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Gas-Solution Processes Gas Solubility Raoult’s Law Edward A. Mottel Department of Chemistry Rose-Hulman Institute of Technology.

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Presentation on theme: "Gas-Solution Processes Gas Solubility Raoult’s Law Edward A. Mottel Department of Chemistry Rose-Hulman Institute of Technology."— Presentation transcript:

1 Gas-Solution Processes Gas Solubility Raoult’s Law Edward A. Mottel Department of Chemistry Rose-Hulman Institute of Technology

2 6/28/2015 Solutions  Reading assignment: Fine, Beall & Stuehr: Chapter 9.1, 9.3  A homogeneous mixture of two or more substances in a single phase. Solute ·The material dissolved in a solution. Solvent ·The medium into which a solute is dissolved to form a solution.

3 6/28/2015 Solutions  What is the solute and what is the solvent? An ounce of isopropanol (rubbing alcohol) is added to a cup of water. A drop of water is added to a cup of isopropanol. A cup of water is added to a cup of isopropanol.

4 Molality One cup of isopropanol is mixed with one cup of water. isopropanol CH 3 -CHOH-CH 3 0.78 g  mL -1 water 1.00 g  mL -1 Molality (m) = moles of solute per kilogram of solvent What is the molality of the solution?

5 6/28/2015 Henry’s Law Assumes that the gas and the solvent don’t react with each other. m dissolved gas = k · P gas

6 6/28/2015 Henry’s Law m dissolved gas = k · P gas  dissolved gas = k’ · P gas The concentration of the gas in solution could also be defined in terms of mole fraction of gas in solution.

7 6/28/2015 Henry’s Law What are the axes? What is the slope? Does the line go through the origin? m dissolved gas = k · P gas Draw a graph that represents Henry’s Law

8 6/28/2015 Henry’s Law Partial Pressure of Gas (torr) 0 0 moles of dissolved gas per kilogram of solvent (m) slope = Henry’s Law constant (k)

9 6/28/2015 Solubility of Gases in Water 1 atm gas pressure above solution CH 4 O2O2 CO N2N2 He Temperature (ºC) Solubility (mmol/kg solvent) 0 0 1 2 102030 What effect does heating have on the amount of dissolved gas?

10 6/28/2015 Solubility of Gases  The solubility of gas in liquids usually decreases with increasing temperature.  However, some gases become more soluble in organic solvents with increasing temperature.

11 6/28/2015 Henry’s Law Applications  Carbonated beverages  Diver’s bends m dissolved gas = k · P gas

12 6/28/2015 Carbonated Beverages m dissolved CO 2 = 0.034 m·atm -1 · P CO 2 Packaged under about 2 atm of CO 2 pressure. Why does a carbonated beverage go "flat" overnight?

13 6/28/2015 Diver's Bends  Caused by dissolved nitrogen in the blood. A diver can go deep for a short period of time, if nitrogen gas does not have a chance to equilibrate in the blood. How do the partial pressures of nitrogen and oxygen in the blood compare at 1 atm pressure and under 33 feet of water?

14 6/28/2015 Physiological Implications Henry’s Law  What are the problems associated with breathing pure oxygen for a sustained period? How do scuba divers avoid breathing high partial pressures of oxygen?  Why is it harder to work and breathe in the mountains? Why do athletes train in the mountains?

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16 Gas-Solution Processes Colligative Properties Edward A. Mottel Department of Chemistry Rose-Hulman Institute of Technology

17 Please Pick Up a Copy of Raoult’s Law and Vapor Pressure problem set Solution Reactions and Concentrations problem set

18 6/28/2015 Temp (°C)P (torr) 109.2 2017.5 3031.8 4055.3 5092.5 Vapor Pressure Water On the graph where is the boiling point located? What does a vapor pressure of water graph look like?

19 Vapor Pressure of Water 0 20 40 60 80 100 01020304050 Temperature (°C) Pressure (torr) What will be the vapor pressure when the temperature is 100 °C?

20 6/28/2015 Vapor Pressure of Water Which of the dozen water molecules shown are evaporating? What factors determine whether a molecule will evaporate?

21 6/28/2015 Which Is More Volatile? Water or Isopropanol Vapor Pressure of Water & Isopropanol

22 6/28/2015 Vapor Pressure of a Solution What factors determine the total vapor pressure of the solution?

23 6/28/2015 Raoult’s Law  The vapor pressure of an ideal solution is equal to the sum of the vapor pressure of each component times its mole fraction. A P   B P   total P  A   B   

24 6/28/2015 P° A is the vapor pressure of pure A. A P   B P   total P  A   B    If the solvent or solute is non-volatile, then P° for that species is zero.

25 6/28/2015  Non-ideal solutions have vapor pressures higher or lower than predicted. A P   B P   total P  A   B    Raoult’s Law

26 6/28/2015 Raoult’s Law Application  Distillation: The separation of one component from a solution by boiling and condensation of the vapor.

27 6/28/2015 Benzene-Toluene Distillation  One mole of benzene is mixed with three moles of toluene.  What will be the vapor pressure of the solution at 88  C. Benzene (C 6 H 6 ) and toluene (C 6 H 5 CH 3 ) form an ideal solution. P  benzene is 960 torr at 88  C P  toluene is 380 torr at 88  C What other questions could be asked?

28 6/28/2015 Benzene-Toluene Distillation  One mole of benzene is mixed with three moles of toluene.  What will be the composition of the vapor that is condensed?

29 6/28/2015 Distillation  Distillation separates one compound from another because the vapor and the solution have different compositions.  How could you make a solution containing two volatile components very pure?

30 6/28/2015 Fractional Distillation

31 6/28/2015 Not All Compounds can be Separated by Distillation  Non-ideal solutions that have the same composition in solution and the vapor are called azeotropes. Isopropanol (BP 82  C) and water (BP 100  C) forms a 80.4  C azeotrope of composition  isopropanol = 0.878,  water = 0.122

32 6/28/2015 Colligative Properties  A physical property of a solution which depends upon the concentration of the solute, but not its nature or identity.  Examples: Vapor pressure lowering Boiling point elevation

33 6/28/2015 Vapor Pressure Lowering Why does the dissolution of a non-volatile solute result in a lower solution vapor pressure than the pure solvent? Describe the cause based on a mathematical argument and a mechanical argument.

34 6/28/2015 Under what conditions does boiling occur? Temperature ( o C) 0 200 400 600 800 020406080100120 Vapor Pressure of Water Solutions Pressure (torr) Water Sugar Water

35 6/28/2015 Temperature ( o C) 0 200 400 600 800 020406080100120 Vapor Pressure of Water Solutions Pressure (torr) Water Sugar Water What is the vapor pressure of a 0.10 mole fraction sugar in water solution at 100 °C?

36 6/28/2015 Temperature ( o C) 0 200 400 600 800 020406080100120 Vapor Pressure of Water Solutions Pressure (torr) Water Sugar Water If the vapor pressure of a solution is lowered, why is the boiling point elevated?

37 6/28/2015 Boiling Point Elevation   T is the increase in boiling point  k b is the boiling point (ebullioscopic) constant ·k b = 0.512 K  m -1 for water  m is the molality of the solution  T k mi b 

38 6/28/2015 Boiling Point Elevation  Sugar water does not conduct electricity.  k b = 0.512 K  m -1 for water  What is the boiling point of 0.10 mole fraction sugar water solution?  T k mi b 

39 6/28/2015 Boiling Point Elevation  i is the multiplicative factor to account for ions if a compound does not ionize then there is one particle per molecule and i = 1. if a compound breaks into two ions then i = 2, etc.  T k mi b 

40 6/28/2015 Rank the following solutions in terms of increasing boiling point water 0.2 m salt water 0.3 m sugar water  T k mi b 

41 6/28/2015 Freezing Point Depression  The dissolution of a solute in a solvent results in a lower freezing point for the solution.  T k mi f 

42 6/28/2015 Freezing Point Depression How does an impurity change the freezing point of the triangular molecules?

43 6/28/2015 Freezing Point Depression   T is the decrease in freezing point  k f is the freezing point (cryoscopic) constant ·k f = 1.86 K  m -1 for water  Applications  T k mi f 

44 6/28/2015 Osmotic Pressure  A pressure differential due to solvent molecules passing through a semipermeable membrane from a region of lower solute concentration to a region of higher solute concentration.  mRT

45 6/28/2015 Osmotic Pressure

46 6/28/2015 Osmotic Pressure

47 6/28/2015 Osmotic Pressure

48 6/28/2015 Osmotic Pressure   mRT

49 6/28/2015 Osmotic Pressure Applications   mRT Molecular weight determination Biochemical Why is physiological 0.9% saline solution important in intravenous transfusions?

50 6/28/2015 Historical Perspective  Colligative properties were important to the early chemists because it provided information about the molecular weight or degree of ionization of the solute.


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