Properties of Solutions and Kinetics. Unit 8 Chapters 4.5, 13 and 14

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1 Properties of Solutions and Kinetics Unit 8 Chapters 4.5, 13 and 14

2 Unit 8.1: Solutions Chapters 4.5,

3 Classification of Matter Solutions are homogeneous mixtures

4 Solute A solute is the dissolved substance in a solution. Salt in salt water Sugar in soda drinks Solvent Carbon dioxide in soda drinks A solvent is the dissolving medium in a solution. Water in salt water Water in soda

5 Heat of Solution The Heat of Solution is the amount of heat energy absorbed (endothermic) or released (exothermic) when a specific amount of solute dissolves in a solvent. Substance Heat of Solution (kj/mol) NaOH NH 4 NO KNO HCl

6 Steps in Solution Formation H 1 Expanding the solute Separating the solute into individual components (requires energy) H 2 Expanding the solvent Overcoming intermolecular forces of the solvent molecules (requires energy) H 3 Interaction of solute and solvent to form the solution (releases energy)

7 Like Dissolves Like Nonpolar solutes dissolve best in nonpolar solvents Fats Steroids Waxes Benzene Hexane Toluene Polar and ionic solutes dissolve best in polar solvents Inorganic Salts Sugars Water Small alcohols Acetic acid

8 Solubility Trends The solubility of MOST solids increases with temperature. The rate at which solids dissolve increases with increasing surface area of the solid, higher temperatures, and agitating. The solubility of gases decreases with increases in temperature. The solubility of gases increases with the pressure above the solution.

9 Therefore Solids tend to dissolve best when: o Heated o Stirred o Ground into small particles gases tend to dissolve best when: o The solution is cold o Pressure is high

10 Saturation of Solutions A solution that contains the maximum amount of solute that may be dissolved under existing conditions is saturated. (equilibrium is reached and additional solute will not dissolve) A solution that contains less solute than a saturated solution under existing conditions is unsaturated. A solution that contains more dissolved solute than a saturated solution under the same conditions is supersaturated.

11 Solubility Chart

12 Calculations of Solution Concentration Mole fraction the ratio of moles of solute to total moles of solution Mole fraction of A = χ A = n A n A + n B Mol fractions do not have a unit 1.00 mol of HCl in 8.00 mol of water X HCl = X H2O = 0.889

13 Calculations of Solution Concentration Parts per million (ppm) mass of solute x 10 6 mass of solution Parts per billion (ppb) mass of solute x 10 9 mass of solution

14 Calculations of Solution Concentration: Molarity is the ratio of moles of solute to liters of solution Molarity = M = moles of solute Liter of solution 5 M or 5 molar means 5 moles of solute are in 1 liter of solution

16 MOLARITY A measurement of the concentration of a solution Molarity (M) is equal to the moles of solute (n) per liter of solution M = n / V = mol / L Calculate the molarity of a solution prepared by mixing 1.5 g of NaCl in ml of solution First calculate the moles of solute: 1.5 g NaCl ( 1 mole NaCl ) = moles of NaCl g NaCl Next convert ml to L: L of solution Last, plug the appropriate values into the correct variables in the equation: M = n / V = moles / L = mol/l

17 MOLARITY M = n / V = mol / L How many grams of LiOH are needed to prepare ml of a 1.25 M solution? First calculate the moles of solute needed: M = n / V, now rearrange to solve for n: n = MV n = (1.25 mol / L ) ( L) = moles of solute needed Next calculate the molar mass of LiOH: g/mol Last, use diminsional analysis to solve for mass: moles (23.95 g LiOH / 1 mol LiOH) = 7.48 g of LiOH

18 Review of electrolytes Strong electrolytes: conduct electricity strongly in solution because there are a lot of ions (soluble ionic, strong acids) Weak electrolytes: conduct electricity weakly in solution because there are a few ions (insoluble ionic, weak acids Nonelectrolyes: Do not conduct electricity in solution because there are no ions (covalent compounds)

19 Concentrations of ions in solution For strong electrolytes K 2 SO 4 2K + + SO 4 2-2M K 2 SO 4 solution is 4M of potassium ions and 2M of sulfate ions

20 MOLARITY & DILUTION M 1 V 1 = M 2 V 2 The act of diluting a solution is to simply add more water (the solvent) thus leaving the amount of solute unchanged. Since the amount or moles of solute before dilution (n i ) and the moles of solute after the dilution (n f ) are the same: n i = n f And the moles for any solution can be calculated by n=mv A relationship can be established such that M i V i = n i = n f = M f V f Or simply : M i V i = M b V b

21 MOLARITY & Dilution Calculate the molarity of a solution prepared by diluting 25.0 ml of 0.05 M potassium iodide with 50.0 ml of water M 1 = 0.05 mol/l M 2 =? V 1 = 25.0 ml V 2 = = 75.0 ml M 1 V 1 = M 2 V 2 M 1 V 1 = M 2 = (0.05 mol/l) (25.0 ml) = M of KI V ml

22 MOLARITY & dilution Given a 6.00 M HCl solution, how would you prepare ml of M HCl? M 1 = 6.00 mol/l M 2 = V 1 =? ml V 2 = ml M 1 V 1 = M 2 V 2 M 2 V 2 = V 1 = (0.150 mol/l) (250.0 ml) = 6.25 ml of 6 M HCl 6.00 mol/l M 1 You would need 6.25 ml of the 6.00 M HCl reagent which would be added to about 100 ml of DI water in a ml graduated cylinder then more water would be added to the mixture until the bottom of the menicus is at ml. Mix well.

23 MOLARITY & Stoichiometry How many grams of calcium carbonate will be precipitated by adding 25.0 ml calcium chloride to 25.0 ml of 0.56 M potassium carbonate? CaCl 2 + K 2 CO 3 CaCO KCl V = 25.0 ml M = 0.56 mol/l m=? V = 25.0 ml First convert volume of A to moles of A: L K 2 CO 3 (0.56 mol/l) = moles of K 2 CO 3 Now convert moles of A to moles of B: mol K 2 CO 3 (1 mol CaCO 3 /1 mol K 2 CO 3 ) =0.014 mol CaCO 3 Next convert moles of B to grams of B:

24 0.028 mol KCl / L = 0.56 M of KCl MOLARITY & Stoichiometry What would be the molarity of the potassium chloride solution from the last problem? CaCl 2 + K 2 CO 3 CaCO KCl V = 25.0 ml M = 0.56 mol/l M =? V = 25.0 ml First convert volume of A to moles of A: L K 2 CO 3 (0.56 mol/l) = moles of K 2 CO 3 Now convert moles of A to moles of B: mol K 2 CO 3 (2 mol KCl /1 mol K 2 CO 3 ) =0.028 mol KCl Next convert moles of B to molarity of B:

25 Suspensions and Colloids Suspensions and colloids are NOT solutions. Suspensions: The particles are so large that they settle out of the solvent if not constantly stirred. Colloids: The particles intermediate in size between those of a suspension and those of a solution.

26 Types of Colloids Examples Dispersing Dispersed Colloid Type Medium Substance Fog, aerosol sprays Gas Liquid Aerosol Smoke, airborn germs Gas Solid Aerosol Whipped cream, soap suds Liquid Gas Foam Milk, mayonnaise Liquid Liquid Emulsion Paint, clays, gelatin Liquid Solid Sol Marshmallow, Styrofoam Solid Gas Solid Foam Butter, cheese Solid Liquid Solid Emulsion Ruby glass Solid Solid Solid sol

27 Colloids scatter light, making a beam visible. Solutions do not scatter light. The Tyndall Effect Which glass contains a colloid? colloid solution

28 Unit 8.2: Rates of reactions Chapter 14

$However, only a small fraction of collisions produces a reaction.$

29 Collision Model Key Idea: Molecules must collide to react. However, only a small fraction of collisions produces a reaction. Why?

30 Collision Model 1. Collisions must have sufficient energy to produce the reaction (must equal or exceed the activation energy). 2. Colliding particles must be correctly oriented to one another in order to produce a reaction.

31 Endothermic Reactions

32 Exothermic Reactions

33 Factors Affecting Rate Increasing temperature always increases the rate of a reaction. Particles collide more frequently Particles collide more energetically Increasing surface area (particle size) increases the rate of a reaction Increasing Concentration (pressure for gases) USUALLY increases the rate of a reaction Presence of Catalysts, which lower the activation energy by providing alternate pathways

34 Catalysis Catalyst: A substance that speeds up a reaction without being consumed Enzyme: A large molecule (usually a protein) that catalyzes biological reactions. Homogeneous catalyst: Present in the same phase as the reacting molecules. Heterogeneous catalyst: Present in a different phase than the reacting molecules.

35 Lowering of Activation Energy by a Catalyst

36 The Arrhenius Equation k = Ae E / RT a k = rate constant at temperature T A = frequency factor E a = activation energy R = Gas constant, J/K mol

37 Reaction Rate The change in concentration of a reactant or product per unit of time Rate = [ A] at timet [ A] at timet t t Rate [ A] = t

38 A B 1 mole 0 mol 20 min mol 0.46 mol 40 min. 0.3 mol 0.7 mol A negative rate just means it s the reactant disappearing and a positive rate means it s the product appearing

39 2 HI H 2 + I 2 What is the rate of appearance of hydrogen if the rate of disappearance of HI is 6.0 x M/s? Answer: 3.0 x M/s -1/2 rate HI = rate H 2 = rate I 2

40 Reaction Rates: 2NO 2 (g) 2NO(g) + O 2 (g) 1. Can measure disappearance of reactants 2. Can measure appearance of products 3. Are proportional stoichiometrically

41 Reaction Rates: 2NO 2 (g) 2NO(g) + O 2 (g) 4. Are equal to the slope tangent to that point [NO 2 ] t 5. Change as the reaction proceeds, if the rate is dependent upon concentration [ NO2 ] t constant

42 Unit 8.3: Rate Laws Chapter 14

43 Rate Laws Differential rate laws express (reveal) the relationship between the concentration of reactants and the rate of the reaction. The differential rate law is usually just called the rate law. Integrated rate laws express (reveal) the relationship between concentration of reactants and time

44 Writing a (differential) Rate Law Problem - Write the rate law, determine the value of the rate constant, k, and the overall order for the following reaction: Experiment 2 NO(g) + Cl 2 (g) 2 NOCl(g) [NO] (mol/l) [Cl 2 ] (mol/l) Rate Mol/L s x x x x 10-6

45 Writing a Rate Law Part 1 Determine the values for the exponents in the rate law: R = k[no] x [Cl 2 ] y Experiment [NO] (mol/l) [Cl 2 ] (mol/l) Rate Mol/L s x x x x 10-5 In experiment 1 and 2, [Cl 2 ] is constant while [NO] doubles. The rate quadruples, so the reaction is second order with respect to [NO] R = k[no] 2 [Cl 2 ] y

46 Writing a Rate Law Part 1 Determine the values for the exponents in the rate law: Experiment R = k[no] 2 [Cl 2 ] y [NO] (mol/l) [Cl 2 ] (mol/l) Rate Mol/L s x x x x 10-5 In experiment 2 and 4, [NO] is constant while [Cl 2 ] doubles. The rate doubles, so the reaction is first order with respect to [Cl 2 ] R = k[no] 2 [Cl 2 ]

47 Writing a Rate Law Part 2 Determine the value for k, the rate constant, by using any set of experimental data: R = k[no] 2 [Cl 2 ] Experiment [NO] [Cl 2 ] Rate (mol/l) (mol/l) Mol/L s x x10 k mol mol mol = L s L L k x 10 mol L x10 L = 3 3 = L s mol mol s

48 Writing a Rate Law Part 3 Determine the overall order for the reaction. R = k[no] 2 [Cl 2 ] = 3 The reaction is 3 rd order Overall order is the sum of the exponents, or orders, of the reactants Most reactions go faster with increasing temperatures. In the rate law equation, k is higher at higher temperatures.

49 Determining Order with Concentration vs. Time data (the Integrated Rate Law) Zero Order: First Order: timevs. concentrationislinear timevs.ln( concentration) islinear Second Order: timevs. 1 concentration islinear

50 Solving an Integrated Rate Law Time (s) [H 2 O 2 ] (mol/l) Problem: Find the integrated rate law and the value for the rate constant, k A graphing calculator with linear regression analysis greatly simplifies this process!!

Time vs. ln[h 2 O 2 ] Time (s) ln[h 2 O 2 ] 0 0 120-0.0943 300-0.

51 Time vs. ln[h 2 O 2 ] Time (s) ln[h 2 O 2 ]

Time vs. [H 2 O 2 ] Time (s) [H 2 O 2 ] 0 1.00 120 0.91 300 0.

52 Time vs. [H 2 O 2 ] Time (s) [H 2 O 2 ]

53 Time vs. 1/[H 2 O 2 ] Time (s) 1/[H 2 O 2 ]

54 And the winner is Time vs. ln[h 2 O 2 ] 1. As a result, the reaction is 1 st order 2. The (differential) rate law is: R = kho [ ] The integrated rate law is: ln[ HO] = kt+ ln[ HO] But what is the rate constant, k?

55 Finding the Rate Constant, k Method #1: Calculate the slope from the Time vs. ln[h 2 O 2 ] table. slope ln[ HO] t 3600 s 2 2 = = slope = 8.32 x10 s Now remember: ln[ HO] = kt+ ln[ HO] k = -slope 4 1 k = 8.32 x 10-4 s -1 Time (s) ln[h 2 O 2 ]

56 Rate Laws Summary Zero Order First Order Second Order Rate Law Rate = k Rate = k[a] Rate = k[a] 2 Integrated Rate Law [A] = -kt + [A] 0 ln[a] = -kt + ln[a] = kt + [ A] [ A] 0 Plot the produces a straight line Relationship of rate constant to slope of straight line [A] versus t ln[a] versus t 1 [ A] versus t Slope = -k Slope = -k Slope = k Half-Life t 1/2 [ A] 0 = t1/2 2k k = 1/2 t 1 = ka [ ] 0

57 Unit 8.4: Kinetics Chapter 14

58 Class starter Dinitrogen pentoxide gas decomposes according to the equation: N 2 O 5 4NO 2 + O 2. This first-order reaction was allowed to proceed at 40 ᴼC and the data was collected. [N2O5] time (min) a. Calculate the rate constant using the values for concentration and time given in the table. Include units. b. After how many minutes will [N 2 O 5 ] be equal to M? c. What will be the concentration of N 2 O 5 after 100 minutes have elapsed? d. Calculate the initial rate of the reaction. Include units with your answer. e. What is the half life of the reaction?

59 Reaction Mechanism The reaction mechanism is the series of elementary steps by which a chemical reaction occurs. The sum of the elementary steps must give the overall balanced equation for the reaction The mechanism must agree with the experimentally determined rate law

60 NO 2 + CO NO + CO 2 Mechanism 1. NO 2 + NO 2 NO 3 + NO 2. NO 3 + CO NO 2 + CO 2 The two steps add up to give the overall rxn. NO 3 is not a reactant or a product in the overall rxn. It was produced and then consumed. NO 3 is the intermediate step.

61 Rate-Determining Step In a multi-step reaction, the slowest step is the rate-determining step. It therefore determines the rate of the reaction. The experimental rate law must agree with the rate-determining step

62 Determining the rate law from the NO 2 + CO NO + CO 2 Mechanism: mechanism 1. NO 2 + NO 2 NO 3 + NO (slow) 2. NO 3 + CO NO 2 + CO 2 (fast) Rate law is determined from the slow step Rate = k [NO 2 ] 2 The intermediate cannot be in the rate law

63 Another example 2 NO + Br 2 2 NOBr Mechanism 1. NO + Br 2 NOBr 2 (fast equilibrium) 2. NOBr 2 + NO 2NOBr (slow) Rate = k [NOBr 2 ][NO] This is a problem because NOBr 2 is an intermediate and cannot be in the rate law, but because the first step is a fast equilibrium we can replace NOBr 2 with NO and Br 2 so Rate = k [NO][Br 2 ][NO] or Rate = k [Br 2 ][NO] 2

64 Identifying Intermediates For the reaction: 2H 2 (g) + 2NO(g) N 2 (g) + 2H 2 O(g) Which species in the reaction mechanism are intermediates? Step #1 Step #2 H 2 (g) + 2NO(g) N 2 O(g) + H 2 O(g) N 2 O(g) + H 2 (g) N 2 (g) + H 2 O(g) 2H 2 (g) + 2NO(g) N 2 (g) + 2H 2 O(g) N 2 O(g) is an intermediate

65 Identifying the Rate-Determining Step For the reaction: 2H 2 (g) + 2NO(g) N 2 (g) + 2H 2 O(g) The experimental rate law is: R = k[no] 2 [H 2 ] Which step in the reaction mechanism is the ratedetermining (slowest) step? Step #1 Step #2 H 2 (g) + 2NO(g) N 2 O(g) + H 2 O(g) N 2 O(g) + H 2 (g) N 2 (g) + H 2 O(g) Step #1 agrees with the experimental rate law

Solvents. Solvents at the hardware store

Solvents. Solvents at the hardware store Solution Chemistry Parts of a Solution Review Solute A solute is the dissolved substance in a solution. Salt in salt water Sugar in soda drinks Solvent Carbon dioxide in soda drinks A solvent is the dissolving