IB Chemistry on Born Haber Cycle and Lattice Enthalpy

1. http://lawrencekok.blogspot.com
Prepared by
Lawrence Kok
Video Tutorial on Born Haber Cycle and Lattice
Enthalpy.

2. Na (s) Na (g)
Born Haber Cycle/BHC
Li+
(g) + CI–
(g) → LiCI (s)
Multi stage Hess’s Cycle Find Lattice enthalpy for IONIC COMPOUND
A → D / A → B → C → D/ ∆H1 = H2 + H3 + H4
∆H1
∆H2 ∆H4
∆H3 Lattice Enthalpy
+∆H (Heat absorb) to convert 1 MOL IONIC
compound to GASEOUS ions
Lattice Enthalpy
-∆H (Heat release) when 1 MOL IONIC
compound form from GASEOUS ions
LiCI (s) → Li+
(g) + CI–
(g)
Li (g) → Li+
(g) + e
2nd Ionization enthalpy
+ ∆H when 1 MOL e removed from
1 MOL unipositive ion in gaseous state
Li+
(g) → Li2+
(g) + e
Formation Enthalpy
-∆H (Heat release) when 1 MOL compound
form from its element under std condition
Na(s) + ½CI2(g) → NaCI(s)
Std Enthalpy Changes ∆Hθ needed for BHC
Ionization Enthalpy
1st Ionization enthalpy
+ ∆H when 1 MOL e removed
from 1 MOL atom in gaseous state
Li Li+
+ e-
+ e-
Li+ Li2+
++ 2+
Electron affinity Enthalpy
+ e-
-
CI-
CI
Electron Affinity enthalpy
-∆H when 1 MOL GASEOUS
atom gain 1 mol electron
CI (g) + e → CI -
(g)
Gaseous
state
Atomization Enthalpy
Atomization enthalpy
+ ∆H when 1 MOL GASEOUS atom form
from its element under STD condition
Formation Enthalpy
Na CI2 NaCI
solid gas
Na(s) → Na(g) ½H2 (g) → H (g) ½O2 (g) → O (g)
½H2 (g) H (g)
1 mol gas
H2 (g) → 2H (g)
O2 (g) → 2O (g)
½O2 (g) O (g)
+∆H -∆H
NOT
atomization
enthalpy
2 mol gas
gas

3. LiCI(s)
Li+
(g) + CI (g)Li+
(g) + CI (g)
Li+
(g) + ½CI2 (g)
Li(g) + ½CI2 (g)
LiCI(s)
∆H atom + ∆H ion + ∆H EA
( Determined
experimentally)
LiCI(s)
∆H lattice = ?????
( Can’t be determined
experimentally)
∆H form = - 409
( Determined
experimentally)
Li(s) + ½CI2 (g)
Born Haber Cycle/BHC
Li+
(g) + CI–
(g) → LiCI (s)
Find Lattice enthalpy for IONIC COMPOUND
Lattice Enthalpy
+∆H (Heat absorb) to convert 1 MOL IONIC
compound to GASEOUS ions
Lattice Enthalpy
-∆H (Heat release) when 1 MOL IONIC
compound form from GASEOUS ions
LiCI (s) → Li+
(g) + CI–
(g)
Li (g) → Li +
(g) + e
Formation Enthalpy
-∆H (Heat release) when 1 MOL compound
form from its element under std condition
Li(s) + ½CI2(g) → LiCI(s)
Std Enthalpy Changes ∆Hθ needed for BHC
Ionization Enthalpy
1st Ionization enthalpy
+ ∆H when 1 MOL e removed
from 1 MOL atom in gaseous state
Li Li+
+ e-
+
Electron affinity Enthalpy
+ e-
-
CI-
CI
Electron Affinity enthalpy
-∆H when 1 MOL GASEOUS
atom gain 1 mol electron
CI (g) + e → CI -
(g)
Gaseous
state
Atomization Enthalpy
Atomization enthalpy
+ ∆H when 1 MOL GASEOUS atom form
from its element under STD condition
Formation Enthalpy
Li CI2 LiCI
½CI2 (g) → CI (g)
+∆H -∆H
½CI2 (g) CI (g)
Find Lattice enthalpy for
IONIC COMPOUND using BHC
LiCI(s)
Li+
(g) + CI–
(g) Li+
(g) + CI–
(g)
∆Hlatt
∆H form = - 409
Li(s) + ½CI2 (g)
∆Hatom = + 159
Li(s) → Li(g)
Li+
(g) + CI-
(g)
∆Hie = + 520
Li(g) → Li+
(g)
∆Hatom = + 121
½CI2(g) → CI(g)
∆He = - 364
CI(g) → CI -
(g)
∆Hlatt = -845

4. NaCI (s)
Na+
(g) + CI (g)Na+
(g) + CI (g)
Na+
(g) + ½CI2 (g)
Na(g) + ½CI2 (g)
NaCI (s)
∆H atom + ∆H ion + ∆H EA
( Determined
experimentally)
NaCI (s)
∆H lattice = ?????
( Can’t be determined
experimentally)
∆H form = - 414
( Determined
experimentally)
Na(s) + ½CI2 (g)
Born Haber Cycle/BHC
Na+
(g) + CI–
(g) → NaCI (s)
Find Lattice enthalpy for IONIC COMPOUND
Lattice Enthalpy
+∆H (Heat absorb) to convert 1 MOL IONIC
compound to GASEOUS ions
Lattice Enthalpy
-∆H (Heat release) when 1 MOL IONIC
compound form from GASEOUS ions
NaCI (s) → Na+
(g) + CI–
(g)
Na (g) → Na +
(g) + e
Formation Enthalpy
-∆H (Heat release) when 1 MOL compound
form from its element under std condition
Na(s) + ½CI2(g) → NaCI(s)
Std Enthalpy Changes ∆Hθ needed for BHC
Ionization Enthalpy
1st Ionization enthalpy
+ ∆H when 1 MOL e removed
from 1 MOL atom in gaseous state
Na Na+
+ e-
+
Electron affinity Enthalpy
+ e-
-
CI-
CI
Electron Affinity enthalpy
-∆H when 1 MOL GASEOUS
atom gain 1 mol electron
CI (g) + e → CI -
(g)
Gaseous
state
Atomization Enthalpy
Atomization enthalpy
+ ∆H when 1 MOL GASEOUS atom form
from its element under STD condition
Formation Enthalpy
Na CI2 NaCI
½CI2 (g) → CI (g)
+∆H -∆H
½CI2 (g) CI (g)
Find Lattice enthalpy for
IONIC COMPOUND using BHC
NaCI (s)
Na+
(g) + CI–
(g) Na+
(g) + CI–
(g)
∆Hlatt
∆H form = - 414
Na(s) + ½CI2 (g)
∆Hatom = + 108
Na(s) → Na(g)
Na+
(g) + CI-
(g)
∆Hie = + 500
Na(g) → Na+
(g)
∆Hatom = + 121
½CI2(g) → CI(g)
∆He = - 364
CI(g) → CI -
(g)
∆Hlatt = -790

5. KCI (s)
K+
(g) + CI (g)K+
(g) + CI (g)
K+
(g) + ½CI2 (g)
K(g) + ½CI2 (g)
KCI (s)
∆H atom + ∆H ion + ∆H EA
( Determined
experimentally)
KCI (s)
∆H lattice = ?????
( Can’t be determined
experimentally)
∆H form = - 436
( Determined
experimentally)
K(s) + ½CI2 (g)
Born Haber Cycle/BHC
K+
(g) + CI–
(g) → KCI (s)
Find Lattice enthalpy for IONIC COMPOUND
Lattice Enthalpy
+∆H (Heat absorb) to convert 1 MOL IONIC
compound to GASEOUS ions
Lattice Enthalpy
-∆H (Heat release) when 1 MOL IONIC
compound form from GASEOUS ions
KCI (s) → K+
(g) + CI–
(g)
K (g) → K +
(g) + e
Formation Enthalpy
-∆H (Heat release) when 1 MOL compound
form from its element under std condition
K(s) + ½CI2(g) → KCI(s)
Std Enthalpy Changes ∆Hθ needed for BHC
Ionization Enthalpy
1st Ionization enthalpy
+ ∆H when 1 MOL e removed
from 1 MOL atom in gaseous state
K K+
+ e-
+
Electron affinity Enthalpy
+ e-
-
CI-
CI
Electron Affinity enthalpy
-∆H when 1 MOL GASEOUS
atom gain 1 mol electron
CI (g) + e → CI -
(g)
Gaseous
state
Atomization Enthalpy
Atomization enthalpy
+ ∆H when 1 MOL GASEOUS atom form
from its element under STD condition
Formation Enthalpy
K CI2 KCI
½CI2 (g) → CI (g)
+∆H -∆H
½CI2 (g) CI (g)
Find Lattice enthalpy for
IONIC COMPOUND using BHC
KCI (s)
K+
(g) + CI–
(g) K+
(g) + CI–
(g)
∆Hlatt
∆H form = - 436
K(s) + ½CI2 (g)
∆Hatom = + 89
K(s) → K(g)
K+
(g) + CI-
(g)
∆Hie = + 425
K(g) → K+
(g)
∆Hatom = + 121
½CI2(g) → CI(g)
∆He = - 364
CI(g) → CI -
(g)
∆Hlatt = -720

6. NaBr(s)
Na+
(g) + Br (g)Na+
(g) + Br (g)
Na+
(g) + ½Br2 (g)
Na(g) + ½Br2 (g)
NaBr(s)
∆H atom + ∆H ion + ∆H EA
( Determined
experimentally)
NaBr(s)
∆H lattice = ?????
( Can’t be determined
experimentally)
∆H form = - 361
( Determined
experimentally)
Na(s) + ½Br2 (g)
Born Haber Cycle/BHC
Na+
(g) + Br–
(g) → NaBr (s)
Find Lattice enthalpy for IONIC COMPOUND
Lattice Enthalpy
+∆H (Heat absorb) to convert 1 MOL IONIC
compound to GASEOUS ions
Lattice Enthalpy
-∆H (Heat release) when 1 MOL IONIC
compound form from GASEOUS ions
NaBr (s) → Na+
(g) + Br–
(g)
Na (g) → Na +
(g) + e
Formation Enthalpy
-∆H (Heat release) when 1 MOL compound
form from its element under std condition
Na(s) + ½Br2(g) → NaBr(s)
Std Enthalpy Changes ∆Hθ needed for BHC
Ionization Enthalpy
1st Ionization enthalpy
+ ∆H when 1 MOL e removed
from 1 MOL atom in gaseous state
Na Na+
+ e-
+
Electron affinity Enthalpy
+ e-
-
Br-
Br
Electron Affinity enthalpy
-∆H when 1 MOL GASEOUS
atom gain 1 mol electron
Br (g) + e → Br -
(g)
Gaseous
state
Atomization Enthalpy
Atomization enthalpy
+ ∆H when 1 MOL GASEOUS atom form
from its element under STD condition
Formation Enthalpy
Na Br2 NaBr
½Br2 (g) → Br(g)
+∆H -∆H
½Br2 (g) Br (g)
Find Lattice enthalpy for
IONIC COMPOUND using BHC
NaBr(s)
Na+
(g) + Br–
(g) Na+
(g) + Br–
(g)
∆Hlatt
∆H form = - 361
Na(s) + ½Br2 (g)
∆Hatom = + 108
Na(s) → Na(g)
Na+
(g) + Br-
(g)
∆Hie = + 500
Na(g) → Na+
(g)
∆Hatom = + 112
½Br2(g) → Br(g)
∆He = - 325
Br(g) → Br -
(g)
∆Hlatt = -750

7. NaF(s)
Na+
(g) + F (g)Na+
(g) + F (g)
Na+
(g) + ½F2 (g)
Na(g) + ½F2 (g)
NaF(s)
∆H atom + ∆H ion + ∆H EA
( Determined
experimentally)
NaF(s)
∆H lattice = ?????
( Can’t be determined
experimentally)
∆H form = - 574
( Determined
experimentally)
Na(s) + ½F2 (g)
Born Haber Cycle/BHC
Na+
(g) + F–
(g) → NaF (s)
Find Lattice enthalpy for IONIC COMPOUND
Lattice Enthalpy
+∆H (Heat absorb) to convert 1 MOL IONIC
compound to GASEOUS ions
Lattice Enthalpy
-∆H (Heat release) when 1 MOL IONIC
compound form from GASEOUS ions
NaF (s) → Na+
(g) + F–
(g)
Na (g) → Na +
(g) + e
Formation Enthalpy
-∆H (Heat release) when 1 MOL compound
form from its element under std condition
Na(s) + ½F2(g) → NaF(s)
Std Enthalpy Changes ∆Hθ needed for BHC
Ionization Enthalpy
1st Ionization enthalpy
+ ∆H when 1 MOL e removed
from 1 MOL atom in gaseous state
Na Na+
+ e-
+
Electron affinity Enthalpy
+ e-
-
F-
F
Electron Affinity enthalpy
-∆H when 1 MOL GASEOUS
atom gain 1 mol electron
F (g) + e → F -
(g)
Gaseous
state
Atomization Enthalpy
Atomization enthalpy
+ ∆H when 1 MOL GASEOUS atom form
from its element under STD condition
Formation Enthalpy
Na F2 NaF
½F2 (g) → F(g)
+∆H -∆H
½F2 (g) F (g)
Find Lattice enthalpy for
IONIC COMPOUND using BHC
NaF(s)
Na+
(g) + F–
(g) Na+
(g) + F–
(g)
∆Hlatt
∆H form = - 574
Na(s) + ½F2 (g)
∆Hatom = + 108
Na(s) → Na(g)
Na+
(g) + F -
(g)
∆Hie = + 500
Na(g) → Na+
(g)
∆Hatom = + 79
½F2(g) → F(g)
∆He = - 328
F(g) → F -
(g)
∆Hlatt = -930

8. NaH(s)
Na+
(g) + H (g)Na+
(g) + H (g)
Na+
(g) + ½H2 (g)
Na(g) + ½H2 (g)
NaH(s)
∆H atom + ∆H ion + ∆H EA
( Determined
experimentally)
NaH(s)
∆H lattice = ?????
( Can’t be determined
experimentally)
∆H form = - 57
( Determined
experimentally)
Na(s) + ½H2 (g)
Born Haber Cycle/BHC
Na+
(g) + H–
(g) → NaH (s)
Find Lattice enthalpy for IONIC COMPOUND
Lattice Enthalpy
+∆H (Heat absorb) to convert 1 MOL IONIC
compound to GASEOUS ions
Lattice Enthalpy
-∆H (Heat release) when 1 MOL IONIC
compound form from GASEOUS ions
NaH (s) → Na+
(g) + H–
(g)
Na (g) → Na +
(g) + e
Formation Enthalpy
-∆H (Heat release) when 1 MOL compound
form from its element under std condition
Na(s) + ½H2(g) → NaH(s)
Std Enthalpy Changes ∆Hθ needed for BHC
Ionization Enthalpy
1st Ionization enthalpy
+ ∆H when 1 MOL e removed
from 1 MOL atom in gaseous state
Na Na+
+ e-
+
Electron affinity Enthalpy
+ e-
-
H-
H
Electron Affinity enthalpy
-∆H when 1 MOL GASEOUS
atom gain 1 mol electron
H (g) + e → H -
(g)
Gaseous
state
Atomization Enthalpy
Atomization enthalpy
+ ∆H when 1 MOL GASEOUS atom form
from its element under STD condition
Formation Enthalpy
Na H2 NaH
½H2 (g) → H (g)
+∆H -∆H
½H2 (g) H (g)
Find Lattice enthalpy for
IONIC COMPOUND using BHC
NaH(s)
Na+
(g) + H–
(g) Na+
(g) + H–
(g)
∆Hlatt
∆H form = - 57
Na(s) + ½H2 (g)
∆Hatom = + 108
Na(s) → Na(g)
Na+
(g) + H -
(g)
∆Hie = + 500
Na(g) → Na+
(g)
∆Hatom = + 218
½H2(g) → H(g)
∆He = - 72
H(g) → H -
(g)
∆Hlatt = -811

9. Mg2+
(g) + O (g)
MgO(s)MgO(s)
Mg2+
(g) + O (g)
Mg2+
(g) + ½O2 (g)
Mg(g) + ½O2 (g)
∆H atom + ∆H ion + ∆H EA
( Determined
experimentally)
MgO(s)
∆H lattice = ?????
( Can’t be determined
experimentally)
∆H form = - 602
( Determined
experimentally)
Mg(s) + ½O2 (g)
Born Haber Cycle/BHC
Mg2+
(g) + O2-
(g) → MgO(s)
Find Lattice enthalpy for IONIC COMPOUND
Lattice Enthalpy
+∆H (Heat absorb) to convert 1 MOL IONIC
compound to GASEOUS ions
Lattice Enthalpy
-∆H (Heat release) when 1 MOL IONIC
compound form from GASEOUS ions
MgO(s) → Mg2+
(g) + O2-
(g)
Mg(g) → Mg2+
(g) + 2e
Formation Enthalpy
-∆H (Heat release) when 1 MOL compound
form from its element under std condition
Mg(s) + ½O2(g) → MgO(s)
Std Enthalpy Changes ∆Hθ needed for BHC
Ionization Enthalpy
1st Ionization enthalpy
+ ∆H when 1 MOL e removed
from 1 MOL atom in gaseous state
Mg Mg2+
+ 2e-
2+
Electron affinity Enthalpy
+ 2e-
2-
O2-
O
Electron Affinity enthalpy
-∆H when 1 MOL GASEOUS
atom gain 1 mol electron
O (g) + e → O2-
(g)
Gaseous
state
Atomization Enthalpy
Atomization enthalpy
+ ∆H when 1 MOL GASEOUS atom form
from its element under STD condition
Formation Enthalpy
Mg O2 MgO
½O2 (g) → O (g)
+∆H -∆H
½O2 (g) O(g)
Find Lattice enthalpy for
IONIC COMPOUND using BHC
MgO(s)
Mg2+
(g) + O2-
(g)
∆Hlatt
∆H form = - 602
Mg(s) + ½O2 (g)
∆Hatom = + 146
Mg(s) → Mg(g)
Mg2+
(g) + O2-
(g)
∆H i 1st/2nd = + 2186
Mg(g) → Mg2+
(g)
∆Hatom = + 249
½O2(g) → O(g)
∆He 1st = - 141
O(g) → O-
(g)
∆Hlatt = -3833
Mg2+
(g) + O2-
(g)
∆He 2nd = + 791
O-
(g) → O2-
(g)

10. Ca2+
(g) + 2CI (g)
CaCI2 (s)CaCI2 (s)
CaCI2 (s)
Ca2+
(g) + 2CI (g)
Ca2+
(g) + CI2 (g)
Ca(g) + CI2 (g)
∆H atom + ∆H ion + ∆H EA
( Determined
experimentally)
∆H lattice = ?????
( Can’t be determined
experimentally)
∆H form = - 795
( Determined
experimentally)
Ca(s) + CI2 (g)
Born Haber Cycle/BHC
Ca2+
(g) + 2CI-
(g) → CaCI2(s)
Find Lattice enthalpy for IONIC COMPOUND
Lattice Enthalpy
+∆H (Heat absorb) to convert 1 MOL IONIC
compound to GASEOUS ions
Lattice Enthalpy
-∆H (Heat release) when 1 MOL IONIC
compound form from GASEOUS ions
CaCI2(s) → Ca2+
(g) + 2CI-
(g)
Ca(g) → Ca2+
(g) + 2e
Formation Enthalpy
-∆H (Heat release) when 1 MOL compound
form from its element under std condition
Ca(s) + CI2(g) → CaCI2(s)
Std Enthalpy Changes ∆Hθ needed for BHC
Ionization Enthalpy
1st Ionization enthalpy
+ ∆H when 1 MOL e removed
from 1 MOL atom in gaseous state
Ca Ca2+
+ 2e-
2+
Electron affinity Enthalpy
+ e-
-
CI-
CI
Electron Affinity enthalpy
-∆H when 1 MOL GASEOUS
atom gain 1 mol electron
CI (g) + e → CI-
(g)
Gaseous
state
Atomization Enthalpy
Atomization enthalpy
+ ∆H when 1 MOL GASEOUS atom form
from its element under STD condition
Formation Enthalpy
Ca CI2 CaCI2
½CI2 (g) → CI (g)
+∆H -∆H
½CI2 (g) CI(g)
Find Lattice enthalpy for
IONIC COMPOUND using BHC
CaCI2 (s)
Ca2+
(g) + 2CI-
(g)
∆Hlatt
∆H form = - 795
Ca(s) + CI2 (g)
∆Hatom = + 190
Ca(s) → Ca(g)
Ca2+
(g) + 2CI-
(g)
∆H i 1st/2nd = + 1730
Ca(g) → Ca2+
(g)
∆Hatom = + 121 x 2
½CI2(g) → CI(g)
∆He 1st = - 354 x 2
CI(g) → CI -
(g)
∆Hlatt = -2249
Ca2+
(g) + 2CI-
(g)

11. Ba2+
(g) + 2CI (g)
BaCI2 (s)BaCI2 (s)
BaCI2 (s)
Ba2+
(g) + 2CI (g)
Ba2+
(g) + CI2 (g)
Ba(g) + CI2 (g)
∆H atom + ∆H ion + ∆H EA
( Determined
experimentally)
∆H lattice = ?????
( Can’t be determined
experimentally)
∆H form = - 860
( Determined
experimentally)
Ba(s) + CI2 (g)
Born Haber Cycle/BHC
Ba2+
(g) + 2CI-
(g) → BaCI2(s)
Find Lattice enthalpy for IONIC COMPOUND
Lattice Enthalpy
+∆H (Heat absorb) to convert 1 MOL IONIC
compound to GASEOUS ions
Lattice Enthalpy
-∆H (Heat release) when 1 MOL IONIC
compound form from GASEOUS ions
BaCI2(s) → Ba2+
(g) + 2CI-
(g)
Ba(g) → Ba2+
(g) + 2e
Formation Enthalpy
-∆H (Heat release) when 1 MOL compound
form from its element under std condition
Ba(s) + CI2(g) → BaCI2(s)
Std Enthalpy Changes ∆Hθ needed for BHC
Ionization Enthalpy
1st Ionization enthalpy
+ ∆H when 1 MOL e removed
from 1 MOL atom in gaseous state
Ba Ba2+
+ 2e-
2+
Electron affinity Enthalpy
+ e-
-
CI-
CI
Electron Affinity enthalpy
-∆H when 1 MOL GASEOUS
atom gain 1 mol electron
CI (g) + e → CI-
(g)
Gaseous
state
Atomization Enthalpy
Atomization enthalpy
+ ∆H when 1 MOL GASEOUS atom form
from its element under STD condition
Formation Enthalpy
Ba CI2 BaCI2
½CI2 (g) → CI (g)
+∆H -∆H
½CI2 (g) CI(g)
Find Lattice enthalpy for
IONIC COMPOUND using BHC
BaCI2 (s)
Ba2+
(g) + 2CI-
(g)
∆Hlatt
∆H form = - 860
Ba(s) + CI2 (g)
∆Hatom = + 175
Ba(s) → Ba(g)
Ba2+
(g) + 2CI-
(g)
∆H i 1st/2nd = + 1500
Ba(g) → Ba2+
(g)
∆Hatom = + 121 x 2
½CI2(g) → CI(g)
∆He 1st = - 354 x 2
CI(g) → CI -
(g)
∆Hlatt = -2049
Ba2+
(g) + 2CI-
(g)

12. K+
(g) + Br (g)
KBr(s)
K+
(g) + Br (g)
K+
(g) + ½Br2 (g)
K(g) + ½Br2 (g)
KBr(s)
∆H atom + ∆H ion + ∆H EA
( Determined
experimentally)
KBr(s)
∆H lattice = ?????
( Can’t be determined
experimentally)
∆H form = - 392
( Determined
experimentally)
Ks) + ½Br2 (g)
Born Haber Cycle/BHC
K+
(g) + Br–
(g) → KBr (s)
Find Lattice enthalpy for IONIC COMPOUND
Lattice Enthalpy
+∆H (Heat absorb) to convert 1 MOL IONIC
compound to GASEOUS ions
Lattice Enthalpy
-∆H (Heat release) when 1 MOL IONIC
compound form from GASEOUS ions
KBr (s) → K+
(g) + Br–
(g)
K (g) → K +
(g) + e
Formation Enthalpy
-∆H (Heat release) when 1 MOL compound
form from its element under std condition
K(s) + ½Br2(g) → KBr(s)
Std Enthalpy Changes ∆Hθ needed for BHC
Ionization Enthalpy
1st Ionization enthalpy
+ ∆H when 1 MOL e removed
from 1 MOL atom in gaseous state
K K+
+ e-
+
Electron affinity Enthalpy
+ e-
-
Br-
Br
Electron Affinity enthalpy
-∆H when 1 MOL GASEOUS
atom gain 1 mol electron
CI (g) + e → CI -
(g)
Gaseous
state
Atomization Enthalpy
Atomization enthalpy
+ ∆H when 1 MOL GASEOUS atom form
from its element under STD condition
Formation Enthalpy
K Br2 KBr
½Br2 (g) → Br(g)
+∆H -∆H
½Br2 (g) Br (g)
Find Lattice enthalpy for
IONIC COMPOUND using BHC
KBr(s)
K+
(g) + Br–
(g) K+
(g) + Br–
(g)
∆Hlatt
∆H form = - 392
K(s) + ½Br2 (g)
∆Hatom = + 89
K(s) → K(g)
K+
(g) + Br-
(g)
∆Hie = + 420
K(g) → K+
(g)
∆Hatom = + 112
½Br2(g) → Br(g)
∆He = - 342
CI(g) → CI -
(g)
∆Hlatt = -671

13. ∆H lattice
NaCI (s)NaCI (s)
Lattice Enthalpy
-∆H
Find Lattice enthalpy using BHC
2
21
r
qq
kF 
Theoretical Lattice Enthalpy
(Calculated using formula)
Lattice enthalpy depend
Find Lattice enthalpy using Coulomb’s Law
Experimental/Actual Lattice Enthalpy
(Calculated using BHC)
Assumption
Ionic compound
Coulomb’s Law
CHARGE on ions
Electrostatic
force
Electric charge
(+) or (-)
Distance
Coulomb
constant
+ -
SIZE of ions
Size increase ↑
↓
Separation bet ions increase ↑
Electrostatic force bet ion decrease ↓
↓
Lattice enthalpy decrease ↓
Charge ↑
↓
Electrostatic forces bet ion increases ↑
↓
Lattice enthalpy increase ↑
Gp1
salt
Lattice
Enthalpy
kJ mol-1
LiCI + 846
NaCI + 771
KCI + 720
Size cation ↑
2
21
r
qq
kF 
Li
Na
K
Gp1 salt Lattice
Enthalpy
kJ mol-1
NaO + 2702
MgO + 3889
AI2O3 + 4020
CI
CI
CI
Charge cation ↑
Na+
Mg2+
AI3+
O
O
O
2
21
r
qq
kF 
Vs
Na+
(g) + CI–
(g)
∆H atom + ∆H ion + ∆H EA
( Determined
experimentally)
∆H form = - 414
( Determined
experimentally)
Na+
(g) + CI–
(g)
Electrostatic forces of attraction
bet opposite charges
Vs
Na(s) + ½CI2 (g)

14. Lattice Enthalpy
2
21
r
qq
kF 
Theoretical Lattice Enthalpy
(Calculated using formula)
Lattice enthalpy depend
Find Lattice enthalpy using Coulomb’s Law
Assumption
Ionic compound
Coulomb’s Law
CHARGE on ions
Electrostatic
force
Electric charge
(+) or (-)
Distance
Coulomb
constant
+ -
SIZE of ions
Size increase ↑
↓
Separation bet ions increase ↑
Electrostatic force bet ion decrease ↓
↓
Lattice enthalpy decrease ↓
Charge ↑
↓
Electrostatic forces bet ion increases ↑
↓
Lattice enthalpy increase ↑
Gp1
salt
Lattice
Enthalpy
kJ mol-1
LiCI + 846
NaCI + 771
KCI + 720
Size cation ↑
2
21
r
qq
kF 
Li
Na
K
Gp1 salt Lattice
Enthalpy
kJ mol-1
NaO + 2702
MgO + 3889
AI2O3 + 4020
CI
CI
CI
Charge cation ↑
Na+
Mg2+
AI3+
O
O
O
2
21
r
qq
kF 
Metal
Halide
Lattice Enthalpy
F CI Br I
Li 1049 864 820 764
Na 930 790 754 705
K 830 720 691 650
Rb 795 695 668 632
Experimental/Actual Lattice Enthalpy
(Calculated using BHC)
Size increase ↑
↓
LE decrease ↓
Li
Na
K
Rb
F CI Br I
Size increase ↑
↓
LE decrease ↓

15. Lattice Enthalpy
2
21
r
qq
kF 
Theoretical Lattice Enthalpy
(Calculated using formula)
Find Lattice enthalpy using Coulomb’s Law
Assumption
Ionic compound
Coulomb’s Law
Electrostatic
force
Electric charge
(+) or (-)
Distance
Coulomb
constant
Metal
Halide
Lattice Enthalpy
F CI Br I
Li 1049 864 820 764
Na 930 790 754 705
K 830 720 691 650
Rb 795 695 668 632
Experimental/Actual Lattice Enthalpy
(Calculated using BHC)
Li
Na
K
Rb
F CI Br I
Size increase ↑
↓
LE decrease ↓
NaF NaCI NaBr NaI
Experimental
Lattice Enthalpy/(BHC)
930 776 740 700
Theoretical
Lattice Enthalpy
910 769 732 682
AgF AgCI AgBr AgI
Experimental
Lattice Enthalpy/(BHC)
974 910 900 865
Theoretical
Lattice Enthalpy
953 770 755 734
Uses of Born Haber Cycle – Determine degree of ionic /covalent character
High Difference in EN value
↓
High degree ionic character (100% ionic bond)
↓
Actual Expt LE (BHC) = Theoretical LE (Assume 100% ionic bond)
↓
Good agreement/Low % diff
Small Difference in EN value
↓
Ionic + Covalent character (NOT 100% ionic bond)
↓
Actual Expt LE (BHC) > Theoretical LE (Assume 100% ionic bond)
↓
Poor agreement/High % diff
Na – F Na - CI Na – Br Na - I
Diff in EN 3.1 2.1 1.9 1.6
Ag – F Ag - CI Ag – Br Ag - I
Diff in EN 2.1 1.1 0.9 0.6

16. NaF NaCI NaBr NaI
Experimental
Lattice Enthalpy/(BHC)
930 776 740 700
Theoretical
Lattice Enthalpy
910 769 732 682
AgF AgCI AgBr AgI
Experimental
Lattice Enthalpy/(BHC)
974 910 900 865
Theoretical
Lattice Enthalpy
953 770 755 734
Uses of Born Haber Cycle – Determine degree of ionic /covalent character
High Difference in EN value
↓
High degree ionic character (100% ionic bond)
↓
Actual Expt LE (BHC) = Theoretical LE (Assume 100% ionic bond)
↓
Good agreement/Low % diff
↓
Small Difference in EN value
↓
Ionic + Covalent character (NOT 100% ionic bond)
↓
Actual Expt LE (BHC) > Theoretical LE (Assume 100% ionic bond)
↓
Poor agreement/High % diff
↓
Na – F Na - CI Na – Br Na - I
Diff in EN 3.1 2.1 1.9 1.6
Ag – F Ag - CI Ag – Br Ag - I
Diff in EN 2.1 1.1 0.9 0.6
Difference in electronegativity
0 0.4 2 4
difference < 0.4
covalent compound
difference > 2
ionic compound
Diff = 2.5 – 2.1
= 0.4
Diff = 3 – 0.9
= 2.1
EN – 2.1EN – 2.5
HC CI-Na+
EN - 3.0EN - 0.9
Click here notes bonding triangle
Polar
covalent
Click here video bonding triangle
Polar covalent
Ionic
Bond
Ionic Bond
Due to high charge density cation (+)
(charge/ionic radius)
↓
Donated electron cloud pull back to cation
to form partial covalent bond
↓
Ionic + covalent character (Polar covalent)
Ag+ CI -
Electron cloud pull
(covalent bond)
Polarization – cause polar covalent
No polarization
(100% ionic)

17. NaF NaCI NaBr NaI
Experimental
Lattice Enthalpy/(BHC)
930 776 740 700
Theoretical
Lattice Enthalpy
910 769 732 682
AgF AgCI AgBr AgI
Experimental
Lattice Enthalpy/(BHC)
974 910 900 865
Theoretical
Lattice Enthalpy
953 770 755 734
Uses of Born Haber Cycle – Determine degree of ionic /covalent character
High Difference in EN value
↓
High degree ionic character (100% ionic bond)
↓
Actual Expt LE (BHC) = Theoretical LE (Assume 100% ionic bond)
↓
Good agreement/Low % diff
↓
Small Difference in EN value
↓
Ionic + Covalent character (NOT 100% ionic bond)
↓
Actual Expt LE (BHC) > Theoretical LE (Assume 100% ionic bond)
↓
Poor agreement/High % diff
↓
Na – F Na - CI Na – Br Na - I
Diff in EN 3.1 2.1 1.9 1.6
Ag – F Ag - CI Ag – Br Ag - I
Diff in EN 2.1 1.1 0.9 0.6
Polar
covalent
Polar covalent
Ionic
Bond
Ionic Bond
Due to high charge density cation (+)
(charge/ionic radius)
↓
Donated electron cloud pull back to cation
to form partial covalent bond
↓
Ionic + covalent character (Polar covalent)
Ag+ CI -
Electron cloud pull
(covalent bond)
Polarization – cause polar covalent
No polarization
(100% ionic)
vs
Lattice enthalpy AgF – AgI > Lattice Enthalpy NaF – NaI
↓ ↓
Size Ag bigger > Size Na smaller
↓ ↓
LE Ag should be lower < LE Na should be higher
↓ ↓
Higher LE Ag due to > Lower LE Na due to
ionic/covalent character only ionic character
Size increase ↑
↓
LE decrease ↓
Ag+
CI - CI -
Electron cloud pull
(covalent bond)
only ionic
BUT BUT

18. Born Haber Cycle/BHC
NaCI (s) → Na+
(g) + CI–
(g)
-∆H
Lattice Enthalpy
-∆H (Heat release) when 1 MOL IONIC
compound form from GASEOUS ions
Find Lattice enthalpy using BHC
2
21
r
qq
kF 
Electrostatic forces of attraction
bet opposite charges
Theoretical Lattice Enthalpy
(Calculated using formula)
Find Lattice enthalpy using Coulomb’s Law
Experimental/Actual Lattice Enthalpy
(Calculated using BHC)
Assume – 100% Ionic Coulomb’s Law
Lattice Enthalpy
Electrostatic
force
Electric charge
(+) or (-)
Distance
Coulomb
constant
+ -
Na CI
Using Born Meyer eqn:







nner
qq
AH
1
1
4
21
A = 1.747
q1 = +1
q2 = -1
n = 8
R = 283 x 10-12
4ƞe = 1.13 x 10 -10
r = Distance n = quantum #
Electric charge
(+) or (-)
A = Madelung constant
Values for NaCI
∆Hlatt = 769
NaCI NaBr NaI
Experimental
Lattice Enthalpy/(BHC)
776 740 700
Theoretical
Lattice Enthalpy (Calculated)
769 732 682
∆Hlatt = 776
Theoretical LE (Assume 100% ionic bond) = Expt LE (BHC)

19. Acknowledgements
Thanks to source of pictures and video used in this presentation
Thanks to Creative Commons for excellent contribution on licenses
http://creativecommons.org/licenses/
Prepared by Lawrence Kok
Check out more video tutorials from my site and hope you enjoy this tutorial
http://lawrencekok.blogspot.com