Thermodynamics Revision
1. Write symbol equations to show the transformation (including state symbols)
associated with the following energy changes
a)
b)
c)
d)
e)
f)
g)
h)
i)
j)
k)
l)
m)
n)
o)
p)
q)
r)
s)
t)
u)
1st I.E. sodium
1st and 2nd E.A. oxygen
∆Hat chlorine
∆Hat carbon
∆Hat calcium
∆Hat bromine
∆Hat oxygen
∆Hat copper
∆Hlat sodium chloride
∆Hlat magnesium oxide
∆Hlat copper (II) fluoride
∆Hlat potassium bromide
∆Hsol potassium bromide
∆Hsol sodium chloride
∆Hsol magnesium iodide
∆Hhyd sodium ion
∆Hhyd chloride ion
∆Hhyd magnesium ion
∆Hf sodium chloride
∆Hf carbon dioxide
∆Hf copper (I) oxide
2.
Using the following values for ∆Hlat, and ∆Hhyd work out the enthalpy change of
solution for following compounds. Comment on how easily you think they will dissolve.
a) CaCl2
b) CsCl
c) NaCl
d) KCl
e) KF
∆Hhyd
Ca2+
Cs+
Na+
K+
FCl-
(kJ.mol-1)
-1577
- 256
- 406
- 322
- 505
- 364
∆Hlat (kJ.mol-1)
CaCl2
-2258
CsCl
-661
NaCl
-771
KCl
-711
KF
-817
(f) Using the ∆Hsoln and ∆Hlat values for the following salts work out the ∆Hhyd values for
the NH4+ ion.
∆Hsoln
NH4Cl 14.8 kJ.mol-1
∆Hlat
NH4Cl -705 kJ.mol-1
(g) Why do salts with a positive value of ∆Hsol still dissolve without heating?
3.
Construct a Born-Haber cycle and using the data supplied
1. Calculate ∆Hf for LiCl(s)
2. Calculate ∆Hlat for MgO(s)
3. Calculate 1st E.A. for F(g) → F-(g)
4. Calculate 1st I.E. for K(g) → K+(g)
5. Calculate ∆Hat for silver.
∆Hf values (kJ.mol-1)
MgO -602
CaF2 -1220
KBr -394
AgCl -127
∆Hlat values (kJ.mol-1)
CaF2 -2630
KBr -679
LiCl -848
AgCl -905
∆Hat values (kJ.mol-1)
Li
159
Mg
150
Ca
178
K
89
Bond enthalpy values (kJ.mol-1)
F-F 158
Cl-Cl 244
Br-Br 194 (note ∆vap = 30)
O=O 498
Ionisation energy (I.E.) values (kJ.mol-1)
Li
520
Mg
738 1455
Ca
590 1145
Ag
731
Electron affinities (E.A.) values (kJ.mol-1)
Cl
-349
Br
-325
O
-141 798
4. Using the following values for the enthalpy of formation (kJ.mol-1) of two related
compounds comment on which is the most stable (least reactive).
Try to write an equation showing the conversion of one into the other and calculate the
energy change for that reaction using Hess’ Law. Is it exothermic or endothermic? Could you
have predicted that beforehand?
H2O2
H2O
-188
-286
LiH
LiOH
-90
-485
AlCl3
Al(OH)3
-704
-1676
5. Do you expect the entropy (∆Ssys) to increase or decrease in these reactions? Why?
a)
b)
c)
d)
e)
f)
3H2(g) + N2(g)
2H2(g) + O2(g)
2H2O2(l)
→
2Na(s) + Cl2(g)
CaCO3(s)
→
3Fe(s) + 2O2(g)
→
2NH3(g)
→
2H2O(l)
2H2O(l)
+ O2(g)
→
2NaCl(s)
CaO(s) + CO2(g)
→
Fe3O4(s)
6. Use the following values for S (J.K-1.mol-1) to calculate actual values of ∆Ssys for the
reactions a) to f) above.
H2 130.6
N2 191.6
O2 205.0
Cl2 165.0
H2O(l) 69.9
H2O2(l) 109.6
Na 51.2
NaCl(s) 72.1
CaO(s) 39.7
CaCO3(s) 92.9
CO2(g) 213.6
Fe 27.3
Fe3O4(s) 146.4
NH3(g) 192.3
7. Using ∆Hf values in the data book calculate ∆Hr values for the above reactions. In some
cases you will need to construct Hess cycles. According to the second law of
thermodynamics entropy always increases, but in reactions a), b) d) and f) above it
decreases. Explain.
8. Calculate ∆G values at 273K for each reaction and comment on its feasibility at that
temperature.
9. What will happen for those reactions with a drop in Ssys, as the temperature increases?
10. If any reactions are not feasible at 273K, what temperature do they become feasible?
11. Find out at what temperature the following reactions become feasible.
a) ZnO(s) + C(s)
→
b) Fe3O4 (s) + 4H2(g) →
c) 2Mg(s) + O2(g)
d) 2Ag2O(s)
→
e) BaCO3(s) →
→
ZnO(s) + CO(g)
∆H= +237 kJ.mol-1, ∆S= + 189.9 J.K-1.mol-1
3Fe(s) + 4H2O(9)
∆H= -144 kJ.mol-1, ∆S= + 226.9 J.K-1.mol-1
2MgO(s)
∆H= -1203 kJ.mol-1, ∆S= - 216.6 J.K-1.mol-1
4Ag(s) + O2(g)
∆H= +62 kJ.mol-1, ∆S= + 132.8 J.K-1.mol-1
BaO(s) + CO2(g)
∆H= +268 kJ.mol-1, ∆S= + 171.9 J.K-1.mol-1
12. Some workers transform a field full of stones into dry stone walls surrounding the field.
How has the entropy changed with respect to the arrangement of stones? How has the
overall entropy of this process increased according to the second law of thermodynamics?
Mark Scheme
c)
d)
e)
f)
g)
h)
i)
j)
k)
l)
m)
n)
o)
p)
q)
r)
s)
t)
u)
Na(g) →
Na(g)
O2(g) + e
→
O-(g) + e→
½ Cl2(g)
→
C(s) →
C(g)
Ca(s)
→
½ Br(l)
→
½ O2(g)
→
Cu(s) →
Cu(g)
Na+(g) + Cl-(g)
Mg2+(g) + O2-(g)
Cu2+(g) + 2F-(g)
K+(g) + Br-(g)
KBr(s)
→
NaCl(s)
→
MgI2(s)
→
+
Na (g)
→
Cl-(g)
→
Mg2+(g)
→
Na(s) + ½ Cl2(g)
C(s) + O2(g)
→
2Cu(s) + ½ O2(g)
2.
a)
a)
b)
+ eO-(g)
O2-(g)
Cl(g)
Ca(g)
Br(g)
O(g)
→
NaCl(s)
→
MgO(s)
→
CuF2(s)
→
KBr(s)
K+(aq) + Br-(aq)
Na+(aq) + Cl-(aq)
Mg2+(aq) + 2I-(aq)
Na+(aq)
Cl-(aq)
Mg2+(aq)
→
NaCl(s)
CO2(g)
→
Cu2O(s)
∆Hsol = ∑∆Hhyd - ∆Hlat
= (-1577) + 2(-364) – (-2258)
= - 47 kJ.mol-1
b)
∆Hsol = (-256) + (-364) – (-661)
= + 41 kJ.mol-1
c)
∆Hsol = (-406) + (-364) – (-771)
= +1 kJ.mol-1
d)
∆Hsol = (-322) + (-364) – (-711)
= + 25 kJ.mol-1
e)
∆Hsol = (-322) + (-505) – (-817)
= -10 kJ.mol-1
Initial thoughts are that those that dissolve with an exotherm a) and e) or very small
endotherm c) would be the most soluble. However CsCl despite its significant endotherm
(+41 kJ.mol-1) is readily soluble.
f)
∆Hsol = ∑∆Hhyd - ∆Hlat
14.8 = (-364) + ∆Hhyd(ammonium) – (-705)
∆Hhyd(ammonium) = - 326.2 kJ.mol-1
g)
This is because formation of solutions from solids always a significant increase in
entropy.
3.
a)
∆Hf
b)
-602 = 150 + 249 + 738 + 1455 – 141 + 798 +∆Hlat
∆Hlat = - 3851 kJ.mol-1
c)
-1220 = 178 + 158 + 178 + 590 + 1145 + 2(E.A.) - 2630
E.A. = - 330.5 kJ.mol-1
d)
- 394 = 89 + 112 + I.E. – 325 – 679
I.E. = + 409 kJ.mol-1
e)
-127 = ∆Hat + 122 + 731 – 349 – 905
∆Hat = + 274 kJ.mol-1
a)
H2O2 →
4.
= 159 + 122 + 520 – 349 – 848
= - 396 kJ.mol-1
H2O
+
½ O2
-286 ↑
-188
Elements in their standard states
b)
c)
∆Hr
= -286-(188) = -98 kJ.mol-1
LiH
+ H2O
∆Hr
= -485-(-90-286) = -109 kJ.mol-1
AlCl3 + 3 H2O
∆Hr
→
LiOH + H2
→ Al(OH)3
+ 3HCl
= -3(286)-704-(-1676-3(93)) = -393 kJ.mol-1
(∆Hf(HCl)= -93 kJ.mol-1)
Compounds with a highly negative value of ∆Hf are more stable or less reactive. This
is illustrated above by the fact that a related more reactive compound can be
converted into a more stable form in an exothermic reaction.
5.
a)
b)
c)
d)
e)
f)
decrease (4 to 2 molecules)
decrease (3 gas to one liquid)
increase (2 to 3 molecules and l to g)
decrease ( g to s )
increase ( s to g)
decrease (g to s)
6.
7.
=
∆S (J.K-1.mol-1)
- 198.8
a)
Sproducts
384.6
-
Sreactants
583.4
b)
139.8
-
466.2
- 326.4
c)
344.8
-
219.2
+ 125.6
d)
144.2
-
267.4
-123.2
e)
253.3
-
92.9
+ 160.4
f)
146.4
-
491.9
- 345.5
a) -92.3 kJ.mol-1
b) -571.6 kJ.mol-1
c) -196.0 kJ.mol-1
d) -822.4 kJ.mol-1
e) +178.3 kJ.mol-1
f) – 1118.4 kJ.mol-1
a), b), d) and f) are exothermic so heat energy is transferred to the surroundings
meaning the entropy of the surroundings will be increased.
8.
a) -38.03 kJ.mol-1
b) -482.5 kJ.mol-1
c) –230.3 kJ.mol-1
d) -788.8 kJ.mol-1
e) +134.2 kJ.mol-1
f) – 1024.1.0 kJ.mol-1
feasible at 273K
feasible at 273K
feasible at 273K
feasible at 273K
not feasible at 273K
feasible at 273K
9. If the products are becoming more ordered (decrease in entropy) the T∆S term
becomes more positive as temperature increases. At sufficiently high temperatures the
reaction will become unfeasible as it will cancel out the exotherm.
10. Becomes feasible at 1110K
11. a) T = 237/0.1899 = 1248K
b) exothermic with increase in entropy – always feasible
c) exothermic with drop in entropy feasible up to 5554K
d) T= 62/0.1328 = 467K
e) T = 268/ 0.01719 = 1559K
12. The stones have fewer ways or being arranged (more ordered) and therefore their
entropy has been reduced. The workers have had to respire to produce this ordering so
glucose molecules have been converted into CO2 and water molecules, and heat has
been generated so the overall entropy of the surroundings has increased.