#### Transcript Entropy

```Chapter 20
Thermodynamics:
Entropy, Free Energy and the
Direction of Chemical Reactions
Limitations of the First Law of Thermodynamics
DE = q + w
Euniverse = Esystem + Esurroundings
DEsystem = -DEsurroundings
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)2.8H2O(s) + 2NH4NO3(s)
Ba2+(aq) + 2NO3-(aq) + 2NH3(aq) + 10H2O(l)
DH0rxn = +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
solid
more order
crystal + liquid
more order
crystal + crystal
less order
liquid
gas
less order
ions in solution
less order
gases + ions in solution
The number of ways to arrange a deck of playing cards
Spontaneous expansion of a gas
stopcock
closed
1 atm
evacuated
stopcock
opened
0.5 atm
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/NA (R = universal gas constant, NA = 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.
DSuniverse = DSsystem + DSsurroundings > 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.
Ssystem = 0 at 0 K
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Predicting Relative S0 Values of a System
1. Temperature changes
S0 increases as the temperature rises.
2. Physical states and phase changes
S0 increases as a more ordered phase changes to a less ordered
phase.
3. Dissolution of a solid or liquid
S0 of a dissolved solid or liquid is usually greater than the S0 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
O2 gas
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Entropy and vibrational motion
NO
NO2
N 2 O4
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 SO2(g) or 1mol of SO3(g)
(b) 1mol of CO2(s) or 1mol of CO2(g)
(c) 3mol of oxygen gas (O2) or 2mol of ozone gas (O3)
(d) 1mol of KBr(s) or 1mol of KBr(aq)
(e) Seawater in midwinter at 20C or in midsummer at 230C
(f) 1mol of CF4(g) or 1mol of CCl4(g)
Sample 2
PROBLEM:
Calculating the Standard Entropy of Reaction, DS0rxn
Calculate DS0rxn for the combustion of 1mol of propane at 250C.
C3H8(g) + 5O2(g)
3CO2(g) + 4H2O(l)
Components of DS0universe for spontaneous reactions
exothermic
endothermic
system becomes more disordered
exothermic
system becomes more disordered
system becomes more ordered
Sample 3
PROBLEM:
Determining Reaction Spontaneity
At 298K, the formation of ammonia has a negative DS0sys;
N2(g) + 3H2(g)
2NH3(g)
DS0sys = -197J/K
Calculate DS0rxn, and state whether the reaction occurs
spontaneously at this temperature. DH0fNH3 = -45.9 kJ/mol
DG0system = DH0system - TDS0system
DG0rxn = S mDG0products - S nDG0reactants
Sample 4
PROBLEM:
Calculating DG0 from Enthalpy and Entropy Values
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
+7
-1
D
4KClO3(s)
3KClO4(s) + KCl(s)
Use DH0f and S0 values to calculate DG0sys at 250C for this reaction.
Sample 5
PROBLEM:
Calculating DG0rxn from DG0f Values
Use DG0f values to calculate DGrxn for the reaction:
4KClO3(s)
D
3KClO4(s) + KCl(s)
Sample 6
PROBLEM:
Determining the Effect of Temperature on DG0
An important reaction in the production of sulfuric acid is the
oxidation of SO2(g) to SO3(g):
2SO2(g) + O2(g)
2SO3(g)
At 298K, DG0 = -141.6kJ; DH0 = -198.4kJ; and DS0 = -187.9J/K
(a) Use the data to decide if this reaction is spontaneous at 250C, and predict
how DG0 will change with increasing T.
(b) Assuming DH0 and DS0 are constant with increasing T, is the reaction
spontaneous at 900.0C?
Sample 7
PROBLEM:
Determining the Effect of Temperature on DG0
A reaction is nonspontaneous at room temperature but is
spontaneous at -40 0C. What can you say about the signs and
relative magnitudes of DH0 DS0 and -TDS0
Reaction Spontaneity and the Signs of DH0, DS0, and DG0
DH0
DS0
-TDS0
DG0
-
+
-
-
Spontaneous at all T
+
-
+
+
Nonspontaneous at all T
+
+
-
+ or -
Spontaneous at higher T;
nonspontaneous at lower T
-
-
+
+ or -
Spontaneous at lower T;
nonspontaneous at higher T
Description
The effect of temperature on reaction spontaneity
DG and the Work a System Can Do
For a spontaneous process, DG is the maximum work obtainable from
the system as the process takes place: DG = workmax
For a nonspontaneous process, DG is the maximum work that must be
done to the system as the process takes place: DG = workmax
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 (DG < 0)
•If Q/K > 1, then ln Q/K > 0; the reaction proceeds to the left (DG > 0)
•If Q/K = 1, then ln Q/K = 0; the reaction is at equilibrium (DG = 0)
DG = RT ln Q/K = RT lnQ - RT lnK
Under standard conditions (1M concentrations, 1atm for gases), Q = 1
and lnQ = 0 so
DG0 = - RT lnK
The Relationship Between DG0 and K at 250C
DG0(kJ)
K
100
3x10-18
50
2x10-9
10
2x10-2
1
7x10-1
0
1
-1
1.5
-10
5x101
-50
6x108
-100
3x1017
-200
1x1035
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
REVERSE REACTION
9x10-36
FORWARD REACTION
200
Significance
Sample Problem 6
Calculating DG at Nonstandard Conditions
PROBLEM: The oxidation of SO2, which we considered in Sample Problem 6
2SO2(g) + O2(g)
2SO3(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. (DG0298 = -141.6kJ/mol of reaction as
written using DH0 and DS0 values at 973K. DG0973 = -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 SO2,
0.0100atm of O2, and 0.100atm of SO3 and kept at 250C and at 700.0C. In
which direction, if any, will the reaction proceed to reach equilibrium at each
temperature?
(c) Calculate DG for the system in part (b) at each temperature.
Sample Problem 7
Calculating DG at Nonstandard Conditions
PROBLEM: At 298 K hypobromous acid (HBrO) dissociates in water with a
Ka of 2.3 x 10-9.
(a) Calculate DG0298
(b) Calculate DG if [H3O+] = 6.0 x 10-4M, [BrO-] = 0.1 M and [HBrO]=0.20 M
The relation between free energy and the extent of reaction
DG0 < 0
K >1
DG0 > 0
K <1
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