ib chemistry on born haber cycle and lattice enthalpy

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

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)


Li+(g) + CI–

(g) → LiCI (s)

Find Lattice enthalpy for IONIC COMPOUND

Lattice Enthalpy +∆H (Heat absorb) to convert 1 MOL IONIC



LiCI (s) → Li+(g) + CI–

(g)

Li (g) → Li +(g) + e


Li(s) + ½CI2(g) → LiCI(s)


Ionization Enthalpy



Li Li+ + e-

+


+ e- -

CI- CI



CI (g) + e → CI -(g)

Gaseous state




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

NaCI (s)

Na+(g) + CI (g) Na+

(g) + CI (g)

Na+(g) + ½CI2 (g)

Na(g) + ½CI2 (g)

NaCI (s)



NaCI (s)



Na(s) + ½CI2 (g)


Na+(g) + CI–

(g) → NaCI (s)





NaCI (s) → Na+(g) + CI–

(g)

Na (g) → Na +(g) + e


Na(s) + ½CI2(g) → NaCI(s)


Ionization Enthalpy



Na Na+ + e-

+


+ e- -

CI- CI



CI (g) + e → CI -(g)

Gaseous state




Formation Enthalpy

Na CI2 NaCI

½CI2 (g) → CI (g)

+∆H -∆H

½CI2 (g) CI (g)


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

KCI (s)

K+(g) + CI (g) K+

(g) + CI (g)

K+(g) + ½CI2 (g)

K(g) + ½CI2 (g)

KCI (s)



KCI (s)



K(s) + ½CI2 (g)


K+(g) + CI–

(g) → KCI (s)





KCI (s) → K+(g) + CI–

(g)

K (g) → K +(g) + e


K(s) + ½CI2(g) → KCI(s)


Ionization Enthalpy



K K+ + e-

+


+ e- -

CI- CI



CI (g) + e → CI -(g)

Gaseous state




Formation Enthalpy

K CI2 KCI

½CI2 (g) → CI (g)

+∆H -∆H

½CI2 (g) CI (g)


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

NaBr (s)

Na+(g) + Br (g) Na+

(g) + Br (g)

Na+(g) + ½Br2 (g)

Na(g) + ½Br2 (g)

NaBr (s)



NaBr (s)



Na(s) + ½Br2 (g)


Na+(g) + Br–

(g) → NaBr (s)





NaBr (s) → Na+(g) + Br–

(g)

Na (g) → Na +(g) + e


Na(s) + ½Br2(g) → NaBr(s)


Ionization Enthalpy



Na Na+ + e-

+


+ e- -

Br- Br



Br (g) + e → Br -(g)

Gaseous state




Formation Enthalpy

Na Br2 NaBr

½Br2 (g) → Br (g)

+∆H -∆H

½Br2 (g) Br (g)


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

NaF (s)

Na+(g) + F (g) Na+

(g) + F (g)

Na+(g) + ½F2 (g)

Na(g) + ½F2 (g)

NaF (s)



NaF (s)



Na(s) + ½F2 (g)


Na+(g) + F–

(g) → NaF (s)





NaF (s) → Na+(g) + F–

(g)

Na (g) → Na +(g) + e


Na(s) + ½F2(g) → NaF(s)


Ionization Enthalpy



Na Na+ + e-

+


+ e- -

F- F



F (g) + e → F -(g)

Gaseous state




Formation Enthalpy

Na F2 NaF

½F2 (g) → F (g)

+∆H -∆H

½F2 (g) F (g)


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

NaH (s)

Na+(g) + H (g) Na+

(g) + H (g)

Na+(g) + ½H2 (g)

Na(g) + ½H2 (g)

NaH (s)



NaH (s)



Na(s) + ½H2 (g)


Na+(g) + H–

(g) → NaH (s)





NaH (s) → Na+(g) + H–

(g)

Na (g) → Na +(g) + e


Na(s) + ½H2(g) → NaH(s)


Ionization Enthalpy



Na Na+ + e-

+


+ e- -

H- H



H (g) + e → H -(g)

Gaseous state




Formation Enthalpy

Na H2 NaH

½H2 (g) → H (g)

+∆H -∆H

½H2 (g) H (g)


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

Mg2+(g) + O (g)

MgO (s) MgO (s)

Mg2+(g) + O (g)

Mg2+(g) + ½O2 (g)

Mg(g) + ½O2 (g)



MgO (s)



Mg(s) + ½O2 (g)


Mg2+(g) + O2-

(g) → MgO(s)





MgO(s) → Mg2+(g) + O2-

(g)

Mg(g) → Mg2+(g) + 2e


Mg(s) + ½O2(g) → MgO(s)


Ionization Enthalpy



Mg Mg2+ + 2e-

2+


+ 2e- 2-

O2- O



O (g) + e → O2-(g)

Gaseous state




Formation Enthalpy

Mg O2 MgO

½O2 (g) → O (g)

+∆H -∆H

½O2 (g) O(g)


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)

Ca2+(g) + 2CI (g)

CaCI2 (s) CaCI2 (s)

CaCI2 (s)

Ca2+(g) + 2CI (g)

Ca2+(g) + CI2 (g)

Ca(g) + CI2 (g)


( Determined experimentally) ∆H lattice = ?????

( Can’t be determined experimentally)


Ca(s) + CI2 (g)


Ca2+(g) + 2CI-

(g) → CaCI2(s)





CaCI2(s) → Ca2+(g) + 2CI-

(g)

Ca(g) → Ca2+(g) + 2e


Ca(s) + CI2(g) → CaCI2(s)


Ionization Enthalpy



Ca Ca2+ + 2e-

2+


+ e- -

CI- CI



CI (g) + e → CI-(g)

Gaseous state




Formation Enthalpy

Ca CI2 CaCI2

½CI2 (g) → CI (g)

+∆H -∆H

½CI2 (g) CI(g)


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)

Ba2+(g) + 2CI (g)

BaCI2 (s) BaCI2 (s)

BaCI2 (s)

Ba2+(g) + 2CI (g)

Ba2+(g) + CI2 (g)

Ba(g) + CI2 (g)


( Determined experimentally) ∆H lattice = ?????

( Can’t be determined experimentally)


Ba(s) + CI2 (g)


Ba2+(g) + 2CI-

(g) → BaCI2(s)





BaCI2(s) → Ba2+(g) + 2CI-

(g)

Ba(g) → Ba2+(g) + 2e


Ba(s) + CI2(g) → BaCI2(s)


Ionization Enthalpy



Ba Ba2+ + 2e-

2+


+ e- -

CI- CI



CI (g) + e → CI-(g)

Gaseous state




Formation Enthalpy

Ba CI2 BaCI2

½CI2 (g) → CI (g)

+∆H -∆H

½CI2 (g) CI(g)


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)

K+(g) + Br (g)

KBr (s)

K+(g) + Br (g)

K+(g) + ½Br2 (g)

K(g) + ½Br2 (g)

KBr (s)



KBr (s)



Ks) + ½Br2 (g)


K+(g) + Br–

(g) → KBr (s)





KBr (s) → K+(g) + Br–

(g)

K (g) → K +(g) + e


K(s) + ½Br2(g) → KBr(s)


Ionization Enthalpy



K K+ + e-

+


+ e- -

Br- Br



CI (g) + e → CI -(g)

Gaseous state




Formation Enthalpy

K Br2 KBr

½Br2 (g) → Br (g)

+∆H -∆H

½Br2 (g) Br (g)


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

∆H lattice

NaCI (s) NaCI (s)

Lattice Enthalpy

-∆H

Find Lattice enthalpy using BHC

2

21

r

qqkF

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

qqkF

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

qqkF

Vs

Na+ (g) + CI–

(g)




Na+ (g) + CI–

(g) Electrostatic forces of attraction bet opposite charges

Vs Na(s) + ½CI2 (g)

Lattice Enthalpy

2

21

r

qqkF


Lattice enthalpy depend



Coulomb’s Law

CHARGE on ions

Electrostatic force



+ -

SIZE of ions


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

qqkF

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

qqkF

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



LE decrease ↓

Li

Na

K

Rb

F CI Br I Size increase ↑

↓ LE decrease ↓

Lattice Enthalpy

2

21

r

qqkF




Coulomb’s Law

Electrostatic force



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


Li

Na

K

Rb

F CI Br I


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


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

NaF NaCI NaBr NaI

Experimental Lattice Enthalpy/(BHC)

930 776 740 700

Theoretical

Lattice Enthalpy 910 769 732 682

AgF AgCI AgBr AgI


974 910 900 865


953 770 755 734



↓




↓


↓




↓


Diff in EN 3.1 2.1 1.9 1.6


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 – 2.1 EN – 2.5

H C CI- Na+

EN - 3.0 EN - 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)

http://www.meta-synthesis.com/webbook/37_ak/triangles.html

https://www.youtube.com/watch?v=ZNGThjz82Xo

https://www.youtube.com/watch?v=ZNGThjz82Xo

NaF NaCI NaBr NaI


930 776 740 700

Theoretical

Lattice Enthalpy 910 769 732 682

AgF AgCI AgBr AgI


974 910 900 865


953 770 755 734



↓




↓


↓




↓


Diff in EN 3.1 2.1 1.9 1.6


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 -


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


LE decrease ↓

Ag+ CI - CI -


only ionic

BUT BUT


NaCI (s) → Na+(g) + CI–

(g)

-∆H


Find Lattice enthalpy using BHC

2

21

r

qqkF

Electrostatic forces of attraction bet opposite charges




Assume – 100% Ionic Coulomb’s Law

Lattice Enthalpy

Electrostatic force



+ - Na CI

Using Born Meyer eqn:

nner

qqAH

11

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 #


A = Madelung constant

Values for NaCI

∆Hlatt = 769

NaCI NaBr NaI


776 740 700

Theoretical

Lattice Enthalpy (Calculated) 769 732 682

∆Hlatt = 776

Theoretical LE (Assume 100% ionic bond) = Expt LE (BHC)

ib chemistry on born haber cycle and lattice enthalpy

Education

h g o2 g

o g o2 g o g h h

nas ci2 g

lis lig li g ci g hie

mol gas h2 g

na s na g

ionization enthalpy

lig li g hatom