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Chemical Kinetics

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  1. Chemical Kinetics • Chemical kinetics study reaction rates in a chemical reaction. When a chemical reaction takes place, a collision happens and the collision has to be strong enough to break bonds. • The rate at which a chemical reaction occurs is often very important to us. A common task of chemists is to find ways to change the rates of chemical reactions - we want our cars and buildings to rust more slowly and our food to spoil more slowly as well. But we need the reactions that produce economically important substances, such as ammonia and medical pharmaceuticals to occur quickly.

  2. Kinetic looks at the key factors that determine whether a reaction will be fast or slow, and how the reaction rate may be changed. In order to understand how these factors affect reaction rates, you will also learn about the collision theory, the theory that helps explain what happens at a molecular level during a chemical reaction.

  3. In thermodynamics one key concept was determining whether or not a reaction would occur spontaneously. • We learned that two thermodynamic factors - enthalpy and entropy - were key determinants of whether a reaction would actually occur or not. • The speed of the reaction was not important - just whether the reaction would occur.

  4. Here we are concerned with how fast the reaction occurs.  The key questions are: • How do you measure the speed, or rate, of a reaction? • What factors influence how fast a reaction occurs?

  5. Reaction rate - the speed at which a reaction occurs; typically measured as the change in concentration of either reactants or products over time. • Reaction rate is a measure of how fast a reaction occurs, or how something changes during a given time period.

  6. Look at the oxidation of glucose, C6H12O6 : • C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(g) • One of the things that happens during this reaction is simply that glucose gets used up as it reacts with oxygen in the air, and carbon dioxide and water start to form. • How could we measure how fast this occurs?

  7. In this example, we might want to measure how quickly the mass of solid glucose decreases, or how quickly the gases (carbon dioxide and water) form. • We might want to measure the volume of these gases.

  8. C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(g) • For other reaction the following could be measured as they change: • change in conductivity • change in pH • color change • change in pressure as gases are formed or used up

  9. C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(g) • The most common measure of reaction rate is to express how the concentration of a reaction participant changes over time. It could be how the concentration of a reactant decreases, or how the concentration of a product increases. This is what we will be using. [C6H12O6]

  10. C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(g) • Now that we have something that changes to measure, we must consider the second key aspect of determining rate - time. Rate is a measure of how something changes over time.  You measure your rate of speed when driving by determining how many kilometers you travel in an hour. Rate = [ ] =change in concentration t change in time

  11. If we wanted to express the rate of the following reaction: 2CO2(g) ) → O2(g) + CO(g) we could either measure the change in concentration of a reactant or product and might use either of the following expressions to calculate our rate: Rate = [ CO2 ] or [ O2 ] min sec

  12. We can choose a point and determine the rate for that instant, using a tangential line. This works for this point. This is the instantaneous rate. But the rate of decomposition is changing. • -If we measured the rate at the start of the reaction, it would be different than if we measured the rate near the end of the reaction. Rate is not a constant - it changes during the course of the reaction.

  13. Generally we do the calculations using a reactant, ie the rate of decomposition of the reactant as the product can decompose and reform the product. • Using this method we find the rate = [ CO2 ] / min • CO2 is a reactant so this is decreasing, but rate is generally expressed as a positive number so rate =-[ CO2 ] / min

  14. Reaction Rate The change in concentration of a reactant or product per unit of time

  15. Reaction Rates: 2NO2(g)  2NO(g) + O2(g) 4. Are equal to the slope tangent to that point 5. Change as the reaction proceeds, if the rate is dependent upon concentration [NO2] t

  16. Reaction Rates: 2NO2(g)  2NO(g) + O2(g) 1. Can measure disappearance of reactants 2. Can measure appearance of products 3. Are proportional stoichiometrically

  17. rate =-[ CO2 ] / min Lets think about what could affect the rate of a reaction A better way of showing how the concentration of CO2 affects the rate is known as the rate law Rate = k[CO2]n k = rate constant – don’t know what it is n= order of reactions

  18. Determining Reaction Order: Here are four ways to learn the order of reaction from easiest to hardest: 1. They tell you in the problem. "In the first order reaction of ….”

  19. Determining Reaction Order: 2. You are given units for the rate constant. For example, if a reaction is first order the units are reciprocaltime: Rate = k [A]1 and rearranging, k = rate/M = (M/sec)/M = 1/sec = sec-1 In other words, the order of a reaction with k= 1.24 x 10-2 min-1 is first order. Do factor labeling to cancel the units to see for your self Remember, this only tells you the total order for the reaction, not the individual orders.

  20. 3. Method of initial rates. The favorite of every kinetics exam in general chemistry, you will be given a series of varying concentrations and a rate and from this asked to determine the individual and reaction rates. This shows up in the multiple choice questions a lot. In each case, you set up a ratio of concentrations and rates to fit the expression: rate = k[A]x or rate = k[B]y and ask, what does the order x or y have to be to make the equation true?

  21. Quick visual method • n= order of reactions • if n =1 • if you double CO2 then the rate will double. • If n=2 • if you double CO2 and then the rate will quadruples

  22. n=3 • If you triple CO2 and then the rate is cubed what order do you have? • n=0 • If you triple CO2 and then the rate doesn’t change what order do you have?

  23. What if you need to do a little math?

  24. Rules of logarithms ln (1) = 0 ln (e) = 1 ln (ex) = x ln Ax = xlnA • log (1) = 0 • log (10) = 1 • log (100) = 2 • Log (10x) = x • log Ax = x log A log(Ax/Bx) = log (A/B)x = x log (A/B) log(AB) = log A + log B log(A/B) = log A - log B

  25. Hold [A] in trial 1 and 2 constant, then look at the rate when [B] is varied. In this case there is no change so it is a 0 order reaction  do the math rate2 /rate1 = k([B2]/[B1])y 1x10-4/1x10-4 = k (.2/.1)y  y = 0.

  26. For A now hold [B] in trial 1 and 3 constant, then look at the rate when the [A] is tripled so does the rate  the reaction is 1 order Or rate3 /rate1 = k([A3]/[A1])x 3x10-4/1x10-4 = k (.3/.1)x  x = 1.

  27. For B now hold [A] in trial 1 and 2 constant, then look at the rate when the [B] is tripled the rate goes up to the 3 times Hold [A] in trial 1 and 2 constant, rate2 /rate1 = k([B2]/[B1])y 54 x 10-3/2x10-3 = k (.3/.1)y  y = 3. This is a 3rd order reaction for B.

  28. Hold [B] in trial 1 and 3 constant, then look at the rate when the [A] is tripled so does the rate  the reaction is 1 order Or rate3 /rate1 = k([A3]/[A1])x 3x10-4/1x10-4 = k (.3/.1)x  x = 1.

  29. More complicated Consider a reaction A + B + C-> Products Several initial conditions of this reaction are investigated and the following data are obtained:

  30. First write the rate law. Rate = k [A]a[B]b[C]c Look to see if we can determine any orders easily Yes C can be because 1 and 3 have the same [A] & [B]

  31. The ratio of the initial rates of runs 1 and 3 is

  32. This reduces to 0.480M/s = (0.398M)c 1.102 M/s (0.525)c

  33. Notice that the units have all been eliminated • Now, to solve for c. • Take the natural log (i.e. ln) of both sides of the equation ln (0.4356) = ln ((0.7581)c) = c * ln (0.7581) c = ln(0.4356)/ ln (0.7581) =3

  34. So now we know the order of C we pick another pair of runs that have a change in the initial concentration of C and just one other reactant. Let use 1 and 4. The ratio of the initial rates of these is: 0.480 M/s = (0.151 M)a(0.213)b(0.398)3 1.102 M/s = (0.151 M)a(0.250)b(0. 480)3

  35. 0.480 M/s = (0.151 M)a(0.213)b(0.398)3 1.102 M/s = (0.151 M)a(0.250)b(0. 480)3 • Note that everything cancels, is known, or is the order b. So: 0.4868 = (.8520)b (0.5701) 0.4868/ (0.5701) = (.8520)b ln (.8521) = b ln (0.8520) b= 1

  36. Now that we know the order of Reactants B and C, we can use another pair of experimental runs to determine the remaining unknown order. We must use a pair of runs where the initial concentration of A changes, so we pick runs 1 and 2. 0.480 M/s = (0.151M)a (0.213M)1(0.398 M)3 0.358 M/s (0.251M)a (0.105M)1(0..325 M)3

  37. 0.480 M/s = (0.151M)a (0.213M)1(0.398 M)3 0.358 M/s (0.251M)a (0.105M)1(0.325 M)3 1.3483 = (0.6016)a (2.0286)1 (1.2246)3 ln (0.3619) = a ln(0.6016) ln(0.3619)/ln (0.6016) =a a=2

  38. The rate constant for the reaction may now be evaluated from any of the experimental runs: • Rate = k [A]2[B]1[C]3

  39. Rate = k [A]2[B]1[C]3 • 0.480 M/s =k (0.151M)2[0.213M]1[0.398M]3 • k = 1.57 x 103   M5/ .s

  40. 4. Determining order from integrated rate equation. The final method for determining orders is indirect and it really shows whether you know your kinetics. (This should up on the 2007 AP) Note that the integrated rate law solutions for zero, first and second order expressions are different functions but all can be written in the form of a straight line. This means that if I plot the concentration, [A], as a function of time for each expression below, the correct order should yield a straight line function.

  41. Graphs are very useful in kinetics. They can tell us a lot about the reaction.

  42. [A]0 t½ = D[A] 2k rate = - Dt Zero-Order Reactions rate = k [A]0 = k [A] is the concentration of A at any time t [A] - [A]0 = kt [A]0 is the concentration of A at time t=0 Half life for zero order t½ = t when [A] = [A]0/2

  43. D[A] rate = - Dt First-Order Reactions [A] = [A]0e-kt rate = k [A] [A] is the concentration of A at any time t ln[A] - ln[A]0 = - kt [A]0 is the concentration of A at time t=0

  44. 1 1 - = - kt [A] [A]0 t½ = D[A] rate = - Dt 1 k[A]0 Second-Order Reactions [A] is the concentration of A at any time t rate = k [A]2 [A]0 is the concentration of A at time t=0 Half life for second order t½ = t when [A] = [A]0/2

  45. Reaction Mechanism The series of steps by which a chemical reaction occurs. A chemical equation does not tell us how reactants become products - it is a summary of the overall process.

  46. Often Used Terms • Intermediate: formed in one step and used up in a subsequent step and so is never seen as a product. • Molecularity: the number of species that must collide to produce the reaction indicated by that step. • Elementary Step: A reaction whose rate law can be written from its molecularity. • uni, bi and termolecular

  47. Reaction Mechanism We can define a reaction mechanism. It is a series of elementary steps that must satisfy two requirements: • The sum of the elementary steps must give the overall balanced equation for the reactions. • The mechanism must agree with the experimentally determined rate law.

  48. Reaction Mechanism • The reaction NO2(g) + CO(g)  NO(g) + CO2(g) • has many steps in the reaction mechanism. • Rate = k[NO2]2 NO2(g) + NO2(g)  NO3(g) + NO(g) NO3(g) + CO(g)  NO2(g) + CO2(g) NO2(g) + NO2(g) + NO3(g) + CO(g)  NO3(g) + NO(g) + NO2(g) + CO2(g) Overall reaction: NO2(g) + CO(g)  NO(g) + CO2(g)

  49. Rate-Determining Step • In a multistep reaction, it is the slowest step. It therefore determines the rate of reaction. – Slow (rate determining) NO2(g) + NO2(g)  NO3(g) + NO(g) – Fast NO3(g) + CO(g)  NO2(g) + CO2(g) Rate of formation of NO3 = D[NO3]/Dt = K1[NO2]2 Overall rate = D[NO3]/Dt = k1[NO2]2