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Chemical Kinetics: Measuring Reaction Rates and Factors Affecting Rate

This chapter explores the measurement of reaction rates and the factors that affect them, such as reactant concentrations, temperature, and the presence of catalysts. The method of initial rates is also discussed.

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Chemical Kinetics: Measuring Reaction Rates and Factors Affecting Rate

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  1. Chapter 14Chemical Kinetics

  2. Kinetics – The speed (or rate) at which a reaction takes place 2 H2O2 (aq) 2 H2O () + O2 (g) 2 ways to measure the rate of any reaction 1. Measure the Increase in [Products] over time 2. Measure the Decrease in [Reactants] over time How do you measure this ? Measure a change in concentration per change in time

  3. y x [ ] t C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq) slope = Rate = slope = Rate

  4. [ Products ] t [ Reactants ] t [ Reactants ] t [ Products ] t slope = -# slope = + # when when Rate = - Rate = * By convention, all rates are reported as positive values

  5. Instantaneous Rate – rate of rxn at a particular instant of time = slope of tangent to the line at that instant of time

  6. Initial Rate – instantaneous rate at t = 0 (very beginning) – All reactions slow down over time. – The best indicator of the rate of a reaction is the instantaneous rate near the beginning of the reaction. – At t = 0, what is the rate?

  7. In this reaction, the ratio of C4H9Cl to C4H9OH is 1:1. Thus, the rate of disappearance of C4H9Cl is the same as the rate of appearance of C4H9OH. [C4H9Cl] t [C4H9OH] t Rate = = Reaction Rates and Stoichiometry C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq)

  8.  [H2O2 ]  t Consider Rate = - 2 H2O2 (aq) 2 H2O () + O2 (g) If slope of tangent line at t = 0 is -2.0 x 10-4M/sec Rate = = - (-2.0 x 10-4M/sec) 2.0 x 10-4M/sec Then,

  9. Can the rate of H2O2 being consumed be related to the rate of products being generated ? 2 H2O2 (aq) 2 H2O () + O2 (g) [H2O] increases at a rate of 2.0 x 10-4M/sec as [H2O2] decreases at a rate of 2.0 x 10-4M/sec What is the rate that O2 forms when the [H2O2] is decreasing at a rate of 2.0 x 10-4M/sec ? [O2] increases at a rate of 1.0 x 10-4M/sec as [H2O2] decreases at a rate of 2.0 x 10-4M/sec

  10. Reaction Rates and Stoichiometry To generalize, for the reaction a A + b B cC + d D [A] t [B] t [C] t [D] t 1 b 1 c 1 a 1 d Rate =  =  = =

  11. Reaction Rates and Stoichiometry How is the rate at which ozone disappears related to the rate at which oxygen appears in the reaction 2 O3(g) 3 O2(g)? If the rate at which O2 appears, D[O2]/Dt, is 6.0 X 10-5 M/s at a particular instant, at what rate is O3 disappearing at this time, - D[O3]/Dt?

  12. Factors That Affect Reaction Rates • Physical state of the reactants (usually means surface area) • Reactant concentrations • Reaction temperature • Presence of a catalyst

  13. Physical State of the Reactants • In order to react, molecules must come in contact with each other. • Homogeneous reactions (all gases or liquids) are often faster. • Heterogeneous reactions that involve solids are slower; the faster of these reactions occurs if the surface area is increased; that is, a fine powder reacts faster than a pellet or tablet.

  14. Reactant Concentrations • Increasing reactant concentration generally increases reaction rate. • Since there are more molecules, more collisions occur.

  15. Temperature • Reaction rate generally increases with increased temperature. • Kinetic energy of molecules is related to temperature. • At higher temperatures, molecules move more quickly, increasing numbers of collisions and the energy the molecules possess during the collisions.

  16. Presence of a Catalyst • Catalysts speed up reactions by changing the mechanism of the reaction. • Catalysts are not consumed during the course of the reaction. • Do not appear in balanced equation.

  17. Reaction Rate vs. Concentration 2 H2O2 (aq) 2 H2O () + O2 (g) Rate law expression takes the form: Rate = k [H2O2]x Rate = rate of the reaction k = rate constant [H2O2] = concentration of reactant x = order of the reactant (usually an integer)

  18. What if the rxn has multiple reactants ? 2 HgCl2+ C2O42- Hg2Cl2+ 2 CO2+ 2 Cl- Rate = k [HgCl2]x [C2O42-]y k = rate constant indicative of this rxn x = order wrt[HgCl2] x and y must be determined experimentally y = order wrt [C2O42-]

  19. Determining Concentration Effect on Rate • How do we determine what effect the concentration of each reactant has on the rate of the reaction? • We keep every concentration constant except for one reactant and see what happens to the rate. Then, we change a different reactant. We do this until we have seen how each reactant has affected the rate.

  20. Method of Initial Rates To determine the numerical value for x and y, examine the data and find 2 trials where [reactant] of interest is varied, while all other reactants are held constant. Compare rate law expressions for these two trials to determine the order of that reactant.

  21. Rate = k [HgCl2]x [C2O42-]y Examine trials 1 and 2 [HgCl2] varies while the [C2O42-] is held constant Use trials 1 and 2 to obtain the numerical value for the order wrt[HgCl2] (which is x) x = 1 unitless

  22. Rate = k [HgCl2]x [C2O42-]y Examine trials 2 and 3 [C2O42-] varies while the [HgCl2] is held constant Use trials 2 and 3 to obtain the numerical value for the order wrt[C2O42-] (which is y) y = 2 unitless

  23. rxn is 1st order wrt HgCl2 rxn is 2nd order wrt C2O42- x = 1 y = 2 Rate = k [HgCl2]x [C2O42-]y Rate = k [HgCl2]1 [C2O42-]2 or Rate = k [HgCl2] [C2O42-]2 x and y do NOT vary under any circumstances !

  24. The experimentally determined rate law for the reaction 2 NO(g) + 2 H2(g) N2(g) + 2 H2O(g) is rate = k[NO]2[H2]. What are the reaction orders in this rate law? What is the effect of doubling the concentration of H2 on the rate? What is the effect of doubling the concentration of NO? Rate Laws

  25. Determine the Numerical Value for the Rate Constant, k Rate = k [HgCl2] [C2O42-]2 Rate[HgCl2] [C2O42-]2 = k units are very important 1 M2·sec k = 8.5 x 10-3 k is constant and does NOT vary with [ ] k does vary with temperature

  26. Rate constants can be used to compare the relative rates of reactions. the larger the value of k, the faster the reaction. the smaller the value of k, the slower the reaction. Rate Constants • The units of the rate constant depend on the overall reaction order of the rate law. • For a second order rxn: • For a first order rxn: eg. Rate = k[H2][I2] Units of rate constant = M-1s-1 eg. Rate = k[N2O5] Units of rate constant = s-1

  27. Rate Constant k Is a constant for a particular reaction. K varies with temperature.

  28. Overall Reaction Order – sum of the individual reaction orders Rate = k [HgCl2] [C2O42-]2 x + y = 1 + 2 = 3 reaction is overall 3rd order

  29. aA bB + jJ [A]o [A]t ln = kt [ ] vs. time Δ if Rate = k[A]1 For rxn if overall 1storder: [A]o = initial [A] at t=0 [A]t = [A] at some time later, t k = rate constant t = time from t=0

  30. CH3NC CH3CN [ ] vs. time is NOT linear slope = - k ln [ ] vs. time is linear only occurs when the rxn is overall 1st order

  31. SO2Cl2 decomposes as follows: SO2Cl2SO2 + Cl2 The rate constant, k = 2.2 x 10-5 s-1 at 300 C. Suppose 0.30 M SO2Cl2 is placed in a container at 300 C and allowed to decompose. What is the [SO2Cl2] after 75 minutes ? 1 t Whenever the rate constant, k has units of 1/time ( or time-1), the reaction is overall 1st order !

  32. Half-life – The amount of time it takes for one-half of a reactant to be used up in a chemical reaction. Abbreviation for half-life is t1/2 For rxnif overall 1storder: 0.693 k t1/2 =

  33. Prozac is a prescription drug which is metabolically eliminated from the body by overall first order kinetics. Prozac has a body clearance half-life of 1.835 x 105 sec. If a patient is given a 20.0 mg dose, how many hours until only 13.8 mg Prozac remains in the body ?

  34. aA uU + kK 1 [A]t = kt [ ] vs. time Δ if Rate = k[A]2 For rxn if overall 2ndorder: 1 [A]o 1 k [A]o t1/2 =

  35. NO2 NO + ½ O2 This rxn is NOT1st order slope = k 1 [ ] vs. time is linear only occurs when the rxn is overall 2nd order

  36. Zero-Order Reactions • Occasionally, rate is independent of the concentration of the reactant: Rate = k • These are zero-orderreactions. • These reactions are linear in concentration. • [A]t = – k t + [A]0 Slope = - k

  37. Factors That Affect Reaction Rate • Temperature • Frequency of collisions • Orientation of molecules • Energy needed for the reaction to take place (activation energy)

  38. Temperature and Rate • An increase in temperature results in an increase in the rate of a reaction • The rate constant is temperature dependent: It increases as temperature increases. • Rate constant doubles (approximately) with every 10 °C rise.

  39. Frequency of Collisions • Molecules must collide to react. • If there are more collisions, more reactions can occur. • So, if there are more molecules, the reaction rate is faster. • In a chemical reaction, bonds are broken and new bonds are formed. Molecules can only react if they collide with each other.

  40. Orientation of Molecules • Molecules can often collide without forming products. • Aligning molecules properly can lead to chemical reactions. • Bonds must be broken and made, and atoms need to be in proper positions. Cl + NOClNO + Cl2

  41. There is a minimum amount of energy required for reaction: the activation energy, Ea. Just as a ball cannot get over a hill if it does not roll up the hill with enough energy, a reaction cannot occur unless the molecules possess sufficient energy to get over the activation-energy barrier. Eais usually the most important in determining whether a particular collision results in a reaction. Energy Needed for a Reaction to Take Place (Activation Energy)

  42. Activation Energy, Ea

  43. Relating Ea to Speeds of Rxn Consider these energy profiles: Rank the reactions according to rate.

  44. How is Activation Energy, Ea, Determined? In lab, keep [reactants] constant and vary the temperature. Calculate values of the rate constant, k at each temperature. Svante Arrhenius1859 – 1927

  45. -Ea RT Arrhenius Equation k = (A) (e(−Ea/RT)) take ln of both sides ln k = + ln A y = mx + b 1 T a plot of ln k vs. gives a straight line -Ea R slope =

  46. ln k vs. slope = -19,000 K 1 T 1 T K-1 K-1

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