born haber l4
DESCRIPTION
Born Haber L4TRANSCRIPT
Born-Haber Cycle
Born-Haber Cycles
• The Born–Haber cycle is an approach to analyzing reaction energies.
• It was named after and developed by the two German scientists Max Born and Fritz Haber.
• Born–Haber cycles are used primarily for calculating lattice enthalpies which cannot otherwise be measured directly.
• The lattice enthalpy is the enthalpy change involved in the formation of an ionic compound from gaseous ions.
For an ionic compound the lattice enthalpy is the enthalpy change when one mole of solid in its standard state is formed from its ions in the gaseous state.
The lattice enthalpy cannot be measured directly and so we make use of other known enthalpies and link them together with an enthalpy cycle.
This enthalpy cycle is the Born-Haber cycle.
What do we mean by lattice enthalpy?
- -
-
--
-
-
-
For an ionic compound the lattice enthalpy is the enthalpy change when one mole of solid in its standard state is formed from its ions in the gaseous state.
http://www.tes.co.uk/teaching-resource/Lattice-Enthalpy-6354465/
1 Sublimation of Sodium
Na(s) + 1/2 Cl2(g)
Na(g) + 1/2 Cl2(g)
0
+100
+200
+300
+400
+500
+600
+700
+800
kJmol-1
-400
-300
-200
-100
H = +107kJmol-1 θS
Born-Haber Cycle for Sodium Chloride
2 Bond Dissociation of Chlorine
Na(s) + 1/2 Cl2(g)
Na(g) + 1/2 Cl2(g)
Na(g) + Cl(g)
0
+100
+200
+300
+400
+500
+600
+700
+800
-400
-300
-200
-100
H = +121kJmol-1θD½
Born-Haber Cycle for Sodium ChloridekJmol-1
e-e-
e-
e-e-
Na(s) + 1/2 Cl2(g)
Na(g) + 1/2 Cl2(g)
Na(g) + Cl(g)
0
+100
+200
+300
+400
+500
+600
+700
+800
-400
-300
-200
-100
Na+(g) + Cl(g)
H = +502kJmol-1θI
Born-Haber Cycle for Sodium Chloride
3 First Ionisation of Sodium
kJmol-1
4 Electron Affinity of Chlorine
e-Na(s) + 1/2 Cl2(g)
Na(g) + 1/2 Cl2(g)
Na(g) + Cl(g)
0
+100
+200
+300
+400
+500
+600
+700
+800
-400
-300
-200
-100
Na(s) + 1/2 Cl2(g)
Na(g) + 1/2 Cl2(g)
Na(g) + Cl(g)
0
+100
+200
+300
+400
+500
+600
+700
+800
-400
-300
-200
-100
Na(s) + 1/2 Cl2(g)
Na(g) + 1/2 Cl2(g)
Na(g) + Cl(g)
0
+100
+200
+300
+400
+500
+600
+700
+800
-400
-300
-200
-100
Na+(g) + Cl(g)
Na+(g) + Cl-(g)
H = -355kJmol-1θE
Born-Haber Cycle for Sodium ChloridekJmol-1
-
Na(s) + 1/2 Cl2(g)
Na(g) + 1/2 Cl2(g)
Na(g) + Cl(g)
0
+100
+200
+300
+400
+500
+600
+700
+800
-400
-300
-200
-100
Na(s) + 1/2 Cl2(g)
Na(g) + 1/2 Cl2(g)
Na(g) + Cl(g)
0
+100
+200
+300
+400
+500
+600
+700
+800
-400
-300
-200
-100
Na(s) + 1/2 Cl2(g)
Na(g) + 1/2 Cl2(g)
Na(g) + Cl(g)
0
+100
+200
+300
+400
+500
+600
+700
+800
-400
-300
-200
-100
NaCl(s)
Born-Haber Cycle for Sodium Chloride
H = -411kJmol-1θF
Na+(g) + Cl(g)
kJmol-1
- -
-
--
5 Formation of Sodium Chloride
Na+(g) + Cl-(g)
Na(s) + 1/2 Cl2(g)
Na(g) + 1/2 Cl2(g)
Na(g) + Cl(g)
0
+100
+200
+300
+400
+500
+600
+700
+800
-400
-300
-200
-100
Na(s) + 1/2 Cl2(g)
Na(g) + 1/2 Cl2(g)
Na(g) + Cl(g)
0
+100
+200
+300
+400
+500
+600
+700
+800
-400
-300
-200
-100
Na(s) + 1/2 Cl2(g)
Na(g) + 1/2 Cl2(g)
Na(g) + Cl(g)
0
+100
+200
+300
+400
+500
+600
+700
+800
-400
-300
-200
-100
NaCl(s)
Lattice Enthalpy for Sodium Chloride
Na+(g) + Cl-(g)
H = -786 kJmol-1θL
Na+(g) + Cl(g)
Born-Haber Cycle for Sodium Chloride
- -
-
--
kJmol-1
Na(s) + 1/2 Cl2(g)
Na(g) + 1/2 Cl2(g)
Na(g) + Cl(g)
0
+100
+200
+300
+400
+500
+600
+700
+800
-400
-300
-200
-100
Na(s) + 1/2 Cl2(g)
Na(g) + 1/2 Cl2(g)
Na(g) + Cl(g)
0
+100
+200
+300
+400
+500
+600
+700
+800
-400
-300
-200
-100
Na(s) + 1/2 Cl2(g)
Na(g) + 1/2 Cl2(g)
Na(g) + Cl(g)
0
+100
+200
+300
+400
+500
+600
+700
+800
-400
-300
-200
-100
NaCl(s)
Na+(g) + Cl-(g)
H = -786 kJmol-1θL
Na+(g) + Cl(g)
Born-Haber Cycle for Sodium Chloride
- -
-
--
kJmol-1
H = -411kJmol-1θF
H = -355kJmol-1θEH = +502kJmol-1
H = +121kJmol-1
H = +107kJmol-1
= -1141kJmol-1
Born-Haber Cycle
Max Born 1882-1970 German physicist and mathematician who was instrumental in the development of quantum mechanics
Fritz Haber 1868-1934 German chemist. Nobel Prize winner in Chemistry in 1918 for synthesis of ammonia.
Born-Haber Cycle - NaCBorn-Haber Cycle - NaCll
1
6
54
3
2
Na(s) + ½Cl2(g)
NaCl(s)
Na(g) + ½Cl2(g)
Na(g) + Cl(g)
Na+(g) + Cl(g)
Na+(g) + Cl–(g)
Enthalpy of formation of NaCl
Na(s) + ½Cl2(g) ——> NaCl(s)
Enthalpy of sublimation of sodium
Na(s) ——> Na(g)
Enthalpy of atomisation of chlorine
½Cl2(g) ——> Cl(g)
Ist Ionisation Energy of sodium
Na(g) ——> Na+(g) + e¯
Electron Affinity of chlorine
Cl(g) + e¯ ——> Cl¯(g)
Lattice Enthalpy of NaCl
Na+(g) + Cl¯(g) ——> NaCl(s)
1
2
3
4
5
6
Lattice Enthalpy is exothermic. Oppositely charged ions are attracted to each other.
Lattice Enthalpy is exothermic. Oppositely charged ions are attracted to each other.
Born-Haber Cycle - NaCBorn-Haber Cycle - NaCll
1
6
54
3
2
Na(s) + ½Cl2(g)
NaCl(s)
Na(g) + ½Cl2(g)
Na(g) + Cl(g)
Na+(g) + Cl(g)
Na+(g) + Cl–(g)
CALCULATING THE LATTICE ENTHALPY CALCULATING THE LATTICE ENTHALPY
Apply Hess’s Law
16 5 4 3 2 = - - - - +
The minus shows you are going in the opposite direction to the definition
= - (-355) - (+502) - (+121) - (+107) + (-411)= - 786 kJ mol-1
Born-Haber Cycle - NaCBorn-Haber Cycle - NaCll
1
6
54
3
2
Na(s) + ½Cl2(g)
NaCl(s)
Na(g) + ½Cl2(g)
Na(g) + Cl(g)
Na+(g) + Cl(g)
Na+(g) + Cl–(g)
CALCULATING THE LATTICE ENTHALPY CALCULATING THE LATTICE ENTHALPY
Apply Hess’s Law
16 5 4 3 2 = - - - - +
The minus shows you are going in the opposite direction to the definition
= - (-364) - (+500) - (+121) - (+108) + (-411)= - 786 kJ mol-1
OR…
Ignore the signs and just use the values;
If you go up you add, if you come down you subtract the value
= - - - -
= (355) - (502) - (121) - (107) - (411)= - 786 kJ mol-1
16 5 4 3 2
Your Task: Your Task: Draw the Born-Haber cycle for - MgCl2
Enthalpy of formation of MgCl2
Mg(s) + Cl2(g) ——> MgCl2(s)
Enthalpy of sublimation of magnesium
Mg(s) ——> Mg(g)
Enthalpy of atomisation of chlorine
½Cl2(g) ——> Cl(g) x2
Ist Ionisation Energy of magnesium
Mg(g) ——> Mg+(g) + e¯
2nd Ionisation Energy of magnesium
Mg+(g) ——> Mg2+(g) + e¯
Electron Affinity of chlorine
Cl(g) + e¯ ——> Cl¯(g) x2
Lattice Enthalpy of MgCl2
Mg2+(g) + 2Cl¯(g) ——> MgCl2(s)
1
65
4
3
2
Mg(s) + Cl2(g)
MgCl2(s)
Mg(g) + Cl2(g)
Mg(g) + 2Cl(g)
Mg2+(g) + 2Cl–(g)
7
Mg+(g) + 2Cl(g)
Mg2+(g) + 2Cl(g)
Enthalpy of formation of MgCl2
Mg(s) + Cl2(g) ——> MgCl2(s)
Enthalpy of sublimation of magnesium
Mg(s) ——> Mg(g)
Enthalpy of atomisation of chlorine
½Cl2(g) ——> Cl(g) x2
Ist Ionisation Energy of magnesium
Mg(g) ——> Mg+(g) + e¯
2nd Ionisation Energy of magnesium
Mg+(g) ——> Mg2+(g) + e¯
Electron Affinity of chlorine
Cl(g) + e¯ ——> Cl¯(g) x2
Lattice Enthalpy of MgCl2
Mg2+(g) + 2Cl¯(g) ——> MgCl2(s)
1
2
3
4
5
7
6
Born-Haber Cycle - MgCBorn-Haber Cycle - MgCll22
Value ∆Ho (kJmol-1)
Enthalpy of sublimation of Magnesium 148
Ist Ionisation energy of magnesium 738
2nd Ionisation energy of magnesium 1451
Enthalpy of atomisation of chlorine gas 244
1st electron affinity of chlorine -364
Lattice energy of MgCl2 -2526
Determine the enthalpy of formation of MgCl2
1
65
4
3
2
Mg(s) + Cl2(g)
MgCl2(s)
Mg(g) + Cl2(g)
Mg(g) + 2Cl(g) Mg2+(g) + 2Cl–(g)
7
Mg+(g) + 2Cl(g)
Mg2+(g) + 2Cl(g)
Enthalpy of formation of MgCl2
Mg(s) + Cl2(g) ——> MgCl2(s)
Enthalpy of sublimation of magnesium
Mg(s) ——> Mg(g)
Enthalpy of atomisation of chlorine
½Cl2(g) ——> Cl(g) x2?
Ist Ionisation Energy of magnesium
Mg(g) ——> Mg+(g) + e¯
2nd Ionisation Energy of magnesium
Mg+(g) ——> Mg2+(g) + e¯
Electron Affinity of chlorine
Cl(g) + e¯ ——> Cl¯(g) x2
Lattice Enthalpy of MgCl2
Mg2+(g) + 2Cl¯(g) ——> MgCl2(s)
1
2
3
4
5
7
6
Born-Haber Cycle - MgCBorn-Haber Cycle - MgCll22
148
244
738
1451 -364 x 2
-2526
-673
http://www.tes.co.uk/teaching-resource/Born-Haber-Cycle-6354470/