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Kinetics

Kinetics. “How fast a reaction occurs” Factors Concentration Temperature Surface area Catalyst. Measuring the Rate of a Reaction. Disappearance of a reactant C 4 H 9 Cl(aq) + H 2 O(l)  C 4 H 9 OH(aq) + HCl(aq). Draw a graph of the following data (x=time). Rate is the Slope.

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Kinetics

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  1. Kinetics “How fast a reaction occurs” Factors • Concentration • Temperature • Surface area • Catalyst

  2. Measuring the Rate of a Reaction Disappearance of a reactant C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)

  3. Draw a graph of the following data (x=time)

  4. Rate is the Slope y = m x + b [C4H9Cl]= (Rate)(t) Rate = D [C4H9Cl] D t Rate = d[C4H9Cl] dt (Note how rate decreased with decreasing concentration)

  5. The problems on this page refer to the data table for C4H9Cl • Using the data in the table, calculate the rate of disappearance of C4H9Cl from 0 to 50 s. • Using the data in the table, calculate the rate of disappearance of C4H9Cl from 50 to 100 s. Now graph the data • Using the graph, calculate the rate at 100 s (-1.64 X 10-4 M/s) • Using the graph, estimate the rate at 0 seconds and 300 s. (-2.0 X 10-4 M/s, -1.1 X 10-4 M/s)

  6. Rates and Stoichiometry C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq) Rate = -D[C4H9Cl] = D[C4H9OH] Dt Dt Rate of disappearance = Rate of appearance (only for 1:1 stoichiometry)

  7. 2HI(g)  H2(g) + I2(g) Rate = -1D[HI] = D[H2] = D[I2] 2 Dt Dt Dt aA + bB  cC + dD Rate = -1D[A] = -1D[B] = 1D[C] = 1D[D ] a Dt b Dt c Dt d Dt

  8. Stoichiometry: Ex 1 How is the rate of the disappearance of ozone related to the appearance of oxygen according to: 2O3(g)  3O2(g) Rate = -1D[O3] = D[O2] 2 Dt 3 Dt -D[O3] = 2D[O2] or RateO3 = -2/3 RateO2 Dt 3 Dt

  9. Stoichiometry: Ex 2 If the rate of appearance of oxygen at some instant is 6.0 X 10-5 M/s, calculate the rate of disappearance of ozone. D[O3] = -2D[O2] Dt 3 Dt D[O3] = -2 (6.0 X 10-5 M/s) Dt 3 D[O3] = -4.0 X 10-5 M/s Dt

  10. Stoichiometry: Ex 3 The rate of decomposition of N2O5 at a particular instant is 4.2 X 10-7 M/s. What is the rate of appearance of NO2 and O2? 2N2O5(g)  4NO2(g) + O2(g) Rate = -1D[N2O5] = 1D[NO2] = D[O2] 2 Dt 4 Dt Dt -D[N2O5] = 1D[NO2] 2 Dt 4 Dt

  11. D[NO2] = – 4D[N2O5] Dt 2 Dt D[NO2] = – 2D[N2O5] = (-2)(4.2 X 10-7 M/s) Dt Dt D[NO2] = 8.4 X 10-7 M/s Dt -D[N2O5] = D[O2] 2 Dt Dt -D[O2]= (½)(4.2 X 10-7 M/s) = 2.1 X 10-7 M/s Dt

  12. The Rate Law aA + bB  cC Rate = k[A]m[B]n k = Rate constant [A] and [B] = Initial concentrations m and n = exponents k, m, and n must be determined experimentally

  13. Rate Law: Ex 1 What is the rate law for the following reaction, given the following rate data: NH4+(aq) + NO2-(aq)  N2(g) + 2H2O(l)

  14. Rate = k[NH4+]m[NO2-]n Divide Experiment 1 and 2 Rate2 = k[NH4+]m[NO2-]n Rate1 = k[NH4+]m[NO2-]n 10.8 X 10-7 = k[0.0200]m[0.200]n 5.4 X 10-7 = k[0.0100]m[0.200]n 2= [0.0200]mm = 1 1 =[0.0100]m

  15. Rate = k[NH4+]m[NO2-]n Divide Experiment 3 and 4 Rate6 = k[NH4+]m[NO2-]n Rate5 = k[NH4+]m[NO2-]n 21.6 X 10-7 = k[0.200]1[0.0404]n 10.8 X 10-7 = k[0.200]1[0.0202]n 2= [0.0404]nn = 1 1 =[0.0202]n

  16. Calculating k Rate = k[NH4+]1[NO2-]1 Consider Experiment 1 (can pick anyone) Rate1 = k[NH4+]1[NO2-]1 5.4 X 10-7 = k[0.0100]1[0.200]1 k = 2.7 X 10-4 M-1s-1

  17. Rate Law: Ex 2 Calculate the rate law for the following general reaction: S + O2 SO2

  18. ANS: Rate = (4.0 X 10-3 M-1s-1)[S]2 What is the rate of the reaction when [S] = 0.050 M and [O2] = 0.100 M? ANS: 1.0 X 10-5 M/s

  19. Rate Law: Ex 3 A particular reaction varies with [H+] as follows. Calculate the Rate Law:

  20. Rate = 3.2 X 10-8 M2s-1[H+]-1

  21. Rate Law: Ex 4 Calculate the rate if the [H+] = 0.400 M Rate = 8 X 10-8 M/s

  22. Calculate the rate law for the following reaction: BrO3-(aq) + 5 Br-(aq) + 8 H+(aq) ----> 3 Br2(l) + H2O(l) 

  23. Reaction Order • m and n are called reaction orders • m + n + …. = overall reaction order • Units of k • k must always have units that allow rate to have a unit of M/s

  24. Reaction Order: Ex 1 What is the overall reaction order for the following reaction. What unit will the rate constant (k) have? CHCl3 + Cl2 CCl4 + HCl Rate=k[CHCl3][Cl2]1/2

  25. Zeroth Order • Reaction order can be zero • Concentration does not affect the rate • 2NH3 (g)  N2(g) + 3H2(g) Rate = k[NH3]0 Rate = k

  26. Recognizing Zeroth Order

  27. Concentration and Time: First Order • First Order Reaction – reaction whose rate varies with the concentration of a single reactant to the first power • Often the decay(chemical or nuclear) or decomposition of one substance ln[A]t = -kt + ln[A]0 y = mx + b First order eqns - linear ln[Conc] vs time graph

  28. Example of a First Order Reaction • First order in CH3NC • ln[CH3NC]t = -kt + ln[CH3NC]0 • Can use this equation to calculate the concentration at any time.

  29. This graph is not linear This graph is [A]t = [A]0e-kt ln[A]t = -kt + ln[A]0

  30. First Order: Ex 1 The first order rate constant for the decomposition of an insecticide is 1.45 yr-1. If the starting concentration of the compound in a lake is 5.0 X 10-7 g/cm3, what will be the concentration the following year? In two years?

  31. ln[A]t = -kt + ln[A]0 ln[A]t = -(1.45 yr-1 )(1 yr) + ln(5.0 X 10-7 g/cm3) ln[A]t = -15.96 [A]t = e-15.96 [A]t = 1.2 X 10-7 g/cm3 or [A]t = [A]0e-kt [A]t = (5.0 X 10-7 g/cm3)e-(1.45 yr-1 )(1yr) [A]t = 1.2 X 10-7 g/cm3 ([A]2yr = 2.8 X 10-8 g/cm3)

  32. First Order: Ex 2 How long will it take the concentration to drop to 3.0 X 10-7 g/cm3? ln[A]t = -kt + ln[A]0

  33. First Order: Ex 3 The decomposition of dimethyl ether is a first order process with k=6.8 X10-4s-1. If the initial pressure is 135 torr, what is the partial pressure after 1420s? (CH3)2O(g)  CH4(g) + H2(g) + CO(g) (ANS: 51 torr)

  34. First Order Half-life • Half-life – time required for the concentration of a reactant to drop to one half the initial value • Medicine in the body • Nuclear decay • [A]t = ½[A]0 t½ = 0.693 k

  35. Half-Life: Ex 1 From the figure below, calculate the half-life and k for the reaction of C4H9Cl with water.

  36. Sodium-24 (used in some medical tests) has a half-life of 14.8 hours. What is the rate constant for its decay? (ANS: 0.0468 hr-1)

  37. Half-Life • Half-life - The time during which one-half of a radioactive sample decays • Ranges from fraction of a second to billions of years. • You can’t hurry half-life.

  38. Carbon-14 dating • 14C in traces of once-living organisms • E.g., a 5730 years after death, only half of the 14C remains. • ~ 50,000 years. • 15% margin of error • Mummies, the Dead Sea Scrolls, Shroud of Turin

  39. Half-Life

  40. Half-life: Example 1 Carbon-14 has a half-life of 5730 years and is used to date artifacts. How much of a 26 g sample will exist after 3 half-lives? How long is that?

  41. Half-life: Example 2 Tritium undergoes beta decay and has a half life of 12.33 years. How much of a 3.0 g sample of tritium remains after 24.66 years?

  42. Half-life: Example 3 Cesium-137 has a half-life of 30 years. If you start with a 200 gram sample, and you now have 25 grams left, how much time has passed?

  43. Rate Law: Ex 1 Uranium-238 has a half-life of 4.5 X 109 yr. If 1.000 mg of a 1.257 mg sample of uranium-238 remains, how old is the sample?

  44. Rate Law: Ex 2 A wooden object is found to have a carbon-14 activity of 11.6 disintegrations per second. Fresh wood has 15.2 disintegrations per second. If the half-life of 14C is 5715 yr, how old is the object? ANS: 2230 yr

  45. After 2.00 yr, 0.953 g of a 1.000 g sample of strontium-90 remains. • Calculate k (0.0241 y-1) • Calculate how much remains after 5.00 years. (0.886 g)

  46. Ex 4 A sample for medical imaging contains 18F (1/2 life = 110 minutes). What percentage of the original sample remains after 300 minutes? ANS: 15.1%

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