5-born haber cycle
DESCRIPTION
Examples on Lattice EnergyTRANSCRIPT
1
Born Haber Cycle Lattice Enthalpy ( )lattH∆ : The enthalpy of formation of 1 mole of a compound
from its gaseous ions under standard conditions ( )1atm (298K, All latt∆H are exothermic
⇒ ( )( )∑∑ ∆+∆+∆−∆=∆ eoi
oat
of
olat HHHHH
∆Hf = ∆H1 + ∆H2 + ∆H3+ ∆H4 + ∆H6
For diatomic molecules ( )N and H ,O ,I ,Br ,Cl ,F 2222222
Energy Bond2
1=∆ natomizatioH
)()(2
12 gg ClCl → )C-E(C)]([ 2
122
1 llgClH natomizatio =∆
)()(2
1OO2 gg → O)-E(O)](O[ 2
122
1 =∆ gH natomizatio
)()( gsNa Na→ natomizatioH∆
atoms Gaseous(g)(g) Cl Na +
)()( gs ClNa 221
Elements
+
(g)(g) Cl Na
atoms Gaseous−+ +
(s)Cl Na
Compound Solid
Ato
miz
ati
on
En
erg
y
Ion
iza
tio
n E
ne
rgy
Ele
ctro
n A
ffin
ity
Enthalpy Lattice
formation ofEnthalpy
2
The Born - Haber cycle
⋅ (s)CaCl (g)2Cl (g)Ca 22 →+ −+ 0,)]([∆H 2latt <sCaCl
• Factors that affect the lattice energy are:
- Sizes (radii) of the ions - Charges on the ions
The smaller the ions and the higher the charges on these ions, the more negative the lattice enthalpy The larger the ions and the smaller the charges on these ions, the less negative the lattice enthalpy Explanation:
- In smaller ions the outer electrons are closer to the positive nucleus; hence the attractive force of nucleus holds them more tightly.
∴lattice energy becomes more negative (larger)
2MgF nmMgr 072.0)( 2 =+ latH∆ =
-2957 1. −molkJ )]([ 2 sMgFH lat∆ > )]([ 2 sCaFH lat∆
Because )()( 22 ++ < CarMgr 2CaF nmCar 10.0)( 2 =+ latH∆ =
-2630 1. −molkJ
- the higher the charges on one or both of the opposite ions the stronger the attractive force between them
∴lattice energy becomes more negative (larger)
Ca(g)
)()(
Elements
2 gClsCa +
)(Ca g+
IE1
2 E
A1
lattH∆
Θ
f∆H
IE2
(g)2C (g)Ca 2 l++
( )ClClE
gCl
−=∆ )]([H2 22
1at
)]([Hat sCa∆
(g)2C (g)Ca 2 l++
(g)2C (g)Ca -2 l++
(s)CaCl2
3
The table shows how increasing the charge on either of the ions increases the lattice enthalpy
Charge on CATION
is increasing
latH∆ / 1. −molkJ is increasing
Charge on ANION
is increasing
latH∆ / 1. −molkJ is increasing
NaCl -771 2MgCl -2493
2MgCl -2493 MgO -3889
3AlCl - 4500
Exercise: Which of the following has the largest lattice enthalpy?
a) 32 or AlFNaF, CaF
b) 322 O or AllNaCl, CaC
c) , MgFCaF 22
Answer
a) 3 AlF ( +3 Al ion is smaller and has higher charge than + Na and +2Ca )
- the attractive force between their opposite charges is stronger and the nucleus of +3 Al holds the electrons on F − more tightly.
b) 32O Al ( +3 Al and −2 O ions are smaller and has higher charges)
c) MgF2 ( − F ion is the same in both compounds, but +2 Mg ion is smaller than +2 Ca )
• Polarizing power of small cations having high charge and polarizability of large anions
and their effect on the nature of the bond
Polarization is due to - Size and charge of the cation - Only size of the anion
∼ Small cation, with large positive charge have high polarizing power ∼ The larger the anion, the more easily it is polarized
Small cations with high positive charge polarize large anions more and DISTORT the ionic bond and result in a considerable degree of covalent character
4
Theoretical )(AgClH lat∆ is less negative than the Experimental )(AgClH lat∆
silver ion is small ( has high polarizing power), it polarized chloride ion and caused distortion of the ionic bond. This results in a considerable degree of covalent character (sharing electrons)
• Predicting how ionic a bond is from LATTICE enthalpy DIFFERENCE
Enthalpy DIFFERENCE: The difference between the EXPERIMENTAL value (calculated by
Born Haber Cycle) and the THEORETICAL value (determined using the ionic sizes and charges)
⋅ When the enthalpy difference is small (less than 5 %), the bond is almost pure ionic
⋅ When the enthalpy difference is big (more than 5 %), the bond has a considerable degree of
covalency , that is sharing of electrons
Therefore the difference is due to the covalent character which is not considered in theoretical calculation
• Predicting stability of a compound over others from ofH∆ calculated from
Born – Haber Cycle
The more negative the o
fH∆ of a compound the more stable it is and the more likely to be formed
The more positive ofH∆ of a compound the less likely to be formed as it requires energy to be
formed. Therefore, less stable it is
• From Comparison between THERRETICAL and EXPERIMENTAL lattices , to
degree of covalency , stability and solubility
tionNopolarizaonpolarizati
pole pole
−
+
+
−
+
+
+ −
−
−
5
( )( )∑∑ ∆+∆+∆−∆=∆ eoi
oat
of
olat HHHHH
The lattice Enthalpy is very highly exothermic the first electron affinity of the very electronegative elements such as nitrogen, oxygen and halogens are endothermic
(g)2Al
)()(2
Elements
22
3 gOsAl +
)(2A gl +
lattH∆
Θ
f∆H
3O(g) (g)2Al 3 ++
( )OO
)](O[H3
2
3
22
1at
−
=∆
E
g
)]([H2 at sAl∆
(s)OA 32l
)]([H2 i1 gAl∆
)(2A 2 gl +
)]([H2 i2 gAl∆
)]([H2 2i3 gAl +∆
3O(g) (g)2Al 3 ++
(g)3O (g)2Al -23 ++
(g)3O (g)2Al -3 ++
)]([OeH3 -2 g∆
)]([OeH3 1 g∆
6
Question 1: Paper 4 Jun 2008 / Q 4 (a) The following data were collected in a Born – Haber cycle for silver fluoride, AgF.
value 1/ .kJ mol −
Enthalpy of atomization of silver +285
First ionization energy of silver +731
enthalpy of atomization of fluorine +79
Enthalpy of formation of silver fluoride -205
Lattice energy of silver fluoride -958
(i) On the following outline of Born – Haber cycle, complete the boxes A and B by adding the
formula and state symbol for the appropriate species. Write the name of enthalpy change C.
[3]
(ii) Use the data to calculate the first electron affinity of fluorine.
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[2]
122
(s) (g)Ag + F
( )A gg+ +
( )gF −
BoxB
BoxA
(s) AgF
.....................................
.....................................
C
7
(b) Hlatt∆ (Theoretical) is the lattice energy calculated assuming the crystal lattice is completely ionic.
Hlatt∆ (Experimental) is the lattice energy determined experimentally using the Born – Haber cycle
The values for silver halides are listed below.
Formula of
halide
Hlatt∆
(Theoretical)
1/ .kJ mol −
Hlatt∆
(Experimental)
1/ .kJ mol −
Hlatt∆ (Theoretical)
minus Hlatt∆ (Experimental)
1/ .kJ mol −
AgF -920 -958 38
AgCl -833 -905 72
AgBr -816 -891 75
AgI -778 -889 111
(i) Explain why the theoretical lattice energies become less exothermic from AgF to AgI.
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[3] (ii) Explain why the values of the theoretical and experimental lattice energies are different.
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[2]
(iii)Explain why the difference between the theoretical and experimental lattice energies increases from AgF to AgI. …………………………………………………………………………………………………………………………
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[2]
- anion (X-) size increases down group (1)
- Charges (on anions) are the same (1)
- so forces of attraction between ions become weaker(from AgF to AgI) (1)
- Theoretical value assumes 100% ionic, no covalent character (1) - (Experimental value is different) due to covalent character OR polarisation
of anion by the cations / distortion of the anion(1)
- (as) size of anion increases down group (1) - so anions become more easily polarised (down) group OR more distortion
of anion (1)
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Question 2: Paper 4 Jan 2008 / Q 3
The Born – Haber cycle below represents the enthalpy changes when calcium hydride CaH2 , is formed from its elements.
(a) Write down in terms of one of the symbols1H∆ to 6H∆ .
(i) The lattice energy of calcium hydride ………………..………………………………………… [1]
(ii) *The first electron affinity of hydrogen. ……………………………………………………… [1]
(b) Use the data in the table below to calculate the enthalpy of formation of calcium hydride, CaH2(s)
value 1/ .kJ mol −
Enthalpy of atomization of calcium +178
First plus second ionization energies of calcium +1735
enthalpy of atomization of hydrogen +218
First electron affinity of hydrogen -73
Lattice energy of calcium hydride -2389
Calculation: ………………………………………………………………………………………………………………………… …………………………………………………………………………………………………………………………
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[2]
(c) Explain why the lattice energy of magnesium hydride, MgH2(s), is more exothermic than the lattice energy of calcium hydride, CaH2(s). …………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………… …………………………………………………………………………………………………………………………
[3]
2(s) (g) Ca + H
1H∆
2 (s) CaH
2(g) (g) Ca + H
22(g) 2e (g) Ca + + H+ −
2(g) 2e (g) Ca + + 2H+ −
2(g) (g) Ca + 2H+ −
2H∆
3H∆
4H∆ 5H∆
6H∆
- Mg2+ ion is smaller than Ca2+ ion (1) - but charges are the same (1) - so stronger (forces of) attraction between ions in MgH2(s) (1)
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Question 3: Paper 4 Jan 2009 / Q 2
(a) Define the term lattice energy.
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[2]
(b) Use the energy cycle and the data to calculate the lattice energy of barium chloride.
value -1/kJ.mol
Enthalpy of formation of BaCl2(s) -859
enthalpy of atomization of barium +180
enthalpy of atomization of chlorine +122
1st + 2nd ionization energies of barium +1468
First electron affinity of chlorine -349
Calculation: …………………………………………………………………………………………………………
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[3]
2(s) (g) Ba + Cl
2(g) (g) Ba + Cl
2(g) (g) 2e Ba + 2C +l+ − 2
(g) (g) Ba + 2Cl+ −
(g) (g) Ba + 2Cl
(s) BaCl
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(c) (i) Explain how comparison of lattice energies from the Bor – Haber cycle with theoretical lattice energies provodes evidence for the nature of the bonding in the chlorides of the group 2 metals.
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[2]
(ii) Explain the difference in bonding between beryllium and chlorine and that between barium and chlorine.
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[2]
Theoretical model is based on 100% ionic bonding (1) If experimental Born Haber value is different or more exothermic / bigger this is due to some covalency or some covalent character in the bonding (1)
- Be2+ ion or beryllium ion is smaller (than the Ba2+ion (and have the same charge) (1)
- Be2+ ion polarizes /distorts the chloride ion more (than Ba2+does), leading to
covalency / covalent character (1)
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Question 4: Paper 4 Jan 2007 / Q 5
(a) A Born-Haber cycle for the formation of magnesium chloride is shown below.
value -1/kJ.mol
enthalpy of atomization of magnesium +150
1st ionization energy of magnesium +736
2nd ionization energy of magnesium +1450
enthalpy of atomization of chlorine +122
Enthalpy of formation of magnesium chloride -642
Lattice energy of magnesium chloride -2526
(i) Use the data to calculate the first electron affinity of fluorine.
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[2]
2(s) (g) Mg + Cl
2 (s) MgC l
2(g) (g) Mg + Cl
2(g) (g) Mg + C + 1el+ −
(g) (g) + 2e Mg + 2Cl −
2(g) (g)Mg + 2Cl+ −2
2(g) (g) Mg + C + 1el+ −
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(ii) The theoretically calculated value for the lattice energy of magnesium chloride is –2326 kJ.mol–1 Explain, in terms of the bonding in magnesium chloride, why the experimentally determined value of –2526 kJ.mol–1 is significantly different from the theoretical value.
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(b) The table shows values for the lattice energies of the metal chlorides of some Group 2 metals.
Group 2 metal chloride
MgCl2 CaCl2 SrCl2 BaCl2
Lattice energy/ -1kJ.mol
-2526 -2237 -2112 -2018
Explain why these lattice energies become less exothermic from MgCl2 to BaCl2.
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[2]
MgCl2 has (a degree of ) covalent character (1) due to polarisation of the anion by the small Mg
2+ cation (1)
- radius of M 2+ (ion) increases down the group (1)
- Charge on ions remains the same/2+ (1) - so down the group forces of attraction between ions become weaker (1)