# Unit 5 Part 1: Thermodynamics

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27 Οκτ 2013 (πριν από 4 χρόνια και 8 μήνες)

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Unit 5

Part 1: Thermodynamics

Entropy and the Second Law of
Thermodynamics

Gibbs Free Energy

Free Energy and Equilibrium Constants

Entropy and the 2nd Law of
Thermodynamics

Industrial chemists are responsible for
designing cost
-
effective manufacturing
processes.

2 NH
3

(g) + CO
2

(g) NH
2
CONH
2

(aq) + H
2
O (l)

urea

Commercial uses of urea:

nitrogen fertilizer for plants

used to manufacture certain plastics and

Entropy and the 2nd Law of
Thermodynamics

Some of the questions a chemist must
consider:

Does the reaction need to be heated?

How much product will be present at
equilibrium?

Does the reaction naturally proceed in this
direction?

Entropy and the 2nd Law of
Thermodynamics

In order to determine if a reaction proceeds
naturally in the direction written, we need to
know if it is
spontaneous.

Capable of proceeding in the direction
written without needing to be driven by an
outside source of energy

Entropy and the 2nd Law of
Thermodynamics

Examples of spontaneous processes:

an egg breaking when dropped

a ball rolling down a hill

ice

Examples of non
-
spontaneous processes:

a ball rolling up a hill

water

i捥 at room temperature

Entropy and the 2nd Law of
Thermodynamics

Examples of spontaneous reactions:

CH
4
(g) + 2 O
2

(g)

2

(g) + 2 H
2
O (g)

2 N (g)

N
2

(g)

Examples of non
-
spontaneous reactions:

2 H
2
O
(g)

2 H
2

(g)

+ O
2

(g)

O
2
(g)

2 传(g)

Entropy and the 2nd Law of
Thermodynamics

Many
but not all

spontaneous processes are
exothermic.

Enthalpy (
D
H) alone 捡nnot be u獥d to predi捴
whether or not a reaction is spontaneous.

To predict whether a reaction is spontaneous,
we need to use the
second law of
thermodynamics

and a thermodynamic quantity
called
entropy.

Entropy and the 2nd Law of
Thermodynamics

Entropy (S):

a thermodynamic quantity related to the
disorder or randomness of a system.

The more disordered or random the system
is, the larger its entropy is.

A state function

not path dependent

Entropy and the 2nd Law of
Thermodynamics

Every chemical has an entropy associated with it
that depends on its physical state, temperature,
and pressure:

H
2
O (l)

69.91 J/mol
.
K at 25
o
C/1 atm

H
2
O (g)

188.83 J/mol
.
K at 100
o
C/1 atm

Appendix C:

table of thermodynamic properties including S

Entropy and the 2nd Law of
Thermodynamics

The
entropy change

(
D

any process:

D
S = S
final

-

S
initial

D
S = S
products

-

S
reactants

Sign conventions for
D

D
S = po獩ti癥

more di獯rdered

D
S = negati癥

le獳 di獯rdered

Entropy and the 2nd Law of
Thermodynamics

Example:

The following process occurs at 0
o
C
and 1 atm pressure. Does the system become
more ordered or more disordered?

Entropy and the 2nd Law of
Thermodynamics

The sign of
D
S 捡n be predi捴ed:

In general, any change that increases the
overall disorder or randomness will result in
a positive value for
D

In general, the overall entropy increases when:

a molecule (or anything else) is broken into
two or more smaller molecules

there is an increase in the number of moles
of a gas

a solid changes to a liquid or gas

a liquid changes to a gas

Entropy and the 2nd Law of
Thermodynamics

Example:

Without doing any calculations, predict
whether
D
S 睩ll be po獩si癥 or negati癥.

Breaking an egg

N
2
(g)

+ 3 H
2

(g)

2 NH
3

(g)

2NH
3
(g)

+ CO
2
(g)

NH
2
CONH
2

(aq)

+ H
2
O
(l)

Entropy and the 2nd Law of
Thermodynamics

The entropy change

(D
S) for a rea捴ion or
pro捥獳 捡n be 捡l捵lated u獩ng the follo睩ng
equation:

D
S
o

=
S
n S
o
products

-

S

m S
o
reactants

where S
o

= the

standard molar entropy

Note: This is similar to the method used to
calculate
D
H
o

for a reaction!

Entropy and the 2nd Law of
Thermodynamics

Standard molar entropy (S
o
) :

the entropy value for one mole of a chemical
species in its standard state

1 atm pressure

1 M (for those in solution)

NOTE:

Unlike
D
H
o
f
, the standard molar
entropy of a pure element is NOT zero.

Entropy and the 2nd Law of
Thermodynamics

Example:

Predict whether the entropy will
increase or decrease for the following reaction.
Calculate
D
S
o
.

C
6
H
12
O
6

(s) 2 C
2
H
5
OH (l) + 2 CO
2

(g)

Entropy and the 2nd Law of
Thermodynamics

The
Second Law of Thermodynamics

can be
used to predict whether a reaction will
occur spontaneously.

The total entropy of a system and its
surroundings always increases for a
spontaneous process.

How does the change in entropy relate to
the spontaneity of a chemical reaction or
process?

Gibbs Free Energy

Simply looking at the sign of
D
S for a 捨emi捡l
reaction or process does not tell you if the
reaction is spontaneous.

Spontaneous reactions involve an overall
increase in the entropy of the
universe
.

Reactions that have a large, negative
D
H tend to
be spontaneous:

Gibbs Free Energy

The
Gibbs free energy (G)

is used to relate
both the enthalpy change and the entropy
change of a reaction to its spontaneity.

G = H
-

TS

where G = Gibbs free energy (“free energy”)

H = enthalpy

T = temperature (K)

S = entropy

Gibbs Free Energy

Free energy is a state function.

The
change in free energy (
D

of a 獹獴em

process or reaction.

For a process occurring at constant
temperature:

D
G =
D

-

T
D

Gibbs Free Energy

For a reaction occurring at
constant temperature
and pressure,

the sign of
D
G 捡n be u獥d to
determine if a reaction is spontaneous in the
direction written:

D
G = negati癥

reaction is spontaneous in the forward
direction

D
G = zero

reaction is at equilibrium

Gibbs Free Energy

The sign of
D
G 捡n be u獥d to determine if a
reaction is spontaneous in the direction written
(cont):

D
G = po獩ti癥

reaction is not spontaneous in the direction
written

work must be supplied by the surroundings
to make the reaction occur in the direction
written

reaction is spontaneous in the
reverse

direction

Gibbs Free Energy

Example:

Using the definition of
D
G, 捡l捵late
the
D
G for the follo睩ng rea捴ion at 35
o
C:

2 H
+

(aq)

+ S
2
-

(aq)

H
2
S

(g)

D
H =
-
61.9 kJ

D
S = + 183.S J⽋

Gibbs Free Energy

The
standard free energy of formation

(
D
G
o
f
)

has been tabulated for many different
substances. (see Appendix C)

the change in free energy associated with the
formation of 1 mole of a substance from its
elements under standard conditions

pure solid

pure liquid

gas at 1 atm pressure

solution with 1 M concentration

Gibbs Free Energy

There is not a standard temperature for
determining
D
G
o
f
.

25
o
C is often used for tables of data

values can be calculated at other
temperatures as well

D
G
o
f

for an element in its standard state is
zero.

Gibbs Free Energy

The
standard free energy change for a
chemical process

can be calculated using the
following expression:

D
G
o

=
S

D
G
o
f

(products)

-

S

D
G
o
f

(reactants)

Note: This is similar to the way we
calculated
D
H
o

and
D
S
o

Gibbs Free Energy

Example:

Calculate
D
G
o

for the following
reaction using the standard free energies of
formation.

2 KClO
3

(s)

2 KCl (猩 + 3 O
2

(g)