IB Chemistry on Born Haber Cycle and Lattice Enthalpy

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Prepared by Lawrence Kok

Video Tutorial on Born Haber Cycle and Lattice Enthalpy.

http://lawrencekok.blogspot.com/


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

enthalpy2 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)


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

KCI (s)

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

(g) + CI (g)

K+(g) + ½CI2

(g)

K(g) + ½CI2 (g)

KCI (s)



KCI (s)


∆H form = - 436( Determined experimentally)

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)


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

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)


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

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)


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

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)


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

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)


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)

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)


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)

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)


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)

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)


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

∆H lattice

NaCI

(s) NaCI

(s)

Lattice Enthalpy

-∆H

Find Lattice enthalpy using BHC

221

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 (-)

DistanceCoulomb 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

LatticeEnthalp

ykJ mol-1

LiCI + 846

NaCI

+ 771

KCI + 720Size cation ↑

221

r

qqkF

Li

Na

K

Gp1 salt

LatticeEnthalp

ykJ mol-1

NaO + 2702

MgO + 3889

AI2O3 + 4020

CI

CI

CI

Charge cation ↑

Na+

Mg2+

AI3+

O

O

O

221

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

221

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

LatticeEnthalp

ykJ mol-1

LiCI + 846

NaCI

+ 771

KCI + 720Size cation ↑

221

r

qqkF

Li

Na

K

Gp1 salt

LatticeEnthalp

ykJ mol-1

NaO + 2702

MgO + 3889

AI2O3 + 4020

CI

CI

CI

Charge cation ↑

Na+

Mg2+

AI3+

O

O

O

221

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 ISize increase ↑

↓ LE decrease ↓

Lattice Enthalpy

221

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 ISize increase ↑

↓ LE decrease ↓

NaF NaCI NaBr NaI

ExperimentalLattice Enthalpy/(BHC)

930 776 740 700


910 769 732 682

AgF AgCI AgBr AgI


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


930 776 740 700


910 769 732 682

AgF AgCI AgBr AgI


974 910 900 865


953 770 755 734






↓





↓

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 < 0.4covalent compound

difference > 2ionic compound

Diff = 2.5 – 2.1 = 0.4

Diff = 3 – 0.9 = 2.1

EN – 2.1EN – 2.5

H C CI-Na+

EN - 3.0EN - 0.9

Click here notes bonding triangle

Polarcovalent

Click here video bonding triangle

Polar covalent

IonicBond

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


910 769 732 682

AgF AgCI AgBr AgI


974 910 900 865


953 770 755 734






↓





↓

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 Polarcovalent

Polar covalent

IonicBond

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

221

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

421

A = 1.747q1 = +1q2 = -1n = 8R = 283 x 10-12

4ƞe = 1.13 x 10 -10r = Distance n = quantum #


A = Madelung constant

Values for NaCI

∆Hlatt = 769

NaCI NaBr NaI


776 740 700


(Calculated)

769 732 682

∆Hlatt = 776

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

Acknowledgements

Thanks to source of pictures and video used in this presentation

Thanks to Creative Commons for excellent contribution on licenseshttp://creativecommons.org/licenses/

Prepared by Lawrence Kok

Check out more video tutorials from my site and hope you enjoy this tutorialhttp://lawrencekok.blogspot.com

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IB Chemistry on Born Haber Cycle and Lattice Enthalpy

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Transcript of IB Chemistry on Born Haber Cycle and Lattice Enthalpy