Chapter 19
CHEMICAL THERMODYNAMICS
19.1 SPONTANEOUS PROCESSES
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.
For example, Drop an egg above a hard surface.
An egg falling and breaking isspontaneous. The reverse process is nonspontaneous, even though energy is conserved inboth processes.
Another example: Gas expansion into vacuum. See Fig. 19.2
In general, processes that arespontaneous 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 atmosphericpressure, when the temperature of the surroundings is above 0 C, ice melts spontaneously, and the reverse process— liquid water turning into iceis 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 necessarilymean that it will occur at an observable rate.A chemical reaction is spontaneous ifit occurs on its own accord, regardless of its speed. A spontaneous reaction can be veryfast, 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.
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 ofbenzene, 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 equilibriumstate 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 systemand surroundings can be restored to their original states by exactly reversing thechange.
An irreversible process is onethat cannot simply be reversed to restore the system and its surroundings to theiroriginal states.
A French engineer named Sadi Carnot (1796–1832) concludedthat: a reversible change produces the maximum amount of work that can bedone by a system on its surroundings.
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 partitionالحاجز is removed, the gas expands spontaneously to fill theevacuated space.
Since the gas expands into a vacuum with no external pressure, itdoes 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 theinitial and final states of the system and not on the path taken from one state to theother:
S = Sfinal –Sinitial[19.1]
For the special case of an isothermal process,S is equal to the heat that would betransferred if the process were reversible, qrev, divided by the absolute temperature atwhich the process occurs:
between states, not just thereversible 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:
Example 19.2
Elemental mercury (Hg) is a silver liquid at room temperature. Its normal freezing point is -38.9 C and itsmolar enthalpy of fusion is Hfusion = 2.29 kJ/mol.What is the entropy change of the system when 50.0 gof 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
The Second Law of Thermodynamics
The first law of thermodynamics states that energy is conserved in anyprocess.
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 roughlythe 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 differenttemperatures.
We can calculate the entropy changeof the system as follows:
Heat lost = - Heat gained
Q (lost by your hand) = - Q(heat gained by the ice)
Hence, the entropy change of the surroundings is
Assuming, the body temperature = 37 C
Thus, the total entropy change is positive:
Stotal = Ssystem + Ssurr = 22.0 -19.4 = 2.6 J/K The process is nonspontaneous.
If the temperature of the surroundings were not 310 K but rather some temperatureinfinitesimally above 273 K, the melting would be reversible instead of irreversible. In thatcase the entropy change of the surroundings would equal and wouldbe 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 spontaneouschangeit 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.
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
S = klogW
Or , S = k ln W19.5
,
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:
Any change in the system that leads to an increase in the number of microstates (WfinalWinitial)
leads to a positive value of S : Entropy increases with the number ofmicrostates 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 entropychanges:
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 themolecule move periodically toward and away from one another.
(iii)Rotational motion: in which the molecule spins about an axis.
FIGURE 19.8 shows the vibrationalmotionsand one of the rotational motions possible for the water molecule. These differentforms of motion are ways in which a molecule can store energy, and we refer to the variousforms 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 ofthese 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.
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 readilymove about relative to one another (translation) and tumble around (rotation). Duringmelting, therefore, the number of possible microstates increases and so does theentropy. In water vapor, the molecules are essentially independent of one another andhave their full range of translational, vibrational, and rotational motions. Thus, watervapor has an even greater number of possible microstates and therefore a higher entropythan 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 adisordering process (the ions become less confined) and an ordering process (somewater 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) 2NO2(g) [19.7]
which results in a decrease in the number of moleculesthree molecules of gaseousreactants form two molecules of gaseous products ( FIGURE 19.11). The formation ofnew N-O bonds reduces the motions of the atoms in the system. The formationof new bonds decreases the number of degrees of freedom, or forms of motion, availableto the atoms. That is, the atoms are less free to move in random fashion because of theformation of new bonds. The decrease in the number of molecules and the resultant decreasein motion result in fewer possible microstates and therefore a decrease in theentropy of the system.
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1
Sample Exercise 19.3:
Predict whetherS is positive or negative for each process, assuming each occurs at constant
temperature:
a) H2O(l)H2O(g)S is positive
b) Ag+(aq) + Cl-(aq) AgCl(s)S is negative
c) 4 Fe(s)+ 3O2(g)2 Fe2O(s)S is negative
d) N2(g)+ O2(g)2NO(g)Szero
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 25C , (b) 2 mol of HCl(g) or 1 mol of HCl(g) at 25C , (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 moleculesin 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 =KlnW = zero, when W = 1
As the temperature is increased from absolute zero, the atoms or molecules in the crystal gainenergy 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 tomove 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 canunderstand this increase as resulting from the increased volume available to theatomsor molecules as they enter the gaseous state.When the gas is heated further, the entropyincreases steadily as more energy is stored in the translational motion of the gas atomsor 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 referencetemperature 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 inFigure 19.13.
3. Standard molar entropies generally increase with increasing molar mass.
4. Standard molar entropies generally increase with an increasing number of atoms inthe 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 andPr Ex. 19.5
Energy Changes in the Surroundings
Because in a constant-pressure process, qsys is simply the enthalpy change for the reaction,H
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, whichmeans an increase in the entropy of the surroundings:
q sys is the enthalpy change for the reactionunder standard conditions,H , so the changes in entropy will be standard entropychanges,S.