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Droplet size and boiling point. Estimate the boiling point of a cloud droplet 2 µ m in diameter. A small droplet has in its free energy an additional term due to its surface tension Δ surf G = γA. r = 1 µ m = 10 –6 m. V = (4/3) πr 3 = 4.19 × 10 –18 m 3.
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Droplet size and boiling point Estimate the boiling point of a cloud droplet 2 µm in diameter A small droplet has in its free energy an additional term due to its surface tension ΔsurfG = γA. r = 1 µm = 10–6 m. V = (4/3)πr3 = 4.19 × 10–18 m3. m = Vρ = 4.19 × 10–18 m3 × 958.3 kg m–3 = 4.01 × 10–15 kg n = m/M = 4.01 × 10–15 kg/(0.018015 kg mol–1) = 2.23 × 10–13 mol ΔsurfG = γA = 4πr2γ = 4 × 3.142 × 10–12 m2 × 0.05891 N m–1 = 7.40 × 10–13 J ΔsurfGm = 7.40 × 10–13 J/(2.23 × 10–13 mol) = 3.32 J mol–1. Thus, at the normal boiling point of 372.86 K, the droplet is unstable by 3.32 J/mol. (∂ΔvG/∂T)p = –ΔvS; at 372.86 ΔvS = 109.1 J mol–1 K–1 CHEM 471: Physical Chemistry We need to lower ΔvGm by 3.32 J mol–1 to get back to equilibrium (ΔG = 0) ΔT ~ –ΔvGm/ΔvSm = –3.32 J mol–1/(109.1 J mol–1 K–1) = –0.030 K So the boiling point of the droplet is 372.83 K
Depression of freezing point For the pure solvent, the solid and liquid phases intersect at point d on the graph. At Tf° µs° = µl° The solute reduces the activity of the solvent, and thence the chemical potential. This reduction is the length of line segment ab. µl = µl° + RT ln a CHEM 471: Physical Chemistry ab = µl – µl° = RT ln a The slope of the solid curve is just the negative of the entropy. –Ss° = bc/cd The slope of the liquid curve is likewise: –Sl° = ac/cd
Depression of freezing point But cd is the depression of freezing point –Ss° = bc/cd = bc/ΔT bc= –Ss°ΔT –Sl° = ac/cd = ac/ΔT ac= –Sl°ΔT ab= ac – bc = –Sl°ΔT + Ss°ΔT = RT ln a ΔT(Sl°–Ss°) = –RT ln a ⇒ ΔTΔfS° = –RT ln a CHEM 471: Physical Chemistry But we know the entropy of fusion is just the enthalpy of fusion divided by temperature ΔfH°ΔT/T = –RT ln a ⇒ ΔT = –RT2ln a/ΔfH°
Depression of freezing point If the solution obeys Raoult’s law and is binary, then aA = XA= 1 – XB ln aA = ln (1 – XB) Power series expansion of ln (1 – x) about x = 0 ln (1 – x) = ln 1 – x [1/(1–x)x = 0] + … = –x +... ln aA = ln (1 – XB) ~ –XB XB = [nB/(nA + nB)] ~ [nB/nA] = [nBMA/mA] But nB/mA is the molality mB XB = [nBMA/mA] = MAmB CHEM 471: Physical Chemistry ΔT = –RT2ln a/ΔfH° = –RT2MAmB/ΔfH° This says the freezing point depression is proportion to the molality, with a constant of proportionality kf given by kf= RT2MA/ΔfH° ΔT = – kfmB
Elevation of boiling point kf= RT2MA/ΔfH° For water: Tf= 273.15, MA = 0.018 kg/mol, ΔfH° = 6007 J/mol kf= RT2MA/ΔfH° = (8.31447 J/(mol.K) × 273.152 K2 × 0.018017 kg/mol)/(6007 J/mol) = 1.86 K kg/mol = 1.86 K/molal Exactly the same logic (but a slightly different diagram) is used for elevation of boiling point, but the constants are different For water: Tb= 373.15, MA = 0.018 kg/mol, ΔfH° = 40667 J/mol kb= RT2MA/ΔbH° = (8.31447 J/(mol.K) × 373.152 K2 × 0.018017 kg/mol)/(40667 J/mol) = 0.513 K kg/mol = 0.513 K/molal Note that because the solution boiling point is higher than the pure solvent boiling point, the sign is different: ΔT = kbmB CHEM 471: Physical Chemistry
Example Determination of molal mass: 5g of naphthalene are dissolved in 100 g of benzene, and the freezing point is depressed by 2.00°C. The freezing point depression coefficient kffor benzene is 5.12 K/molal. What is the molecular weight of naphthalene? Answer: the depression of freezing point ΔT = –kf mBso mB= ΔT/kf So mB = (2.00 K)/(5.12 K/molal) = 0.3906 molal. 5 g naphthalene in 100 g of benzene is 50 g/kg solvent, so 50 g = 0.3906 mol This means M = 50 g/(0.3906 mol) = 128 g/mol. CHEM 471: Physical Chemistry
Osmotic pressure Pure solvent on the right, solution on the left μA, right = μA° μA, left = μA° + RT ln aA ΔμA = RT ln aA ln aA ≅ ln XA = ln(1 – XB) ≅ – XB ≅ –nB/nA Difference in chemical potential due to concentration: ΔμA, conc = –RT nB/nA Difference in chemical potential due to pressure: CHEM 471: Physical Chemistry is called the partial molar volume. Assume it is pressure independent, and integrate both sides
Osmotic pressure The chemical potential difference due to concentration balances the chemical potential difference due to osmotic pressure π = ρgh In dilute solution: π= RTnB /V = cRT Example: 1 g/L solution of lysozyme (14300 g mol–1), Molarity = 7.0 × 10–5 M ≡ 7.0 × 10–2 mol m–3 π = (7.0 × 10–2 mol m–3 × 298 K × 8.3145 J mol–1 K–1)= 173.4 Pa Solvent density ρ = 998.3 kg m–3 h = π/ρg = 173.4/(998.3 × 9.81) = 0.018 m = 1.8 cm CHEM 471: Physical Chemistry Caveats (1) measures number average molecular weight (2) dissociation: do experiment near pI
Weight and number average molecular masses Question: Mix 1 g of glucose (M = 180 g/mol) with 100 g of lysozyme (M = 14,313 g/mol). What is the average molecular mass? Answer: it depends. Averaging by the mass: (1 g × 180 g/mol + 100 g × 14,313 g/mol)/101 g = 14,173 g/mol But we can also average by the number of moles. {(1 g/180 g/mol) × 180 g/mol + (100 g /14,313 g/mol) × 14,313 g/mol}/ {(1 g/180 g/mol} + (100 g /14,313 g/mol)} = (1 + 100)g/{{(1 g/180 g/mol} + (100 g /14,313 g/mol)} = 8,052 g/mol CHEM 471: Physical Chemistry Some methods for m.w. determination (e.g. analytical ultracentrifugation) measure mass-averaged molecular masses. Colligative methods measure number averaged molecular masses, and are therefor very vulnerable to low molecular mass impurities.
