Chapter16-Thermodyna.. - cllenes-chem-corner

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Oct 27, 2013 (4 years and 16 days ago)

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Chapter 16


Thermodynamics

Entropy, Free Energy, and
Equilibrium

Enthalpy, Entropy and
Spontaneous Processes



Although an exothermic process (
-
∆H
) favors the spontaneity
of a chemical reaction, it doesn’t guarantee that it happens.




The melting of ice is an endothermic process yet it occurs
spontaneously at certain temperatures.




The other thermodynamic quantity that affects the spontaneity
of a process is the change in entropy (
∆S
). A positive entropy
favors a spontaneous process.




Entropy is a measure of the randomness of a system. The more
random the system, the higher the entropy.




Ice melts spontaneously because of the transition from a highly
ordered state of a crystal to a less ordered state of a liquid.

Spontaneous Processes



A spontaneous process once started proceeds without
any external influence.




An example would be a match once lit continues to
burn even out any outside influence.




A low activation energy favors a spontaneous process.

Activation

Energy

Energy
Released

Reactant

Product



The reverse of a spontaneous reaction is always non
-
spontaneous.




Some spontaneous reactions occur extremely slowly.

Problems

16.1 Which of the following processes are spontaneous? Which are
non
-
spontaneous?

(a) Diffusion of perfume molecules from one side of the


room to the other.

(b) Heat flow from a cold object to a hot object.

(c) Decomposition of rust (Fe
2
O
3
.H
2
O) to iron metal,


oxygen, and water.

(d) Decomposition of solid CaCO
3

to solid
CaO

and



gaseous CO
2

at 25
o
C and 1
atm

pressure (
K
p

= 1.4 x 10
-
23
)


16.2 Predict the sign of

S in the system for each of the following
processes:

(a)
H
2
O(g)


H
2
O(l)

(b)
I
2
(g)


2I(g)

(c)
CaCO
3
(s)


CaO
(s) + CO
2
(g)

(d)
Ag
+
(
aq
) + Br
-
(
aq
)


AgBr
(s)

Entropy and Probability



The tendency of a process to increase in entropy is due to the
fact that a more disordered state is more
probable
and
consequently more achievable.




Austrian physicist Ludwig Boltzmann quantified the relationship
between probability and entropy with the equation:

S =
k

ln W


where S is the entropy of the system, W is the number of possible
ways a system can be arranged, and k is the Boltzmann constant
equal to the gas constant divided by Avogadro’s number (R/N
A
).




For gases expanding to a larger volume at constant temperature
and given the initial and final volume ∆S is equal to:



V
final

∆S = nR ln
----------


V
initial

Problem

16.4 Which state has the higher entropy? Explain
in terms of probability.


(a)
A perfectly ordered crystal of solid nitrous
oxide or a disordered crystal in which the
molecules are oriented randomly.


(b)

Quartz glass or quartz crystal.


(c)
1 mol of N
2

gas at STP or 1 mol of N
2

gas at
273 K in a volume of 11.2 L.


(d)
1 mol of N
2

gas at STP or 1 mol of N
2

gas at
273 K and 0.25 atm.

Entropy and Temperature



The higher the temperature the greater the kinetic energy
of particles, the higher the degree of randomness, the
higher the entropy.




The 3
rd

law of thermodynamics states that the entropy of
a perfectly ordered crystalline solid at absolute 0 (0
o
K) is
zero.




As temperature of a solid rises, entropy steadily increases
as the vibrational energy can be distributed in many more
ways.




At the melting point, entropy jumps vertically since the
randomness in liquids are much higher than solids.




The same changes occur going from liquid to gaseous
state.

Standard Molar Entropies and Standard
Entropies of Reaction



Standard Molar Entropy is defined as:

The entropy of 1 mole of pure substance at a pressure
of 1 atm and at a specified temperature usually 25
o
C.




The standard molar entropy is used as basis for the
calculation of the standard entropy of reactions.


Problem

16.5 Calculate the standard entropy at 25
o
C for the
decomposition of calcium carbonate:


CaCO
3
(s)


CaO(s) + CO
2
(g)

Entropy and the 2
nd

Law of Thermodynamics



The 2
nd

law of thermodynamics states that in any spontaneous
process the total entropy of the system and its surrounding
always

increases.




Though the 1
st

law of thermodynamics tells us which direction of
a chemical reaction is favored energy wise, it doesn’t tell us
whether the reaction is spontaneous. The 2
nd

law does. It states
that the total entropy change is equal to the change in entropy of
the system and the change of entropy of the surroundings

∆S
total

= ∆S
system

+ ∆S
surroundings




For a spontaneous process,
∆S
total

is
+
, for a non
-
spontaneous it’s
-
, and for a process in equilibrium
it’s
equal to
0
.




The change in entropy of the surroundings is proportional to
-
∆H
but is inversely proportional to the absolute temperature.



∆H


∆S
surroundings

=
-

-----


T

Problem

16.6 By determining the sign of ∆S
total
, show
whether the decomposition of CaCO
3

is
spontaneous under the standard conditions at
25
o
C.


CaCO
3
(s)


CaO(s) + CO
2
(g)




Gibb’s Free Energy



To calculate for the thermodynamic properties of a system, the
∆S
surroundings
have to be determined which can be complicated.




A way around this is by using the Gibb’s free energy (
G
):
G = H
-

TS
.
Since
H

and
S

are state functions,
G

IS

also a state function.
So the change in free energy is:

G =

H
-

T


S


If

G

< 0, the reaction is spontaneous.

If

G

> 0, the reaction is non
-
spontaneous.

If

G

= 0, the reaction is in equilibrium.


H


S


G

Results

Examples

-

+

-

Spontaneous at all temperatures.

2NO
2
(g)



N
2
(g)

+ 2O
2
(g)

-

-

-

or +

Spontaneous at low

temperatures,

Non
-
spontaneous at high temperatures

N
2
(g) + 3
H
2
(g)


2NH
3
(g)



+

-

+

Non
-
spontaneous at all temperatures.

3O
2
(g)


2O
3
(g)

+

+

-

or +

Non
-
spontaneous at low

temperatures,

Spontaneous at high temperatures

2HgO
(s)



2
Hg
(l) +
O
2
(g)



Problem

16.7 Consider the decomposition of N
2
O
4
:


N
2
O
4
(g)


2
NO
2
(g) ∆H
o

= 55.3kJ, ∆S
o

= +175.7 J/K


(a)
Is the reaction spontaneous under the standard
conditions at 25
o
C?


(b)
Estimate the temperature at which the reaction
becomes spontaneous.


16.9 What are the signs (+,
-
, or 0) of ∆H, ∆S, ∆G for
the following spontaneous reaction of A atoms (red)
and B atoms (black)?

Standard Free
-
Energy Changes for Reactions



Just like the other thermodynamic quantities, free
energy is temperature, state and concentration
dependent.




The standard conditions are for:

1. Solids
, liquids
, and gases in pure form and at 1 atm.

2. Solutions at 1M concentrations

3. Temperature at 25
o
C.




Just like enthalpy, the value of ∆G depends on the
stoichiometry of the reaction.




In the calculation of the standard free energy change,
the standard enthalpy and standard entropy must be
used.





G
o

=

H
o

-

T


S
o

Problem

16.10 Consider the thermal decomposition of CaCO
3
:


CaCO
3
(s)


CaO
(s) + CO
2
(g)


(a)
Use the data in Appendix B to calculate the standard free
-
energy change
for this reaction at 25
o
C?


(b)
Will a mixture of solid CaCO
3
, solid
CaO
, and gaseous CO
2

at 1
atm

pressure react spontaneously at 25
o
C to produce more
CaO

and CO
2
?


(c)
Assuming that ∆H
o

and ∆S
o

are independent of temperature, estimate the
temperature at which the reaction becomes spontaneous.


16.11 Consider the following endothermic reaction of AB
2

molecules:






(a)
What is the sign (+,
-
, or 0) of ∆S
o

for the reaction?

(b)
Is the reaction more likely to be spontaneous at high or low temperatures?

Standard Free Energies of Formation



The standard free energies of formation is the change in free
energy when 1 mole of substance is formed from its elements in
their standard states.




Calculations with
∆G
o
f

is similar to
∆H
o
f

. The most stable form
of the element has a
∆G
o
f

=
0
.




Substances with a
∆G
o
f

< 0
or (
-
) are stable and have little
tendency to decompose. Conversely substances with a
∆G
o
f

> 0
or
(+) are unstable and may spontaneously decompose.




The
∆G
o

of substances can be calculated from given
∆G
o
f

using
the equation:

∆G
o


= ∆G
o
f

products

-

∆G
o
f

reactants




A high (+)
∆G
o

indicates that the substance can not be easily
synthesized at standard temperatures and pressures.

Problem

16.12 (a) Using values of
∆G
o
f

in Appendix B,
calculate the standard free energy change for the
reaction of calcium carbide (CaC
2
) with water. Might
this reaction be used for the synthesis of acetylene
(HC≡CH, or C
2
H
2
)?


CaC
2
(s) + 2H
2
O(l)


C
2
H
2
(g) + Ca(OH)
2
(s)


(b) Is it possible to synthesize acetylene from solid
graphite and gaseous H
2

at 25
o
C and 1 atm pressure?

Free
-
Energy Changes and Composition of
the Reaction Mixture



The free energy at non
-
standard conditions depends on the
composition of the reaction mixture.




The following equation is used for the calculation of
∆G

at non
-
standard pressures and concentrations:

∆G


= ∆G
o


+ RT ln
Q


where
∆G


is the free energy at non
-
standard conditions,
∆G
o


is
the standard free energy,
R

is the gas constant,
T

is the
temperature in
o
K, and
Q

is the concentration quotient Q
C

or Q
P
.




For the reaction: A(g) + B(g) ↔
C(g) + D(g)







[C][D] P
C

P
D



Q
C

=
---------

Q
P

=
---------


[A][B]
P
A

P
B







Problems

16.13 Calculate ∆G

for the formation of ethylene (C
2
H
4
)
from carbon and hydrogen at 25
o
C when the partial
pressures are 100 atm H
2

and 0.10 atm C
2
H
4
.


2C(s) + 2H
2
(g)


C
2
H
4
(g) ∆G
o

= 68.1 kJ


Is the reaction spontaneous in the forward or reverse
direction?

Free
-
Energy and Chemical Equilibrium



The value of the expression
RT ln Q
determines whether a
reaction is spontaneous or not.




A large Q (much more products than reactants) means that RT
ln Q is a very large (+) value and no matter what the value of

G
o
,

G will be (+) so the forward reaction will be non
-
spontaneous.




At equilibrium Q = K and

G = 0 so the free energy chemical
equilibrium equation becomes:

0 = ∆G
o

+ RT ln K
or
∆G
o

=
-
RT ln K

where K can be K
c

or K
p

∆G
o

ln

K

K

∆G
o

< 0

ln

K > 0

K > 1

Mainly products

∆G
o

> 0

ln

K < 0

K < 1

Mainly reactants

∆G
o

= 0

ln

K = 0

K = 1

Products and reactants have

comparable amounts

Problems

16.15 Given the data in Appendix B, calculate K
p

at
25
o
C for the reaction:


CaCO
3
(s) ↔ CaO(s) + CO
2
(g)


16.16 Use data in Appendix B to calculate the vapor
pressure of water at 25
o
C.


16.17 At 25
o
C, K
w

for the dissociation of water is
1.0 x 10
-
14
. Calculate ∆G
o

for the reaction:


2H
2
O(l) ↔ H
3
O
+
(aq) + OH
-
(aq).


Table of Thermodynamic Properties

Table of Thermodynamic Properties