DOCSLIB.ORG

Page 1 of 6 This Is Henry's Law. It Says That at Equilibrium the Ratio of Dissolved NH3 to the Partial Pressure of NH3 Gas In

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
Page 1 of 6 This Is Henry's Law. It Says That at Equilibrium the Ratio of Dissolved NH3 to the Partial Pressure of NH3 Gas In CHMY 361 HANDOUT#6 October 28, 2012 HOMEWORK #4 Key Was due Friday, Oct. 26 1. Using only data from Table A5, what is the boiling point of water deep in a mine that is so far below sea level that the atmospheric pressure is 1.17 atm? 0 ΔH vap = +44.02 kJ/mol H20(l) --> H2O(g) Q= PH2O /XH2O = K, at ⎛ P2 ⎞ ⎛ K 2 ⎞ ΔH vap ⎛ 1 1 ⎞ ln⎜ ⎟ = ln⎜ ⎟ − ⎜ − ⎟ equilibrium, i.e., the Vapor Pressure ⎝ P1 ⎠ ⎝ K1 ⎠ R ⎝ T2 T1 ⎠ for the pure liquid. ⎛1.17 ⎞ 44,020 ⎛ 1 1 ⎞ ln⎜ ⎟ = − ⎜ − ⎟ = 1 8.3145 ⎜ T 373 ⎟ ⎝ ⎠ ⎝ 2 ⎠ ⎡1.17⎤ − 8.3145ln 1 ⎢ 1 ⎥ 1 = ⎣ ⎦ + = .002651 T2 44,020 373 T2 = 377 2. From table A5, calculate the Henry’s Law constant (i.e., equilibrium constant) for dissolving of NH3(g) in water at 298 K and 340 K. It should have units of Matm-1;What would it be in atm per mole fraction, as in Table 5.1 at 298 K? o For NH3(g) ----> NH3(aq) ΔG = -26.5 - (-16.45) = -10.05 kJ/mol ΔG0 − [NH (aq)] K = e RT = 0.0173 = 3 This is Henry’s Law. It says that at equilibrium the ratio of dissolved P NH3 NH3 to the partial pressure of NH3 gas in contact with the liquid is a constant = 0.0173 (Henry’s Law Constant). This also says [NH3(aq)] =0.0173PNH3 or -1 PNH3 = 0.0173 [NH3(aq)] = 57.8 atm/M x [NH3(aq)] The latter form is like Table 5.1 except it has NH3 concentration in M instead of XNH3. Need the convert [NH3(aq)] to XNH3(aq) Mole fraction of a 1 M solution in water is 1/(55 +1) = 1/56 = 0.018 moles NH3 / total moles. -1 -1 So, the conversion factor is: XNH3 = molL NH3 x 0.018 mol fraction/molL -1 -1 57.8 atmM /0.018 M (mol fraction)M = 3200 atm (per mol fraction dissolved NH3 ) Note: this is similar to that for CO2 in Table 5.1, another fairly soluble gas. Page 1 of 6 3. (a) How many moles of N2 gas are dissolved in the blood of a diver before entering the water. Assume 2.0 L of blood. (Use Table 5.1) (b) How many liters of gas at 1 atm and 298 K does this number of moles equal? (c) 1 atmosphere = 760 mm of Hg. What is 1 atm in feet of water? (d) If a diver stays at a depth in the water that is the answer to (c) until equilibrium is reached, what volume of N2 gas would be released from the blood if the diver came to the surface quickly? [Note, you don't have to find the answer to (c) in order to know the answer to (d) ] a. 2.0 L blood (assume it is pure water) has close to 2.0 kg / .018kg mol-1 = 111 mol water PN 0.78 atm at 1 atm air and 298 K, X dissolved = 2 = = 9.18 ×10−6 N2 k 85000 atm N2 −6 mols N 2 dissolved = 9.18 ×10 x 111 = 0.00102 nRT 0.00102 x 0.0821 x 298 PV = nRT V = = = 0.025 L = 25 mL P 1 nRT 0.00102 x 0.0821 x 298 b. PV = nRT V = = = 0.025 L = 25 mL P 1 c. 0.76 m Hg x 3.28 ft/m = 2.49 ft Hg. Hg is 13.6 time more dense, so 2.49 ft Hg x 13.6 ft H2O/ft Hg =33.9 ft H20 = 1 atm d. At this depth the total pressure would be 2 atm: 1 atm from the atmosphere and 1atm from the water, meaning that exactly twice the amount of N2 would be dissolved than at the surface. Thus 25 ml of N2 gas would be released as bubbles into the blood. The bubbles clog capillaries, causing low oxygen delivery to tissue. 4. Consider a 1x10-4 M polymer solution inside a dialysis bag that is permeable to the ligand, A, but not to the polymer. The following measurements of the free ligand outside the bag and total ligand inside the bag are measured at 3 equilibrium values. The units are M. From a Scatchard plot find Kd for the binding of A to its binding sites on the polymer and the number of binding sites per polymer molecule. (the notation 5E-07 means 5 x 10-7, etc.) [A]out [A]in tot [SA] [SA]/[A] 1.100E‐05 7.443E‐04 7.33E-04 6.67E+01 9.100E‐07 3.821E‐04 3.81E-04 4.19E+02 8.300E‐08 6.139E‐05 6.13E-05 7.39E+02 -4 Plot of [SA]/[A] vs. [SA]. The x intercept is 8 x 10 M = Stotal ; y intercept is 800 = Stotal/Kd. -4 -4 Therefore number of binding sites per polymer is Stotal/ [polymer] = 8 x 10 / 1 x 10 = 8 sites -4 -6 The value of Kd is Stotal /800 = 8 x 10 / 800 = 1 x 10 M 8.00E+02 6.00E+02 4.00E+02 Series1 2.00E+02 0.00E+00 0.00E+00 2.00E‐04 4.00E‐04 6.00E‐04 8.00E‐04 Page 2 of 6 5. If the partition coefficient for solute B = [Bmethanol]/ [Bwater] = 0.010, and the solubility of B in water is 0.10 M at 25o C, answer the following: (a)What is ΔGo for the process B(methanol) --> B(aq)? (b) What is ΔGo for the process B(methanol) --> B(s)? (c) What is ΔGo for the process B(aq) --> B(s)? a. ΔGo for B(aq) --> B(methanol) = -RTln K = -8.3145 x 298 x ln( 0.01) = 11,400 J/mol (could have been written by inspection as 2 x 5,700). Therefore ΔGo for B(methanol) --> B(aq) = -11,400 J/mol o c. K for B(s) --> B(aq) is given as 0.1 M , so by inspection ΔG for this 5,700 J/mol Therefore ΔGo for B(aq) --> B(s) = -5,700 J/mol b. Adding the processes of a. and c. gives B(methanol) --> B(s). Therefore adding the ΔGo values of a. and c. gives the answer = -11,400 -5,700 = -22,100 J/mol 6. In certain mitochondria, the oxidation of glucose products pumps protons into the intermembrane space between the double membrane to reach a pH of 4.8. In the interior of the mitochondrian, the pH is 7.2. In addition, the intermembrane space has an electric potential difference of +80 mV relative to the interior. (“membrane potential” is -80 mV) This problem is very much like Prob. 5 of Chap. 5, with solution given in the Appendix. (a) ATP is synthesized by the combined free energy change of the protons flowing down the concentration gradient from the intermembrane space to the interior and the change of electrical free energy coming from the voltage difference. Calculate the ΔG per mole of protons transferred. ⎛ a + ()outside ⎞ ⎜ H ⎟ Δμ = RT ln + ZF()φout − φin , where in means interior ⎜ a + inside ⎟ ⎝ H ()⎠ J ⎛1×10−7.2 ⎞ ⎛ C ⎞⎛ J J ⎞ = 8.314 ()298K ln⎜ ⎟ + ()1 ⎜96485 ⎟⎜0 − 0.080 ⎟ ⎜ −4.8 ⎟ molK ⎝1×10 ⎠ ⎝ mol ⎠⎝ C C ⎠ kJ = −13700 − 7700 = −21400 mol (b) How many molecules of ATP from ADP and P are obtained per proton if [ATP]/[ADP] = 3 and [P] 0.003 M? ⎛ ATP ⎞ ⎛⎛ ATP ⎞⎛ 1 ⎞⎞ o [] o ⎜ [ ] ⎟ ΔG = ΔG + RT ln⎜ ⎟ = ΔG + RT ln⎜⎜ ⎟⎜ ⎟⎟ = ⎝ [][]ADP P ⎠ ⎝⎝ []ADP ⎠⎝ []P ⎠⎠ kJ J ⎛ ⎛ 1 ⎞⎞ kJ 31.0 + 8.314 ()()298K ln⎜ 3 ⎜ ⎟⎟ = 48.1 mol molK ⎝ ⎝ 0.003 ⎠⎠ mol kJ 21.4 mol = 0.5 ; Therefore it requires about 2 protons to make 1 ATP kJ 48.1 mol Page 3 of 6 + + + + 7. Consider Fig. 5.22 but with [Na ]out/[Na ]in = 20.0 and [K ]out/[K ]in = 0.040 and φout - φin = 60 mV. (a) What is Δμ for Na+(in) --> Na+(out)? (b) What is Δμ for K+(out) --> K+(in) ? (c) What is ΔG for the process: 3Na+ (in) + 2K+ (out)--> 3Na+ (out) 2K+(in)? (d) If [ATP] = 0.08 [ADP] and [P] = 0.02 M will the process in (c) be spontaneous if the hydrolysis of ATP is coupled perfectly to it? a. a + ()outside Na = 20.0 aNa+ ()inside C ⎛ 1V ⎞ Z = 1 F = 96485 ΔΦ = 60mV ⎜ ⎟ = 0.06V mol ⎝1000mV ⎠ ⎛ a + ()outside ⎞ J ⎛ C ⎞ Δμ = RT ln⎜ Na ⎟ + ZFΔΦ = 8.314 ()()()298K ln 20.0 + 1 ⎜96485 ⎟()0.06V ⎜ a + inside ⎟ molK mol ⎝ Na ()⎠ ⎝ ⎠ kJ Δμ = 13.2 mol a + ()outside a + ()inside K = 0.04; K = 25. aK + ()inside aK + ()outside C ⎛ 1V ⎞ Z = 1 F = 96485 ΔΦ = −60mV ⎜ ⎟ = − 0.06V mol ⎝1000mV ⎠ b. ⎛ a + ()inside ⎞ J ⎛ C ⎞ Δμ = RT ln⎜ K ⎟ + ZFΔΦ = 8.314 ()()()298K ln 25. + 1 ⎜96485 ⎟()− 0.06V ⎜ a + outside ⎟ molK mol ⎝ K ()⎠ ⎝ ⎠ kJ Δμ = 7.98 − 5.79 = 2.19 mol c. 3Na + ()in + 2K + ()out → 3Na + ()out + 2K + ()in ⎛ kJ ⎞ kJ 3Na + ()in → 3Na + (out )3Δμ = 3⎜13.2 ⎟ = 39.6 ⎝ mol ⎠ mol ⎛ kJ ⎞ kJ 2K + ()out → 2K + ()in 2Δμ = 2⎜2.19 ⎟ = 4.38 ⎝ mol ⎠ mol kJ 3Na + ()in + 2K + ()out → 3Na + ()out + 2K + ()in ΔG = 44.0 mol o ⎛ [][]ADP P ⎞ J J ⎛ (0.02)⎞ kJ d. Δμ = Δμ + RT ln⎜ ⎟ = −31000 + 8.314 ()298K ln⎜ ⎟ = −33.4 ⎝ []ATP ⎠ mol molK ⎝ 0.08 ⎠ mol Page 4 of 6 8. For a 0.3 g/L aqueous solution of a protein with molecular weight of 5 x 104 g/mol, what is the osmotic pressure in mm of liquid water at 298 K, using the relationship that reminds one of the ideal gas law? ΠV=nRT approximately, where V is volume of solution, and n = moles solute ⎛ g ⎞ ⎜ 0.3 ⎟ w ⎛ Latm ⎞ Π = RT = ⎜ L ⎟⎜0.08205 ⎟()298K =1.47 ×10−4 atm M ⎜ g ⎟⎝ Kmol ⎠ ⎜ 50000 ⎟ ⎝ mol ⎠ −4 ⎛ 760mmHg ⎞⎛13.6mmWater ⎞ 1.47 ×10 atm⎜ ⎟⎜ ⎟ =1.52 mm water ⎝ 1atm ⎠⎝ mmHg ⎠ 9.
Load more

Recommended publications
  • Solutes and Solution
    Solutes and Solution The first rule of solubility is “likes dissolve likes” Polar or ionic substances are soluble in polar solvents Non-polar substances are soluble in non- polar solvents Solutes and Solution There must be a reason why a substance is soluble in a solvent: either the solution process lowers the overall enthalpy of the system (Hrxn < 0) Or the solution process increases the overall entropy of the system (Srxn > 0) Entropy is a measure of the amount of disorder in a system—entropy must increase for any spontaneous change 1 Solutes and Solution The forces that drive the dissolution of a solute usually involve both enthalpy and entropy terms Hsoln < 0 for most species The creation of a solution takes a more ordered system (solid phase or pure liquid phase) and makes more disordered system (solute molecules are more randomly distributed throughout the solution) Saturation and Equilibrium If we have enough solute available, a solution can become saturated—the point when no more solute may be accepted into the solvent Saturation indicates an equilibrium between the pure solute and solvent and the solution solute + solvent solution KC 2 Saturation and Equilibrium solute + solvent solution KC The magnitude of KC indicates how soluble a solute is in that particular solvent If KC is large, the solute is very soluble If KC is small, the solute is only slightly soluble Saturation and Equilibrium Examples: + - NaCl(s) + H2O(l) Na (aq) + Cl (aq) KC = 37.3 A saturated solution of NaCl has a [Na+] = 6.11 M and [Cl-] =
    [Show full text]
  • THE SOLUBILITY of GASES in LIQUIDS Introductory Information C
    THE SOLUBILITY OF GASES IN LIQUIDS Introductory Information C. L. Young, R. Battino, and H. L. Clever INTRODUCTION The Solubility Data Project aims to make a comprehensive search of the literature for data on the solubility of gases, liquids and solids in liquids. Data of suitable accuracy are compiled into data sheets set out in a uniform format. The data for each system are evaluated and where data of sufficient accuracy are available values are recommended and in some cases a smoothing equation is given to represent the variation of solubility with pressure and/or temperature. A text giving an evaluation and recommended values and the compiled data sheets are published on consecutive pages. The following paper by E. Wilhelm gives a rigorous thermodynamic treatment on the solubility of gases in liquids. DEFINITION OF GAS SOLUBILITY The distinction between vapor-liquid equilibria and the solubility of gases in liquids is arbitrary. It is generally accepted that the equilibrium set up at 300K between a typical gas such as argon and a liquid such as water is gas-liquid solubility whereas the equilibrium set up between hexane and cyclohexane at 350K is an example of vapor-liquid equilibrium. However, the distinction between gas-liquid solubility and vapor-liquid equilibrium is often not so clear. The equilibria set up between methane and propane above the critical temperature of methane and below the criti­ cal temperature of propane may be classed as vapor-liquid equilibrium or as gas-liquid solubility depending on the particular range of pressure considered and the particular worker concerned.
    [Show full text]
  • Introduction to the Solubility of Liquids in Liquids
    INTRODUCTION TO THE SOLUBILITY OF LIQUIDS IN LIQUIDS The Solubility Data Series is made up of volumes of comprehensive and critically evaluated solubility data on chemical systems in clearly defined areas. Data of suitable precision are presented on data sheets in a uniform format, preceded for each system by a critical evaluation if more than one set of data is available. In those systems where data from different sources agree sufficiently, recommended values are pro­ posed. In other cases, values may be described as "tentative", "doubtful" or "rejected". This volume is primarily concerned with liquid-liquid systems, but related gas-liquid and solid-liquid systems are included when it is logical and convenient to do so. Solubilities at elevated and low 'temperatures and at elevated pressures may be included, as it is considered inappropriate to establish artificial limits on the data presented. For some systems the two components are miscible in all proportions at certain temperatures or pressures, and data on miscibility gap regions and upper and lower critical solution temperatures are included where appropriate and if available. TERMINOLOGY In this volume a mixture (1,2) or a solution (1,2) refers to a single liquid phase containing components 1 and 2, with no distinction being made between solvent and solute. The solubility of a substance 1 is the relative proportion of 1 in a mixture which is saturated with respect to component 1 at a specified temperature and pressure. (The term "saturated" implies the existence of equilibrium with respect to the processes of mass transfer between phases) • QUANTITIES USED AS MEASURES OF SOLUBILITY Mole fraction of component 1, Xl or x(l): ml/Ml nl/~ni = r(m.IM.) '/.
    [Show full text]
  • Producing Nitrogen Via Pressure Swing Adsorption
    Reactions and Separations Producing Nitrogen via Pressure Swing Adsorption Svetlana Ivanova Pressure swing adsorption (PSA) can be a Robert Lewis Air Products cost-effective method of onsite nitrogen generation for a wide range of purity and flow requirements. itrogen gas is a staple of the chemical industry. effective, and convenient for chemical processors. Multiple Because it is an inert gas, nitrogen is suitable for a nitrogen technologies and supply modes now exist to meet a Nwide range of applications covering various aspects range of specifications, including purity, usage pattern, por- of chemical manufacturing, processing, handling, and tability, footprint, and power consumption. Choosing among shipping. Due to its low reactivity, nitrogen is an excellent supply options can be a challenge. Onsite nitrogen genera- blanketing and purging gas that can be used to protect valu- tors, such as pressure swing adsorption (PSA) or membrane able products from harmful contaminants. It also enables the systems, can be more cost-effective than traditional cryo- safe storage and use of flammable compounds, and can help genic distillation or stored liquid nitrogen, particularly if an prevent combustible dust explosions. Nitrogen gas can be extremely high purity (e.g., 99.9999%) is not required. used to remove contaminants from process streams through methods such as stripping and sparging. Generating nitrogen gas Because of the widespread and growing use of nitrogen Industrial nitrogen gas can be produced by either in the chemical process industries (CPI), industrial gas com- cryogenic fractional distillation of liquefied air, or separa- panies have been continually improving methods of nitrogen tion of gaseous air using adsorption or permeation.
    [Show full text]
  • Chapter 15: Solutions
    452-487_Ch15-866418 5/10/06 10:51 AM Page 452 CHAPTER 15 Solutions Chemistry 6.b, 6.c, 6.d, 6.e, 7.b I&E 1.a, 1.b, 1.c, 1.d, 1.j, 1.m What You’ll Learn ▲ You will describe and cate- gorize solutions. ▲ You will calculate concen- trations of solutions. ▲ You will analyze the colliga- tive properties of solutions. ▲ You will compare and con- trast heterogeneous mixtures. Why It’s Important The air you breathe, the fluids in your body, and some of the foods you ingest are solu- tions. Because solutions are so common, learning about their behavior is fundamental to understanding chemistry. Visit the Chemistry Web site at chemistrymc.com to find links about solutions. Though it isn’t apparent, there are at least three different solu- tions in this photo; the air, the lake in the foreground, and the steel used in the construction of the buildings are all solutions. 452 Chapter 15 452-487_Ch15-866418 5/10/06 10:52 AM Page 453 DISCOVERY LAB Solution Formation Chemistry 6.b, 7.b I&E 1.d he intermolecular forces among dissolving particles and the Tattractive forces between solute and solvent particles result in an overall energy change. Can this change be observed? Safety Precautions Dispose of solutions by flushing them down a drain with excess water. Procedure 1. Measure 10 g of ammonium chloride (NH4Cl) and place it in a Materials 100-mL beaker. balance 2. Add 30 mL of water to the NH4Cl, stirring with your stirring rod.
    [Show full text]
  • THE SOLUBILITY of GASES in LIQUIDS INTRODUCTION the Solubility Data Project Aims to Make a Comprehensive Search of the Lit- Erat
    THE SOLUBILITY OF GASES IN LIQUIDS R. Battino, H. L. Clever and C. L. Young INTRODUCTION The Solubility Data Project aims to make a comprehensive search of the lit­ erature for data on the solubility of gases, liquids and solids in liquids. Data of suitable accuracy are compiled into data sheets set out in a uni­ form format. The data for each system are evaluated and where data of suf­ ficient accuracy are available values recommended and in some cases a smoothing equation suggested to represent the variation of solubility with pressure and/or temperature. A text giving an evaluation and recommended values and the compiled data sheets are pUblished on consecutive pages. DEFINITION OF GAS SOLUBILITY The distinction between vapor-liquid equilibria and the solUbility of gases in liquids is arbitrary. It is generally accepted that the equilibrium set up at 300K between a typical gas such as argon and a liquid such as water is gas liquid solubility whereas the equilibrium set up between hexane and cyclohexane at 350K is an example of vapor-liquid equilibrium. However, the distinction between gas-liquid solUbility and vapor-liquid equilibrium is often not so clear. The equilibria set up between methane and propane above the critical temperature of methane and below the critical temperature of propane may be classed as vapor-liquid equilibrium or as gas-liquid solu­ bility depending on the particular range of pressure considered and the par­ ticular worker concerned. The difficulty partly stems from our inability to rigorously distinguish between a gas, a vapor, and a liquid, which has been discussed in numerous textbooks.
    [Show full text]
  • Lecture 3. the Basic Properties of the Natural Atmosphere 1. Composition
    Lecture 3. The basic properties of the natural atmosphere Objectives: 1. Composition of air. 2. Pressure. 3. Temperature. 4. Density. 5. Concentration. Mole. Mixing ratio. 6. Gas laws. 7. Dry air and moist air. Readings: Turco: p.11-27, 38-43, 366-367, 490-492; Brimblecombe: p. 1-5 1. Composition of air. The word atmosphere derives from the Greek atmo (vapor) and spherios (sphere). The Earth’s atmosphere is a mixture of gases that we call air. Air usually contains a number of small particles (atmospheric aerosols), clouds of condensed water, and ice cloud. NOTE : The atmosphere is a thin veil of gases; if our planet were the size of an apple, its atmosphere would be thick as the apple peel. Some 80% of the mass of the atmosphere is within 10 km of the surface of the Earth, which has a diameter of about 12,742 km. The Earth’s atmosphere as a mixture of gases is characterized by pressure, temperature, and density which vary with altitude (will be discussed in Lecture 4). The atmosphere below about 100 km is called Homosphere. This part of the atmosphere consists of uniform mixtures of gases as illustrated in Table 3.1. 1 Table 3.1. The composition of air. Gases Fraction of air Constant gases Nitrogen, N2 78.08% Oxygen, O2 20.95% Argon, Ar 0.93% Neon, Ne 0.0018% Helium, He 0.0005% Krypton, Kr 0.00011% Xenon, Xe 0.000009% Variable gases Water vapor, H2O 4.0% (maximum, in the tropics) 0.00001% (minimum, at the South Pole) Carbon dioxide, CO2 0.0365% (increasing ~0.4% per year) Methane, CH4 ~0.00018% (increases due to agriculture) Hydrogen, H2 ~0.00006% Nitrous oxide, N2O ~0.00003% Carbon monoxide, CO ~0.000009% Ozone, O3 ~0.000001% - 0.0004% Fluorocarbon 12, CF2Cl2 ~0.00000005% Other gases 1% Oxygen 21% Nitrogen 78% 2 • Some gases in Table 3.1 are called constant gases because the ratio of the number of molecules for each gas and the total number of molecules of air do not change substantially from time to time or place to place.
    [Show full text]
  • Dissolved Oxygen in the Blood
    Chapter 1 – Dissolved Oxygen in the Blood Say we have a volume of blood, which we’ll represent as a beaker of fluid. Now let’s include oxygen in the gas above the blood (represented by the green circles). The oxygen exerts a certain amount of partial pressure, which is a measure of the concentration of oxygen in the gas (represented by the pink arrows). This pressure causes some of the oxygen to become dissolved in the blood. If we raise the concentration of oxygen in the gas, it will have a higher partial pressure, and consequently more oxygen will become dissolved in the blood. Keep in mind that what we are describing is a dynamic process, with oxygen coming in and out of the blood all the time, in order to maintain a certain concentration of dissolved oxygen. This is known as a) low partial pressure of oxygen b) high partial pressure of oxygen dynamic equilibrium. As you might expect, lowering the oxygen concentration in the gas would lower its partial pressure and a new equilibrium would be established with a lower dissolved oxygen concentration. In fact, the concentration of DISSOLVED oxygen in the blood (the CdO2) is directly proportional to the partial pressure of oxygen (the PO2) in the gas. This is known as Henry's Law. In this equation, the constant of proportionality is called the solubility coefficient of oxygen in blood (aO2). It is equal to 0.0031 mL / mmHg of oxygen / dL of blood. With these units, the dissolved oxygen concentration must be measured in mL / dL of blood, and the partial pressure of oxygen must be measured in mmHg.
    [Show full text]
  • Chemistry C3102-2006: Polymers Section Dr. Edie Sevick, Research School of Chemistry, ANU 5.0 Thermodynamics of Polymer Solution
    Chemistry C3102-2006: Polymers Section Dr. Edie Sevick, Research School of Chemistry, ANU 5.0 Thermodynamics of Polymer Solutions In this section, we investigate the solubility of polymers in small molecule solvents. Solubility, whether a chain goes “into solution”, i.e. is dissolved in solvent, is an important property. Full solubility is advantageous in processing of polymers; but it is also important for polymers to be fully insoluble - think of plastic shoe soles on a rainy day! So far, we have briefly touched upon thermodynamic solubility of a single chain- a “good” solvent swells a chain, or mixes with the monomers, while a“poor” solvent “de-mixes” the chain, causing it to collapse upon itself. Whether two components mix to form a homogeneous solution or not is determined by minimisation of a free energy. Here we will express free energy in terms of canonical variables T,P,N , i.e., temperature, pressure, and number (of moles) of molecules. The free energy { } expressed in these variables is the Gibbs free energy G G(T,P,N). (1) ≡ In previous sections, we referred to the Helmholtz free energy, F , the free energy in terms of the variables T,V,N . Let ∆Gm denote the free eneregy change upon homogeneous mix- { } ing. For a 2-component system, i.e. a solute-solvent system, this is simply the difference in the free energies of the solute-solvent mixture and pure quantities of solute and solvent: ∆Gm G(T,P,N , N ) (G0(T,P,N )+ G0(T,P,N )), where the superscript 0 denotes the ≡ 1 2 − 1 2 pure component.
    [Show full text]
  • CEE 370 Environmental Engineering Principles Henry's
    CEE 370 Lecture #7 9/18/2019 Updated: 18 September 2019 Print version CEE 370 Environmental Engineering Principles Lecture #7 Environmental Chemistry V: Thermodynamics, Henry’s Law, Acids-bases II Reading: Mihelcic & Zimmerman, Chapter 3 Davis & Masten, Chapter 2 Mihelcic, Chapt 3 David Reckhow CEE 370 L#7 1 Henry’s Law Henry's Law states that the amount of a gas that dissolves into a liquid is proportional to the partial pressure that gas exerts on the surface of the liquid. In equation form, that is: C AH = K p A where, CA = concentration of A, [mol/L] or [mg/L] KH = equilibrium constant (often called Henry's Law constant), [mol/L-atm] or [mg/L-atm] pA = partial pressure of A, [atm] David Reckhow CEE 370 L#7 2 Lecture #7 Dave Reckhow 1 CEE 370 Lecture #7 9/18/2019 Henry’s Law Constants Reaction Name Kh, mol/L-atm pKh = -log Kh -2 CO2(g) _ CO2(aq) Carbon 3.41 x 10 1.47 dioxide NH3(g) _ NH3(aq) Ammonia 57.6 -1.76 -1 H2S(g) _ H2S(aq) Hydrogen 1.02 x 10 0.99 sulfide -3 CH4(g) _ CH4(aq) Methane 1.50 x 10 2.82 -3 O2(g) _ O2(aq) Oxygen 1.26 x 10 2.90 David Reckhow CEE 370 L#7 3 Example: Solubility of O2 in Water Background Although the atmosphere we breathe is comprised of approximately 20.9 percent oxygen, oxygen is only slightly soluble in water. In addition, the solubility decreases as the temperature increases.
    [Show full text]
  • Gp-Cpc-01 Units – Composition – Basic Ideas
    GP-CPC-01 UNITS – BASIC IDEAS – COMPOSITION 11-06-2020 Prof.G.Prabhakar Chem Engg, SVU GP-CPC-01 UNITS – CONVERSION (1) ➢ A two term system is followed. A base unit is chosen and the number of base units that represent the quantity is added ahead of the base unit. Number Base unit Eg : 2 kg, 4 meters , 60 seconds ➢ Manipulations Possible : • If the nature & base unit are the same, direct addition / subtraction is permitted 2 m + 4 m = 6m ; 5 kg – 2.5 kg = 2.5 kg • If the nature is the same but the base unit is different , say, 1 m + 10 c m both m and the cm are length units but do not represent identical quantity, Equivalence considered 2 options are available. 1 m is equivalent to 100 cm So, 100 cm + 10 cm = 110 cm 0.01 m is equivalent to 1 cm 1 m + 10 (0.01) m = 1. 1 m • If the nature of the quantity is different, addition / subtraction is NOT possible. Factors used to check equivalence are known as Conversion Factors. GP-CPC-01 UNITS – CONVERSION (2) • For multiplication / division, there are no such restrictions. They give rise to a set called derived units Even if there is divergence in the nature, multiplication / division can be carried out. Eg : Velocity ( length divided by time ) Mass flow rate (Mass divided by time) Mass Flux ( Mass divided by area (Length 2) – time). Force (Mass * Acceleration = Mass * Length / time 2) In derived units, each unit is to be individually converted to suit the requirement Density = 500 kg / m3 .
    [Show full text]
  • Topic720 Composition: Mole Fraction: Molality: Concentration a Solution Comprises at Least Two Different Chemical Substances
    Topic720 Composition: Mole Fraction: Molality: Concentration A solution comprises at least two different chemical substances where at least one substance is in vast molar excess. The term ‘solution’ is used to describe both solids and liquids. Nevertheless the term ‘solution’ in the absence of the word ‘solid’ refers to a liquid. Chemists are particularly expert at identifying the number and chemical formulae of chemical substances present in a given closed system. Here we explore how the chemical composition of a given system is expressed. We consider a simple system prepared using water()l and urea(s) at ambient temperature and pressure. We designate water as chemical substance 1 and urea as chemical substance j, so that the closed system contains an aqueous solution. The amounts of the two substances are given by n1 = = ()wM11 and nj ()wMjj where w1 and wj are masses; M1 and Mj are the molar masses of the two chemical substances. In these terms, n1 and nj are extensive variables. = ⋅ + ⋅ Mass of solution, w n1 M1 n j M j (a) = ⋅ Mass of solvent, w1 n1 M1 (b) = -1 For water, M1 0.018 kg mol . However in reviewing the properties of solutions, chemists prefer intensive composition variables. Mole Fraction The mole fractions of the two substances x1 and xj are given by the following two equations: =+ =+ xnnn111()j xnnnjj()1 j (c) += Here x1 x j 10. In general terms for a system comprising i - chemical substances, the mole fraction of substance k is given by equation (d). ji= = xnkk/ ∑ n j (d) j=1 ji= = Hence ∑ x j 10.
    [Show full text]