Chemistry 11 (HL)
Unit 6 / IB Topic 15.2
Energetics Cycle 6 Practice Problems
Born Haber Cycles and Lattice Enthalpies
1.
2.
Identify the type of energy change (ionization energy, energy of atomization, etc)
shown by each of these steps in the Born Haber cycle for potassium chloride.
State with a reason whether the step is exothermic or endothermic.
a)
1/2 Cl2(g) → Cl(g)
b)
Cl(g) → Cl (g)
first electron affinity; exo; energy is released
when an atom gains one electron
c)
K(s) → K(g)
enthalpy of atomization / enthalpy of
vapourization; endo; energy absorbed to
break metallic bonds in the solid
d)
K(g) → K (g) + e
e)
K(s) + 1/2Cl2(g) → KCl(s)
f)
KCl(s)  K (g) + Cl (g)
½ bond enthalpy for the Cl-Cl bond; endo;
bond breaking absorbs energy
–
+
+
–
first ionization energy; endo; energy is
absorbed when removing electrons from an
atom
–
enthalpy of formation; exo; formation of
ionic compounds from their elements
releases energy overall
lattice enthalpy; endo; energy is required to
break ionic bonds
Define the following terms and then using MgO as an example write equations to
represent each term. Include states.
State whether the process is exothermic or endothermic.
See your notes for DEFINITIONS.
a)
standard enthalpy of formation Mg(s) + ½ O2(g)  MgO
b)
standard lattice enthalpy
c)
standard enthalpy of atomization for a metal and non metal
metal
nonmetal
2+
(exo)
2-
MgO(s)  Mg (g) + O (g)
(endo)
Mg(s)  Mg(g)
(endo)
½ O2(g)  O(g)
(endo)
1+
–
d)
standard enthalpy of 1st ionization
Mg(g)  Mg (g) + e
e)
standard enthalpy of 2nd ionization
Mg (g)  Mg (g) + e
f)
standard 1st electron affinity
O(g) + e  O (g)
g)
standard 2nd electron affinity
O (g) + e  O (g)
1+
–
–
–
2+
(endo)
–
(endo)
–
(exo)
2–
(endo)
p. 1
Chemistry 11 (HL)
Unit 6 / IB Topic 15.2
3.
a)
Draw and label a Born-Haber cycle for the formation of calcium oxide.
3.
b)
Calculate the lattice enthalpy of calcium oxide from the following data,
data found in the IB Chemistry Data Booklet and the Born Haber cycle drawn
in part (a).
-1
enthalpy of atomisation of Ca(s) = +178 kJ mol
-1
second ionization energy of Ca(g) = +1150 kJ mol
-1
enthalpy of formation of calcium oxide = -635 kJ mol
[∆Hat Ca + IE1 Ca + IE2 Ca + ½ BE O2 + Ea1 O + Ea2 O] = ∆Hºf CaO + ∆Hºlattice CaO
[ 178 + 590 + 1150 + ½ (498) + (-141) + 78 ] = -635 + ∆Hºlattice
2104 = -635 + ∆Hºlattice
∆Hºlattice = +2708 kJ mol
4.
-1
This diagram shows an alternative way of representing a Born Haber enthalpy cycle
for potassium oxide, K2O:
+
2–
2K (g) + O (g)
+
–
Y
2K (g) + 2e + O(g)
X
- ∆Hºlattice(K2O)
2K(g) + O(g)
W
2K(g) + ½ O2(g)
2(89.2 kJ)
2K(s) + ½ O2(g)
Z (-361 kJ)
K2O(s)
p. 2
Chemistry 11 (HL)
a)
Unit 6 / IB Topic 15.2
Identify (by name) the enthalpy changes identified by the letters W, X, Y and
Z.
W = enthalpy of atomization (or ½ bond enthalpy) for O2
X = 2 x (IE1 for potassium)
Y = EA1 + EA2 for oxygen
Z = enthalpy of formation of K2O
b)
Use the energy cycle and information from Tables 7 and 10 of the Data
Booklet to calculate an experimental value for the lattice enthalpy of
potassium oxide.
Note: In this enthalpy cycle, the negative value of the lattice enthalpy is
used. But the clockwise arrows still add up to equal the counterclockwise
arrows.
2(89.2) + W + X + Y + (–∆Hlattice) = Z
2(89.2) + ½ (498) + 2(419) + (-141) + 798 + (–∆Hlattice) = -361
-1
∆Hlattice = 2283.4 kJ mol
5.
Consider the following theoretical and experimental lattice enthalpies and then
deduce the order of increasing covalent character.
compound
KF
CaO
LiO
experimental lattice
-1
enthalpy (kJ mol )
801
3513
744
theoretical lattice
-1
enthalpy (kJ mol )
795
3477
728
Remember: The closer the agreement (lower % difference) between the
experimental and theoretical values, the greater the ionic character. The lower the
agreement (greater % difference) between the two values, the greater the covalent
character.
order of increasing covalent character:
least covalent (most ionic)
most covalent
KF
CaO
LiO
% difference = 0.75%
% difference = 1.02%
% difference = 2.15%
p. 3
Chemistry 11 (HL)
6.
Unit 6 / IB Topic 15.2
Construct a Born Haber cycle to calculate the first electron affinity of iodine from the
following data:
-1
enthalpy of atomisation of Ag(s) = 284 kJ mol
-1
first ionization energy of Ag(g) = 731 kJ mol
-1
enthalpy of vapourization of I2(s) = 42 kJ mol
-1
bond enthalpy of I2(g) x ½ = 76 kJ mol
-1
lattice enthalpy of AgI(s) = 889 kJ mol
-1
enthalpy of formation of AgI(s) -62 kJ mol
Ag+(g) + I–(g)
EA1 I = x
Ag+(g) + I(g)
1/2 BE I2 = +76
Ag+(g) + 1/2 I2 (g)
∆Hvap I2 = +42
∆Hºlat AgI = +889
Ag+(g) + 1/2 I2(s)
IE1 Ag = +731
Ag(g) + 1/2 I2(s)
∆Hat Ag(s) = +284
Ag(s) + 1/2 I2(s)
AgI(s)
∆Hºf AgI = -62
∆Hat Ag + IE1 Ag + ∆Hvap I2 + ½ BE I2 + EA1 I(g) = ∆Hºf AgI + ∆Hºlattice AgI
284 + 731 + 42 + 76 + x = (-62) + 889
-1
x = -306 kJ mol
Multiple Choice Questions
7.
The lattice enthalpy of calcium bromide is the energy change for the reaction
A.
C.
E.
8.
Ca(g) + 2Br(g) → CaBr2(g)
Ca(s) + Br2(l) → CaBr2(s)
–
2+
CaBr2(s)→ Ca (g) + 2Br (g)
B.
D.
CaBr2(s) → Ca(g) + 2Br(g)
CaBr2(s) → Ca(g) + Br2(g)
The standard enthalpy of formation of KCl(s) is -437 kJmol-1. In a Born-Haber cycle
for the formation of KCl(s) which other enthalpy change(s) are exothermic?
A.
B.
C.
D.
E.
the lattice enthalpy and the electron affinity of chlorine
the electron affinity of chlorine
the formation of Cl(g) from Cl2(g)
the enthalpy of atomization of K(s) and the first ionization energy of K(g)
lattice enthalpy
Note: B is the answer and not A because we using the IB definition for lattice
enthalpy (turning the solid ionic compound into gaseous ions, not the other way
around).
p. 4
Chemistry 11 (HL)
9.
Unit 6 / IB Topic 15.2
The Born-Haber cycle for the formation of potassium chloride includes the steps
below:
+
I.
K(g) → K (g)
II.
1/2 Cl2(s) → Cl(g)
–
–
III.
Cl(g) + e → Cl (g)
+
–
IV.
K (g) + Cl (g) → KCl(s)
Which of these steps are exothermic?
A.
C.
10.
I and II
I, II and III
B.
D.
III and IV
I, II and IV only
Which of the following has the largest lattice enthalpy?
A.
D.
MgCl2(s)
NaCl(s)
B.
E.
KCl(s)
AgCl(s)
C.
LiCl(s)
p. 5