Second Law of Thermodynamics

flinkexistenceMechanics

Oct 27, 2013 (3 years and 7 months ago)

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Second Law of Thermodynamics

Law
of Disorder


the disorder (or entropy) of a system tends to
increase

ENTROPY (S)


Entropy is a measure of disorder



Low entropy (S) = low disorder


High entropy (S) = greater disorder


Operates at the level of atoms and molecules



hot metal block tends to cool



gas spreads out as much as possible

Factors affecting Entropy

A. Entropy increase as matter moves from a solid to a liquid to a gas

Increasing Entropy

B. Entropy increases when a substance is divided into parts

Increasing Entropy

C. Entropy tends to increase in reactions in which the number of
molecules increases

Increasing Entropy

D. Entropy increase with an increase in temperature

Entropy Changes in the System
(∆
S
sys
)

a
A

+
b
B


c
C

+
d
D

D
S
0

rxn

d
S
0
(D)

c
S
0
(C)

=

[

+

]

-

b
S
0
(B)

a
S
0
(A)

[

+

]

D
S
0

rxn

n
S
0
(products
)

=

S

m
S
0
(reactants
)

S

-

The
standard entropy of reaction

(∆ S
0
) is the entropy change for a
reaction carried out at 1
atm

and 25
0
C.

rxn

What is the standard entropy change for the following reaction at 25
0
C?
2CO
(
g
)

+ O
2

(
g
)

2CO
2

(
g
)

S
0
(CO) = 197.9 J/
K
•mol

S
0
(O
2
) = 205.0 J/
K
•mol

S
0
(CO
2
) = 213.6 J/
K
•mol

D
S
0

rxn

= 2 x S
0
(CO
2
)


[2 x S
0
(CO) + S
0

(O
2
)]

D
S
0

rxn

= 427.2


[395.8 + 205.0] =
-
173.6 J/
K
•mol


Entropy Changes in the System
(∆
S
sys
)

When gases are produced (or consumed)



If a reaction produces more gas molecules than
it consumes,

S
0

> 0.


If the total number of gas molecules diminishes,

S
0

< 0.


If there is no net change in the total number of
gas molecules, then

S
0

may be positive or
negative BUT

S
0

w
ill be a small number.

What is the sign of the entropy change for the following reaction? 2Zn
(
s
)

+ O
2

(
g
)


2ZnO
(
s
)

The total number of gas molecules goes down,

S

is negative.

D
S
univ

=
D
S
sys

+
D
S
surr

> 0

Spontaneous process:

D
S
univ

=
D
S
sys

+
D
S
surr

= 0

Equilibrium process:

Gibbs Free Energy

For a constant
-
temperature process:

D
G =
D
H
sys

-
T
D
S
sys

Gibbs free energy
(G)

D
G < 0
The reaction is spontaneous in the forward direction.

D
G > 0

reaction is spontaneous in the reverse
direction.

The reaction is
non
-
spontaneous
as
written. The

D
G = 0
The reaction is at equilibrium
.

D
G

=
D
H

-

T
D
S

a
A

+
b
B


c
C

+
d
D

D
G
0

rxn

d
D
G
0

(D)

f

c
D
G
0

(C)

f

=

[

+

]

-

b
D
G
0

(B)

f

a
D
G
0

(A)

f

[

+

]

D
G
0

rxn

n
D
G
0

(products)

f

=

S

m
D
G
0

(reactants)

f

S

-

The
standard free
-
energy of reaction
(

G
0

)

is the free
-
energy change
for a reaction when it occurs under standard
-
state conditions.

rxn

Standard free energy of formation
(∆ G
0
)

i
s the free
-
energy change that occurs when
1 mole

of the compound is formed from its
elements in their standard states.

f

D
G
0

of any element in its stable form is
zero.

f

f

2C
6
H
6

(
l
)

+ 15O
2

(
g
)


12CO
2

(
g
)

+ 6H
2
O
(
l
)

D
G
0

rxn

n
D
G
0

(products)

f

=

S

m
D
G
0

(reactants)

f

S

-

What is the standard free
-
energy change for the
following reaction at 25
0
C?

D
G
0

rxn

6
D
G
0

(H
2
O)

f

12
D
G
0

(CO
2
)

f

=

[

+

]

-

2
D
G
0

(C
6
H
6
)

f

[

]

D
G
0

rxn

=

[ 12x

394.4 + 6x

237.2 ]


[ 2x124.5 ] =
-
6405 kJ

Is the reaction spontaneous at 25
0
C?

D
G
0

=
-
6405 kJ

< 0

spontaneous

Recap: Signs of Thermodynamic Values

Negative

Positive

Enthalpy
(
Δ
H)

Exothermic

Endothermic

Entropy (
Δ
S)


Less disorder

More
disorder

Gibbs Free
Energy (
Δ
G)


Spontaneous

Not
spontaneous

Gibbs Free Energy and Chemical Equilibrium

D
G

=
D
G
0

+
RT
ln
Q

R

is the gas constant (8.314 J/
K
•mol
)

T

is the absolute temperature (K)

Q

is the reaction quotient

At Equilibrium

D
G

= 0

Q

=
K

0 =
D
G
0

+
RT
ln
K

D
G
0

=
-

RT
ln
K

D
G
0

=
-

RT
ln
K