Physics 622. T.R. Lemberger. Jan. 2003

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1 Physics 622. T.R. Lemberger. Jan Connections between thermodynamic quantities: enthalpy, entropy, energy, and microscopic quantities: kinetic and bonding energies. Useful formulas for one mole of ideal gas in the classical high-t limit: n Q [(kt/4π)/(ħ 2 /2M)] 3/2. P = nk B T, n = density of molecules. S = R[ln(n Q /n) + 5/2] + R{ln(2IkT/ħ 2 σ) + 1} + R{ln(kT/ħω 0 ) + 1} + Rln(2S+1). U = 1.5RT {+RT} {+RT} {+ bonding energy, as appropriate). C P = T S/ T P,N = 2.5R {+R} {+R}. µ = -k B Tln(n Q V/N) {k B Tln[kT/(σħ 2 /2I)] - k B Tln(kT/ħω 0 ) - k B Tln(2S+1)}. The terms in curly brackets are present in the classical limit, i.e., when kt is large enough that the arguments of the logarithms in S are much greater than unity. The σ in entropy of rotations is a symmetry parameter.[see Wiberg] σ = 2 for homonuclear diatomic molecules like O 2, H 2, Cl 2, and for H 2 O; σ = 1 for heteronuclear molecules like HCl; and σ = 4 for methane, CH 4. Useful numbers: n Q,H = Å -3 for H atoms (1 amu); 2.76 Å -3 for H 2 (2 amu); 74.5 Å -3 for H 2 O (18 amu); 176 Å -3 for O 2 (32 amu), 583 Å -3 for Cl 2 (71 amu), 4086 for I 2 (260 amu). n 0 = P 0 /k B T 0 = 1/(34.4 Å) 3 = Å 298 K & J/m 3. n Q,H /n 0 = ; ln(n Q,H /n 0 )= 10.6; ln(n Q,Cl2 /n 0 ) = 17.0; ln(n Q,I2 /n 0 ) = Density of water = 1/(3.1 Å) 3 = Å -3. RT 0 = 2.48 kj at T = 298 K; N A 1eV = 96.5 kj.

2 Rules of Thumb: 1. Energies and enthalpies are generally dominated by bond energies at room temperature. Molecular bond energies are typically a few electron volts; kinetic energies and work done due to volume changes are both typically a few fortieths of an ev. 2. Bond energy associated with dissolving a gas, e.g., oxygen, in water is typically a few tenths of an ev. 3. Tabulated enthalpies involve the change in bonding energy in creating a certain molecule from elements in their standard states, not the absolute bonding energy that one might calculate in a quantum mechanics class. 4. Vibrational modes for gas molecules are typically frozen out at room temperature. Exceptions occur when heavy atoms are involved, like in I 2, where each I atom has a mass of 130 amu. 5. Rotational modes are typically classical at room temperature. One must include the symmetry number, σ, properly to get quantitative agreement between ideal gas formulas and tabulated entropies. 6. For gaseous systems, translational entropy is typically 80% of the total entropy, and it can be calculated to high accuracy. Contributions from rotations and spin contribute about 20%.

3 Problem: Deduce the H-H bond energy from thermodynamic data on the dissociation reaction: ½ H 2 H, at 298 K and 1 atm. pressure ( J/m 3 ). 1. Consider one mole of H gas. Schroeder (p. 405) gives thermodynamic data on entropy, heat capacity at fixed pressure, enthalpy of formation, and Gibbs free energy of formation at 298 K and 1 atm. pressure: S = J/K; C P = J/K; f H = kj; f G = kj. We notice that C P = 5R/2, consistent with a monatomic ideal gas. Check the tabulated entropy against the ideal gas formula. The translational contribution is: S trans = J/K [ ] = J/K. The spin-1/2 of the electron in H leads to entropy: S spin = Rln(2) = 5.76 J/K. Spin entropy of the proton is not included, by convention. Total calculated entropy: S calc = J/K, which agrees well with the tabulated value J/K. We conclude that for T = 298 K, H gas is described by: U(T,V,N) = Nε B /2 + 3Nk B T/2. S(T,V,N) = Nk B [ln(n Q,H V/N) + 5/2]. F(T,V,N) = Nε B /2 - Nk B T - Nk B Tln(n Q,H V/N). H(T,V,N) = Nε B /2 + 5Nk B T/2. G(T,V,N) = Nε B /2 - Nk B Tln(n Q,H2 V/N). µ H (T,V,N) = ε B /2 - k B Tln(n Q,H V/N) = ε B /2 - k B Tln(n Q,H /n 0 ) + k B Tln(n/n 0 ). n 0 is the density at 298 K and 1 atm. pressure, and ε B (= ev) is the bonding energy of an H 2 molecule.

4 2. Consider 1 mole of H 2 gas molecules. Schroeder gives: S = J/K; C P = J/K; f H = 0 kj; f G = 0 kj. Enthalpy of formation, f H, and Gibbs free energy of formation, f G, are defined to be zero because H 2 gas is the standard state of hydrogen at STP. We note that C P =28.82 J/K = 7R/2, which suggests that rotations are classical. The interproton distance in H 2 is 0.74 Å, so its moment of inertia is: I = 2M H (0.37 Å) 2 = kg m 2 and kt/(ħ 2 /2I) = 3.38 > 1. Calculating, we find: S = J/K + R[ln(3.38/2) + 1] = J/K. The first term is from translation, the second from rotations, including σ = 2. This entropy is essentially equal to the tabulated value. We conclude that for T near 298 K, H 2 gas is described by: U(T,V,N) = 5Nk B T/2. S(T,V,N)=Nk B [ln(n Q,H2 V/N)+ln(Ik B T/ħ 2 )+7/2]. F(T,V,N) = -Nk B T - Nk B T[ln(n Q,H2 V/N) +ln(ik B T/ħ 2 )]. H(T,V,N) = 7Nk B T/2. G(T,V,N) = -Nk B T[ln(n Q,H2 V/N) +ln(ik B T/ħ 2 )]. µ H2 (T,V,N) = -k B T[ln(n Q,H2 V/N) + ln(k B TI/ħ 2 )]. Of course, N refers to the number of H 2 molecules.

5 3. Consider the reaction in which ½ mole of gaseous H 2 dissociates into 1 mole of H, at 298 K, 1 atm. We expect the change in energy to be: U = 3RT/2 ½ [5RT/2] + [ ½ mole of H 2 ε B = 4.47 ev each] = RT/ kj = kj. We have anticipated the bond energy, 4.47 ev. During the reaction, the gas would expand to twice its initial volume. The work done by the expanding gas would be P V = RT/2 = 1.24 kj. Thus, we expect that the heat needed to fuel this reaction would be the increase in energy plus energy lost via work: U + P V = kj. This is close to the tabulated heat of formation of one mole of H gas, kj. The tabulated Gibbs free energy of formation, G f = U + P V T S, is T times the net entropy change of the universe during the reaction, if it goes to completion. The heat needed to fuel the reaction, f H = U + P V, was withdrawn from a thermal reservoir thereby reducing its entropy by f H/T. In other words, S reservoir = - f H/T. The increase in entropy of the hydrogen system is S = loss of ½ mole of H 2 + gain of 1 mole of H: S = -½ J/K J/K = 49.4 J/K. The decrease in entropy of the reservoir is 218 kj/298 K = 732 J/K. Thus, the reaction will not go to completion at room temperature and pressure.

6 4. How far does the dissociation reaction actually go? The number of hydrogen atoms, N H, increases until µ H2 = 2µ H : -k B T[ln(n Q,H2 V/N H2 ) + ln(k B TI/ħ 2 )] = ε B - 2k B Tln(n Q,H V/N H ). ln(n Q,H2 V/N H2 ) + ln(k B TI/ħ 2 ) = -ε B /k B T + 2ln(n Q,H V/N H ) Rearrange and exponentiate to get: N H 2 /VN H2 = (n Q,H /2 3/2 )( ħ 2 /Ik B T) exp(-ε B /k B T) K(T). We have used the result: n Q,H2 = 2 3/2 n Q,H. Note that K(T) has dimensions of density. Using numbers from above: n Q,H = m -3 ; k B T/(ħ 2 /I) = 1.69; ε B = 4.47 ev; k B T = 1/40 ev, we find: K(298K) (1/2 3/2 1.69) e m (1/1.7) m m -3. Since the reaction takes place at fixed pressure, volume V changes as the reaction proceeds. We must use the ideal gas result: V(T,p,N tot ) = (N H + N H2 )k B T/p. If we start with ½ mole of H 2, then conservation of hydrogen requires: N H2 = ½ N A ½ N H. Finally, our equilibrium condition can be expressed in terms of the number of free hydrogen atoms, N H : N H 2 /(N A 2 - N H 2 ) = K(T)k B T/4p. We can solve for N H 2 : N H 2 = N A 2 /[1 + 4p/K(T)k B T]. The fraction of molecules that dissociate is N H /N A. Putting in numbers, we have 4p/Kk B T = 4/[K(T) 24.2 liters] = , so N H /N A = This means that less than 1 molecule dissociates; in other words, for 1 millisecond in each second, one molecule is dissociated.

7 5. Just to see that we understand the microscopics behind tabulated entropies, consider 1 mole of O 2 gas at T = 298 K, p = 1 atm. Schroeder gives: S = J/K; C P = J/K; f H = 0 kj; f G = 0 kj. Since C P = 7R/2, it seems that rotations are classical. Internuclear distance in O 2 is 1.21 Å, so: I = 2M O (0.6 Å) 2 = kg m 2 ; ħ 2 /2I = mev = 2.11 K, and kt/(ħ 2 /2I) = 141. Also, n Q /n = = A subtlety: O 2 has a net electron spin of unity! Thus: S = R[ln(n Q /n) + 5/2] + R[ln(IkT/ħ 2 ) + 1] + Rln(2S+1) = J/K J/K J/K = J/K. The terms are from translation, rotation (including σ = 2), and spin. Without the spin term, we would have been way off. Apparently vibrations are frozen out at room temperature. 6. Consider 1 mole of heteronuclear diatomic molecules, carbon monoxide, CO, gas, at T = 298 K, p = 1 atm. Schroeder gives: S = J/K; C P = J/K; f H = kj; f G = kj. Since C P = 7R/2, we expect that rotations are classical. The internuclear distance in CO is 1.21 Å; its effective mass for rotations is 16 12/28 amu. Its moment of inertia is: I = 6.86 amu (1.21 Å) 2 = kg m 2, so ħ 2 /2I = mev = 2.41 K, and kt/(ħ 2 /2I) = 124. Also, n Q /n = = Thus: S = R[ln(n Q /n)+5/2] +R[ln(2IkT/ħ 2 ) +1]= J/K J/K= J/K. The first term is from translation, the second from rotations, without the -ln(2).

8 7. Consider the triatomic molecule, H 2 O(gas), at T = 298 K, p = 1 atm. [See Wiberg, p. 223, for his treatment of water.] Schroeder gives: S = J/K; C P = J/K; f H = kj; f G = kj. A nonlinear polyatomic molecule has three principle moments of inertia, and three rotational degrees of freedom. Given that the classical hamiltonian is: L x 2 /2I x + L y 2 /2I y + L z 2 /2I z, the classical partition function involves three gaussian integrals, and results in: Z rot (2kT/ħ 2 ) 3/2 (I x I y I z ) 1/2, hence: S rot Rlog[(2kT/ħ 2 ) 3/2 (I x I y I z ) 1/2 ] + 3R/2. According to Wiberg, the rotational entropy for a nonlinear polyatomic molecule is, in the classical limit: S rot = R{3/2 ln[2(πi 1 I 2 I 3 ) 1/3 kt/σħ 2 ] + 3/2}, where σ is called the symmetry number. Wiberg says that σ is 2 for water, since there is one rotation that exchanges one H with the other. (For methane, σ = 4.) C P should be 8R/2 instead of 7R/2, due to the extra rotational degree of freedom. Since C P = 8R/2, we expect that rotations are classical. The H-O bond distance is about 1 Å, so the principle moments of inertia are all about 2 amu (1 Å) 2 = kg m 2. Wiberg gives (I 1 I 2 I 3 ) 1/3 = (1 2 3) 1/ kg m 2 = kg m 2. Thence, ħ 2 /2(πI 1 I 2 I 3 ) 1/3 = 1.32 mev = 15.3 K, and kt2(πi 1 I 2 I 3 ) 1/3 /ħ 2 = Also, n Q /n = = Thus: S = R[ln(n Q /n) + 5/2] + R{3/2 ln[kt2(πi 1 I 2 I 3 ) 1/3 /2ħ 2 ] + 3/2} = J/K J/K = J/K. Looks like Wiberg s formula is accurate. We are missing just about R/3 of entropy from somewhere, possibly from slight errors in moments of inertia, which change with increasing rotation rate. Creating one mole of gaseous water requires breaking ½ mole of O 2 bonds, at a cost of kj. Combining free oxygen atoms with H 2 molecules releases kj = 491 kj.

9 References. Thermodynamic data can be found in: Introduction to Thermal Physics, by Daniel V. Schroeder, p General Chemistry, by Linus Pauling, App. XV, p. 939, provides data on many elements and compounds; App. VIII, p. 914 provides data on monatomic gases of the elements. CRC Handbook of Chemistry and Physics, Sec. D, pp. D32 et seq. Bond energies can be found in: (note: 1 calorie = J) General Chemistry, by Linus Pauling, App. VIII, p Chemistry, a Study of Matter, by Garrett, Lippincott, and Verhoek, Ch. 6, p CRC Handbook of Chemistry and Physics, Sec. F, pp. F The difference in rotational entropy between homo- and hetero-nuclear diatomic molecules is discussed in: General Chemistry, by Linus Pauling, Sec , p Physical Organic Chemistry, by Wiberg, Sec. 2-3, p Bond lengths can be found in: General Chemistry, by L. Pauling, Sec 6-13, p (Single-bond covalent radii.) CRC Handbook of Chemistry and Physics, Sec. F, pp. F

Physics 213. Practice Final Exam Spring The next two questions pertain to the following situation:

Physics 213. Practice Final Exam Spring The next two questions pertain to the following situation: The next two questions pertain to the following situation: Consider the following two systems: A: three interacting harmonic oscillators with total energy 6ε. B: two interacting harmonic oscillators, with