of the factor (n − 1) in the exponent of the above equation, where n can be greater than 103 ; this resulted in explosions in the early days of polymer chemistry. (c) The van’t Hoﬀ relation gives (approximately) ∂ ln K ∂T = Δh◦ kT 2 = (n − 1)Δε0 , kT 2 so the slope of ln K versus T gets steeper as n gets larger. Hence, the formation of longer chains is more sensitive to temperature than that of shorter chains. 6. Hydrogen ionization. A hydrogen atom can ionize in the gas phase: K H −→ H+ + e− . Calculate the equilibrium constant K for temperature T = 5000 K. There is no rotational or vibrational partition function for H, H+ , or e− , but there are spin partition functions: qs = 1 for H+ and qs = 2 for e− . ΔD = −311 kcal mol−1 . Kp = qH+ qe− −Δε0 /kT e qH = kT 2πmH+ kT h2 3/2 3/2 2πme kT e−Δε0 /kT h2 2πm kT 3/2 H h2 = kT = 2πme kT h2 3/2 e−Δε0 /kT (8.21 × 10−5 m3 atm K−1 mol−1 )(5000 K) 6.02 × 1023 molecules mol−1 (2π)(9.11 × 10−31 kg)(1.38 × 10−23 J K−1 )(5000 K) (6.626 × 10−34 J s)2 3/2 = (6.82 × 10−25 m3 atm)(8.53 × 1026 m−3 )(2.54 × 10−14 ) = 1.48 × 10−11 atm. 164 e−311,000/9935 8. Pressure denaturation of proteins. For a typical protein, folding can be regarded as involving two states, native (N ) and denatured (D): K N −→ D. At T = 300 K, Δμ◦ = 10 kcal mol−1 . Applying about 10,000 atm of pressure can denature a protein at T = 300 K. What is the volume change Δv for the unfolding process? We assume here that with the added 10,000 atm pressure, K2 = 1 =⇒ Δμ◦2 = 0. Therefore, following Example 13.8, Δμ◦2 Δμ◦1 RT − − RT (ln K2 − ln K1 ) RT RT =− Δv = vD − vN = − Δp Δp (0 + 10,000) cal mol−1 = − 10,000 atm 1010 ˚ A × 1m 3 8.21 × 10−5 m3 atm K−1 mol−1 2 cal K−1 mol−1 1 mol 6.02 × 1023 molecules 3 = 68 ˚ A molecule−1 . Since a protein may have about 100 amino acids, and each amino acid has a volume of about 100 ˚ A3 , the volume of a protein is about 104 ˚ A3 . Hence, Δv is a small change. 9. Clusters. Sometimes isolated molecules of type A can be in a two-state equilibrium with a cluster of K m monomers, mA −→ Am , where Am represents an m-mer cluster. (a) At equilibrium, what is the relationship between μ1 , the chemical potential of the monomer, and μm , the chemical potential of A in the cluster? (b) Express the equilibrium constant K in terms of the partition functions. (a) Use Equation (13.15) of the text with a = m, b = 0, c = 1 to get μm = mμ1 (b) Equation (13.17) gives K = qm −Δε0 /kT e . q1m 167 Chapter 14 Phase Equilibria 1. Applying the Clausius–Clapeyron equation. (a) The vapor pressure of water is 23 mmHg at T = 300 K and 760 mmHg at T = 373 K. Calculate the enthalpy of vaporization Δhvap . (b) Assuming that each water has z = 4 nearest neighbors, calculate the interaction energy wAA . (a) p2 ln p1 =− Δhvap R =⇒ ln 1 373 760 23 − 1 1 − T2 T1 1 300 1.987 cal mol−1 = −Δhvap =⇒ Δhvap = 10.653 kcal mol−1 (b) Since − zw = Δhvap , 2 Δhvap ≈ −5.3 kcal mol−1 . w = − 2 170 8. Sublimation of graphite. The heat of sublimation of graphite is Δhsub = 716.7 kJ mol−1 . Use this number to estimate the strength of a carbon–carbon bond. Equation (14.22) of the text applies to sublimation as well as to vaporization, so Δhsub = − zwAA . 2 In graphite, carbon is in the sp2 hybridization state, with three bonds and a delocalized π bond. For simplicity, we take z = 4 neighbors, so wAA = − Δhsub = −358.35 kJ mol−1 = −85.6 kcal mol−1 2 The experimental value of a C−C single bond is −342 kJ mol−1 , but it can typically vary from about −300 to −400 kJ mol−1 , depending on neighboring bond substituent eﬀects. 9. Surface tension of mercury. The surface tension of water is 72 erg cm−2 and that of liquid mercury is 487 erg cm−2 . If the water–water attraction is about 5 kcal mol−1 , what is the mercury–mercury attraction? wAA2 γ2 = wAA1 γ1 =⇒ wAAHg 487 = (5 kcal mol−1 ) = 33 kcal mol−1 . 72 10. Squeezing ice. Use the Clapeyron relation to compute the pressure that is required to lower the melting temperature of ice by 10 K. For water, Δh = 6.008 kJ mol−1 and Δv = −1.64 cm3 mol−1 . P2 P1 Δh T2 dT , dP = Δv T1 T P2 = P1 + T2 Δh ln . Δv T1 The melting temperature of water at 1 atm is 273.15 K. 263.15 6.008 × 103 J ln , −6 3 −1.64 × 10 m 273.15 1 atm = 1348 atm = 0.137 kJ cm−3 . P2 = 1.366 × 108 Pa 1.013 × 105 Pa P2 = 1.013 × 105 Pa + 175

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