-



y para el O-






Hatm

EI1

EI2= 4567

Hatm

AE1

Hordmred=

___________________________________________





Uo = Uc + UB


4πε0r 0 r0n


4πε0r 0 4πε0nr0

factorizando


4πε0r0 n





NA = 6023 x 10 23


0 = 8854188 x 10 ndash12 C2Jm


2 2

0 0 0

14

AN Az eU

r r

= minus minus

Born-Mayer

UB = NAB e (-r)




z+ y z- = 1

NA = 6023 x 10 23


0 = 8854188 x 10 ndash12 C2Jm




31309x10-20 8 (C2Jm) m


Uo = ndash 863192692 (78)


1000J


8

Ec Born-Lande


U0 para el NaCl

Uc =

ndash86319Hordmred

-788 kJmol 1001095

9585






_____________________________________________________________________________________________




AgCl


Cl + AE)



Uo = Uc + UB





3 2933x10-20

Uo = -71804 [kJmol]

Ec Born-Lande


2 2

6 8exp

4 r

AA A A


r r


Energiacutea de

London (UL)

6

C





D

r





J E Mayer (1933)

1( )


1 2

1 2

3

2 ( )

E EC

E E +minus =

+

23

4C E ++ + += 23


1 2

1 2

3

2 ( )

E EC

E E +minus =

+

23

4C E ++ + += 23


Donde





cristal


energiacuteas


r0 = 2772 x 10-10 m

Entonces

6

0

C


Para el AgClDatos

Polarizabilidad


αCl- = 345 x 10-30 m3




NA NA




dipolo-cuadrupolo D



dipolo-dipolo



AgCl es


UL=

8

D

r

2 2

6 8exp

4 r

AA A A


r r


2 2

6 8exp

4 r

AA A A


r r



4πε0r0


Uo = Uc + UB + UL




3 2933x10-20



Ec Born-Lande







_____________________________________________________________________________________________


Hordmred = -9108


-71804 KJmol -84304 KJmol -9108

788 956 100





α+

Li+ Be

2+ Cu

+ Zn

2+ Al

3+

0029 0008 043 029 0054

Na+ Mg

2+ Ag

+ Cd

2+ Ga

3+

018 0094 172 109 02

k+ Ca

2+ Au

+ Hg

2+ In

3+

084 047 188 125 073

Rb+ Sr

2+ Tl

3+

142 086 087

Cs+ Ba

2+

244 156

α-

H- F

- O

2-

102 105 392

Cl- S

2-

369 103

Br- Se

2-

481 106

I- Te

2-

716 141





max

1

2Z AU N hv=



(calc)

ΔH 298

(ciclo Born)

NaCl -766 -863 114 -25 8 -776 -787 CsI -580 -619 90 -54 3 -592 -602

AgI -795 -808 138 -128 3 -807 -891

CuBr -870 -925 114 -64 4 -882 -977

108772


CsI -580 es del 964

AgI -795 es del 892

CuBr -870 es del 89


788

AgCl -840 -820 1026 -125 3 -9108


U0 = - NAA z+z-e2 (1-1n)

40 (r0)


(r+ + r-)

1pm = 1x10-12 m = 10-2 Ả


(r+ + r-)






2

0

11

4 2

c aA

c a

Z ZN Ae vU

r r n

= minus

+

2

0

14 2

c aA

c a c a

Z ZN Ae vU

r r r r

= minus

+ +




2 2

0 0

11

4

AN AZ eU

r n

= minus minus

2 2

0 0 0

14

AN Az eU

r r

= minus minus

Born-Landeacute

Born-Mayer

ndash

ndash




bull n asymp 8

n2 =22 =1

bull2

04

AN e

108 c a

c a

Z ZU v KJ mol

r r=

+-

2

0

11

4 2

c aA

c a

Z ZN Ae vU

r r n

= minus

+

ndash


Ejemplo MgF2

kJmol

Mg(s) rarr Hatm 149

rarr EI1 +EI2 7375 y 1450


rarr -2AE -339 X2

rarr Hordmred= _____________________________________________________________________________________________


-29395



0 0

11

4 r

AN Az eU

n

= minus



Madelung A= 2408


Mg2+(6 86 pm) F- (3116pm) = 202 pm

0202 nm y 202 x 10-10 m

z = 1 2

n=8

U = -28996 kJmol

Zc = +2 Za = -1

v = 3


bull0211nm

U =- 307109 kJmol

Ejemplo MgF2

U0 = - 139 x 10-4 A z+z- (1-1n) Jmol

(r+ + r-) (m)

108 c a

c a

Z ZU v KJ mol

r r=

+-

-

Hordmred= -29395 100986 1045

Uc = -33139 kJmol y UB = 4142






(g)]





kJmol

2ΔH(Na+) 1220

iquestProceso real

AE 1+2 (SO4-2

) = ΔH(SO4-2)

Hordmred

____________________________________________________________________________________



-2)


-2)

kJmol

2ΔH(K+) 1028

iquestProceso real

AE 1+2 (SO4-2

) = ΔH(SO4-2)

Hordmred

____________________________________________________________________________________



-2)


-2)


r(Na+) = 116 pm = 0116 nm r(K+) = 152 pm = 0152 nm

Regla de los radios





6 oct entre 0414 y 0732


12 dodecaedro 1








Reglas de Fajans











EjemploNaX

F Cl Br I kJmol

ΔHatm Na

EI

ΔHatm X

-AE

ΔHred

-569 -410 -360 -288 ΔHf

Ejemplo MI

Li Na K Rb Cs kJmol

161

526

108

502

90

424

82

408

78

381

ΔHatm M

EI

106

-301

106

-301

106

-301

106

-301

106

-301

ΔHatm I

-AE

ΔHred

-271 -288 -328 -329 -337 ΔHf





F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________







F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf


T = 298 K










Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf





Gf = -531- 298(-105) kJmol

1000

Gf = -531 + 313 = -4997 kJmol









(g) AE2

Mg 2+(g) + O2minus


___________________________________________


149 kJmol

738

1450

249

-142

+844

-610

Extra

GASTO

De ENERGIacuteA








____________________________________________


149 kJmol

738

249

-142

108 c a

c a

Z ZU v KJ mol

r r=

+-



y para el O-






Hatm

EI1

EI2= 4567

Hatm

AE1

Hordmred=

___________________________________________








z+ y z- = 1

NA = 6023 x 10 23


0 = 8854188 x 10 ndash12 C2Jm




31309x10-20 8 (C2Jm) m


Uo = ndash 863192692 (78)


1000J


8

Ec Born-Lande


U0 para el NaCl

Uc =

ndash86319Hordmred

-788 kJmol 1001095

9585






_____________________________________________________________________________________________




AgCl


Cl + AE)



Uo = Uc + UB





3 2933x10-20

Uo = -71804 [kJmol]

Ec Born-Lande


2 2

6 8exp

4 r

AA A A


r r


Energiacutea de

London (UL)

6

C





D

r





J E Mayer (1933)

1( )


1 2

1 2

3

2 ( )

E EC

E E +minus =

+

23

4C E ++ + += 23


1 2

1 2

3

2 ( )

E EC

E E +minus =

+

23

4C E ++ + += 23


Donde





cristal


energiacuteas


r0 = 2772 x 10-10 m

Entonces

6

0

C


Para el AgClDatos

Polarizabilidad


αCl- = 345 x 10-30 m3




NA NA




dipolo-cuadrupolo D



dipolo-dipolo



AgCl es


UL=

8

D

r

2 2

6 8exp

4 r

AA A A


r r


2 2

6 8exp

4 r

AA A A


r r



4πε0r0


Uo = Uc + UB + UL




3 2933x10-20



Ec Born-Lande







_____________________________________________________________________________________________


Hordmred = -9108


-71804 KJmol -84304 KJmol -9108

788 956 100





α+

Li+ Be

2+ Cu

+ Zn

2+ Al

3+

0029 0008 043 029 0054

Na+ Mg

2+ Ag

+ Cd

2+ Ga

3+

018 0094 172 109 02

k+ Ca

2+ Au

+ Hg

2+ In

3+

084 047 188 125 073

Rb+ Sr

2+ Tl

3+

142 086 087

Cs+ Ba

2+

244 156

α-

H- F

- O

2-

102 105 392

Cl- S

2-

369 103

Br- Se

2-

481 106

I- Te

2-

716 141





max

1

2Z AU N hv=



(calc)

ΔH 298

(ciclo Born)

NaCl -766 -863 114 -25 8 -776 -787 CsI -580 -619 90 -54 3 -592 -602

AgI -795 -808 138 -128 3 -807 -891

CuBr -870 -925 114 -64 4 -882 -977

108772


CsI -580 es del 964

AgI -795 es del 892

CuBr -870 es del 89


788

AgCl -840 -820 1026 -125 3 -9108


U0 = - NAA z+z-e2 (1-1n)

40 (r0)


(r+ + r-)

1pm = 1x10-12 m = 10-2 Ả


(r+ + r-)






2

0

11

4 2

c aA

c a

Z ZN Ae vU

r r n

= minus

+

2

0

14 2

c aA

c a c a

Z ZN Ae vU

r r r r

= minus

+ +




2 2

0 0

11

4

AN AZ eU

r n

= minus minus

2 2

0 0 0

14

AN Az eU

r r

= minus minus

Born-Landeacute

Born-Mayer

ndash

ndash




bull n asymp 8

n2 =22 =1

bull2

04

AN e

108 c a

c a

Z ZU v KJ mol

r r=

+-

2

0

11

4 2

c aA

c a

Z ZN Ae vU

r r n

= minus

+

ndash


Ejemplo MgF2

kJmol

Mg(s) rarr Hatm 149

rarr EI1 +EI2 7375 y 1450


rarr -2AE -339 X2

rarr Hordmred= _____________________________________________________________________________________________


-29395



0 0

11

4 r

AN Az eU

n

= minus



Madelung A= 2408


Mg2+(6 86 pm) F- (3116pm) = 202 pm

0202 nm y 202 x 10-10 m

z = 1 2

n=8

U = -28996 kJmol

Zc = +2 Za = -1

v = 3


bull0211nm

U =- 307109 kJmol

Ejemplo MgF2

U0 = - 139 x 10-4 A z+z- (1-1n) Jmol

(r+ + r-) (m)

108 c a

c a

Z ZU v KJ mol

r r=

+-

-

Hordmred= -29395 100986 1045

Uc = -33139 kJmol y UB = 4142






(g)]





kJmol

2ΔH(Na+) 1220

iquestProceso real

AE 1+2 (SO4-2

) = ΔH(SO4-2)

Hordmred

____________________________________________________________________________________



-2)


-2)

kJmol

2ΔH(K+) 1028

iquestProceso real

AE 1+2 (SO4-2

) = ΔH(SO4-2)

Hordmred

____________________________________________________________________________________



-2)


-2)


r(Na+) = 116 pm = 0116 nm r(K+) = 152 pm = 0152 nm

Regla de los radios





6 oct entre 0414 y 0732


12 dodecaedro 1








Reglas de Fajans











EjemploNaX

F Cl Br I kJmol

ΔHatm Na

EI

ΔHatm X

-AE

ΔHred

-569 -410 -360 -288 ΔHf

Ejemplo MI

Li Na K Rb Cs kJmol

161

526

108

502

90

424

82

408

78

381

ΔHatm M

EI

106

-301

106

-301

106

-301

106

-301

106

-301

ΔHatm I

-AE

ΔHred

-271 -288 -328 -329 -337 ΔHf





F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________







F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf


T = 298 K










Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf





Gf = -531- 298(-105) kJmol

1000

Gf = -531 + 313 = -4997 kJmol









(g) AE2

Mg 2+(g) + O2minus


___________________________________________


149 kJmol

738

1450

249

-142

+844

-610

Extra

GASTO

De ENERGIacuteA








____________________________________________


149 kJmol

738

249

-142

108 c a

c a

Z ZU v KJ mol

r r=

+-



y para el O-






Hatm

EI1

EI2= 4567

Hatm

AE1

Hordmred=

___________________________________________





Ec Born-Lande


U0 para el NaCl

Uc =

ndash86319Hordmred

-788 kJmol 1001095

9585






_____________________________________________________________________________________________




AgCl


Cl + AE)



Uo = Uc + UB





3 2933x10-20

Uo = -71804 [kJmol]

Ec Born-Lande


2 2

6 8exp

4 r

AA A A


r r


Energiacutea de

London (UL)

6

C





D

r





J E Mayer (1933)

1( )


1 2

1 2

3

2 ( )

E EC

E E +minus =

+

23

4C E ++ + += 23


1 2

1 2

3

2 ( )

E EC

E E +minus =

+

23

4C E ++ + += 23


Donde





cristal


energiacuteas


r0 = 2772 x 10-10 m

Entonces

6

0

C


Para el AgClDatos

Polarizabilidad


αCl- = 345 x 10-30 m3




NA NA




dipolo-cuadrupolo D



dipolo-dipolo



AgCl es


UL=

8

D

r

2 2

6 8exp

4 r

AA A A


r r


2 2

6 8exp

4 r

AA A A


r r



4πε0r0


Uo = Uc + UB + UL




3 2933x10-20



Ec Born-Lande







_____________________________________________________________________________________________


Hordmred = -9108


-71804 KJmol -84304 KJmol -9108

788 956 100





α+

Li+ Be

2+ Cu

+ Zn

2+ Al

3+

0029 0008 043 029 0054

Na+ Mg

2+ Ag

+ Cd

2+ Ga

3+

018 0094 172 109 02

k+ Ca

2+ Au

+ Hg

2+ In

3+

084 047 188 125 073

Rb+ Sr

2+ Tl

3+

142 086 087

Cs+ Ba

2+

244 156

α-

H- F

- O

2-

102 105 392

Cl- S

2-

369 103

Br- Se

2-

481 106

I- Te

2-

716 141





max

1

2Z AU N hv=



(calc)

ΔH 298

(ciclo Born)

NaCl -766 -863 114 -25 8 -776 -787 CsI -580 -619 90 -54 3 -592 -602

AgI -795 -808 138 -128 3 -807 -891

CuBr -870 -925 114 -64 4 -882 -977

108772


CsI -580 es del 964

AgI -795 es del 892

CuBr -870 es del 89


788

AgCl -840 -820 1026 -125 3 -9108


U0 = - NAA z+z-e2 (1-1n)

40 (r0)


(r+ + r-)

1pm = 1x10-12 m = 10-2 Ả


(r+ + r-)






2

0

11

4 2

c aA

c a

Z ZN Ae vU

r r n

= minus

+

2

0

14 2

c aA

c a c a

Z ZN Ae vU

r r r r

= minus

+ +




2 2

0 0

11

4

AN AZ eU

r n

= minus minus

2 2

0 0 0

14

AN Az eU

r r

= minus minus

Born-Landeacute

Born-Mayer

ndash

ndash




bull n asymp 8

n2 =22 =1

bull2

04

AN e

108 c a

c a

Z ZU v KJ mol

r r=

+-

2

0

11

4 2

c aA

c a

Z ZN Ae vU

r r n

= minus

+

ndash


Ejemplo MgF2

kJmol

Mg(s) rarr Hatm 149

rarr EI1 +EI2 7375 y 1450


rarr -2AE -339 X2

rarr Hordmred= _____________________________________________________________________________________________


-29395



0 0

11

4 r

AN Az eU

n

= minus



Madelung A= 2408


Mg2+(6 86 pm) F- (3116pm) = 202 pm

0202 nm y 202 x 10-10 m

z = 1 2

n=8

U = -28996 kJmol

Zc = +2 Za = -1

v = 3


bull0211nm

U =- 307109 kJmol

Ejemplo MgF2

U0 = - 139 x 10-4 A z+z- (1-1n) Jmol

(r+ + r-) (m)

108 c a

c a

Z ZU v KJ mol

r r=

+-

-

Hordmred= -29395 100986 1045

Uc = -33139 kJmol y UB = 4142






(g)]





kJmol

2ΔH(Na+) 1220

iquestProceso real

AE 1+2 (SO4-2

) = ΔH(SO4-2)

Hordmred

____________________________________________________________________________________



-2)


-2)

kJmol

2ΔH(K+) 1028

iquestProceso real

AE 1+2 (SO4-2

) = ΔH(SO4-2)

Hordmred

____________________________________________________________________________________



-2)


-2)


r(Na+) = 116 pm = 0116 nm r(K+) = 152 pm = 0152 nm

Regla de los radios





6 oct entre 0414 y 0732


12 dodecaedro 1








Reglas de Fajans











EjemploNaX

F Cl Br I kJmol

ΔHatm Na

EI

ΔHatm X

-AE

ΔHred

-569 -410 -360 -288 ΔHf

Ejemplo MI

Li Na K Rb Cs kJmol

161

526

108

502

90

424

82

408

78

381

ΔHatm M

EI

106

-301

106

-301

106

-301

106

-301

106

-301

ΔHatm I

-AE

ΔHred

-271 -288 -328 -329 -337 ΔHf





F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________







F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf


T = 298 K










Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf





Gf = -531- 298(-105) kJmol

1000

Gf = -531 + 313 = -4997 kJmol









(g) AE2

Mg 2+(g) + O2minus


___________________________________________


149 kJmol

738

1450

249

-142

+844

-610

Extra

GASTO

De ENERGIacuteA








____________________________________________


149 kJmol

738

249

-142

108 c a

c a

Z ZU v KJ mol

r r=

+-



y para el O-






Hatm

EI1

EI2= 4567

Hatm

AE1

Hordmred=

___________________________________________










_____________________________________________________________________________________________




AgCl


Cl + AE)



Uo = Uc + UB





3 2933x10-20

Uo = -71804 [kJmol]

Ec Born-Lande


2 2

6 8exp

4 r

AA A A


r r


Energiacutea de

London (UL)

6

C





D

r





J E Mayer (1933)

1( )


1 2

1 2

3

2 ( )

E EC

E E +minus =

+

23

4C E ++ + += 23


1 2

1 2

3

2 ( )

E EC

E E +minus =

+

23

4C E ++ + += 23


Donde





cristal


energiacuteas


r0 = 2772 x 10-10 m

Entonces

6

0

C


Para el AgClDatos

Polarizabilidad


αCl- = 345 x 10-30 m3




NA NA




dipolo-cuadrupolo D



dipolo-dipolo



AgCl es


UL=

8

D

r

2 2

6 8exp

4 r

AA A A


r r


2 2

6 8exp

4 r

AA A A


r r



4πε0r0


Uo = Uc + UB + UL




3 2933x10-20



Ec Born-Lande







_____________________________________________________________________________________________


Hordmred = -9108


-71804 KJmol -84304 KJmol -9108

788 956 100





α+

Li+ Be

2+ Cu

+ Zn

2+ Al

3+

0029 0008 043 029 0054

Na+ Mg

2+ Ag

+ Cd

2+ Ga

3+

018 0094 172 109 02

k+ Ca

2+ Au

+ Hg

2+ In

3+

084 047 188 125 073

Rb+ Sr

2+ Tl

3+

142 086 087

Cs+ Ba

2+

244 156

α-

H- F

- O

2-

102 105 392

Cl- S

2-

369 103

Br- Se

2-

481 106

I- Te

2-

716 141





max

1

2Z AU N hv=



(calc)

ΔH 298

(ciclo Born)

NaCl -766 -863 114 -25 8 -776 -787 CsI -580 -619 90 -54 3 -592 -602

AgI -795 -808 138 -128 3 -807 -891

CuBr -870 -925 114 -64 4 -882 -977

108772


CsI -580 es del 964

AgI -795 es del 892

CuBr -870 es del 89


788

AgCl -840 -820 1026 -125 3 -9108


U0 = - NAA z+z-e2 (1-1n)

40 (r0)


(r+ + r-)

1pm = 1x10-12 m = 10-2 Ả


(r+ + r-)






2

0

11

4 2

c aA

c a

Z ZN Ae vU

r r n

= minus

+

2

0

14 2

c aA

c a c a

Z ZN Ae vU

r r r r

= minus

+ +




2 2

0 0

11

4

AN AZ eU

r n

= minus minus

2 2

0 0 0

14

AN Az eU

r r

= minus minus

Born-Landeacute

Born-Mayer

ndash

ndash




bull n asymp 8

n2 =22 =1

bull2

04

AN e

108 c a

c a

Z ZU v KJ mol

r r=

+-

2

0

11

4 2

c aA

c a

Z ZN Ae vU

r r n

= minus

+

ndash


Ejemplo MgF2

kJmol

Mg(s) rarr Hatm 149

rarr EI1 +EI2 7375 y 1450


rarr -2AE -339 X2

rarr Hordmred= _____________________________________________________________________________________________


-29395



0 0

11

4 r

AN Az eU

n

= minus



Madelung A= 2408


Mg2+(6 86 pm) F- (3116pm) = 202 pm

0202 nm y 202 x 10-10 m

z = 1 2

n=8

U = -28996 kJmol

Zc = +2 Za = -1

v = 3


bull0211nm

U =- 307109 kJmol

Ejemplo MgF2

U0 = - 139 x 10-4 A z+z- (1-1n) Jmol

(r+ + r-) (m)

108 c a

c a

Z ZU v KJ mol

r r=

+-

-

Hordmred= -29395 100986 1045

Uc = -33139 kJmol y UB = 4142






(g)]





kJmol

2ΔH(Na+) 1220

iquestProceso real

AE 1+2 (SO4-2

) = ΔH(SO4-2)

Hordmred

____________________________________________________________________________________



-2)


-2)

kJmol

2ΔH(K+) 1028

iquestProceso real

AE 1+2 (SO4-2

) = ΔH(SO4-2)

Hordmred

____________________________________________________________________________________



-2)


-2)


r(Na+) = 116 pm = 0116 nm r(K+) = 152 pm = 0152 nm

Regla de los radios





6 oct entre 0414 y 0732


12 dodecaedro 1








Reglas de Fajans











EjemploNaX

F Cl Br I kJmol

ΔHatm Na

EI

ΔHatm X

-AE

ΔHred

-569 -410 -360 -288 ΔHf

Ejemplo MI

Li Na K Rb Cs kJmol

161

526

108

502

90

424

82

408

78

381

ΔHatm M

EI

106

-301

106

-301

106

-301

106

-301

106

-301

ΔHatm I

-AE

ΔHred

-271 -288 -328 -329 -337 ΔHf





F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________







F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf


T = 298 K










Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf





Gf = -531- 298(-105) kJmol

1000

Gf = -531 + 313 = -4997 kJmol









(g) AE2

Mg 2+(g) + O2minus


___________________________________________


149 kJmol

738

1450

249

-142

+844

-610

Extra

GASTO

De ENERGIacuteA








____________________________________________


149 kJmol

738

249

-142

108 c a

c a

Z ZU v KJ mol

r r=

+-



y para el O-






Hatm

EI1

EI2= 4567

Hatm

AE1

Hordmred=

___________________________________________







Uo = Uc + UB





3 2933x10-20

Uo = -71804 [kJmol]

Ec Born-Lande


2 2

6 8exp

4 r

AA A A


r r


Energiacutea de

London (UL)

6

C





D

r





J E Mayer (1933)

1( )


1 2

1 2

3

2 ( )

E EC

E E +minus =

+

23

4C E ++ + += 23


1 2

1 2

3

2 ( )

E EC

E E +minus =

+

23

4C E ++ + += 23


Donde





cristal


energiacuteas


r0 = 2772 x 10-10 m

Entonces

6

0

C


Para el AgClDatos

Polarizabilidad


αCl- = 345 x 10-30 m3




NA NA




dipolo-cuadrupolo D



dipolo-dipolo



AgCl es


UL=

8

D

r

2 2

6 8exp

4 r

AA A A


r r


2 2

6 8exp

4 r

AA A A


r r



4πε0r0


Uo = Uc + UB + UL




3 2933x10-20



Ec Born-Lande







_____________________________________________________________________________________________


Hordmred = -9108


-71804 KJmol -84304 KJmol -9108

788 956 100





α+

Li+ Be

2+ Cu

+ Zn

2+ Al

3+

0029 0008 043 029 0054

Na+ Mg

2+ Ag

+ Cd

2+ Ga

3+

018 0094 172 109 02

k+ Ca

2+ Au

+ Hg

2+ In

3+

084 047 188 125 073

Rb+ Sr

2+ Tl

3+

142 086 087

Cs+ Ba

2+

244 156

α-

H- F

- O

2-

102 105 392

Cl- S

2-

369 103

Br- Se

2-

481 106

I- Te

2-

716 141





max

1

2Z AU N hv=



(calc)

ΔH 298

(ciclo Born)

NaCl -766 -863 114 -25 8 -776 -787 CsI -580 -619 90 -54 3 -592 -602

AgI -795 -808 138 -128 3 -807 -891

CuBr -870 -925 114 -64 4 -882 -977

108772


CsI -580 es del 964

AgI -795 es del 892

CuBr -870 es del 89


788

AgCl -840 -820 1026 -125 3 -9108


U0 = - NAA z+z-e2 (1-1n)

40 (r0)


(r+ + r-)

1pm = 1x10-12 m = 10-2 Ả


(r+ + r-)






2

0

11

4 2

c aA

c a

Z ZN Ae vU

r r n

= minus

+

2

0

14 2

c aA

c a c a

Z ZN Ae vU

r r r r

= minus

+ +




2 2

0 0

11

4

AN AZ eU

r n

= minus minus

2 2

0 0 0

14

AN Az eU

r r

= minus minus

Born-Landeacute

Born-Mayer

ndash

ndash




bull n asymp 8

n2 =22 =1

bull2

04

AN e

108 c a

c a

Z ZU v KJ mol

r r=

+-

2

0

11

4 2

c aA

c a

Z ZN Ae vU

r r n

= minus

+

ndash


Ejemplo MgF2

kJmol

Mg(s) rarr Hatm 149

rarr EI1 +EI2 7375 y 1450


rarr -2AE -339 X2

rarr Hordmred= _____________________________________________________________________________________________


-29395



0 0

11

4 r

AN Az eU

n

= minus



Madelung A= 2408


Mg2+(6 86 pm) F- (3116pm) = 202 pm

0202 nm y 202 x 10-10 m

z = 1 2

n=8

U = -28996 kJmol

Zc = +2 Za = -1

v = 3


bull0211nm

U =- 307109 kJmol

Ejemplo MgF2

U0 = - 139 x 10-4 A z+z- (1-1n) Jmol

(r+ + r-) (m)

108 c a

c a

Z ZU v KJ mol

r r=

+-

-

Hordmred= -29395 100986 1045

Uc = -33139 kJmol y UB = 4142






(g)]





kJmol

2ΔH(Na+) 1220

iquestProceso real

AE 1+2 (SO4-2

) = ΔH(SO4-2)

Hordmred

____________________________________________________________________________________



-2)


-2)

kJmol

2ΔH(K+) 1028

iquestProceso real

AE 1+2 (SO4-2

) = ΔH(SO4-2)

Hordmred

____________________________________________________________________________________



-2)


-2)


r(Na+) = 116 pm = 0116 nm r(K+) = 152 pm = 0152 nm

Regla de los radios





6 oct entre 0414 y 0732


12 dodecaedro 1








Reglas de Fajans











EjemploNaX

F Cl Br I kJmol

ΔHatm Na

EI

ΔHatm X

-AE

ΔHred

-569 -410 -360 -288 ΔHf

Ejemplo MI

Li Na K Rb Cs kJmol

161

526

108

502

90

424

82

408

78

381

ΔHatm M

EI

106

-301

106

-301

106

-301

106

-301

106

-301

ΔHatm I

-AE

ΔHred

-271 -288 -328 -329 -337 ΔHf





F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________







F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf


T = 298 K










Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf





Gf = -531- 298(-105) kJmol

1000

Gf = -531 + 313 = -4997 kJmol









(g) AE2

Mg 2+(g) + O2minus


___________________________________________


149 kJmol

738

1450

249

-142

+844

-610

Extra

GASTO

De ENERGIacuteA








____________________________________________


149 kJmol

738

249

-142

108 c a

c a

Z ZU v KJ mol

r r=

+-



y para el O-






Hatm

EI1

EI2= 4567

Hatm

AE1

Hordmred=

___________________________________________






2 2

6 8exp

4 r

AA A A


r r


Energiacutea de

London (UL)

6

C





D

r





J E Mayer (1933)

1( )


1 2

1 2

3

2 ( )

E EC

E E +minus =

+

23

4C E ++ + += 23


1 2

1 2

3

2 ( )

E EC

E E +minus =

+

23

4C E ++ + += 23


Donde





cristal


energiacuteas


r0 = 2772 x 10-10 m

Entonces

6

0

C


Para el AgClDatos

Polarizabilidad


αCl- = 345 x 10-30 m3




NA NA




dipolo-cuadrupolo D



dipolo-dipolo



AgCl es


UL=

8

D

r

2 2

6 8exp

4 r

AA A A


r r


2 2

6 8exp

4 r

AA A A


r r



4πε0r0


Uo = Uc + UB + UL




3 2933x10-20



Ec Born-Lande







_____________________________________________________________________________________________


Hordmred = -9108


-71804 KJmol -84304 KJmol -9108

788 956 100





α+

Li+ Be

2+ Cu

+ Zn

2+ Al

3+

0029 0008 043 029 0054

Na+ Mg

2+ Ag

+ Cd

2+ Ga

3+

018 0094 172 109 02

k+ Ca

2+ Au

+ Hg

2+ In

3+

084 047 188 125 073

Rb+ Sr

2+ Tl

3+

142 086 087

Cs+ Ba

2+

244 156

α-

H- F

- O

2-

102 105 392

Cl- S

2-

369 103

Br- Se

2-

481 106

I- Te

2-

716 141





max

1

2Z AU N hv=



(calc)

ΔH 298

(ciclo Born)

NaCl -766 -863 114 -25 8 -776 -787 CsI -580 -619 90 -54 3 -592 -602

AgI -795 -808 138 -128 3 -807 -891

CuBr -870 -925 114 -64 4 -882 -977

108772


CsI -580 es del 964

AgI -795 es del 892

CuBr -870 es del 89


788

AgCl -840 -820 1026 -125 3 -9108


U0 = - NAA z+z-e2 (1-1n)

40 (r0)


(r+ + r-)

1pm = 1x10-12 m = 10-2 Ả


(r+ + r-)






2

0

11

4 2

c aA

c a

Z ZN Ae vU

r r n

= minus

+

2

0

14 2

c aA

c a c a

Z ZN Ae vU

r r r r

= minus

+ +




2 2

0 0

11

4

AN AZ eU

r n

= minus minus

2 2

0 0 0

14

AN Az eU

r r

= minus minus

Born-Landeacute

Born-Mayer

ndash

ndash




bull n asymp 8

n2 =22 =1

bull2

04

AN e

108 c a

c a

Z ZU v KJ mol

r r=

+-

2

0

11

4 2

c aA

c a

Z ZN Ae vU

r r n

= minus

+

ndash


Ejemplo MgF2

kJmol

Mg(s) rarr Hatm 149

rarr EI1 +EI2 7375 y 1450


rarr -2AE -339 X2

rarr Hordmred= _____________________________________________________________________________________________


-29395



0 0

11

4 r

AN Az eU

n

= minus



Madelung A= 2408


Mg2+(6 86 pm) F- (3116pm) = 202 pm

0202 nm y 202 x 10-10 m

z = 1 2

n=8

U = -28996 kJmol

Zc = +2 Za = -1

v = 3


bull0211nm

U =- 307109 kJmol

Ejemplo MgF2

U0 = - 139 x 10-4 A z+z- (1-1n) Jmol

(r+ + r-) (m)

108 c a

c a

Z ZU v KJ mol

r r=

+-

-

Hordmred= -29395 100986 1045

Uc = -33139 kJmol y UB = 4142






(g)]





kJmol

2ΔH(Na+) 1220

iquestProceso real

AE 1+2 (SO4-2

) = ΔH(SO4-2)

Hordmred

____________________________________________________________________________________



-2)


-2)

kJmol

2ΔH(K+) 1028

iquestProceso real

AE 1+2 (SO4-2

) = ΔH(SO4-2)

Hordmred

____________________________________________________________________________________



-2)


-2)


r(Na+) = 116 pm = 0116 nm r(K+) = 152 pm = 0152 nm

Regla de los radios





6 oct entre 0414 y 0732


12 dodecaedro 1








Reglas de Fajans











EjemploNaX

F Cl Br I kJmol

ΔHatm Na

EI

ΔHatm X

-AE

ΔHred

-569 -410 -360 -288 ΔHf

Ejemplo MI

Li Na K Rb Cs kJmol

161

526

108

502

90

424

82

408

78

381

ΔHatm M

EI

106

-301

106

-301

106

-301

106

-301

106

-301

ΔHatm I

-AE

ΔHred

-271 -288 -328 -329 -337 ΔHf





F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________







F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf


T = 298 K










Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf





Gf = -531- 298(-105) kJmol

1000

Gf = -531 + 313 = -4997 kJmol









(g) AE2

Mg 2+(g) + O2minus


___________________________________________


149 kJmol

738

1450

249

-142

+844

-610

Extra

GASTO

De ENERGIacuteA








____________________________________________


149 kJmol

738

249

-142

108 c a

c a

Z ZU v KJ mol

r r=

+-



y para el O-






Hatm

EI1

EI2= 4567

Hatm

AE1

Hordmred=

___________________________________________







J E Mayer (1933)

1( )


1 2

1 2

3

2 ( )

E EC

E E +minus =

+

23

4C E ++ + += 23


1 2

1 2

3

2 ( )

E EC

E E +minus =

+

23

4C E ++ + += 23


Donde





cristal


energiacuteas


r0 = 2772 x 10-10 m

Entonces

6

0

C


Para el AgClDatos

Polarizabilidad


αCl- = 345 x 10-30 m3




NA NA




dipolo-cuadrupolo D



dipolo-dipolo



AgCl es


UL=

8

D

r

2 2

6 8exp

4 r

AA A A


r r


2 2

6 8exp

4 r

AA A A


r r



4πε0r0


Uo = Uc + UB + UL




3 2933x10-20



Ec Born-Lande







_____________________________________________________________________________________________


Hordmred = -9108


-71804 KJmol -84304 KJmol -9108

788 956 100





α+

Li+ Be

2+ Cu

+ Zn

2+ Al

3+

0029 0008 043 029 0054

Na+ Mg

2+ Ag

+ Cd

2+ Ga

3+

018 0094 172 109 02

k+ Ca

2+ Au

+ Hg

2+ In

3+

084 047 188 125 073

Rb+ Sr

2+ Tl

3+

142 086 087

Cs+ Ba

2+

244 156

α-

H- F

- O

2-

102 105 392

Cl- S

2-

369 103

Br- Se

2-

481 106

I- Te

2-

716 141





max

1

2Z AU N hv=



(calc)

ΔH 298

(ciclo Born)

NaCl -766 -863 114 -25 8 -776 -787 CsI -580 -619 90 -54 3 -592 -602

AgI -795 -808 138 -128 3 -807 -891

CuBr -870 -925 114 -64 4 -882 -977

108772


CsI -580 es del 964

AgI -795 es del 892

CuBr -870 es del 89


788

AgCl -840 -820 1026 -125 3 -9108


U0 = - NAA z+z-e2 (1-1n)

40 (r0)


(r+ + r-)

1pm = 1x10-12 m = 10-2 Ả


(r+ + r-)






2

0

11

4 2

c aA

c a

Z ZN Ae vU

r r n

= minus

+

2

0

14 2

c aA

c a c a

Z ZN Ae vU

r r r r

= minus

+ +




2 2

0 0

11

4

AN AZ eU

r n

= minus minus

2 2

0 0 0

14

AN Az eU

r r

= minus minus

Born-Landeacute

Born-Mayer

ndash

ndash




bull n asymp 8

n2 =22 =1

bull2

04

AN e

108 c a

c a

Z ZU v KJ mol

r r=

+-

2

0

11

4 2

c aA

c a

Z ZN Ae vU

r r n

= minus

+

ndash


Ejemplo MgF2

kJmol

Mg(s) rarr Hatm 149

rarr EI1 +EI2 7375 y 1450


rarr -2AE -339 X2

rarr Hordmred= _____________________________________________________________________________________________


-29395



0 0

11

4 r

AN Az eU

n

= minus



Madelung A= 2408


Mg2+(6 86 pm) F- (3116pm) = 202 pm

0202 nm y 202 x 10-10 m

z = 1 2

n=8

U = -28996 kJmol

Zc = +2 Za = -1

v = 3


bull0211nm

U =- 307109 kJmol

Ejemplo MgF2

U0 = - 139 x 10-4 A z+z- (1-1n) Jmol

(r+ + r-) (m)

108 c a

c a

Z ZU v KJ mol

r r=

+-

-

Hordmred= -29395 100986 1045

Uc = -33139 kJmol y UB = 4142






(g)]





kJmol

2ΔH(Na+) 1220

iquestProceso real

AE 1+2 (SO4-2

) = ΔH(SO4-2)

Hordmred

____________________________________________________________________________________



-2)


-2)

kJmol

2ΔH(K+) 1028

iquestProceso real

AE 1+2 (SO4-2

) = ΔH(SO4-2)

Hordmred

____________________________________________________________________________________



-2)


-2)


r(Na+) = 116 pm = 0116 nm r(K+) = 152 pm = 0152 nm

Regla de los radios





6 oct entre 0414 y 0732


12 dodecaedro 1








Reglas de Fajans











EjemploNaX

F Cl Br I kJmol

ΔHatm Na

EI

ΔHatm X

-AE

ΔHred

-569 -410 -360 -288 ΔHf

Ejemplo MI

Li Na K Rb Cs kJmol

161

526

108

502

90

424

82

408

78

381

ΔHatm M

EI

106

-301

106

-301

106

-301

106

-301

106

-301

ΔHatm I

-AE

ΔHred

-271 -288 -328 -329 -337 ΔHf





F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________







F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf


T = 298 K










Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf





Gf = -531- 298(-105) kJmol

1000

Gf = -531 + 313 = -4997 kJmol









(g) AE2

Mg 2+(g) + O2minus


___________________________________________


149 kJmol

738

1450

249

-142

+844

-610

Extra

GASTO

De ENERGIacuteA








____________________________________________


149 kJmol

738

249

-142

108 c a

c a

Z ZU v KJ mol

r r=

+-



y para el O-






Hatm

EI1

EI2= 4567

Hatm

AE1

Hordmred=

___________________________________________





1 2

1 2

3

2 ( )

E EC

E E +minus =

+

23

4C E ++ + += 23


Donde





cristal


energiacuteas


r0 = 2772 x 10-10 m

Entonces

6

0

C


Para el AgClDatos

Polarizabilidad


αCl- = 345 x 10-30 m3




NA NA




dipolo-cuadrupolo D



dipolo-dipolo



AgCl es


UL=

8

D

r

2 2

6 8exp

4 r

AA A A


r r


2 2

6 8exp

4 r

AA A A


r r



4πε0r0


Uo = Uc + UB + UL




3 2933x10-20



Ec Born-Lande







_____________________________________________________________________________________________


Hordmred = -9108


-71804 KJmol -84304 KJmol -9108

788 956 100





α+

Li+ Be

2+ Cu

+ Zn

2+ Al

3+

0029 0008 043 029 0054

Na+ Mg

2+ Ag

+ Cd

2+ Ga

3+

018 0094 172 109 02

k+ Ca

2+ Au

+ Hg

2+ In

3+

084 047 188 125 073

Rb+ Sr

2+ Tl

3+

142 086 087

Cs+ Ba

2+

244 156

α-

H- F

- O

2-

102 105 392

Cl- S

2-

369 103

Br- Se

2-

481 106

I- Te

2-

716 141





max

1

2Z AU N hv=



(calc)

ΔH 298

(ciclo Born)

NaCl -766 -863 114 -25 8 -776 -787 CsI -580 -619 90 -54 3 -592 -602

AgI -795 -808 138 -128 3 -807 -891

CuBr -870 -925 114 -64 4 -882 -977

108772


CsI -580 es del 964

AgI -795 es del 892

CuBr -870 es del 89


788

AgCl -840 -820 1026 -125 3 -9108


U0 = - NAA z+z-e2 (1-1n)

40 (r0)


(r+ + r-)

1pm = 1x10-12 m = 10-2 Ả


(r+ + r-)






2

0

11

4 2

c aA

c a

Z ZN Ae vU

r r n

= minus

+

2

0

14 2

c aA

c a c a

Z ZN Ae vU

r r r r

= minus

+ +




2 2

0 0

11

4

AN AZ eU

r n

= minus minus

2 2

0 0 0

14

AN Az eU

r r

= minus minus

Born-Landeacute

Born-Mayer

ndash

ndash




bull n asymp 8

n2 =22 =1

bull2

04

AN e

108 c a

c a

Z ZU v KJ mol

r r=

+-

2

0

11

4 2

c aA

c a

Z ZN Ae vU

r r n

= minus

+

ndash


Ejemplo MgF2

kJmol

Mg(s) rarr Hatm 149

rarr EI1 +EI2 7375 y 1450


rarr -2AE -339 X2

rarr Hordmred= _____________________________________________________________________________________________


-29395



0 0

11

4 r

AN Az eU

n

= minus



Madelung A= 2408


Mg2+(6 86 pm) F- (3116pm) = 202 pm

0202 nm y 202 x 10-10 m

z = 1 2

n=8

U = -28996 kJmol

Zc = +2 Za = -1

v = 3


bull0211nm

U =- 307109 kJmol

Ejemplo MgF2

U0 = - 139 x 10-4 A z+z- (1-1n) Jmol

(r+ + r-) (m)

108 c a

c a

Z ZU v KJ mol

r r=

+-

-

Hordmred= -29395 100986 1045

Uc = -33139 kJmol y UB = 4142






(g)]





kJmol

2ΔH(Na+) 1220

iquestProceso real

AE 1+2 (SO4-2

) = ΔH(SO4-2)

Hordmred

____________________________________________________________________________________



-2)


-2)

kJmol

2ΔH(K+) 1028

iquestProceso real

AE 1+2 (SO4-2

) = ΔH(SO4-2)

Hordmred

____________________________________________________________________________________



-2)


-2)


r(Na+) = 116 pm = 0116 nm r(K+) = 152 pm = 0152 nm

Regla de los radios





6 oct entre 0414 y 0732


12 dodecaedro 1








Reglas de Fajans











EjemploNaX

F Cl Br I kJmol

ΔHatm Na

EI

ΔHatm X

-AE

ΔHred

-569 -410 -360 -288 ΔHf

Ejemplo MI

Li Na K Rb Cs kJmol

161

526

108

502

90

424

82

408

78

381

ΔHatm M

EI

106

-301

106

-301

106

-301

106

-301

106

-301

ΔHatm I

-AE

ΔHred

-271 -288 -328 -329 -337 ΔHf





F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________







F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf


T = 298 K










Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf





Gf = -531- 298(-105) kJmol

1000

Gf = -531 + 313 = -4997 kJmol









(g) AE2

Mg 2+(g) + O2minus


___________________________________________


149 kJmol

738

1450

249

-142

+844

-610

Extra

GASTO

De ENERGIacuteA








____________________________________________


149 kJmol

738

249

-142

108 c a

c a

Z ZU v KJ mol

r r=

+-



y para el O-






Hatm

EI1

EI2= 4567

Hatm

AE1

Hordmred=

___________________________________________






r0 = 2772 x 10-10 m

Entonces

6

0

C


Para el AgClDatos

Polarizabilidad


αCl- = 345 x 10-30 m3




NA NA




dipolo-cuadrupolo D



dipolo-dipolo



AgCl es


UL=

8

D

r

2 2

6 8exp

4 r

AA A A


r r


2 2

6 8exp

4 r

AA A A


r r



4πε0r0


Uo = Uc + UB + UL




3 2933x10-20



Ec Born-Lande







_____________________________________________________________________________________________


Hordmred = -9108


-71804 KJmol -84304 KJmol -9108

788 956 100





α+

Li+ Be

2+ Cu

+ Zn

2+ Al

3+

0029 0008 043 029 0054

Na+ Mg

2+ Ag

+ Cd

2+ Ga

3+

018 0094 172 109 02

k+ Ca

2+ Au

+ Hg

2+ In

3+

084 047 188 125 073

Rb+ Sr

2+ Tl

3+

142 086 087

Cs+ Ba

2+

244 156

α-

H- F

- O

2-

102 105 392

Cl- S

2-

369 103

Br- Se

2-

481 106

I- Te

2-

716 141





max

1

2Z AU N hv=



(calc)

ΔH 298

(ciclo Born)

NaCl -766 -863 114 -25 8 -776 -787 CsI -580 -619 90 -54 3 -592 -602

AgI -795 -808 138 -128 3 -807 -891

CuBr -870 -925 114 -64 4 -882 -977

108772


CsI -580 es del 964

AgI -795 es del 892

CuBr -870 es del 89


788

AgCl -840 -820 1026 -125 3 -9108


U0 = - NAA z+z-e2 (1-1n)

40 (r0)


(r+ + r-)

1pm = 1x10-12 m = 10-2 Ả


(r+ + r-)






2

0

11

4 2

c aA

c a

Z ZN Ae vU

r r n

= minus

+

2

0

14 2

c aA

c a c a

Z ZN Ae vU

r r r r

= minus

+ +




2 2

0 0

11

4

AN AZ eU

r n

= minus minus

2 2

0 0 0

14

AN Az eU

r r

= minus minus

Born-Landeacute

Born-Mayer

ndash

ndash




bull n asymp 8

n2 =22 =1

bull2

04

AN e

108 c a

c a

Z ZU v KJ mol

r r=

+-

2

0

11

4 2

c aA

c a

Z ZN Ae vU

r r n

= minus

+

ndash


Ejemplo MgF2

kJmol

Mg(s) rarr Hatm 149

rarr EI1 +EI2 7375 y 1450


rarr -2AE -339 X2

rarr Hordmred= _____________________________________________________________________________________________


-29395



0 0

11

4 r

AN Az eU

n

= minus



Madelung A= 2408


Mg2+(6 86 pm) F- (3116pm) = 202 pm

0202 nm y 202 x 10-10 m

z = 1 2

n=8

U = -28996 kJmol

Zc = +2 Za = -1

v = 3


bull0211nm

U =- 307109 kJmol

Ejemplo MgF2

U0 = - 139 x 10-4 A z+z- (1-1n) Jmol

(r+ + r-) (m)

108 c a

c a

Z ZU v KJ mol

r r=

+-

-

Hordmred= -29395 100986 1045

Uc = -33139 kJmol y UB = 4142






(g)]





kJmol

2ΔH(Na+) 1220

iquestProceso real

AE 1+2 (SO4-2

) = ΔH(SO4-2)

Hordmred

____________________________________________________________________________________



-2)


-2)

kJmol

2ΔH(K+) 1028

iquestProceso real

AE 1+2 (SO4-2

) = ΔH(SO4-2)

Hordmred

____________________________________________________________________________________



-2)


-2)


r(Na+) = 116 pm = 0116 nm r(K+) = 152 pm = 0152 nm

Regla de los radios





6 oct entre 0414 y 0732


12 dodecaedro 1








Reglas de Fajans











EjemploNaX

F Cl Br I kJmol

ΔHatm Na

EI

ΔHatm X

-AE

ΔHred

-569 -410 -360 -288 ΔHf

Ejemplo MI

Li Na K Rb Cs kJmol

161

526

108

502

90

424

82

408

78

381

ΔHatm M

EI

106

-301

106

-301

106

-301

106

-301

106

-301

ΔHatm I

-AE

ΔHred

-271 -288 -328 -329 -337 ΔHf





F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________







F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf


T = 298 K










Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf





Gf = -531- 298(-105) kJmol

1000

Gf = -531 + 313 = -4997 kJmol









(g) AE2

Mg 2+(g) + O2minus


___________________________________________


149 kJmol

738

1450

249

-142

+844

-610

Extra

GASTO

De ENERGIacuteA








____________________________________________


149 kJmol

738

249

-142

108 c a

c a

Z ZU v KJ mol

r r=

+-



y para el O-






Hatm

EI1

EI2= 4567

Hatm

AE1

Hordmred=

___________________________________________





dipolo-cuadrupolo D



dipolo-dipolo



AgCl es


UL=

8

D

r

2 2

6 8exp

4 r

AA A A


r r


2 2

6 8exp

4 r

AA A A


r r



4πε0r0


Uo = Uc + UB + UL




3 2933x10-20



Ec Born-Lande







_____________________________________________________________________________________________


Hordmred = -9108


-71804 KJmol -84304 KJmol -9108

788 956 100





α+

Li+ Be

2+ Cu

+ Zn

2+ Al

3+

0029 0008 043 029 0054

Na+ Mg

2+ Ag

+ Cd

2+ Ga

3+

018 0094 172 109 02

k+ Ca

2+ Au

+ Hg

2+ In

3+

084 047 188 125 073

Rb+ Sr

2+ Tl

3+

142 086 087

Cs+ Ba

2+

244 156

α-

H- F

- O

2-

102 105 392

Cl- S

2-

369 103

Br- Se

2-

481 106

I- Te

2-

716 141





max

1

2Z AU N hv=



(calc)

ΔH 298

(ciclo Born)

NaCl -766 -863 114 -25 8 -776 -787 CsI -580 -619 90 -54 3 -592 -602

AgI -795 -808 138 -128 3 -807 -891

CuBr -870 -925 114 -64 4 -882 -977

108772


CsI -580 es del 964

AgI -795 es del 892

CuBr -870 es del 89


788

AgCl -840 -820 1026 -125 3 -9108


U0 = - NAA z+z-e2 (1-1n)

40 (r0)


(r+ + r-)

1pm = 1x10-12 m = 10-2 Ả


(r+ + r-)






2

0

11

4 2

c aA

c a

Z ZN Ae vU

r r n

= minus

+

2

0

14 2

c aA

c a c a

Z ZN Ae vU

r r r r

= minus

+ +




2 2

0 0

11

4

AN AZ eU

r n

= minus minus

2 2

0 0 0

14

AN Az eU

r r

= minus minus

Born-Landeacute

Born-Mayer

ndash

ndash




bull n asymp 8

n2 =22 =1

bull2

04

AN e

108 c a

c a

Z ZU v KJ mol

r r=

+-

2

0

11

4 2

c aA

c a

Z ZN Ae vU

r r n

= minus

+

ndash


Ejemplo MgF2

kJmol

Mg(s) rarr Hatm 149

rarr EI1 +EI2 7375 y 1450


rarr -2AE -339 X2

rarr Hordmred= _____________________________________________________________________________________________


-29395



0 0

11

4 r

AN Az eU

n

= minus



Madelung A= 2408


Mg2+(6 86 pm) F- (3116pm) = 202 pm

0202 nm y 202 x 10-10 m

z = 1 2

n=8

U = -28996 kJmol

Zc = +2 Za = -1

v = 3


bull0211nm

U =- 307109 kJmol

Ejemplo MgF2

U0 = - 139 x 10-4 A z+z- (1-1n) Jmol

(r+ + r-) (m)

108 c a

c a

Z ZU v KJ mol

r r=

+-

-

Hordmred= -29395 100986 1045

Uc = -33139 kJmol y UB = 4142






(g)]





kJmol

2ΔH(Na+) 1220

iquestProceso real

AE 1+2 (SO4-2

) = ΔH(SO4-2)

Hordmred

____________________________________________________________________________________



-2)


-2)

kJmol

2ΔH(K+) 1028

iquestProceso real

AE 1+2 (SO4-2

) = ΔH(SO4-2)

Hordmred

____________________________________________________________________________________



-2)


-2)


r(Na+) = 116 pm = 0116 nm r(K+) = 152 pm = 0152 nm

Regla de los radios





6 oct entre 0414 y 0732


12 dodecaedro 1








Reglas de Fajans











EjemploNaX

F Cl Br I kJmol

ΔHatm Na

EI

ΔHatm X

-AE

ΔHred

-569 -410 -360 -288 ΔHf

Ejemplo MI

Li Na K Rb Cs kJmol

161

526

108

502

90

424

82

408

78

381

ΔHatm M

EI

106

-301

106

-301

106

-301

106

-301

106

-301

ΔHatm I

-AE

ΔHred

-271 -288 -328 -329 -337 ΔHf





F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________







F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf


T = 298 K










Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf





Gf = -531- 298(-105) kJmol

1000

Gf = -531 + 313 = -4997 kJmol









(g) AE2

Mg 2+(g) + O2minus


___________________________________________


149 kJmol

738

1450

249

-142

+844

-610

Extra

GASTO

De ENERGIacuteA








____________________________________________


149 kJmol

738

249

-142

108 c a

c a

Z ZU v KJ mol

r r=

+-



y para el O-






Hatm

EI1

EI2= 4567

Hatm

AE1

Hordmred=

___________________________________________





2 2

6 8exp

4 r

AA A A


r r



4πε0r0


Uo = Uc + UB + UL




3 2933x10-20



Ec Born-Lande







_____________________________________________________________________________________________


Hordmred = -9108


-71804 KJmol -84304 KJmol -9108

788 956 100





α+

Li+ Be

2+ Cu

+ Zn

2+ Al

3+

0029 0008 043 029 0054

Na+ Mg

2+ Ag

+ Cd

2+ Ga

3+

018 0094 172 109 02

k+ Ca

2+ Au

+ Hg

2+ In

3+

084 047 188 125 073

Rb+ Sr

2+ Tl

3+

142 086 087

Cs+ Ba

2+

244 156

α-

H- F

- O

2-

102 105 392

Cl- S

2-

369 103

Br- Se

2-

481 106

I- Te

2-

716 141





max

1

2Z AU N hv=



(calc)

ΔH 298

(ciclo Born)

NaCl -766 -863 114 -25 8 -776 -787 CsI -580 -619 90 -54 3 -592 -602

AgI -795 -808 138 -128 3 -807 -891

CuBr -870 -925 114 -64 4 -882 -977

108772


CsI -580 es del 964

AgI -795 es del 892

CuBr -870 es del 89


788

AgCl -840 -820 1026 -125 3 -9108


U0 = - NAA z+z-e2 (1-1n)

40 (r0)


(r+ + r-)

1pm = 1x10-12 m = 10-2 Ả


(r+ + r-)






2

0

11

4 2

c aA

c a

Z ZN Ae vU

r r n

= minus

+

2

0

14 2

c aA

c a c a

Z ZN Ae vU

r r r r

= minus

+ +




2 2

0 0

11

4

AN AZ eU

r n

= minus minus

2 2

0 0 0

14

AN Az eU

r r

= minus minus

Born-Landeacute

Born-Mayer

ndash

ndash




bull n asymp 8

n2 =22 =1

bull2

04

AN e

108 c a

c a

Z ZU v KJ mol

r r=

+-

2

0

11

4 2

c aA

c a

Z ZN Ae vU

r r n

= minus

+

ndash


Ejemplo MgF2

kJmol

Mg(s) rarr Hatm 149

rarr EI1 +EI2 7375 y 1450


rarr -2AE -339 X2

rarr Hordmred= _____________________________________________________________________________________________


-29395



0 0

11

4 r

AN Az eU

n

= minus



Madelung A= 2408


Mg2+(6 86 pm) F- (3116pm) = 202 pm

0202 nm y 202 x 10-10 m

z = 1 2

n=8

U = -28996 kJmol

Zc = +2 Za = -1

v = 3


bull0211nm

U =- 307109 kJmol

Ejemplo MgF2

U0 = - 139 x 10-4 A z+z- (1-1n) Jmol

(r+ + r-) (m)

108 c a

c a

Z ZU v KJ mol

r r=

+-

-

Hordmred= -29395 100986 1045

Uc = -33139 kJmol y UB = 4142






(g)]





kJmol

2ΔH(Na+) 1220

iquestProceso real

AE 1+2 (SO4-2

) = ΔH(SO4-2)

Hordmred

____________________________________________________________________________________



-2)


-2)

kJmol

2ΔH(K+) 1028

iquestProceso real

AE 1+2 (SO4-2

) = ΔH(SO4-2)

Hordmred

____________________________________________________________________________________



-2)


-2)


r(Na+) = 116 pm = 0116 nm r(K+) = 152 pm = 0152 nm

Regla de los radios





6 oct entre 0414 y 0732


12 dodecaedro 1








Reglas de Fajans











EjemploNaX

F Cl Br I kJmol

ΔHatm Na

EI

ΔHatm X

-AE

ΔHred

-569 -410 -360 -288 ΔHf

Ejemplo MI

Li Na K Rb Cs kJmol

161

526

108

502

90

424

82

408

78

381

ΔHatm M

EI

106

-301

106

-301

106

-301

106

-301

106

-301

ΔHatm I

-AE

ΔHred

-271 -288 -328 -329 -337 ΔHf





F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________







F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf


T = 298 K










Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf





Gf = -531- 298(-105) kJmol

1000

Gf = -531 + 313 = -4997 kJmol









(g) AE2

Mg 2+(g) + O2minus


___________________________________________


149 kJmol

738

1450

249

-142

+844

-610

Extra

GASTO

De ENERGIacuteA








____________________________________________


149 kJmol

738

249

-142

108 c a

c a

Z ZU v KJ mol

r r=

+-



y para el O-






Hatm

EI1

EI2= 4567

Hatm

AE1

Hordmred=

___________________________________________











_____________________________________________________________________________________________


Hordmred = -9108


-71804 KJmol -84304 KJmol -9108

788 956 100





α+

Li+ Be

2+ Cu

+ Zn

2+ Al

3+

0029 0008 043 029 0054

Na+ Mg

2+ Ag

+ Cd

2+ Ga

3+

018 0094 172 109 02

k+ Ca

2+ Au

+ Hg

2+ In

3+

084 047 188 125 073

Rb+ Sr

2+ Tl

3+

142 086 087

Cs+ Ba

2+

244 156

α-

H- F

- O

2-

102 105 392

Cl- S

2-

369 103

Br- Se

2-

481 106

I- Te

2-

716 141





max

1

2Z AU N hv=



(calc)

ΔH 298

(ciclo Born)

NaCl -766 -863 114 -25 8 -776 -787 CsI -580 -619 90 -54 3 -592 -602

AgI -795 -808 138 -128 3 -807 -891

CuBr -870 -925 114 -64 4 -882 -977

108772


CsI -580 es del 964

AgI -795 es del 892

CuBr -870 es del 89


788

AgCl -840 -820 1026 -125 3 -9108


U0 = - NAA z+z-e2 (1-1n)

40 (r0)


(r+ + r-)

1pm = 1x10-12 m = 10-2 Ả


(r+ + r-)






2

0

11

4 2

c aA

c a

Z ZN Ae vU

r r n

= minus

+

2

0

14 2

c aA

c a c a

Z ZN Ae vU

r r r r

= minus

+ +




2 2

0 0

11

4

AN AZ eU

r n

= minus minus

2 2

0 0 0

14

AN Az eU

r r

= minus minus

Born-Landeacute

Born-Mayer

ndash

ndash




bull n asymp 8

n2 =22 =1

bull2

04

AN e

108 c a

c a

Z ZU v KJ mol

r r=

+-

2

0

11

4 2

c aA

c a

Z ZN Ae vU

r r n

= minus

+

ndash


Ejemplo MgF2

kJmol

Mg(s) rarr Hatm 149

rarr EI1 +EI2 7375 y 1450


rarr -2AE -339 X2

rarr Hordmred= _____________________________________________________________________________________________


-29395



0 0

11

4 r

AN Az eU

n

= minus



Madelung A= 2408


Mg2+(6 86 pm) F- (3116pm) = 202 pm

0202 nm y 202 x 10-10 m

z = 1 2

n=8

U = -28996 kJmol

Zc = +2 Za = -1

v = 3


bull0211nm

U =- 307109 kJmol

Ejemplo MgF2

U0 = - 139 x 10-4 A z+z- (1-1n) Jmol

(r+ + r-) (m)

108 c a

c a

Z ZU v KJ mol

r r=

+-

-

Hordmred= -29395 100986 1045

Uc = -33139 kJmol y UB = 4142






(g)]





kJmol

2ΔH(Na+) 1220

iquestProceso real

AE 1+2 (SO4-2

) = ΔH(SO4-2)

Hordmred

____________________________________________________________________________________



-2)


-2)

kJmol

2ΔH(K+) 1028

iquestProceso real

AE 1+2 (SO4-2

) = ΔH(SO4-2)

Hordmred

____________________________________________________________________________________



-2)


-2)


r(Na+) = 116 pm = 0116 nm r(K+) = 152 pm = 0152 nm

Regla de los radios





6 oct entre 0414 y 0732


12 dodecaedro 1








Reglas de Fajans











EjemploNaX

F Cl Br I kJmol

ΔHatm Na

EI

ΔHatm X

-AE

ΔHred

-569 -410 -360 -288 ΔHf

Ejemplo MI

Li Na K Rb Cs kJmol

161

526

108

502

90

424

82

408

78

381

ΔHatm M

EI

106

-301

106

-301

106

-301

106

-301

106

-301

ΔHatm I

-AE

ΔHred

-271 -288 -328 -329 -337 ΔHf





F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________







F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf


T = 298 K










Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf





Gf = -531- 298(-105) kJmol

1000

Gf = -531 + 313 = -4997 kJmol









(g) AE2

Mg 2+(g) + O2minus


___________________________________________


149 kJmol

738

1450

249

-142

+844

-610

Extra

GASTO

De ENERGIacuteA








____________________________________________


149 kJmol

738

249

-142

108 c a

c a

Z ZU v KJ mol

r r=

+-



y para el O-






Hatm

EI1

EI2= 4567

Hatm

AE1

Hordmred=

___________________________________________









α+

Li+ Be

2+ Cu

+ Zn

2+ Al

3+

0029 0008 043 029 0054

Na+ Mg

2+ Ag

+ Cd

2+ Ga

3+

018 0094 172 109 02

k+ Ca

2+ Au

+ Hg

2+ In

3+

084 047 188 125 073

Rb+ Sr

2+ Tl

3+

142 086 087

Cs+ Ba

2+

244 156

α-

H- F

- O

2-

102 105 392

Cl- S

2-

369 103

Br- Se

2-

481 106

I- Te

2-

716 141





max

1

2Z AU N hv=



(calc)

ΔH 298

(ciclo Born)

NaCl -766 -863 114 -25 8 -776 -787 CsI -580 -619 90 -54 3 -592 -602

AgI -795 -808 138 -128 3 -807 -891

CuBr -870 -925 114 -64 4 -882 -977

108772


CsI -580 es del 964

AgI -795 es del 892

CuBr -870 es del 89


788

AgCl -840 -820 1026 -125 3 -9108


U0 = - NAA z+z-e2 (1-1n)

40 (r0)


(r+ + r-)

1pm = 1x10-12 m = 10-2 Ả


(r+ + r-)






2

0

11

4 2

c aA

c a

Z ZN Ae vU

r r n

= minus

+

2

0

14 2

c aA

c a c a

Z ZN Ae vU

r r r r

= minus

+ +




2 2

0 0

11

4

AN AZ eU

r n

= minus minus

2 2

0 0 0

14

AN Az eU

r r

= minus minus

Born-Landeacute

Born-Mayer

ndash

ndash




bull n asymp 8

n2 =22 =1

bull2

04

AN e

108 c a

c a

Z ZU v KJ mol

r r=

+-

2

0

11

4 2

c aA

c a

Z ZN Ae vU

r r n

= minus

+

ndash


Ejemplo MgF2

kJmol

Mg(s) rarr Hatm 149

rarr EI1 +EI2 7375 y 1450


rarr -2AE -339 X2

rarr Hordmred= _____________________________________________________________________________________________


-29395



0 0

11

4 r

AN Az eU

n

= minus



Madelung A= 2408


Mg2+(6 86 pm) F- (3116pm) = 202 pm

0202 nm y 202 x 10-10 m

z = 1 2

n=8

U = -28996 kJmol

Zc = +2 Za = -1

v = 3


bull0211nm

U =- 307109 kJmol

Ejemplo MgF2

U0 = - 139 x 10-4 A z+z- (1-1n) Jmol

(r+ + r-) (m)

108 c a

c a

Z ZU v KJ mol

r r=

+-

-

Hordmred= -29395 100986 1045

Uc = -33139 kJmol y UB = 4142






(g)]





kJmol

2ΔH(Na+) 1220

iquestProceso real

AE 1+2 (SO4-2

) = ΔH(SO4-2)

Hordmred

____________________________________________________________________________________



-2)


-2)

kJmol

2ΔH(K+) 1028

iquestProceso real

AE 1+2 (SO4-2

) = ΔH(SO4-2)

Hordmred

____________________________________________________________________________________



-2)


-2)


r(Na+) = 116 pm = 0116 nm r(K+) = 152 pm = 0152 nm

Regla de los radios





6 oct entre 0414 y 0732


12 dodecaedro 1








Reglas de Fajans











EjemploNaX

F Cl Br I kJmol

ΔHatm Na

EI

ΔHatm X

-AE

ΔHred

-569 -410 -360 -288 ΔHf

Ejemplo MI

Li Na K Rb Cs kJmol

161

526

108

502

90

424

82

408

78

381

ΔHatm M

EI

106

-301

106

-301

106

-301

106

-301

106

-301

ΔHatm I

-AE

ΔHred

-271 -288 -328 -329 -337 ΔHf





F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________







F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf


T = 298 K










Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf





Gf = -531- 298(-105) kJmol

1000

Gf = -531 + 313 = -4997 kJmol









(g) AE2

Mg 2+(g) + O2minus


___________________________________________


149 kJmol

738

1450

249

-142

+844

-610

Extra

GASTO

De ENERGIacuteA








____________________________________________


149 kJmol

738

249

-142

108 c a

c a

Z ZU v KJ mol

r r=

+-



y para el O-






Hatm

EI1

EI2= 4567

Hatm

AE1

Hordmred=

___________________________________________







max

1

2Z AU N hv=



(calc)

ΔH 298

(ciclo Born)

NaCl -766 -863 114 -25 8 -776 -787 CsI -580 -619 90 -54 3 -592 -602

AgI -795 -808 138 -128 3 -807 -891

CuBr -870 -925 114 -64 4 -882 -977

108772


CsI -580 es del 964

AgI -795 es del 892

CuBr -870 es del 89


788

AgCl -840 -820 1026 -125 3 -9108


U0 = - NAA z+z-e2 (1-1n)

40 (r0)


(r+ + r-)

1pm = 1x10-12 m = 10-2 Ả


(r+ + r-)






2

0

11

4 2

c aA

c a

Z ZN Ae vU

r r n

= minus

+

2

0

14 2

c aA

c a c a

Z ZN Ae vU

r r r r

= minus

+ +




2 2

0 0

11

4

AN AZ eU

r n

= minus minus

2 2

0 0 0

14

AN Az eU

r r

= minus minus

Born-Landeacute

Born-Mayer

ndash

ndash




bull n asymp 8

n2 =22 =1

bull2

04

AN e

108 c a

c a

Z ZU v KJ mol

r r=

+-

2

0

11

4 2

c aA

c a

Z ZN Ae vU

r r n

= minus

+

ndash


Ejemplo MgF2

kJmol

Mg(s) rarr Hatm 149

rarr EI1 +EI2 7375 y 1450


rarr -2AE -339 X2

rarr Hordmred= _____________________________________________________________________________________________


-29395



0 0

11

4 r

AN Az eU

n

= minus



Madelung A= 2408


Mg2+(6 86 pm) F- (3116pm) = 202 pm

0202 nm y 202 x 10-10 m

z = 1 2

n=8

U = -28996 kJmol

Zc = +2 Za = -1

v = 3


bull0211nm

U =- 307109 kJmol

Ejemplo MgF2

U0 = - 139 x 10-4 A z+z- (1-1n) Jmol

(r+ + r-) (m)

108 c a

c a

Z ZU v KJ mol

r r=

+-

-

Hordmred= -29395 100986 1045

Uc = -33139 kJmol y UB = 4142






(g)]





kJmol

2ΔH(Na+) 1220

iquestProceso real

AE 1+2 (SO4-2

) = ΔH(SO4-2)

Hordmred

____________________________________________________________________________________



-2)


-2)

kJmol

2ΔH(K+) 1028

iquestProceso real

AE 1+2 (SO4-2

) = ΔH(SO4-2)

Hordmred

____________________________________________________________________________________



-2)


-2)


r(Na+) = 116 pm = 0116 nm r(K+) = 152 pm = 0152 nm

Regla de los radios





6 oct entre 0414 y 0732


12 dodecaedro 1








Reglas de Fajans











EjemploNaX

F Cl Br I kJmol

ΔHatm Na

EI

ΔHatm X

-AE

ΔHred

-569 -410 -360 -288 ΔHf

Ejemplo MI

Li Na K Rb Cs kJmol

161

526

108

502

90

424

82

408

78

381

ΔHatm M

EI

106

-301

106

-301

106

-301

106

-301

106

-301

ΔHatm I

-AE

ΔHred

-271 -288 -328 -329 -337 ΔHf





F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________







F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf


T = 298 K










Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf





Gf = -531- 298(-105) kJmol

1000

Gf = -531 + 313 = -4997 kJmol









(g) AE2

Mg 2+(g) + O2minus


___________________________________________


149 kJmol

738

1450

249

-142

+844

-610

Extra

GASTO

De ENERGIacuteA








____________________________________________


149 kJmol

738

249

-142

108 c a

c a

Z ZU v KJ mol

r r=

+-



y para el O-






Hatm

EI1

EI2= 4567

Hatm

AE1

Hordmred=

___________________________________________







(calc)

ΔH 298

(ciclo Born)

NaCl -766 -863 114 -25 8 -776 -787 CsI -580 -619 90 -54 3 -592 -602

AgI -795 -808 138 -128 3 -807 -891

CuBr -870 -925 114 -64 4 -882 -977

108772


CsI -580 es del 964

AgI -795 es del 892

CuBr -870 es del 89


788

AgCl -840 -820 1026 -125 3 -9108


U0 = - NAA z+z-e2 (1-1n)

40 (r0)


(r+ + r-)

1pm = 1x10-12 m = 10-2 Ả


(r+ + r-)






2

0

11

4 2

c aA

c a

Z ZN Ae vU

r r n

= minus

+

2

0

14 2

c aA

c a c a

Z ZN Ae vU

r r r r

= minus

+ +




2 2

0 0

11

4

AN AZ eU

r n

= minus minus

2 2

0 0 0

14

AN Az eU

r r

= minus minus

Born-Landeacute

Born-Mayer

ndash

ndash




bull n asymp 8

n2 =22 =1

bull2

04

AN e

108 c a

c a

Z ZU v KJ mol

r r=

+-

2

0

11

4 2

c aA

c a

Z ZN Ae vU

r r n

= minus

+

ndash


Ejemplo MgF2

kJmol

Mg(s) rarr Hatm 149

rarr EI1 +EI2 7375 y 1450


rarr -2AE -339 X2

rarr Hordmred= _____________________________________________________________________________________________


-29395



0 0

11

4 r

AN Az eU

n

= minus



Madelung A= 2408


Mg2+(6 86 pm) F- (3116pm) = 202 pm

0202 nm y 202 x 10-10 m

z = 1 2

n=8

U = -28996 kJmol

Zc = +2 Za = -1

v = 3


bull0211nm

U =- 307109 kJmol

Ejemplo MgF2

U0 = - 139 x 10-4 A z+z- (1-1n) Jmol

(r+ + r-) (m)

108 c a

c a

Z ZU v KJ mol

r r=

+-

-

Hordmred= -29395 100986 1045

Uc = -33139 kJmol y UB = 4142






(g)]





kJmol

2ΔH(Na+) 1220

iquestProceso real

AE 1+2 (SO4-2

) = ΔH(SO4-2)

Hordmred

____________________________________________________________________________________



-2)


-2)

kJmol

2ΔH(K+) 1028

iquestProceso real

AE 1+2 (SO4-2

) = ΔH(SO4-2)

Hordmred

____________________________________________________________________________________



-2)


-2)


r(Na+) = 116 pm = 0116 nm r(K+) = 152 pm = 0152 nm

Regla de los radios





6 oct entre 0414 y 0732


12 dodecaedro 1








Reglas de Fajans











EjemploNaX

F Cl Br I kJmol

ΔHatm Na

EI

ΔHatm X

-AE

ΔHred

-569 -410 -360 -288 ΔHf

Ejemplo MI

Li Na K Rb Cs kJmol

161

526

108

502

90

424

82

408

78

381

ΔHatm M

EI

106

-301

106

-301

106

-301

106

-301

106

-301

ΔHatm I

-AE

ΔHred

-271 -288 -328 -329 -337 ΔHf





F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________







F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf


T = 298 K










Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf





Gf = -531- 298(-105) kJmol

1000

Gf = -531 + 313 = -4997 kJmol









(g) AE2

Mg 2+(g) + O2minus


___________________________________________


149 kJmol

738

1450

249

-142

+844

-610

Extra

GASTO

De ENERGIacuteA








____________________________________________


149 kJmol

738

249

-142

108 c a

c a

Z ZU v KJ mol

r r=

+-



y para el O-






Hatm

EI1

EI2= 4567

Hatm

AE1

Hordmred=

___________________________________________






U0 = - NAA z+z-e2 (1-1n)

40 (r0)


(r+ + r-)

1pm = 1x10-12 m = 10-2 Ả


(r+ + r-)






2

0

11

4 2

c aA

c a

Z ZN Ae vU

r r n

= minus

+

2

0

14 2

c aA

c a c a

Z ZN Ae vU

r r r r

= minus

+ +




2 2

0 0

11

4

AN AZ eU

r n

= minus minus

2 2

0 0 0

14

AN Az eU

r r

= minus minus

Born-Landeacute

Born-Mayer

ndash

ndash




bull n asymp 8

n2 =22 =1

bull2

04

AN e

108 c a

c a

Z ZU v KJ mol

r r=

+-

2

0

11

4 2

c aA

c a

Z ZN Ae vU

r r n

= minus

+

ndash


Ejemplo MgF2

kJmol

Mg(s) rarr Hatm 149

rarr EI1 +EI2 7375 y 1450


rarr -2AE -339 X2

rarr Hordmred= _____________________________________________________________________________________________


-29395



0 0

11

4 r

AN Az eU

n

= minus



Madelung A= 2408


Mg2+(6 86 pm) F- (3116pm) = 202 pm

0202 nm y 202 x 10-10 m

z = 1 2

n=8

U = -28996 kJmol

Zc = +2 Za = -1

v = 3


bull0211nm

U =- 307109 kJmol

Ejemplo MgF2

U0 = - 139 x 10-4 A z+z- (1-1n) Jmol

(r+ + r-) (m)

108 c a

c a

Z ZU v KJ mol

r r=

+-

-

Hordmred= -29395 100986 1045

Uc = -33139 kJmol y UB = 4142






(g)]





kJmol

2ΔH(Na+) 1220

iquestProceso real

AE 1+2 (SO4-2

) = ΔH(SO4-2)

Hordmred

____________________________________________________________________________________



-2)


-2)

kJmol

2ΔH(K+) 1028

iquestProceso real

AE 1+2 (SO4-2

) = ΔH(SO4-2)

Hordmred

____________________________________________________________________________________



-2)


-2)


r(Na+) = 116 pm = 0116 nm r(K+) = 152 pm = 0152 nm

Regla de los radios





6 oct entre 0414 y 0732


12 dodecaedro 1








Reglas de Fajans











EjemploNaX

F Cl Br I kJmol

ΔHatm Na

EI

ΔHatm X

-AE

ΔHred

-569 -410 -360 -288 ΔHf

Ejemplo MI

Li Na K Rb Cs kJmol

161

526

108

502

90

424

82

408

78

381

ΔHatm M

EI

106

-301

106

-301

106

-301

106

-301

106

-301

ΔHatm I

-AE

ΔHred

-271 -288 -328 -329 -337 ΔHf





F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________







F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf


T = 298 K










Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf





Gf = -531- 298(-105) kJmol

1000

Gf = -531 + 313 = -4997 kJmol









(g) AE2

Mg 2+(g) + O2minus


___________________________________________


149 kJmol

738

1450

249

-142

+844

-610

Extra

GASTO

De ENERGIacuteA








____________________________________________


149 kJmol

738

249

-142

108 c a

c a

Z ZU v KJ mol

r r=

+-



y para el O-






Hatm

EI1

EI2= 4567

Hatm

AE1

Hordmred=

___________________________________________










2

0

11

4 2

c aA

c a

Z ZN Ae vU

r r n

= minus

+

2

0

14 2

c aA

c a c a

Z ZN Ae vU

r r r r

= minus

+ +




2 2

0 0

11

4

AN AZ eU

r n

= minus minus

2 2

0 0 0

14

AN Az eU

r r

= minus minus

Born-Landeacute

Born-Mayer

ndash

ndash




bull n asymp 8

n2 =22 =1

bull2

04

AN e

108 c a

c a

Z ZU v KJ mol

r r=

+-

2

0

11

4 2

c aA

c a

Z ZN Ae vU

r r n

= minus

+

ndash


Ejemplo MgF2

kJmol

Mg(s) rarr Hatm 149

rarr EI1 +EI2 7375 y 1450


rarr -2AE -339 X2

rarr Hordmred= _____________________________________________________________________________________________


-29395



0 0

11

4 r

AN Az eU

n

= minus



Madelung A= 2408


Mg2+(6 86 pm) F- (3116pm) = 202 pm

0202 nm y 202 x 10-10 m

z = 1 2

n=8

U = -28996 kJmol

Zc = +2 Za = -1

v = 3


bull0211nm

U =- 307109 kJmol

Ejemplo MgF2

U0 = - 139 x 10-4 A z+z- (1-1n) Jmol

(r+ + r-) (m)

108 c a

c a

Z ZU v KJ mol

r r=

+-

-

Hordmred= -29395 100986 1045

Uc = -33139 kJmol y UB = 4142






(g)]





kJmol

2ΔH(Na+) 1220

iquestProceso real

AE 1+2 (SO4-2

) = ΔH(SO4-2)

Hordmred

____________________________________________________________________________________



-2)


-2)

kJmol

2ΔH(K+) 1028

iquestProceso real

AE 1+2 (SO4-2

) = ΔH(SO4-2)

Hordmred

____________________________________________________________________________________



-2)


-2)


r(Na+) = 116 pm = 0116 nm r(K+) = 152 pm = 0152 nm

Regla de los radios





6 oct entre 0414 y 0732


12 dodecaedro 1








Reglas de Fajans











EjemploNaX

F Cl Br I kJmol

ΔHatm Na

EI

ΔHatm X

-AE

ΔHred

-569 -410 -360 -288 ΔHf

Ejemplo MI

Li Na K Rb Cs kJmol

161

526

108

502

90

424

82

408

78

381

ΔHatm M

EI

106

-301

106

-301

106

-301

106

-301

106

-301

ΔHatm I

-AE

ΔHred

-271 -288 -328 -329 -337 ΔHf





F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________







F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf


T = 298 K










Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf





Gf = -531- 298(-105) kJmol

1000

Gf = -531 + 313 = -4997 kJmol









(g) AE2

Mg 2+(g) + O2minus


___________________________________________


149 kJmol

738

1450

249

-142

+844

-610

Extra

GASTO

De ENERGIacuteA








____________________________________________


149 kJmol

738

249

-142

108 c a

c a

Z ZU v KJ mol

r r=

+-



y para el O-






Hatm

EI1

EI2= 4567

Hatm

AE1

Hordmred=

___________________________________________








bull n asymp 8

n2 =22 =1

bull2

04

AN e

108 c a

c a

Z ZU v KJ mol

r r=

+-

2

0

11

4 2

c aA

c a

Z ZN Ae vU

r r n

= minus

+

ndash


Ejemplo MgF2

kJmol

Mg(s) rarr Hatm 149

rarr EI1 +EI2 7375 y 1450


rarr -2AE -339 X2

rarr Hordmred= _____________________________________________________________________________________________


-29395



0 0

11

4 r

AN Az eU

n

= minus



Madelung A= 2408


Mg2+(6 86 pm) F- (3116pm) = 202 pm

0202 nm y 202 x 10-10 m

z = 1 2

n=8

U = -28996 kJmol

Zc = +2 Za = -1

v = 3


bull0211nm

U =- 307109 kJmol

Ejemplo MgF2

U0 = - 139 x 10-4 A z+z- (1-1n) Jmol

(r+ + r-) (m)

108 c a

c a

Z ZU v KJ mol

r r=

+-

-

Hordmred= -29395 100986 1045

Uc = -33139 kJmol y UB = 4142






(g)]





kJmol

2ΔH(Na+) 1220

iquestProceso real

AE 1+2 (SO4-2

) = ΔH(SO4-2)

Hordmred

____________________________________________________________________________________



-2)


-2)

kJmol

2ΔH(K+) 1028

iquestProceso real

AE 1+2 (SO4-2

) = ΔH(SO4-2)

Hordmred

____________________________________________________________________________________



-2)


-2)


r(Na+) = 116 pm = 0116 nm r(K+) = 152 pm = 0152 nm

Regla de los radios





6 oct entre 0414 y 0732


12 dodecaedro 1








Reglas de Fajans











EjemploNaX

F Cl Br I kJmol

ΔHatm Na

EI

ΔHatm X

-AE

ΔHred

-569 -410 -360 -288 ΔHf

Ejemplo MI

Li Na K Rb Cs kJmol

161

526

108

502

90

424

82

408

78

381

ΔHatm M

EI

106

-301

106

-301

106

-301

106

-301

106

-301

ΔHatm I

-AE

ΔHred

-271 -288 -328 -329 -337 ΔHf





F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________







F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf


T = 298 K










Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf





Gf = -531- 298(-105) kJmol

1000

Gf = -531 + 313 = -4997 kJmol









(g) AE2

Mg 2+(g) + O2minus


___________________________________________


149 kJmol

738

1450

249

-142

+844

-610

Extra

GASTO

De ENERGIacuteA








____________________________________________


149 kJmol

738

249

-142

108 c a

c a

Z ZU v KJ mol

r r=

+-



y para el O-






Hatm

EI1

EI2= 4567

Hatm

AE1

Hordmred=

___________________________________________






Ejemplo MgF2

kJmol

Mg(s) rarr Hatm 149

rarr EI1 +EI2 7375 y 1450


rarr -2AE -339 X2

rarr Hordmred= _____________________________________________________________________________________________


-29395



0 0

11

4 r

AN Az eU

n

= minus



Madelung A= 2408


Mg2+(6 86 pm) F- (3116pm) = 202 pm

0202 nm y 202 x 10-10 m

z = 1 2

n=8

U = -28996 kJmol

Zc = +2 Za = -1

v = 3


bull0211nm

U =- 307109 kJmol

Ejemplo MgF2

U0 = - 139 x 10-4 A z+z- (1-1n) Jmol

(r+ + r-) (m)

108 c a

c a

Z ZU v KJ mol

r r=

+-

-

Hordmred= -29395 100986 1045

Uc = -33139 kJmol y UB = 4142






(g)]





kJmol

2ΔH(Na+) 1220

iquestProceso real

AE 1+2 (SO4-2

) = ΔH(SO4-2)

Hordmred

____________________________________________________________________________________



-2)


-2)

kJmol

2ΔH(K+) 1028

iquestProceso real

AE 1+2 (SO4-2

) = ΔH(SO4-2)

Hordmred

____________________________________________________________________________________



-2)


-2)


r(Na+) = 116 pm = 0116 nm r(K+) = 152 pm = 0152 nm

Regla de los radios





6 oct entre 0414 y 0732


12 dodecaedro 1








Reglas de Fajans











EjemploNaX

F Cl Br I kJmol

ΔHatm Na

EI

ΔHatm X

-AE

ΔHred

-569 -410 -360 -288 ΔHf

Ejemplo MI

Li Na K Rb Cs kJmol

161

526

108

502

90

424

82

408

78

381

ΔHatm M

EI

106

-301

106

-301

106

-301

106

-301

106

-301

ΔHatm I

-AE

ΔHred

-271 -288 -328 -329 -337 ΔHf





F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________







F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf


T = 298 K










Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf





Gf = -531- 298(-105) kJmol

1000

Gf = -531 + 313 = -4997 kJmol









(g) AE2

Mg 2+(g) + O2minus


___________________________________________


149 kJmol

738

1450

249

-142

+844

-610

Extra

GASTO

De ENERGIacuteA








____________________________________________


149 kJmol

738

249

-142

108 c a

c a

Z ZU v KJ mol

r r=

+-



y para el O-






Hatm

EI1

EI2= 4567

Hatm

AE1

Hordmred=

___________________________________________







0 0

11

4 r

AN Az eU

n

= minus



Madelung A= 2408


Mg2+(6 86 pm) F- (3116pm) = 202 pm

0202 nm y 202 x 10-10 m

z = 1 2

n=8

U = -28996 kJmol

Zc = +2 Za = -1

v = 3


bull0211nm

U =- 307109 kJmol

Ejemplo MgF2

U0 = - 139 x 10-4 A z+z- (1-1n) Jmol

(r+ + r-) (m)

108 c a

c a

Z ZU v KJ mol

r r=

+-

-

Hordmred= -29395 100986 1045

Uc = -33139 kJmol y UB = 4142






(g)]





kJmol

2ΔH(Na+) 1220

iquestProceso real

AE 1+2 (SO4-2

) = ΔH(SO4-2)

Hordmred

____________________________________________________________________________________



-2)


-2)

kJmol

2ΔH(K+) 1028

iquestProceso real

AE 1+2 (SO4-2

) = ΔH(SO4-2)

Hordmred

____________________________________________________________________________________



-2)


-2)


r(Na+) = 116 pm = 0116 nm r(K+) = 152 pm = 0152 nm

Regla de los radios





6 oct entre 0414 y 0732


12 dodecaedro 1








Reglas de Fajans











EjemploNaX

F Cl Br I kJmol

ΔHatm Na

EI

ΔHatm X

-AE

ΔHred

-569 -410 -360 -288 ΔHf

Ejemplo MI

Li Na K Rb Cs kJmol

161

526

108

502

90

424

82

408

78

381

ΔHatm M

EI

106

-301

106

-301

106

-301

106

-301

106

-301

ΔHatm I

-AE

ΔHred

-271 -288 -328 -329 -337 ΔHf





F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________







F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf


T = 298 K










Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf





Gf = -531- 298(-105) kJmol

1000

Gf = -531 + 313 = -4997 kJmol









(g) AE2

Mg 2+(g) + O2minus


___________________________________________


149 kJmol

738

1450

249

-142

+844

-610

Extra

GASTO

De ENERGIacuteA








____________________________________________


149 kJmol

738

249

-142

108 c a

c a

Z ZU v KJ mol

r r=

+-



y para el O-






Hatm

EI1

EI2= 4567

Hatm

AE1

Hordmred=

___________________________________________










(g)]





kJmol

2ΔH(Na+) 1220

iquestProceso real

AE 1+2 (SO4-2

) = ΔH(SO4-2)

Hordmred

____________________________________________________________________________________



-2)


-2)

kJmol

2ΔH(K+) 1028

iquestProceso real

AE 1+2 (SO4-2

) = ΔH(SO4-2)

Hordmred

____________________________________________________________________________________



-2)


-2)


r(Na+) = 116 pm = 0116 nm r(K+) = 152 pm = 0152 nm

Regla de los radios





6 oct entre 0414 y 0732


12 dodecaedro 1








Reglas de Fajans











EjemploNaX

F Cl Br I kJmol

ΔHatm Na

EI

ΔHatm X

-AE

ΔHred

-569 -410 -360 -288 ΔHf

Ejemplo MI

Li Na K Rb Cs kJmol

161

526

108

502

90

424

82

408

78

381

ΔHatm M

EI

106

-301

106

-301

106

-301

106

-301

106

-301

ΔHatm I

-AE

ΔHred

-271 -288 -328 -329 -337 ΔHf





F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________







F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf


T = 298 K










Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf





Gf = -531- 298(-105) kJmol

1000

Gf = -531 + 313 = -4997 kJmol









(g) AE2

Mg 2+(g) + O2minus


___________________________________________


149 kJmol

738

1450

249

-142

+844

-610

Extra

GASTO

De ENERGIacuteA








____________________________________________


149 kJmol

738

249

-142

108 c a

c a

Z ZU v KJ mol

r r=

+-



y para el O-






Hatm

EI1

EI2= 4567

Hatm

AE1

Hordmred=

___________________________________________





kJmol

2ΔH(Na+) 1220

iquestProceso real

AE 1+2 (SO4-2

) = ΔH(SO4-2)

Hordmred

____________________________________________________________________________________



-2)


-2)

kJmol

2ΔH(K+) 1028

iquestProceso real

AE 1+2 (SO4-2

) = ΔH(SO4-2)

Hordmred

____________________________________________________________________________________



-2)


-2)


r(Na+) = 116 pm = 0116 nm r(K+) = 152 pm = 0152 nm

Regla de los radios





6 oct entre 0414 y 0732


12 dodecaedro 1








Reglas de Fajans











EjemploNaX

F Cl Br I kJmol

ΔHatm Na

EI

ΔHatm X

-AE

ΔHred

-569 -410 -360 -288 ΔHf

Ejemplo MI

Li Na K Rb Cs kJmol

161

526

108

502

90

424

82

408

78

381

ΔHatm M

EI

106

-301

106

-301

106

-301

106

-301

106

-301

ΔHatm I

-AE

ΔHred

-271 -288 -328 -329 -337 ΔHf





F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________







F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf


T = 298 K










Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf





Gf = -531- 298(-105) kJmol

1000

Gf = -531 + 313 = -4997 kJmol









(g) AE2

Mg 2+(g) + O2minus


___________________________________________


149 kJmol

738

1450

249

-142

+844

-610

Extra

GASTO

De ENERGIacuteA








____________________________________________


149 kJmol

738

249

-142

108 c a

c a

Z ZU v KJ mol

r r=

+-



y para el O-






Hatm

EI1

EI2= 4567

Hatm

AE1

Hordmred=

___________________________________________






-2)

kJmol

2ΔH(K+) 1028

iquestProceso real

AE 1+2 (SO4-2

) = ΔH(SO4-2)

Hordmred

____________________________________________________________________________________



-2)


-2)


r(Na+) = 116 pm = 0116 nm r(K+) = 152 pm = 0152 nm

Regla de los radios





6 oct entre 0414 y 0732


12 dodecaedro 1








Reglas de Fajans











EjemploNaX

F Cl Br I kJmol

ΔHatm Na

EI

ΔHatm X

-AE

ΔHred

-569 -410 -360 -288 ΔHf

Ejemplo MI

Li Na K Rb Cs kJmol

161

526

108

502

90

424

82

408

78

381

ΔHatm M

EI

106

-301

106

-301

106

-301

106

-301

106

-301

ΔHatm I

-AE

ΔHred

-271 -288 -328 -329 -337 ΔHf





F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________







F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf


T = 298 K










Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf





Gf = -531- 298(-105) kJmol

1000

Gf = -531 + 313 = -4997 kJmol









(g) AE2

Mg 2+(g) + O2minus


___________________________________________


149 kJmol

738

1450

249

-142

+844

-610

Extra

GASTO

De ENERGIacuteA








____________________________________________


149 kJmol

738

249

-142

108 c a

c a

Z ZU v KJ mol

r r=

+-



y para el O-






Hatm

EI1

EI2= 4567

Hatm

AE1

Hordmred=

___________________________________________






-2)


-2)


r(Na+) = 116 pm = 0116 nm r(K+) = 152 pm = 0152 nm

Regla de los radios





6 oct entre 0414 y 0732


12 dodecaedro 1








Reglas de Fajans











EjemploNaX

F Cl Br I kJmol

ΔHatm Na

EI

ΔHatm X

-AE

ΔHred

-569 -410 -360 -288 ΔHf

Ejemplo MI

Li Na K Rb Cs kJmol

161

526

108

502

90

424

82

408

78

381

ΔHatm M

EI

106

-301

106

-301

106

-301

106

-301

106

-301

ΔHatm I

-AE

ΔHred

-271 -288 -328 -329 -337 ΔHf





F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________







F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf


T = 298 K










Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf





Gf = -531- 298(-105) kJmol

1000

Gf = -531 + 313 = -4997 kJmol









(g) AE2

Mg 2+(g) + O2minus


___________________________________________


149 kJmol

738

1450

249

-142

+844

-610

Extra

GASTO

De ENERGIacuteA








____________________________________________


149 kJmol

738

249

-142

108 c a

c a

Z ZU v KJ mol

r r=

+-



y para el O-






Hatm

EI1

EI2= 4567

Hatm

AE1

Hordmred=

___________________________________________





Regla de los radios





6 oct entre 0414 y 0732


12 dodecaedro 1








Reglas de Fajans











EjemploNaX

F Cl Br I kJmol

ΔHatm Na

EI

ΔHatm X

-AE

ΔHred

-569 -410 -360 -288 ΔHf

Ejemplo MI

Li Na K Rb Cs kJmol

161

526

108

502

90

424

82

408

78

381

ΔHatm M

EI

106

-301

106

-301

106

-301

106

-301

106

-301

ΔHatm I

-AE

ΔHred

-271 -288 -328 -329 -337 ΔHf





F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________







F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf


T = 298 K










Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf





Gf = -531- 298(-105) kJmol

1000

Gf = -531 + 313 = -4997 kJmol









(g) AE2

Mg 2+(g) + O2minus


___________________________________________


149 kJmol

738

1450

249

-142

+844

-610

Extra

GASTO

De ENERGIacuteA








____________________________________________


149 kJmol

738

249

-142

108 c a

c a

Z ZU v KJ mol

r r=

+-



y para el O-






Hatm

EI1

EI2= 4567

Hatm

AE1

Hordmred=

___________________________________________





Reglas de Fajans











EjemploNaX

F Cl Br I kJmol

ΔHatm Na

EI

ΔHatm X

-AE

ΔHred

-569 -410 -360 -288 ΔHf

Ejemplo MI

Li Na K Rb Cs kJmol

161

526

108

502

90

424

82

408

78

381

ΔHatm M

EI

106

-301

106

-301

106

-301

106

-301

106

-301

ΔHatm I

-AE

ΔHred

-271 -288 -328 -329 -337 ΔHf





F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________







F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf


T = 298 K










Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf





Gf = -531- 298(-105) kJmol

1000

Gf = -531 + 313 = -4997 kJmol









(g) AE2

Mg 2+(g) + O2minus


___________________________________________


149 kJmol

738

1450

249

-142

+844

-610

Extra

GASTO

De ENERGIacuteA








____________________________________________


149 kJmol

738

249

-142

108 c a

c a

Z ZU v KJ mol

r r=

+-



y para el O-






Hatm

EI1

EI2= 4567

Hatm

AE1

Hordmred=

___________________________________________





EjemploNaX

F Cl Br I kJmol

ΔHatm Na

EI

ΔHatm X

-AE

ΔHred

-569 -410 -360 -288 ΔHf

Ejemplo MI

Li Na K Rb Cs kJmol

161

526

108

502

90

424

82

408

78

381

ΔHatm M

EI

106

-301

106

-301

106

-301

106

-301

106

-301

ΔHatm I

-AE

ΔHred

-271 -288 -328 -329 -337 ΔHf





F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________







F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf


T = 298 K










Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf





Gf = -531- 298(-105) kJmol

1000

Gf = -531 + 313 = -4997 kJmol









(g) AE2

Mg 2+(g) + O2minus


___________________________________________


149 kJmol

738

1450

249

-142

+844

-610

Extra

GASTO

De ENERGIacuteA








____________________________________________


149 kJmol

738

249

-142

108 c a

c a

Z ZU v KJ mol

r r=

+-



y para el O-






Hatm

EI1

EI2= 4567

Hatm

AE1

Hordmred=

___________________________________________





Ejemplo MI

Li Na K Rb Cs kJmol

161

526

108

502

90

424

82

408

78

381

ΔHatm M

EI

106

-301

106

-301

106

-301

106

-301

106

-301

ΔHatm I

-AE

ΔHred

-271 -288 -328 -329 -337 ΔHf





F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________







F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf


T = 298 K










Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf





Gf = -531- 298(-105) kJmol

1000

Gf = -531 + 313 = -4997 kJmol









(g) AE2

Mg 2+(g) + O2minus


___________________________________________


149 kJmol

738

1450

249

-142

+844

-610

Extra

GASTO

De ENERGIacuteA








____________________________________________


149 kJmol

738

249

-142

108 c a

c a

Z ZU v KJ mol

r r=

+-



y para el O-






Hatm

EI1

EI2= 4567

Hatm

AE1

Hordmred=

___________________________________________









F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________







F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf


T = 298 K










Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf





Gf = -531- 298(-105) kJmol

1000

Gf = -531 + 313 = -4997 kJmol









(g) AE2

Mg 2+(g) + O2minus


___________________________________________


149 kJmol

738

1450

249

-142

+844

-610

Extra

GASTO

De ENERGIacuteA








____________________________________________


149 kJmol

738

249

-142

108 c a

c a

Z ZU v KJ mol

r r=

+-



y para el O-






Hatm

EI1

EI2= 4567

Hatm

AE1

Hordmred=

___________________________________________








F(g) + e rarr F-(g) -AE -333

Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf


T = 298 K










Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf





Gf = -531- 298(-105) kJmol

1000

Gf = -531 + 313 = -4997 kJmol









(g) AE2

Mg 2+(g) + O2minus


___________________________________________


149 kJmol

738

1450

249

-142

+844

-610

Extra

GASTO

De ENERGIacuteA








____________________________________________


149 kJmol

738

249

-142

108 c a

c a

Z ZU v KJ mol

r r=

+-



y para el O-






Hatm

EI1

EI2= 4567

Hatm

AE1

Hordmred=

___________________________________________













Cs +(g) + F-


________________________________________


Gf = Hf ndash T Sf





Gf = -531- 298(-105) kJmol

1000

Gf = -531 + 313 = -4997 kJmol









(g) AE2

Mg 2+(g) + O2minus


___________________________________________


149 kJmol

738

1450

249

-142

+844

-610

Extra

GASTO

De ENERGIacuteA








____________________________________________


149 kJmol

738

249

-142

108 c a

c a

Z ZU v KJ mol

r r=

+-



y para el O-






Hatm

EI1

EI2= 4567

Hatm

AE1

Hordmred=

___________________________________________













(g) AE2

Mg 2+(g) + O2minus


___________________________________________


149 kJmol

738

1450

249

-142

+844

-610

Extra

GASTO

De ENERGIacuteA








____________________________________________


149 kJmol

738

249

-142

108 c a

c a

Z ZU v KJ mol

r r=

+-



y para el O-






Hatm

EI1

EI2= 4567

Hatm

AE1

Hordmred=

___________________________________________












____________________________________________


149 kJmol

738

249

-142

108 c a

c a

Z ZU v KJ mol

r r=

+-



y para el O-






Hatm

EI1

EI2= 4567

Hatm

AE1

Hordmred=

___________________________________________








Hatm

EI1

EI2= 4567

Hatm

AE1

Hordmred=

___________________________________________





Cristal iónicodepa.fquim.unam.mx/amyd/archivero/Diapo2_35658.pdf · 2019-02-20 · Cristal...

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