Thermodynamics: Enthalpy, Entropy & Gibbs Free Energy

technicianlibrarianMechanics

Oct 27, 2013 (3 years and 9 months ago)

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Chapter 15

(not much on




Thermodynamics:
Enthalpy, Entropy
& Gibbs Free
Energy





Thermo
2

Thermodynamics:

thermo

= heat (energy)







dynamics

= movement, motion


Some thermodynamic terms chemists use:

System:

the portion of the universe that we are
considering


open system:



energy & matter can transfer


closed system:


energy transfers only


isolated system:


no

transfers

Surroundings:

everything else besides the system

Isothermal:

a system that is kept at a
constant
temperature

by adding or subtracting heat from the
surroundings.

Heat Capacity:

the amount of heat energy required to
raise the

temperature of a certain amount of material by
1°C (or 1 K).


Specific Heat Capacity:

1 g by 1°C


Molar Heat Capacity:

1 mole by 1°C

Thermo
3

Calorie:

the amount of heat required to raise the
temperature of 1g of water by 1°C.


specific heat of water = 1 cal
/g °C

1 calorie = 4.18 joules


Specific Heats and Molar Heat Capacities

Substance

Specific Heat (J/°C

g)

Molar Heat (J/°C

mol)

Al

0.90

24.3

Cu

0.38

24.4

Fe

0.45

25.1

CaCO
3

0.84

83.8

Ethanol

2.43

112.0

Water

4.18

75.3

Air

1.00

~ 29

important to: engineers



chemists

EXAMPLE:

How many joules of energy are needed to raise
the temperature of an iron nail (7.0 g) from 25°C to 125°C?
The specific heat of iron is 0.45 J/°C




䡥H琠敮敲杹

=

獰散s晩f⁨慴

m慳s


T
)


䡥H琠
敮敲杹

=

〮0㔠䨯5C

g

㜮〠7

㄰グ0
⤠)㴠″=㔠5

Note that

T

捡渠扥nº䌠C爠K,b畴⁎佔ºF.坨敮畳琠
T

楳i扥楮朠畳敤e
楮⁡i獣s敮瑩晩c景牭畬愠tw楬l畳畡汬礠扥b
k
敬v楮
䬩K

Thermo
4

Problem:

How much energy does it take to raise
the body temperature 2.5ºC (a fever of just over
103ºF) for someone who weighs 110 pounds (50 kg).
Assume an average body specific heat capacity of

3 J/ºC.g.












Problem:

What would be more effective at melt
ing
a frozen pipe


hot water or a hair dryer (hot air
gun). Why?

Thermo
5

State Functions

System properties, such as
pressure

(
P
),
volume

(
V
),
and
temperature

(
T
) are called
state functions
.

The value of a state function depends only on the state
of the system and not on the way in which the system
came to be in that state.

A
change

in a state function

describes a difference
between the two states.
It is independent of the
process or pathway
by which the change occurs.


For example, if we heat a block of iron from room
temperature to 100
°
C, it is not important exactly how
we did it. Just on the initial state and the final state
conditions. For example, we could heat it to 150
°
C,
then cool it
to 100
°
C. The path we take is unimportant,
so long as the final temperature is 100
°
C.

Miles per gallon for a car, is NOT a state function. It
depends highly on the path: acceleration, speed, wind,
tire inflation, hills, etc.

Most of the thermodynamic values we will discuss in
this chapter are state functions.

Thermo
6

Energy:

"The capacity to do work
and/or transfer heat"

Forms of Energy:

Kinetic

(E
kinetic

= ½
m
v
2
)

Heat

Light (
&
Electromagnetic)

Electricity

Sound

Potential

Gravitational

Chemical

Nuclear

-

Matter (E =
m
c
2
)



WORK

Thermo
7

First Law of Thermodynamics:

The total amount of energy (and
mass) in the universe is constant.

In any process energy can be
changed from one form to
another; but it can
never be
created nor
destroyed
.




You can't get something for
nothing





Thermo
8

Enthalpy (Heats) of Reaction

The amount of heat released or absorbed by a
chemical reaction at
constant pressure

(as one would
do in a laboratory) is called the
enthalpy

or
heat or
reaction
. We use the symbol

H

瑯⁩湤楣慴攠
敮瑨慬灹
⸠.

卩杮潴慴楯渠S
EXTREMELY IMPORTANT!!
):

+

H

indicates that
heat

is being
absorbed

in the
reaction (it gets
cold
)
endo
thermic



H

indicates that
heat

is being
given off

in the
r
eaction (it gets
hot
)
exo
thermic


Standard

Enthalpy


=


H
° (° is called a “
not
”)






Occurring under
Standard

Conditions:

Pressure

1 atm (760 torr)

Concentration

1.0 M

Temperature

is
not

defined or part of
Standard
Conditions
, but is often measured at 298 K (25°C).


Thermo
9

Standard Enthalpy of Formation
--


The amount of heat absorbed (
endothermic
) or
released (
exothermic
) in a reaction in which
one
mole

of a substance is formed from its
elements

in
th
eir
standard states
, usually at
298 K

(25°C).

Also called
heat of formation
.


=
0

for any
element

in its
standard state

(the
natural elemental form at 1 atm or 1 M) at
298 K
.


EXAMPLES:


Thermo
10


Note that
I

usually will
not

have you calculate

H
f
º on homeworks or tests


so you generally don’t have to worry about normalizing your answer to a
per mole basis.

Hess's Law
--

Adding Reactions

The overall
heat of reaction

(

H
rxn
) is equal to the
sum of the

H
f

(
products
)

minus the sum of the

H
f

(
reactants
)
:


Therefore, by knowing

H
f

of the
reactants

and
products
, we can determine the

H
rxn

for any
reaction that involves these
reactants

and
products
.

Thermo
11

EXAMPLE:

CO
2

is used in certain kinds of fire extinguishers
to put out simple fires. It works by smothering the fire with
"heavier"

CO
2

that replaces oxygen needed to maintain a fire.
CO
2

is not good, however, for more exotic electrical and
chemical fires.



Therefore, Mg will
"burn"

CO
2

!


Thermo
12

You can also add two reactions together to get the

H
rxn


for
another new reaction:


If we have to multiply one (or more) of the reactions by some
constant to get them to add correctly, then we also would have
to multiply

H
rxn

for that reaction by the same amount.



Thermo
13

Chemists use
bomb calorimeters

to measure
Enthalpies of formation or reaction.



In order to use this effectively one must know the
heat capacity

of the bomb (inner part) and water
bath. By measuring the temperature increase of the
water one can calcul
ate the amount of heat given off
during the combustion process.

Thermo
14

Problem:

Calculate

H
rxn

for the following
reactions given the following
values:


(SO
2
,
g
) =

㈹2
kJ/mol




(SO
3
,
g
) =

㌹3
kJ/mol

(H
2
SO
4
,
l
) =

8ㄴ1
kJ/mol


(H
2
SO
4
,
aq
) =

㤰㠠
kJ/mol

(H
2
O,
l
) =

㈸2
kJ/mol



(H
2
S,
g
) =

㈰
kJ/mol


a)

S(
s
) + O
2
(
g
)
SO
2
(
g
)




b)
2SO
2
(
g
) + O
2
(
g
)
2SO
3
(
g
)




c)

SO
3
(
g
) + H
2
O(
l
)

H
2
SO
4
(
l
)




d)

2H
2
S(g) + 3O
2
(
g
)
2SO
2
(
g
) + 2H
2
O(
l
)





Thermo
15

Internal Energy
--


E


The
internal energy
,
E
, represents all the energy
contained within a material. It includes kinetic
energy (heat), intra
-

and intermolecular forces
(bond energies, electrostatic forces, van der Waals),
and any other forms of energy present. As with
enthalpy,
H
, the absolute va
lue can’t (or is extremely
difficult) to define.

What we can track is the change in

E
:


E

㴠=
E
final



E
inital

=
E
products



E
reactants

A key relationship is:


E

㴠=
q

+
w

Where
q

= heat and
w

= work performed on or by
the system. Sign notations:

q

= positive = heat added to system (adds energy)

q

= negative = heat removed (removes energy)

w

= positive = work done on system (adds energy)

w

= negative = work done by system (removes energy)

Thermo
16

The most common type of work involves
pressure/volume changes: e.g., explosion of gasoline
vapors in an internal combustion engine. The
explosion creates a dramatic pressure and volume
increase that pushes the piston and creates work.

If one has a const
ant volume situation, then no
pressure/volume work will be done and
w

= 0.

So under
constant volume conditions
:


E

㴠=
q

The change in internal energy is, therefore, equal to
the amount of heat added or removed from the
material (system).


Thermo
17

Relationship

Between


E



H


E

=

q

(at constant volume and temperature)


H

㴠=

q

(at constant pressure and temperature)

The difference is that volume changes occur for

H

慮搠
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慲攠楮癯汶敤e†漠睥w汹湥敤⁴n⁢攠捯湣敲湥搠慢o畴
癯汵浥⁷潲欠睨敮⁴e敲e⁩猠⁣桡湧攠e渠n桥⁡浯畮琠t映f慳a
灲潤畣敤⁩⁡⁣桥浩慬⁲敡捴楯渮

T
he relationship between

H

慮搠

E

楳⁤敦楮敤Ⱐ
瑨敲敦潲攬⁡猺


H

㴠=

E

+


n
)RT

Where R = gas constant; T = temperature in kelvin,
and:


n

= equivalents (moles) of product gas


equivalents
(moles)
of reactant gas

If

n

= 0, then

H



E


But even when

n

≠ 0, the PV work component is usually
small. See example in textbook (pages 574
-
575).

Thermo
18

Entropy

The final state of a system
is more energetically
favorable
i
f
:


1.

Energy can be dispersed over a greater number
and variety of molecules.

2.

The particles

of the system can be more
dispersed (more disordered).

The dispersal of energy and matter is described by
the thermodynamic state function
entropy
,

S
.


T
he greater the dispersal
of
energy
or matter
in a
system,

the
higher

is its
entropy
.

The

greater the

disorder

(dispersal of
energy and
matter, both in

space and in variety) the
higher the entropy
.

Adding heat to a material increases the
disorder
.


Thermo
19

Unlike

H

敮瑲潰o

捡渠扥⁤敦c湥搠數慣e汹⁢散慵獥l
潦⁴桥h
周楲i⁌w映
T桥牭潤祮慭楣s
:

Third Law of Thermodynamics:

Any
pure crystalline substance

at a
temperature of absolute zero (0.0 K) has
an
entropy

of zero (
S

= 0.0 J/K

浯氩.

Sign notation (
EXTREMELY IMPORTANT!!
):

+

S

indicates that
entropy

is
increasing

in the
reaction or transformation (it's getting
more

disordered

--

mother nature likes)



S

indicates that
entropy

is
decreasing

in the
reaction or transformation (it's getting
less

disordered

{more ordered}
--

mother nature doesn't
like, but it does happ
en)


Thermo
20

Qualitative
"Rules"

About Entropy:

1)

Entropy
increases

as one goes from a solid to a
liquid
, or more dramatically, a
liquid

to a
gas
.


2)

Entropy
increases

if a solid or
liquid

is dissolved
in a solvent.

3)

Entropy
increases

as the
number of particles

(molecules) in a system

increases:

N
2
O
4
(
g
)
2
NO
2
(
g
)



S
°

= 304 J/K

(1 mole)



S
°

= 480 J/K (
2

moles)

Th
ese

first 3 above are
most important

for
evaluating

S
rxn
.


Thermo
21

The rules below are mainly for comparing the
entropy

of individual molecules or materials.

4)

The Entropy of any material
increases

with
increasing temperature

5)

Entropy
increases

as the
mass of a molecule

increases

S
°
(Cl
2
(
g
)) >
S
°
(F
2
(
g
))





S
°

= 165
J/K

mol

S
°

= 158 J/K

mol

6)

Entropy is
higher

for
weakly bonded compounds

than for compounds with
very strong covalent
bonds

S
°
(graphite) >
S
°
(diamond)




S
°

= 5.7 J/K

mol

S
°

= 2.4 J/K

mol


Note that for
individual molecules

(materials) the
higher the entropy, the more likely the molecule will
want to

fall apart


to produce a number of smaller
molecules.

Thermo
22

7)

Entropy
increases

as the
complexity

(# of atoms,
# of heavier atoms, etc.) of a molecule
increases

Entropy of a Series of
Gaseous

Hydrocarbons



S
° = 186 J/K

mol



S
° = 201 J/K

mol



S
° = 220 J/K

mol



S
° = 230 J/K

mol



S
° = 270 J/K

mol


Thermo
23

What are the biggest factors for
evaluating

S
rxn


for
a chemical rxn?

1) phase change 2) change in # of molecules

Problem:

For the following reactions, is the entropy
of the reaction
increasing

or
decreasing
?

a)

Ag
+
(
aq
) + Cl
-
(
aq
)
AgCl(
s
)

b)

H
2
CO
3
(
aq
)

H
2
O + CO
2
(
g
)

c)

Ni(
s
) + 4CO(
g
)

Ni(CO)
4
(
l
)

d)

H
2
O(
s
)

H
2
O(
l
)

e)

graphite


diamond

f)

2Na(
s
) + 2H
2
O


2Na
+
(
aq
) + 2OH
-
(
aq
) + H
2
(g)

g)

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

H
2
O(
l
) + SO(
g
)

h)

2H
2
O(
l
)

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

i)

CO
2
(
g
) + CaO(
s
)

CaCO
3
(
s
)

j)

CaCl
2
(
s
) + 6H
2
O(
l
)

CaCl
2

6H
2
O(
s
)

k)

2NO
2
(
g
)

N
2
O
4
(
g
)

Thermo
24

Just as with
enthalpies
, one can calculate
entropies

of reaction
.


EXAMPLE:



Thermo
25

Spontaneous Processes

A process that takes place without the
net

input of
energy from an external source is said to be
spontaneous

(
not

instantaneous
).

1) Rxn of sodium metal with water:

2Na(
s
) + 2H
2
O


2Na
+
(
aq
) + 2OH
-
(
aq
) + H
2
(g)


2)

Combustion rxns:

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

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

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

2H
2
O(
l
)


3)

Expansion of a gas into a vacuum

x
CO
2
(
g
)

y
CO
2
(
s
) +
z
CO
2
(
g
) (
x

=
y

+
z
)


4)

A salt dissolving into solution:

NH
4
NO
3
(
s
) + H
2
O(
l
)
NH
4
+
(
aq
) + NO
3
-
(
aq
)


Thermo
26

Second Law of Thermodynamics:


In
any
spontaneous

process the
entropy

of the
universe

increases



S
universe

=

S
system

+

S
surroundings


Second Law (variant):

in trying to do
work, you
always

lose energy to the
surroundings.

You can't even break
even!


Neither
entropy

(

S
) or
enthalpy

(

H

慬潮攠捡渠瑥汬⁵猠睨整桥爠愠捨敭楣慬
牥慣瑩潮⁷楬r⁢攠獰潮瑡湥潵猠潲潴s

䅮A潢癩潵o

?
) conclusion is that one
needs to use some combination of the two.

Thermo
27

Gibbs Free Energy

The combination of entropy, temperature
and enthalpy explains whether a reaction is
going to be
spontaneous

or not. The symbol

G

is used to define the
Free Energy

of a
system. Since this was discovered by J.
Willard Gibbs it is also called the
Gibbs
Fre
e Energy.

"Free"

energy refers to the
amount of energy available to do work once
you have paid your price to entropy. Note
that this is not given simply by

H
Ⱐ瑨,⁨敡琠
敮敲杹牥l敡獥搠楮⁡⁲a捴楯c.




=





T



When

G

楳i
negative
, it indicates that a
reaction or process is
spontaneous
. A
positive


G

楮摩捡c敳⁡e
non
-
spontaneous
reaction.

Thermo
28


G

=

H



T

S



Spontaneous = exoergic (energy releasing)

Non
-
spontaneous = endoergic (energy releasing)
Thermo
29

Remember that
entropies

are
given in units of
J/K

浯m

睨楬攠
敮瑨慬灩敳

慮搠
晲敥⁥湥牧楥f

慲攠楮a
歊⽭潬
.

DON'T forget to convert all units to
kJ or J when using both

匠慮搠


楮⁴桥⁳慭攠i畡瑩潮℡



Thermo
30




vs.

G
:

Standard vs. Non
-
Standard Conditions

Remember that the º (“not”) on



indicates that the
numerical value of



楳⁢i獥搠s渠n桥⁲敡捴楯渠n琠
獴慮摡牤⁣潮摩瑩潮s

(
ㄠ䴠獯汵瑩潮⁣潮捥湴牡瑩c測nㄠ慴洠
条猠灲敳s畲e
⤮)⁔敭灥牡瑵牥⁩猠t佔O灡牴映獴f湤n牤r
捯湤楴楯湳n

䅳⁳潯渠n猠s湥⁨n猠s
捯湣敮瑲c瑩t渠n楦晥牥湴⁴桡渠1

M or
1 atm pressure, the º “not” goes away and one has

G
.

Consider the reaction:

Initial:

1 atm 1 atm




1 atm




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

2SO
3
(
g
)





rxn

=

ㄴ1

k䨯浯m

The


rxn

of

ㄴ1

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睩瑨w愠捯湣敮瑲c瑩t渠n映
ㄠ慴1
⸠.周楳⁩T摩捡瑥猠瑨t琠瑨攠
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灲潤畣瑳
獰潮瑡湥潵猩⸠s

䅳⁴桥⁲敡捴慮瑳⁳瑡牴t牥r捴楮cⰠ,潷敶敲Ⱐ瑨敩爠
捯湣敮瑲c瑩t湳⁤散牥慳
攠e

2

twice as fast as
O
2
) and



瑵牮猠楮瑯

G

慮搠扥d潭敳敳猠湥条瑩t攮†

坨敮W

G

㴠〠瑨攠牥慣瑩t渠n慳a牥慣桥搠
敱畩汩扲極b
⸠.
䅬瑨A畧栠h潲⁴桩猠牸測n

2

is probably the limiting
reagent (not enough present to complete the rxn).

Thermo
31

Example:

Calculate


f

for CO
2

at 298 K.

H
f
°
(CO
2
) =

3㤳⁋䨯浯氬m
S
뀠⡏
2
) = 205 J/mol

K,
S
°
(C) = 6 J/mol

K,
S
° (CO
2
) = 213 J/mol

K


C(
graphite
) + O
2
(
g
)

CO
2
(
g
)




f

=


f



T


f



f

=


prod





react



f

= (213 J/mol

K)


⠲〵‫‶⁊⽭潬

K)



f

= 2 J/mol

K)



f

= (

㌹㌠䭊3浯氩m


⠲㤸⁋⤨〮)〲⁋䨯浯m

K)



f

= (

㌹㌠䭊3浯氩m


⠱⁋䨯浯m)



f

=

㌹㐠䭊4浯m


Problem:

Calculate


f

for CO at 298 K.

H
f
°
(CO) =

ㄱ〠䭊0浯氬m
S
뀨°
2
) = 205 J/mol

K,

S
°(C) = 6 J/mol

K,
S
°(CO) = 198 J/mol

K


2C(
graphite
) + O
2
(
g
)

2CO(
g
)

Note change in
units


J to KJ

DANGER!!

Common
mistake!!

Thermo
32

Just as with
enthalpies
and
entropies
, one can
calculate
free energies of reaction
.


EXAMPLE:

Thermo
33

Example:


To make iron, a steel mill takes Fe
2
O
3

(rust
or iron ore) and reacts it with coke (a complex, impure
form of carbon) to make iron and CO
2
. Based on the
data below, this is a

non
-
spontaneous

reaction at room
temperature, but it becomes
spontaneous

at higher
temperatures. Assuming that

H
° and

S
° do not
change much with temperature, calculate the temp
-
erature above which the reaction becomes spontaneous
(i.e.,

G
°
rxn

= 0).


H
°
rxn

= +465 kJ/mol


S
°
rxn

= +552 J/molK (
or

0.552 kJ/molK)


G
°
rxn

= +301 kJ/mol (
at 298 K
)


G
°
rxn

=

H
°
rxn



T

S
°
rxn


as we raise the temperature,

G
뀠wil氠ev敮瑵慬ty⁲慣栠〠慮搠
瑨敮t杯敧慴iv攠☠獰潮瑡湥潵猬t獯整e

G
뀠°‰⁡搠獯sve⁦爠
听⁴攠
瑥t灥牡瑵牥⁡琠w桩捨⁴楳iwi汬⁨慰l敮e

0 =

H
°
rxn



T

S
°
rxn

rearranging to solve for T gives:

T = (

H
°
rxn
)
/

(

S
°
rxn
)

T = (465 kJ/mol)
/
(0.552 kJ/molK)

T = 842 K

(above this temperature


rxn

will be
negative



we will have
a
spontaneous

reaction)

Thermo
34

Problem:

Calculate


rxn

for the following.




f

(SO
2
,
g
) =

㌰3
kJ/mol




f

(SO
3
,
g
) =

㌷3
kJ/mol



f

(H
2
SO
4
,
l
) =

㘹〠
kJ/mol



f

(H
2
SO
4
,
aq
) =

㜴㈠
kJ/mol



f

(H
2
O,
l
) =

㈳2

kJ/mol




f

(H
2
S,
g
) =



kJ/mol


a)

S(
s
) + O
2
(
g
)
SO
2
(
g
)




b)
2SO
2
(
g
) + O
2
(
g
)
2SO
3
(
g
)




c)

SO
3
(
g
) + H
2
O(
l
)

H
2
SO
4
(
l
)




d)

2H
2
S(g) + 3O
2
(
g
)
2SO
2
(
g
) + 2H
2
O(
l
)






Thermo
35

Comparisons of

H
º
rxn

and

G
º
rxn



S(
s
) + O
2
(
g
)
SO
2
(
g
)



H
º
rxn

=

㈹㜠2䨯浯m


G
º
rxn

=

㌰3

k䨯浯m


㉓2
2
(
g
) + O
2
(
g
)
2SO
3
(
g
)



H
º
rxn

=

ㄹ1

k䨯浯m


G
º
rxn

=

ㄴ1

k䨯浯m



3
(
g
) + H
2
O(
l
)

H
2
SO
4
(
l
)



H
º
rxn

=

ㄳ1

k䨯浯m


G
º
rxn

=




k䨯浯m



2
S(g) + 3O
2
(
g
)
2SO
2
(
g
) + 2H
2
O(
l
)



H
º
rxn

=

ㄱ㈶

k䨯浯m


G
º
rxn

=

㄰〶

k䨯浯m


S
º
rxn

=

1ㄠ
䨯J潬

K

Thermo
36

Entropy of Fusion and Vaporization

While the entropy of a substance increases steadily
with increasing temperature, there is a considerable
jump in the entropy at a
phase transition
:


This jump at the melting point is called the
entropy
of fusion
,

S
fusion
, an
d as you might expect, it is
related to the enthalpy (or heat) of fusion,

H
fusion
:


Thermo
37

The jump in entropy at the boiling point is called the
entropy of vaporization
,

S
vaporization
, and it is
related to the
enthalpy of vaporization
,

H
vaporization
:


EXAMPLE:

What is the boiling point of bromine
(Br
2
)?

S
vapor

= 93.2 J/K

浯氠慮搠

H
vapor

= 30.9
kJ/K

浯m⸠




but we want to solve for
T
b
, the boiling point
temperature, so we need to rearrange the
formula:




Note how we have to
convert kJ/mol to J/mol in
order for the units on

H
to be the same as the units
on

S!! To convert kJ to
J we need to multiply the
kJ value by 1000.

Thermo
38

Problem:

Calculate the boiling point for etha
nol
(CH
3
CH
2
OH) from the data in the thermodynamic
tables.

















(literature value = 78.5°C)
Thermo
39


Thermo
40


From “General Chemistry”, 7
th

Ed, by Whitten, Davis, Peck & Stanley. Thomson Brooks/Cole Publisher.