Chapter 6. Heat capacity, enthalpy, & entropy

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1 Chapter 6 Heat capacity, enthalpy, & entropy 1

2 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 (TT 2, PP) and (TT 1, PP) gives the difference between the molar enthalpies of the two states as (6.1) 2

3 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, εε ii = kt, E = 3N 0 kt = 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 hvv (6.2) a system of 3n linear harmonic oscillators (due to vibration in the x, y, and z directions) (6.3) The Energy of Einstein crystal 3

4 6.2 THEORETICAL CALCULATION OF THE HEAT CAPACITY Using, εε ii = ii hvv & eq Into 4

5 6.2 THEORETICAL CALCULATION OF THE HEAT CAPACITY Taking hνν where xx = ee kkkk, gives and in which case (6.4) Differentiation of eq. with respect to temperature at constant volume 5

6 6.2 THEORETICAL CALCULATION OF THE HEAT CAPACITY Defining hυυ kk = θθ EE : Einstein characteristic temperature CC VV RR aaaa TT CC VV 0 aaaa TT 0 (6.5) the Einstein equation good at higher T, the theoretical values approach zero more rapidly than do the actual values. 6

7 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 EE nn = hvv EE 2 + nnnvv 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. : the maximum frequency of vibration of an oscillator 7

8 6.2 THEORETICAL CALCULATION OF THE HEAT CAPACITY Integration Einstein s equation in the range, 0 vv vv mmmmmm 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 = θθ DD kk h Debye s equation gives an excellent fit to the experimental data at lower T. 8

9 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 TT 3 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. 9

10 6.3 THE EMPIRICAL REPRESENTATION OF HEAT CAPACITIES By experimental measurements, : Normally fitted 10

11 6.4 ENTHALPY AS A FUNCTION OF TEMPERATURE AND COMPOSITION For a closed system of fixed composition, with a change in T from T 1 to T 2 at the const. P T ⅰ) H = H T 2, P H T 1, P = 2 T1 C p dt (6.1) : H is the area under a plot of CC PP vvvv 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 (6.8) : Hess law H < 0 exothermic H > 0 endothermic 11

12 6.4 ENTHALPY AS A FUNCTION OF TEMPERATURE AND COMPOSITION 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 TT 1 to TT 2 at constant pressure. (ii) 12

13 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 2 g = MO s at 298K HH 298 = HH MMMM ss,298 HH MM ss, HH OO 2 gg,298 = HH MMMM ss,298 as HH MM ss,298 & HH OO2 gg,298=0 by convention 13

14 6.4 ENTHALPY AS A FUNCTION OF TEMPERATURE AND COMPOSITION Fig 6.7 : For the oxidation Pb + 1 O 2 2 = PbO with H of 1 mole of 2 O 2 gas, 1mole of Pb (s) at 298K (=0 by convention) T ab : 298 T 600K, where H Pb(s) = 298 Cp,Pb(s) dt T Cp,O2 (g)dt ; ac : 298 T 3000K, where H1 2 O 2(g) = H PbO s,298k = -219,000 J T de : 298 T 1159K where H PbO s,t = 219, Cp,PbO(s) dt J 14

15 6.4 ENTHALPY AS A FUNCTION OF TEMPERATURE AND COMPOSITION With H of 1 mole of O 2 2(g) and 1mole of Pb (s) 298K(=0 by convention) at f : H of 1 2 mole of O 2(g) and 1mole of Pb (s) at T. g : H of 1mole of PbO (s) at T. Thus where 15

16 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 600 to 1200K, given as 16

17 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 17

18 6.4 ENTHALPY AS A FUNCTION OF TEMPERATURE AND COMPOSITION Thus This gives HH 1000 = 216,700 JJ at TT =1000K 18

19 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

20 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: 20

21 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

22 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, At const. P, (3.8) As T increased, (6.12) the molar S of the system at any T is given by (6.13) 22

23 6.5 THE DEPENDENCE OF ENTROPY ON TEMPERATURE AND THE THIRD LAW OF THERMODYNAMICS T. W. Richards (1902) found experimentally that ΔS 0 and ΔC p 0 as T 0. (Clue for the 3 rd 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

24 6.5 THE DEPENDENCE OF ENTROPY ON TEMPERATURE AND THE THIRD LAW OF THERMODYNAMICS (i) ΔC p = Σν i C pi 0 means that each C pi 0 (solutions) by Einstein & Debye (T 0, C v 0) (ii) ΔS = Σν i S i 0 means that each S i 0 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

25 6.5 THE DEPENDENCE OF ENTROPY ON TEMPERATURE AND THE THIRD LAW OF THERMODYNAMICS 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 BB = 0 aaaa TT = 0 and if SS AA and SS BB 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

26 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 - SS 0 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. 26

27 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 35 Cl 37 4 Point defects - entropy of mixing with vacancy Ex) Solid CO Structure 27

28 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

29 6.6 EXPERIMENTAL VERIFICATION OF THE THIRD LAW 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: For the cycle shown in Fig For the Third Law to be obeyed, SS Ⅳ =0, which requires that where 29

30 6.6 EXPERIMENTAL VERIFICATION OF THE THIRD LAW In Fig 6.11, a monoclinic form which is stable above K and an orthorhombic form which is stable below K The measured heat capacities give 30

31 6.6 EXPERIMENTAL VERIFICATION OF THE THIRD LAW Assigning a value of zero to S 0 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>TT mm 31

32 6.6 EXPERIMENTAL VERIFICATION OF THE THIRD LAW Richard s rule (generally metal) HH mm TTmm SS mm 9.6J/K(FCC), 8.3J/K(BCC) From FCC Trouton s rule (generally metal)-more useful!! SS bb 88J/K(for both FCC and BCC) HH VV TTbb From BCC 32

33 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 1 2 SS TT,OO 2 at 298K S is of similar magnitude to that caused by the disappearance of the gas, i.e., of 1 2 mole of O 2(g) 33

34 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 PP ) TT = ( VV TT ) PP and Thus 34

35 6.7 THE INFLUENCE OF PRESSURE ON ENTHALPY AND ENTROPY The change in molar enthalpy caused by the change in state from (P 1, T) to (P 2, T) is thus (6.14) For an ideal gas, αα = 1 TT an Eq. (6.14) = 0, H of an ideal gas is independent of P. The molar V and α of Fe are, respectively, 7.1cccc 3 and KK 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 to 301 K at 1 atm P. 35

36 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 PP ) TT = ( VV TT ) PP & Thus, for the change of state from (P 1, T) to (P 2, T) (6.15) For an ideal gas, as αα=1/t, Eq. (6.15) simplifies to Same as decreasing temperature 36

37 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 = J/K Al: from 1 to 100 atm at 298K ΔS = J/K - For same ΔS, how much is the temperature change? Fe 0.29K required Al 0.09K required very insignificant effect 37

38 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. 38

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