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1 Kinetics: Rates and Mechanisms of Reactions Chemical Kinetics tells us: …how fast a reaction will occur …how molecules react (MECHANISM) a mechanism.

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Presentation on theme: "1 Kinetics: Rates and Mechanisms of Reactions Chemical Kinetics tells us: …how fast a reaction will occur …how molecules react (MECHANISM) a mechanism."— Presentation transcript:

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2 1 Kinetics: Rates and Mechanisms of Reactions Chemical Kinetics tells us: …how fast a reaction will occur …how molecules react (MECHANISM) a mechanism is a sequence of steps that lead to the product

3 2 Factors Affecting the rate: 1. concentration: 2. temperature: generally a 10 o C increase will double the rate 3. nature of the reactant: i.e. surface area 4. catalyst: (two types: homogenous and heterogeneous) 5. mechanism: (orientation, shape, & order) COLLISION THEORY = CAPPING A MARKER

4 3 Rate of RXN = The increase in concentration of a product per unit time. or The decrease in concentration of a reactant per unit time.

5 4 Conc. is usually measured in M (Molarity= mol/L) for solutions. Rate =  M  time = mol L s or molL -1 s -1 or Ms -1 Since many reactions involve gases,  P is often used for concentration. Square brackets [ ] are often used to express molarity (i.e. [HCl] means Molarity of HCl) moles/L

6 5 3ClO - (aq)  2Cl - (aq) + ClO 3 - (aq) Consider the reaction (net ionic eq.): Rate could be defined in at least 3 ways: (3 coefficients and ions) 1. appearance of ClO appearance of Cl - 3. disappearance of ClO - Disappearance

7 6 3ClO - (aq)  2Cl - (aq) + ClO 3 - (aq) Consider the reaction (net ionic): 1. appearance of ClO appearance of Cl - 3. disappearance of ClO - Question: Are these three rates equal? Disappearance

8 7 3ClO - (aq)  2Cl - (aq) + ClO 3 - (aq) Consider the reaction (net ionic): Let’s make these three rates equal. Note the use of coefficients and the - sign

9 8 General Form: aA + bB  cC + dD rate =  [C] c  time =  [D] d  time = -  [A] a  time = -  [B] b  time “PRODUCTS” “REACTANTS”

10 9 Average rate = slope (over time period) negative sign

11 10 instantaneous rate = tangent slope (changing) WHY? Collision Theory!

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13 12 IF we can now somehow get a linear plot in the form of: y = mx + b. The slope would be a constant independent of concentration!

14 13 We could call the slope the rate constant and assign it the letter k! rate constant = k  instantaneous rate or rate “call in the mathematicians”

15 14 rate constant: k is conc. independent Is still temperature dependent and mechanism dependent! General form of rate law : for RXN: A  products m = RXN order according to A. Determined by experiment only! conc. of A rate constant

16 15 General form of rate law : RXN orders (w, x, y, and z)must be determined by exp. only! Total (overall) order =  individual orders

17 16 General Equation Forms: 0 order: rate = k 1st order: rate = k[A] 2nd order: rate = k[A] 2 or rate = k[A][B] 3rd order: rate = k[A] 3 or Simple experiments are done by doubling 1 conc. at a time and looking at the effect.

18 17 General Equation Forms: 0 order: rate = k 1st order: rate = k[A] 2nd order: rate = k[A] 2 or rate = k[A][B] 3rd order: rate = k[A] 3 or Simple experiments are done by doubling 1 conc. at a time and looking at the effect. order doubling effect on rate 0[2] 0 = 1none 1/2[2] 1/2 = increase by [2] 1 = 2doubles 2[2] 2 = 4quadruples Question: suppose rate = k[A] 2 [B] what is the effect of doubling both A and B?

19 18 Let’s look a some rate data for the RXN: NH 4 + (aq) + NO 2 - (aq)  N 2 (g) + 2H 2 O(l)

20 19 NH 4 + (aq) + NO 2 - (aq)  N 2 (g) + 2H 2 O(l) doubles double

21 20 NH 4 + (aq) + NO 2 - (aq)  N 2 (g) + 2H 2 O(l) 2 2 for NH 4 + : [2] x = 2 x = 1 (1st Order) for NO 2 - : [2] y = 2 y = 1 (1st Order) 2 2

22 21 first order with respect to each reactant, 2nd order overall. orders usually have integer values, but can be fractional. Can also be (-) (inhibitors). Since rate has units, k must also have units. so units of k must work with [ ] to match units.

23 22 Determine the units of k in each of the following: 1. rate = k[A] 2. rate = k[A] 2 Since rate has units, k must also have units. (so units of k must work with [ ] to match units.) M ? k units must = time - 1 M2M2 ? k units must = time - 1 M - 1 so k must have units of M -1 t -1

24 23 4. rate = k[A][B] 2 [C] 5. rate = k[A] 0 k units =M - 3 time - 1

25 24 Ebbing 4th ed. P In a kinetic study of the reaction: 2NO(g) + O 2 (g)  2NO 2 (g) the following data were obtained for the initial rates disap- pearance of NO: Obtain the rate law. What is the value of the rate constant?.

26 25 Ebbing 4th ed. P In a kinetic study of the reaction: 2NO(g) + O 2 (g)  2NO 2 (g) the following data were obtained for the initial rates disap- pearance of NO: Obtain the rate law. What is the value of the rate constant? Write overall rate equation: rate = k[NO] x [O 2 ] y For NO: select a pair of experiments in which the conc. of [NO] is changed, but other concentrations are unchanged. Let’s use Exp. 1 and 2

27 26 Ebbing 4th ed. P In a kinetic study of the reaction: 2NO(g) + O 2 (g)  2NO(g) Obtain the rate law. What is the value of the rate constant? overall rate equation:rate = k[NO] x [O 2 ] y “Let’s divide Exp.1 by Exp.2 to allow us to cancel terms.

28 27 Ebbing 4th ed. P In a kinetic study of the reaction: 2NO(g) + O 2 (g)  2NO(g) Obtain the rate law. What is the value of the rate constant? overall rate equation:rate = k[NO] x [O 2 ] y

29 28 Ebbing 4th ed. P In a kinetic study of the reaction: 2NO(g) + O 2 (g)  2NO(g) overall rate equation:rate = k[NO] x [O 2 ] y How do we solve for x? Use logarithms

30 29 Ebbing 4th ed. P In a kinetic study of the reaction: 2NO(g) + O 2 (g)  2NO(g) overall rate equation:rate = k[NO] x [O 2 ] y How do we solve for x? Use logarithms Therefore the rate equation becomes:rate = k[NO] 2 [O 2 ] y

31 30 Now let’s determine the reaction order with respect to [O 2 ]

32 31 Ebbing 4th ed. P In a kinetic study of the reaction: 2NO(g) + O 2 (g)  2NO 2 (g) the following data were obtained for the initial rates disap- pearance of NO: Write overall rate equation:rate = k[NO] 2 [O 2 ] y For O 2 : select a pair of experiments in which the conc. of [O 2 ] is changed, but all other concentrations are unchanged. Let’s use Exp. 1 and 3

33 32 Ebbing 4th ed. P In a kinetic study of the reaction: 2NO(g) + O 2 (g)  2NO 2 (g) the following data were obtained for the initial rates disap- pearance of NO: Write overall rate equation:rate = k[NO] 2 [O 2 ] y Let’s use Exp. 1 and 3

34 33 Ebbing 4th ed. P In a kinetic study of the reaction: 2NO(g) + O 2 (g)  2NO 2 (g) Write overall rate equation:rate = k[NO] 2 [O 2 ] y

35 34 Ebbing 4th ed. P In a kinetic study of the reaction: 2NO(g) + O 2 (g)  2NO 2 (g) Write overall rate equation:rate = k[NO] 2 [O 2 ] y y = 1 Therefore: rate = k[NO] 2 [O 2 ] 1 Now solve for k.

36 35 Ebbing 4th ed. P In a kinetic study of the reaction: 2NO(g) + O 2 (g)  2NO 2 (g) Overall rate equation: rate = k[NO] 2 [O 2 ] 1 Choose any Exp. and substitute in experimental to obtain k. i.e. Exp1. : rate = k[NO] 2 [O 2 ] 1 so : = k[0.0125] 2 [0.0253] k = 7100 M - 2 s - 1

37 36 Ebbing 4th ed. P Iodine ion is oxidized to hypoiodite ion, IO -, by hypochlorite ion, ClO -, in basic solution. the following initial rate experiments were run: Obtain the rate law. What is the value of the rate constant?.

38 37 k = 6.1 s -1

39 38 This data isn’t linear! What can we do? Integrated Rate Laws

40 39 IF we can now somehow get a linear plot in the form of: y = mx + b. The slope would be a constant, independent of concentration!

41 40 We could call the slope the rate constant and assign it the letter k! rate constant = k  rate “call in the mathematicians”

42 41 Key Equations: * Since the units of rate are concentration/time, the units of k (the rate constant) must dimensionally agree. So for each order, k will have different units and those units can tell one which equation to use. [ ] means the concentration of the enclosed species in Molarity (M).

43 42 The data below was collected for the reaction: NOCl(g)  NO(g) + 1/2Cl 2 (g) Time (s) [NOCl] (M) Prepare THREE graphs to determine if the RXN is ZERO, 1st, or 2nd order. Then determine the value and units of the rate constant k.

44 43 Zero Order Plot [A] t vs. time rate = k [A] t = - kt + [A] 0 y = mx + b

45 44 First Order Plot ln[A] t vs. time rate =k[A] ln[A] t = - kt + ln[A] 0 y = mx + b

46 45 2nd Order Plot 1/[A] t vs. time Plot is linear so 2nd Order k = slope = M -1 s -1 rate = k[A] 2 1/[A] t = kt + 1/[A] 0 y = mx + b

47 46 Zero Order: rate = k[A] 0 = k rate = k integrated gives: [A] t = -kt + [A] 0 y = mx + b slope = - k If a RXN is zero order, a plot of [A] vs. time should be linear and the slope = -k. Integrated rate laws:

48 47 Integrated rate laws: 1st order rate laws: rate = k[A] integrated gives: rearranged to : y = mx + b gives: ln[A] t = -kt + ln[A] 0 slope = - k If reaction data is 1st order, a plot of ln[A] vs. time should be linear.

49 48 2nd Order Integrated Rate Equations: rate = k[A] 2 integrated gives: y = mx + b slope = k If a RXN is 2nd order, a plot of 1/[A] vs. time should be linear and the slope = k.

50 49 slope = - k RXN is first order with respect to CH 3 NC ln[CH 3 NC] t = -kt + ln[CH 3 NC] 0 Zero Order Plot1st Order Plot

51 50 2nd order Plot Slope = k First order plot RXN is 2nd order with respect to [NO 2 ]

52 51 General form 1st order: ln[A] t = -kt + ln[A] 0 Note: This is a formula that can be used to solve (1st order) problems. (If all but one of the variables are given) 1. Given the RXN: C 3 H 6  CH 2 =CHCH 3 Where k = 6.0 x s o C. Looking at the units of k, determine the order. Problem: if [C 3 H 6 ] 0 = M, find [C 3 H s. ln[A] t = -kt + ln[A] 0 ln[A] t = [A] t = M

53 52 The other equations can be used in a similar fashion.

54 53 Half-life: the time it takes to decrease the concentration to 1/2 its initial value. “fold paper to view subsequent half-lives”

55 54 Formulas: 1st order: t 1/2 = 0.693/k 1/2 : and: so: Half-lives:

56 55 Half-life formulas: Zero OrderFirst Order 2nd Order These 2 are conc. dependent (and not very useful).

57 56 Energy Diagrams: All chemical and physical changes are accompanied by energy changes. e n e r g y  time  reactants products  E Question: What keeps the reactants from rolling down the hill? E a = Activation energy Exothermic

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63 62 rate constant (k) varies with temperature. but not with concentration

64 63 The Arrhenius Equation: Temp in K 8.31 J/molK activation energy base e (natural ln) frequency factor (1/time), fraction of collisions with correct geometry. rate constant - Ea/RT is always <1 and refers to the fraction of molecules having minimum energy for a RXN.

65 64 How can we make this a linear equation in the form of y = mx + b? Take the ln of each side. ln k = ln A - E a /RT or: y = b - mx A plot of ln k vs. 1/T gives a straight line with the slope = -E a /R (E a = x slope)

66 65 At two diff. temps. we get:

67 66 Reaction Mechanisms: Reaction is broken into steps with intermediates being formed. “some RXNS occur in one step, but most occur in in multiple steps.” Each Step is called an elementary step, and the number of molecules involved in each step defines the molecularity of the step. uni-molecular: = 1 i.e. O 3 *  O 2 + O bi-molecular: = 2 (these are the most common) i.e. HI + HI  activated complex  H 2 + I 2 ter-molecular: = 3 (rare, due to probability of orientation and energy both being correct.) i.e. O(g) + O 2 (g) + N 2 (g)  O 3 (g) + “energetic” N 2 (g)

68 67 The Raschig process for the preparation of hydrazine (N 2 H 4 ) Overall RXN: 2NH 3 (g) + NaOCl(aq)  N 2 H 4 (aq) + NaCl(aq) + H 2 O(l) Proposed Mechanism: (Only from experiment) Step 1: NH 3 (aq) + OCl - (aq)  NH 2 Cl(aq) ‡ + OH - ‡ (aq) Step 2: NH 2 Cl(aq) ‡ + NH 3 (aq)  N 2 H 5 +‡ + Cl - (aq) Step 3: N 2 H 5 + (aq) ‡ + OH - ‡ (aq)  N 2 H 4 (aq) + H 2 O(l) “Cancel intermediates and “add steps” to give overall RXN.” 2NH 3 (g) + OCl - (aq)  N 2 H 4 (aq) + Cl - (aq) + H 2 O(l) The overall rate law, mechanism, and the total order can’t be predicted from the stoichiometry, only by experiment.

69 68 The following is only true for individual steps: The rate law of an elementary step is given by the product of a rate constant and the conc. of the reactants in the step. Step Molecularityrate law A  Product(s) uni rate = k[A] A + B  Product(s) bi rate = k[A][B] A + A  Product(s) bi rate = k[A] 2 2A + B  Product(s) ter rate = k[A] 2 [B] The overall mechanism must match the observed rate law. Usually one STEP is assumed to be the rate determining step.

70 69 Example: Overall RXN: 2NO 2 (g) + F 2 (g)  2NO 2 F(g) Observed Experimental rate law: rate = k[NO 2 ][F 2 ] Question:Why does this rule out a single step RXN? Answer: rate law for single step process would be: rate = k[NO 2 ] 2 [F 2 ] “Let’s try to work out a Mechanism that matches the observed rate law.”

71 70 Example: Overall RXN: 2NO 2 (g) + F 2 (g)  2NO 2 F(g) Observed Experimental rate law: rate = k[NO 2 ][F 2 ] slow:NO 2 (g) + F 2 (g)  NO 2 F(g) + F(g) ‡ k1k1 fast: NO 2 (g) + F(g) ‡  NO 2 F(g) k2k2 The rate law is dependent upon the slow step. rate = k[NO 2 ][F 2 ] 2NO 2 (g) + F 2 (g)  2NO 2 F(g)

72 71 Let’s try making the 2 nd RXN slow and the first fast Overall RXN: 2NO 2 (g) + F 2 (g)  2NO 2 F(g) Observed Experimental rate law: rate = k[NO 2 ][F 2 ] fast:NO 2 (g) + F 2 (g)  NO 2 F(g) + F(g) ‡ k1k1 slow: NO 2 (g) + F(g) ‡  NO 2 F(g) k2k2 The rate law is dependent upon the slow step. 2NO 2 (g) + F 2 (g)  2NO 2 F(g) rate = k[NO 2 ][F ‡ ]

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74 73 1. At low temperatures, the rate law for the reaction: CO(g) + NO 2 (g)  CO 2 (g) + NO(g) is:rate = k[NO 2 ] 2 Which mechanism is consistent with the rate law? a. CO + NO 2  CO 2 + NO b. 2NO 2  N 2 O 4 ‡ (fast) N 2 O 4 ‡ + 2CO  2CO 2 + 2NO (slow) c. 2NO 2  NO 3 ‡ + NO (slow) NO 3 ‡ + CO  NO 2 + CO 2 (fast) d. 2NO 2  2NO + O 2 ‡ (slow) 2CO + O 2 ‡  2CO 2 (fast) rate = k[CO][NO 2 ]

75 74 For the following mechanism: 2NO  N 2 O 2 (fast) N 2 O 2 + H 2  H 2 O + N 2 0(slow) N 2 O + H 2  N 2 + H 2 O(fast) a. Determine the overall reaction b. Does the mechanism agree with the rate law: rate = k[NO] 2 [H 2 ]

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