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Unit 5 Exam Analysis

Kinetics Why are some chemical reactions slow (rusting of iron) and some reactions fast (burning of coal)?. Unit 5 Exam Analysis. 20 minute individual review….for half credit! 10 minute peer review Learning is a PROCESS! **Reflections NEED to show learning!

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Unit 5 Exam Analysis

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  1. KineticsWhy are some chemical reactions slow (rusting of iron) and some reactions fast (burning of coal)?

  2. Unit 5 Exam Analysis • 20 minute individual review….for half credit! • 10 minute peer review • Learning is a PROCESS! **Reflections NEED to show learning! • For credit you must correct your answer and provide a brief reflection of why your answer choice was incorrect and how you corrected it. (3 to 4 sentences per question)

  3. Unit 5 Exam Analysis • Next Unit – Chemical Kinetics and Equilibrium • Topic Checklist • Keep up with READINGS!! Complete Ch. 12 if you have not already! • Begin working through book practice and unit packet practice. • Quiz will take place Tuesday!

  4. Collision Theory of Reaction Rates • Atoms, ions, and molecules must collide in order to react. • Reacting substances must collide with the correct orientation. • Reacting substances must collide with sufficient energy (activation energy) to form the activated complex.

  5. Reaction Rate • The change in concentration of a reactant or product per unit of time. General Form: aA + bB cC + dD Rate = Units =

  6. Factors that Influence Reaction Rate • The nature of the reactants • Physical State • Concentration • Temperature • Catalysts • Activated complex • Surface Area • Video

  7. Ways to view the reaction rate. • 1. The rate of disappearance of a reactant. • 2. The rate of appearance of a product. • 3. The rate at which the overall reaction proceeds. From the data on the left, calculate the average rate at which A disappears over the time interval from 20. s to 40. s. Think about signs of reaction rates. What do they indicate?

  8. Reaction Rate Example For the reaction 2 H2(g) + O2(g)  2 H2O(g); when [O2] is decreasing at 0.23 mol/L∙s, at what rate is [H2O] increasing? (A: 0.46 mol/L∙s)

  9. Reaction Rate Example For the reaction: 2A + nBqC + rD If –Δ[A]/Δt = -0.050 mol/L∙s –Δ[B]/Δt= -0.150 mol/L∙s –Δ[C]/Δt = 0.075 mol/L∙s –Δ[D]/Δt = 0.025 mol/L∙s What are coefficients n, q, and r? (Hint: Use stoichiometry.)

  10. Rate Laws • Inquiry Activity • Partners • Be prepared to share out responses. • You may write out on the handout • Time: _________minutes • Big Ideas!!

  11. Rate Laws • The rate law for a reaction expresses the rate as a function of the reactant concentrations. For a reaction of the form aA + bB products the rate law has the form: Rate = k[A]m[B]n k is a proportionality constant called the rate constant, with units that when multiplied by [A]m[B]nproduce units of M/time for the rate. The exponents are used to describe the order of the reaction with respect to each reactant: 0 zero order 2 second order 1 first order 3 third order

  12. Rate Laws Continued… • The overall order of a reaction is the sum of all the exponents. • There is NO relationship between the exponents of the rate law and the coefficients of the balanced chemical equation!! The exponents MUST be determined by experiment!!

  13. Reaction Orders – Method of Initial Rates A + B  products

  14. Rate Law Example 1 2 NO(g) + O2(g)  2 NO2(g) What is the rate law and the value of the rate constant, k?

  15. Reaction Orders and Units for Rate Constants • Zero- order • First-order • Second-order • Third-order

  16. Rate Law Example 2 Initial rate data for the decomposition of gaseous N2O5 at 55°C are shown in the table below. What is the rate law and the value of the rate constant, k?

  17. Individual Practice

  18. Class Practice • Isoprene (C5H8) forms a dimer called dipentene (C10H16). 2 C5H8 C10H16 What is the rate law for this reaction given the following data for the initial rate of formation of dipentene? What is the value of the rate constant?

  19. Differential Rate Laws • Express the relationship between the concentration of reactants and the rate of the reaction. • The differential rate law is usually just called “the rate law”.

  20. The oxidation of iodide ion by hydrogen peroxide in an acidic solution is described by the equation H2O2(aq) + 3 I-(g) + 2 H+(aq)  I3-(aq) + 2 H2O The rate of formation of the red-colored triiodide ion, Δ[I3-]/ Δt, can be determined by measuring the rate of appearance of the color. Following are initial rate data at 25°C. a. What is the rate law for the formation of I3-? b. What is the value of the rate constant? c. What is the rate of formation of I3- when the concentrations are [H2O2] = 0.300 M and [I-] = 0.400 M?

  21. Integrated Rate Laws • Express the relationship between concentration of reactants and time • Take into account the passage of time in a chemical reaction. • For a general first-order reaction aAproducts Rate = k[A] • The integrated rate law is ln([A]t/[A]0) =-kt [A]t=[A]0e-kt

  22. First Order Reactions • Rearrange the first order integrated rate law: ln[A]t=-kt + ln[A]0 (this is the equation of a line…) The graph of ln[A]t (y-axis) vs. time (x-axis) will give a straight line (with a slope of –k) for a first order reaction, but not for a second order reaction. (linear relationship: time vs. ln(concentrations))

  23. Half-life of a first-order reaction • The time required for a reactant to reach half its original concentration. • Natural reactions are almost always 1st order reactions with a half-life of • Equation to use for completions other than 50% ln([A]t/[A]0) =-kt

  24. Half-life Example • A certain first-order reaction has a half-life of 25.0 minutes. What is the rate constant?

  25. Second Order Reaction • Ex: For a general second-order reaction aA products Rate = k[A]2 The integrated rate law is 1/[A]t=kt + 1/[A]0 The graph of 1/[A]t (y-axis) vs. time (x-axis) will give a straight line (with a slope of –k) for a second order reaction, but not for a first order reaction. (linear relationship: time vs. (1/concentration))

  26. Half-life of a second-order reaction

  27. Zeroth-Order Reactions • Ex: For a general zeroth-order reaction aA products Rate =k[A]0 =k The integrated rate law is [A]t=-kt + [A]0 The graph of [A]t (y-axis) vs. time (x-axis) will give a straight line (with a slope of –k) for a zeroth-order reaction. (linear relationship: time vs. concentrations) Half-life of a zeroth-order reaction:

  28. Orders and Graphs

  29. Rate Laws Summary

  30. Temperature and Reaction Rates • The rates of chemical reactions are sensitive to temperature: most reactions slow down at lower temperatures and speed up at higher temperatures (molecular theory). • This temperature dependence is contained in the rate constant, k. Rate = k[A] • Increasing the value of k increases the rate of reaction. • The temperature of dependence of k is given by the Arrhenius equation: k= Ae-Ea/RT

  31. k= Ae-Ea/RT • k is the rate constant • A is the frequency factor (orientation) • R is 8.314 J/mol K • T is temperature in Kelvin • Ea is the activation energy

  32. Slope of linear plot for a rate law is equal to :

  33. Example Problem The gas-phase reaction between methane and diatomic sulfur is given by the equation CH4(g) + 2 S2(g)  CS2(g) + 2 H2S(g) At 550°C the rate constant for this reaction is 1.1 L/mol∙s and at 625°C the rate constant is 6.4 L/mol∙s. Using these values, calculate Ea for this reaction.

  34. Reaction Mechanism • 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 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.

  35. Reaction Mechanism

  36. 2 A + 2 B  C + D (Rate = k[A]2[B]) The reaction takes place by the following three-step mechanism: I. A + A  X (fast) II. X + B  C + Y (slow) – rate determining step III. Y + B  D (fast) Elementary Reaction: Overall Reaction:

  37. Reaction: 2 O3(g)  3 O2(g) Mechanism: Step 1 O3(g)  O2(g) + O(g) unimolecular Step 2 O3(g) + O  2 O2(g) bimolecular Reaction Intermediate • Species in a reaction mechanism that does not show up in the final balanced equation. The species is produced and also consumed during the reaction.

  38. Example The following two-step mechanism has been proposed for the gas-phase decomposition of nitrous oxide (N2O). Step 1 N2O(g)  N2 (g) + O(g) slow Step 2 N2O(g) + O(g)  N2(g) + O2(g) fast a. Write the equation for the overall reaction. b. Identify any reaction intermediates. c. What are the molecularities (unimolecular or bimolecular) of the elementary reactions? d. What is the molecularity of the overall reaction?

  39. For an elementary reaction, the stoichiometric coefficients are equal to the reaction orders in the rate law. (This is true ONLY for elementary reactions!!) A  products Rate = k[A] 2A  products Rate = k[A]2 A + B  products Rate = k[A][B] 2A + B  products Rate = k[A]2[B] Rate-Determining Step

  40. Rate Laws for Overall Reactions: • The elementary steps in the mechanism must sum to the overall reaction. (All reaction intermediates must cancel out) • The rate law predicted by the mechanism must be consistent with the experimentally observed rate law. (That is, the rate law for the slow, rate-determining step in the mechanism must match the experimental rate law.)

  41. Example The reaction NO2(g) + CO(g) → NO(g) + CO2(g) can be thought of as occurring in two elementary steps: NO2+ NO2 → NO + NO3(slow step) NO3+ CO → NO2 + CO2(fast step) • What is the overall balanced equation? • What is the rate law?

  42. Example 2 • The reaction 2 NO2(g) + F2(g)  2 NO2F(g) has an experimental rate law of Rate = k[NO2][F2]. Propose a possible reaction mechanism and identify the rate-determining (slow) step.

  43. Catalysts A substance that speeds up a reaction without being consumed. Stabilizes a transition state which lowers the activation energy (surface catalysis) Forms a new reaction intermediate which provide a new pathway or mechanism (enzyme catalysis). 2A + 2B  D Step1: A + B  AB Step 2: AB + C  ABC Step 3: A + B  AB Step 4: AB + ABC  C + D • What are intermediates? • Which is the catalyst?

  44. Types of Catalysts • 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.

  45. How does a catalyst work?

  46. Kinetics Lab • Pre-lab: READ THROUGH THE LAB HANDOUT!! • Black Binder • State purpose • Create abstract of procedure • Construct the appropriate data tables • Quantitative (see suggested in handout) and Qualitative

  47. Kinetics Lab Partners • Group 1: Mackenzie and Austin • Group 2: Paige and Cass • Group 3: Cassie and Libby • Group 4: Kyli and Shae and Manu (Madi) • Group 5: Chetha and Odessa

  48. Individual Practice • Unit 6 Review Packet Kinetics • Kinetics Quiz –Review • Next we will begin Equilibrium (Ch. 13) • Begin reading!

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