Gases & colligative properties. Ch.14

Transcription

1 Gases & colligative properties Ch.14

2 Gases dissolving in liquids Pressure and temperature influence gas solubility Solubility directly proportional to gas pressure Henry s s Law: S g k H P g S g gas solubility (M mol/l) k H Henry s s law constant (unique to each gas; M/mm Hg) P g partial pressure of gaseous solute (mm Hg)

3 Increase partial pressure increase solubility

4 Example 27.0 g of acetylene gas dissolves in 1.00 L of acetone at 1.00 atm partial pressure of acetylene. If the partial pressure of acetylene is increased to 6.00 atm,, what is the solubility of acetylene in acetone in mol/l? MW of acetylene g/mol g x (mol/ g) x (1/1.00 L) 1.04 M 2. S g k H P g M k H x 1.00 atm 1. k H 1.04 M/atm 4. S g (1.04 M/atm atm) ) x 6.00 atm M Could also solve this by: (S g1 /P g1 ) (S g2 /P g2 ) How did I come up with this?

5 Problem The partial pressure of oxygen gas, O 2, in air at sea level is 0.21 atm. Using Henry s Law, calculate the molar concentration of oxygen gas in the surface water (at 20 C) of a lake saturated with air given that the solubility of O 2 at 20 C and 1.0 atm pressure is M.

6 Solution M 1.0atm S atm 2 4 S M

7 They call it pop in the Midwest Drinks carbonated under high pressure Above 90 atm Under CO 2 atmosphere Once bottle opened, partial pressure of gas above soda plummets CO 2 solubility decreases drastically Gas bubbles out of soln Once the fizz is gone, it can never be regained Truly, one of the existential tragedies of this universe

8 The bends Deeper diving has higher pressures Must use breathing tank If it contains N 2 then higher pressure forces N 2 to dissolve in higher amounts in blood If ascension too fast, lower pressure causes N 2 to start bubbling out of blood too quickly Rupturing of arteries Excruciatingly painful death Must be rushed to hyperbaric chamber Tanks now don t t use N 2, but He Why?

9 Effects of temp on solubility Obviously, as temp increases, solubility decreases Since increasing heat causes gases to dissolve out (endothermic) dissolving gases is an exothermic process

10 Another look at gas solubility: Le Châtelier s Principle Explains temperature relevance of solubility For systems in equilibrium, change in one side causes system to counteract on other side: Gas + liquid solvent sat. soln + heat So add heat, rxn goes to left by kicking out gas Add gas, rxn goes to right by saturating soln & giving off heat

11 Solubility of solids based on In general, solubility increases w/ increasing temp But exceptions No general behavior pattern noted temperature

12 Crystals One can separate impure dissolved salts by reducing temperature Impurity or desired product crystallizes out at specific temp as solubility collapses

13 Colligative properties Vapor and osmotic pressures, bp,, and mp are colligative properties Depend on relative # of solute and solvent particles

14 Remember: Vapor Pressure Equilibrium vapor pressure Pressure of vapor when liq and vapor in equilibrium at specific temp Vapor pressure of soln lower than pure solvent vapor pressure Vapor pressure of solvent relative # of solvent molecules in soln i.e., solvent vapor pressure solvent mole fraction

15 Raoult s Law P solution X solvent P solvent So if 75% of molecules in soln are solvent molecules (0.75 X solvent ) Vapor pressure of solvent (P( solvent ) 75% of P solvent

16 Problem The vapor pressure of pure acetone (CH 3 COCH 3 ) at 30 C is atm. Suppose 15.0 g of benzophenone, C 13 H 10 O (MW g/mol), is dissolved in 50.0 g of acetone (MW g/mol). Calculate the vapor pressure of acetone above the resulting solution.

17 Solution solute :15.0g solvent :50.0g X P solvent solution mol g mol 58.09g mol 0.861mol 0.861mol mol mol X P solvent atm solvent atm

18 Problem The vapor pressure of pure liquid CS 2 is atm at 20 C. When 40.0 g of rhombic sulfur (a naturally occurring form of sulfur) is dissolved in 1.00 kg of CS 2, the vapor pressure falls to atm. Determine the molecular formula of rhombic sulfur.

19 Solution Solution 8 rhombic sulfur rhombic sulfur 3 solvent solvent solvent solvent solution S g sulfur mol mol 256g mol 256g 0.156mol 40.0g mol mol 13.1mol 13.1mol mol g mol g :1.00kg solvent X atm X atm P X P +

20 Limitations of Raoult s Law Doesn t t take into consideration attractive forces in solns For ideal soln (to right), forces between solute/solvent molecules forces w/in pure solvent Thus, P tot P A + P B Like graph to right Fine for similarly constructed molecules (hydrocarbons) London dispersion forces are weakest

21 Solute-solvent > solv-solv Decreases vapor pressure decreased volatility Get lower vapor pressure than calculated Ex: CHCl 3 & C 2 H 5 OC 2 H 5 H on former H-bonds H to latter Does it increase or decrease the latter s IMF?

22 Solute-solvent < solv-solv Increases vapor pressure increased volatility Get higher vapor pressure than calculated Ex: C 2 H 5 OH and H 2 O Former disrupts H-H bonding of latter Does it increase or decrease the latter s IMF?

23 Salts Nonvolatile solute added to solvent Lower vapor pressure of solvent Make solvent less volatile

24 Nonvolatile solute added to Raises bp Lowers mp Why? Adding more nonvolatile solute or increasing solute molality decreases vapor pressure even more Phase diagram to right Pure water (black) Adulterated water (pink) solvent

25 Bp and molality relationship T bp K bp m solute K bp molal boiling pt elevation constant for solvent ( C/ C/m) Bp elevation, T bp, directly proportional to solute molality

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27 Antifreeze Propylene glycol 1,2-propanediol Formerly used ethylene glycol Phased out Poisonous Lowers melting pt Increases boiling pt Reduces risk of radiator boiling over Appreciated during the summer months in the desert

28 Example Pure toluene (C 7 H 8 ) has a normal boiling point of C. A solution of 7.80 g of anthracene (C 14 H 10 ) in g of toluene has a boiling point of C. Calculate K b for toluene. 1. T bp K bp m solute bp C - 2. T bp C 1.46 C g x (mol/178.23g) 4.38 x 10-2 mol 4. (4.38 x 10-2 mol/ kg) m C/0.438 m 3.33 C/m

29 Freezing point depression Similarly, T fp K fp m solute K fp molal fp depression constant ( C/( C/m) Antifreeze & CaCl 2

30 Problem Barium chloride has a freezing point of 962 C and a K f of 108 C/m. A solution of 12.0 g of an unknown substance dissolved in 562 g of barium chloride gives a freezing point of 937 C. Determine the molecular weight of the unknown substance.

31 T T m moles solute 12.0g 0.13mol Solution C C 108 m moles solute 0.562g solvent 92g mol 937 C 0.13 C m 25 C

32 Solutions containing ions: their colligative properties Colligative properties based on amount of solute/solvent Molality of ions depend on number of constituents in cmpd Different for ionic vs. covalent cmpds Ex: 1. NaCl ionizes into two ions So 0.5 m NaCl has 0.5 x 2 m 1 m tot 2. Benzene doesn t t ionize So 0.5 m benzene 0.5 m tot Using equation w/out above factor will lead to values that are off

33 How to correct for it: the van t Hoff factor i the number of solute particles after dissolving Colligative properties are larger for electrolytes than for nonelectrolytes of the same molality Why? (Hint: solve the below) Give the i-values for: methanol, CaSO 4, BaCl 2 T fp (measured) K fp m i

34 Problem How many grams of Al(NO 3 ) 3 must be added to 1.00 kg of water to raise the boiling point to C K b 0.51 C/ C/m MW g/mol

35 solution T 5.0 m C 2.5 moles solute 2.5mol 0.51 m C C m 4 moles solute 1.00kg solvent g mol C 5.0 C 530g Al(NO 3 ) 3 needed

36 Osmosis Net movement of water (solvent) from area of lower solute concentration to area of higher solute concentration across a semi-permeable membrane Bio101

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38 More Pressure of column of soln pressure of water moving through membrane Osmotic pressure pressure made by column of soln diff of heights Π crt c mol/l M R L atm/(mol K) ideal gas law T in Kelvin Π atm Useful for measuring MM of biochemical macromolecules Proteins and carbs