Physical Equilibria PDF

Physical Equilibria Physical Equilibria • Biological organisms are inhomogenous; different regions of each cell have different concentrations of molecules and different biological functions. Applications • Membranes and Transport Applications • Membranes and Transport • Ligand Binding Applic...

Physical Equilibria Physical Equilibria • Biological organisms are inhomogenous; different regions of each cell have different concentrations of molecules and different biological functions. Applications • Membranes and Transport Applications • Membranes and Transport • Ligand Binding Applications • Membranes and Transport • Ligand Binding • Colligative Properties Phase Equilibria • For living systems to exist they must be out of equilibrium; dynamic processes need to occur to maintain the living state. When an organism dies, it approaches closer to equilibrium. Phase Equilibria • The usefulness of considering equilibrium in connection with living organisms is that it helps define the direction of dynamic processes. • For an open system, spontaneous processes are accompanied by a decrease in chemical potential, ∆𝞵T,P < 0. At equilibrium, ∆𝞵T,P = 0. One-Component Systems The Exact Clapeyron Equation • The master equation that allows us to understand phase changes: 𝒅𝑮 = 𝑽𝒅𝑷 − 𝑺𝒅𝑻 Molar free-energy surface of (a) water and (b) ice as a function of temperature and pressure relative to Gm for H2O at 273 K and 1 bar. One-Component Systems ∆𝝓 𝑮𝒎 = 𝑮𝒎,𝒘𝒂𝒕𝒆𝒓 − 𝑮𝒎.𝒊𝒄𝒆 ∆𝝓 𝑽𝒎 = 𝑽𝒎,𝒘𝒂𝒕𝒆𝒓 − 𝑽𝒎.𝒊𝒄𝒆 < 𝟎 One-Component Systems 𝒅𝑮𝒘𝒂𝒕𝒆𝒓 = 𝑽𝒘𝒂𝒕𝒆𝒓 𝒅𝑷 − 𝑺𝒘𝒂𝒕𝒆𝒓 𝒅𝑻 𝒅𝑮𝒊𝒄𝒆 = 𝑽𝒊𝒄𝒆 𝒅𝑷 − 𝑺𝒊𝒄𝒆 𝒅𝑻 ∆𝝓 𝑮𝒎 = ∆𝝓 𝑽𝒎 𝒅𝑷 − ∆𝝓 𝑺𝒎 𝒅𝑻 𝐆𝐢𝐯𝐞𝐧 ∆𝝓 𝑮𝒎 = 𝟎 ∆𝝓 𝑽𝒎 𝒅𝑷 = ∆𝝓 𝑺𝒎 𝒅𝑻 𝐀𝐭 𝐚 𝐫𝐞𝐯𝐞𝐫𝐬𝐢𝐛𝐥𝐞 𝐩𝐡𝐚𝐬𝐞 𝐭𝐫𝐚𝐧𝐬𝐢𝐭𝐢𝐨𝐧 ∆𝝓 𝑺𝒎 = ∆𝝓 𝑯𝒎 /𝑻𝝓 𝑻𝝓 𝐢𝐬 𝐭𝐡𝐞 𝐞𝐪𝐮𝐢𝐥𝐢𝐛𝐫𝐢𝐮𝐦 𝐩𝐡𝐚𝐬𝐞 𝐭𝐫𝐚𝐧𝐬𝐢𝐭𝐢𝐨𝐧 𝐭𝐞𝐦𝐩𝐞𝐫𝐚𝐭𝐮𝐫𝐞 𝝏𝑻 𝝏𝑷 𝝓 𝑻𝝓 ∆𝝓 𝑽𝒎 = ∆𝝓 𝑯 𝒎 Using the exact Clapeyron equation The enthalpy of melting ice at 1 bar is 6.007 kJ/mol; the density of water at 0°C is 999.9 kg m-3, while that of ice is 915.0 kg m-3. Assuming ∆fusVm and ∆fusHm (the molar volume and enthalpy change on fusion) are constant, determine the freezing point of water at 100 bar. Using the exact Clapeyron equation The enthalpy of melting ice at 1 bar is 6.007 kJ/mol; the density of water at 0°C is 999.9 kg m-3, while that of ice is 915.0 kg m-3. Assuming ∆fusVm and ∆fusHm (the molar volume and enthalpy change on fusion) are constant, determine the freezing point of water at 100 bar. From the densities and the molar mass we can obtain the molar volumes: 𝑴 𝒎𝒐𝒍&𝟏 &𝟓 𝒎𝟑 𝒎𝒐𝒍&𝟏 𝑽𝒎 = ; 𝑽𝒎,𝒊𝒄𝒆 = 𝟎. 𝟎𝟏𝟖𝟎𝟏𝟓 𝒌𝒈 = 𝟏. 𝟗𝟔𝟗 × 𝟏𝟎 𝝆 𝟗𝟏𝟓. 𝟎 𝒌𝒈 𝒎&𝟑 𝒎𝒐𝒍&𝟏 𝑽𝒎,𝒘𝒂𝒕𝒆𝒓 = 𝟎. 𝟎𝟏𝟖𝟎𝟏𝟓 𝒌𝒈 = 𝟏. 𝟖𝟎𝟐 × 𝟏𝟎&𝟓 𝒎𝟑 𝒎𝒐𝒍&𝟏 &𝟑 𝟗𝟗𝟗. 𝟗 𝒌𝒈 𝒎 ∆𝒇𝒖𝒔 = 𝑽𝒎,𝒘𝒂𝒕𝒆𝒓 − 𝑽𝒎,𝒊𝒄𝒆 = −𝟏. 𝟔𝟕 × 𝟏𝟎&𝟔 𝒎𝟑 𝒎𝒐𝒍&𝟏 𝝏𝑻 𝝏𝑷 𝝏𝑻 𝝏𝑷 𝒇𝒖𝒔 𝒇𝒖𝒔 𝑻𝒇𝒖𝒔 ∆𝒇𝒖𝒔 𝑽𝒎 𝟐𝟕𝟑. 𝟏𝟓𝑲 × −𝟏. 𝟔𝟕×𝟏𝟎/𝟔 𝒎𝟑 𝒎𝒐𝒍/𝟏 = = ∆𝒇𝒖𝒔 𝑯𝒎 𝟔𝟎𝟎𝟕 𝑱 𝒎𝒐𝒍/𝟏 = −𝟕. 𝟔𝟎 × 𝟏𝟎/𝟖 𝑲 𝑷𝒂/𝟏 Using the exact Clapeyron equation The enthalpy of melting ice at 1 bar is 6.007 kJ/mol; the density of water at 0°C is 999.9 kg m-3, while that of ice is 915.0 kg m-3. Assuming ∆fusVm and ∆fusHm (the molar volume and enthalpy change on fusion) are constant, determine the freezing point of water at 100 bar. From the densities and the molar mass we can obtain the molar volumes: 𝝏𝑻 𝝏𝑷 𝒇𝒖𝒔 = −𝟕. 𝟔𝟎 × 𝟏𝟎/𝟖 𝑲 𝑷𝒂/𝟏 Assumption: ∆𝑻 obtain ∆𝑷 𝒇𝒖𝒔 𝝏𝑻 ≅ 𝝏𝑷 ; since ∆P = 99 bar = 9.9 x 106 Pa, we 𝒇𝒖𝒔 ∆𝐓 = 𝟗. 𝟗 × 𝟏𝟎𝟔 𝑷𝒂 × −𝟕. 𝟔𝟎×𝟏𝟎/𝟖 𝑲 𝑷𝒂/𝟏 = −𝟎. 𝟕𝟓 𝑲 So the 100 bar freezing point is -0.75 K or 272.40°C. The Gibbs Phase Rule All pure substances have an equation of state that, given a particular T and P, defines a Vm. They thus have three variables of state, and one constraint, leading to 2 degrees of freedom. If two phases are present at equilibrium, there is an additional constraint: ∆𝜙Gm=0. This leaves only a single degree of freedom. 𝑭=𝑪−𝑷+𝟐 F: number of degrees of freedom C: number of components P: number of phases The Gibbs Phase Rule F: number of degrees of freedom C: number of components P: number of phases Vapor-Liquid Equilibria and the Clausius-Clapeyron Equation The pressure dependence of the vapor-liquid equilibrium temperature has some special properties. At pressures close to ambient, three simplifying assumptions can be made: • ∆𝒗𝒂𝒑𝑽 = 𝑽𝒎,𝒈𝒂𝒔 − 𝑽𝒎,𝒍𝒊𝒒𝒖𝒊𝒅 ~ 𝑽𝒎,𝒈𝒂𝒔 • 𝑽𝒎,𝒈𝒂𝒔 = 𝑹𝑻/𝑷 • Enthalpy of vaporization is independent of pressure and temperature. 𝑻𝝓 ∆𝝓 𝑽𝒎 = ∆𝝓 𝑯 𝒎 𝝓 𝑻𝒗𝒂𝒑 𝑹𝑻/𝑷 𝝏𝑻 𝑹𝑻𝟐 = = 𝝏𝑷 𝒗𝒂𝒑 ∆𝒗𝒂𝒑 𝑯𝒎 𝑷∆𝒗𝒂𝒑 𝑯𝒎 𝒅𝑷𝒗𝒂𝒑 ∆𝒗𝒂𝒑 𝑯𝒎 𝒅𝑻 = 𝑷𝒗𝒂𝒑 𝑹 𝑻𝟐 ∆𝒗𝒂𝒑 𝑯𝒎 𝐥𝐧 𝑷𝒗𝒂𝒑 = − +𝑪 𝑹𝑻 𝝏𝑻 𝝏𝑷 Vapor-Liquid Equilibria and the Clausius-Clapeyron Equation At Pvap = 1 bar pressure, ln Pvap = 0, and so, since the vaporization is under reversible conditions ∆𝒗𝒂𝒑 𝑯𝒎 ∆𝒗𝒂𝒑 𝑺𝒎 𝑪= = 𝑹𝑻 𝑹 ∆𝒗𝒂𝒑 𝑯𝒎 ∆𝒗𝒂𝒑 𝑺𝒎 𝐥𝐧 𝑷𝒗𝒂𝒑 = − + 𝑹𝑻 𝑹 Plot of Clausius-Clapeyron plot for the water/water vapor equilibrium. The fit gives an intercept ∆vapSm/R = 13.27 and a slope −∆vapHm/R = 4968 K. Vapor-Liquid Equilibria and the Clausius-Clapeyron Equation At Pvap = 1 bar pressure, ln Pvap = 0, and so, since the vaporization is under reversible conditions ∆𝒗𝒂𝒑 𝑯𝒎 ∆𝒗𝒂𝒑 𝑺𝒎 𝑪= = 𝑹𝑻 𝑹 ∆𝒗𝒂𝒑 𝑯𝒎 ∆𝒗𝒂𝒑 𝑺𝒎 𝐥𝐧 𝑷𝒗𝒂𝒑 = − + 𝑹𝑻 𝑹 𝑷𝒗𝒂𝒑,𝟐 ∆𝒗𝒂𝒑 𝑯𝒎 𝟏 𝟏 𝐥𝐧 =− − 𝑷𝒗𝒂𝒑,𝟏 𝑹 𝑻𝟐 𝑻𝟏 Clausius-Clapeyron Equation; Inexact Clapeyron Equation Vapor-Liquid Equilibria and the Clausius-Clapeyron Equation We can make a minor modification to the Clausius-Clapeyron equation that substantially increases its accuracy. ∆𝒗𝒂𝒑 𝑯 = ∆𝒗𝒂𝒑 𝑯𝒐 − ∆𝒗𝒂𝒑 𝑪𝒐𝑷 𝑻 − 𝑻𝒐 ∆𝒗𝒂𝒑𝑪𝒐𝑷 = 𝑪𝒐𝑷 𝒈𝒂𝒔 − 𝑪𝒐𝑷 𝒍𝒊𝒒𝒖𝒊𝒅 ; 𝑻𝒐𝒊𝒔 𝒔𝒐𝒎𝒆 𝒔𝒕𝒂𝒏𝒅𝒂𝒓𝒅 𝒕𝒆𝒎𝒑𝒆𝒓𝒂𝒕𝒖𝒓𝒆 𝒘𝒉𝒆𝒓𝒆∆𝒗𝒂𝒑𝑪𝒐𝑷 𝒂𝒏𝒅 ∆𝒗𝒂𝒑𝑯𝒐 𝒂𝒓𝒆 𝒌𝒏𝒐𝒘𝒏 𝒅𝑷𝒗𝒂𝒑 ∆𝒗𝒂𝒑 𝑯𝒎 𝒅𝑻 = 𝑷𝒗𝒂𝒑 𝑹 𝑻𝟐 𝒅𝑷𝒗𝒂𝒑 ∆𝒗𝒂𝒑 𝑯𝒐 − ∆𝒗𝒂𝒑 𝑪𝒐𝑷 𝑻𝒐 𝒅𝑻 ∆𝒗𝒂𝒑 𝑪𝒐𝑷 𝒅𝑻 = + 𝟐 𝑷𝒗𝒂𝒑 𝑹𝑻 𝑹𝑻 𝑷𝒗𝒂𝒑 ∆𝒗𝒂𝒑 𝑯𝒐 − ∆𝒗𝒂𝒑 𝑪𝒐𝑷 𝑻𝒐 𝐥𝐧 𝒐 =− 𝑷𝒗𝒂𝒑 𝑹 ∆𝒗𝒂𝒑 𝑪𝒐𝑷 𝟏 𝟏 𝑻 − 𝒐 + 𝐥𝐧 𝒐 𝑻 𝑻 𝑹 𝑻 Vapor-Liquid Equilibria and the Clausius-Clapeyron Equation 𝑷𝒗𝒂𝒑 ∆𝒗𝒂𝒑 𝑯𝒐 − ∆𝒗𝒂𝒑 𝑪𝒐𝑷 𝑻𝒐 𝐥𝐧 𝒐 =− 𝑷𝒗𝒂𝒑 𝑹 ∆𝒗𝒂𝒑 𝑪𝒐𝑷 𝟏 𝟏 𝑻 − 𝒐 + 𝐥𝐧 𝒐 𝑻 𝑻 𝑹 𝑻 Vapor pressure of water as a function of temperature. Blue line is a fit assuming constant enthalpy of vaporization; gray line is a fit assuming linear dependence of temperature. Solutions of Two or More Components Let’s consider a 2-component system consisting gasoline and water. 𝒅𝑮 = 𝝁𝑨,𝟏 𝒅𝒏𝑨,𝟏 + 𝝁𝑩,𝟏 𝒅𝒏𝑩,𝟏 + 𝝁𝑨,𝟐 𝒅𝒏𝑨,𝟐 + 𝝁𝑩,𝟐 𝒅𝒏𝑩,𝟐 Assume equilibrium; number of moles of B is constant in both phases 𝒅𝑮 = 𝝁𝑨,𝟏 𝒅𝒏𝑨,𝟏 + 𝝁𝑨,𝟐 𝒅𝒏𝑨,𝟐 Allow A to change only by moving from phase 1 to phase 2 𝒅𝑮 = 𝝁𝑨,𝟏 𝒅𝒏𝑨,𝟏 + 𝝁𝑨,𝟐 𝒅𝒏𝑨,𝟏 = 𝝁𝑨,𝟏 − 𝝁𝑨,𝟐 𝒅𝒏𝑨,𝟏 Negative sign means a decrease of free energy 𝝁𝑨,𝟏 − 𝝁𝑨,𝟐 = 𝟎 At equilibrium, not only does the total free energy of each phase have to be identical – the chemical potential of each component must also be identical. Vapor Pressure Raoult’s Law governs the vapor-liquid equilibrium. lim <! →> 𝑷𝒗𝒂𝒑 = 𝝌𝑨 𝒐 𝑷𝒗𝒂𝒑 𝑷𝒗𝒂𝒑: Vapor pressure of the solvent in the solution 𝑷𝒐𝒗𝒂𝒑: Vapor pressure of the pure solvent 𝝌𝑨: Mole fraction of solvent in solution Chemical potential of the vapor, assuming ideal gas 𝝁 = 𝝁𝒐 + 𝑹𝑻 𝐥𝐧 𝑷𝒗𝒂𝒑 𝝁𝒐 : chemical potential of the real gas; standard state of 1 bar 𝑷𝒗𝒂𝒑 𝝁 = 𝝁 + 𝑹𝑻 𝐥𝐧 𝑷𝒐 𝒐 𝝁 = 𝝁𝒐 + 𝑹𝑻 𝐥𝐧 𝝌𝑨 + 𝑹𝑻 𝐥𝐧 𝑷𝒐𝒗𝒂𝒑 Vapor Pressure 𝝁 = 𝝁𝒐 + 𝑹𝑻 𝐥𝐧 𝝌𝑨 + 𝑹𝑻 𝐥𝐧 𝑷𝒐𝒗𝒂𝒑 𝝁𝒐𝑨 = 𝝁𝒐 + 𝑹𝑻 𝐥𝐧 𝑷𝒐𝒗𝒂𝒑 𝝁𝑨 = 𝝁𝒐𝑨 + 𝑹𝑻 𝐥𝐧 𝝌𝑨 𝝁𝑨 = 𝝁𝒐𝑨 𝑷𝒗𝒂𝒑 + 𝑹𝑻 𝐥𝐧 𝒐 𝑷𝒗𝒂𝒑 Henry’s Law 𝑶𝟐 𝒈 ↔ 𝑶𝟐 𝒂𝒒 𝑷𝒗𝒂𝒑,𝑩 = 𝒌𝑩 𝝌𝑩 Henry’s Law 𝑷𝒗𝒂𝒑,𝑩 = 𝒌𝑩 𝝌𝑩 Partition Equilibria Partitioning: the distribution of a solute between two immiscible phases, often liquid phases, but sometimes (as in chromatography) a liquid phase and the surface of a solid. 𝝁𝑩𝟏 = 𝝁𝑩𝟐 𝝁𝑩𝟏 = 𝝁𝒐𝑩𝟏 + 𝑹𝑻 𝐥𝐧 𝒂𝑩𝟏 𝝁𝑩𝟐 = 𝝁𝒐𝑩𝟐 + 𝑹𝑻 𝐥𝐧 𝒂𝑩𝟐 𝝁𝒐𝑩𝟏 + 𝑹𝑻 𝐥𝐧 𝒂𝑩𝟏 = 𝝁𝒐𝑩𝟐 + 𝑹𝑻 𝐥𝐧 𝒂𝑩𝟐 𝑹𝑻 𝐥𝐧 𝒂𝑩𝟐 − 𝐥𝐧 𝒂𝑩𝟏 = − 𝝁𝒐𝑩𝟐 − 𝝁𝒐𝑩𝟏 𝒐 𝒐 𝝁 /𝝁 𝒂𝑩𝟐 / 𝑩𝟐 𝑩𝟏 / 𝑹𝑻 =𝒆 =𝒆 𝒂𝑩𝟏 ∆𝝁𝒐𝑩 𝑹𝑻 Partition coefficient Partition Equilibria CnH2n+1COOH with n > 4 E.g. palmitic acid (hexadecanoic acid, C15H31COOH) Solubility measurements indicate that 𝜇°heptane - 𝜇°water = -38 kJ mol-1 Solubility data for fatty acids with numbers of carbon atoms nC, where nC = 8 (octanoic), nC = 10 (decanoic), nC = 12 (dodecanoic) to nC = 22 (doeicosanoic) can be used to generate the relation: 𝝁𝒐𝒉𝒆𝒑𝒕𝒂𝒏𝒆 − 𝝁𝒐𝒘𝒂𝒕𝒆𝒓 = 𝟏𝟕. 𝟖𝟐 − 𝟑. 𝟒𝟓𝒏𝑪 𝒌𝑱 𝒎𝒐𝒍/𝟏 Partition Equilibria Seatwork: 1. The Henry’s Law coefficient 𝒌𝑵𝟐 for nitrogen gas in water at 25°C is 8.4 x 104 bar. Calculate the number of moles of N2 dissolved in 1 mole of water at N2 partial pressures of 0.8 bar and 4 bar pressure, and from this, calculate what volume of N2 at 0.8 bar pressure will come bubbling out of 1 L of water that has previously been equilibrated with N2 at 4 bar pressure. Seatwork: 2. Use Raoult’s Law to calculate the vapor pressure of water saturated with air at 1 bar pressure and 25°C. For this purpose, we can dry air to be a mixture of 78.08% N2 (by volume), 20.95% O2, and 0.93% Ar. Seatwork: 3. The hexane/water partition coefficient of phenol at 298.15 K is 0.20; calculate the free energy of transfer of phenol from water to hexane. 4. The solubility of phenol in water is 83 g/L. What is its solubility in hexane? 5. 1-octanol and water are immiscible solvents, and octanol is often thought to have solvent properties similar to biological lipids. The partition equilibria between water and octanol have therefore been extensively tabulated to indicate the tendency of chemical compounds to concentrate in lipids. Benzoic acid, a common preservative, MM = 122.12 g mol-1, has a partition coefficient between octanol and water ao/aw = 76, and the solubility of benzoic acid in water is 3.5 g/kg water. What is the solubility in octanol? Binding of Small Molecules by a Polymer 𝑨 + 𝑩 ⇋ 𝑨𝑩 𝑨𝑩 𝑲= 𝑨 𝑩 ∆𝑮𝒐 = −𝑹𝑻 𝒍𝒏 𝑲 Binding of Small Molecules by a Polymer Binding of Small Molecules by a Polymer Binding of Small Molecules by a Polymer 𝑨 + 𝑩 ⇋ 𝑨𝑩 𝑨𝑩 𝑲= 𝑨 𝑩 ∆𝑮𝒐 = −𝑹𝑻 𝒍𝒏 𝑲 Identical-and-Independent-Sites Model Polymer P has a number of identical and independent sites for binding a smaller molecule A Identical: each has same affinity for A Independent: the occupation of one site has no effect on the binding to another site Identical-and-Independent-Sites Model (𝟏) 𝑷𝑨(𝟏) 𝑲= 𝑷 𝑨 𝑷𝑨(𝟐) 𝑷𝑨(𝟐) 𝑲= 𝑷 𝑨 (𝟑) 𝑷𝑨(𝟑) 𝑲= 𝑷 𝑨 𝑷𝑨(𝟒) 𝑷𝑨(𝟒) 𝑲= 𝑷 𝑨 𝑷 + 𝑨 ⟺ 𝑷𝑨 𝑷+𝑨⟺ 𝑷 + 𝑨 ⟺ 𝑷𝑨 𝑷+𝑨⟺ The superscript (1, 2, 3, or 4) in each reaction specifies which site of the polymer is involved. 𝑷𝑨 = 𝑷𝑨(𝟏) + 𝑷𝑨(𝟐) + 𝑷𝑨(𝟑) + 𝑷𝑨(𝟒) 𝑷𝑨 = 𝟒𝑲 𝑷 𝑨 Identical-and-Independent-Sites Model 𝑲𝟏 𝑷 + 𝑨 ⇔ ⇋ 𝑷𝑨 𝒂𝒏𝒚 𝒔𝒊𝒕𝒆 𝑷𝑨 𝑲𝟏 = = 𝟒𝑲 𝑷 𝑨 Identical-and-Independent-Sites Model 𝑷𝑨 𝑲𝟏 = = 𝟒𝑲 𝑷 𝑨 𝑲𝟏 𝑷 + 𝑨 ⇔ ⇋ 𝑷𝑨 𝒂𝒏𝒚 𝒔𝒊𝒕𝒆 𝑲𝟐 𝑷𝑨 𝒂𝒏𝒚 𝒔𝒊𝒕𝒆 + 𝑨 ⇔ ⇋ 𝑷𝑨𝟐 𝒂𝒏𝒚 𝒕𝒘𝒐 𝒔𝒊𝒕𝒆𝒔 𝑷𝑨𝟐 𝟒−𝟏 𝟑 𝑲𝟐 = = 𝑲= 𝑲 𝑷𝑨 𝑨 𝟐 𝟐 Identical-and-Independent-Sites Model 𝑷𝑨 𝑲𝟏 = = 𝟒𝑲 𝑷 𝑨 𝑷𝑨𝟐 𝟒−𝟏 𝟑 𝑲𝟐 = = 𝑲= 𝑲 𝑷𝑨 𝑨 𝟐 𝟐 𝑷𝑨𝟑 𝟒−𝟐 𝟐 𝑲𝟑 = = 𝑲= 𝑲 𝑷𝑨𝟐 𝑨 𝟑 𝟑 𝑷𝑨𝟒 𝟒−𝟑 𝟏 𝑲𝟒 = = 𝑲= 𝑲 𝑷𝑨𝟑 𝑨 𝟒 𝟒 Identical-and-Independent-Sites Model 𝑲𝟏 = 𝑷𝑨 𝑷𝑨𝟐 𝟒−𝟏 𝟑 𝑷𝑨𝟑 𝟒−𝟐 𝟐 = 𝑲= 𝑲 = 𝟒𝑲 𝑲𝟐 = = 𝑲 = 𝑲 𝑲𝟑 = 𝑷𝑨𝟐 𝑨 𝟑 𝟑 𝑷 𝑨 𝑷𝑨 𝑨 𝟐 𝟐 𝑲𝟒 = 𝑷𝑨𝟒 𝟒−𝟑 𝟏 = 𝑲= 𝑲 𝑷𝑨𝟑 𝑨 𝟒 𝟒 𝑷𝑨 = 𝟒𝑲 𝑷 𝑨 = 𝟒 𝑷 𝑲 𝑨 𝑷𝑨𝟐 𝑷𝑨𝟑 𝑷𝑨𝟑 𝟑 𝟒o𝟑 𝟐 = 𝑲 𝑷𝑨 𝑨 = 𝑲 𝑷 𝑨 𝟐 𝟐 𝟐 =𝟔𝑷 𝑲𝑨 𝟐 𝟐 𝟒o𝟑o𝟐 𝟑 = 𝑲 𝑷𝑨𝟐 𝑨 = 𝑲 𝑷 𝑨𝟑=𝟒𝑷 𝑲𝑨 𝟑 𝟑o𝟐 𝟏 = 𝑲 𝑷𝑨𝟑 𝑨 = 𝑲𝟒 𝑷 𝑨 𝟒 = 𝑷 𝑲 𝑨 𝟒 𝟒 𝟑 Identical-and-Independent-Sites Model 𝑷𝑨 = 𝟒𝑲 𝑷 𝑨 = 𝟒 𝑷 𝑲 𝑨 𝟑 𝟒8𝟑 𝟐 𝑷𝑨𝟐 = 𝑲 𝑷𝑨 𝑨 = 𝑲 𝑷 𝑨𝟐=𝟔𝑷 𝑲𝑨 𝟐 𝟐 𝟐 𝟐 𝟒8𝟑8𝟐 𝟑 𝑷𝑨𝟑 = 𝑲 𝑷𝑨𝟐 𝑨 = 𝑲 𝑷 𝑨𝟑=𝟒𝑷 𝑲𝑨 𝟑 𝟑8𝟐 𝟏 𝑷𝑨𝟑 = 𝑲 𝑷𝑨𝟑 𝑨 = 𝑲𝟒 𝑷 𝑨 𝟒 = 𝑷 𝑲 𝑨 𝟒 𝟒 𝟑 Let K[A] = S. The total concentration of the polymer is 𝑷 𝑻 = 𝑷 + 𝑷𝑨 + 𝑷𝑨𝟐 + 𝑷𝑨𝟑 + 𝑷𝑨𝟒 𝑷 𝑻 = 𝑷 𝟏 + 𝟒𝑺 + 𝟔𝑺𝟐 + 𝟒𝑺𝟑 + 𝑺𝟒 𝑷 𝑻 = 𝑷 𝟏+𝑺 𝟒 Identical-and-Independent-Sites Model The total concentration of A molecules that are bound to the polymer molecules is 𝑨 𝒃𝒐𝒖𝒏𝒅 = 𝑷𝑨 + 𝟐 𝑷𝑨𝟐 + 𝟑 𝑷𝑨𝟑 + 𝟒 𝑷𝑨𝟒 𝑨 𝒃𝒐𝒖𝒏𝒅 = 𝑷 𝟒𝑺 + 𝟏𝟐𝑺𝟐 + 𝟏𝟐𝑺𝟑 + 𝟒𝑺𝟒 𝑨 𝒃𝒐𝒖𝒏𝒅 = 𝟒 𝑷 𝑺 𝑺 + 𝟑𝑺 + 𝟑𝑺𝟐 + 𝑺𝟑 𝑨 𝒃𝒐𝒖𝒏𝒅 =𝟒 𝑷 𝑺 𝟏+𝑺 𝟑 Identical-and-Independent-Sites Model The number of bound A molecules per polymer molecule is 𝜈: 𝑷 𝑨 𝑻 𝒃𝒐𝒖𝒏𝒅 = 𝑷 𝟏+𝑺 𝟒 =𝟒 𝑷 𝑺 𝟏+𝑺 𝟑 𝑨 𝒃𝒐𝒖𝒏𝒅 𝟒 𝑷 𝑺 𝟏 + 𝑺 𝟑 𝟒𝑺 𝝂= = = 𝑷 𝒕𝒐𝒕𝒂𝒍 𝑷 𝟏+𝑺 𝟒 𝟏+𝑺 In general, if there are N identical and independent sites: 𝑵𝑺 𝝂= 𝟏+𝑺 𝑵𝑲 𝑨 𝝂= 𝟏+𝑲 𝑨 Identical-and-Independent-Sites Model 𝑵𝑲 𝑨 𝝂= 𝟏+𝑲 𝑨 𝝂 𝟏+𝑲 𝑨 = 𝑵𝑲 𝑨 𝝂= 𝑵−𝝂 𝑲 𝑨 𝝂 =𝑲 𝑵−𝝂 𝑨 SCATCHARD EQUATION Equilibrium Dialysis Equilibrium Dialysis Equilibrium Dialysis Dialysis Tubing Equilibrium Dialysis Micro-dialyzers, Dialysis cups Equilibrium Dialysis Scatchard Equation A convenient way of treating the data to extract these values is through the Scatchard equation. There are three experimentally measurable quantities: Concentration of the small molecule (ligand), A, outside Concentration of A inside Concentration of the macromolecule inside 𝝂 =𝑲 𝑵−𝝂 𝑨 𝑨 = 𝑨 𝑨 𝒃𝒐𝒖𝒏𝒅,𝒊𝒏𝒔𝒊𝒅𝒆 𝝂= 𝑨 = 𝑨 𝒊𝒏𝒔𝒊𝒅𝒆 𝒐𝒖𝒕𝒔𝒊𝒅𝒆 𝒊𝒏𝒔𝒊𝒅𝒆 − 𝑨 𝑴 − 𝑨 𝒐𝒖𝒕𝒔𝒊𝒅𝒆 𝒐𝒖𝒕𝒔𝒊𝒅𝒆 Binding of O2 to human myoglobin Data obtained at 30°C are plotted as the fraction of binding sites occupied, 𝞶, against the pressure of O2. Scatchard plots of the binding of O2 to Mb at several temperatures. Ligand Binding: Example The formyltetrahydrofolate synthetases can utilize the energy of hydrolysis of ATP (to ADP and phosphate) for the formation of a carbon-nitrogen bond between (l)tetrahydrofolate and formate. Binding of the Mg complex of ATP to such an enzyme from C. cyclindrosporum has been measured. The data are plotted below. Ligand Binding 𝞶: average number of molecules of A bound to the macromolecule N: number of binding sites on the macromolecule 𝞶: average number of molecules of A bound to the macromolecule N: number of binding sites on the macromolecule 𝒓 =𝑲 𝒏−𝒓 𝑨 Trypanosoma brucei Nearest-neighbor Interactions and Statistical Weights Frequently, the binding of a molecule to one site affects the binding of molecules to other sites. Use of statistical weights to simplify: 𝒄𝒊 𝝎𝒊 = 𝒄𝒓𝒆𝒇 Statistical weight, 𝟂i of species I is defined as the concentration relative to a reference concentration. 𝝎𝒊 𝝌𝒊 = ∑𝒊 𝝎 𝒊 Partition function in statistical mechanics: sum over states 𝓠 = x 𝝎𝒊 𝒊 Nearest-neighbor Interactions and Statistical Weights Ising model: Consider a special case with N identical but interacting sites arranged in a linear array. The identical sites form a one-dimensional lattice. 1 0 1 1 1 0 0 0 1 K: equilibrium constant for the binding of an A molecule to a site with no occupied nearest neighbors 𝝉K: equilibrium constant for the binding of an A molecule to a site with one adjacent site occupied. The parameter 𝝉 accounts for the interaction between the two adjacent occupied sites. If 𝝉 < 1, the binding is anticooperative; if 𝝉 > 1, the binding is cooperative; if 𝝉 = 1? Binding to a trimer For a simple case with N = 3: The ratio of concentrations of [001] to [000], the statistical weight of [001] is 𝟎𝟎𝟏 𝝎𝒊 = 𝟎𝟎𝟎 =𝑲𝑨 The statistical weights of [101], [011], and [111] are (K[A])2, 𝝉(K[A])2, and 𝝉2(K[A])3, respectively. The free species [000] has a statistical weight of 1 by convention. Defining the product K[A] as S, Binding to a trimer Where K[A] = S: General formula: 𝝉iSj Where j is the number of 1’s (occupied sites) in the state and i is the number of 1’s following 1 (number of nearest-neighbor interactions) Binding to a trimer Summing the statistical weights gives 𝓠 = x 𝝎𝒊 = 𝟏 + 𝟑𝑺 + 𝑺𝟐 + 𝟐𝝉𝑺𝟐 + 𝝉𝟐 𝑺𝟑 𝒊 Binding to a trimer 𝓠 = x 𝝎𝒊 = 𝟏 + 𝟑𝑺 + 𝑺𝟐 + 𝟐𝝉𝑺𝟐 + 𝝉𝟐 𝑺𝟑 𝒊 The average number of bound A per P is 𝟏 𝑷𝑨 + 𝟐 𝑷𝑨𝟐 + 𝟑 𝑷𝑨𝟑 𝝂= 𝑷 + 𝑷𝑨 + 𝑷𝑨𝟐 + 𝑷𝑨𝟑 𝟑𝑺 + 𝟐 𝑺𝟐 + 𝟐𝝉𝑺𝟐 + 𝟑 𝝉𝟐 𝑺𝟑 𝝂= 𝟏 + 𝟑𝑺 + 𝑺𝟐 + 𝟐𝝉𝑺𝟐 + 𝝉𝟐 𝑺𝟑 𝟑𝑺 + 𝟐 𝑺𝟐 + 𝟐𝝉𝑺𝟐 + 𝟑 𝝉𝟐 𝑺𝟑 𝝂= 𝓠 Binding to a trimer 𝓠 = x 𝝎𝒊 = 𝟏 + 𝟑𝑺 + 𝑺𝟐 + 𝟐𝝉𝑺𝟐 + 𝝉𝟐 𝑺𝟑 𝒊 The average number of bound A per P is 𝓠 = 𝟏 + 𝟑𝑺 + 𝑺𝟐 + 𝟐𝝉𝑺𝟐 + 𝝉𝟐 𝑺𝟑 𝟏 𝑷𝑨 + 𝟐 𝑷𝑨𝟐 + 𝟑 𝑷𝑨𝟑 𝝂= 𝑷 + 𝑷𝑨 + 𝑷𝑨𝟐 + 𝑷𝑨𝟑 𝟑𝑺 + 𝟐 𝑺𝟐 + 𝟐𝝉𝑺𝟐 + 𝟑 𝝉𝟐 𝑺𝟑 𝝂= 𝟏 + 𝟑𝑺 + 𝑺𝟐 + 𝟐𝝉𝑺𝟐 + 𝝉𝟐 𝑺𝟑 𝟑𝑺 + 𝟐 𝑺𝟐 + 𝟐𝝉𝑺𝟐 + 𝟑 𝝉𝟐 𝑺𝟑 𝝂= 𝓠 𝝏𝓠/𝝏𝑺 𝝂=𝑺 𝓠 𝝏𝓠/𝓠 𝝂= 𝝏𝑺/𝑺 𝝉 𝝉 𝝏 𝐥𝐧 𝓠 𝝂= 𝝏 𝐥𝐧 𝑺 𝝉 𝝏𝓠 𝝏𝑺 𝝏𝓠 𝑺 𝝏𝑺 𝝉 𝝉 = 𝟑 + 𝟐𝑺 + 𝟒𝝉𝑺 + 𝟑𝝉𝟐 𝑺𝟐 = 𝑺𝟑 + 𝟐 𝑺𝟐 + 𝟐𝝉𝑺𝟐 + 𝟑𝝉𝟐 𝑺𝟐 SEATWORK A certain macromolecule P has four identical sites in a linear array for the binding of a small molecule A. Prepare a table listing all species with half of the sites occupied (species with two occupied sites and two unoccupied sites) and their statistical weights for the following cases: a. The sites are independent b. Nearest-neighbor interactions are present Fraction of sites occupied, 𝑓 Independent sites model, the spatial distribution of the sites has no effect on the final binding equation: 𝝂= 𝑵−𝝂 𝑲 𝑨 𝝂z 𝑵 =𝑲𝑨 𝟏 − 𝝂z𝑵 𝝂 : number of sites occupied per polymer molecule (macromolecule) N : total number of sites per polymer molecule 𝒇 : fraction of sites occupied, 𝝂/N 𝒇 =𝑲𝑨 𝟏−𝒇 Cooperative, Anticooperative, Excluded-Site Binding For 𝜏 > 1, there is cooperative binding; binding of A to one site makes it easier to bind another A to an adjacent site. A sigmoidal plot of 𝑓 versus S is indicative of cooperative binding. Cooperative, Anticooperative, Excluded-Site Binding 𝒇 =𝑲𝑨 𝟏−𝒇 Since N does not appear in this binding equation, the shape of the 𝑓 versus S curve is not affected by the number of sites N per polymer molecule. → Independent-sites model For cooperative binding, the curve will be dependent on the value of N. 𝓠 = x 𝝎𝒊 = 𝟏 + 𝟑𝑺 + 𝑺𝟐 + 𝟐𝝉𝑺𝟐 + 𝝉𝟐 𝑺𝟑 𝒊 𝝏 𝐥𝐧 𝓠 𝝂= 𝝏 𝐥𝐧 𝑺 𝝉 For the same value of 𝜏, the difference in S for a given change in 𝑓 is less for larger N. → abrupt cooperative binding Cooperative, Anticooperative, Excluded-Site Binding If N is very large, it can be shown that 𝑓 changes from essentially zero to 1 in the range 𝟐 𝟐 𝟏− < 𝝉𝑺 < 𝟏 + 𝝉 𝝉 At 𝜏S = 1 and for large values of 𝜏, a sharp transition can be observed. Recall that S = K[A]. At low concentrations of [A], 𝜏S = 𝜏K[A] is less than 1, and very little binding occurs. As [A] increases, 𝜏S = 𝜏K[A] becomes slightly larger than 1, the cooperative binding is triggered and nearly all sites are filled. Cooperative, Anticooperative, Excluded-Site Binding We can also compare the relative amounts of various species between noncooperative and cooperative binding. For N = 3, and 𝜈 = 3/2 (half of the sites are occupied): 𝝎𝒊 𝝌𝒊 = ∑𝒊 𝝎 𝒊 The mol fractions were evaluated for 𝜏 = 1 (𝜈 = 3/2 at S =1) and 𝜏 = 20 (𝜈 = 3/2 at S =0.131) and are plotted. Cooperative, Anticooperative, Excluded-Site Binding 𝜏 = 1 (𝜈 = 3/2 at S =1) 𝜏 = 20 (𝜈 = 3/2 at S =0.131) All-or-none limit In the all-or-none limit (N = 3), we can simply consider the reaction as 𝑷 + 𝟑𝑨 → 𝑷𝑨𝟑 𝑷𝑨𝟑 = 𝑷 𝑨𝟑 𝝂 𝟑𝑲: 𝑨 𝟑 =𝒇 = : 𝟑 𝟑 𝟏+𝑲 𝑨 𝒇 = 𝑲: 𝑨 𝟑 𝟏−𝒇 𝑲: 𝟑 𝑷𝑨𝟑 𝝂= 𝑷 + 𝑷𝑨𝟑 Fraction of sites occupied 𝒇 𝒍𝒐𝒈 = 𝟑 𝒍𝒐𝒈 𝑨 + 𝟑 𝒍𝒐𝒈𝑲: 𝟏−𝒇 𝒇 𝒍𝒐𝒈 = 𝑵 𝒍𝒐𝒈 𝑨 + 𝑵 𝒍𝒐𝒈𝑲: 𝟏−𝒇 All-or-none limit: Hill plot 𝒇 𝒍𝒐𝒈 = 𝑵 𝒍𝒐𝒈 𝑨 + 𝑵 𝒍𝒐𝒈𝑲: 𝟏−𝒇 Hill plot Hill constant, n Cooperative binding in hemoglobin Fraction of hemes of Mb or Hb occupied by oxygen, 𝑓, as a function of the pressure of oxygen. Cooperative binding in hemoglobin Scatchard plot of the hemoglobin data presented previously. Lipids, Bilayers, and Membranes (a) (b) (c) (d) (e) Lipid monolayer Solubilization of grease Micelle Lipid bilayer Vesicle or liposome Lipids, Bilayers, and Membranes Assembly is entropy-driven: release of water molecules from the solvation shell Liposomes are good model system for the cell membrane. Liposomes are also used for drug delivery and cosmetic preparations. Liposomes can also form as multilayer vesicles (MLVs). MLVs are like an onion with concentric spheres of phospholipid bilayers with diameters between 1000 and 50,000 Å. Phase Transitions in Lipids, Bilayers, and Membranes Cooperative Unit, CU Gel to liquid-crystalline phase transition is highly “cooperative.” Tm: the temperature at which phase transition occurs Near the Tm, the distance range of this cooperation increases. CU (Cooperative Unit): Number of molecules in disordered “islands”; the larger the CU the narrower the phase transition range Differential Thermal Analysis, DTA A technique in which the difference in temperature between the sample and a reference material is monitored against time or temperature while the temperature of the sample, in a specified atmosphere, is programmed. Differential Thermal Analysis, DTA (a) Temperature change of the furnace, the reference, and the sample against time. (b) Change in temperature difference (∆T) against time detected with the differential thermocouple. Differential Scanning Calorimetry, DSC Simple schematic of a DSC set-up Differential Scanning Calorimetry, DSC Differential scanning calorimetry of a suspensin of dipalmitoylphosphatidylcholine in water DSC of phospholipids DSC of phospholipids DSC of phospholipids