Kinetics

1. Kinetics Chapter 12 3 out of 75 M/C questions Free Response—almost every year

2. Chemical Kinetics Study of • Reaction rates • Reaction mechanism—series of steps by which a reaction takes place

3. Rate of Reaction • the change in concentration of a reactant or product per unit time Rate = [A]t2 – [A]t1 / (t2 – t1) OR Rate = D[A] / Dt • Usually considered positive even if concentration is decreasing. The square brackets indicate concentration of a substance.

4. Example Problem Nitrogen dioxide decomposes to nitrogen monoxide and oxygen gas. At the beginning of the reaction, a pressurized 1-L flask contains 2.0 mol of nitrogen dioxide. At the end of 20 minutes, the flask contains 1.5 mol of nitrogen dioxide and 0.5 mol of nitrogen monoxide. What is the rate of the reaction in terms of M/s of nitrogen dioxide?

5. Example Problem Solution • Important information: • 2.0 mol/ L nitrogen dioxide initially • After 20 minutes, 1.5 mol/L of nitrogen dioxide • Change in concentration: 2.0 mol – 1.5 mol = 0.5 mol • Change in time: 20 min • Rate: 0.5 mol / 20 min = 0.025 mol/min

6. Average vs. Instantaneous Rate • Rate is usually not constant, but decreases over time • Average rate—total change over total time (like the previous example) • Instantaneous rate—rate at a particular instant in time

7. Instantaneous Rate Think about a slope: change in "y” over change in “x”—It’s the same thing as the average rate if y is the concentration and x is the time. • Calculated by line’s slope tangent to the point in time on the rate curve Page 558

8. Rate Law • A mathematical expression which describes the dependence of reaction rate on concentration of reactants • Rate = k[A]n • k = proportionality constant This is a very important concept! The [A] means concentration of substance A. The exponent “n” is known as the order of the reaction. The order must be determined experimentally from data. The lower case k is the rate constant. It must be lower case because upper case K is an equilibrium constant.

9. Rate Law • [A] = concentration of reactant(s) • n = order of reactant • Both k and n must be experimentally determined • Cannot determine from balanced equation!!!

10. Write the rate law Rate law always begins with “rate = k….” Next, put the formula for each reactant in brackets and raise it to the power of the order. • 2NO2 2NO + O2 if NO2 is second order • Rate = k [NO2]2 • 2H2 + 2NO N2 + 2H2O if NO is second order and H2 is first order You don’t need to include zero order reactants since anything raised to the zero power is equal to one.

11. Combining Rate Laws Since both of these are equal to the rate, they are equal to each other. Rate = D[A] / Dt Rate = k[A]n D[A] / Dt = k[A]n More product terms can be added Use this equation to solve for either rate, the rate constant or a concentration.

12. Example Rate = k[A]n[B]m[C]p (Greater value of exponent means greater effect on rate.) To find exponents, determine what happens if concentration of a reactant doubles.

13. Determining orders of reactants • Find two experiments in which one concentration remains the same while the other changes. Rate 2 Rate 1 = (D concentration)n

14. Example Calculation To determine orders of reactions, by method of initial rates: Look at the data and find two trials that start with the same concentrations in all but one reactant. See trials 1 and 2. The concentration of reactant C was doubled while A and B stayed the same. There was no effect on the rate, so reactant C is zero order. Leave it out of the rate law. Continued on next slide

15. Example Calculation Compare trials 1 and 3. While the concentrations of A and C stayed the same, the concentration of B was doubled and the rate doubled. Since the effect was the same, B is a first order reactant. (2 x B = 2 x rate; 21 = 2)—The exponent is the order. Compare trials 3 and 4. The concentrations of B and C are the same while the concentration of A was doubled. The rate increased by a factor of 4 so A is a second order reactant. (2 x A = 4 x rate; 22 = 4)—The exponent is the order. Rate = k [A]2 [B]1

16. Example • Overall Rate: Rate = k[A]2[B]1[C]0 • Overall order of reaction: 2 + 1 + 0 = 3

17. Applying Rate Laws • Once the rate law is established, use data to solve for k and to find rates at different conditions.

18. Example • Calculate the rate constant k for the previous reaction. Rate = k[A]2[B]

19. Example • Since [C] has no effect on rate, we can leave it out when solving for k. Rate = k[A]2[B] • k = Rate / [A]2[B] • k = .01M/sec / (.10M)2(.10M) • k = 10 M-2sec-1 To find the rate constant, k, choose any of the trials and substitute the concentrations and initial rate into the equation. Solve for k. Pay attention to units and include them for k.

20. Write the Rate Law Equation • Look at data sets from different trials: • NH4+ + NO2- N2 + 2 H2O

21. Answer • Both reactants are first order, so Rate = k [NH4+]1[NO2-]1 • Solve for k.

22. Answer • Rate = k [NH4+]1[NO2-]1 • Using information from trial #1: • 1.35 x 10-7 M/s = k (0.100 M)(0.0050 M) • k = 2.7 x 10-4 M-1s-1

23. 2 ClO2+ 2 OH- ClO3-+ H2O Write the rate law equation and find the rate constant.

24. Answer • Rate = k [ClO2]2 [OH-] • k = 230 M-2 s-1

25. Types of Rate Laws • Differential Rate Law—expression of how rate depends on concentration (previous problems) • Integrated Rate Law—expression of how concentration depends on time • When one is determined, the other can be calculated.

26. Integrated Rate Law • Shows how concentration of A depends on time • Usually given [A]0 and k • Equation (whichever form) is in the form y = mx + b so it will be a straight line when graphed.

27. Calculating Integrated Rate Law—First Order • ln[A] = -kt + ln[A]0 • ln = natural logarithm • t = time • [A] = conc. at time t • [A]0 = conc. at time 0

28. Integrated Rate Law Equation • Equation can also be expressed as a ratio: ln ([A]0 / [A]) = kt

29. Half Life • The time required for a reactant to reach half its original concentration • t1/2

30. Calculating Half Life • When t = t1/2, then [A] = [A]0/2 • Use equation ln ([A]0 / [A]) = kt • ln ([A]0/ [A]0/2) = kt1/2 • So…ln(2) = kt1/2 • Notice, half life does not depend on concentration.

31. Calculating Integrated Rate Law—Second Order • Rate = k[A]2 • Integrated Rate Law 1/[A] = kt + 1/[A]0

32. Integrated Second Order • Plot of 1/[A] versus time is a straight line with slope = k • Half life for second order: t1/2 = 1/k[A]0 For 2nd order, half life does depend on initial concentration.

33. Zero Order Rate Laws • Often encountered when a reaction rate is limited because a surface (such as a platinum catalytic converter) is required. • When surface is covered, increasing concentration of a reactant has no effect.

34. If the reaction takes place on a surface, increasing concentration will not affect reaction rate.

35. Zero Order Rate Laws • Rate = k[A]0 = k(2) = k • Integrated Rate Law: [A] = -kt + [A]0 • Half Life: t1/2 = [A]0 / 2k

36. More Complex Rate Laws • Many reactions have several reactants—each affects rate • To study them, high concentrations of all but one reactant are used. • Highly concentrated reactants stay practically constant, so the remaining reactant can be studied.

37. Pseudo-First-Order • If conc. of B & C are large, Rate = k[A]n[B]m[C]p = k’[A] • Calculate k’ and solve for k

38. Good Summary—Table 12.6

39. Importance • Helps determine reaction mechanism • Find rate determining step in a series of reactions so total process can be sped up

40. Reaction Mechanisms • Sum of elementary steps must give overall balanced equation • Must agree with experimentally determined rate law

41. Reaction Mechanisms • Most reactions are much more complicated that they appear from their equations. • Intermediate—a species that is neither a reactant nor a product but that appears and then is consumed in the course of a reaction

43. Reaction Mechanisms • Each reaction is called an elementary step. • Rate of each elementary step can be written from its molecularity. • Molecularity = the number of species that must collide to cause the reaction

44. Molecularity • Unimolecular—depends on one molecule; Rate = k[A] • Bimolecular—depends on two molecules; Rate = k[A]2or Rate = k[A][B] • Termolecular—depends on 3 molecules; Rate = k[A]2[B] (etc)

45. Rate Determining Step • The slowest step in a reaction mechanism • Determines overall rate much like pouring water through a funnel—limiting factor

46. Rate Law • Comes from the slow step and every step prior to it. • Only consider species that are in the overall reaction. • Only consider reactants.

47. Species Not in Overall Reaction • Intermediate—a species that is produced and then consumed in a reaction—appears in mechanism but not in overall reaction—appears as a product and then a reactant • Catalyst—a species that is used during one step of a mechanism but is reproduced later—appears as a reactant and then a product

48. Determining Mechanism • See if rates of elementary steps agree with observed rate law • If yes, it is an acceptable mechanism • Never proven—only possibly correct

49. Example Problem • 2NO2 + F2 2NO2F • Rate = k[NO2][F2] • Possible Mechanism?? • NO2 + F2  NO2F + F (slow) • F + NO2  NO2F (fast)

50. Example • Overall Reaction: I2 + H2 2HI • Step 1: I2  2I • Step 2: I + H2  H2I • Step 3: H2I + I  2HI • Does this give the overall equation? • If step one is rate determining, what is the rate law? Step 2? • Identify any catalysts or intermediates.