Chapter 6

Chapter 6

Heat capacity, enthalpy, & entropy

• By eq. 2.6 & 2.7

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.

(2.6) (2.7)

(2.6a)

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

(classical theory: avg. E for 1-D oscillator, 𝜀𝜀_𝑖𝑖= 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 𝜀𝜀_𝑖𝑖 = 𝑖𝑖 + ¹₂ ℎ𝑣𝑣

– a system of 3n linear harmonic oscillators

6.2 THEORETICAL CALCULATION OF THE HEAT CAPACITY

The Energy of Einstein crystal

(6.2)

(6.3)

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

6.2 THEORETICAL CALCULATION OF THE HEAT CAPACITY

Taking

6.2 THEORETICAL CALCULATION OF THE HEAT CAPACITY

where 𝑥𝑥 = 𝑒𝑒^{−ℎ𝜈𝜈 𝑘𝑘𝑘𝑘⁄} , gives and

in which case (6.4)

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

• Defining ℎ𝜐𝜐 𝑘𝑘 = 𝜃𝜃⁄ _𝐸𝐸 : Einstein characteristic temperature

6.2 THEORETICAL CALCULATION OF THE HEAT CAPACITY

𝐶𝐶_𝑉𝑉 ≈ 𝑅𝑅 𝑎𝑎𝑎𝑎 𝑇𝑇 → ∞ 𝐶𝐶_𝑉𝑉 ≈ 0 𝑎𝑎𝑎𝑎 𝑇𝑇 → 0

(6.5)

• 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 𝐸𝐸_𝑛𝑛 = ^ℎ𝑣𝑣₂^𝐸𝐸 + 𝑛𝑛ℎ𝑣𝑣_𝐸𝐸 (n = 0,1,2,….)

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

• 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.

6.2 THEORETICAL CALCULATION OF THE HEAT CAPACITY

• Integration Einstein’s equation in the range, 0 ≤ 𝑣𝑣 ≤ 𝑣𝑣_{𝑚𝑚𝑚𝑚𝑚𝑚}

6.2 THEORETICAL CALCULATION OF THE HEAT CAPACITY

obtained the heat capacity of the solid

which, with x=hυ/kT, gives (6.6)

• Defining 𝜃𝜃_𝐷𝐷 = ℎ𝜐𝜐_{𝑚𝑚𝑚𝑚𝑚𝑚}⁄𝑘𝑘 = ℎ𝜐𝜐_𝐷𝐷⁄𝑘𝑘 : Debye characteristic T

• 𝑉𝑉_𝐷𝐷(Debye frequency)=𝑉𝑉_{𝑚𝑚𝑚𝑚𝑚𝑚} = ^{𝜃𝜃𝐷𝐷}_ℎ^�𝑘𝑘

• Debye’s equation gives an excellent fit to the experimental data at

• 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

6.2 THEORETICAL CALCULATION OF THE HEAT CAPACITY

(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 Cv should vary with T as

in which bT is the electronic contribution.

• By experimental measurements,

6.3 THE EMPIRICAL REPRESENTATION OF HEAT CAPACITIES

: Normally fitted

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 = ∫_T^T₁²C_pdT : ∆H is the area under a plot of 𝐶𝐶_𝑃𝑃 𝑣𝑣𝑎𝑎 𝑇𝑇

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

∆H T, P = H_AB T, P − H_A T, P − H_B T, P : Hess′ law

∆H < 0 exothermic

∆H > 0 endothermic

6.4 ENTHALPY AS A FUNCTION OF TEMPERATURE AND COMPOSITION

(6.1)

(6.8)

• Enthalpy change

Consider the change of state

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

(𝒾𝒾)

6.4 ENTHALPY AS A FUNCTION OF TEMPERATURE AND COMPOSITION

∴ or

where

(6.9)

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

6.4 ENTHALPY AS A FUNCTION OF TEMPERATURE AND COMPOSITION

ex) M(s) + 1/2O₂ g = MO s at 298K

∆𝐻𝐻₂₉₈ = 𝐻𝐻𝑀𝑀𝑀𝑀 𝑠𝑠 ,298 − 𝐻𝐻_{𝑀𝑀 𝑠𝑠 ,298} − ¹₂𝐻𝐻_{𝑀𝑀2 𝑔𝑔 ,298}

= 𝐻𝐻𝑀𝑀𝑀𝑀 𝑠𝑠 ,298 as 𝐻𝐻_{𝑀𝑀 𝑠𝑠 ,298} &

𝐻𝐻_{𝑀𝑀2 𝑔𝑔 ,298}=0 by convention

Fig 6.7 : For the oxidation Pb + ¹₂O₂ = PbO with H of ¹₂ mole of O₂ gas , 1mole of Pb_(s) at 298K (=0 by convention)

ab : 298 ≤ T ≤ 600K, where H_Pb(s) = ∫₂₉₈^T C_p,Pb(s)dT

�

ac : 298 ≤ T ≤ 3000K, where H¹

2O₂(g) = ¹

2∫₂₉₈^T C_p,O2(g)dT ;

∆HPbO s ,298K = -219,000 J

de : 298 ≤ T ≤ 1159K where H_{PbO s ,T} = 219,000 + ∫₂₉₈^T C_p,PbO(s)dT J

6.4 ENTHALPY AS A FUNCTION OF TEMPERATURE AND COMPOSITION

With H of ¹

2mole of O_2(g) and 1mole of Pb_(s) at 298K(=0 by convention)

f : H of ¹

2mole of O_2(g) and 1mole of Pb_(s) at T.

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

Thus

where

6.4 ENTHALPY AS A FUNCTION OF TEMPERATURE AND COMPOSITION

From the data in Table 6.1,

and, thus, from 298 to 600 K (𝑇𝑇_{𝑚𝑚,𝑃𝑃𝑃𝑃})

With T=500K, ∆H_500K = −217,800 J

In Fig. 6.7a, h: H of 1 mole of 𝑃𝑃𝑏𝑏_(𝑙𝑙)at 𝑇𝑇_𝑚𝑚 of 600K and

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_2(g), and hence ∆H_T^′ is calculated from the cycle

where

6.4 ENTHALPY AS A FUNCTION OF TEMPERATURE AND COMPOSITION

Thus

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

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.

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_m is always positive. The general Eq. (6.9) can be obtained as follows:

6.4 ENTHALPY AS A FUNCTION OF TEMPERATURE AND COMPOSITION

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.

6.4 ENTHALPY AS A FUNCTION OF TEMPERATURE AND COMPOSITION

• 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,

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

(3.8) At const. P,

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

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 ΔC_p → 0 as T → 0. (Clue for the 3^rd law)

→ 0

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

(i) ΔC_p = Σν_iC_pi → 0 means that each C_pi → 0 (solutions)

by Einstein & Debye (T → 0, C_v → 0) (ii) ΔS = Σν_iSi → 0 means that each Si → 0

thus, Ω = Ω = 1

• If ( ⁄^𝜕𝜕∆G _𝜕𝜕T)_P and (^𝜕𝜕∆H⁄_𝜕𝜕T)_P → 0 as T → 0, ∆S & ∆C_P → 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 = 𝑆𝑆_{𝐴𝐴𝐴𝐴} − 𝑆𝑆_𝐴𝐴 − 𝑆𝑆_𝐴𝐴 = 0 𝑎𝑎𝑎𝑎 𝑇𝑇 = 0 and if 𝑆𝑆_𝐴𝐴 and 𝑆𝑆_𝐴𝐴 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.”

6.5 THE DEPENDENCE OF ENTROPY ON TEMPERATURE AND THE

THIRD LAW OF THERMODYNAMICS

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

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

the substance be in complete internal equilibrium:

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

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,

6.5 THE DEPENDENCE OF ENTROPY ON TEMPERATURE AND THE

THIRD LAW OF THERMODYNAMICS

• The Third Law can be verified by considering a phase transition in an element such as α → β where 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, 𝑆𝑆^Ⅳ=0, which requires that

where

• 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

6.6 EXPERIMENTAL VERIFICATION OF THE THIRD LAW

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

6.6 EXPERIMENTAL VERIFICATION OF THE THIRD LAW

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

Richard’s rule (generally metal) Trouton’s rule (generally metal)-more useful!!

6.6 EXPERIMENTAL VERIFICATION OF THE THIRD LAW

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 −¹₂𝑆𝑆_{𝑘𝑘,𝑀𝑀2} at 298K

∆S is of similar magnitude to that caused by the disappearance of the gas, i.e., of ¹mole of O_2(g)

(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

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

6.7 THE INFLUENCE OF PRESSURE ON ENTHALPY AND ENTROPY

(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 298

(ii) 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

Maxwell’s equation (5.34) gives ( ⁄^{𝜕𝜕𝑆𝑆} _{𝜕𝜕𝑃𝑃})_𝑘𝑘 = −( ⁄^{𝜕𝜕𝑉𝑉} _{𝜕𝜕𝑘𝑘})_𝑃𝑃 &

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

(6.15)

- 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

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

6.7 THE INFLUENCE OF PRESSURE ON ENTHALPY AND ENTROPY

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

(6.17)