Weight and number average molecular masses

1. 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.

2. Electrochemistry: basic electricity The most fundamental relationship in electricity is Ohm’s law I = V/R I is the current; the SI unit of current is the Ampere, or Coulomb per second I = Q/t, so Coulombs are the SI unit of charge. V is voltage, or electrical potential, similar to the gravitational potential gh The SI units of electrical potential are volts. Moving a mass m against a gravitational potential gh does work w = mgh. Moving a charge Q against an electrical potential V does work w = QV, or ∫VdQ if the voltage changes.Obviously 1 volt = 1 Joule/1 Coulomb 1 Ampere x 1 volt x 1 second = 1 C s–1 x 1 J/C x 1 s = 1 J W (electrical power) = VI : 1 Ampere x 1 volt = 1 J s–1 = 1 Watt First Law including electrical work: ΔU = q + wpV + wE CHEM 471: Physical Chemistry ΔH = ΔU + Δ(pV) = q + wpV + wE + Δ(pV) ΔG = ΔH–TΔS = q + wpV + wE + Δ(pV) –TΔS At constant p: Δ(pV) = pΔV = –wpV At constant T, reversible conditions: q = TΔS ΔGT,p, rev = wE

3. Electrochemistry: capacitance and electrical neutrality The quantity that determines the potential on an object per unit of charge it carries is the capacitance (C): C = Q/V The SI unit of capacitance is the farad (F) 1 F = 1 C/ 1 V. Capacitance (actually, the self capacitance) of a conducting sphere: C = 4πε0r ε0 = 8.854187817620 × 10–12 F m–1 Capacitance of 1 L of electrolyte solution in a round-bottom flask C = 4 x 3.14159 x 8.85418 × 10–12 F m–1 × 0.062 m = 6.9 × 10–12 F To charge that solution to 1 V requires = 6.9 × 10–12 F × 1 V = 6.9 × 10–12 C 6.9 × 10–12 C = (6.9 × 10–12 C)/F mol of electrons or ions ~ 7.2 × 10–17 mol ~ 43,000,000 ions. CHEM 471: Physical Chemistry NaCl solution containing 1 m Cl– and a 1 V potential has (1 + 7.2 × 10–17)m Na+ = 1.000000000000000072m Na+ Even highly charged solutions are electrically neutral for chemical purposes!

4. The Faraday Let’s say we have a chemical process that involves 1 mole of electrons (e.g. 1 mol Fe2+ -> 1 mol Fe3+ + 1 mol e– ) How much charge is involved? Charge on 1 electron is e = –1.60217649 × 10–19 C A mole is NA particles: NA= 6.0221418 × 1023 mol–1 q = NAe = 6.0221418 × 1023 mol–1 × –1.60217649 × 10–19 C = 96485.3 C mol–1 = 1 F (Faraday) ΔGE = wE= qV = –nFV ΔGE,m = –FV for a 1 electron/mole process CHEM 471: Physical Chemistry ΔGE,m = –NFV for a N electron/mole process

5. Electrochemical potentials We derived the equivalence between Gibbs Free Energy, electrical work, and the product qV. Consider an electrochemical cell under reversible conditions, with an external voltage source set to exactly balance the cell potential ΔGE,m,ext = –NFV for the external process ΔGE,m,electrochem. = –NFE for the electrochemical cell V ΔGE,m = –NFE E° CHEM 471: Physical Chemistry

6. The Nernst Equation ΔGE,m = ΔG°E,m + RT ln Q This is the Nernst Equation: it shows the dependence of cell potential on concentration –NFE = –NFE° + RT ln Q E = E° – (RT/NF) ln Q Problem: find the half-cell potential of a zinc cell in equilibrium with 0.01 M ZnSO4: E0 (Zn2+/Zn) = –0.7628 V at 298 K half-cell reaction: Zn2+ + 2e– → Zn: E° = –0.7628 V Q = aZn/aZn2+ ~ 1/cZn2+ = 1/0.01 = 100 CHEM 471: Physical Chemistry = –0.8219 V Useful numbers: RT/F = 0.025680 V at 298 K (RT/F)ln 10 = 0.059130 V at 298 K

7. Chemists’ and biochemists’ standard potential The chemists’ standard potential is defined at pH 0 (aH+ = 1 m) The biochemists’ standard potential is defined at pH 7 (aH+ = 10–7m) Let’s consider a reaction involving N electrons and m protons Yadda + N e– + m H+ → Yaddayadda E° = x V Q = cYaddayadda /[cYadda(cH+)m] E = E° – (RT/NF) ln Q E°′ is just E° with the Q adjusted for pH 7, and everything else at standard concentration (cYaddayadda = cYadda = 1 m) CHEM 471: Physical Chemistry = E° – 0.4145 (m/N) The most common processes are 2 proton, 2 electron, and so the difference between E°′ and E° is –0.4145 V at 298.15 K. See the table for examples

8. CHEM 471: Physical Chemistry