# THERMODYNAMICS

Mechanics

Oct 28, 2013 (4 years and 8 months ago)

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THERMODYNAMICS

Chapter 19

A process that occurs without ongoing

outside intervention
.

Examples

Nails rusting outdoors

SPONTANEOUS PROCESS

Ice melting at room temperature

Formation of water from O
2
(g) and H
2
(g):

Expansion of gas into an evacuated space

2H
2
(g) + O
2
(g) 2H
2
O(g)

Why are some processes spontaneous and
others not?

We know that

temperature

has an effect on the
spontaneity of a process.

e.g.
T>0
o
C

ice浥lts

spntaneusatthiste浰.

H
2
O (s)

H
2
O (l)

T<0
o
C

waterfreees

spntaneusatthiste浰.

H
2
O (l)

H
2
O (s)

T=0
o
C

waterandiceineuilibrium

H
2
O (l)

H
2
O (s)

Exothermic processes tend to be spontaneous.

Rusting of nail
-

SPONTANEOUS!

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

2
O
3
(s)

H=
-

822.2kJ.浯l
-
1

Example

Formation of water
-

SPONTANEOUS!

2H
2
(g) + O
2
(g) 2H
2
O(l)

H=
-

285.8kJ.浯l
-
1

Pt cat

However
, the dissolution of ammonium
nitrate is also
spontaneous
, but it is also
endothermic
.

NH
4
NO
3
(s)

4
+
(aq) + NO
3
-
(aq)

H=+25.7kJ.浯l
-
1

So is:

2N
2
O
5
(s)

4NO
2
(g) + 2O
2
(g)

H=+109.5kJ.浯l
-
1

a process does not have to be exothermic
to be spontaneous.

s浥thingelsebesidessignf

H

cntributetdeter浩ningwhetheraprcessis
spntaneusrnt.

That something else is:

ENTROPY (S)

extent of disorder!

More disordered

largerentrpy

S = S
final

-

S
initial

Entropy is a state function

Units: J K
-
1
mol
-
1

Entropy’s effects on the mind

NH
4
NO
3
(s)

4
+
(aq) + NO
3
-
(aq)

H‽+25⸷kJ⹭l
-
1

2N
2
O
5
(s)

4NO
2
(g) + 2O
2
(g)

H‽+109⸵kJ⹭l
-
1

Examples of spontaneous processes where
entropy increases:

Dissolution of ammonium nitrate:

Decomposition of dinitrogen pentoxide:

Spontaneous

Non
-
spontaneous

However, entropy does not always increase for
a s
pontaneous process

At room temperature:

SECOND LAW OF THERMODYNAMICS

The entropy of the universe increases in any
spontaneous process.

S
universe

=

S
system

+

S
surroundings

Spontaneous process:

S
universe

> 0

Process at equilibrium:

S
universe

= 0

Thus

S
univ

is continually increasing!

S
univ

must increase during a spontaneous
process, even if

S
syst

decreases.

Q: What is the connection between sausages and the
second law of thermo?

A: Because of the 2nd law, you can put a
pig into a machine and get sausage, but
you can't put sausage into the machine
and get the pig back.

For example:

Rusting nail =
spontaneous process

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

㉆2
2
O
3
(s)

S
syst
<0

BUT reaction is
exothermic
,

entrpyfsurrundingsincreases

asheatis
evlvedbythesyste洠therebyincreasing
motion of molecules in the surroundings.

S
univ
>0

S
surr
>0

Thus for

S
univ

=

S
syst

+

S
surr

>0


S
surr

>

S
syst

-
ve

+ve

+ve

Special circumstance =
Isolated system
:

Does not exchange energy nor matter with

surroundings


S
surr

= 0

Spontaneous process:

S
syst

> 0

Process at equilibrium:

S
syst

= 0

Spontaneous

Ther浯dyna浩callyfavurable

(Ntnecessarilyccuratbservablerate.)

Thermodynamics

directinandextentf
reactin,ntspeed.

EXAMPLE

State whether the processes below are
spontaneous, non
-
spontaneous or in equilibrium:

CO
2

decomposes to form diamond and O
2
(g)

Water boiling at 100
o
C to produce steam

in a
closed container

Sodium chloride dissolves in water

NON
-
SPONTANEOUS

SPONTANEOUS

EQUILIBRIUM

MOLECULAR INTEPRETATION OF S

Decrease in number of gaseous molecules

decreasein

S

e.g.2NO(g)+O
2
(g)

2NO
2
(g)

3 moles gas

2 moles gas

Molecules have 3 types of motion:

Translational motion

-

Entire molecule moves in
a direction (gas > liquid > solid)

Vibrational motion

within a molecule

Rotational motion

“spinning”

Greater the number of degrees of freedom

greaterentrpy

Decrease in temperature

decreaseinther浡lenergy

decreaseintranslatinal,vibratinal

andrtatinal浯tin

decreaseinentrpy

Asthete浰eraturekeepsdecreasing,these
motions “shut down”

reaches a point of
perfect order.

EXAMPLE

Which substance has the great entropy in each
pair? Explain.

C
2
H
5
OH(l) or C
2
H
5
OH(g)

2 moles of NO(g) or 1.5 moles of NO(g)

1 mole O
2
(g) at STP or 1 mole NO
2
(g) at STP

THIRD LAW OF THERMODYNAMICS

The entropy of a pure crystalline substance at
absolute zero is zero.

S(0 K) = 0

perfectrder

Ginsberg's Theorem

(The modern statement of the three laws of
thermodynamics)

1. You can't win.

2. You can't even break even.

3. You can't get out of the game.

Entropy increases for s

g

EXAMPLE

Predict whether the entropy change of the
system in each reaction is positive or negative.

CaCO
3
(s)

CaO(s)+CO
2
(g)

2SO
2
(g) + O
2
(g)

2SO
3
(g)

N
2
(g) + O
2
(g)

2NO(g)

H
2
O(l) at 25
o
C

H
2
O(l) at 55
o
C

+ve

-
ve

?

+ve

3 mol gas

2浯lgas

2 mol gas

2浯lgas

Can’t predict, but it
is close to zero

Increase thermal energy

Standard molar entropy (S
o
)

= molar entropy for substances in their

standard state

NOTE

S
o

0frele浥ntsintheirstandardstate

S
o
(gas) > S
o
(liquid) > S
o
(solid)

S
o

generally increases with increasing molar

mass

S
o

generally increases with increasing

number of atoms in the formula of the

substance

Calculation of

(S
o

from tabulated data)

Stoichiometric coefficients

EXAMPLE

Calculate

S
o

for the synthesis of ammonia
from N
2
(g) and H
2
(g):

N
2
(g) + 3H
2
(g)

2NH
3
(g)

S
o
/J.K
-
1
.mol
-
1

N
2
(g)

191.5

H
2
(g)

130.6

NH
3
(g)

192.5

Calculation of

For a process that occurs at constant
temperature and pressure, the entropy change
of the surroundings is:

(T &P constant)

GIBB’S FREE ENERGY (G)

Defined as:
G = H

TS

-

state function

-

extensive property

S
univ

=

S
sys

+

S
surr

At constant T and P:

-

T

S
univ

=
-

T

S
sys

+

H
sys

(at constant T & P)

S
univ

=

S
sys

-

H
sys

T

G=

T

S

Spontaneous process:

S
univ

> 0

Process at equilibrium:

S
univ

= 0

G
syst

=
-
T

S
univ

We know:

Therefore:

Spontaneous process:

Process at equilibrium:

< 0

= 0

-
T

S
univ

-
T

S
univ

Spontaneity involves

S

H

G =

T

S

Spontaneity is favoured by
increasing

S

Hislargeandnegative
.

T

G allows us to predict whether a process is
spontaneous or not (under constant
temperature and pressure conditions):

G<0

spntaneus

infrwarddirectin

G>0

-
spntaneus

infrward

directin/spntaneusinreverse

directin

G = 0

at
euilibrium

But nothing about rate

Standard free energy (

G
o
)

G
o

=

H
o

T

S
o

Standard states:

Gas

-

1 atm

Solid

-

pure substance

Liquid

-

pure liquid

Solution

-

Concentration = 1M

G
f
o

= 0 kJ/mol for elements in their standard

states

Tabulated data of

G
f
o

can be used to calculate
standard free energy change for a reaction as
follows:

Stoichiometric coefficients

Substance

H

(kJ mol
-
1
)

S
(J K
-
1

mol
-
1
)

G
(kJ mol
-
1
)

Substance

H

(kJ mol
-
1
)

S
(J K
-
1

mol
-
1
)

G

(kJ mol
-
1
)

AgS

0

+42.6

0

I
2
(s)

0

+116.1

0

AgCl(s)

127.1

+96.2

-
109.8

I
2
(g)

+62.4

+260.6

+19.4

Al(s)

0

28.32

0

MgO(s)

-
601.5

+27.0

-
569.2

AlCl
3
(s)

-
704.2

+110.7

-
628.8

MnO
2
(s)

-
520.0

+53.1

-
465.2

Al
2
O
3
(s)

-
1669.8

+51.0

-
1576.5

N
2
(g)

0

+191.5

0

Br
2
(

)

0

+152.2

0

N
2
O
4
(g)

+9.3

+304.2

+97.8

BrF
3
(g)

-
255.6

+292.4

-
229.5

Na(s)

0

+51.3

0

C(g)

+716.7

+158.0

+671.3

NaF(s)

-
569.0

+51.3

-
546.3

C(graphite)

0

+5.8

0

NaCl(s)

-
411.1

+72.4

-
384.3

C(diamond)

+1.9

+2.4

+2.9

NaBr(s)

-
361.1

+87.2

-
349.1

CO(g)

-
110.5

+197.6

-
137.2

NaI(s)

-
287.8

+98.5

-
282.4

CO
2
(g)

-
393.5

+213.7

-
394.4

NaOH(s)

-
425.6

+64.5

-
379.5

CH
4
(g)

-
74.5

+186.1

-
50.8

NH
3
(g)

-
46.2

+192.7

-
16.4

C
3
H
8
(g)

-
103.8

+269.9

-
23.4

N
2
H
4
(

)

+50.6

+121.2

+149.2

Ca(s)

0

41.4

0

NO(g)

+90.3

+210.6

+86.6

CaO(s)

-
635.1

+38.1

-
603.5

NO
2
(g)

+33.2

+240.0

+51.3

CaCO
3
(s)(calcite)

-
1206.9

+92.9

-
1128.8

HNO
3
(

)

-
174.1

+155.6

-
80.8

Cl
2
(g)

0

+223.0

0

O
2
(g)

0

+205.0

0

Cu(s)

0

+33.2

0

O
3
(g)

+142.7

+238.8

+163.2

F
2
(g)

0

+202.7

0

P(s)(white)

0

+41.1

0

Fe(s)

0

+27.3

0

P
4
O
10
(s)

-
3010.0

+231.0

-
2724.0

Fe
2
O
3
(s)(hematite)

-
824.2

+87.4

-
742.2

PCl
3
(g)

-
287.0

+311.7

-
267.8

H(g)

+218.0

+114.6

+203.3

PCl
5
(g)

-
374.9

+364.5

-
305.0

H
2
(g)

0

+130.6

0

PbO
2
(s)

-
277.4

+68.6

-
217.4

HCl(g)

-
92.3

+186.8

-
95.3

S(s)(orthorhombic)

0

+32.0

0

HF(g)

-
271.1

+173.8

-
273.2

H
2
S(g)

-
20.6

+205.6

-
33.4

HI(g)

+26.4

+206.5

+1.6

SiO
2
(s)(quartz)

-
910.7

+41.5

-
856.3

HBr(g)

-
36.4

+198.6

-
53.5

SiCl
4
(

)

-
687.0

+239.7

-
619.9

HCN(g)

+135.1

+201.7

+124.7

SO
2
(g)

-
296.8

+248.1

-
300.2

H
2
O(g)

-
241.8

+188.7

-
228.6

SO
3
(g)

-
395.7

+256.6

-
371.1

H
2
O(

)

-
285.8

+70.0

-
237.2

Zn(s)

0

+41.6

0

H
2
O
2
(

)

-
187.8

+109.6

-
120.4

ZnO(s)

-
350.5

+43.6

-
320.5

Hg(

)

0

+75.9

0

EXAMPLE

The combustion of propane gas occurs as
follows:

C
3
H
8
(g) + 5O
2
(g)

3CO
2
(g) + 4H
2
O(l)

Using thermodynamic data for

G
o
, calculate
the standard free energy change for the
reaction at 298 K.

G
f
o
/kJ.mol
-
1

C
3
H
8
(g)

-
23.47

CO
2
(g)

-
394.4

H
2
O(g)

-
228.57

H
2
O(l)

-
237.13

C
3
H
8
(g) + 5O
2
(g)

3CO
2
(g) + 4H
2
O(l)

G
f
o
/kJ.mol
-
1

C
3
H
8
(g)

-
23.47

CO
2
(g)

-
394.4

H
2
O(g)

-
228.57

H
2
O(l)

-
237.13

G
o

= [3(
-
394.4) + 4(
-
237.13)]

[(
-
23.47)

5(0)]

G
o

=
-
2108 kJ

Free Energy and Temperature

How is change in free energy affected by
change in temperature?

G=

H

T

S

H

S

-

G=

-

+

+

+

+

+

+

-

-

-

-

-

-

-

+

at all temp

-

at all temp

-

at high temp

+

at low temp

+

at high temp

-

at low temp

Note
:

For a spontaneous process the maximum
useful work that can be done by the system:

w
max

=

G

“free energy” = energy available to do work

EXAMPLE

The combustion of propane gas occurs as
follows at 298K:

C
3
H
8
(g) + 5O
2
(g)

3CO
2
(g) + 4H
2
O(l)

H
o

=
-
2220 kJ.mol
-
1

a)
Without using thermodynamic data tables,

predict whether

G
o
, for this reaction is

more or less negative than

H
o
.

b)
Given that

S
o

=
-
374.46 J.K
-
1
.mol
-
1

at 298 K

for the above reaction, calculate

G
o
. Was

C
3
H
8
(g) + 5O
2
(g)

3䍏
2
(g) + 4H
2
O(l)

H
o

=
-
2220 kJ.mol
-
1

a)

Without using thermodynamic data tables,

predict whether

G
o
, for this reaction is more or

less negative than

H
o
.

G
o

=

H
o

T

S
o

-
ve

-
ve

6 moles gas

3⁭lesgas

T

S
o

> 0

H
o

T

S
o

will be less negative than

H
o

i.e.

G
o

will be less negative than

H
o

C
3
H
8
(g) + 5O
2
(g)

3䍏
2
(g) + 4H
2
O(l)

H
o

=
-
2220 kJ.mol
-
1

b)
Given that

S
o

=
-
374.46 J.K
-
1
.mol
-
1

at 298 K for the

above reaction, calculate

G
o
.

Was

G
o

=

H
o

T

S
o

G
o

=
(
-
2220 kJ)

(298 K)(
-
374.46
x10
-
3

kJ
.K
-
1
.mol
-
1
)

G
o

=
-
2108 kJ.mol
-
1

Predictinwascrrect.

EXAMPLE (TUT no. 5a)

At what temperature is the reaction below
spontaneous?

AI
2
O
3
(s) + 2Fe(s)

2AI(s)+Fe
2
O
3
(s)

H
o

= 851.5 kJ;

S
o

= 38.5 J K
-
1

At what temperature is the reaction below
spontaneous?

AI
2
O
3
(s) + 2Fe(s)

2AI(s)+⁆e
2
O
3
(s)

H
o

= 851.5 kJ;

S
o

= 38.5 J K
-
1

G
o

=

H
o

T

S
o

Assume

Hand

Sdntvarythat

0 = (851.5 kJ)

T(38.5x10
-
3

kJ.K
-
1
)

T = 22117 K

equilibrium

Set

G=0

ateuilibrium

Fr
spntaneus

reactin:

G<0

T>22117K

Free Energy and the equilibrium constant

Recall:

G

= Change in Gibb’s free energy under standard
conditions.

G

can⁢e⁣alculatedfrm⁴abulatedvalues.

BUTmst⁲eactinsdnt⁯ccur⁵nderstandard
cnditins.

Calculate

䜠undernn
-
standardcnditins:

Q = reaction quotient

R = gas constant = 8.314 J.K
-
1
.mol
-
1

Under standard conditions:

(1 M, 1 atm)

Q = 1

††
lnQ=0††

††

䜠=

G
o

At equilibrium
:

䜠=‰††and††⁑‽⁋
eq

If

G
o

< 0

††
lnK
eq

> 0

††
K
eq

> 1

i.e. the more negative

G
o
, the larger K etc.

G
o

< 0


K
eq

> 1

G
o

> 0


K
eq

< 1

G
o

= 0


K
eq

= 1

Also

EXAMPLE

Calculate K for the following reaction at 25
o
C:

2H
2
O(
l
) 2H
2
(
g
) + O
2
(
g
)

G
f
o
/kJ.mol
-
1

H
2
O(g)

-
228.57

H
2
O(l)

-
237.13

G
o

= [2(0) + (0)]

[2(
-
237.13)]

G
o

= 474.26 kJ.mol
-
1

474.26x10
3

J.mol
-
1

=
-
(
8.314 J.K
-
1
.mol
-
1
)(298 K) lnK

lnK =
-
191.4

K = 7.36x10
-
84