Unit 5 Part 1: Thermodynamics

draweryaleΜηχανική

27 Οκτ 2013 (πριν από 4 χρόνια και 2 μήνες)

83 εμφανίσεις

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
adhesives

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


睡ter at room temperature




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


捡n be 捡l捵lated for
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
捡n be u獥d to determine the 獰ontaneity of a
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)