1 / 63

Osmotic Pressure: a fascinating behavior. Yet it is the result of a very simple tendency to

Osmotic Pressure: a fascinating behavior. Yet it is the result of a very simple tendency to equalize the concentrations of solutions. The critical part is the membrane!!!. A VERY practical application/consequence of Osmotic Pressure. Hypertonic soln crenation (shrivels).

Download Presentation

Osmotic Pressure: a fascinating behavior. Yet it is the result of a very simple tendency to

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Osmotic Pressure: a fascinating behavior. Yet it is the result of a very simple tendency to equalize the concentrations of solutions.

  2. The critical part is the membrane!!!

  3. A VERY practical application/consequence of Osmotic Pressure.

  4. Hypertonic soln • crenation (shrivels) • Hypotonic soln • hemolysis (bursts)

  5. Cucumber placed in NaCl solution loses water to shrivel up and become a pickle. • Limp carrot placed in water becomes firm because water enters via osmosis. • Salty food causes retention of water and swelling of tissues (edema). • Water moves into plants through osmosis. • Salt added to meat or sugar to fruit prevents bacterial infection (a bacterium placed on the salt honey will lose water through osmosis and die).

  6. Osmotic Pressure π V = n R T or π = (n/V) R T or π = M R T π = ρ g h, where ρ = density of solution g = 9.807 m s-2 h= height of column

  7. π = ρ g h, where ρ = density of solution g = 9.807 m s-2 h= height of column If h = 0.17 m of a dilute aqueous soln with ρ = 1.00 g/cm3 π = [(1.00g/cm3)(10-3 kg/g)(106 cm3/m3)](9.807 m/s2)(0.17 m) = 1.7 x 103 kg m-1 s-1 = 1.7 x 103 Pa or = (1.7 x 103 Pa) / (1.013 x 105 Pa/atm) = 0.016 atm

  8. A chemist dissolves 2.00 g of protein in 0.100 L of water. The observed osmotic pressure is 0.021 atm at 25 oC. What is the MW of the protein?

  9. Colloids

  10. Hydrophilic and Hydrophobic Colloids • Focus on colloids in water. • “Water loving” colloids: hydrophilic. • “Water hating” colloids: hydrophobic. • Molecules arrange themselves so that hydrophobic portions are oriented towards each other. • If a large hydrophobic macromolecule (giant molecule) needs to exist in water (e.g. in a biological cell), hydrophobic molecules embed themselves into the macromolecule leaving the hydrophilic ends to interact with water.

  11. Hydrophilic and Hydrophobic Colloids…….. • Typical hydrophilic groups are polar (containing C-O, O-H, N-H bonds) or charged. • Hydrophobic colloids need to be stabilized in water. • Adsorption: when something sticks to a surface we say that it is adsorbed. • If ions are adsorbed onto the surface of a colloid, the colloids appears hydrophilic and is stabilized in water. • Consider a small drop of oil in water. • Add to the water sodium stearate.

  12. Hydrophilic and Hydrophobic Colloids

  13. Hydrophilic and Hydrophobic Colloids • Sodium stearate has a long hydrophobic tail (CH3(CH2)16-) and a small hydrophobic head (-CO2-Na+). • The hydrophobic tail can be absorbed into the oil drop, leaving the hydrophilic head on the surface. • The hydrophilic heads then interact with the water and the oil drop is stabilized in water.

  14. Removal of Colloidal Particles • Colloid particles are too small to be separated by physical means (e.g. filtration). • Colloid particles may be coagulated (enlarged) until they can be removed by filtration. • Methods of coagulation: • heating (colloid particles move and are attracted to each other when they collide); • adding an electrolyte (neutralize the surface charges on the colloid particles). • Dialysis: using a semipermeable membranes separate ions from colloidal particles

  15. Chapter 14 Chemical Kinetics 14.1 Factors that Affect Reaction Rates 14.2 Reaction Rates Changes of Rate with Time Reaction Rates and Stoichiometry 14.3 Concentration and Rate Exponents in the Rate Law Units of Rate Constants Using Initial Rates to Determine Rate Laws 14.4 The Change of Concentration with Time First-Order Reactions Second-Order Reactions Half-Life 14.5 Temperature and Rate 14.5 Reaction Mechanisms 14.7 Catalysis

  16. C4H9Cl(aq) + H2O (l)  C4H9OH (aq) + HCl (aq) Note the signs!

  17. In fact, the instantaneous rate corresponds to d[A]/dt

  18. Consider the reaction 2 HI(g)  H2(g) + I2(g) It’s convenient to define the rate as And, in general for aA + bB  cC + dD

  19. Sample exercise 14.2 The decomposition of N2O5 proceeds according to the equation 2 N2O5 (g)  4 NO2 (g) + O2 (g) If the rate of decomposition of of N2O5 at a particular instant in a vessel is 4.2 X 10-7 M/s, what is the rate of appearance of (a) NO2; (b) O2 ? i.e. the rate of the reaction is 2.1 x 10-7 M/s the rate of appearance of NO2 is 8.4 x 10-7 M/s and the rate of appearance of O2 is 2.1 x 10-7 M/s

  20. 2 N2O5 = 4 NO2 + O2 (g) at T = 45 oC in carbon tetrachloride as a solvent Time ∆t {N2O5] ∆[N2O5] - ∆[N2O5]/ ∆t min min mol/L mol/L mol/L-min 0 2.33 184 2.08 319 1.91 526 1.67 867 1.35 1198 1.11 1877 0.72

  21. 2 N2O5 = 4 NO2 + O2 (g) at T = 45 oC in carbon tetrachloride as a solvent Time ∆t {N2O5] ∆[N2O5] - ∆[N2O5]/ ∆t min min mol/L mol/L mol/L-min 0 2.33 184 0.25 1.36 x 10-3 184 2.08 135 0.17 1.26 x 10-3 319 1.91 207 0.24 1.16 x 10-3 526 1.67 341 0.32 0.94 x 10-3 867 1.35 331 0.24 0.72 x 10-3 1198 1.11 679 0.39 0.57 x 10-3 1877 0.72

  22. The information on the previous slide is a bit of a nuisance, since the instantaneous rate keeps changing —and you know how much we like constant values or linear relationships! So let’s try something rather arbitrary at this point. Let’s divide the instantaneous, average rate by [N2O5] and/or [N2O5]2

  23. 2 N2O5 = 4 NO2 + O2 (g) at T = 45 oC in carbon tetrachloride as a solvent {N2O5] [N2O5] ∆[N2O5] - ∆[N2O5]/ ∆t Avg rate Avg rate mol/L avg mol/L mol/L-min /[N2O5]av /[N2O5]av2 2.33 2.21 0.25 1.36 x 10-36.2 x 10-4 2.8 x 10-4 2.08 2.00 0.17 1.26 x 10-36.3 x 10-4 3.2 x 10-4 1.91 1.79 0.24 1.16 x 10-36.5 x 10-4 3.6 x 10-4 1.67 1.51 0.32 0.94 x 10-36.2 x 10-4 4.1 x 10-4 1.35 1.23 0.24 0.72 x 10-35.9 x 10-4 4.8 x 10-4 1.11 0.92 0.39 0.57 x 10-36.2 x 10-4 6.7 x 10-4 0.72 Notice the nice constant value!!!

  24. It’s convenient to write this result in symbolic form: Rate = k [N2O5] where the value of k is about 6.2 x 10-4 so that when [N2O5] = 0.221, Rate = (6.2 x 10-4 )(0.221) = 1.37 x 10-4 which is the ‘average rate’ we started with In fact, we really should take into account the 2 in front of the N2O5, in accordance with the rule we developed earlier.

  25. This leads us to the general concept of Reaction Order When Rate = k [reactant 1]m [reactant 2]n we say the reaction is m-th order in reactant 1 n-th order in reactant 2 and (m + n)-th order overall. Be careful—because these orders are NOT related necessarily to the stoichiometry of the reaction!!!

  26. Other reactions and their observed reaction orders 2 N2O5 = 4 NO2 + O2 (g) Rate = k [N2O5] !!! CHCl3 (g) + Cl2 (g)  CCl4 (g) + HCl(g) Rate = k[CHCl3][Cl2]1/2 H2 (g) + I2 (g)  2 HI (g) Rate = k[H2][I2] The order must be determined experimentally!!! We’ll see later that it depends on the Reaction Mechanism, rather than the overall stoichiometry. Be careful: the measurement of the rate will always depend on observations of the reactants or products and involves stoichiometry, but the part on the right, the order, does not depend on the stoichiometry.

  27. Let’s explore the results for the result Rate = k [N2O5] This can be expressed as Rate = - (Δ[N2O5] / Δ t = - d[N2O5] / dt = k [N2O5] or, in general for A  products Rate = - Δ[A] / Δt = d[A] / dt = k [A] rearrangement and integration from time = 0 to t = t gives the result ln[A]t - ln[A]o = -kt or ln [A]t = -kt + ln [A]o or ln ([A]t/[A]o = - kt This is the expression of concentration vs time for a First-Order Reaction

  28. High: 175/175 Lo: 10/175 Mean 106 (60.6%) DISTRIBUTION OF SCORES 0.00 - 9.99 0 10.00 - 19.99 1 20.00 - 29.99 0 30.00 - 39.99 4 40.00 - 49.99 4 50.00 - 59.99 22 60.00 - 69.99 32 70.00 - 79.99 37 80.00 - 89.99 48 90.00 - 99.99 61 100.00 - 109.99 61 110.00 - 119.99 51 120.00 - 129.99 52 130.00 - 139.99 31 140.00 - 149.99 37 150.00 - 159.99 25 160.00 - 169.99 12 170.00 - 175.00 2 BCDCC DBCEB CDBBE CAACA AEEDD ECBBC EBC These people need to talk with Dr. Mathews

  29. Chapter 14 Chemical Kinetics 14.1 Factors that Affect Reaction Rates 14.2 Reaction Rates Changes of Rate with Time Reaction Rates and Stoichiometry 14.3 Concentration and Rate Exponents in the Rate Law Units of Rate Constants Using Initial Rates to Determine Rate Laws 14.4 The Change of Concentration with Time First-Order Reactions Second-Order Reactions Half-Life 14.5 Temperature and Rate 14.5 Reaction Mechanisms 14.7 Catalysis

  30. Consider First-Order Reactions To give these forms of the “integrated rate law”:

  31. An example of the plots of concentration vs time for a First-Order Reaction

  32. The Change of Concentration with Time Half-Life • Half-life is the time taken for the concentration of a reactant to drop to half its original value. • That is, half life, t1/2 is the time taken for [A]0 to reach ½[A]0. • Mathematically,

  33. The Change of Concentration with Time For a First-Order Reaction The identical length of the first and second half-life is a SPECIFIC characteristic of First-Order reactions

  34. Consider now Second-Order Reactions

  35. Second-Order Reactions • We can show that the half life • A reaction can have rate constant expression of the form rate = k[A][B], i.e., is second order overall, but has first order dependence on A and B.

More Related