Chapter 8. THERMODYNAMICS: THE SECOND AND THIRD LAW

2013 General Chemistry I 1

Chapter 8. THERMODYNAMICS: THE SECOND AND THIRD LAW

2012 General Chemistry I

ENTROPY

8.1 Spontaneous Change8.2 Entropy and Disorder8.3 Changes in Entropy8.4 Entropy Changes Accompanying Changes in Physical State8.5 A Molecular Interpretation of Entropy8.6 The Equivalence of Statistical and Thermodynamic Entropies8.7 Standard Molar Entropies8.8 Standard Reaction Entropies


ENTROPY (Sections 8.1-8.8)

- The 1st law of thermodynamics says if a reaction takes place, then the total energy of the universe remains unchanged. It cannot be used to predict the directionality of a process.

- The natural progression of a system and its surroundings (or “the universe”) is from order to disorder, from organized to random.

- A new thermodynamic state function is needed topredict both directionality and extent of disorder.


Spontaneous change is a change that has a tendency to occur without needing to be driven by an external influence.

- Spontaneous changes need not be fast: e.g. C(diamond) C(graphite); H2(g) + 1/2O2(g) H2O(l)

8.1 Spontaneous Change8.1 Spontaneous Change

Heat flowMixing of gases


8.2 Entropy and Disorder8.2 Entropy and Disorder

- Energy and matter tend to disperse in a disorderly fashion.

Entropy, S is defined as a measure of disorder.

The second law of thermodynamics:

unit: J·K-1

- Entropy is a state function; the change in entropy of a system is independent of the path between its initial and final states.

The entropy of an isolated system increases in any

spontaneous change.

At constant temperature


Self-Test 8.1A

Calculate the change in entropy of a large block of icewhen 50. J of energy is removed from it as heat at 0 oCin the freezer.

Solution

S =50. J

273.15 K=

__ 0.18 J K-1


8.3 Changes in Entropy8.3 Changes in Entropy

- Thermal disorder: arising from the thermal motion of the molecules- Positional disorder: related to the locations of the molecules

S for a process with changing temperature:

→

(CV if V is constant, CP if P is constant)


S for a reversible, isothermal expansion of an ideal gas

S =qrev

T( wrev)

T

nRT ln(V2/V1)

T

= nR ln V2

V1

_=

S

Hence for the isothermal expansion of an ideal gas,

=

or = nR ln P1

P2

S


Self-Test 8.2A

The temperature of 1.00 mol He(g) is increased from25.0 oC to 300. oC at constant volume. What is thechange in entropy of the helium? Assume idealbehavior and use Cv,m = 3/2R.

Solution

25.0 oC = 298 K; 300. oC = 573 K

S = nCv,m ln(T2/T1)

= (1.00 mol) x 3 x (8.3145 J mol-1 K-1)2

ln 573 K298 K

x

= +8.15 J K-1


Self-Test 8.3A

Calculate the change in molar entropy of an ideal gaswhen it is compressed isothermally to one-third itsinitial volume.

Solution

S = nR ln(V2/V1)

= (1.00 mol) x (8.3145 J K-1 mol-1) x ln 0.33

= 9.13 J K-1 mol-1_


Self-Test 8.4A

Calculate the change in entropy when the pressureof 1.50 mol Ne(g) is decreased isothermally from20.00 bar to 5.00 bar. Assume ideal behavior.

Solution

S = nR ln(P1/P2)

= (1.50 mol) x (8.3145 J K-1 mol-1) x ln20.00 bar

5.00 bar

= + 17.3 J K-1


EXAMPLE 8.5 In an experiment, 1.00 mol Ar(g) was compressedsuddenly (and irreversibly) from 5.00 L to 1.00 L by driving in a piston.and in the process its temperature was increased from 20.0 oC to 25.2 oC.What is the change in entropy of the gas?

To solve this problem, we consider tworeversible stages between initial and final states. Then S(irrev) = S(rev 1) +S(rev 2).


8.4 Entropy Changes Accompanying Changes 8.4 Entropy Changes Accompanying Changes in Physical Statein Physical State

- Phase transition: solid → liquid, Tf (fusion or melting point)liquid → solid, Tb (boiling point)

- At the transition temperature (such as Tb),

The temperature remains constant as heat is supplied.

The transfer of heat is reversible.

The heat supplied is equal to the enthalpy change due to the

constant pressure (at 1 atm).

Entropy of vaporization, Svap

qrev = Hvap

> 0 in all cases


Some Standard entropies of vaporization at Tb

(Table 8.1)

- Standard entropy of vaporization, Svapo :

Svap at 1 bar


Trouton’s rule: Svapo = ~85 J·K-1mol-1

There is approximately the same increase in positional disorder for most liquids when evaporating.

- Exceptions; water, methanol, ethanol, ··· due to extensive hydrogen bonding in liquid phases

- Standard entropy of fusion, Sfuso

> 0 in all cases


Temperature dependence of Svapo

-To determine the entropy of vaporization of water at 25 oC(not at Tb), we can use an entropy change cycle:

Svap(25 oC) = S1(liq. heating; 25 oC Tb)+ Svap(Tb)+ S2(vap. condensing: Tb 25 oC)

Liquid (Tb = 100 oC) Vapor (Tb)

Liquid (25 oC) Vapor (25 oC)

S1 = nCp,m(liq.) ln S2 = nCp,m(vap.) ln

Svap(Tb) = Hvap/Tb

Svap(25 oC)o

373.15 K

298.15 K 373.15 K

298.15 K


Calculate the standard entropy of vaporization of water at 85 oC, given

that its standard entropy of vaporization at 100 oC is 109.0 J·K-1·mol-1 and

the molar heat capacities at constant pressure of liquid water and water

vapor are 75.3 J·K-1·mol-1 and 33.6 J·K-1·mol-1, respectively, in this range.

297s

Exercise 8.43

Solution


8.5 A Molecular Interpretation of Entropy8.5 A Molecular Interpretation of Entropy

The third law of thermodynamics

S → 0 as T → 0

Boltzmann formula

- statistical entropy: S = kB ln W kB = 1.381 × 10-23 J·K-1 = R/NA

- W : the number of microstates the number of ways that the atoms or molecules in the sample can be arranged and yet still give rise to the same total energy

- When we measure the bulk properties of a system, we are measuring an average taken over the many microstates (ensemble) that the system has occupied during the measurement.


EXAMPLE 8.7

Calculate the entropy of a tiny solid made up of four diatomic molecules

of a compound such as carbon monoxide, CO, at T = 0 when (a) the four

molecules have formed a perfectly ordered crystal in which all molecules

are aligned with their C atoms on the left and (b) the four molecules lie

in random orientations, but parallel.

(a) 4 CO molecules perfectly ordered:

(b) 4 CO in random, but parallel:

(c) 1 mol CO in random, but parallel:


- Residual entropy at T = 0, arising from positional disorder

4.6 J·K-1 for the entropy of 1 mol CO < 5.76 J·K-1

Nearly random arrangement due to a small electric dipole moment

- Solid HCl; S ~ 0 at T = 0 due to the bigger dipole momentleading strict head-to-tail arrangement


EXAMPLE 8.8

The entropy of 1.00 mol FClO3(s) at T = 0 is 10.1 J·K-1. Interpret it.

4 orientations possible

nearly random arrangement


8.6 The Equivalence of Statistical and 8.6 The Equivalence of Statistical and Thermodynamic EntropiesThermodynamic Entropies

Thermodynamic entropy

S = qrev/T

behavior of bulk matter

Statistical entropy

S = k ln W

behavior of molecules=

- Consider a one-dimensional box, for the statistical entropy,

• At T = 0, only the lowest energy level occupied → W = 1 and S = 0

• At T > 0, W > 1 and S > 0

• When the box length is increased at constant T, the molecules are distributed across more levels. → W and S increase.


- W = constant × V

- For N molecules,

- The change when a sample expands isothermally from V1 to V2 is,

= nR lnV2

V1

- By raising the temperature,

The molecules have access to larger number ofenergy levels → W and S increase.


- The equations used to calculate changes in the statistical entropy and the thermal entropy lead to the same result.

1. Both are state functions.

3. Both increase in a spontaneous change.

2 × no. of molecules = entropy changes from k ln W to 2k ln W

Number of microstates depends only on its current state.

2. Both are extensive (dependent on “extent”) properties.

4. Both increase with temperature.

In any irreversible change, the overall disorder increases→ no. of microstates increases.

When T increases, more microstates become accessible.


If SO2F2 adopts a disordered arrangement in its crystal form, what

would its residual molar entropy be?

302s

Exercise 8.27

Solution


8.29 Which substance in each of the following pairs has the

higher molar entropy at 298 K: (a) HBr(g) or HF(g); (b) NH3(g)

or Ne(g); (c) I2(s) or I2(l); (d) 1.0 mol Ar(g) at 1.00 atm or 1.0

mol Ar(g) at 2.00 atm?

302sExercise 8.29

Solution


List the following substances in order of increasing molar

entropy at 298 K: H2O(l), H2O(g), H2O(s), C(s, diamond). Explain

your reasoning.

302sExercise 8.31

Solution


Which substance in each of the following pairs would you expect to

have the higher standard molar entropy at 298 K? Explain your

reasoning: (a) Iodine vapor or bromine vapor;

(b) the two liquids cyclopentane and 1-pentene;

(c) ethene (ethylene) or an equivalent mass of polyethylene, a

substance formed by the polymerization of ethylene.

302sExercise 8.33

Solution


Without performing any calculations, predict whether there is an

Increase or a decrease in entropy for each of the following

processes:

(a)Cl2(g) + H2O(l) → HCl(aq) + HClO(aq);

(b)Cu3(PO4)2(s) → 3 Cu2+(aq) + 2 PO43-(aq);

(c)SO2(g) + Br2(g) + 2 H2O(l) → H2SO4(aq) + 2 HBr(aq).

302sExercise 8.35

Solution


8.7 Standard Molar Entropies8.7 Standard Molar Entropies

For heating at constant P,

CP and CP/T → 0as T → 0

Molar entropy, S(T), can be determined from measurement of Cp at different temperatures.


Standard molar entropy, Smo is the molar entropy of the

pure substance at 1 bar.


- Diamond (2.4 J·K-1) vs. lead (64.8 J·K-1):

rigid bonds vs. vibrational energy levels

- H2 (130.7 J·K-1) vs. N2 (191.6 J·K-1):

the greater the mass, the closer energy levels

light heavy

- CaCO3 (92.9 J·K-1) vs. CaO (39.8 J·K-1):

large, complex vs. smaller, simpler

- In general, Smo: gases >> liquids > solids

Related to freedom of movement and disordered state

Standard molar entropy, Smo :


8.8 Standard Reaction Entropies8.8 Standard Reaction Entropies

Standard reaction entropy, So, is the difference

between the standard molar entropies of the products

and those of the reactants, taking into account their

stoichiometric coefficients.


EXAMPLE 8.9 Calculate So for N2(g) + 3H2(g) → 2NH3(g) at 25 oC.


307s

Use data in Table 8.3 or Appendix 2A to calculate the standard reaction

entropy for each of the following reactions at 25 C. Interpret the sign and

magnitude of the reaction entropy. (a) The formation of 1.00 mol H2O(l)

from the elements in their most stable state at 298 K. (b) The oxidation of

1.00 mol CO(g) to carbon dioxide. (c) The decomposition of 1.00 mol

calcite, CaCO3(s), to carbon dioxide gas and solid calcium oxide. (d) The

decomposition of potassium chlorate: 4 KClO3 (s) 3 KClO4 (s) + KCl (s)

Exercise 8.35


307sSolutions


307s


Chapter 8. THERMODYNAMICS: THE SECOND AND THIRD LAW

2012 General Chemistry I

GLOBAL CHANGES IN ENTROPY8.9 The Surroundings8.10 The Overall Change in Entropy8.11 Equilibrium

GIBBS-FREE ENERGY

8.12 Focusing on the System8.13 Gibbs Free Energy of Reaction8.14 The Gibbs Free Energy and Nonexpansion Work8.15 The Effect of Temperature8.16 Impact on Biology: Gibbs Free Energy Changes in Biological Systems


GLOBAL CHANGES IN ENTROPY(Sections 8.9-8.11)

8.9 The Surroundings8.9 The Surroundings

- The second law refers to an isolated system (system + surroundings = universe).

- Only if the total entropy change is positive will theprocess be spontaneous.


Sometimes Ssurr can be difficult to compute, but in generalit can be obtained from the enthalpy change for the process(that is, of the system).


Self-Test 8.14A

Calculate the entropy change of the surroundingswhen 1.00 mol H2O(l) vaporizes at 90 oC and 1 bar.Take the enthalpy of vaporization of water as +40.7kJ mol-1.

Solution

Ssurr = H/T

=

_

_ ( +40 700 J mol-1)

90 oC = 363 K

363 K

= _ 112 J K-1


8.10 The Overall Change in Entropy8.10 The Overall Change in Entropy

- To use the entropy to

judge the direction of

spontaneous change, we

must consider the

change in the entropy of

the system plus the

entropy change in the

surroundings:

-Spontaneous exothermic

(H<0) reactions:


EXAMPLE 8.11 Is the reaction spontaneous at 298 K?

2 Mg(s) + O2(g) → 2 MgO(s) So = -217 J·K-1 Ho = -1202 kJ

The reaction is spontaneous


-Spontaneous endothermic (H>0) reactions:

There can still be an overall increase in entropyif the disorder of the system increases enough.

Summary


- A process produces maximum work if it takes place reversibly.

Clausius inequality

S = > S

- For an isolated system (universe), q = 0

The entropy of an isolated system cannot decrease.

- For two given states of the system,ΔS is a state

function (path-independent) but ΔStot is not.

(See EXAMPLE 8.12)


EXAMPLE 8.12

Calculate S, Ssurr, and Stot for (a) the isothermal, reversible expansion and (b) the isothermal, free expansion of 1.00 mol of ideal gas molecules from 8.00 L to 20.00 L at 292 K. Explain any differences between the two paths.

(a) Isothermal reversible expansion at 292 K


(b) Isothermal free expansion 292 K


312s

Initially a sample of ideal gas at 323 K occupies 1.67 L at 4.95 atm.

The gas is allowed to expand to 7.33 L by two pathways:

(a) isothermal, reversible expansion (b) isothermal, irreversible free

expansion.

Calculate ΔStot, ΔS, ΔSsurr for these pathways.

Exercise 8.49


8.11 Equilibrium8.11 Equilibrium

Dynamic equilibrium is one where there is no net tendency to change but microscopic forward and reverse processes occur at matching rates.

Thermal equilibrium: no net flow of energy as heat

Mechanical equilibrium: no tendency to expand or contract

Chemical equilibrium: no net change in composition

at thermodynamic equilibrium


GIBBS FREE ENERGY(Sections 8.12-8.16)

8.12 Focusing on the System8.12 Focusing on the System

- at constant T and P

Gibbs free energy, G G = H - TS

- Change in Gibbs free energy

at constant T and P-

The direction of spontaneous change is the direction ofdecreasing Gibbs free energy.


Gsys < 0 Spontaneous, irreversible

Gsys = 0 Reversible

Gsys > 0 Nonspontaneous

at constant T and P

- the condition for equilibrium, Stot = 0, and

G = 0 at constant T and P


EXAMPLE 8.13

Calculate the change in molar Gibbs free energy, Gm, for the processH2O(s) → H2O(l) at 1 atm and (a) 10 oC; (b) 0 oC. Decide for eachtemperature whether melting is spontaneous or not. Treat Hfus andSfus as independent of temperature.

(a) At 10 oC,

= -0.22 kJ·mol-1 < 0; spontaneous melting


(b) At 0 oC,

equilibrium (reversible)


- G decreases as its T is raised at constant P.

G↓ = H – T↑S ; H and S vary little with T, S > 0

- Decreasing rate of Gm: vapor >> liquid > solid

Sm(vapor) >> Sm(liquid) > Sm(solid)

Variation of G with temperature: phasetransitions

In most cases (opposite), heatingleads to melting, then boiling.


In some cases, at certain pressures,G for the liquid may never be lowerthan those of the other two phases.The liquid phase is unstable and thephase transition is solid to vapor (sublimation).


8.13 Gibbs Free Energy of Reaction8.13 Gibbs Free Energy of Reaction

Gibbs free energy of reaction

Standard Gibbs free energy of reaction (standard state: pure form at 1 bar)

- Go is fixed for a given reaction and temperature.

- G depends on the composition of the reaction mixture and so it varies – and might even change sign as the reaction proceeds.


Standard Gibbs free energy of formation, Gfo , is the

standard Gibbs free energy of reaction per mole for the formation of a compound from its elements in their most stable form.

- For the most stable forms of elements, Gfo = 0

E.g. Gfo(I2, s) = 0; Gf

o(I2, g) > 0

Examples of most stable forms of elements(Table 8.6)


Some Standard Gibbs free Energies of Formation at25 oC (kJ mol-1) (Table 8.7)


EXAMPLE 8.14

Calculate the standard Gibbs free energy of formation of HI(g) at 25 oCfrom its standard molar entropy and standard enthalpy of formation.


- Thermodynamically stable compound;

- Thermodynamically unstable compound;

• Stable and unstable: thermodynamic tendency to

decompose into its elements• Labile, nonlabile, and inert: the rate at which a

thermodynamic tendency to react is realized

but nonlabile or even inert

- Another solution for standard Gibbs free energies of reaction:

Thermodynamic stability and reactivity


EXAMPLE 8.15

Calculate the standard Gibbs free energy of the reaction

4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g)

and decide whether the reaction is spontaneous under standardconditions at 25 oC.

The reaction isspontaneous


8.14 The Gibbs Free Energy and Nonexpansion 8.14 The Gibbs Free Energy and Nonexpansion WorkWork

- The Gibbs free energy is a measure of the energy free to do nonexpansion work.


E.g. Bioenergetics of glucose oxidation

- The maximum nonexpansion work obtainable from 1 mol of glucose is +2879 kJ at 1 bar.

- 180 g of glucose can be used to build 170 (= 2879/17) mole of peptide links.

In practice, only about 10 moles of peptide links can be built.

- If we know the change in Gibbs free energy of a process taking place at constant T and P, then we immediately know how much nonexpansion work it can do.


8.15 The Effect of Temperature8.15 The Effect of Temperature

- The values of Ho and So do not change much with temperature.

- However, Go does depend much on temperature.

Go = Ho - TSo

1) For an exothermic reaction (Ho<0) with So<0 (disorder decrease L-R),

Go<0 at low Tbut it may become >0 at high T.


3) For an endothermic reaction (Ho>0) with So<0,

Go>0 at all Tand the reaction is never spontaneous.

2) For an endothermic reaction (Ho>0) with So>0 (disorder increase L-R),

Go>0 at low Tbut it may become <0 at high T.


4) For an exothermic reaction (Ho<0) with So>0,

Go<0 at all Tand the reaction is always spontaneous.

The Gibbs free energy increases with T for reactions with a negative So and decreases with T for reactions with a positive So.


EXAMPLE 8.16

Estimate T at which it is thermodynamically possible for carbon toreduce iron(III) oxide to iron under standard conditions by theendothermic reaction, 2Fe2O3(s) + 3C(s) 4Fe(s) + 3CO2(g)

, above 565 oC


8.16 Impact on Biology: Gibbs Free Energy 8.16 Impact on Biology: Gibbs Free Energy Changes in Biological SystemsChanges in Biological Systems

- A reaction that produces a lot of entropy can drive another nonspontaneous reaction forward.

- A process may be driven uphill in Gibbs free energy by another reaction that rolls downhill.


Hydrolysis of adenosine triphosphate (ATP) to adenosine

diphosphate (ADP): the reaction used for driving

nonspontaneous biochemical reactions

- The nonspontaneous reaction restoring of ATP from ADP is driven by the food we eat.


The End!


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Chapter 8. THERMODYNAMICS: THE SECOND AND THIRD LAW

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