Limitations of the First Law of Thermodynamics
D
E = q + w
E universe
= E system
+ E surroundings
D
E system
= -
D
E surroundings
The total energy-mass of the universe is constant.
However, this does not tell us anything about the direction of change in the universe.

A spontaneous endothermic chemical reaction water
Ba(OH)
2
.
8H
2
O( s ) + 2NH
4
NO
3
( s ) Ba 2+ ( aq ) + 2NO
3
( aq ) + 2NH
3
( aq ) + 10H
2
O( l )
D
H 0 rxn
= +62.3 kJ

The Concept of Entropy (S)
Entropy refers to the state of order.
A change in order is a change in the number of ways of arranging the particles, and it is a key factor in determining the direction of a spontaneous process.
more order less order solid liquid gas more order less order crystal + liquid ions in solution more order less order crystal + crystal gases + ions in solution

The number of ways to arrange a deck of playing cards

1 atm
Spontaneous expansion of a gas stopcock closed evacuated
0.5 atm stopcock opened
0.5 atm

1877 Ludwig Boltzman S = k ln W where S is entropy, W is the number of ways of arranging the components of a system, and k is a constant (the Boltzman constant),
R/N
A
(R = universal gas constant, N
A
= Avogadro’s number.
•A system with relatively few equivalent ways to arrange its components (smaller W) has relatively less disorder and low entropy.
•A system with many equivalent ways to arrange its components
(larger W) has relatively more disorder and high entropy.
D
S universe
=
D
S system
+
D
S surroundings
> 0
This is the second law of thermodynamics.

Random motion in a crystal
The third law of thermodynamics .
A perfect crystal has zero entropy at a temperature of absolute zero.
S system
= 0 at 0 K
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Predicting Relative S 0 Values of a System
1. Temperature changes
S 0 increases as the temperature rises.
2. Physical states and phase changes
S 0 increases as a more ordered phase changes to a less ordered phase.
3. Dissolution of a solid or liquid
S 0 of a dissolved solid or liquid is usually greater than the S 0 of the pure solute. However, the extent depends upon the nature of the solute and solvent.
4. Dissolution of a gas
A gas becomes more ordered when it dissolves in a liquid or solid.
5. Atomic size or molecular complexity
In similar substances, increases in mass relate directly to entropy.
In allotropic substances, increases in complexity (e.g. bond flexibility) relate directly to entropy.

The increase in entropy from solid to liquid to gas
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The entropy change accompanying the dissolution of a salt pure solid
MIX pure liquid solution
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The small increase in entropy when ethanol dissolves in water
Ethanol Water
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Solution of ethanol and water

The large decrease in entropy when a gas dissolves in a liquid
O
2 gas
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

NO
NO
2
Entropy and vibrational motion
N
2
O
4

Sample 1 Predicting Relative Entropy Values
PROBLEM: Choose the member with the higher entropy in each of the following pairs, and justify your choice [assume constant temperature, except in part (e)]:
(a) 1mol of SO
2
( g ) or 1mol of SO
3
( g )
(b) 1mol of CO
2
( s ) or 1mol of CO
2
( g )
(c) 3mol of oxygen gas (O
2
) or 2mol of ozone gas (O
(d) 1mol of KBr( s ) or 1mol of KBr( aq )
3
)
(e) Seawater in midwinter at 2 0 C or in midsummer at 23 0 C
(f) 1mol of CF
4
( g ) or 1mol of CCl
4
( g )

Sample 2 Calculating the Standard Entropy of Reaction,
D
S 0 rxn
PROBLEM: Calculate
D
S 0 rxn for the combustion of 1mol of propane at 25 0 C.
C
3
H
8
( g ) + 5O
2
( g ) 3CO
2
( g ) + 4H
2
O( l )

Components of
D
S 0 universe for spontaneous reactions exothermic endothermic system becomes more disordered exothermic system becomes more disordered system becomes more ordered

Sample 3 Determining Reaction Spontaneity
PROBLEM: At 298K, the formation of ammonia has a negative
D
S 0 sys
;
N
2
( g ) + 3H
2
( g ) 2NH
3
( g )
D
S 0 sys
= -197J/K
Calculate
D
S 0 rxn
, and state whether the reaction occurs spontaneously at this temperature.
D
H 0 f
NH
3
= -45.9 kJ/mol

D
G 0 system
=
D
H 0 system
- T
D
S 0 system
D
G 0 rxn
=
S m
D
G 0 products
-
S n
D
G 0 reactants

Sample 4 Calculating
D
G 0 from Enthalpy and Entropy Values
PROBLEM: Potassium chlorate, one of the common oxidizing agents in explosives, fireworks, and matchheads, undergoes a solid-state redox reaction when heated. In this reaction, note that the oxidation number of Cl in the reactant is higher in one of the products and lower in the other (disproportionation):
+5
4KClO
3
( s
D
+7
) 3KClO
4
-1
( s ) + KCl( s )
Use
D
H 0 f and S 0 values to calculate
D
G 0 sys at 25 0 C for this reaction.

Sample 5 Calculating
D
G 0 rxn from
D
G 0 f
Values
PROBLEM:
Use
D
G 0 f values to calculate
D
G rxn for the reaction:
4KClO
3
( s
D
) 3KClO
4
( s ) + KCl( s )

Sample 6 Determining the Effect of Temperature on
D
G 0
PROBLEM: An important reaction in the production of sulfuric acid is the oxidation of SO
2
( g ) to SO
3
( g ):
2SO
2
( g ) + O
2
( g ) 2SO
3
( g )
At 298K,
D
G 0 = -141.6kJ;
D
H 0 = -198.4kJ; and
D
S 0 = -187.9J/K
(a) Use the data to decide if this reaction is spontaneous at 25 0 C, and predict how
D
G 0 will change with increasing T.
(b) Assuming
D
H 0 and
D
S 0 are constant with increasing T, is the reaction spontaneous at 900.
0 C?

Sample 7 Determining the Effect of Temperature on
D
G 0
PROBLEM: A reaction is nonspontaneous at room temperature but is spontaneous at -40 0 C. What can you say about the signs and relative magnitudes of
D
H 0
D
S 0 and -T
D
S 0

Reaction Spontaneity and the Signs of
D
H 0 ,
D
S 0 , and
D
G 0
D
H 0
D
S 0 -T
D
S 0
D
G 0 Description
-
+
+
-
+
-
+
-
-
+
-
+
-
+
+ or -
+ or -
Spontaneous at all T
Nonspontaneous at all T
Spontaneous at higher T; nonspontaneous at lower T
Spontaneous at lower T; nonspontaneous at higher T

The effect of temperature on reaction spontaneity

D
G and the Work a System Can Do
For a spontaneous process,
D
G is the maximum work obtainable from the system as the process takes place:
D
G = work max
For a nonspontaneous process,
D
G is the maximum work that must be done to the system as the process takes place:
D
G = work max
An example

The coupling of a nonspontaneous reaction to the hydrolysis of ATP.

The cycling of metabolic free enery through ATP

Free Energy, Equilibrium and Reaction Direction
•If Q/K < 1, then ln Q/K < 0; the reaction proceeds to the right ( D
G < 0)
•If Q/K > 1, then ln Q/K > 0; the reaction proceeds to the left ( D
G > 0)
•If Q/K = 1, then ln Q/K = 0; the reaction is at equilibrium ( D
G = 0)
D
G = RT ln Q/K = RT lnQ - RT lnK
Under standard conditions (1M concentrations, 1atm for gases), Q = 1 and lnQ = 0 so
D
G 0 = - RT lnK

The Relationship Between
D
G 0 and K at 25 0 C
D
G 0 (kJ)
200
100
-10
-50
-100
-200
50
10
1
0
-1
K
9x10 -36
3x10 -18
2x10 -9
2x10 -2
7x10 -1
1
1.5
5x10 1
6x10 8
3x10 17
1x10 35
Significance
Essentially no forward reaction; reverse reaction goes to completion
Forward and reverse reactions proceed to same extent
Forward reaction goes to completion; essentially no reverse reaction

Sample Problem 6 Calculating
D
G at Nonstandard Conditions
PROBLEM: The oxidation of SO
2
, which we considered in Sample Problem 6
2SO
2
( g ) + O
2
( g ) 2SO
3
( g ) is too slow at 298K to be useful in the manufacture of sulfuric acid. To overcome this low rate, the process is conducted at an elevated temperature.
(a) Calculate K at 298K and at 973K. (
D
G 0
298
= -141.6kJ/mol of reaction as written using
D
H 0 and
D
S 0 values at 973K.
D
G 0
973
= -12.12kJ/mol of reaction as written.)
(b) In experiments to determine the effect of temperature on reaction spontaneity, two sealed containers are filled with 0.500atm of SO
2
,
0.0100atm of O
2
, and 0.100atm of SO
3 and kept at 25 0 C and at 700.
0 C. In which direction, if any, will the reaction proceed to reach equilibrium at each temperature?
(c) Calculate
D
G for the system in part (b) at each temperature.

Sample Problem 7 Calculating
D
G at Nonstandard Conditions
PROBLEM: At 298 K hypobromous acid (HBrO) dissociates in water with a
K a of 2.3 x 10 -9 .
(a) Calculate
D
G 0
298
(b) Calculate
D
G if [H
3
O+] = 6.0 x 10 -4 M, [BrO-] = 0.1 M and [HBrO]=0.20 M

The relation between free energy and the extent of reaction
D
G 0 < 0
K >1
D
G 0 > 0
K <1