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Chapter 6 Heat Capacity, Enthalpy, & Entropy

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Chapter 6 Heat Capacity, Enthalpy, & Entropy Chapter 6 Heat capacity, enthalpy, & entropy 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) 2 1 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) - 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 2 ℎ (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 2 ℎ 4 6.2 THEORETICAL CALCULATION OF THE HEAT CAPACITY Taking where = , gives −ℎ⁄ and 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 ℎ⁄ (6.5) 0 0 ≈ → ∞ ≈ → 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,….) ℎ Einstein assumed2 thatℎ 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. : 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 ≤ ≤ obtained the heat capacity of the solid which, with x=hυ/kT, gives (6.6) • Defining = = : Debye characteristic T ℎ⁄ ℎ⁄ • (Debye frequency)= = � 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 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. 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 ∆ 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 ∆to → at constant pressure. 1 2 ( ) 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 ∆298 298 − 298 − 2 298 , =0 by convention 298 298 2 298 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. s 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 ( , ) 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 − 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 ′ ∆1000 − 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. ∆ − − • 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 0 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.
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