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18.4 Entropy Changes and the Third Law of Thermodynamics


Learning Objective

To use thermodynamic cycles to calculate transforms in entropy.

The atoms, molecules, or ions that write a chemical device deserve to undergo several kinds of molecular activity, including translation, rotation, and also vibration (Figure 18.13 "Molecular Motions"). The better the molecular movement of a device, the higher the number of feasible microsays and also the greater the entropy. A perfectly ordered device with just a single microstate accessible to it would certainly have an entropy of zero. The only mechanism that meets this criterion is a perfect crystal at a temperature of absolute zero (0 K), in which each component atom, molecule, or ion is addressed in place within a crystal lattice and also exhibits no activity. Such a state of perfect order (or, conversely, zero disorder) corresponds to zero entropy. In practice, absolute zero is a perfect temperature that is unobtainable, and a perfect single crystal is also a perfect that cannot be accomplished. Nonethemuch less, the combination of these two ideals constitutes the basis for the third legislation of thermodynamicsThe entropy of any kind of perfectly ordered, crystalline substance at absolute zero is zero.: the entropy of any type of perfectly ordered, crystalline substance at absolute zero is zero.

You are watching: Which substance has the lower molar entropy?


Figure 18.13 Molecular Motions

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Vibrational, rotational, and also translational motions of a carbon dioxide molecule are illustrated right here. Only a perfectly ordered, crystalline substance at absolute zero would exhilittle bit no molecular motion and also have zero entropy. In practice, this is an unattainable appropriate.


The third legislation of thermodynamics has actually 2 crucial consequences: it defines the sign of the entropy of any kind of substance at temperatures over absolute zero as positive, and it offers a fixed reference point that permits us to meacertain the absolute entropy of any substance at any temperature.In practice, chemists recognize the absolute entropy of a substance by measuring the molar warmth capacity (Cp) as a role of temperature and also then plotting the amount Cp/T versus T. The area under the curve between 0 K and also any type of temperature T is the absolute entropy of the substance at T. In contrast, various other thermodynamic properties, such as interior energy and also enthalpy, deserve to be evaluated in only relative terms, not absolute terms. In this area, we research 2 various means to calculate ΔS for a reaction or a physical change. The first, based on the meaning of absolute entropy offered by the 3rd regulation of thermodynamics, provides tabulated values of absolute entropies of substances. The second, based on the truth that entropy is a state feature, supplies a thermodynamic cycle equivalent to those we first encountered in Chapter 5 "Energy Changes in Chemical Reactions".


Calculating ΔS from Standard Molar Entropy Values

One method of calculating ΔS for a reactivity is to use tabulated worths of the conventional molar entropy (S°)The entropy of 1 mol of a substance at a standard temperature of 298 K., which is the entropy of 1 mol of a substance at a conventional temperature of 298 K; the systems of S° are J/(mol·K). Unprefer enthalpy or inner power, it is possible to obtain absolute entropy worths by measuring the entropy readjust that occurs in between the referral point of 0 K S = 0 J/(mol·K)> and also 298 K.

As displayed in Table 18.1 "Standard Molar Entropy Values of Selected Substances at 25°C", for substances via approximately the very same molar mass and also variety of atoms, S° values loss in the order S°(gas) > S°(liquid) > S°(solid). For instance, S° for liquid water is 70.0 J/(mol·K), whereas S° for water vapor is 188.8 J/(mol·K). Likewise, S° is 260.7 J/(mol·K) for gaseous I2 and also 116.1 J/(mol·K) for solid I2. This order renders qualitative sense based on the kinds and also extents of activity obtainable to atoms and molecules in the 3 phases. The correlation in between physical state and absolute entropy is depicted in Figure 18.14 "A Generalized Plot of Entropy versus Temperature for a Single Substance", which is a generalized plot of the entropy of a substance versus temperature.


Table 18.1 Standard Molar Entropy Values of Schosen Substances at 25°C

Substance S°
Gases
He 126.2
H2 130.7
Ne 146.3
Ar 154.8
Kr 164.1
Xe 169.7
H2O 188.8
N2 191.6
O2 205.2
CO2 213.8
I2 260.7
Liquids
H2O 70.0
CH3OH 126.8
Br2 152.2
CH3CH2OH 160.7
C6H6 173.4
CH3COCl 200.8
C6H12 (cyclohexane) 204.4
C8H18 (isooctane) 329.3
Solids
C (diamond) 2.4
C (graphite) 5.7
LiF 35.7
SiO2 (quartz) 41.5
Ca 41.6
Na 51.3
MgF2 57.2
K 64.7
NaCl 72.1
KCl 82.6
I2 116.1

Figure 18.14 A Generalized Plot of Entropy versus Temperature for a Single Substance

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Absolute entropy increases steadily with increasing temperature until the melting suggest is got to, where it jumps unexpectedly as the substance undergoes a phase adjust from a extremely ordered solid to a disordered liquid (ΔSfus). The entropy again increases steadily through enhancing temperature till the boiling allude is got to, where it jumps all of a sudden as the liquid undergoes a phase readjust to a highly disordered gas (ΔSvap).


A closer examicountry of Table 18.1 "Standard Molar Entropy Values of Selected Substances at 25°C" additionally reveals that substances with equivalent molecular structures tfinish to have equivalent S° worths. Amongst crystalline materials, those through the lowest entropies tend to be rigid crystals composed of small atoms connected by solid, extremely directional bonds, such as diamond <S° = 2.4 J/(mol·K)>. In comparison, graphite, the softer, less rigid allotrope of carbon, has actually a greater S° <5.7 J/(mol·K)> as a result of even more disorder in the crystal. Soft crystalline substances and also those with bigger atoms tend to have actually greater entropies bereason of increased molecular activity and disorder. Similarly, the absolute entropy of a substance tends to boost via raising molecular intricacy bereason the number of obtainable microsays rises with molecular complexity. For instance, compare the S° values for CH3OH(l) and also CH3CH2OH(l). Finally, substances through solid hydrogen bonds have actually lower worths of S°, which reflects a much more ordered framework.

To calculate ΔS° for a chemical reactivity from typical molar entropies, we usage the acquainted “commodities minus reactants” dominion, in which the absolute entropy of each reactant and product is multiplied by its stoichiometric coeffective in the balanced chemical equation. Example 7 illustprices this procedure for the burning of the liquid hydrocarbon isooctane (C8H18; 2,2,4-trimethylpentane).


Example 7

Use the information in Table 18.1 "Standard Molar Entropy Values of Selected Substances at 25°C" to calculate ΔS° for the reactivity of liquid isooctane with O2(g) to give CO2(g) and H2O(g) at 298 K.

Given: traditional molar entropies, reactants, and also products

Asked for: ΔS°

Strategy:

Write the well balanced chemical equation for the reactivity and recognize the proper quantities in Table 18.1 "Standard Molar Entropy Values of Schosen Substances at 25°C". Subtract the sum of the absolute entropies of the reactants from the sum of the absolute entropies of the commodities, each multiplied by their proper stoichiometric coefficients, to achieve ΔS° for the reactivity.

Solution:

The balanced chemical equation for the complete combustion of isooctane (C8H18) is as follows:

C 8 H 18 (l) + 25 2 O 2 (g) → 8 CO 2 (g) + 9 H 2 O(g)

We calculate ΔS° for the reactivity making use of the “products minus reactants” rule, wright here m and also n are the stoichiometric coefficients of each product and also each reactant:

Δ S ° rxn = ∑ m S ° (products) − ∑ n S ° (reactants) = < 8 S ° ( CO 2 ) + 9 S ° ( H 2 O ) > − < S ° ( C 8 H 18 ) + 25 2 S ° ( O 2 ) > = < 8   mol  CO 2 × 213.8  J/ ( mol ⋅ K ) > + < 9   mol  H 2 O × 188.8  J/ ( mol ⋅ K ) >   − < 1 mol  C 8 H 18 × 329.3  J/ ( mol ⋅ K ) > + < 25 2 mol  O 2 × 205.2  J/ ( mol ⋅ K ) > = 515.3  J/K

ΔS° is positive, as meant for a combustion reactivity in which one large hydrocarbon molecule is converted to many kind of molecules of gaseous commodities.

Exercise

Use the data in Table 18.1 "Standard Molar Entropy Values of Schosen Substances at 25°C" to calculate ΔS° for the reaction of H2(g) via liquid benzene (C6H6) to offer cyclohexane (C6H12).

Answer: −361.1 J/K


Keep in mind the Pattern

Entropy increases through softer, much less rigid solids, solids that contain bigger atoms, and also solids with complex molecular structures.


Keep in mind the Pattern

ΔS° for a reactivity can be calculated from absolute entropy values making use of the very same “commodities minus reactants” rule provided to calculate ΔH°.


Calculating ΔS from Thermodynamic Cycles

We can likewise calculate a adjust in entropy making use of a thermodynamic cycle. As you learned in Chapter 5 "Energy Changes in Chemical Reactions", the molar heat capacity (Cp) is the amount of heat needed to raise the temperature of 1 mol of a substance by 1°C at consistent pressure. Similarly, Cv is the amount of warmth necessary to raise the temperature of 1 mol of a substance by 1°C at continuous volume. The boost in entropy through boosting temperature in Figure 18.14 "A Generalized Plot of Entropy versus Temperature for a Single Substance" is around proportional to the heat capacity of the substance.

Respeak to that the entropy change (ΔS) is related to heat circulation (qrev) by ΔS = qrev/T. Because qrev = nCT at continuous pressure or nCT at constant volume, where n is the variety of moles of substance existing, the readjust in entropy for a substance whose temperature transforms from T1 to T2 is as follows:

Δ S = q r e v T = n C p Δ T T   ( constant pressure )

As you will certainly discover in more progressed math courses than is required below, it have the right to be shown that this is equal to the following:For a review of herbal logarithms, check out Essential Skills 6 in Chapter 11 "Liquids".


Equation 18.20

Δ S = n C p ln  T 2 T 1   ( constant push )

Similarly,


Equation 18.21

Δ S = n C v ln  T 2 T 1   ( constant volume )

Thus we deserve to use a combination of warmth capacity measurements (Equation 18.20 or Equation 18.21) and also experimentally measured worths of enthalpies of fusion or vaporization if a phase readjust is affiliated (Equation 18.18) to calculate the entropy adjust corresponding to a readjust in the temperature of a sample.

We deserve to usage a thermodynamic cycle to calculate the entropy adjust as soon as the phase change for a substance such as sulfur cannot be measured straight. As listed in the exercise in Example 6, elepsychological sulhair exists in 2 forms (component (a) in Figure 18.15 "Two Forms of Elepsychological Sulfur and also a Thermodynamic Cycle Showing the Transition from One to the Other"): an orthorhombic develop with a highly ordered framework (Sα) and also a less-ordered monoclinic develop (Sβ). The orthorhombic (α) develop is even more steady at room temperature however undergoes a phase change to the monoclinic (β) form at temperatures greater than 95.3°C (368.5 K). The shift from Sα to Sβ deserve to be defined by the thermodynamic cycle presented in component (b) in Figure 18.15 "Two Forms of Elemental Sulfur and also a Thermodynamic Cycle Showing the Transition from One to the Other", in which liquid sulfur is an intermediate. The change in entropy that acproviders the conversion of liquid sulhair to Sβ (−ΔSfus(β) = ΔS3 in the cycle) cannot be measured straight. Due to the fact that entropy is a state feature, yet, ΔS3 can be calculated from the all at once entropy readjust (ΔSt) for the Sα–Sβ change, which equates to the sum of the ΔS values for the steps in the thermodynamic cycle, utilizing Equation 18.20 and tabulated thermodynamic parameters (the warm capacities of Sα and also Sβ, ΔHfus(α), and also the melting suggest of Sα.)


Figure 18.15 Two Forms of Elepsychological Sulfur and also a Thermodynamic Cycle Showing the Transition from One to the Other

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(a) Orthorhombic sulhair (Sα) has a very ordered structure in which the S8 rings are stacked in a “crankshaft” setup. Monoclinic sulfur (Sβ) is additionally written of S8 rings however has actually a less-ordered framework. (b) At 368.5 K, Sα undergoes a phase shift to Sβ. Although ΔS3 cannot be measured directly, it have the right to be calculated using the worths shown in this thermodynamic cycle.


If we know the melting allude of Sα (Tm = 115.2°C = 388.4 K) and ΔSt for the as a whole phase shift , we deserve to calculate ΔS3 from the values provided in component (b) in Figure 18.15 "Two Forms of Elemental Sulhair and a Thermodynamic Cycle Showing the Transition from One to the Other" wbelow Cp(α) = 22.70 J/mol·K and also Cp(β) = 24.77 J/mol·K (subscripts on ΔS refer to steps in the cycle):

Δ S t = Δ S 1 + Δ S 2 + Δ S 3 + Δ S 4 1.09  J/ ( mol ⋅ K ) = C p ( α ) ln ( T 2 T 1 ) + Δ H fus T m + Δ S 3 + C p ( β ) ln ( T 4 T 3 ) = 22.70  J/ ( mol ⋅ K ) ln ( 388.4 368.5 ) + ( 1.722   kJ /mol 388.4  K × 1000  J/ kJ ) + Δ S 3 + 24.77  J/ ( mol ⋅ K ) ln ( 368.5 388.4 ) = < 1.194  J/ ( mol ⋅ K ) > + < 4.434  J/ ( mol ⋅ K ) > + Δ S 3 + < − 1.303  J/ ( mol ⋅ K ) >

Solving for ΔS3 gives a value of −3.24 J/(mol·K). As expected for the convariation of a less ordered state (a liquid) to a more ordered one (a crystal), ΔS3 is negative.


Summary

The third regulation of thermodynamics claims that the entropy of any type of perfectly ordered, crystalline substance at absolute zero is zero. At temperatures better than absolute zero, entropy has actually a positive value, which allows us to measure the absolute entropy of a substance. Measurements of the warm capacity of a substance and also the enthalpies of fusion or vaporization deserve to be provided to calculate the alters in entropy that acagency a physical change. The entropy of 1 mol of a substance at a conventional temperature of 298 K is its typical molar entropy (S°). We deserve to usage the “products minus reactants” dominance to calculate the traditional entropy readjust (ΔS°) for a reaction utilizing tabulated worths of S° for the reactants and the products.

See more: Why 3-Phase Voltage Is 440 Volts, Is 440 Volt The Same As 480 Volt


Key Takeaway

Entropy transforms have the right to be calculated using the “assets minus reactants” dominion or from a mix of warmth capacity dimensions and measured worths of enthalpies of fusion or vaporization.

Key Equations

Temperature dependence of entropy at constant pressure

Equation 18.20: ΔS=nCpln T2T1

Temperature dependence of entropy at continuous volume

Equation 18.21: ΔS=nCvln T2T1


Conceptual Problems


Crystalline MgCl2 has actually S° = 89.63 J/(mol·K), whereas aqueous MgCl2 has actually S° = −25.1 J/(mol·K). Is this continual with the third regulation of thermodynamics? Exsimple your answer.


Why is it possible to measure absolute entropies but not absolute enthalpies?


How many microsays are easily accessible to a system at absolute zero? How many are obtainable to a substance in its liquid state?


Substance A has actually a greater heat capacity than substance B. Do you intend the absolute entropy of substance A to be much less than, comparable to, or greater than that of substance B? Why? As the 2 substances are heated, for which substance carry out you predict the entropy to increase more rapidly?