chapter_18_entropy_free_energy_and_equilibrium.ppt

chapter_18_entropy_free_energy_and_equilibrium.ppt

1. Entropy, Free Energy,
and Equilibrium
Chapter 18
Copyright
 © The McGraw-Hill Companies, Inc.  Permission required for reproduction or display.

2. Spontaneous Physical and Chemical Processes
• A waterfall runs downhill
• A lump of sugar dissolves in a cup of coffee
• At 1 atm, water freezes below 0
0
C and ice melts above 0
0
C
• Heat flows from a hotter object to a colder object
• A gas expands in an evacuated bulb
• Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
18.2

3. spontaneous
nonspontaneous
18.2

4. Does a decrease in enthalpy mean a reaction proceeds
spontaneously?
CH
4 (g) + 2O
2 (g) CO
2 (g) + 2H
2O (l) H
0
= -890.4 kJ
H
+
(aq) + OH
-
(aq) H
2
O (l) H
0
= -56.2 kJ
H
2O (s) H
2O (l) H
0
= 6.01 kJ
NH
4
NO
3
(s) NH
4
+
(aq) + NO
3
-
(aq) H
0
= 25 kJ
H
2O
Spontaneous reactions
18.2

5. Entropy (S) is a measure of the randomness or disorder of a
system.
order
SdisorderS
S = S
f
- S
i
If the change from initial to final results in an increase in randomness
S
f > S
i S > 0
For any substance, the solid state is more ordered than the
liquid state and the liquid state is more ordered than gas state
S
solid
< S
liquid
<< S
gas
H
2
O (s) H
2
O (l) S > 0
18.3

6. W = 1
W = 4
W = 6
W = number of microstates
S = k ln W
S = S
f - S
i
S = k ln
W
f
W
i
W
f
> W
i
then S > 0
W
f < W
i then S < 0
Entropy
18.3

7. Processes that
lead to an
increase in
entropy (S > 0)
18.2

8. How does the entropy of a system change for each of the
following processes?
(a) Condensing water vapor
Randomness decreases Entropy decreases (S < 0)
(b) Forming sucrose crystals from a supersaturated solution
Randomness decreases Entropy decreases (S < 0)
(c) Heating hydrogen gas from 60
0
C to 80
0
C
Randomness increases Entropy increases (S > 0)
(d) Subliming dry ice
Randomness increases Entropy increases (S > 0)
18.3

9. Entropy
State functions are properties that are determined by the state
of the system, regardless of how that condition was achieved.
Potential energy of hiker 1 and hiker 2
is the same even though they took
different paths.
energy, enthalpy, pressure, volume, temperature, entropy
18.3

10. First Law of Thermodynamics
Energy can be converted from one form to another but
energy cannot be created or destroyed.
Second Law of Thermodynamics
The entropy of the universe increases in a spontaneous
process and remains unchanged in an equilibrium process.
S
univ = S
sys + S
surr > 0Spontaneous process:
S
univ
= S
sys
+ S
surr
= 0Equilibrium process:
18.4

11. Entropy Changes in the System (S
sys
)
aA + bB cC + dD
S
0
rxn dS
0
(D)cS
0
(C)=[ + ]- bS
0
(B)aS
0
(A)[ + ]
S
0
rxnnS
0
(products)= mS
0
(reactants)-
The standard entropy of reaction (S
0
) is the entropy
change for a reaction carried out at 1 atm and 25
0
C.
rxn
18.4
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
S
0
rxn= 2 x S
0
(CO
2
) – [2 x S
0
(CO) + S
0
(O
2
)]
S
0
rxn= 427.2 – [395.8 + 205.0] = -173.6 J/K•mol

12. Entropy Changes in the System (S
sys
)
18.4
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
will 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.

13. Entropy Changes in the Surroundings (S
surr
)
Exothermic Process
S
surr
> 0
Endothermic Process
S
surr
< 0
18.4

14. Third Law of Thermodynamics
The entropy of a perfect crystalline substance is zero at the
absolute zero of temperature.
18.3
S = k ln W
W = 1
S = 0

15. S
univ = S
sys + S
surr > 0Spontaneous process:
S
univ
= S
sys
+ S
surr
= 0Equilibrium process:
Gibbs Free Energy
For a constant-temperature process:
G = H
sys
-TS
sys
Gibbs free
energy (G)
G < 0 The reaction is spontaneous in the forward direction.
G > 0 The reaction is nonspontaneous as written. The
reaction is spontaneous in the reverse direction.
G = 0 The reaction is at equilibrium.
18.5

16. 18.5
aA + bB cC + dD
G
0
rxn
dG
0
(D)
f
cG
0
(C)
f=[ + ]- bG
0
(B)
f
aG
0
(A)
f
[ + ]
G
0
rxn
nG
0
(products)
f= mG
0
(reactants)
f
-
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
) is the free-energy
change that occurs when 1 mole
of the compound is formed from its
elements in their standard states.
f
G
0
of any element in its stable
form is zero.
f

17. 2C
6
H
6
(l) + 15O
2
(g) 12CO
2
(g) + 6H
2
O (l)
G
0
rxn
nG
0
(products)
f= mG
0
(reactants)
f
-
What is the standard free-energy change for the following
reaction at 25
0
C?
G
0
rxn 6G
0
(H
2O)
f
12G
0
(CO
2)
f=[ + ]-2G
0
(C
6H
6)
f
[ ]
G
0
rxn=[ 12x–394.4 + 6x–237.2 ] – [ 2x124.5 ] = -6405 kJ
Is the reaction spontaneous at 25
0
C?
G
0
= -6405 kJ< 0
spontaneous
18.5

18. G = H - TS
18.5

19. CaCO
3 (s) CaO (s) + CO
2 (g)
H
0
= 177.8 kJ
S
0
= 160.5 J/K
G
0
= H
0
– TS
0
At 25
0
C, G
0
= 130.0 kJ
G
0
= 0 at 835
0
C
18.5
Temperature and Spontaneity of Chemical Reactions
Equilibrium Pressure of CO
2

20. Gibbs Free Energy and Phase Transitions
H
2
O (l) H
2
O (g)
G
0
= 0= H
0
– TS
0
S =
T
H
=
40.79 kJ
373 K
= 109 J/K
18.5

21. Efficiency = X 100%
T
h - T
c
T
c
Chemistry In Action: The Efficiency of Heat Engines
A Simple Heat Engine

22. Gibbs Free Energy and Chemical Equilibrium
G = G
0
+ RT lnQ
R is the gas constant (8.314 J/K•mol)
T is the absolute temperature (K)
Q is the reaction quotient
At Equilibrium
G = 0 Q = K
0 = G
0
+ RT lnK
G
0
=  RT lnK
18.6

23. 18.6
Free Energy Versus Extent of Reaction
G
0
< 0 G
0
> 0

24. G
0
=  RT lnK
18.6

25. 18.7
ATP + H
2
O + Alanine + Glycine ADP + H
3
PO
4
+ Alanylglycine
Alanine + Glycine Alanylglycine
G
0
= +29 kJ
G
0
= -2 kJ
K < 1
K > 1

26. 18.7
The Structure of ATP and ADP in Ionized Forms

27. High Entropy Low Entropy
TS = H - G
Chemistry In Action: The Thermodynamics of a Rubber Band