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Chemical Kinetics

Chemical Kinetics. Unit 11. Factors affecting reactions Collision Model Activation energy, activated complex Exothermic/endothermic reactions Energy of reactions ∆E Reaction Rate Average reaction rate Instantaneous reaction rate Kinetics Stoichiometry Reaction Rate and Concentration

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Chemical Kinetics

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  1. Chemical Kinetics Unit 11

  2. Factors affecting reactions Collision Model Activation energy, activated complex Exothermic/endothermic reactions Energy of reactions ∆E Reaction Rate Average reaction rate Instantaneous reaction rate Kinetics Stoichiometry Reaction Rate and Concentration (Differential) Rate laws Specific rate constant (k) Writing rate laws Reaction Order Rate laws Integrated Rate Laws Graphing Solving (0, 1st, 2nd order) Half life equations Rate Constant and Temperature Arrhenius equation Catalysis Homogeneous, heterogeneous, enzymes Reaction Mechanisms Identifying intermediates Rate determining step Overview

  3. Chemical Kinetics • Thermochemistry = does a reaction take place? • Kinetics = how fast does the reaction happen? • Kinetics: The area of chemistry concerned with the speeds (rates) of reactions

  4. Factors Affecting Reaction Rates • Physical state of reactants (surface area) • Concentration of reactants • Temperature at which reaction occurs • Presence of a catalyst • Pressure of gaseous reactants or products

  5. Molecules must collide to react Only a small fraction of collisions produces a reaction. Why? Collision Model

  6. Collision Model • Collisions must have enough energy to produce the reaction (must equal or exceed the activation energy). • Reactants must have proper orientation to allow the formation of new bonds.

  7. Collision Model

  8. Activation Energy • Activation Energy (Ea)– minimum energy required to transform reactants into the activated complex • (The minimum energy required to produce an effective collision) • Example: flame, spark, high temperature, radiation • The lower the Ea, the faster the reaction

  9. Activation Energy • Activated Complex – transitional structure • All bonds have been broken but no new bonds have formed • Point in reaction of highest energy but lowest stability

  10. Exothermic Processes Processes in which energy is released as it proceeds, and surroundings become warmer Reactants  Products + energy

  11. Endothermic Processes Processes in which energy is absorbed as it proceeds, and surroundings become colder Reactants + energy  Products

  12. Energy of Reactions (∆E) • Same as ∆H • “Heat” and “energy” can be used interchangeably • ∆E = ∆Eproducts- ∆Ereactants • - ∆E = exothermic (energy lost) • +∆E = endothermic (energy gained)

  13. A B rate = D[A] D[B] rate = - Dt Dt Reaction Rate Reaction Rate: the change in concentration of reactants per unit time as a reaction proceeds (M/s) D[A] = change in concentration of A over time period Dt D[B] = change in concentration of B over time period Dt Because [A] decreases with time, D[A] is negative.

  14. A B time rate = D[A] D[B] rate = - Dt Dt 13.1

  15. Reaction Rate • Average reaction rate: average rate throughout the course of the reaction • Instantaneous reaction rate: rate of reaction at a specific point in time during the reaction • Think of a car trip where you drove 100 miles in 2 hours • Average speed = 50 mi/hr • Instantaneous speed is speed at specific point – you don’t go exactly 50 mi/hr the entire trip

  16. Br2(aq) + HCOOH (aq) 2Br-(aq) + 2H+(aq) + CO2(g) slope of tangent slope of tangent slope of tangent [Br2]final – [Br2]initial D[Br2] average rate= - = - Dt tfinal - tinitial instantaneous rate= rate for specific instance in time (slope of tangent)

  17. Kinetics Stoichiometry • When mole ratios are not 1:1, stoichiometry is used to compare rates of reactions aA + bB  cC + dD Rate = - = - = = 1 ∆[A] a ∆t 1 ∆[B] b ∆t 1 ∆[C] c ∆t 1 ∆[D] d ∆t

  18. Kinetics Stoichiometry 2HI  H2 + I2 If the rate at which H2 appears, ∆[H2]/∆t, is 6.0×10-5 M/s at a particular instant, at what rate is HI disappearing at this same time, - ∆[HI]/∆t? Rate = - = Therefore, - = (6.0×10-5 M/s ) - = 1.2×10-4 M/s 1 ∆[HI] 2 ∆t 1 ∆[H2] 1 ∆t 1 1 1 ∆[HI] 2 ∆t ∆[HI] ∆t

  19. Reaction Rate and Concentration • The initial rate of a reaction depends on the concentration of the reactants • Rate changes based on the concentration of reactants • (Differential) Rate Law: an equation that relates reaction rate and concentrations of reactants Rate a[reactants] ameans “proportional to”

  20. Rate Law R = k[A]n[B]m R = reaction rate k= specific rate constant [A] and [B] = concentrations of reactants n and m = the order of the reactant (usually 0, 1, 2) (not related to moles of reactants from the equation) Overall order of reaction = n + m

  21. F2(g) + 2ClO2(g) 2FClO2(g) 1 Rate Laws • Rate laws are always determined experimentally. • Reaction order is always defined in terms of reactant (not product) concentrations. • The order of a reactant is not related to the stoichiometric coefficient of the reactant in the balanced chemical equation. rate = k [F2][ClO2] 13.2

  22. Specific Rate Constant (k) • Once the reaction orders are known, the value of k must be determined from experimental data • Value of k is for a specific reaction; k has a different value for other reactions, even at the same conditions • Units of k depend on the overall order of the reaction • Value of k does not change for different concentrations of reactants or products • Value of k is for reaction at a specific temperature; if temperature increases, k increases • Value of k changes if a catalyst is present

  23. 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: 2 NO(g) + Cl2(g)  2 NOCl(g)

  24. Writing a Rate Law Step 1 – Determine the values for the exponents in the rate law: R = k[NO]n[Cl2]m In experiment 1 and 2, [Cl2] is constant while [NO] doubles. The rate quadruples, so the reaction is second order with respect to [NO]  R = k[NO]2[Cl2]m

  25. Writing a Rate Law R = k[NO]2[Cl2]m In experiment 2 and 4, [NO] is constant while [Cl2] doubles. The rate doubles, so the reaction is first order with respect to [Cl2]  R = k[NO]2[Cl2]

  26. 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[Cl2]

  27. Reaction Order Given the rate law R = k[NO]2[Cl2] We say that the reaction is • 2nd order with respect to NO • 1st order with respect to Cl2 Overall the order of the reaction is a 3rd order reaction • 2 + 1 = 3

  28. Example 2 What is the order with respect to A? What is the order with respect to B? What is the overall order of the reaction? 0 1 1

  29. Example 3 What is the order with respect to Cl2? What is the order with respect to NO? What is the overall order of the reaction? 1 2 3

  30. Reaction Order

  31. Concentration VS Time • Integrated Rate Law • Graph data to solve for order of reaction Zero Order = time vs concentration is linear 1st Order = time vs ln[concentration] is linear 2nd Order = time vs is linear 1 concentration

  32. Zero Order Reactions 13.3

  33. First Order Reactions 13.3

  34. Second Order Reactions

  35. Solving an Integrated Rate Law Problem: Find the integrated rate law and the value for the rate constant, k

  36. Time vs. [H2O2] Time Regression results: [H2O2] y = ax + b a = -2.64 x 10-4 b = 0.841 r2 = 0.8891 r = -0.9429

  37. Time vs. ln[H2O2] ln[H2O2] Time Regression results: y = ax + b a = -8.35 x 10-4 b = -.005 r2 = 0.99978 r = -0.9999

  38. Time vs. 1/[H2O2] Time 1/[H2O2] Regression results: y = ax + b a = 0.00460 b = -0.847 r2 = 0.8723 r = 0.9340

  39. Integrated Rate Law • The graph is linear for “time vs ln[H2O2]” • The reaction is 1st order VS VS

  40. Concentration VS Time Rate laws for concentration of reactants versus time are based on the linear equation y = ax + b • Zero Order Reactions (time vs concentration) [A] = -kt + [A]0 • 1st Order (time vsln[concentration]) ln[A] = -kt + ln[A]0 • 2nd Order (time vs 1/concentration)

  41. The reaction 2A B is first order in A with a rate constant of 2.8 x 10-2 s-1 at 800C. How long will it take for A to decrease from 0.88 M to 0.14 M ? 0.88 M ln 0.14 M = 2.8 x 10-2 s-1 ln ln[A]0 – ln[A] = k k [A]0 [A] ln[A] = -kt + ln[A]0 [A]0 = 0.88 M ln[A] - ln[A]0 = - kt [A] = 0.14 M ln[A]0 - ln[A] = kt = 66 s t =

  42. Half-life t½ • Time it takes for concentration of reactant to equal ½ of its initial value • A fast reaction will have a short half-life • In a first order reaction, concentration of reactant decreases by ½ each time interval of t½ [A] = ½ [A]0

  43. Half-life t½ • For zero and second order reactions, half-life changes based on concentration of the reactant • To find equation for half-life, place ½[A] 0 in for [A] in each integrated rate law and put t½ in for t When [A] = ½[A]0 and t = t½ … Zero Order: [A] = -kt + [A]0 1st Order ln[A] = -kt + ln[A]0 2nd Order

  44. What is the half-life of N2O5 if it decomposes with a rate constant of 5.7 x 10-4 s-1? t½ 0.693 = 5.7 x 10-4 s-1 Half-life t½ = 1200 s = 20 minutes units of k are (s-1) How do you know decomposition is first order?

  45. Concentration-Time Equation Order Rate Law Half-Life 1 1 - = kt [A] [A]0 = [A]0 t½ = t½ t½ = Ln 2 2k k 1 k[A]0 SUMMARY [A] - [A]0 = - kt rate = k 0 ln[A] - ln[A]0 = - kt 1 rate = k [A] 2 rate = k [A]2

  46. Rate Constant and Temperature • Reaction rate increases with temperature • Molecules move faster • More molecules collide to cause a reaction in a time period • Molecules have greater kinetic energy at higher temperatures • More molecules will have enough energy to reach the activation energy of the reaction at higher temperatures

  47. Boltzmann Distribution Curve

  48. Rate Constant and Temperature • At higher temperatures (T2), more molecules are able to reach the activation energy

  49. Lnk= - + lnA -Ea 1 T R Rate Constant and Temperature • Arrhenius Equation: relationship between k and T Eais the activation energy (J/mol) R is the gas constant (8.314 J/K•mol) T is the absolute temperature (K) A is the frequency factor Takes into account collision frequency and orientation

  50. Graphing Arrhenius • A plot of lnkversus 1 / T produces a straight line with the familiar form y = - mx + b, where x = 1 / T y = lnk m = - E a / R b = lnA • The activation energy E a can be determined from the slope m of this line: E a = - m · R

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