# Thermodynamics Revision - Watford Boys

Mechanics

Oct 27, 2013 (4 years and 8 months ago)

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Thermodynamics Revision

1.
Write
symbol
equat
ions to show the transformation
(including state symbols)
associated with the following energy changes

a)

1
st

I.E. sodium

b)

1
st

and 2
nd

E.A. oxygen

c)

∆H
at

chlorine

d)

∆H
at

carbon

e)

∆H
at

calcium

f)

∆H
at

bromine

g)

∆H
at

oxygen

h)

∆H
at

copper

i)

∆H
lat
sodium chloride

j)

∆H
lat

magnesium oxide

k)

∆H
lat

copper

(II)

fluoride

l)

∆H
lat

potassium bromide

m)

∆H
sol
potassium bromide

n)

∆H
sol
sodium chloride

o)

∆H
sol

magnesium iodide

p)

∆H
hyd

sodium ion

q)

∆H
hyd
chloride ion

r)

∆H
hyd

magnesium ion

s)

∆H
f
sodium chloride

t)

∆H
f
ca
rbon dioxide

u)

∆H
f
copper
(I)
oxide

2.

Using the following values for ∆H
lat
, and ∆H
hyd
work out the enthalpy change of
solution for following compounds.

Comment on how easily you think they will dissolve.

a)

CaCl
2

b)

CsCl

c)

NaCl

d)

KCl

e)

KF

∆H
hyd

(kJ.mol
-
1
)

∆H
lat

(kJ.mol
-
1
)

Ca
2+

-
1577

CaCl
2

-
2258

Cs
+

-

256

CsCl

-
661

Na
+

-

406

NaCl

-
771

K
+

-

322

KCl

-
711

F
-

-

505

KF

-
817

Cl
-

-

364

(f)
Using the ∆H
soln

and ∆H
lat

values for the following salts work out the ∆H
hyd

values for
t
he NH
4
+
ion.

∆H
soln

∆H
lat

NH
4
Cl 14.8 kJ.mol
-
1

NH
4
Cl
-
705 kJ.mol
-
1

(g) Why do salts with a positive value of ∆H
sol

still dissolve without heating?

3.

Construct a Born
-
Haber cycle and using
the data supplied

1. Calculate ∆H
f

for LiCl
(s)

2. Calcu
late ∆H
lat

for MgO
(s)

3. Calculate 1
st

E.A. for F
(g)

→ F
-
(g)

4. Calculate 1
st

I.E. for K
(g)

→ K
+
(g)

5. Calculate ∆H
at

for silver.

∆H
f

values (kJ.mol
-
1
)

MgO

-
602

CaF
2

-
1220

KBr

-
394

AgCl

-
127

∆H
lat

values (kJ.mol
-
1
)

CaF
2

-
2630

KBr

-
679

LiCl

-
848

AgCl

-
905

∆H
at

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.)

v
alues (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 in
to the other and calculate the
energy change for that reaction using Hess’ Law. Is it exothermic or endothermic? Could you
have predicted that beforehand?

H
2
O
2

-
188

H
2
O

-
286

LiH

-
90

LiOH

-
485

AlCl
3

-
704

Al(OH)
3

-
1676

5. Do you expect the entro
py (∆S
sys
) to increase or decrease in these reactions? Why?

a)

3H
2(g)

+ N
2(g)

2NH
3(g)

b)

2H
2(g)

+ O
2(g)

2H
2
O
(l)

c)

2H
2
O
2(l
)

2H
2
O
(l
)

+ O
2(g)

d)

2Na
(s)

+ Cl
2(g)

2NaCl
(s)

e)

CaCO
3(s)

Ca
O
(s)

+ CO
2
(g)

f)

3Fe
(s
)

+ 2O
2(g)

Fe
3
O
4(s
)

6. Us
e
the following
values

for S (J.K
-
1
.mol
-
1
)

to calculate actual values of ∆S
sys

for the
reactions a) to f) above.

H
2
130.6

N
2

191.6

O
2

205.0

Cl
2

165.0

H
2
O
(l)

69.9

H
2
O
2(l)

109.6

Na
51.2

NaCl
(s)

72.1

CaO
(s)

39.7

CaCO
3(s)

92.9

CO
2(g)

213.6

Fe
27.3

F
e
3
O
4(s)

146.4

NH
3(g)

192.3

7. Using
∆H
f

values in the data book calculate ∆H
r

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 S
sys
, as the temperature increases?

10
. If any rea
ctions 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)

ZnO
(s)

+
CO
(g)

∆H= +237 kJ.mol
-
1
, ∆S= + 189.9 J.K
-
1
.mol
-
1

b)
Fe
3
O
4 (s)
+ 4H
2(g)

3Fe
(s
)

+ 4H
2
O
(9)

∆H=
-
144

kJ.mol
-
1
, ∆S= + 226
.9 J.K
-
1
.mol
-
1

c)
2Mg
(s)
+ O
2(g
)

2MgO
(s
)

∆H=
-
1203

kJ.mol
-
1
, ∆S=
-

216.6

J.K
-
1
.mol
-
1

d) 2Ag
2
O
(s)

4Ag
(s)

+
O
2
(g)

∆H= +62

kJ.mol
-
1
, ∆S= + 1
32.8

J.K
-
1
.mol
-
1

e) BaC
O
3(s)

BaO
(s)

+
CO
2
(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 t
o the second law of thermodynamics?

Mark Scheme

a)

Na
(g)

Na
(
g
)

+ e
-

b)

O
2(g)
+ e
-

O
-
(g)

O
-
(g)

+ e
-

O
2
-
(g)

c)

½ Cl
2(g)

Cl
(g)

d)

C
(s)

C
(g)

e)

Ca
(s)

Ca
(g)

f)

½ Br
(l)

Br
(g)

g)

½ O
2(g)

O
(g)

h)

Cu
(s)

Cu
(g)

i)

Na
+
(g)

+ Cl
-
(g)

NaCl
(s)

j)

Mg
2+
(g)

+ O
2
-
(g)

MgO
(s)

k)

Cu
2+
(g)

+ 2F
-
(g)

CuF
2(s)

l)

K
+
(g)

+ Br
-
(g)

KBr
(s)

m)

KBr
(s)

K
+
(aq)

+ Br
-
(aq)

n)

NaCl
(s)

Na
+
(aq)

+ Cl
-
(aq)

o)

MgI
2(s)

Mg
2+
(aq)

+ 2I
-
(aq)

p)

Na
+
(g)

Na
+
(aq)

q)

Cl
-
(g)

Cl
-
(aq)

r)

Mg
2+
(g)

Mg
2+
(
aq)

s)

Na
(s)

+ ½ Cl
2(g)

NaCl
(s)

t)

C
(s)

+ O
2(g)

CO
2(g)

u)

2Cu
(s)

+ ½ O
2(g)

Cu
2
O
(s)

2.

a)

∆H
sol

= ∑
∆H
hyd

-

∆H
lat

= (
-
1577) + 2(
-
364)

(
-
2258)

=

-

47 kJ.mol
-
1

b)

∆H
sol

= (
-
256) + (
-
364)

(
-
661)

= + 41 kJ.mol
-
1

c)

∆H
sol

= (
-
406) +

(
-
364)

(
-
771)

= +1 kJ.mol
-
1

d)

∆H
sol

= (
-
322) + (
-
364)

(
-
711)

= + 25 kJ.mol
-
1

e)

∆H
sol

= (
-
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 b
e the most soluble. However CsCl despite its significant endotherm
(+41 kJ.mol
-
1
) is readily soluble.

f)

∆H
sol

= ∑∆H
hyd

-

∆H
lat

14.8

= (
-
364) + ∆H
hyd(ammonium)

(
-
705)

∆H
hyd(ammonium)

=
-

326.2 kJ.mol
-
1

g)

This is because formation of solutions from
solids always a significant increase in
entropy.

3.

a)

∆H
f

= 159 + 122

+ 520

349

848

=
-

396 kJ.mol
-
1

b)

-
602

= 150 + 249 + 738 + 1455

141 + 798 +∆H
lat

∆H
lat

=
-

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

= ∆H
at

+ 122 + 731

349

905

∆H
at

= + 274

kJ.mol
-
1

4
.

a)

H
2
O
2

H
2
O

+

½ O
2

-
286

-
188

Elements in their standard states

∆H
r

=
-
286
-
(188) =
-
98 kJ.mol
-
1

b)

LiH

+ H
2
O

LiOH

+ H
2

∆H
r

=
-
485
-
(
-
90
-
286) =
-
109 kJ.mol
-
1

c)

AlCl
3

+ 3 H
2
O

→ Al(OH)
3

+ 3HCl

∆H
r

=
-
3(286)
-
704
-
(
-
1676
-
3(93)) =
-
393 kJ.mol
-
1

(∆H
f(HCl)
=
-
93 kJ.mol
-
1
)

Compounds with a highly negative value of ∆H
f

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)

decrease (4 to 2 molecules)

b)

decrease (3 gas to one liquid)

c)

increase (2 to 3
molecules and l to g)

d)

decrease ( g to s )

e)

increase ( s to g)

f)

decrease

(g to s)

6.

S
products

-

S
reactants

=

∆S (J.K
-
1
.mol
-
1
)

a)

384.6

-

583.4

-

198.8

b)

139.8

-

466.2

-

326.4

c)

344.8

-

219.2

+ 125.6

d)

144.2

-

267.4

-
1
23.2

e)

253.3

-

92.9

+ 160.4

f)

146.4

-

491.9

-

345.5

7.

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 th
e surroundings
meaning the entropy of the surroundings will be increased.

8.

a)
-
38.03

kJ.mol
-
1

feasible

at 273K

b)
-
482
.
5

kJ.mol
-
1

feasible at 273K

c)

230.3

kJ.mol
-
1

feasible at 273K

d)
-
788.8

kJ.mol
-
1

feasible at 273K

e) +134.2

kJ.mol
-
1

not feasible
at 273K

f
)

1024.1
.0 kJ.mol
-
1

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 wil
l cancel out the exotherm.

10.
Becomes feasible at 1110
K

11.
a) T = 237/0.1899 = 1248K

b) exothermic with increase in entropy

always feasible

c) exothermic with d
rop in entropy feasible up to 5554
K

d) T= 62/0.1328 = 4
67K

e) T = 268/ 0.01719 = 15
59K

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 CO
2

and water molecules, and heat has

been generated so the overall entropy of the surroundings has increased.