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KINETICS Chapter 14

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  1. KINETICSChapter 14

  2. 14.1 Factors that Affect Reaction Rates Chemical kinetics is the study of how fast chemical reactions occur. • Generally, the more frequently collisions between reaction particles occur, the faster the reaction.

  3. There are several important factors which affect rates of reactions: Physical state of the reactants • Homogeneous mixtures react faster than heterogeneous mixtures. • When reactants are in different phases: more surface area  faster reaction Concentration of the reactants • more reactant molecules so they collide more frequently  faster reaction Temperature of the reaction • higher temperature molecules move faster and collide more frequently  faster reaction Presence or absence of a catalyst • use a catalyst usually changes mechanism faster reaction

  4. 14.2 Reaction Rates • The speed of a reaction is defined as the change that occurs per unit time. • It is often determined by measuring the change in concentration of a reactant or product with time [M/sec] (could also use change in moles but if volume is constant [most reactions], then molarity and moles are directly proportional – concentration is easier to directly measure than moles. Pressure is sometimes used for gases.) • The speed of the chemical reaction is its reaction rate.

  5. Suppose A reacts to form B. A  B • Let us begin with [A] = 1.00 M • At t = 0 (time zero) there is 1.00 M A and no B present. • At t = 20 sec, there is 0.54 M A and 0.46 M B. • At t = 40 sec, there is 0.30 M A and 0.70 M B. • We can uses this information to find the average rate of the reaction.

  6. For the reaction A  B, there are 2 ways of measuring rate: • The rate of appearance of product B (i.e., change in concentration of B per unit time) as in the preceding example. • The rate of disappearance of reactant A (i.e., the change in concentration of A per unit time): • Note the negative sign! • Reaction rates are always positive!

  7. Change of Rate with Time • The average rate generally decreases with time • The rate at any instant in time is called the instantaneous rate. • • It is the slope of the straight line tangent to the curve at that instant. • • Instantaneous rate is different from average rate. • • We usually call the "instantaneous rate" the rate.

  8. Example: Using the data in table above, calculate the average rate of disappearance of C4H9Cl over the time interval from 50.0 to 150.0 seconds. (b) Using the figure, estimate the instantaneous rate of disappearance of C4H9Cl at t = 0 (the initial rate).

  9. the initial rate is given by the slope of the dashed line (y / x) pick two points on the tangent line: (0, .100) and (200, .060)

  10. Reaction Rates and Stoichiometry • For the reaction: C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq) • The rate of appearance of C4H9OH must equal the rate of disappearance of C4H9Cl.

  11. What if the stoichiometric relationships are not one-to-one? For the reaction: 2HI(g)  H2(g) + I2(g) The rate may be expressed as: We can generalize this equation a bit. For the reaction: aA + bBcC + dD The rate may be expressed as:

  12. The isomerization of methyl isonitrile CH3NC to acetonitrile, CH3CN, was studied in the gas phase at 215oC and the following data was obtained: Time (s) 0 2000 5000 8000 12000 15000 [CH3NC] (M) 0.0165 0.0110 0.00591 0.00314 0.00137 0.00074 (a) Calculate the average rate of reaction in M/s, for the time interval between each measurement. (a) Time Time Interval [CH3NC]  M Rate (sec) (Δt) (sec) (M) (ΔM/ Δt) 0 0.0165 2000 2000 0.0110 - 0.0055 2.8  10-6 5000 3000 0.00591 - 0.0051 1.7  10-6 8000 3000 0.00314 - 0.00277 .923  10-6 12000 4000 0.00137 - 0.00177 .443  10-6 15000 3000 0.00074 - 0.00063 .21  10-6 - ( - 0.0055 M / 2000 s)

  13. Consider the hypothetical reaction: A(aq)  B(aq). A flask is charged with 0.065 mol of A and allowed to react to form B in a total volume of 100.0 mL. The following data are collected: Time (min) 0 10 20 30 40 Mol of A 0.065 0.051 0.042 0.036 0.031 (a) Calculate the number of moles of B at each time in the table, assuming there are no molecules of B at time zero (b) Calculate the average rate of disappearance of A for each 10 min interval, in units of M/s. (c) Between t = 10 min and t = 30 min, what is the average rate of appearance of B in units of M/s? Assume the volume of the solution is constant. Time(min) Mol A (a)Mol B [A]  [A] (b) Rate -( [A]/time) 0 0.065 0 0.65 10 0.051 0.014 0.51- 0.142.3  10-4 20 0.042 0.023 0.42 - 0.09 2  10-4 30 0.036 0.029 0.36 - 0.06 1  10-4 40 0.031 0.034 0.31 - 0.05 0.8  10-4 (c) - ( - 0.14 M / 600 s)

  14. For each of the following gas-phase reactions, indicate how the rate of disappearance of each reactant is related to the rate of appearance of each product: (a) H2O2(g)  H2(g) + O2(g) (b) 2 N2O(g)  2 N2(g) + O2(g) (c) N2(g) + 3 H2(g)  2 NH3(g) (a) (b) (c)

  15. Consider the combustion of H2(g), 2 H2(g) + O2(g)  2 H2O(g). If hydrogen is burning at the rate of 0.85 mol/s, what is the rate of consumption of oxygen? What is the rate of formation of water vapor?

  16. The reaction 2 NO(g) + Cl2(g)  2 NOCl(g) is carried out in a closed vessel. If the partial pressure of NO is decreasing at the rate of 23 torr/min, what is the rate of change of the total pressure in the vessel? The change in total pressure is the sum of the changes of each partial pressure. NO and Cl2 are disappearing and NOCl is appearing. ΔPNO/Δt = -23 torr/min (given) ΔPCl/Δt = ½ (ΔPNO/Δt ) = ½ (-23) = -11.5 torr/min ΔPNOCl /Δt = - ΔPNO/Δt = -(-23) = +23 torr/min PTOT = -23 – 11.5 + 23 = -11.5 = -12 torr/min Pressure is decreasing at a rate of 12 torr/min

  17. 14.3 Concentration and Rate In general, rates: • Increase when reactant concentration is increased. • Decrease when reactant concentration is decreased. • We often examine the effect of concentration on reaction rate by measuring the way in which reaction rate at the beginning of a reaction depends on different starting conditions.

  18. Consider the reaction: NH4+(aq) + NO2– (aq)  N2(g) + 2H2O(l) • We measure initial reaction rates. • The initial rate is the instantaneous rate at time t = 0. • We find this at various initial concentrations of each reactant.

  19. As [NH4+] doubles with [NO2–] constant the rate doubles. • We conclude the rate is proportional to [NH4+]. • As [NO2–] doubles with [NH4+] constant the rate doubles • We conclude that the rate is proportional to [NO2–].

  20. Rate Law • The overall concentration dependence of reaction rate is given in a rate law or rate expression. • For a general reaction the rate law is: Rate = k [A]m [B]n The proportionality constant k is called the rate constant. Once we have determined the rate law and the rate constant, we can use them to calculate initial reaction rates under any set of initial concentrations.

  21. Exponents in the Rate Law The exponents m and n are called reaction orders The rate is “m” order in A The rate is “n” order in B We commonly encounter reaction orders of 0, 1 or 2. But, fractional or negative values are possible. The overall reaction order is the sum of the reaction orders. The overall order of a reaction is m + n + …. Note that reaction orders must be determined experimentally. They DO NOT necessarily correspond to the stoichiometric coefficients in the balanced chemical equation!(but they can and sometimes do)

  22. Using Initial Rates to Determine Rate Laws • To determine the rate law, we observe the effect of changing initial concentrations. • A reaction is nth order if multiplying the concentration by x causes a xn increase in rate. x increase in concentration = xn increase in rate (n is the order) For our example – the concentration was multiplied by 2 and the rate was also multiplied by 2 (21) so the rate is 1st order with respect to both NH4+(aq) and NO2– (aq) 2 (increase in concentration) = 21 (increase in rate) (1st order)

  23. Another way to look at order of reaction vs. change in concentration: Zero order rxn: change in concentration = no change in rate 1st order rxn: change in concentration = change in rate 2nd order rxn: change in concentration = change in rate is the squareof change in concentration • Most Common Scenarios: change in conc. of reactantchange in rate order doubled no change - rate is multiplied by 1 or 20 doubled doubled - rate is multiplied by 2 or 21 doubled quadrupled - rate is multiplied by 4 or 22 tripled no change - rate is multiplied by 1 or 30 tripled tripled - rate is multiplied by 3 or 31 tripled rate is multiplied by 9 or 32 0 1 2 0 1 2

  24. For the reaction in our example: NH4+(aq) + NO2– (aq)  N2(g) + 2H2O(l) The rate law is: Rate = k [NH4+] [ NO2–] • The reaction is said to be first order in [NH4+], first order in [NO2–], and second order overall.

  25. Units of Rate Constants • Units of the rate constant depend on the overall reaction order. • Second order overall: • Rate = k [A]2 • First order overall: • Rate = k [A]

  26. Note that the rate, not the rate constant, depends on concentration. • The rate constant IS affected by temperature and by the presence of a catalyst (more on this later).

  27. Examples of rate laws What are the individual and overall reaction orders for the reactions described in the following equations: (1) 2 N2O5(g)  4 NO2(g) + O2(g) Rate = k [N2O5] (2) CHCl3(g) + Cl2(g)  CCl4(g) + HCl (g) Rate = k [CHCl3][Cl2]1/2 • Reaction (1) is first order in N2O5 and first order overall. • Reaction (2) is first order in CHCl3 and ½ order in Cl2 and 3/2 order overall (add the exponents).

  28. A reaction A + B  C, obeys the following rate law: rate = k [B]2. (a) if [A] is doubled, how will the rate change? Will the rate constant change? (b) What are the reaction orders for A and B? What is the overall reaction order? (c) What are the units of the rate constant? (a) if [A] is doubled, there will be no change in the rate or the rate constant (b) the reaction is zero order in A, second order in B and second order overall (c)

  29. The decomposition of N2O5 in carbon tetrachloride proceeds as follows: 2 N2O5 2 NO2 + O2. The rate law is first order in N2O5. At 64oC, the rate constant is 4.82  10-3 s-1. (a) Write the rate law for the reaction. (b) What is the rate of reaction when [N2O5] = 0.0240 M? (c) What happens to the rate when the concentration of N2O5 is doubled to 0.0480 M? (a) rate = k [N2O5] = (4.82  10-3 s-1) [N2O5] (b) rate = (4.82  10-3 s-1) (0.0240 M) = 1.16  10-4 M/s (c) one way: rate = 4.82  10-3 s-1 (0.0480 M) = 2.31  10-4 M/s quicker way: We know that the rxn is 1st order in N2O5, so doubling the conc. will double the rate

  30. Consider the following reaction: CH3Br + OH-1 CH3OH + Br-1 . The rate law for this reaction is first order in CH3Br and first order in OH-1. When [CH3Br] is 0.0050 M and [OH-1] is 0.050 M, the reaction rate at 298 K is 0.0432 M/s. (a) What is the value of the rate constant? (b) What are the units of the rate constant? (c) What would happen to the rate if the concentration of OH-1 were tripled? (a,b) rate = k [CH3Br] [OH-1] (c) Since the rate law is first order in [OH-1], if [OH-1] is tripled, the rate triples.

  31. The iodide ion reacts with hypochlorite ion (the active ingredient in bleach) in the following way: OCl-1 + I-1 OI-1 + Cl-1. This rapid reaction gives the following rate data: Experiment [OCl-1] (M) [I-1] (M) Rate (M/s) 1 0.0015 0.0015 1.36  10-4 2 0.0030 0.0015 2.72  10-4 3 0.0015 0.0030 2.72  10-4 (a) What is the rate law for the reaction? (b) What is the value of the rate constant? (c) Calculate the rate when [OCl-1] = 0.0020 M and [I-1] = 0.00050 M (a) from exp. 1&2, doubling [OCl-1] while keeping [I-1] constant doubles the rate – rxn is 1st order in OCl-1 from exp. 1&3, doubling [I-1] while keeping [OCl-1] constant doubles the rate – rxn is 1st order in I-1 rate = k [OCl-1][I-1] (b) solving the rate law for k and using the data of experiment 1: (c) using the rate law (from a) and the value of k (from b):

  32. Consider the gas-phase reaction between nitric oxide and bromine at 273oC: 2 NO + Br2 2 NOBr. The following data for the initial rate of NOBr were obtained: Experiment [NO] (M) [Br2] (M) Initial Rate (M/s) 1 0.10 0.20 24 2 0.25 0.20 150 3 0.10 0.50 60 4 0.35 0.50 735 (a) Determine the rate law. (b) Calculate the value of the rate constant for the appearance of NOBr. (c) How is the rate of appearance of NOBr related to the rate of disappearance of Br2? (d) What is the rate of disappearance of Br2 when [NO] = 0.075 M and [Br2] = 0.25? (a) From exp. 1&2, multiplying [NO] by 2.5 multiplies the rate by 2.52 – rxn is 2nd order in NO From exp. 1&3, multiplying [Br2] by 2.5 multiplies the rate by 2.5 – rxn is 1st order in Br2 The rate law for the appearance of NOBr is : rate = k [NO]2[Br2] (b)

  33. Consider the gas-phase reaction between nitric oxide and bromine at 273oC: 2 NO + Br2 2 NOBr. The following data for the initial rate of NOBr were obtained: Experiment [NO] (M) [Br2] (M) Initial Rate (M/s) 1 0.10 0.20 24 2 0.25 0.20 150 3 0.10 0.50 60 4 0.35 0.50 735 (a) Determine the rate law. (b) Calculate the value of the rate constant for the appearance of NOBr. (c) How is the rate of appearance of NOBr related to the rate of disappearance of Br2? (d) What is the rate of disappearance of Br2 when [NO] = 0.075 M and [Br2] = 0.25? (c) (d) The rate law for the appearance of NOBr is : rate = k [NO]2[Br2] The rate of disappearance of Br2 is ½ the rate of appearance of NOBr.

  34. Consider the reaction of peroxydisulfate ion, S2O8-2 , with iodide ion, I-1 , in aqueous solution: S2O8-2 + 3 I-1 2 SO4-2 + I3-1 At a particular temperature, the rate of disappearance of S2O8-2 varies with reactant concentrations in the following manner: Experiment [S2O8-2] (M) [I-1] (M) Initial Rate (M/s) 1 0.018 0.036 2.6  10-6 2 0.027 0.036 3.9  10-6 3 0.036 0.054 7.8  10-6 4 0.050 0.072 1.4  10-5 • Determine the rate law for the reaction. (a) from experiments 1 and 2, increasing [S2O8-2] by 1.5 increases the rate by 1.5 so [S2O8-2] is 1st order

  35. Experiment [S2O8-2] (M) [I-1] (M) Initial Rate (M/s) 1 0.018 0.036 2.6  10-6 2 0.027 0.036 3.9  10-6 3 0.036 0.054 7.8  10-6 4 0.050 0.072 1.4  10-5 ALTERNATE SOLUTION In exp 1 and 2, keeping [I-1] constant and increasing [S2O8-2] by 1.5 increases the rate by 1.5 meaning that the reaction is first order in [S2O8-2] So we know that the rate law is: rate = k [S2O8-2] [I-1]x and we need to find x Looking at exp 1 and 3, the rate inc by a factor of 3 or Rateexp 3 / Rateexp1 = 3  Rate 3 k [S2O8-2] [I-1]x k [.036] [.054]x 2 [.054]x -------- = -------------------- = -------------------- = ------------- = 3 Rate 1 k [S2O8-2] [I-1]x k [.018] [.036]x [.036]x [.054]x 3 -------- = ----- (or 1.5) [.036]x 2 (1.5)x = 1.5 x = 1, so the reaction is 1st order in [I-1] rate = k [S2O8-2] [I-1]

  36. Kinetics Part 2 14.4

  37. Goal: Convert the rate law into a convenient equation that gives concentration as a function of time.

  38. (slope= −k) ln[A]t Time (s) First-Order Reactions • Rate must depend on only one reactant and be raised to the 1st power. • For a first-order reaction, the rate doubles as the concentration of a reactant doubles. Rate = k[A] After some calculus……… y = mx + b •A plot of ln[A]t versus t is a straight line with slope -k and intercept ln[A]0

  39. (slope= k) 1/[A]t Time (s) Second-Order Reactions • A second-order reaction is one whose rate depends on the reactant concentration to the second power or on the concentration of two reactants, each raised to the first power. • For a 2nd order reaction with just one reactant: calculus y = mx + b A plot of 1/[A]t versus t is a straight line with slope k and intercept 1/[A]0.

  40. Half-life • Half-life, t½, is the time required for the concentration of a reactant to decrease to half its original value. • That is, half life, t½, is the time taken for [A]0 to reach ½ [A]0.

  41. Mathematically, the half life of a first-order reaction is: Radioactive decay is 1st order. Note that the half-life of a first-order reaction is independent of the initial concentration of the reactant.

  42. We can show that the half-life of a second order reaction is: • Note that the half-life of a second-order reaction is dependent on the initial concentration of reactant.

  43. Sample Exercise #1 • The first order rate constant for the decomposition of a certain insecticide in water at 12oC is 1.45 yr-1. A quantity of this insecticide is washed into a lake on June 1, leading to a concentration of 5.0 x 10-7 g/cm3 of water. Assume that the effective temperature of the lake is 12oC. (a) What is the concentration of the insecticide on June 1 of the following year? (b) How long for the conc. of the insecticide to drop to 3.0 x 10-7 g/cm3?

  44. (a) ln [A]t = - k t + ln [A]0 ln[A] t= - (1.45 yr-1)(1.00 yr) + ln(5.0 x 10-7) eln [A] t = e-15.96 [A] t = 1.2 x 10-7 g/cm3 (note: conc. units for [A] and [A]0 must be the same)

  45. (b) ln[A]t = - k t + ln [A]0 ln (3.0  10-7) = - (1.45)(t) + ln (5.0  10-7) t = 0.35 yr

  46. Using the figure to the right, estimate the half-life of the reaction of C4H9Cl with water. • The initial value of [C3H9Cl] is 0.100 M. • The half-life for this reaction is the time for [C3H9Cl] to be 0.050 M. • This point occurs at approx. 340 s. At the end of the second half-life, about 680 s, the concentration should decrease by another factor of 2 to about 0.025 M. The graph indicates that this is true.

  47. The reaction SO2Cl2 SO2 + Cl2 is first order in SO2Cl2. Using the following kinetic data, determine the magnitude of the first order rate constant: Time (s) Pressure SO2Cl2 (atm) ln Pressure SO2Cl2 0 1.000 0 2500 0.947 - 0.0545 5000 0.895 - 0.111 7500 0.848 - 0.165 10000 0.803 - 0.219 Graph ln P vs. time (first order) The graph is linear with slope of about - 2.19  10-5 s-1

  48. Sample Exercise #4 The following data was obtained for the gas phase decomposition of NO2(g) at 300oC: 2 NO2(g)  2 NO(g) + O2(g) Time (s) [NO2] (M) 0.0 0.01000 50.0 0.00787 100.0 0.00649 200.0 0.00481 300.0 0.00380 What is the rate law for this reaction?

  49. To test whether the reaction is first or second order, we can construct plots of ln [NO2] and 1 / [NO2] against time. Time (s) [NO2] ln [NO2] 1/[NO2] 0.0 0.01000 50.0 0.00787 100.0 0.00649 200.0 0.00481 300.0 0.00380 -4.610 -4.845 -5.038 -5.337 -5.573 100 127 • 154 • 208 • 263