Download Chapter 19 CHEMICAL THERMODYNAMICS 19.1 SPONTANEOUS

Chapter 19
CHEMICAL THERMODYNAMICS
19.1 SPONTANEOUS PROCESSES
Engineering Chemistry
A spontaneous process is one that proceeds on its own without any outside
assistance.
A spontaneous process occurs in one direction only, and the reverse of any spontaneous process is
always nonspontaneous.
Generally, processes that are spontaneous in one direction are nonspontaneous in the opposite direction.
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For example, Drop an egg above a hard surface.
An egg falling and breaking is spontaneous. The reverse process is nonspontaneous, even though
energy is conserved in both processes.
Another example: Gas expansion into vacuum. See Fig. 19.2
In general, processes that are spontaneous in one direction are nonspontaneous in the opposite
direction.
Experimental conditions, such as temperature and pressure, are often important in determining
whether a process is spontaneous. Consider, for example, ice melting. At atmospheric pressure,
when the temperature of the surroundings is above 0 C, ice melts spontaneously, and the reverse
process— liquid water turning into ice is not spontaneous. However, when the temperature of the
surroundings is below 0 C, the opposite is true—liquid water turns to ice spontaneously, but the
reverse process is not spontaneous.
It is important to realize that the fact that a process is spontaneous does not necessarily mean that it
will occur at an observable rate. A chemical reaction is spontaneous if it occurs on its own accord,
regardless of its speed. A spontaneous reaction can be very fast, as in the case of acid–base
neutralization, or very slow, as in the rusting of iron.
Thermodynamics tells us the direction and extent of a reaction but nothing about the
speed.
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Example 19.1
Predict whether each process is spontaneous as described, spontaneous in the reverse direction,
or in equilibrium:
(a ) Water at 40 C gets hotter when a piece of metal heated to 150 C is added.
(b) Water at room temperature decomposes into H2(g) and O2(g).
(c) Benzene vapor, C6H6 (g), at a pressure of 1 atm condenses to liquid benzene at the normal
boiling point of benzene, 80.1 C .
Solution:
(a) The process is spontaneous. (transfer of heat from the hotter object to the colder one).
(b) Nonspontaneous.
(c) The process is spontaneous in which an equilibrium state is established.
Reversible and Irreversible processes
A reversible process is a specific way in which a system changes its state, and the change occurs in
such a way that the system and surroundings can be restored to their original states by exactly
reversing the change.
An irreversible process is one that cannot simply be reversed to restore the system and its
surroundings to their original states.
A French engineer named Sadi Carnot (1796–1832) concluded that: a reversible change produces
the maximum amount of work that can be done by a system on its surroundings.
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Another example, the expansion of an ideal gas at constant temperature (referred to as an
isothermal process). In the cylinder-piston arrangement of ( FIGURE 19.5, when the
partition ‫ الحاجز‬is removed, the gas expands spontaneously to fill the evacuated space.
Since the gas expands into a vacuum with no external pressure, it does no work on the
surroundings. Thus, for the expansion, w = 0.
We can use the piston to compress the gas back to its original state, but doing so requires
that the surroundings do work on the system, meaning that w>0 for the compression.
In other words, the path that restores the system to its original state requires a different value of w
(and, by the first law, a different value of q) than the path by which the system was first changed.
The fact that the same path can’t be followed to restore the system to its original state indicates that
the process is irreversible.
ENTROPY AND THE SECOND LAW OF THERMODYNAMICS:
Entropy: is associated either with the extent of randomness in a system or with the extent to
which energy is distributed among the various motions of the molecules of the system.
Entropy Change:
The entropy, S, of a system is a state function just like internal energy, E, and enthalpy, H.
As with these other quantities, the value of S is a characteristic of the state of a system.
Thus, the change in entropy,S , in a system depends only on the initial and final states of the
system and not on the path taken from one state to the other:
S = Sfinal –Sinitial
[19.1]
For the special case of an isothermal process, S is equal to the heat that would be transferred if the
process were reversible, qrev, divided by the absolute temperature at which the process occurs:
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S 
qrev
T
(cons tan t T )
[19.2]
between states, not just the reversible one.
S for phase Changes
Consider for example, the melting of ice. At 1 atm pressure, ice and liquid water are in equilibrium
at 0 C . Imagine melting 1 mol
of ice at 0 C, 1 atm to form 1 mol of liquid water at 0 C, 1 atm. We can achieve this
change by adding a certain amount of heat to the system from the surroundings q = Hfusion:
For the melting of ice at 0 C, qrev = Hfusion,Enthalpy of fusion of ice = 6.01 kJ/mol
From Equation 19.2, we can calculate Sfusion for melting 1 mol of ice at 273 K:
S fusion 
q rev H fusion (1 mol)(6.01  10 3 J / mol)


 22.0 J / K
T
T
273 K
Example 19.2
Elemental mercury (Hg) is a silver liquid at room temperature. Its normal freezing point is -38.9 C
and its molar enthalpy of fusion is Hfusion = 2.29 kJ/mol. What is the entropy change of the system
when 50.0 g of Hg(l) freezes at the normal freezing point?
Note that freezing is the reverse of melting, the enthalpy change that accompanies ‫ يصاحب عملية‬the
freezing of 1 mol of Hg is -2.29 kJ/mol.So, - Hfusion = -2.29 kJ/mol = qrev (for 1 mol Hg).
The heat evolved during freezing a quantity of 50.0 g of Hg is (-2290 J/200.59 g/mol)×50 g = -571 J
So, qrev = -571 J
 - 571 J 
q
   2.44 J / K
S system   rev  
T
 234.3 K 
The Second Law of Thermodynamics
The first law of thermodynamics states that energy is conserved in any process.
Entropy, however, is not conserved. For any spontaneous
process, the total change in entropy, which is the sum of the entropy change of the system
plus the entropy change of the surroundings, is greater than zero.
Consider for example, a system of 1 mol of ice (a piece roughly the size of an ice cube) melting in
the palm of your hand, which is part of the surroundings. The process is not reversible because the
system and surroundings are at different temperatures.
We can calculate the entropy change of the system as follows:
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Heat lost = - Heat gained
Q (lost by your hand) = - Q(heat gained by the ice)
Hence, the entropy change of the surroundings is
S sys 
q rev (1 mol )(6.01  10 3 J / mol )

 22.0 J / K
T
273 K
Assuming, the body temperature = 37 C
S Surr 
q rev (1 mol)(  6.01  10 3 J / mol)

  19.4 J / K
T
(273  37) K
Thus, the total entropy change is positive:
Stotal = Ssystem + Ssurr = 22.0 - 19.4 = 2.6 J/K The process is non spontaneous.
If the temperature of the surroundings were not 310 K but rather some temperature infinitesimally
above 273 K, the melting would be reversible instead of irreversible. In that case the entropy change
of the surroundings would equal and would be zero.
Statement of the second law of thermodynamics:
Any irreversible process results in an increase in total entropy, whereas any reversible process
results in no overall change in entropy.
The sum of the entropy of a system plus the entropy of the surroundings is everything
there is, and so we refer to the total entropy change as the entropy change of the
universe, Suniv .
Mathematically, the second law of thermodynamics:
Reversible Process: Suniv = Ssystem + Ssurr = 0
Irreversible Process: Suniv = Ssystem + Ssurr > 0
Another form for the second law of thermodynamics: the entropy of the universe increases in any
spontaneous process.
The second law of thermodynamics tells us the essential character of any spontaneous change it is
always accompanied by an increase in the entropy of the universe.
We can use this criterion to predict whether a given process is spontaneous or not.
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19.3 Molecular interpretation of entropy
Expansion of a Gas at the Molecular Level
It is mentioned before that expansion of a gas into a vacuum is a spontaneous process. it is an
irreversible process and that the entropy of the universe increases during the expansion.
Tracking two of the gas molecules as they move around.
Before the stopcock is opened, both molecules are confined to the left
flask, as shown in Fig. 19.6(a).
After the stopcock is opened, the molecules travel randomly throughout the entire apparatus.
From Figure 19.6(b): There are four possible arrangements for the two molecules once both flasks
are available to them. (Because the molecular motion is random) all four arrangements are equally
likely.
Note that now only one arrangement corresponds to the situation before the stopcock was opened:
both molecules in the left flask.
MICROSTATE
A microstate is a single possible arrangement of the positions and kinetic energies of the gas
molecules when the gas is in a specific thermodynamic state.
we can use the tools of statistics and probability to determine the total number of microstates for the
thermodynamic state. (That is where the statistical part of the name statistical thermodynamics
comes in.) Each thermodynamic state has a characteristic number of microstates
associated with it, and we will use the symbol W for that number.
A microstate is a particular microscopic arrangement of the atoms or molecules of the system that
corresponds to the given state of the system.
The connection between the number of microstates of a system, W, and the entropy
of the system, S, is expressed in a beautifully simple equation developed by Boltzmann
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S = k log W
Or , S = k ln W
19.5
,
Where, S is the entropy, k is Boltzmann’s constant =R/NA = 1.38 ×10-23 (R = 8.314 J mol-1 K-1)and
W is the number of microstates. Thus, entropy(s) is a measure of how many microstates are
associated with a particular macroscopic state.
From Equation 19.5, we see that the entropy change accompanying any process is:
S  k ln W final  k ln Winitial  k ln
W final
Winitial
19.6
Any change in the system that leads to an increase in the number of microstates (Wfinal > Winitial)
leads to a positive value of S : Entropy increases with the number of microstates of the system.
When the entropy increase?
Generally, entropy will increase as the number of microstates of the system increases.
There are two modifications to our ideal-gas sample showing how the entropy changes:
First: Increasing Volume
suppose we increase the volume of the system, which is analogous
to allowing the gas to expand isothermally. A greater volume means a greater
number of positions available to the gas atoms and therefore a greater number of microstates.
The entropy therefore increases as the volume increases.
Second: Increasing volume
suppose we keep the volume fixed but increase the temperature.
An increase in temperature increases the most probable speed of the molecules and also broadens
the distribution of speeds. Hence, the molecules have a greater number of possible kinetic energies,
and the number of microstates increases. Thus, the entropy of the system increases with increasing
temperature.
Molecular motions and Energy:
Any real molecule can undergo ‫ يعاني من‬three kinds of more complex motion.
(i) Translational motion: in which the entire molecule can move in one direction.
The molecules in a gas have more freedom of translational motion than those
in a liquid, which have more freedom of translational motion than the molecules of
a solid.
(ii) Vibrational motion: in which the atoms in the molecule move periodically toward and away
from one another.
(iii)Rotational motion: in which the molecule spins about an axis.
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FIGURE 19.8 shows the vibrational motions and one of the rotational motions possible for the
water molecule. These different forms of motion are ways in which a molecule can store energy,
and we refer to the various forms collectively as the motional energy of the molecule.
Factors affecting the number of possible microstates:
There are four factors which can directly increase the number of microstates of a substance:
1- Temperature.
2- Volume.
3- Number of moles (or molecules).
4- Complexity or the number of atoms per molecule.
The number of microstates possible for a system increases with an increase in
(i) volume,(ii) an increase in temperature, or (iii) an increase in the number of molecules because
any of these changes increases the possible positions and kinetic energies of the molecules making
up the system. (iv) an increase in the complexity of the molecule (No. of atoms), which increases
the available vibrational motions.
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In ice, hydrogen bonding leads to the rigid structure shown in FIGURE 19.9. Each molecule in the
ice is free to vibrate, but its translational and rotational motions are much more restricted than in
liquid water.
Although there are hydrogen bonds in liquid water, the molecules can more readily move
about relative to one another (translation) and tumble around (rotation). During melting, therefore,
the number of possible microstates increases and so does the entropy. In water vapor, the molecules
are essentially independent of one another and have their full range of translational, vibrational, and
rotational motions. Thus, water vapor has an even greater number of possible microstates and
therefore a higher entropy than liquid water or ice.
Entropy change of dissolution processes:
When an ionic solid dissolves in water, a mixture of water and ions replaces the
pure solid and pure water, as shown for KCl in FIGURE 19.10. The ions in the liquid
move in a volume that is larger than the volume in which they were able to move in the
crystal lattice and so possess more motional energy.
The water molecules becomes more ordered than before because they are now confined to the
immediate environment of the ions. Therefore, the dissolving of a salt involves both a disordering
process (the ions become less confined) and an ordering process (some water molecules become
more confined).
The disordering processes are usually dominant, and so the overall effect is an increase in the
randomness of the system.
The same ideas apply to chemical reactions. Consider the reaction between nitric
oxide gas and oxygen gas to form nitrogen dioxide gas:
2 NO(g) + O2(g)  2 NO2(g)
[19.7]
which results in a decrease in the number of molecules three molecules of gaseous reactants form
two molecules of gaseous products ( FIGURE 19.11). The formation of new N-O bonds reduces
the motions of the atoms in the system. The formation of new bonds decreases the number of
degrees of freedom, or forms of motion, available to the atoms. That is, the atoms are less free to
move in random fashion because of the formation of new bonds. The decrease in the number of
molecules and the resultant decrease in motion result in fewer possible microstates and therefore a
decrease in the entropy of the system.
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Sample Exercise 19.3:
Predict whether S is positive or negative for each process, assuming each occurs at constant
temperature:
a) H2O(l)  H2O(g)
b) Ag+(aq) + Cl-(aq) AgCl (s)
c) 4 Fe(s) + 3O2(g) 2 Fe2O (s)
d) N2(g) + O2 (g) 2NO(g)
S is positive
S is negative
S is negative
S  zero
See Practice ex 19.3
Sample Exercise 19.4:
In each pair, choose the system that has greater entropy and explain your choice: (a) 1 mol of
NaCl(s) or 1 mol of HCl(g) at 25C , (b) 2 mol of HCl(g) or 1 mol of HCl(g) at 25C , (c) 1 mol
of HCl(g) or 1 mol of Ar(g) at 298 K.
Answer:
a) 1 mol of HCl(g): the particles in gases are more disordered.
b) 2 mol of HCl(g): the 2-mol sample has twice the number of microstates.
c) 1 mol of HCl(g): more complexity.
The Third Law of Thermodynamics
The entropy of a pure crystalline substance at absolute zero (0 K) is zero: S(0)= 0.
At absolute zero, the individual atoms or molecules in the lattice would be perfectly ordered and as
well defined in position as they could be.
Because none of them would have thermal motion, there is only one possible microstate.
As a result, Equation 19.5 becomes S = K ln W = zero, when W = 1
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As the temperature is increased from absolute zero, the atoms or molecules in the crystal gain
energy in the form of vibrational motion about their lattice positions. This means that the degrees of
freedom and the entropy both increase.
Entropy Changes in Chemical Reactions
According to the third law of thermodynamics, the entropy of a pure crystalline solid at 0 K is zero.
The entropy increases as the temperature of the crystal is increased.
FIGURE 19.13 shows that the entropy of the solid increases steadily with increasing temperature
up to the melting point of the solid. When the solid melts, the atoms or molecules are free to move
about the entire volume of the sample. The added degrees of freedom increase the randomness of
the substance, thereby increasing its entropy. We therefore see a sharp (or an abrupt) increase in
the entropy at the melting point.
At the boiling point of the liquid, another abrupt increase in entropy occurs. We can understand
this increase as resulting from the increased volume available to the atoms or molecules as they
enter the gaseous state. When the gas is heated further, the entropy increases steadily as more
energy is stored in the translational motion of the gas atoms or molecules.
TABLE 19.1 lists the values of for a number of substances at 298 K; Appendix C gives a more
extensive list.
We can make several observations about the values of S in Table 19.1
1. Unlike enthalpies of formation, standard molar entropies of elements at the reference temperature
of 298 K are not zero.
2. The standard molar entropies of gases are greater than those of liquids and solids, consistent with
our interpretation of experimental observations, as represented in Figure 19.13.
3. Standard molar entropies generally increase with increasing molar mass.
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4. Standard molar entropies generally increase with an increasing number of atoms in the formula of
a substance. (See Figure 19.14)
The entropy change in a chemical reaction equals the sum of the entropies of the
products minus the sum of the entropies of the reactants:
S° =  nS° (products)-  mS° (reactants)
[19.8]
As in Equation 5.31, the coefficients n and m are the coefficients in the balanced chemical equation.
See Ex 19.5 and Pr Ex. 19.5
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Energy Changes in the Surroundings
S surroundings  
q sys
T
Because in a constant-pressure process, qsys is simply the enthalpy change for the reaction,H
S surroundings  
H sys
T
19.9
For the reaction in Sample Exercise 19.5,

N2(g) + 3H2(g)
2NH3(g)
S = -198.3 J/K
H=-92.38 kJ
The negative value of H tells us that at 298 K the formation of ammonia from H2(g) and
N2(g) is exothermic. The surroundings absorb the heat given off by the system, which means an
increase in the entropy of the surroundings:
q sys is the enthalpy change for the reaction under standard conditions, H , so the changes in
entropy will be standard entropy changes, S.
S surroundings  
q sys
T

92.38
 0.310kJ / K  310 J / K
298
Notice that the magnitude of the entropy gained by the surroundings is greater than
that lost by the system, calculated as in Sample Exercise 19.5.
The overall entropy change for the reaction is:
S°univ = S°sys + S°surr = -198.3 J/K + 310 J/K = 112 J/K.
So, the reaction given in ex 19.5 moves spontaneously toward formation of NH3(g).
19.5 GIBBS FREE ENERGY
How can we use H and S to predict whether a given reaction occurring at constant
temperature and pressure will be spontaneous?
The means for doing so was first developed by the American mathematician J. Willard Gibbs
(1839–1903).
Gibbs proposed a new state function, called the Gibbs free energy (or just free energy), G,
and defined as:
G = H – TS
[19.10]
the change in the free energy of the system,G , is: G = H – TS
[19.11]
Under standard conditions, this equation becomes: G = H – TS
[19.12]
 H sys
S°univ = S°sys + S°surr = S°sys +  
T





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Multiplying both sides by (–T) gives us:
-TSuniv =Hsys -T Ssys
[19.13]
Comparing Equations 19.11 and 19.13, we see that in a process occurring at constant temperature
and pressure, the free-energy change, G , is equal to -TSuniv. We know that for spontaneous
processes, Suniv is always positive and, therefore, -TSuniv is always negative. Thus, the sign of G
provides us with extremely valuable information
about the spontaneity of processes that occur at constant temperature and pressure.
If both T and P are constant, the relationship between the sign of G and the spontaneity
of a reaction is as follows:
1. If G , the reaction is spontaneous in the forward direction.
2. If G , the reaction is at equilibrium.
3. If G , the reaction in the forward direction is nonspontaneous (work must be
done to make it occur) but the reverse reaction is spontaneous.
See EX 19.6
Standard Free Energy of Formation: (Gf) :
Standard free energies of formation are useful in calculating the standard free energy
change for chemical processes. The procedure is analogous to the calculation of H
(Equation 5.31) and S (Equation 19.8):
G   n G f (products) -  m G f (Reactants )
[19.14]
SAMPLE EXERCISE 19.7
(a) Use data from Appendix C to calculate the standard free-energy change for the reaction:
P4(g) + 6 Cl2 (g)  4 PCl3 (g). run at 298 K.
(b) What is G for the reverse of this reaction?
Solution:
(a) G f (rxn) = 4 G f [PCl3 (g)] - G f [P4(g)] - 6 G f [Cl2(g)]
= (4 mol)(-269.6 kJ/mol) - (1 mol)(24.4 kJ/mol) – 0
= -1102.8 kJ  Spontaneous reaction in the forward direction to form more PCl3.
(b) When we reverse the reaction, we reverse the roles of the reactants and products.
4 PCl3 (g)

P4(g) + 6 Cl2 (g)

we reverse the sign of G f (rxn). Hence, using the result from part (a), we have G f (for the
reverse direction) = + 1102 kJ.
See EX. 19.8
19.6 FREE ENERGY AND TEMPERATURE
To see how
G is affected by temperature, let’s look again at Equation 19.11:
14
G = H – TS
[19.11]
Or,
Notice that we have written the expression for G as a sum of two contributions, an enthalpy
term, (H) , and an entropy term, (– TS) . Because the value of depends directly on the absolute
temperature T, G varies with temperature. We know that the enthalpy term, (H), can be either
positive or negative and that T is positive at all temperatures other than absolute zero. The entropy
term, (– TS), can also be positive or negative. When S is positive, which means the final state
has greater randomness (a greater number of microstates) than the initial state, the term (– TS) is
negative. When S is negative, the term (– TS) is positive.
The sign of G , which tells us whether a process is spontaneous, depends on the signs and
magnitudes of S and H . See TABLE 19.3.
SAMPLE EXERCISE 19.9
The Haber process for the production of ammonia involves the equilibrium
N2(g) + 3 H2(g)
2 NH3(g)
Assume that H and S for this reaction do not change with temperature.
(a) Predict the direction in which G for the reaction changes with increasing temperature.
(b) Calculate G° at 25°C and at 500 °C.
Solution:
(a) We expect S for this reaction to be negative because the number of NH3(g)
is smaller in the products. Because S is negative,the term (-TS) is positive and increases
with increasing temperature. So, G becomes less negative (or more positive) with increasing
temperature.
(b) The calculated values of H and S are H= -92.38 kJ and S = -198.3 J/K
We can calculate the value of G from equation 19.12.
G = -92.38 kJ - (298 K)(-0.1983 kJ/K) = -33.3 kJ
Similarly, at 500 C:
G = -92.38 kJ - (773 K)(-0.1983 kJ/K) = 61 kJ
15
Comment:
Increasing the temperature from 298 K to 773 K changes G from -33.3 kJ to + 61 kJ. (i.e) it
makes the reaction nonspontaneous in the forward direction. (the result agrees with our prediction
in part (a).
See Practice Ex. 19.9
FREE ENERGY AND EQUILIBRIUM CONSTANT
The relationship between G (for a system at equilibrium) and equilibrium constant is given by:
G = G + RT ln Q
[19.19]
Where, Q is the reaction quotient for the reaction mixture of interest.
Under standard conditions, the concentrations of all the reactants and products
are equal to 1. Thus, under standard conditions Q=1, and ln Q = 0, and Equation 19.19
reduces to G = G under standard conditions, as it should.
SAMPLE EXERCISE 19.10
(a) Write the chemical equation that defines the normal boiling point of liquid carbon tetrachloride,
CCl4(l).
(b) What is the value of G for the equilibrium in part (a)?
(c) Use data from Appendix C and Equation 19.12
to estimate the normal boiling point of CCl4.
Answer:
(a) CCl4(l)

CCl4(g)
P = 1 atm
(b) At equilibrium, G = 0 . In any normal boiling-point equilibrium, both liquid and vapor are in
their standard state of pure liquid and vapor at 1 atm (Table 19.2). Consequently, Q=1 , ln Q = 0 ,
and G = G for this process. We conclude that G = 0 for the equilibrium representing the
normal boiling point of any liquid.
(We would also find that G = 0 for the equilibria relevant to normal melting points and normal
sublimation points.
(c)
Relation between G and K
16
We can now use Equation 19.19 to derive the relationship between and the equilibrium constant, K.
At equilibrium, G = 0 and Q = K . Thus, at equilibrium, Equation 19.19 transforms as follows:
By solving Equation 19.20 for K, we obtain an expression that allows us to calculate
K if we know the value of G :
See EX. 19.12 and Pr Ex 19.12
Important sections in chapter 19:
Page 786: definitions, P787: Ex. 19.1,
PP790-793, P794-798: Summary of entropy increase and Ex.19.3, PP800-803.
PP804-816: Table 19.3, examples and equations.
Selected problems: 19.11, 13, 16, 24-27, 31, 35, 41-43, 45, 47, 48-50, 53, 54, 57, 59, 61, 64, 66,
69, 71, 73, 77-80, 82-84.
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