Part 1 (18 points) Define three out of six terms, given below.

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1 Chemistry 45/456 3 August End-of-term Examination Professor G. Drobny Useful Constants and Conversions Universal Gas Constant= R = 8.31 J / mole K =.81 L atm / mole K Faraday s Constant= I= 96,458 Coulombs / mole gravitational constant=g=9.8m/s Avagadro s Number=N A =6.x1 3 molecules/mole 1 atm=11,35 Nt/m =76 torr 1 bar=1 5 Nt/m = atm=75.5 torr 1 Nt/m =1Pa 1 torr=1 mmhg 1L=1m 3 Part 1 (18 points) Define three out of six terms, given below. 1.1) Donnan Effect. Define and explain how the Donnan effect perturbs osmotic pressure measurements. Answer: The unequal equilibrium distribution of small diffusible ions on two sides of a membrane that is permeable to these ions but impermeable to a macromolecular ion, that is on one side of the membrane. The unequal ion distribution produces two opposing osmotic pressures. The net pressure deviates from the van t Hoff Law. 1.) The Gibbs Phase Rule. Define and give the equation for the degrees of freedom in a multi-component, multi-phase system. What is the physical meaning of degrees of freedom, as the term is used in the phase rule? Answer: The Gibbs Phase Rule gives a relationship between the degrees of freedom f of the system, the number of components c, and the number of phases p. The relationship is f=c-p+. The number of degrees of freedom f is the number of intensive variables required to specify the state of the system. Their values can be altered independently without changing the number of phases. 1.3) Fugacity. Define and describe two methods for measuring the fugacity coefficient. Answer: The fugacity is a measurement of general escape tendency from a system and is related to the pressure P by f = γ f P where γ f is the fugacity coefficient, which measures the deviation of a gas from ideal behavior. If PV=nRT then γ f =1. The fugacity coefficient P 1 RT γ f = exp αdp whereα V RT =. The quantity a can therefore be obtained by P independent measurements of V, P, and T or can be calculated from the equation of state. 1.4) Cooperativity. Define as the term as it is used in ligand binding. Give the binding isotherm equation for complete cooperativity.

2 Answer: A mechanism whereby binding of one ligand to a target macromolecule promotes binding of other ligands. For complete cooperativity the isotherm equation is N ν K[ A] θ = f = = where f (or θ ) is the fraction of sites bound, K is the affinity N N 1 + K [ A] constant, N is the number of binding sites, and [A] is the concentration of free ligand. 1.5) The B.E.T. Isotherm Equation. Give the equation, state the assumptions behind the equation and describe the utility of B.E.T. isotherm data. Answer: This is the Brunauer, Emmett, and Teller isotherm equation V cz P = where Z = Vm ( 1 Z) ( 1 ( 1 c) Z) P V is the volume of adsorbed gas, referenced to a standard temperature and pressure, V m is the volume of gas adsorbed into a monolayer, similarly references, P is the pressure the gas exerts on the surface at equilibrium, P is the pressure exerted on a layer of adsorbed gas greater than a monolayer and approaching bulk liquid. C is a constant dependent on the enthalpy of adsorption. This isotherm equation assumes adsorbed layers of gas can act as a substrate for subsequent adsorption, which can proceed to indefinite limits. B.E.T. isotherm data is used to estimate the surface area of solid substrates. 1.6) Henry s Law Solution. Explain the Henry s Law Solution Model. In particular explain the manner in which the physical behavior of the Henry s Law solute differs from the components of an ideal solution. Explain how the Henry s Law solute standard state differs from the standard state of a component of an ideal solution. Solution: In a Henry s Law solution, the partial pressure of the solute vapor is related to µ soln µ vap the solution composition by P = kh χ where k H = exp. In an ideal solution RT the standard state of a component is the pure liquid form of the component. The solute standard state is defined as a situation where the solute is energetically equivalent to being surrounded by solvent molecules. Part ( points total) Answer two out of four questions, given below..1) Although the surface tension of water opposes expansion of the surface area of water, certain insects are observed to support their weight on small bubbles at the water surface, rather than directly on the water surface itself. Explain. What is the surface tension force for a bubble relative to a droplet of water of the same shape and size? Explain. Answer: The insect s weight is supported on the water surface by the surface tension, which resists the tendency to expand the surface area of the liquid. The bubble supports the insect s weight because it has two surfaces, an inner surface and an outer surface.

3 .) Hemoglobin is a protein that transports oxygen from the lungs to actively metabolizing tissue. There are four oxygen binding sites on each hemoglobin molecule. Carbon monoxide poisoning occurs when a person is exposed to CO and as a result, CO binds to hemoglobin instead of oxygen. Death can occur when 5% of the total binding sites on hemoglobin are occupied by CO. On the other hand, a person with sickle cell anemia can have up to half of their hemoglobin inactive and still live. Explain. Answer: Sickle cell anemia inactivates half the hemoglobin molecules, but the other half can function normally. On the other hand, oxygen binding to hemoglobin is cooperative. Binding to a given site enhances binding to other sites on the same protein. CO will not do this and so if half the hemoglobin sites are saturated with CO, the other half do not have normal oxygen binding affinities..3) Time release drugs have the advantage of releasing the drug to the body at a constant rate so that the drug concentration at any time is not high enough to have harmful side effects or so low as to be ineffective. A schematic diagram of a timerelease capsule is shown below. Explain how it works, thermodynamically..4) Workers in deep sea diving suits necessarily breath air at greater than normal pressure. At a point 4 meters below the surface air is breathed at 6 atm of pressure, and if the diver returns to the surface too quickly, small gas bubbles called emboli form in the blood stream and can cause decompression sickness and death. Explain why divers breath oxygen in a gas mixture other than air. What gas would you mix with oxygen in order to avoid decompression sickness? Use the concept of Henry s Law in your explanation. A table of Henry s Law constants K H are given below. Gas Hydrogen Helium Nitrogen Oxygen Carbon Dioxide K H (torr) 5.54x x x x x1 6

4 Solution: Henry s Law states that the partial pressure of a vapor above a solution is related to the concentration in the solution by P = KH χ. This means that at a given a pressure, the larger the Henry s Law constant the lower the solubility of the gas and the less tendency to form emboli during decompression. For this reason helium, with the largest Henry s Law constant, is mixed with oxygen. Part 3 (3 points total) Calculate three out of five problems 3.1) Calculate the mean activity coefficient for a.1m solution of copper sulfate (CuSO 4 ) at T=98K. Solution: logγ ± =.59zz + µ 1 whereµ = zc i i i 1 µ = (( + ) (.1) + ( ) (.1) ) =.4 1/ logγ ± =.59zz + µ = (.59)( )( )(.4) = γ = 1 =.39 ± 3.) Consider the two compartments, separated by a membrane that is permeable to the passage of sodium and chloride ions, but impermeable to the passage of a protein P. Assume the protein exists in solution as a monovalent anion P -1. Initially in the left compartment [P - ]=[Na + ]=.1M, and in the right compartment [Na + ]=[Cl - ]=.5M (see diagram). Calculate the net osmotic pressure at T=3K when the system reaches equilibrium. Solution: x is the amount of chloride ion that moves from right to left. Cl right, initial (.5) (.5) x = = = =.3M P + Cl.1+ (.5).11 left, final right, initial + Na =.5 + x = =.73 Na + = Cl =.5 x =.7 M right, final right, final 3.3) The concentration of sodium ion in a nerve cell is [Na + ] in =.1 M and the concentration outside the cell is [Na + ] out =.14 M. The difference in

5 electrostatic potential across the membrane is ψ =ψ in -ψ out =-.8V. Calculate the work required to move reversibly one mole of sodium ion from inside the membrane to outside the membrane. Assume the standard chemical potential of sodium inside the membrane is the same as the chemical potential outside the membrane. Assume T=31K and assume all activity coefficients equal 1. Solution: The reversible work per mole is the change in chemical potential required to move the sodium. Therefore a out w= µ = µ out µ in = RTln + zi( ψout ψin ) ain.14 = 1mole 8.31 J / mole K 31K ln mole 96,485 C / mole.8v.1 = 6798J J = 14,517J ( )( )( ) ( )( )( )( ) 3.4) The data show, as a function of temperature, the pressures of carbon monoxide required to adsorb 1cm 3 (corrected to P=1 atm and T=73K) of gas onto charcoal. Calculate the differential heat of adsorption for CO on charcoal. A graph is useful but not necessary. T(K) P(kPa) lnp /T ln P H a ln P Ha Solution: The equation is = or = T θ RT ( 1/ T) R. This means θ the slope of the graph of lnp versus 1/T gives the heat of adsorption divided by R. If the student choses to do a graph it should appear as above. Note the x axis is reversed

6 so the slope is actually negative. The slope is H a 1 1 = 9K Ha = ( 8.31 J / mole K)( 9K ) = 7496J : R 1 1 If a graph is not used the correct integrated equation is ln P Ha = P1 R T T1 Part 4 (3 points total). Perform one of the two multi-step calculations, given below. 4.1) Heptane and -methylpentane are liquids at T=315K which form ideal solutions when mixed. Suppose you mix one mole of heptane and three moles of -methylpentane at T=315K. Calculate the entropy of mixing, the Gibbs free energy of mixing, and the enthalpy of mixing for this process. Also, calculate the vapor pressures of heptane and -methylpentane above the resulting solution. At 315K the vapor pressure of pure heptane is.13 atm and the vapor pressure of pure -methylpentane is.53 atm. What are the mole fractions of heptane and -methylpentane in the vapor phase? Solution: For an ideal solution =. H mix nhept 1 χhept = = =.5 nhept + nmp 1+ 3 nmp 3 χmp = = =.75 nhept + nmp 1+ 3 or χmp = 1 χhept =.75 S = R χ ln χ + χ ln χ ( ) i ( ) i ( ) mix hept hept MP MP ( 8.31 J / mole K).5ln (.5).75ln (.75) ( 8.31 J / mole K) (.5)( 1.38) (.75)(.8) ( 8.31 J / moleik)(.345.1) ( 8.31 J / moleik)(.56) = + = + = = = 4.65 J / moleik There are 4 moles total so Smix = ( 4moles) Smix = ( 4)( 4.65 J / K) = J / K Gmix = T Smix = ( 315K)( J / K) = 586J Phept = χheptphept = (.5)(.13atm) =.35atm P = χ P =.75.53atm =.4atm MP MP MP ( )( )

7 Dalton s Law of Partial Pressure states PT = Pheptan e + P MP = =.435atm Phep tan e.35 χhep tan e = = =.75 PT.435 χ = 1 χ = 1.75 =.95 MP hep tane 4.) I. Langmuir measured the following data for the adsorption of methane onto 1 gram of mica at T=9K: P (bars) V (mm 3 ) Where P is the pressure of the methane above the mica surface at equilibrium, and V is the volume of gas adsorbed onto the surface, measured at T=93K and at a pressure of 1 atm. a) If adsorption of methane onto mica at T=9K occurs without interactions between methane molecules on the surface, all surface sites are equivalent, and adsorption P 1 P cannot proceed beyond monolayer coverage, show that = +, where V VmK Vm V m is the volume of gas adsorbed at T=93K and P=1 atm to form a monolayer, and K is the equilibrium constant for adsorption. Solution: From the conditions mentioned the appropriate isotherm equation is the Langmuir isotherm. Then VKP m V VP V = V + VKP = VmKP P = + 1+ KP VmK Vm P 1 P = + V VmK Vm b) From an appropriate graph of the data in the Table, determine V m and K. Note: You do not have to graph all the data but you must graph enough data to establish linear behavior (i.e. more than two points). Express V m in units of meters 3.

8 P/V (bars/mm^3) P(bars) Solution: Slope=1/V m =.8mm -3. So V m =14mm 3 =1.4x1-7 m 3. The y intercept is.484 bars/mm 3. Therefore 1 1 =.484 bars / mm K = =.167 bars 3 3 VK.484 bars / mm 14mm m 3 1 ( )( ) c) From the information determined in part b, determine the pressure of methane at which the surface is half covered at T=9K. Gives you answer in bars. Hint: Use the equation from part a. Solution: P 1 P Vm P 1 P 1 = + V = = + P = V VmK Vm Vm VmK Vm K 1 P = = 6 bars 1.167bars d) Suppose the cross sectional area of a methane molecule is σ=1-19 m. Use the information obtained in part b to determine the surface area of one gram of mica. Assume the methane vapor behaves ideally. Solution: You need the molar volume at T=93K and P=1atm. This is V RT (.81 L atm/ moles K)( 93K) 3 = V = = = 4L=.4m n P 1atm

9 Then the surface area per gram is A = σ NV V m A m = =.311 / gm ( 19 )( m 6. 1 mole )( m 3 / gm).4m 3