1 / 56

Topic # 15 Chemical Kinetics

Topic # 15 Chemical Kinetics. Conditions of Reaction Rates Rate of a reaction General Rate Law Orders of Reactions Half-Life Reaction Mechanisms. (Zumdahl Chapter 12). Collision Theory. A reaction will only take place if three conditions are met:.

galia
Download Presentation

Topic # 15 Chemical Kinetics

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Topic # 15 Chemical Kinetics Conditions of Reaction Rates Rate of a reaction General Rate Law Orders of Reactions Half-Life Reaction Mechanisms (Zumdahl Chapter 12)

  2. Collision Theory A reaction will only take place if three conditions are met: 1. the reactants come into contact (collide) 2. the collision happens with enough energy, (activation energy) 3. the reactants hit at the best possible orientation to facilitate a reaction.

  3. Factors that control the rate of a reaction A. Concentration putting more reactants in the same space will increase the collision frequency, producing a faster reaction rate. Same effect occurs by increasing the pressure in a gaseous environment. B. Temperature Increasing temperature will increase the rate of a reaction. GENERALLY, a 10oC rise will result in approx. doubling the rate.

  4. As the system is heated, the position of the average kinetic energy must shift to the right in order to reflect that change. As the system cools, the average kinetic energy must respond by shifting position to the left. At higher temperatures, the particles are more evenly distributed over a range of kinetic energy values.

  5. Rate and temperature Reaction rates are temperature dependent. Here are rate constants for N2O5 decomposition at various temperatures. T, oC k x 104, s-1 20 0.235 25 0.469 30 0.933 35 1.82 40 3.62 45 6.29 k x 104 (s-1) Temperature (oC)

  6. C. Solid particle size increase in surface area of the solid will increase the reaction rate D. Catalysis Catalysts are substances that increase the the rate of a reaction while remaining unchanged chemically. They work by providing an different reaction route that needs a lower activation energy.

  7. concentrationlater - concentrationearlier timelater - timeearlier D [ ] Dt Rates of reactions Rate of a chemical reaction. The change in the quantity of a reactant or product that takes place in a period of time. rate = = Molarity

  8. Rates of reactions To study rates of reaction, you must: • Identify the reactants and products. • Carry out the reaction. • Measure the concentrations of one of the reactants or products at known intervals. There MUST be DATA! There needs to be a way to measure at least one of the species involved.

  9. The example reaction Decomposition of N2O5 Dinitrogen pentoxide is known to decompose completely by the following reaction. 2N2O5 (g) 2N2O4 (g) + O2 (g) This reaction can be conducted in an inert solvent like carbon tetrachloride. When N2O5 decomposes, N2O4 remains in solution and O2 escapes and can be measured.

  10. An example reaction We can easily measure the oxygen as dinitrogen pentoxide decomposes. • Temperature must be maintained to within +0.01 oC. • The reaction flask must be shaken to keep oxygen from forming a supersaturated solution. • It is found that the reaction initially occurs very rapidly but gradually slows down.

  11. An example reaction Gas buret Constant temperature bath

  12. An example reaction Time (s) STP O2 in mL 0 0 300 1.15 600 2.18 900 3.11 1200 3.95 1800 5.36 2400 6.50 3000 7.42 4200 8.75 5400 9.62 6600 10.17 7800 10.53 Here are the results for an experiment.

  13. An example reaction Volume, mL O2 The rate of O2 production slows down with time. Time, s

  14. DV O2 D t Time, s Rate O2* 0 300 0.0038 600 0.0034 900 0.0031 1200 0.0028 1800 0.0024 2400 0.0019 3000 0.0015 Calculations with Average rates We can calculate the average rate of oxygen formation during any time interval as: Average rate of O2 formation = *The rates shown here have units O2 mL(at STP)/s. Notice how the rate decreases with time.

  15. D[N2O5] Dt D[O2] Dt Looking at the N2O5 reaction again….. Since we know the stoichiometry for our reaction, we can calculate the concentration of N2O5 during the reaction. 2N2O5 (g) 2N2O4 (g) + O2 (g) For each mole of O2 produced, two moles of N2O5 will have decomposed. The rate of reaction will be: rate of reaction = = - 1 2

  16. This method can be used for any equation……. Using the equation: 2 A + B + 3 C + D The rate of reaction can be expressed in terms of D: d[D] dt 1 d[A] 2 dt - - • d[C] • 3 dt d[B] dt = = rate = = Note the use of coefficients and choosing a component to base the rates on!

  17. Use the Equation to solve for concentrations: 2 A + B +  3 C + D d[C] dt = 0.35 mole/liter-min d[A] dt Find value of 2 3 Relationship of [A] will be negative and ratio - 2 [0.35] 3 1 = 0.23 mole/liter-min is being lost

  18. General Rate Equation A + B  C + D For an equation: rate = k [A]x[B]y x & y are ORDER of reaction for that component they are NOT coefficients!! k is the rate constant Called the rate law for the equation d[C] dt Actual Rate k [A]x[B]y = Collected data is used to answer questions on rate orders.

  19. Finding rate order Method of initial rates The order for each reactant is found by completing an experiment and using the data: • Change the initial concentration of a reactant. • Hold all other initial concentrations and conditions constant. • Measure the initial (or starting) rates of reaction The change in rate is used to determine the order for that specific reactant. The process is repeated for each reactant.

  20. N2O5 example 2N2O5(g) 2N2O4(g) + O2(g) The following data was obtained for the decomposition of N2O5. Experiment [N2O5] Initial rate, M/s 1 0.100 3.62 x 10-5 2 0.200 7.29 x 10-5 We know that the rate expression is: rate = k [N2O5]x Our goal is to determine what x (the order) is.

  21. N2O5 example For exp. 2 7.29 x 10-5 M/s = k (0.200 M)x For exp. 1 3.62 x 10-5 M/s = k (0.100 M)x We can now divide the equation for exp. two by the one for exp. one. 7.29 x 10-5 M/s k (0.200 M)x 3.62 x 10-5 M/s k (0.100 M)x which give 2.01 = (2.00)x and x = 1 (first order reaction) = rate = k [N2O5]1

  22. The data below for rate of reaction was obtained for the following reaction: A + B + C . . . Exp. [A] [B] [C] Initial rate, M/s 1 0.030 0.010 0.050 1.7 x 10-8 2 0.060 0.010 0.050 6.8 x 10-8 3 0.030 0.020 0.050 4.9 x 10-8 4 0.030 0.010 0.100 1.7 x 10-8 Note that the concentrations double in some experiments. Notice the products are not needed!

  23. Equation: A + B + C  ……… Rate Law: rate = k[A]x [B]y [C]z Pick data so calculations for two of the reactants will cancel out……..

  24. Order for A Use exps one and two since [B] and [C] are the same and would cancel out. 6.8 x 10-8 M/s k(0.060 M)x 1.7 x 10-8 M/s k(0.030 M)x 4.0 = (2.0)x x = 2[A] is second order =

  25. Order for B Use experiments one and three. 4.9 x 10-8 M/s (0.020 M)y 1.7 x 10-8 M/s (0.010 M)y 2.9 = (2.0)y The order is not obvious by inspection. You must take the natural logarithm of both sides and solve for y. ln 2.9 = y (ln 2.0) y = 1.54 or = 3 2

  26. Order for C Use experiments one and four. Experiment [C] Initial Rate 1 0.050 1.7 x 10-8 4 0.100 1.7 x 10-8 Here the rate did not change when [C] was doubled. This is an example of a zero order reaction. z = 0

  27. A more complex example We can now write the overall rate law. rate = [A]2 [B]3/2 [C]0 since [C] has no effect on the rate: rate = [A]2 [B]3/2 The overall order for the reaction is: x + y + z = 2 + 3/2 + 0 = 3 1/2

  28. 1 [A]0 1 [A]t 1 [A]t Graphing rate laws Graphical method (some form of [ ] versus time!) Using integrated rate law, one can produce straight line plots. The order for a reactant is assigned if the data produces a straight line. Rate integrated Graph Slope Order law rate law vs. time 0 rate = k[A]t = -kt + [A]0 [A]t -k 1 rate = k[A] ln[A]t = -kt + ln[A]0 ln[A]t -k 2 rate=k[A]2= kt + k

  29. Graphing rate laws 0 order plot 2nd order plot [N2O5] 1/[N2O5] Time (s) Time (s) Time (s) As you can see from these plots of the N2O5 data, only a first order plot results in a straight line. So N2O5 is 1st order! 1st order plot ln[N2O5]

  30. Time (s) [NO2] (mol/L) 0 0.500 1.2 x103 0.444 3.0 x 103 0.381 4.5 x 103 0.340 9.0 x103 0.250 1.8 x 104 0.174 Practice Graphing: Graph: [NO2] vs. time Zero order ln[NO2] vs. time First Order 1/[NO2] vs. time Second Order Straight line tells the order !

  31. Special First order reactions Reactions that are first order with respect to a reactant are of great importance. • Describe how many drugs pass into the blood stream or used by the body. • Often useful in geochemistry • Radioactive decay Half-life (t1/2) The time required for one-half of the quantity of reactant originally present to react.

  32. Half-life From our N2O5 data, we can see that it takes about 1900 seconds for the concentration to be reduced in half. It takes another 1900 seconds to reduce the concentration in half again. [N2O5] Time (s)

  33. Half-life The half-life and the rate constant are related. t1/2 = Half-life (time) can be used to calculate the first order rate constant. For our N2O5 example, the reaction took 1900 seconds to react half way so: k = = = 3.65 x 10-4 s-1 0.693 k 0.693 t1/2 0.693 1900 s

  34. . Fluorine -21 has a half-life of 5 seconds. What fraction of the original nuclei remain after 1 minute? The answer is solved by creating the fraction Where n = number of ½ lives. In 1 minute there will be 12 half-lives. Answer is…….

  35. Reaction mechanisms A detailed molecular-level picture of how a reaction might take place. activated complex = bonds in the process of breaking or being formed

  36. Could also look like: R + S  D but in reality… R B B + S  D Or even: A + B  C + D is really: A  X X + B  C + D Never see the activated complex in the equation!

  37. Reaction mechanisms Molecularity The number of particles that come together to form the activated complex in an elementary process (each step). 1 - unimolecular 2 - bimolecular 3 - termolecular

  38. Reaction mechanisms For some elementary processes, the exponents for each species in the rate law are the same as the coefficients in the equation for the step. For our earlier example, Bimolecular…. the rate law is: rate = k [NO] [O3] (Bimolecular mechanism)

  39. Reaction mechanisms In general, the rate law gives the composition of the activated complex, because that is what comes together to form the complex. • The power of a species in the rate law is the same as the number of particles of the species in the activated complex. • If the exponents in the rate law are not the same as the coefficients of the equation for the reaction, the overall reaction must consist of more than one step. Lets look at N2O5 - again!

  40. Reaction mechanisms Earlier we found that for: 2N2O5 2N2O4 + O2 The rate law was: rate = k [N2O5] • According to the equation, it should be second order (coefficient of 2 in front) but the data shows it to be first order. • The reaction must involve more than one step.

  41. Reaction Mechanisms Consider the following reaction. 2NO2 (g) + F2 (g) 2NO2F (g) If the reaction took place in a single step the rate law might be: rate = k [NO2]2 [F2] However, the experimentally observed rate law is: rate = k [NO2] [F2]

  42. Reaction Mechanisms Since the observed rate law is not the same as if the reaction took place in a single step, we know two things. • More than one step must be involved • The activated complex must be produced from two species. A possible reaction mechanism might be: Step one NO2 + F2 NO2F + F Step two NO2 + F NO2F Overall 2NO2 + F2 2NO2F

  43. Reaction Mechanisms Rate-determining step. When a reaction occurs in a series of steps, with one slow step, it is the slow step that determines the overall rate. Step one NO2 + F2 NO2F + F Expected to be slow. It involves breaking an F-F bond. Step two NO2 + F NO2F Expected to be fast. A fluorine atom is very reactive.

  44. Reaction Mechanisms Since step one is slow, we can expect this step to be the determiner of the overall rate of the reaction. NO2 + F2 NO2F + F This would give a rate expression of: rate = k1 [NO2] [ F2] This agrees with the experimentally observed results.

  45. Let’s look again: R + S  D but in reality… R B slow B + S  D very fast The rate only depends on the concentration of R so the rate law only contains R! Rate = k[R]x The Slow process determines the rate law!

  46. A + B  C + D is really: A  X Fast and equilibrium X + B  C + D SLOW Here the slow step contains X and B, but the formation of X depends on A. Since X is an intermediate it isn’t in the rate law. X is substituted with A…. Rate = k[A][B] No intermediates are written in the rate law!

  47. Take note: the sum of all steps needs to tally up to the reaction equation. R + S  D R B slow B + S  D very fast

More Related