THERMOCHEMISTRY Energy Enthalpy Specific Heat Calorimetry Phase Changes.
Chapter 6 Heat capacity, enthalpy, & entropyenthalpy, entropy, and Gibbs free energy, as functions...
Transcript of Chapter 6 Heat capacity, enthalpy, & entropyenthalpy, entropy, and Gibbs free energy, as functions...
Chapter 6
Heat capacity, enthalpy, & entropy
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โข By eq. 2.6 & 2.7
6.1 Introduction
Integration of Eq. (2.7a) between the states (๐๐2,๐๐) and (๐๐1,๐๐) 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 โ 3R
(classical theory: avg. E for 1-D oscillator, ๐๐๐๐= 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 โ a crystal containing n atoms, each of which
behaves as a harmonic oscillator vibrating independently
discrete energy ๐๐๐๐ = ๐๐ + 12โ๐ฃ๐ฃ
โ 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, ๐๐๐๐ = ๐๐ + 12โ๐ฃ๐ฃ & eq. 4.13 Into
6.2 THEORETICAL CALCULATION OF THE HEAT CAPACITY
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Taking
6.2 THEORETICAL CALCULATION OF THE HEAT CAPACITY
where ๐ฅ๐ฅ = ๐๐ โโโ๐๐ ๐๐๐๐, gives
and
in which case (6.4)
โข Differentiation of eq. with respect to temperature at constant volume
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โข Defining โโ๐๐ ๐๐ = ๐๐๐ธ๐ธ : Einstein characteristic temperature
6.2 THEORETICAL CALCULATION OF THE HEAT CAPACITY
๐ถ๐ถ๐๐ โ ๐ ๐ ๐๐๐๐ ๐๐ โ โ๐ถ๐ถ๐๐ โ 0 ๐๐๐๐ ๐๐ โ 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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โข Problem: although the Einstein equation adequately represents actual heat capacities at highertemperatures, the theoretical values approach zero more rapidly than do the actual values.
โข This discrepancy is caused by the fact that the oscillators do not vibrate with a singlefrequency.
โข In a crystal lattice as a harmonic oscillator, energy is expressed as
๐ธ๐ธ๐๐ = โ๐ฃ๐ฃ๐ธ๐ธ2
+ ๐๐โ๐ฃ๐ฃ๐ธ๐ธ (n = 0,1,2,โฆ.)
Einstein assumed that ๐ฃ๐ฃ๐ธ๐ธ is const. for all the same atoms in the oscillator.
โข Debyeโs 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.
6.2 THEORETICAL CALCULATION OF THE HEAT CAPACITY
: the maximum frequency of vibration of an oscillator
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โข Integration Einsteinโs equation in the range, 0 โค ๐ฃ๐ฃ โค ๐ฃ๐ฃ๐๐๐๐๐๐
6.2 THEORETICAL CALCULATION OF THE HEAT CAPACITY
obtained the heat capacity of the solid
which, with x=hฯ /kT, gives (6.6)
โข Defining ๐๐๐ท๐ท = โโ๐๐๐๐๐๐๐๐ ๐๐ = โโ๐๐๐ท๐ท ๐๐ : Debye characteristic T
โข ๐๐๐ท๐ท(Debye frequency)=๐๐๐๐๐๐๐๐ = ๐๐๐ท๐ท๐๐โ
โข Debyeโs equation gives an excellent fit to the experimental data atlower T.
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โข 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 ๐๐3 law for low-temperature heat capacities.
Debyeโs theory: No consideration on the contribution made to the heat capacity by the uptake of
energy by electrons (โ absolute temperature)
โข 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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โข 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
โ ฐ) โH = H T2, P โ H T1, P = โซT1T2 CpdT : โH is the area under a plot of ๐ถ๐ถ๐๐ ๐ฃ๐ฃ๐๐ ๐๐
โ ฑ) A + B = AB chem. rxn or phase change at const. T, P
โH T, P = HAB T, P โ HA T, P โ HB T, P : Hessโฒ law
โH < 0 exothermicโH > 0 endothermic
6.4 ENTHALPY AS A FUNCTION OF TEMPERATURE AND COMPOSITION
(6.1)
(6.8)
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โข Enthalpy changeConsider the change of state
where โ๐ป๐ป(๐๐ โ ๐๐) is the heat required toincrease the temperature of one mole of solidA from ๐๐1 to ๐๐2 at constant pressure.
(๐พ๐พ)
6.4 ENTHALPY AS A FUNCTION OF TEMPERATURE AND COMPOSITION
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orโด
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 298Kโ๐ป๐ป298 = ๐ป๐ป๐๐๐๐ ๐ ๐ ,298 โ ๐ป๐ป๐๐ ๐ ๐ ,298 โ
12๐ป๐ป๐๐2 ๐๐ ,298
= ๐ป๐ป๐๐๐๐ ๐ ๐ ,298 as ๐ป๐ป๐๐ ๐ ๐ ,298 & ๐ป๐ป๐๐2 ๐๐ ,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 โค T โค 600K, where HPb(s) = โซ298T Cp,Pb(s)dT
ac : 298 โค T โค 3000K, where H12O2(g) = 1
2 โซ298T Cp,O2(g)dT ;
โHPbO s ,298K = -219,000 J
de : 298 โค T โค 1159K where HPbO s ,T = 219,000 + โซ298T Cp,PbO(s)dT J
6.4 ENTHALPY AS A FUNCTION OF TEMPERATURE AND COMPOSITION
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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
6.4 ENTHALPY AS A FUNCTION OF TEMPERATURE AND COMPOSITION
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From the data in Table 6.1,
and, thus, from 298 to 600 K (๐๐๐๐,๐๐๐๐)
With T=500K, โH500K = โ217,800 JIn Fig. 6.7a, h: H of 1 mole of ๐๐๐๐(๐๐)at ๐๐๐๐ of 600K and600 to 1200K, given as
6.4 ENTHALPY AS A FUNCTION OF TEMPERATURE AND COMPOSITION
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โข In Fig. 6.7b, ajkl: H of 1 mole of Pb and 1 mole of O2(g), and
hence โHTโฒ is calculated from the cycle
where
6.4 ENTHALPY AS A FUNCTION OF TEMPERATURE AND COMPOSITION
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Thus
This gives โ๐ป๐ป1000 = โ216,700 ๐ฝ๐ฝ at ๐๐โฒ=1000K
6.4 ENTHALPY AS A FUNCTION OF TEMPERATURE AND COMPOSITION
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If the T of interest is higher than the Tm of both themetal and its oxide, then both latent heats ofmelting must be considered.
6.4 ENTHALPY AS A FUNCTION OF TEMPERATURE AND COMPOSITION
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โข 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 Chatelierโs principle.
โข 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 โH for the change is always a positive quantity. ThusโHm is always positive. The general Eq. (6.9) can be obtained as follows:
6.4 ENTHALPY AS A FUNCTION OF TEMPERATURE AND COMPOSITION
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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 Kirchhoffโs Law.
6.4 ENTHALPY AS A FUNCTION OF TEMPERATURE AND COMPOSITION
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โข The 3rd law of thermodynamics: Entropy of homogeneous substance at complete internalequilibrium state is โ0โ 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
6.5 THE DEPENDENCE OF ENTROPY ON TEMPERATURE AND THETHIRD LAW OF THERMODYNAMICS
Why? by differentiating Eq. (5.2) G = H โ TS with respect to T at constant P:
From Eq. (5.12)
thus
dG = -SdT + VdP
โ 0 as T โ 0.Nernst (1906)
T. W. Richards (1902) found experimentally that ฮS โ 0 and ฮCp โ 0 as T โ 0. (Clue for the 3rd law)
โ 0
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6.5 THE DEPENDENCE OF ENTROPY ON TEMPERATURE AND THETHIRD LAW OF THERMODYNAMICS
(i) ฮCp = ฮฃฮฝiCpi โ 0 means that each Cpi โ 0 (solutions)
by Einstein & Debye (T โ 0, Cv โ 0)
(ii) ฮS = ฮฃฮฝiSi โ 0 means that each Si โ 0
i.e., every particles should be at ground state at 0 K, (ฮฉ th = 1)
every particles should be uniform in concentration (ฮฉ conf = 1).
Thus, it should be at internal equilibrium. Plank statement
thus, ฮฉ th = ฮฉ conf = 1
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โข If ( โ๐๐โG๐๐T)P and ( โ๐๐โH
๐๐T)P โ 0 as T โ 0, โS & โCP โ 0 as T โ 0
โข Nernstโs heat theorem states that โfor all reactions involving substances in the condensed
state, ฮS is zero at the absolute zero of temperatureโ
โข Thus, for the general reaction A + B = AB,
โ๐๐ = ๐๐๐ด๐ด๐ด๐ด โ ๐๐๐ด๐ด โ ๐๐๐ด๐ด = 0 ๐๐๐๐ ๐๐ = 0 and if ๐๐๐ด๐ด and ๐๐๐ด๐ด are assigned the value of zero at 0 K,
then the compound AB also has zero entropy at 0 K.
โข The incompleteness of Nernstโs theorem was pointed out by Planck, who stated that โthe
entropy of any homogeneous substance, which is in complete internal equilibrium, may be
taken to be zero at 0 K.โ
6.5 THE DEPENDENCE OF ENTROPY ON TEMPERATURE AND THETHIRD LAW OF THERMODYNAMICS
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โ Glasses
- noncrystalline, supercooled liquids
liquid-like disordered atom arrangements
โ frozen into solid glassy state โ metastable
- ๐๐0 โ 0, depending on degree of atomic order
โก 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.
6.5 THE DEPENDENCE OF ENTROPY ON TEMPERATURE AND THETHIRD LAW OF THERMODYNAMICS
the substance be in complete internal equilibrium:
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โข Even chemically pure elements
- mixtures of isotopes โ entropy of mixing
ex)Cl35 โ Cl37
โฃ Point defects
- entropy of mixing with vacancy
Ex) Solid CO Structure
6.5 THE DEPENDENCE OF ENTROPY ON TEMPERATURE AND THETHIRD LAW OF THERMODYNAMICS
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โข 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 Stirlingโs approximation,
measured value: 4.2 J/mole K : requires complete internal equilibrium
6.5 THE DEPENDENCE OF ENTROPY ON TEMPERATURE AND THETHIRD LAW OF THERMODYNAMICS
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โข The Third Law can be verified by considering a phase transition in an element such as ฮฑ โ ฮฒwhere ฮฑ & ฮฒ 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, ๐๐โ ฃ=0, which requires that
where
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โข 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
โข The measured heat capacities give
6.6 EXPERIMENTAL VERIFICATION OF THE THIRD LAW
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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
When T>๐๐๐๐
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Richardโs rule (generally metal)โ๐ป๐ป๐๐๐๐๐๐ โ โ๐๐๐๐ โ 9.6J/K(FCC) ,
8.3J/K(BCC)
Troutonโs rule (generally metal)-more useful!!โ๐ป๐ป๐๐๐๐๐๐ โ โ๐๐๐๐ โ 88J/K(for both FCC and BCC)
From FCC From BCC
6.6 EXPERIMENTAL VERIFICATION OF THE THIRD LAW
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6.6 EXPERIMENTAL VERIFICATION OF THE THIRD LAW
Because of the similar molar S of the condensedphases Pb and PbO, it is seen that โS for the reaction,
is very nearly equal to โ12๐๐๐๐,๐๐2 at 298K
โS 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)
Maxwellโs equation (5.34) gives ( โ๐๐๐๐๐๐๐๐)๐๐ = โ( โ๐๐๐๐
๐๐๐๐)๐๐
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
6.7 THE INFLUENCE OF PRESSURE ON ENTHALPY AND ENTROPY
(6.14)
For an ideal gas, ๐ผ๐ผ = โ1 ๐๐ an Eq. (6.14) = 0, H of an ideal gas is independent of P.
โข The molar V and ฮฑ of Fe are, respectively, 7.1๐๐๐๐3 and 0.3 ร 10โ4๐พ๐พโ1 .
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.
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(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
Maxwellโs equation (5.34) gives ( โ๐๐๐๐๐๐๐๐)๐๐ = โ( โ๐๐๐๐
๐๐๐๐)๐๐ &
Thus, for the change of state from (P1, T) to (P2, T)
For an ideal gas, as ๐ผ๐ผ=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
โ ฮS = -0.0022 J/K
Al: from 1 to 100 atm at 298K
โ ฮS = -0.007 J/K
- For same ฮS, how much is the temperature change?
Fe โ 0.29K required
Al โ 0.09K required
โด very insignificant effect
6.7 THE INFLUENCE OF PRESSURE ON ENTHALPY AND ENTROPY
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(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)
For condensed phases over small ranges of P, these P dependencies can be ignored.
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