Chapter 19. Thermodynamics:. Entropy, Free Energy and the Direction of Chemical Reactions. Thermodynamics: Entropy, Free Energy, and the Direction of Chemical Reactions. The Second Law of Thermodynamics: Predicting Spontaneous Change. Calculating the Change in Entropy of a Reaction.
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Entropy, Free Energy, and the Direction of Chemical Reactions
The Second Law of Thermodynamics:
Predicting Spontaneous Change
Calculating the Change in Entropy of a Reaction
Entropy, Free Energy, and Work
Free Energy, Equilibrium, and Reaction Direction
Limitations of the First Law of Thermodynamics
Euniverse= Esystem + Esurroundings
DEsystem = DEsurroundings
DEsystem + DEsurroundings = 0 = DEuniverse
The total energymass of the universe is constant.
However, this does not tell us anything about the direction of change in the universe.
.
Ba(OH)2 8H2O(s) + 2NH4NO3(s) Ba2+(aq) + 2NO3(aq) + 2NH3(aq) + 10H2O(l)
Figure20.1
A spontaneous endothermic chemical reaction.
DH0rxn = +62.3 kJ
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 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.
1877 Ludwig Boltzman
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.
DSuniverse = DSsystem + DSsurroundings > 0
This is the second law of thermodynamics.
Figure 20.4
The third law of thermodynamics.
A perfect crystal has zero entropy at a temperature of absolute zero.
Ssystem = 0 at 0 K
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.
solution
Figure 20.6
The entropy change accompanying the dissolution of a salt.
pure solid
pure liquid
Ethanol
Water
Figure 20.7
The small increase in entropy when ethanol dissolves in water.
Choose the member with the higher entropy in each of the following pairs, and justify your choice [assume constant temperature, except in part (e)]:
PLAN:
In general less ordered systems have higher entropy than ordered systems and entropy increases with an increase in temperature.
Sample Problem 20.1:
Predicting Relative Entropy Values
(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)
SOLUTION:
(a) 1mol of SO3(g)  more atoms
(d) 1mol of KBr(aq)  solution > solid
(b) 1mol of CO2(g)  gas > solid
(e) 230C  higher temperature
(c) 3mol of O2(g)  larger #mols
(f) CCl4  larger with more orbitals/e
Ssurroundings = 
T
Entropy Changes in the System
S0rxn  the entropy change that occurs when all reactants
and products are in their standard states.
S0rxn = S0products  S0reactants
The change in entropy of the surroundings is directly
related to an opposite change in the heat of the system
and inversely related to the temperature at which the
heat is transferred.
Calculate DS0rxn for the combustion of 1mol of propane at 250C.
C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l)
PLAN:
Use summation equations. It is obvious that entropy is being lost because the reaction goes from 6 mols of gas to 3 mols of gas.
SOLUTION:
Find standard entropy values in the Appendix or other table.
Sample Problem 20.2:
Calculating the Standard Entropy of Reaction, DS0rxn
DS = [(3 mol)(S0 CO2) + (4 mol)(S0 H2O)]  [(1 mol)(S0 C3H8) + (5 mol)(S0 O2)]
DS = [(3 mol)(213.7J/mol*K) + (4 mol)(69.9J/mol*K)]  [(1 mol)(269.9J/mol*K) + (5 mol)(205.0J/mol*K)]
Unrealistic – entropy for gaseous water = 188.7
DS =  374 J/K
At 298K, the formation of ammonia has a negative DS0sys;
N2(g) + 3H2(g) 2NH3(g) DS0sys = 197 J/K
PLAN:
DS0universe must be > 0 in order for this reaction to be spontaneous, so DS0surroundings must be > 197 J/K. To find DS0surr, first find DHsys; DHsys = DHrxn which can be calculated using DH0f values from tables. DS0universe = DS0surr + DS0sys.
Sample Problem 20.3:
Determining Reaction Spontaneity
Calculate DH0rxn, and state whether the reaction occurs spontaneously at this temperature.
SOLUTION:
DH0rxn = [(2 mol)(DH0fNH3)]  [(1 mol)(DH0fN2) + (3 mol)(DH0fH2)]
DH0rxn = 91.8 kJ
DS0surr = DH0sys/T =
(91.8x103J/298K)
= 308 J/K
DS0universe = DS0surr + DS0sys
= 308 J/K + (197 J/K) = 111 J/K
DS0universe > 0 so the reaction is spontaneous.
G, the change in the free energy of a system, is a measure of the spontaneity of the process and of the useful energy available from it.
DG0system = DH0system  TDS0system
G < 0 for a spontaneous process
G > 0 for a nonspontaneous process
G = 0 for a process at equilibrium
DG0rxn = S mDG0products  SnDG0reactants
Components of DS0universe for spontaneous reactions
Figure 20.10
exothermic
endothermic
system becomes more disordered
exothermic
system becomes more disordered
system becomes more ordered
+


+

+
+
+
+

+ or 


+
+ or 
Table 20.1 Reaction Spontaneity and the Signs of DH0, DS0, and DG0
DH0
DS0
TDS0
DG0
Description
Spontaneous at all T
Nonspontaneous at all T
Spontaneous at higher T;
nonspontaneous at lower T
Spontaneous at lower T;
nonspontaneous at higher T
Potassium chlorate, a common oxidizing agent in fireworks and matchheads, undergoes a solidstate disproportionation reaction when heated.
Note that the oxidation number of Cl in the reactant is higher in one of the products and lower in the other (disproportionation).
D
4KClO3(s) 3KClO4(s) + KCl(s)
PLAN:
Use Appendix B values for thermodynamic entities; place them into the Gibbs Free Energy equation and solve.
Sample Problem 20.4:
Calculating DG0 from Enthalpy and Entropy Values
+5
+7
1
Use DH0f and S0 values to calculate DG0sys (DG0rxn) at 250C for this reaction.
SOLUTION:
DH0rxn= SmDH0products  SnDH0reactants
DH0rxn= (3 mol)(432.8 kJ/mol) + (1 mol)(436.7 kJ/mol) 
(4 mol)(397.7 kJ/mol)
DH0rxn= 144 kJ
Calculating DG0 from Enthalpy and Entropy Values
continued
DS0rxn= SmDS0products  SnDS0reactants
DS0rxn= (3 mol)(151 J/mol*K) + (1 mol)(82.6 J/mol*K) 
(4 mol)(143.1 J/mol*K)
DS0rxn= 36.8 J/K
DG0rxn= DH0rxn  T DS0rxn
DG0rxn= 144 kJ  (298K)(36.8 J/K)(kJ/103 J)
DG0rxn= 133 kJ
Sample Problem 20.6:
Determining the Effect of Temperature on DG0
PROBLEM:
An important reaction in the production of sulfuric acid is the oxidation of SO2(g) to SO3(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?
PLAN:
The sign of DG0 tells us whether the reaction is spontaneous and the signs of DH0 and DS0 will be indicative of the T effect. Use the Gibbs free energy equation for part (b).
SOLUTION:
(a) The reaction is spontaneous at 250C because DG0 is (). Since DH0 is () but DS0 is also (), DG0 will become less spontaneous as the temperature increases.
Determining the Effect of Temperature on DG0
continued
(b) DG0rxn= DH0rxn  T DS0rxn
DG0rxn= 198.4kJ  (1173K)(187.9J/mol*K)(kJ/103J)
DG0rxn= 22.0 kJ; the reaction will be nonspontaneous at 900.0C
DG and the Work a System Can Do
For a spontaneous process, DG is the maximum work obtainablefrom the system as the process takes place: DG = workmax
For a nonspontaneous process, DG is the minimum work that must be done to the system as the process takes place: DG = workmin
An example
work = pΔv
4KClO3(s) 3KClO4(s) + KCl(s)
PLAN:
Use the DG summation equation.
SOLUTION:
DG0rxn= SmDG0products  SnDG0reactants
Sample Problem 20.5:
Calculating DG0rxn from DG0f Values
PROBLEM:
Use DG0f values to calculate DGrxn for the reaction in Sample Problem 20.4:
DG0rxn= (3mol)(303.2kJ/mol) + (1mol)(409.2kJ/mol) 
(4mol)(296.3kJ/mol)
DG0rxn= 134kJ
The coupling of a nonspontaneous reaction to the hydrolysis of ATP.
The cycling of metabolic free enery through ATP
Figure B20.4
Why is ATP a highenergy molecule?
Figure B20.5
Free Energy, Equilibrium and Reaction Direction
DG = RT ln Q/K = RT lnQ  RT lnK
Under standard conditions (1M concentrations, 1atm for gases), Q = 1 and ln Q = 0 so
DG0 =  RT lnK
Essentially no forward reaction; reverse reaction goes to completion
Forward and reverse reactions proceed to same extent
FORWARD REACTION
REVERSE REACTION
Forward reaction goes to completion; essentially no reverse reaction
Table 20.2 The Relationship Between DG0 and K at 250C
DG0(kJ)
K
Significance
200
9x1036
100
3x1018
50
2x109
10
2x102
1
7x101
0
1
1
1.5
10
5x101
50
6x108
100
3x1017
200
1x1035
Sample Problem 20.7:
Calculating DG at Nonstandard Conditions
PROBLEM:
The oxidation of SO2, which we considered in Sample Problem 20.6
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.
PLAN:
Use the equations and conditions found on slide .
(141.6kJ/mol)(103J/kJ)
At 298, the exponent is DG0/RT
= 
(8.314J/mol*K)(298K)
(8.314J/mol*K)(973K)
Sample Problem 20.7:
Calculating DG at Nonstandard Conditions
continued (2 of 3)
SOLUTION:
(a) Calculating K at the two temperatures:
DG0 = RTlnK so
= 57.2
= e57.2
= 7x1024
At 973, the exponent is DG0/RT
= 1.50
= e1.50
= 4.5
(0.100)2
=
(pSO2)2(pO2)
(0.500)2(0.0100)
Sample Problem 20.7:
Calculating DG at Nonstandard Conditions
continued (3 of 3)
(b) The value of Q =
= 4.00
Since Q is < K at both temperatures the reaction will shift right; for 298K there will be a dramatic shift while at 973K the shift will be slight.
(c) The nonstandard DG is calculated using DG = DG0 + RTlnQ
DG298 = 141.6kJ/mol + (8.314J/mol*K)(kJ/103J)(298K)(ln4.00)
DG298 = 138.2kJ/mol
DG973 = 12.12kJ/mol + (8.314J/mol*K)(kJ/103J)(973K)(ln4.00)
DG973 = 0.9kJ/mol