Chapter 6 Heat Capacity, Enthalpy, & Entropy

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Chapter 6

1 6.1 Introduction

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.

• By eq. 2.6 & 2.7 (2.6) (2.7) (2.6a) (2.7a)

Integration of Eq. (2.7a) between the states ( , ) and ( , ) gives the difference between the molar enthalpies of the two states as (6.1) 𝑇𝑇2 𝑃𝑃 𝑇𝑇1 𝑃𝑃

2 6.2 THEORETICAL CALCULATION OF THE HEAT CAPACITY

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

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

𝜀𝜀𝑖𝑖 - 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 = + (6.2) 1 𝜀𝜀𝑖𝑖 𝑖𝑖 2 ℎ𝑣𝑣 (6.3) – a system of 3n linear harmonic oscillators The Energy of Einstein crystal (due to vibration in the x, y, and z directions)

3 6.2 THEORETICAL CALCULATION OF THE HEAT CAPACITY

Using, = + & eq. 4.13 Into 1 𝜀𝜀𝑖𝑖 𝑖𝑖 2 ℎ𝑣𝑣

4 6.2 THEORETICAL CALCULATION OF THE HEAT CAPACITY

Taking

where = , gives −ℎ𝜈𝜈⁄𝑘𝑘𝑘𝑘 and 𝑥𝑥 𝑒𝑒

in which case (6.4)

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

5 6.2 THEORETICAL CALCULATION OF THE HEAT CAPACITY

• Defining = : Einstein characteristic temperature

ℎ𝜐𝜐⁄𝑘𝑘 𝜃𝜃𝐸𝐸

(6.5)

0 0 𝐶𝐶𝑉𝑉 ≈ 𝑅𝑅 𝑎𝑎𝑎𝑎 𝑇𝑇 → ∞ 𝐶𝐶𝑉𝑉 ≈ 𝑎𝑎𝑎𝑎 𝑇𝑇 → the Einstein equation good at higher T, the theoretical values approach zero more rapidly than do the actual values.

6 6.2 THEORETICAL CALCULATION OF THE HEAT CAPACITY

• Problem: although the Einstein equation adequately represents actual heat capacities at higher temperatures, 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 single frequency.

• In a crystal lattice as a harmonic oscillator, energy is expressed as = + (n = 0,1,2,….) ℎ𝑣𝑣𝐸𝐸 Einstein𝐸𝐸𝑛𝑛 assumed2 𝑛𝑛that�𝑣𝑣𝐸𝐸 is const. for all the same atoms in the oscillator. 𝑣𝑣𝐸𝐸 • Debye’s assumption (1912) : the range of frequencies of vibration available to the oscillators is the same as that available to the elastic vibrations in a continuous solid.

: the maximum frequency of vibration of an oscillator

7 6.2 THEORETICAL CALCULATION OF THE HEAT CAPACITY

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

𝑚𝑚𝑚𝑚𝑚𝑚 obtained the heat capacity of the solid 𝑣𝑣 𝑣𝑣

which, with x=hυ/kT, gives (6.6)

• Defining = = : Debye characteristic T

𝜃𝜃𝐷𝐷 ℎ𝜐𝜐𝑚𝑚𝑚𝑚𝑚𝑚⁄𝑘𝑘 ℎ𝜐𝜐𝐷𝐷⁄𝑘𝑘 • (Debye frequency)= = 𝜃𝜃𝐷𝐷�𝑘𝑘 𝑉𝑉𝐷𝐷 𝑉𝑉𝑚𝑚𝑚𝑚𝑚𝑚 ℎ • Debye’s equation gives an excellent fit to the experimental data at lower T.

8 6.2 THEORETICAL CALCULATION OF THE HEAT CAPACITY

• 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 law for low-temperature heat capacities. 3 Debye’s theory: No consideration on the contribution𝑇𝑇 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 Cv should vary with T as

in which bT is the electronic contribution.

9 6.3 THE EMPIRICAL REPRESENTATION OF HEAT CAPACITIES

• By experimental measurements,

: Normally fitted

10 6.4 ENTHALPY AS A FUNCTION OF TEMPERATURE AND COMPOSITION

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

1 2 ⅰ) H = H T , P H T , P = C dT (6.1) : H is the area under a plot of T2 ∆ 2 − 1 ∫T1 p ∆ 𝐶𝐶𝑃𝑃 𝑣𝑣𝑣𝑣 𝑇𝑇 ⅱ) A + B = AB chem. rxn or phase change at const. T, P

H T, P = H T, P H T, P H T, P (6.8) : Hess′ law H < 0 exothermic ∆ AB − A − B H > 0 endothermic ∆ ∆

11 6.4 ENTHALPY AS A FUNCTION OF TEMPERATURE AND COMPOSITION

• Enthalpy change Consider the change of state

where ( ) is the heat required to increase the temperature of one mole of solid A from ∆𝐻𝐻to𝑎𝑎 → at𝑑𝑑 constant pressure.

𝑇𝑇1 𝑇𝑇2 ( ) 𝒾𝒾

12 6.4 ENTHALPY AS A FUNCTION OF TEMPERATURE AND COMPOSITION

or (6.9)

∴where

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

ex) M(s) + 1/2O g = MO s at 298K = 2 , , , 1 = , as , & 2 ∆𝐻𝐻298 𝐻𝐻𝑀𝑀𝑀𝑀 𝑠𝑠 298 − 𝐻𝐻𝑀𝑀 𝑠𝑠 298 − 2 𝐻𝐻𝑂𝑂 𝑔𝑔 298 , =0 by convention 𝐻𝐻𝑀𝑀𝑀𝑀 𝑠𝑠 298 𝐻𝐻𝑀𝑀 𝑠𝑠 298 𝐻𝐻𝑂𝑂2 𝑔𝑔 298

13 6.4 ENTHALPY AS A FUNCTION OF TEMPERATURE AND COMPOSITION

Fig 6.7 : For the oxidation Pb + O = PbO with H of mole of 1 1 O gas , 1mole of Pb( ) at 298K (=0 by convention) 2 2 2 2 s ab : 298 T 600K, where H ( ) = C , ( )dT T ≤ ≤ Pb s ∫298 p Pb s ac : 298 T 3000K, where H ( ) = C , ( )dT ; 1 1 T H , = -219p O2,000g J � ≤ ≤ 2O2 g 2 ∫298 ∆ PbO s 298K de : 298 T 1159K where H , = 219,000 + C , ( )dT J T ≤ ≤ PbO s T ∫298 p PbO s

14 6.4 ENTHALPY AS A FUNCTION OF TEMPERATURE AND COMPOSITION

With H of mole of O ( ) and 1mole of Pb( ) at 298K(=0 by 1convention) 2 2 g s

f : H of mole of O ( ) and 1mole of Pb( ) at T. 1 2 2 g s g : H of 1mole of PbO( ) at T.

Thus

where

15 6.4 ENTHALPY AS A FUNCTION OF TEMPERATURE AND COMPOSITION

From the data in Table 6.1,

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

𝑇𝑇𝑚𝑚 𝑃𝑃𝑃𝑃

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

16 6.4 ENTHALPY AS A FUNCTION OF TEMPERATURE AND COMPOSITION

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

hence H is calculated from the cycle 2 g ′ ∆ T

where

17 6.4 ENTHALPY AS A FUNCTION OF TEMPERATURE AND COMPOSITION

Thus

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

18 6.4 ENTHALPY AS A FUNCTION OF TEMPERATURE AND COMPOSITION

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

19 6.4 ENTHALPY AS A FUNCTION OF TEMPERATURE AND COMPOSITION

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

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

20 6.4 ENTHALPY AS A FUNCTION OF TEMPERATURE AND COMPOSITION

Subtraction gives

or (6.10)

and integrating from state 1 to state 2 gives

(6.11)

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

21 6.5 THE DEPENDENCE OF ENTROPY ON TEMPERATURE AND THE THIRD LAW OF THERMODYNAMICS

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

For a closed system undergoing a reversible process,

(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 6.5 THE DEPENDENCE OF ENTROPY ON TEMPERATURE AND THE THIRD LAW OF THERMODYNAMICS

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

Nernst (1906) → 0 as T → 0. Why? by differentiating Eq. (5.2) G = H – TS with respect to T at constant P:

From Eq. (5.12) dG = -SdT + VdP

thus → 0

23 6.5 THE DEPENDENCE OF ENTROPY ON TEMPERATURE AND THE THIRD LAW OF THERMODYNAMICS

(i) ΔC = ΣνiC i → 0 means that each C i → 0 (solutions)

p p by Einsteinp & Debye (T → 0, C → 0)

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

thus, Ω th = Ω conf = 1

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 24 6.5 THE DEPENDENCE OF ENTROPY ON TEMPERATURE AND THE THIRD LAW OF THERMODYNAMICS

• If ( ) and ( ) 0 as T 0, S & C 0 as T 0 Nernst’s𝜕𝜕∆G heat theorem𝜕𝜕∆H states that “for all reactions involving substances in the condensed • ⁄𝜕𝜕T P ⁄𝜕𝜕T P → → ∆ ∆ P → → state, ΔS is zero at the absolute zero of temperature”

• Thus, for the general reaction A + B = AB, = = 0 = 0 and if and are assigned the value of zero at 0 K, then the compound AB also has zero entropy at 0 K. ∆𝑆𝑆 𝑆𝑆𝐴𝐴𝐴𝐴 − 𝑆𝑆𝐴𝐴 − 𝑆𝑆𝐵𝐵 𝑎𝑎𝑎𝑎 𝑇𝑇 𝑆𝑆𝐴𝐴 𝑆𝑆𝐵𝐵

• 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.”

25 6.5 THE DEPENDENCE OF ENTROPY ON TEMPERATURE AND THE THIRD LAW OF THERMODYNAMICS

the substance be in complete internal equilibrium: ① Glasses - noncrystalline, supercooled liquids liquid-like disordered atom arrangements → frozen into solid glassy state → metastable - ≠0, depending on degree of atomic order

𝑆𝑆0 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.

26 6.5 THE DEPENDENCE OF ENTROPY ON TEMPERATURE AND THE THIRD LAW OF THERMODYNAMICS

③ Even chemically pure elements - mixtures of isotopes → entropy of mixing ex)Cl Cl 35 37 − ④ Point defects - entropy of mixing with vacancy Ex) Solid CO Structure

27 6.5 THE DEPENDENCE OF ENTROPY ON TEMPERATURE AND THE THIRD LAW OF THERMODYNAMICS

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

using Stirling’s approximation,

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

28 6.6 EXPERIMENTAL VERIFICATION OF THE THIRD LAW

• The Third Law can be verified by considering a phase transition in an element such as where & are allotropes of the element and this for the case of sulfur: α → β α β For the cycle shown in Fig. 6.11

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

𝑆𝑆

where

29 6.6 EXPERIMENTAL VERIFICATION OF THE THIRD LAW

• 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 6.6 EXPERIMENTAL VERIFICATION OF THE THIRD LAW

Assigning a value of zero to S allows the absolute value of the entropy of any material to be determined as 0 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 31 6.6 EXPERIMENTAL VERIFICATION OF THE THIRD LAW

Richard’s rule (generally metal) Trouton’s rule (generally metal)-more useful!! 9.6J/K(FCC) , 88J/K(for both FCC and BCC) 𝑚𝑚 ∆8.3J/K(BCC)𝐻𝐻 𝑉𝑉 𝑇𝑇𝑚𝑚 𝑚𝑚 ∆𝐻𝐻 � ≈ ∆𝑆𝑆 ≈ �𝑇𝑇𝑏𝑏 ≈ ∆𝑆𝑆𝑏𝑏 ≈ From FCC From BCC

32 6.6 EXPERIMENTAL VERIFICATION OF THE THIRD LAW

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

is very nearly equal to , at 298K 1 2 − 2 𝑆𝑆𝑇𝑇 𝑂𝑂

∆S is of similar magnitude to that caused by the

disappearance of the gas, i.e., of mole of O ( ) 1 2 2 g 33 6.7 THE INFLUENCE OF PRESSURE ON ENTHALPY AND ENTROPY

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

(dH =TdS+VdP)

Maxwell’s equation (5.34) gives ( ) = ( ) 𝜕𝜕𝑆𝑆 𝜕𝜕𝑉𝑉 ⁄𝜕𝜕𝑃𝑃 𝑇𝑇 − ⁄𝜕𝜕𝑇𝑇 𝑃𝑃 and Thus

34 6.7 THE INFLUENCE OF PRESSURE ON ENTHALPY AND ENTROPY

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

1 2 (6.14)

For an ideal gas, = an Eq. (6.14) = 0, H of an ideal gas is independent of P. 1 𝛼𝛼 ⁄𝑇𝑇 • The molar V and α of Fe are, respectively, 7.1 and 0.3 × 10 . the P increase on Fe from 1 to 100 atm at 2983 K causes the−H4 to−1increase by 𝑐𝑐𝑐𝑐 𝐾𝐾

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

35 6.7 THE INFLUENCE OF PRESSURE ON ENTHALPY AND ENTROPY

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

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

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

1 2 (6.15)

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

𝛼𝛼 Same as decreasing temperature 36 6.7 THE INFLUENCE OF PRESSURE ON ENTHALPY AND ENTROPY

- 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 6.7 THE INFLUENCE OF PRESSURE ON ENTHALPY AND ENTROPY

(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.