1. Thermodynamics Change in Entropy, S
CHEM 3310
Thermodynamics Change in Entropy, S

2. Entropy, S
Entropy is the measure of dispersal The natural spontaneous direction of any process is toward
greater dispersal of matter and of energy.
Dispersal of matter: We analyze the constraints on a system. The more the system is constrained, the less dispersed it is.
Dispersal of energy: We analyze the flow of heat to measure how much energy is spread out in a particular process at a specific temperature.
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3. Like E and H, S is a state function.
Entropy, S
Entropy measures the dispersal of matter S = k ln W
where W is the number of ways of describing the system, and k is the Boltzmann constant (1.38 x J K-1).
The more ways that the system can reside, the greater the Entropy, S.
Note: If there is only one way to describe the system when W=1, the system is fully constrained. S=0 because ln (1)=0
The change in entropy is S.
S = S2 - S1
where S1 is the entropy of the initial state, and S2 is the entropy of the final state.
Like E and H, S is a state function.
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4. S = k log W
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5. Change in Entropy, S Entropy measures the dispersal of matter
Example 1: Melting of ice
H2O (s)  H2O (l)
S = S2-S1
S >0
In the ice crystal, all the molecules are constrained to fixed positions.
This is state 1 corresponding to an entropy state, S1.
In liquid water, molecules are free to move around.
This is state 2 corresponding to an entropy state, S2.
Image credit:
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6. Change in Entropy, S Entropy measures the dispersal of matter
Example 2: Rusting of iron
4 Fe (s) + 3 O2 (g) + 3 H2O (l)  2 Fe2O33 H2O (s)
S < 0
Image credit:
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7. Change in Entropy, S Entropy measures the dispersal of matter
Example 3: Combustion of methane
CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O (l)
S < 0
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8. Change in Entropy, S Entropy measures the dispersal of matter
Example 1: Melting of water S > 0
Example 2: Iron rusting S < 0
Example 3: Combustion of methane S < 0
All of the above processes are spontaneous at room temperature and
atmospheric pressure.
Conclusion: When a system changes spontaneously, there may be in increase of entropy (S > 0), or a decrease in entropy (S < 0).
The above entropy changes are looking at the S of the system. But in the real world, there is always heat exchanges with the surroundings. Therefore,
we must consider the entropy change of the universe.
Usually when we write S, it implies S of the system (Ssystem).
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9. Change in Entropy of the Universe, Suniverse
The Second Law of Thermodynamics:
In any spontaneous process, there is always an increase in the entropy of the universe.
Suniverse = Ssystem + Ssurroundings > 0
For a spontaneous process, if there is a decrease in the Ssystem, then the heat is absorbed by the surroundings must have caused an increase in the Ssurroundings such that overall there is an increase in the Suniverse (i.e. Suniverse > 0).
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10. Change in Entropy, S
Entropy change measures the dispersal of energy How much energy is spread out in a particular process, or how widely spread out it becomes at a specific temperature.
Every process has a preferred direction – natural spontaneity.
Spontaneous processes:
- Contraction of a stretched elastic band
- Heat flows from a warm body to a cold body
These processes increases random kinetic energy of the system. As a result, an increase of energy dispersion results.
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11. Change in Entropy, S
Entropy change measures the dispersal of energy We analyze the flow of heat to measure how much energy is spread out in a particular process at a specific temperature.
Every process has a preferred direction – natural spontaneity.
Let’s look at what energy dispersal means on a microscopic scale with micro-states.
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12. Entropy S = k ln W Entropy is a measure of the available microstates
Consider a chamber with 4 particles in it. Where the particle can only have energies 2, 4, 6, 8, 10, … (i.e. the particles energy is quantized)
The lowest energy state is when all 4 particles have energy = 2 or total energy is 8. There is only one way the particles can be arrange in this configuration.
4 particles with a total energy of 8

13. Entropy
This system is now heated so that the total energy of the particles is 10.
There are 4 microstates (arrangements of the particles) that satisfy Et=10.
4 particles with a total energy of 10

14. Entropy 3 particles with E =2 and 1 particle with E = 6
This system is now heated so that the total energy of the particles is 12.
There are 10 microstates (arrangements of the particles) that satisfy Et=12.
4 microstates with 3 particles with E =2 and 1 particle with E = 6
3 particles with E =2 and 1 particle with E = 6

15. Entropy 2 particles with E = 2 and 2 particles with E = 4
This system is now heated so that the total energy of the particles is 12.
There are 10 microstates (arrangements of the particles) that satisfy Et=12.
6 microstates with 2 particles with E =2 and 2 particles with E = 4
2 particles with E = 2 and 2 particles with E = 4

16. Entropy
S = k ln 
This system is now heated so that the total energy of the particles is 12.
There are 10 mircostates (arrangements of the particles) that satisfy Et=12.
4 microstates with 3 particles with E =2 and 1 particles with E = 6
6 microstates with 2 particles with E =2 and 2 particles with E = 4
Total of 10 microstates with total energy = 12

17. Entropy
S = k ln 
This system is now heated so that the total energy of the particles is 14.
There are 20 mircostates (arrangements of the particles) that satisfy Et=14.
4 microstates with 3 particles with E =2 and 1 particle with E = 8
12 microstates with 2 particles with E =2 and 1 particles with E = 4 and 1 particle with E = 6
4 microstates with 1 particle with E =2 and 3 particles with E = 4
Total of 20 microstates with total energy = 14

18. Entropy
As the energy of the system increases the number of microstates increases.
Thus the entropy of the system increases.
For 4 particles with quantized energy levels of E = 2, 4, 6, … E  8 1 10 4 12 14 20 16 35 18 56 84

19. Entropy System 1 Eo = 8 System 2 Eo = 20 System 1 E = 10 System 2
Consider 2 of these 4 particles system, one with Et = 8 and one with Et = 20.
The systems are placed in thermal contact with each other.
Number of microstates x = 84
If E = 2 units of energy are transferred from system 2 to system 1
Number of micro states x = 224
System 1
Eo = 8
System 2
Eo = 20
System 1
E = 10
System 2
E = 18

20. Entropy
Consider 2 of these 4 particles system, one with Et = 8 and one with Et = 20.
Total Energy = 28
The systems are placed in thermal contact with each other.
System 1
E = E1
System 2
E = E2
System 1
System 2 E1 1 E2 2
total 8 1 20 84 10 4 18 56
224 12 16 35
350 14
400

21. Entropy
Consider 2 of these 4 particles system, one with Et = 8 and one with Et = 20.
Total Energy = 28
The probability of a finding a given amount of energy in system 1 is show in the graph
Notice that the most likely outcome is when systems 1 and 2 have equal energy

22. Change in Entropy, S On a molecular scale
The number of microstates and, therefore, the entropy tends to increase with increases in:
Temperature
Volume (gases)
The number of independently moving molecules
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23. Entropy, S
Entropy is related to the various modes of motion in molecules.
Translational: Movement of the entire molecule from one place to another.
Vibrational: Periodic motion of atoms within a molecule.
Rotational: Rotation of the molecule on about an axis or rotation about  bonds.
Image credit: Chemistry, The Central Science, 10th edition Theodore L. Brown, H. Eugene LeMay, Jr., and Bruce E. Bursten
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24. Entropy, S
More particles lead to more energy states, and therefore, more entropy
Higher temperature lead to more energy states, and therefore, more entropy
Less structure (gas is less structured than liquid, which is less structured than solid) lead to more states, and therefore, more entropy
Entropy increases with the freedom of motion of molecules. Therefore,
S(g) > S(l) > S(s)
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25. Change in Entropy, S More spontaneous processes:
- Free expansion of a gas
S > 0
This process neither absorbs nor emits heat.
- Combustion of methane
S < 0
CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O (l)
The process produces heat (exothermic).
- Melting of ice cubes at room temperature
H2O (s)  H2O (l)
S > 0
This process requires heat (endothermic).
Conclusion: Naturally spontaneous processes could be exothermic, endothermic, or not involving heat.
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26. Change in Entropy, S Photosynthesis More spontaneous processes:
Complex and highly-energetic compounds (compared to the starting materials, CO2 and H2O) are formed, it appears that there a decrease in entropy in the process.
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27. Change in Entropy, S A good example is photosynthesis. CHEM 3310
S<0; It appears that there a decrease in entropy in this natural process.
A good example is photosynthesis.
Photosynthesis in a plant does not consist of a isolated system of the plant alone.
Experiments and calculations indicate that the maximum efficiency of photosynthesis in most plants is in the 30% range.
This means:
30% of the sunlight that strikes the plant is absorbed by the plant in the photosynthesis of substances that decreases entropy in the plant.
70% of the sunlight that strikes the plant is dispersed to the environment (an entropy increase in slightly heating the leaf and the atmosphere.
Image credit:
When the system and surroundings are considered, there is a net increase in entropy as a result of photosynthesis!
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28. Change in Entropy, S For any spontaneous process,
Suniverse = Ssystem + Ssurroundings > 0
Spontaneous process,
favoured by a decrease in H (exothermic)
favoured by an increase in S
Nonspontaneous process,
favoured by an increase in H (endothermic)
favoured by an decrease in S
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29. Change in Entropy, S What is qrev?
Entropy measures the dispersal of energy At a specific temperature, the change in entropy, S, is
where qrev is the heat transferred at a constant temperature T in Kelvin.
What is qrev?
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30. Change in Entropy, S What is qrev? wrev wirrev CHEM 3310
Consider a gas in a piston and cylinder configuration. The gas is held in place by a pile of sand. The gas undergoes isothermal expansion.
Recall E = q+w
Since E = 0 for an isothermal process, q is the same magnitude of w but opposite in sign.
Let’s take the PV curve of five different
paths to take the gas from State I to State F.
Work done by the gas (w<0) is the area under the PV curve.
wrev
wirrev
This is the reversible path. The area under the PV curve is the greatest. This path yields the smallest w (most negative), and the largest q. We refer to the heat transfer of a reversible path reversible q, qrev.
Of all the different possible paths to take the gas from State I to F, qrev is the limit of the observed heat transferred.
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31. Entropy, S What is qrev? wrev wirrev CHEM 3310
Likewise, consider a gas in a piston and cylinder configuration. The gas is held in place by a pile of sand. The gas undergoes isothermal compression.
Since E = 0 for an isothermal process, q is the same magnitude of w but opposite in sign.
Let’s take the PV curve of five different
paths to take the gas from State I to State F.
Work done on the gas (w>0) is the area under the PV curve.
wrev
wirrev
This is the reversible path. The area under the PV curve is the smallest. This path yields the smallest w, and the largest q (least negative). We refer to the heat transfer of a reversible path reversible q, qrev.
Of all the different possible paths to take the gas from State I to F, qrev is the limit of observed heat transferred.
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32. Change in Entropy, S What is qrev?
A reversible change is one that is carried out in such a way, such that, when the change is “undone”, both the system and the surroundings remain unchanged. This is achieved when the process proceeds in
infinitesimal steps that would take infinitely long to occur.
Reversibility is an idealization that is unachievable in real process, except when the system is in equilibrium.
A process that is carried out reversibly has the smallest wrev, and the largest qrev.
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33. This is a reversible process.
Change in Entropy, S
To express entropy is to describe the number of arrangements for positions and/or energy levels available to a system — a measurement of dispersion.
Entropy measures the dispersal of energy At a specific temperature, the change in entropy, S, is
where qrev is the heat transferred at a constant temperature T in Kelvin.
Although most real processes are irreversible, there are real reversible processes.
Consider ice melting at 0oC, where Hfusion = cal/g = kJ/mole
This is a reversible process.
If the heat is absorbed in small enough increments such that it does not disturb the equilibrium, this is a reversible process.
There is an increase in the entropy of water when ice melts to form liquid water.
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34. This is a reversible process.
Change in Entropy, S
Similarly, consider water vapourizing at 100oC, where Hvapourization = 40.7 kJ/mole
This is a reversible process.
If the heat is absorbed in small enough increments such that it does not disturb the equilibrium, this is a reversible process.
There is an increase in the entropy of water when water is converted to steam.
Note: S for the vapourization of water is much higher than that of fusion.
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35. Work done by the gas (w<0) is the area under the PV curve.
Change in Entropy, S
Expansion of an Ideal Gas
Recall how we define a reversible process. For the expansion of a gas in a piston and cylinder configuration under constant temperature condition.
Since E = 0 for an isothermal process, q is the same magnitude of w but opposite in sign.
Work done by the gas (w<0) is the area under the PV curve.
It follows that or
wrev
Entropy increases with an increase in volume or a decrease in pressure.
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36. Change in Entropy in the surroundings, Ssurroundings
Entropy Changes in Surroundings
Heat that flows into or out of the system also changes the entropy of the surroundings.
For an isothermal process:
At constant pressure, qsys is H for the system.
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37. Change in Entropy in the surroundings, Ssurroundings
A phase change is isothermal (no change in T).
For H2O:
Hfusion = kJ/mole
Hvapourization = 40.7 kJ/mole
If this is done reversibly: Ssurr = –Ssys
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38. This is true for a reversible process. The system is at equilibrium.
Example:
1.00 mole of a monatomic ideal gas at 273 K is expanded via two paths:
(a) isothermally and reversibly from 22.4 L to 44.8 L. Find w, q, E, H, Ssystem, Ssurroundings, and Suniverse.
External pressure, Pext
Under isothermal condition, T=0, therefore
E = H = 0
(w < 0)
Since E depends only on temperature, E = 0 and H = E + (PV) = E + (nRT) = 0
This is because work of expansion done by the gas equals to the heat flow into the system. Net overall change in internal energy and enthalpy is zero. q
(q > 0)
This is Ssystem. What is Ssurroundings? What is Suniverse?
qrev = ST = 273 = kJ
Since E = 0, w = -q = kJ
Since qsurr = -qsys, qsurr = kJ
This is true for a reversible process. The system is at equilibrium.
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39. This is a spontaneous process. Suniverse > 0
Example:
1.00 mole of a monatomic ideal gas at 273 K is expanded via two paths:
(b) via free expansion from 22.4 L to 44.8 L. Find w, q, E, H, and
Ssystem, Ssurroundings, and Suniverse.
This is an irreversible process with no change in temperature, T=0.
As a result, E = H = 0
w = 0 because Pext = 0
w = -q, q =0
Since E depends only on temperature, E = 0 and H = E + (PV) = E + (nRT) = 0
To calculate S, we need to find a reversible path that will take the gas from the same initial state to the same final state. Part (a) is such a path.
This is Ssystem. It has the same disorder as calculated in Part (a). What is Ssurroundings? What is Suniverse?
This is a spontaneous process. Suniverse > 0
Since the no heat flowed into the surroundings, qsurr = 0,
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40. Change in Entropy, S Entropy associated with temperature change
At a specific
temperature T
Recall, q = n C T
where n = number of moles
C = molar heat capacity
T = temperature difference
For a change in temperature of n moles of substance from T1 to T2 where the heat capacity is constant over the temperature range.
Under constant
volume condition
Under constant
pressure condition
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41. Change in Entropy, S S H Entropy associated with temperature change
At a specific
temperature T
For a change in temperature of n moles of substance from T1 to T2 and the heat capacity is expressed as a power series, a + bT + cT2 + …
T in Kelvin S H
Compare
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42. Change in Entropy, S Calculate S for the change in state S
Increase the volume of the container. The amount of gaseous H2O present is governed by the vapour pressure of water at 25oC.
Given Cp (H2O,liq) = 18 cal mole-1K-1
Cp (H2O, g) = 9.0 cal mole-1K Hvap = 9713 cal/mole (at 373K)
Let’s devise a reversible process to calculate S. S
Warming liquid water, S1
Cooling gaseous water, S3
S2
S = S1 + S2 + S3
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43. Change in Entropy, S Calculate S for the change in state S
Given Cp (H2O,liq) = 18 cal mole-1 K-1
Cp (H2O, g) = 9.0 cal mole-1 K Hvap = 9713 cal/mole (at 373K)
S < 0
S > 0 S
Warming liquid water, S1
Cooling gaseous water, S3
S2
S > 0
S = S1 + S2 + S3 = = 28.1 cal K-1
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44. Change in Entropy, S Trouton’s Rule
A useful generalization that works for many liquids at their normal boiling point (i.e. boiling point at 1 atm) is that, the standard molar entropy of vapourization has a value of about 22 cal mole-1 K-1 or 88 J mole-1 K-1
The idea is, if the degree of disorder associated in the phase transformation of 1 mole of liquid to 1 mole of vapour at 1 atm is the same, then the Sovapourization of different liquids should be similar.
Since
Sovapourization = 22 cal mole-1 K-1 or 88 J mole-1 K-1
Trouton’s rule is useful method for estimating the enthalpy of vapourization of a liquid if its normal boiling point is known.
Example: We can estimate the enthalpy of vapourization of Hg using Trouton’s rule
from the normal boiling point of Hg. The normal boiling point of Hg is 357oC.
Compare: Heat of vaporization of mercury:
59.11 kJ mol-1
Cited value: Wikipedia
Hvapourization = 630 K  88 J mole-1 K-1 = 55 kJ mole-1
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45. Change in Entropy, S Trouton’s Rule
Sovapourization = 22 cal mole-1K-1 or 88 J mole-1K-1
Substance
Hvap
(kcal mole-1) Tb
(oC)
Svap
(cal mole-1 K-1) O2
1.630
18.07
CH4
1.955
17.51
H2O
9.717
100.0
26.04
NH3
5.581
-33.43
23.28
C2H5OH
9.23
78.3
26.27
CHCl3
7.02
61.2
20.8
C6H6
7.36
80.1
20.83
(C2H5)2O ether
6.21
34.5
20.18
(CH3)2CO acetone
7.22
56.2
21.90
In liquid state, hydrogen bonding between molecules produces a greater degree of order than expected. As a result, the degree of disorder produced
in the vapourization process is generally greater than the nominal 22 cal mole-1.
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46. Entropy, S The Third Law of Thermodynamics:
The entropy of a pure perfect crystal is zero when the crystal is at zero Kelvin (0 K)
A pure perfect crystal is one in which every molecule is identical, and the molecular alignment is perfectly even throughout the substance.
For non-pure crystals, or those with less-than perfect alignment, there will be some dispersal of the imperfection, so the entropy cannot be zero.
Image credit: Chemistry, The Central Science, 10th edition Theodore L. Brown, H. Eugene LeMay, Jr., and Bruce E. Bursten

47. Entropy, S
The entropy of the crystal gradually increases with temperature as the average kinetic energy of the particles increases.
The entropy of the liquid gradually increases as the liquid becomes warmer because of the increase in the vibrational, rotational, and translational motion of the particles.
As the crystal warms to temperatures above 0 K, the particles in the crystal start to move, generating some disorder.
At the melting point, Tm, the entropy of the system increases in entropy without a temperature change as the compound is transformed into a liquid, which is not as well ordered as the solid.
At the boiling point, there is another increase in the entropy of the substance without a temperature change as the compound is transformed into a gas.
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48. Standard Molar Entropy, So
Standard entropies (measured at 298K) have been carefully measured for many substances.
For example, the standard molar entropy of some solids: So
J K-1 mole-1
C(diamond)
2.38
C (graphite)
5.74
Sodium
51.3
Phosphorus (white)
41.1
Sulfur (rhombic)
31.8
Silver
42.6
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49. Standard Molar Entropy, So
Standard entropies (measured at 298 K) have been carefully measured for many substances.
For example, the standard molar entropies (standard entropy per mole) for gases are usually higher because heat of melting and heat of vapourization must be included. The standard molar entropies for noble gases are: So
J K-1 mole-1 He
126.0 Ne
146.2 Ar
154.7 Kr
164.0 Xe
169.6
For liquids and gases So are usually higher because heat of melting and heat of vaporization must be included.
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50. Standard Molar Entropy, So
Standard entropies (measured at 298 K) have been carefully measured for many substances.
For example, the standard molar entropies (standard entropy per mole) for diatomic gases. So
J K-1 mole-1 H2
130.6 N2
191.5 O2
205.1
Halogens: F2
203.7
Cl2
222.9
Br2
245.4 I2
260.6
For diatomic gases, they are usually higher than those of monatomic noble gases as there are more degrees of freedom of motion.
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51. Standard Molar Entropy, So
Standard entropies (measured at 298 K) have been carefully measured for many substances.
For example, the standard molar entropies (standard entropy per mole) for more gases.
In general, the more complicate the molecule and the heavier the molar mass, the higher the standard entropy. So
J K-1 mole-1
CH4 (g)
186.3
C2H2 (g)
200.8
C2H4 (g)
219.5
C2H6 (g)
229.6
C3H8 (g)
270.3
Image credit: Chemistry, The Central Science, 10th edition Theodore L. Brown, H. Eugene LeMay, Jr., and Bruce E. Bursten
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52. Standard Molar Entropy, So
For example, the standard molar entropies (standard entropy per mole) for compounds. So
J K-1 mole-1
H2O (l)
69.9
H2O2 (l)
110.0
CH3OH(l)
126.9
CH3Cl(l)
145.3
CHCl3(l)
202.9 So
J K-1 mole-1
H2O (g)
188.7
CO (g)
197.8
CO2 (g)
213.6
NO (g)
210.6
NO2 (g)
240.4
N2O4 (g)
304.3
SO2 (g)
248.4
A NO molecule has only one type of vibrational molecule
A NO2 molecule has three types of vibration.

53. Standard Entropy for Aqueous Ions
These are not ‘third law’ entropies, presumably because the task of working up from absolute zero is too complex a task for aqueous ions.
The hydrogen ion, is arbitrarily chosen to have an entropy of zero and all other ions are compared with it.
Some of these other ions have greater entropy than the hydrogen ions, some lower, hence the negative values.
H+ (aq)
OH- (aq)
H2O (l)
So (J K-1 mole-1)
-10.7
70.0
H+(aq) + OH-(aq)  H2O(l) 
So = So(H2O (l)) – (So(H+(aq)) - So(OH-(aq))
So = 70.0 – ( )
So = 80.7 J K-1
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54. Sreactiono
Calculate the change in Entropy of a reaction from Standard Molar Entropy.
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55. Sreactiono
Calculate the standard entropy change accompanying the burning of ethane, C2H6.
C2H6 (g) + 7/2 O2 (g)  2 CO2 (g) + 3 H2O (g) S = ? So
J K-1 mole-1
C2H6 (g)
229.5
O2 (g)
205.1
CO2 (g)
213.6
H2O (g)
188.7
In general, the more complicate the molecule and the heavier the molar mass, the higher the standard entropy.
So = 2So(CO2 (g)) + 3 So(H2O (g)) – (So(C2H6 (g)) + 7/2 So(O2 (g)))
So = 2(213.6) + 3 (188.7) – ( /2 (205.1))
So = 46.3 J K-1
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56. Summary:
The Second Law of Thermodynamics: In any spontaneous process, there is always an increase in the entropy of the universe.
Suniverse = Ssystem + Ssurroundings > 0
Change in entropy, S, at a specific temperature, T.
Change in entropy, S, measured over a temperature range from T1 to T2.
Assuming that Cv and Cp are constant from T1 to T2.
For an isothermal process expansion or compression of a gas.
The Third Law says the entropy of a pure perfect crystal is zero when the crystal is at zero Kelvin (0 K). For a chemical reaction, So can be calculated from the reactants’ and products’ standard molar entropies.
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