# Introduction to Thermodynamics

Mécanique

27 oct. 2013 (il y a 5 années et 2 mois)

101 vue(s)

Introduction to Thermodynamics

I.
Conservation of Energy

A.
First Law of Thermodynamics = energy cannot be created or destroyed, only
converted between different forms

1.
Example: CH
4
(g) + 2O
2
(g) CO
2
(g) + 2H
2
O + energy

a)
Reaction gives off energy as heat

b)
Potential energy stored in chemical bonds is lowered

c)
Total energy is unchanged

2.
Uses and Shortcomings

a)
Lets us keep track of energy flow in processes

b)
Does not tell us if or why a given process occurs

c)
Does not tell us direction of a chemical reaction

II.
Entropy

A.
The “Heat Tax”

1.
Conversion of energy between forms is inefficient:

2.
Usually, some energy is lost as heat

3.
The fewer energy conversion, the better

B.
Direction of Processes and Reactions

1)
Examples:

a)
Ball at the top of a hill Ball at the bottom of a hill

b)
Steel + H
2
O + O
2

Rust

c)
Gas in one part of a container Gas filling a container

d)
Ice at 5
o
C Water at 5
o
C

2)
What is in common?

a)
Exothermic? Not ice melting or gas expanding

b)
Increased Disorder = Increased Entropy = +
D
S

3)
Entropy = S = driving force of spontaneous reactions = disorder or random

a)
Probability (likelihood): there are many ways for objects/molecules to
be disordered, but only a few to be ordered

b)
Nature proceeds towards the most likely state

= state with greatest number of energetically

equivalent arrangements

4)
Expansion of a Gas

a)
Possible arrangements of 4 gas molecules in a 2
-
bulb system

b)
Microstates = possible configurations of a particular arrangement

c)
Entropy selects most likely arrangement = 2 molecules in each bulb

Microstates

ABCD

---

ABC

D

ABD

C

ACD

B

BCD

A

AB

CD

CD

AB

AC

BD

BD

AC

BC

BC

Order

1 microstate

Ordered

4 microstates

Somewhat

Disordered

6 microstates

Fully disordered

B

5)
Probabilities

Probability of all molecules in left bulb

Number of Molecules

Probability

1

½

2

½ x ½ = ¼

3

½ x ½ x ½ = 1/8

5

1/2
5

= 1/32

10

1/2
10

= 1/1024

n

1/2
n

6.022 x 10
23

1 x 10
-
23

6)
Positional Entropy = entropy depending on configuration in space

a)
Changes in state depend on positional entropy

b)
S
solid

< S
liquid

<< S
gas

c)

Larger volume allows many more available positions for particles

d)

Dissolving a solid provides more volume for particles to occupy

e)

Example
: What has highest positional entropy

i)
1 mole solid CO
2

or 1 mole of gaseous CO
2
?

ii)
1 mole N
2

at 1 atm or 1 mole of N
2

at 0.01 atm?

f)
Example
: Predict the sign of the entropy change for

i)
Dissolving solid sugar into water

ii)
Iodine vapor condensing to crystals on a surface

7)

Second Law of Thermodynamics

= in any spontaneous process, there is
always an increase in the entropy of the universe

a)
Energy is conserved = constant

b)
Entropy is always increasing

c)

D
S
universe

=
D
S
system

+
D
S
surroundings

d)
For a given process: if
D
S
universe

= + the process is spontaneous

if
D
S
universe

=
-

the process is not spontaneous

e)
Life = constant battle against entropy

i)

Large molecules are assembled from smaller ones

ii)
Organizing a cell is
D
S
system

=
-

the process is not spontaneous

iii)
Fortunately, it is
D
S
universe

that must be positive in a process

iii)
D
S
surroundings

= large + for life to occur

III.
Temperature and Spontaneity

A.
Change in state: 1 mol = 18 ml H
2
O(l) 1 mol = 31 L H
2
O(g)

1)
D
S
surrounding

depends on flow of heat into or out of the system

a)
Heat increases the motion (randomness) of particles

b)
Exothermic reactions release heat to surroundings
D
S
surrounding

= +

c)
Endothermic reactions absorb heat from surroundings
D
S
surrounding

=
-

d)
Vaporization of water is endothermic
D
S
surrounding

=
-

2)

D
S
universe

=
D
S
system

+
D
S
surrounding

= (+) + (
-
) = +/
-

?

a)
Depends on the temperature

b)
If T > 100
o
C,
D
S
universe

= +

If T < 100
o
C,
D
S
universe

=
-

B.
Temperature Effects

1)
D
S
surroundings

depends on heat flow

a)
Exothermic reactions usually favors spontaneity

b)
Spontaneity usually lowers the energy of the starting material as it
becomes product

c)
The difference of these energies = heat released to surroundings

2)
Importance of Exothermicity of
D
S
universe

depends on Temperature

a)
Adding heat to hot surroundings has little effect

b)
Adding heat to cold surroundings has a large effect

c)
Heat transfer is more important at low temperatures

3)
In Summary

a)
Sign of
D
S
surr

depends on direction of heat transfer

b)
Magnitude of
D
S
surr

depends on T

c)

d)
The (
-
) is there because
D
H is for the system, which is opposite of
D
H of the surroundings

4)
Example
: Find
D
S
surr

at 25
o
C

a)
Sb
2
S
3
(s) + 3Fe(s) 2Sb(s) + 3FeS(s)
D
H =
-
125 kJ/mol

b)
Sb
4
O
6
(s) + 6C(s) 4Sb(s) + 6CO(g)
D
H = +778 kJ/mol

IV.
Free Energy

A.
Free Energy = G = H

TS

1)
D
G
process

=
D
H

T
D
S

2)
Divide by

T

3)

4)
A process is spontaneous if
D
G =
-

5)
Chemists use
D
G rather than
D
S because we only need to know system

B.
Example
: Predicting Spontaneity using
D
G

1)

H
2
O(s) H
2
O(l)
D
H
o

= 6030 J/mol,
D
S
o

= 22.1 J/K mol

T

T

D
H
o

D
S
o

D
S
surr

D
S
univ

T
D
S
o

D
G
o

o
C

K

J/mol

J/K

mol

J/K

mol

J/K

mol

J/mol

J/mol

-
10

263

6030

22.1

-
22.9

-
0.8

5810

+220

0

273

6030

22.1

-
22.1

0.0

6030

0

10

283

6030

22.1

-
21.3

0.8

6250

-
220

2)
Classifying Processes/Reactions based on
D
H and
D
S

3)

Example
: At what T is Br
2
(l) Br
2
(g) spontaneous (1 atm) given that
D
H = 31.0 kJ/mol and
D
S = 93.0 J/Kmol

a)
Spontaneous when
D
G =
-

b)
Set
D
G = 0 and solve for T

c)
When T > 333K, T
D
S >
D
H and
D
G =
-

(Entropy controlled)

d)
When T < 333K, T
D
S <
D
H and
D
G = + (Enthalpy controlled)

e)
333K is the boiling point of Br
2
(l)