chapter 6 heat capacity, enthalpy, & entropyenthalpy, entropy, and gibbs free energy, as...

Chapter 6

Heat capacity, enthalpy, & entropy

1

• By eq. 2.6 & 2.7

6.1 Introduction

Integration of Eq. (2.7a) between the states (𝑇𝑇2,𝑃𝑃) and (𝑇𝑇1,𝑃𝑃) gives thedifference between the molar enthalpies of the two states as

In this lecture, we examine the heat capacity as a function of temperature, compute the enthalpy, entropy, and Gibbs free energy, as functions of temperature.

We then begin to assess phase equilibria constructing a phase diagram for a single component (unary) system.

(2.6) (2.7)

(2.6a)

(2.7a)

(6.1)

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- Empirical rule by Dulong and Petit (1819) : Cv ≈ 3R

(classical theory: avg. E for 1-D oscillator, 𝜀𝜀𝑖𝑖= kT, E = 3N0kT = 3RT)

- Calculation of Cv of a solid element as a function of

T by the quantum theory: First calculation by Einstein (1907)

- Einstein crystal – a crystal containing n atoms, each of which

behaves as a harmonic oscillator vibrating independently

discrete energy 𝜀𝜀𝑖𝑖 = 𝑖𝑖 + 12ℎ𝑣𝑣

– a system of 3n linear harmonic oscillators

(due to vibration in the x, y, and z directions)

6.2 THEORETICAL CALCULATION OF THE HEAT CAPACITY

The Energy of Einstein crystal

(6.2)

(6.3)

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Using, 𝜀𝜀𝑖𝑖 = 𝑖𝑖 + 12ℎ𝑣𝑣 & eq. 4.13 Into


4

Taking


where 𝑥𝑥 = 𝑒𝑒 ⁄−ℎ𝜈𝜈 𝑘𝑘𝑘𝑘, gives

and

in which case (6.4)

• Differentiation of eq. with respect to temperature at constant volume

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• Defining ⁄ℎ𝜐𝜐 𝑘𝑘 = 𝜃𝜃𝐸𝐸 : Einstein characteristic temperature


𝐶𝐶𝑉𝑉 ≈ 𝑅𝑅 𝑎𝑎𝑎𝑎 𝑇𝑇 → ∞𝐶𝐶𝑉𝑉 ≈ 0 𝑎𝑎𝑎𝑎 𝑇𝑇 → 0

the Einstein equation good at higher T,

the theoretical values approach zero more rapidly than do the actual values.

(6.5)

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• Problem: although the Einstein equation adequately represents actual heat capacities at highertemperatures, the theoretical values approach zero more rapidly than do the actual values.

• This discrepancy is caused by the fact that the oscillators do not vibrate with a singlefrequency.

• In a crystal lattice as a harmonic oscillator, energy is expressed as

𝐸𝐸𝑛𝑛 = ℎ𝑣𝑣𝐸𝐸2

+ 𝑛𝑛ℎ𝑣𝑣𝐸𝐸 (n = 0,1,2,….)

Einstein assumed that 𝑣𝑣𝐸𝐸 is const. for all the same atoms in the oscillator.

• Debye’s assumption (1912) : the range of frequencies of vibration available to the oscillators isthe same as that available to the elastic vibrations in a continuous solid.


: the maximum frequency of vibration of an oscillator

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• Integration Einstein’s equation in the range, 0 ≤ 𝑣𝑣 ≤ 𝑣𝑣𝑚𝑚𝑚𝑚𝑚𝑚


obtained the heat capacity of the solid

which, with x=hυ/kT, gives (6.6)

• Defining 𝜃𝜃𝐷𝐷 = ⁄ℎ𝜐𝜐𝑚𝑚𝑚𝑚𝑚𝑚 𝑘𝑘 = ⁄ℎ𝜐𝜐𝐷𝐷 𝑘𝑘 : Debye characteristic T

• 𝑉𝑉𝐷𝐷(Debye frequency)=𝑉𝑉𝑚𝑚𝑚𝑚𝑚𝑚 = 𝜃𝜃𝐷𝐷𝑘𝑘ℎ

• Debye’s equation gives an excellent fit to the experimental data atlower T.

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• The value of the integral in Eq. (6.6) from 0 to infinity is 25.98, and thus, for very low temperatures, Eq. (6.6) becomes


(6.7) : Debye 𝑇𝑇3 law for low-temperature heat capacities.

Debye’s theory: No consideration on the contribution made to the heat capacity by the uptake of

energy by electrons (∝ absolute temperature)

• At high T, where the lattice contribution approaches the Dulong and Petit value, the molar Cvshould vary with T as

in which bT is the electronic contribution.

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• By experimental measurements,

6.3 THE EMPIRICAL REPRESENTATION OF HEAT CAPACITIES

: Normally fitted

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For a closed system of fixed composition, with a change in T from T1 to T2 at the const. P

ⅰ) ∆H = H T2, P − H T1, P = ∫T1T2 CpdT : ∆H is the area under a plot of 𝐶𝐶𝑃𝑃 𝑣𝑣𝑎𝑎 𝑇𝑇

ⅱ) A + B = AB chem. rxn or phase change at const. T, P

∆H T, P = HAB T, P − HA T, P − HB T, P : Hess′ law

∆H < 0 exothermic∆H > 0 endothermic

6.4 ENTHALPY AS A FUNCTION OF TEMPERATURE AND COMPOSITION

(6.1)

(6.8)

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• Enthalpy changeConsider the change of state

where ∆𝐻𝐻(𝑎𝑎 → 𝑑𝑑) is the heat required toincrease the temperature of one mole of solidA from 𝑇𝑇1 to 𝑇𝑇2 at constant pressure.

(𝒾𝒾)


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or∴

where

(6.9)

convention assigns the value of zero to H of elements in their stable states at 298 K.


ex) M(s) + 1/2O2 g = MO s at 298K∆𝐻𝐻298 = 𝐻𝐻𝑀𝑀𝑀𝑀 𝑠𝑠 ,298 − 𝐻𝐻𝑀𝑀 𝑠𝑠 ,298 −

12𝐻𝐻𝑀𝑀2 𝑔𝑔 ,298

= 𝐻𝐻𝑀𝑀𝑀𝑀 𝑠𝑠 ,298 as 𝐻𝐻𝑀𝑀 𝑠𝑠 ,298 & 𝐻𝐻𝑀𝑀2 𝑔𝑔 ,298=0 by convention

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Fig 6.7 : For the oxidation Pb + 12

O2 = PbO with H of 12mole of

O2 gas , 1mole of Pb(s) at 298K (=0 by convention)

ab : 298 ≤ T ≤ 600K, where HPb(s) = ∫298T Cp,Pb(s)dT

ac : 298 ≤ T ≤ 3000K, where H12O2(g) = 1

2 ∫298T Cp,O2(g)dT ;

∆HPbO s ,298K = -219,000 J

de : 298 ≤ T ≤ 1159K where HPbO s ,T = 219,000 + ∫298T Cp,PbO(s)dT J


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With H of 12mole of O2(g) and 1mole of Pb(s) at

298K(=0 by convention)

f : H of 12mole of O2(g) and 1mole of Pb(s) at T.

g : H of 1mole of PbO(s) at T.

Thus

where


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From the data in Table 6.1,

and, thus, from 298 to 600 K (𝑇𝑇𝑚𝑚,𝑃𝑃𝑃𝑃)

With T=500K, ∆H500K = −217,800 JIn Fig. 6.7a, h: H of 1 mole of 𝑃𝑃𝑏𝑏(𝑙𝑙)at 𝑇𝑇𝑚𝑚 of 600K and600 to 1200K, given as


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• In Fig. 6.7b, ajkl: H of 1 mole of Pb and 1 mole of O2(g), and

hence ∆HT′ is calculated from the cycle

where


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Thus

This gives ∆𝐻𝐻1000 = −216,700 𝐽𝐽 at 𝑇𝑇′=1000K


18

If the T of interest is higher than the Tm of both themetal and its oxide, then both latent heats ofmelting must be considered.


19

• If the system contains a low-temperature phase in equilibrium with a high-temperature phase atthe equilibrium phase transition temperature then introduction of heat to the system (theexternal influence) would be expected to increase the temperature of the system (the effect) byLe Chatelier’s principle.

• However, the system undergoes an endothermic change, which absorbs the heat introduced atconstant temperature, and hence nullifies the effect of the external influence. The endothermicprocess is the melting of some of the solid. A phase change from a low- to a high-temperaturephase is always endothermic, and hence ∆H for the change is always a positive quantity. Thus∆Hm is always positive. The general Eq. (6.9) can be obtained as follows:


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Subtraction gives

or

and integrating from state 1 to state 2 gives

(6.10)

(6.11)

Equations (6.10) and (6.11) are expressions of Kirchhoff’s Law.


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• The 3rd law of thermodynamics: Entropy of homogeneous substance at complete internalequilibrium state is ‘0’ at 0 K.

For a closed system undergoing a reversible process,

6.5 THE DEPENDENCE OF ENTROPY ON TEMPERATURE AND THETHIRD LAW OF THERMODYNAMICS

(3.8)

At const. P,

As T increased, (6.12)

the molar S of the system at any T is given by

(6.13)22


Why? by differentiating Eq. (5.2) G = H – TS with respect to T at constant P:

From Eq. (5.12)

thus

dG = -SdT + VdP

→ 0 as T → 0.Nernst (1906)

T. W. Richards (1902) found experimentally that ΔS → 0 and ΔCp → 0 as T → 0. (Clue for the 3rd law)

→ 0

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(i) ΔCp = ΣνiCpi → 0 means that each Cpi → 0 (solutions)

by Einstein & Debye (T → 0, Cv → 0)

(ii) ΔS = ΣνiSi → 0 means that each Si → 0

i.e., every particles should be at ground state at 0 K, (Ω th = 1)

every particles should be uniform in concentration (Ω conf = 1).

Thus, it should be at internal equilibrium. Plank statement

thus, Ω th = Ω conf = 1

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• If ( ⁄𝜕𝜕∆G𝜕𝜕T)P and ( ⁄𝜕𝜕∆H

𝜕𝜕T)P → 0 as T → 0, ∆S & ∆CP → 0 as T → 0

• Nernst’s heat theorem states that “for all reactions involving substances in the condensed

state, ΔS is zero at the absolute zero of temperature”

• Thus, for the general reaction A + B = AB,

∆𝑆𝑆 = 𝑆𝑆𝐴𝐴𝐴𝐴 − 𝑆𝑆𝐴𝐴 − 𝑆𝑆𝐴𝐴 = 0 𝑎𝑎𝑎𝑎 𝑇𝑇 = 0 and if 𝑆𝑆𝐴𝐴 and 𝑆𝑆𝐴𝐴 are assigned the value of zero at 0 K,

then the compound AB also has zero entropy at 0 K.

• The incompleteness of Nernst’s theorem was pointed out by Planck, who stated that “the

entropy of any homogeneous substance, which is in complete internal equilibrium, may be

taken to be zero at 0 K.”


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① Glasses

- noncrystalline, supercooled liquids

liquid-like disordered atom arrangements

→ frozen into solid glassy state → metastable

- 𝑆𝑆0 ≠0, depending on degree of atomic order

② Solutions

- mixture of atoms, ions or molecules

- entropy of mixing

- atomic randomness of a mixture determines its degree of order

: complete ordering : every A is coordinated only by B atoms and vice versa

: complete randomness : 50% of the neighbors of every atom are A atoms and 50% are B atoms.


the substance be in complete internal equilibrium:

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③ Even chemically pure elements

- mixtures of isotopes → entropy of mixing

ex)Cl35 − Cl37

④ Point defects

- entropy of mixing with vacancy

Ex) Solid CO Structure


27

• Maximum value if equal numbers of molecules were oriented in opposite directions and randommixing of the two orientations occurred. From Eq. (4.18) the molar configurational entropy of mixingwould be

using Stirling’s approximation,

measured value: 4.2 J/mole K : requires complete internal equilibrium


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• The Third Law can be verified by considering a phase transition in an element such as α → βwhere α & β are allotropes of the element and this for the case of sulfur:

6.6 EXPERIMENTAL VERIFICATION OF THE THIRD LAW

For the cycle shown in Fig. 6.11

For the Third Law to be obeyed, 𝑆𝑆Ⅳ=0, which requires that

where

29

• In Fig 6.11, a monoclinic form which is stable above 368.5 K and an orthorhombic form which is

stable below 368.5 K

• The measured heat capacities give


30

Assigning a value of zero to S0 allows the absolute value of the entropy of any material

to be determined as


and molar entropies are normally tabulated at 298 K, where

With the constant-pressure molar heat capacity of the solid expressed in the form

the molar entropy of the solid at the temperature T is obtained as

When T>𝑇𝑇𝑚𝑚

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Richard’s rule (generally metal)∆𝐻𝐻𝑚𝑚𝑘𝑘𝑚𝑚 ≈ ∆𝑆𝑆𝑚𝑚 ≈ 9.6J/K(FCC) ,

8.3J/K(BCC)

Trouton’s rule (generally metal)-more useful!!∆𝐻𝐻𝑉𝑉𝑘𝑘𝑏𝑏 ≈ ∆𝑆𝑆𝑃𝑃 ≈ 88J/K(for both FCC and BCC)

From FCC From BCC


32


Because of the similar molar S of the condensedphases Pb and PbO, it is seen that ∆S for the reaction,

is very nearly equal to −12𝑆𝑆𝑘𝑘,𝑀𝑀2 at 298K

∆S is of similar magnitude to that caused by thedisappearance of the gas, i.e., of 1

2mole of O2(g)

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(i) For a closed system of fixed composition, with a change of P at const. T,

6.7 THE INFLUENCE OF PRESSURE ON ENTHALPY AND ENTROPY

(dH =TdS+VdP)

Maxwell’s equation (5.34) gives ( ⁄𝜕𝜕𝑆𝑆𝜕𝜕𝑃𝑃)𝑘𝑘 = −( ⁄𝜕𝜕𝑉𝑉

𝜕𝜕𝑘𝑘)𝑃𝑃

and Thus

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The change in molar enthalpy caused by the change in state from (P1, T) to (P2, T) is thus


(6.14)

For an ideal gas, 𝛼𝛼 = ⁄1 𝑘𝑘 an Eq. (6.14) = 0, H of an ideal gas is independent of P.

• The molar V and α of Fe are, respectively, 7.1𝑐𝑐𝑐𝑐3 and 0.3 × 10−4𝐾𝐾−1 .

the P increase on Fe from 1 to 100 atm at 298 K causes the H to increase by

The same increase in molar H would be obtained by heating Fe from 298to 301 K at 1 atm P.

35

(ii) For a closed system of fixed composition, with a change of P at const. T,


Maxwell’s equation (5.34) gives ( ⁄𝜕𝜕𝑆𝑆𝜕𝜕𝑃𝑃)𝑘𝑘 = −( ⁄𝜕𝜕𝑉𝑉

𝜕𝜕𝑘𝑘)𝑃𝑃 &

Thus, for the change of state from (P1, T) to (P2, T)

For an ideal gas, as 𝛼𝛼=1/T, Eq. (6.15) simplifies to

Same as decreasing temperature

(6.15)

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- Solid : An increase in the pressure exerted on Fe

Fe: from 1 to 100 atm at 298K

⇒ ΔS = -0.0022 J/K

Al: from 1 to 100 atm at 298K

⇒ ΔS = -0.007 J/K

- For same ΔS, how much is the temperature change?

Fe → 0.29K required

Al → 0.09K required

∴ very insignificant effect


37

(iii) For a closed system of fixed composition with changes in both P and T,

combination of Eqs. (6.1) and (6.14) gives


(6.16)

and combination of Eqs. (6.12) and (6.15) gives

(6.17)

For condensed phases over small ranges of P, these P dependencies can be ignored.

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chapter 6 heat capacity, enthalpy, & entropyenthalpy, entropy, and gibbs free energy, as...

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