Use Born-Haber cycles to calculate the following from the
data given below.
1. The lattice enthalpy of KBr.
2. The electron attachment enthalpy of Cl(g)
3. The ionisation enthalpy of Ag(g).
4. The lattice enthalpy of Li2O.
5. The sublimation enthalpy of Fe(s).
6. The lattice enthalpy of Al2O3.
Answers in random order:
738, -677, -2853, -347, -15301, 434 kJ.mol-1.
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ DATA ^^^^^^^^^^^^^^^^^^^^^^^^^^^
Sublimation enthalpies:
DHsub(Al,s) = 326 kJ mol-1
DHsub(Ag,s) = 289 kJ mol-1
DHsub(K,s) = 89 kJ mol-1
Atomisation enthalpies:
DHatom(Cl) = 121 kJ.mol-1
DHatom(Li,s) = 159 kJ mol-1
N.B. These are the enthalpy change when one mole of gaseous
atoms is formed from the element in its Standard State at
298K and 1 atm (101.325 kPa) pressure. They may be
equivalent to one or more other steps. e.g. for a solid metal
it is the same as the sublimation enthalpy, but for bromine
it is HALF the sum of the enthalpies of vaporisation and
dissociation (half because these are defined per mole of Br2
molecules - which give two moles of atoms).
Bond dissociation enthalpies:
DHbond(Br2) = +224 kJ mol-1
DHbond(O2) = +498 kJ mol-1
DHbond(F2) = +158 kJ mol-1
Ionization enthalpies:
Al(g) --> Al3+(g) + 3e- DH = 5138 kJ mol-1
Fe(g) --> Fe+(g) + e- DH = 762 kJ mol-1
Fe+(g)--> Fe+2(g) + e- DH = 1561 kJ mol-1
K(g) --> K+(g) + e- DH = 418 kJ.mol-1
Li(g) --> Li+(g) + e- DH = 520 kJ mol-1
Zn(g) --> Zn2+(g) + 2e- DH = 2638 kJ mol-1
Electron attachment (affinity) enthalpies:
Br(g) + e-(g) --> Br-(g) DH = -342 kJ mol-1
O(g) + e-(g) --> O-(g) DH = -141 kJ mol-1
O-(g) + e-(g) --> O2-(g) DH = 791 kJ mol-1
S(g) + 2e- ---> S2-(g) DH = 449 kJ mol-1
Lattice enthalpies:
DHlatt(AgBr)= -904 kJ mol-1
DHlatt(FeO) = -3923 kJ mol-1
DHlatt(KCl) = -717 kJ mol-1
Enthalpies of formation:
DHform(Al2O3)= -1676 kJ mol-1
DHform(Li2O) = -596 kJ mol-1
DHform(AgBr) = -99.5 kJ mol-1
DHform(FeO) = -267 kJ mol-1
DHform(ZnS) = -206.4 kJ mol-1
DHform(KBr) = -392 kJ mol-1
DHform(KCl) = -436 kJ.mol-1
Enthalpy of vaporisation:
DXHvap(Br2,l) = 15 kJ.mol-1
-----------------------------------------------------------------
Construct a Born-Haber cycle that includes the unknown quantity as one of the steps in the form of an enthalpy diagram. Make sure each level contains ALL the atoms (and electrons if necessary) in the right numbers - each step is a balanced chemical equation. Put in the values of all the enthalpy changes except for the one sought and an arrow to indicate the direction of the change it applies to (i.e. as in the definition of the quantity.) Remember to include the necessary factors.
Put in the arrow for the unknown quantity and then go round the cycle in the other direction changing the sign for any quantity whose arrow goes the opposite way. The sum of all these is the quantity you want.
N.B. This display needs a full width screen to make sense.
1. K+(g) + e-(g) + Br(g)
^^^^^^^^^^^^^^^^^^^^^
DHea = -342 kJ.mol-1
DHion = 418 kJ.mol-1
K+(g) + Br-(g)
K(g) + Br(g) ^^^^^^^^^^^^^^^^^^^^^
^^^^^^^^^^^^^^^^^^^^^
DHdiss = 224/2 kJ.mol-1
K(g) + 1/2Br2(g) DHlatt = x
^^^^^^^^^^^^^^^^^^^^^
DHvap = 15/2 kJ.mol-1
K(g) + 1/2Br2(l)
^^^^^^^^^^^^^^^^^^^^^
DHsub = 89 kJ.mol-1
K(s) + 1/2Br2(l)
^^^^^^^^^^^^^^^^^^^^^
DHform = -392 kJ.mol-1
KBr(s)
^^^^^^^^^^^^^^^^^^^^^
x = -(-342 + 418 + 224/2 + 15/2 + 89) +(-392) kJ.mol-1
= -676.5 kJ.mol-1
= -677 kJ.mol-1 (decimal places not justified.)
============
-----------------------------------------------------------------
2. K+(g) + e-(g) + Cl(g)
^^^^^^^^^^^^^^^^^^^^^
DHea = y
DHion = 418 kJ.mol-1
K+(g) + Cl-(g)
K(g) + Cl(g) ^^^^^^^^^^^^^^^^^^^^^
^^^^^^^^^^^^^^^^^^^^^
DHatom = 121 kJ.mol-1
K(g) + 1/2Cl2(g) DHlatt = -717
kJ.mol-1
^^^^^^^^^^^^^^^^^^^^^
DHsub = 89 kJ.mol-1
K(s) + 1/2Cl2(l)
^^^^^^^^^^^^^^^^^^^^^
DHform = -436 kJ.mol-1
KCl(s)
^^^^^^^^^^^^^^^^^^^^^
y = -(418 + 121 + 89) + (-436) -(-717) kJ.mol-1
y = -347 kJ.mol-1
=============
-----------------------------------------------------------------
3. Ag+(g) + e-(g) + Br(g)
^^^^^^^^^^^^^^^^^^^^^
DHea = -342 kJ.mol-1
DHion = z
Ag+(g) + Br-(g)
Ag(g) + Br(g) ^^^^^^^^^^^^^^^^^^^^^
^^^^^^^^^^^^^^^^^^^^^
DHdiss = 224/2 kJ.mol-1
Ag(g) + 1/2Br2(g) DHlatt = -904 kJ.mol-1
^^^^^^^^^^^^^^^^^^^^^
DHvap = 15/2 kJ.mol-1
Ag(g) + 1/2Br2(l)
^^^^^^^^^^^^^^^^^^^^^
DHsub = 289 kJ.mol-1
Ag(s) + 1/2Br2(l)
^^^^^^^^^^^^^^^^^^^^^
DHform = -99.5 kJ.mol-1
AgBr(s)
^^^^^^^^^^^^^^^^^^^^^
z = -(112 + 289 +15/2) +(-99.5) -(-904) -(-342) kJ.mol-1
z = 738 kJ.mol-1
============
------------------------------------------------------------------
4. 2Li+(g) + O2-(g)
^^^^^^^^^^^^^^^^^^^^^
/
/
2Li+(g) + 2e-(g) + O(g) DHea(O-) = +791 kJ.mol-1
^^^^^^^^^^^^^^^^^^^^^ /
^ \ DHea(O) = -141 kJ.mol-1
| \ /
| \ 2Li+(g) + e-(g) + O-(g)
| ^^^^^^^^^^^^^^^^^^^^^
2 x DHion = 2 x 520 kJ.mol-1
|
| DHlatt = p
2Li(g) + O(g)
^^^^^^^^^^^^^^^^^^^^^
DHdiss = 498/2 kJ.mol-1
2Li(g) + 1/2O2(g)
^^^^^^^^^^^^^^^^^^^^^
2DHatom = 2 x 159 kJ.mol-1
2Li(s) + 1/2O2(g)
^^^^^^^^^^^^^^^^^^^^^
DHform = -596 kJ.mol-1
Li2O(s)
^^^^^^^^^^^^^^^^^^^^^
p = -(2x159 +498/2 + 2x520 - 141 + 791)-596 kJ.mol-1
p = -2853 kJ.mol-1
=============
-----------------------------------------------------------------
5. Fe2+(g) + O2-(g)
^^^^^^^^^^^^^^^^^^^^^
DHea(O) + DHea(O-) = -141 + 791 kJ.mol-1
Fe2+(g) + 2e-(g) + O(g)
^^^^^^^^^^^^^^^^^^^^^
DHion = 762 + 1561 kJ.mol-1
DHlatt = -3923 kJ.mol-1
Fe(g) + O(g)
^^^^^^^^^^^^^^^^^^^^^
DHdiss = 498/2 kJ.mol-1
Fe(g) + 1/2O2(g)
^^^^^^^^^^^^^^^^^^^^^
DHsub = s
Fe(s) + 1/2O2(g)
^^^^^^^^^^^^^^^^^^^^^
DHform = -267 kJ.mol-1
FeO(s)
^^^^^^^^^^^^^^^^^^^^^
s = -267 -(-3923 +791 -141 +1561 +762 +498/2) kJ.mol-1
s = +434 kJ.mol-1
=============
-----------------------------------------------------------------
6. 2Al3+(g) + 3O2-(g)
^^^^^^^^^^^^^^^^^^^^^
3 x (DHea(O) + DHea(O-)) = 3(-141 + 791) kJ.mol-1
2Al3+(g) + 6e-(g) + 3O(g)
^^^^^^^^^^^^^^^^^^^^^
3 x DHion = 3 x 5138 kJ.mol-1
DHlatt = q
2Al(g) + 3O(g)
^^^^^^^^^^^^^^^^^^^^^
DHdiss = 3 x 498/2 kJ.mol-1
2Al(g) + 3/2O2(g)
^^^^^^^^^^^^^^^^^^^^^
2DHsub = 2 x 326 kJ.mol-1
2Al(s) + 3/2O2(g)
^^^^^^^^^^^^^^^^^^^^^
DHform = -1676 kJ.mol-1
Al2O3(s)
^^^^^^^^^^^^^^^^^^^^^
q = -(3(-141+791) +2x5138 +3/2x498 +2x326)-1676 kJ.mol-1
q = -15301 kJ.mol-1
=============
(C)H.J. Clase 1998