Chem 1711
1.
Born-Haber Cycle, Practice Problems
Given the following information for magnesium, oxygen, and magnesium oxide calculate the second electron gain
enthalpy for oxygen {i.e. for O— (g) + e—  O2— (g)}.
for Mg (s), ΔHsub = +148 kJ/mol
1st ionization energy for Mg = +738 kJ/mol
2nd ionization energy for Mg = +1450 kJ/mol
2.
bond dissociation energy for O2 = +499 kJ/mol
1st electron gain enthalpy for O = –141 kJ/mol
for MgO (s), lattice energy = +3890 kJ/mol
for MgO (s), enthalpy of formation = –602 kJ/mol
Consider an ionic compound MX2 where M is a metal that forms a cation of +2 charge, and X is a nonmetal that forms an
anion of –1 charge. A Born-Haber cycle for MX2 is given below. Each step in this cycle has been assigned a number
(1 – 7).
M (s)
+
3
X2 (g)
6
M (g)
1
MX2 (s)
M+ (g)
7
5
2
(g)
Identify one step (1 - 7) that is endothermic as written.
______________
b.
Which step (1 - 7) corresponds to ΔHosub?
_____________
c.
Which step (1 - 7) corresponds to ΔHof?
_____________
d.
Use the following energy values to calculate the lattice energy (in kJ/mol) for MX2.
ΔHosub = 296 kJ/mol; ΔHof = -421 kJ/mol; 1st ionization energy = 378 kJ/mol;
2nd ionization energy = 555 kJ/mol; bond dissociation enthalpy = 310 kJ/mol;
electron affinity = -427 kJ/mol.
2 X (g)
4
M2+
a.
+
2 X- (g)
Born'Haber CYcle,PracticeProblerns
C)\eM lltl
eh#{B
(iivcrt tho i'oiio*'irrg irti'orrrrttiicrir
t;;iygcil, aiiicl,t l. t-(.l'l,r.l 'L. .. \' r. L. ;r "r "r l. . ' . i . 1 , ' . ' . , l , . r r l , , l r . t l r ! . { , r ' r r t r l
1'ortrlzigircsiiili'l,
+
O2 (S)}
cnthalpl lbr oxygen {i.c. for O (S).' o-
i.
energy1'orO. = +-199kJ/rnol
bonclclissociation
gaincnthalpy1'orO = -l'11 k.l/mcll
l" elcctt'ott
I-orMgO (s),leitticceIrerg)'= +3890k.l/ttrol
1'orMgO (s),cnthalpyof'lbnnation= 602 k.l/rnol
= +148 k.l/lriol
lbr Mg (s), AH.,,.,
l ' ' i o n i z a l i o ne n c r g ) ' f o rM g = + 7 3 8k . l / m o l
lirr Mg = +1-l-50k.l/rnol
cnerg_\'
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for MX. is given bclon,. llacli stcp in tliis cycle hiis bcen assi-9ncdr rtutttbcr
el
a n i o no 1 ' - l c h a r g c .A L l o r n - H a t r c1,clc
(t -1).
2.
i\l (s)
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i
.j"*'
+
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+
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lrt: (g)
+
Xl (g)
"lt
\
*
\
ri.
Idcntil'1'oncstcp (l - 7) that is crldothcrmrcas wrrtten
b.
Which step (1 - 7) corrcspoudsto AHu.,,s'?
3,'l r 5, lo
3
\
zi(st
\
I
NL\: (s)
to AH"t:,'
Which stcp ( I - 7) corlcsl.rotttls
L . l s et h e l b l l o r v i n gc n c r g y ' v a l r r ctso c a l c t t l a t ct h c l a t l i c ec n c r . g r ' ( i nk J / n r o l )f i r r ' l \ ' l X .
AHu.ul,= 296 k.l/nrol; AI'lu,= -'12I lt.l/nrol; l't iottiz'ationetrcrg\'- 378 k'l/rrtol;
. l / n r o l ; b o n cdli s s o c i a t i o pc n t h a l p l ' = . 11 0 k . l / n r o l ;
2 ' ) ( il o r r i z a t i o 'e n c r g y* - 5 - 5k- 5
-127
k.llnoI.
clectron afl'inrt1'=
rl
I
- LE
A H a ; s+
, 2(r4)
-aHl
+z(€il
AHj'^b + l s l j - t + z n A : - ,+Ea l - l a i s s
AHr= aHJ"ub+ l s + J E+ z " d T E+
tV*ol
: $ - q , , +3 - 1 8 + s s s3+1 0 +2 G q a - -r k) L t L ' t
L-
b
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