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Demo today ???

Demo today ???. Colloids. 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.

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Demo today ???

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  1. Demo today ???

  2. Colloids

  3. 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.

  4. 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.

  5. Hydrophilic and Hydrophobic Colloids

  6. 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.

  7. 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

  8. 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

  9. Following the Progress of the Reaction A  B

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

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

  12. 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

  13. 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 disappearance of N2O5 is 4.2 x 10-7 M/s 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

  14. 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

  15. 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

  16. 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

  17. 2 N2O5 = 4 NO2 + O2 (g) at T = 45 oC in carbon tetrachloride as a solvent [N2O5] [N2O5]av∆[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!!! which we’ll call ‘k’

  18. 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.

  19. 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!!!

  20. 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.

  21. Let’s explore the results when 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 where [A]o = conc. at t = 0 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

  22. Same information in a better format: To give these forms of the “integrated rate law”:

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

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

  25. 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,

  26. 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

  27. Consider now Second-Order Reactions Note the 2 !

  28. Second-Order Reactions • We can show that the half life • A reaction also can have rate constant expression of the form rate = k[A][B], i.e., be second order overall, but be first order in A and in B.

  29. Recall for 1st order: And for 2nd order: Is this first or second order in the reactant? What is k?

  30. t1/2 = 1.73 sec (‘2nd half-life’) t1/2 = [(0.4)(0.5)] -1 = 5.0 sec (‘2nd half life’) t1/2 = (k[A]0)-1 = [(0.4)(1.0)] -1 = 2.5 sec (‘1st half-life’) t1/2 = 0.693/0.4 = 1.73 sec

  31. MQ-1 122, WI 07: THE MEAN WAS 112.00 (64.00%) THE MODE WAS 95.00 WITH 10 STUDENTS. High: 175 THE MEDIAN WAS 112.50 Low 38 THE STANDARD DEVIATION WAS 29.35 N = 472 THE POSSIBLE RANGE WAS FROM 0.00 TO 175.00 THE ACTUAL RANGE WAS FROM 38.00 TO 175.00 (21.7% TO 100.0%) CUTS TAKEN AT 72.88 (APROX. 72.50) AND 26.86 (APROX. 27.50) PERCENTILES. THE UPPER CUT CONTAINED 130 STUDENTS; THE LOWER, 130 STUDENTS. TOTAL NUMBER OF DETECTED ERRORS: 0. DISTRIBUTION OF SCORES 0.00 - 9.99 0 10.00 - 19.99 0 20.00 - 29.99 0 30.00 - 39.99 1 40.00 - 49.99 8 50.00 - 59.99 6 60.00 - 69.99 22 70.00 - 79.99 32 80.00 - 89.99 45 90.00 - 99.99 50 100.00 - 109.99 51 110.00 - 119.99 59 120.00 - 129.99 50 130.00 - 139.99 53 140.00 - 149.99 38 150.00 - 159.99 42 160.00 - 169.99 12 170.00 - 175.00 3 Correct answers: BCDAE CCDDE CADBE BDABD AAADC ACDBA A These students are in danger of failing! Please See Me!!!

  32. 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

  33. First Order Reactions Second Order Reactions

  34. Example Exercise 14.8 NO2 (g)  NO (g) + ½ O2 (g) at 300 oC Page 540 Time/s[NO2]ln[NO2]1/[NO2] 0.0 0.0100 -4.610 100 50.0 0.00787 -4.845 127 100.0 0.00649 -5.038 154 200.0 0.00481 -5.337 208 300.0 0.00380 -5.573 263 Is the reaction first or second order in NO2 ? What is the rate constant for the reaction?

  35. NO2(g) NO (g) + 1/2 O2(g) Second-Order Processes The decomposition of NO2 at 300°C is described by the equation and yields data comparable to this:

  36. Second-Order Processes • Graphing ln [NO2] vs.t yields: • The plot is not a straight line, so the process is not first-order in [A].

  37. Second-Order Processes • Graphing ln 1/[NO2] vs. t, however, gives this plot. • Because this is a straight line, the process is second-order in [A].

  38. Example of Second-Order Plots of conc vs time NO2 (g)  NO (g) + ½ O2 (g) at 300 oC (See page 591) Since graph (b) gives a straight line, it is a second-order reaction in NO2 . i.e. rate = k[NO2]2 . And the slope is k, where k = 0.534 M -1 s -1 .

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