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Chapter 12 Chemical Kinetics. Introduction l Consider the following: diamond  graphite ΔG = -2.9 kJ l Thoughts? Questions?  If this is true, why don’t.

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Presentation on theme: "Chapter 12 Chemical Kinetics. Introduction l Consider the following: diamond  graphite ΔG = -2.9 kJ l Thoughts? Questions?  If this is true, why don’t."— Presentation transcript:

1 Chapter 12 Chemical Kinetics

2 Introduction l Consider the following: diamond  graphite ΔG = -2.9 kJ l Thoughts? Questions?  If this is true, why don’t we see diamonds in jewelry turning into graphite?  Hint: the reaction happens, yes, but at what pace?  This happens VERY slowly! Also ΔG is very small, so just barely spontaneous.

3 Introduction l What have we learned about thermochemical values and what they do/don’t tell us? l Reactions can be spontaneous but this DOES NOT mean they occur fast. l Chemical kinetics: branch of chemistry that studies reaction rates.

4 What Affects Rates? l Brainstorm- what do you think? Remember- in order to react, two particles must collide! (1) Identity of reactant- large & complex = fewer chances that the collision will take place at the correct location (rate slows). (2) Temperature- greater KE means more energetic and frequent collisions. (3) Concentration- larger amounts mean increased likelihood of collisions.

5 O N Br O N O N O N O N O N O N O N No Reaction O N Br O N Molecular Orientation

6 What Affects Rates? (4) States of matter- reactants in the same state are more able to collide & react; liquids and gases are more able to collide due to greater surface area. (5) Adding a catalyst- adding this increases the rate. *Is recoverable at the end of the rxn. *Reduces activation energy.

7 Unit Essential Question l What role do rates play in chemical reactions?

8 REACTION RATES Section 12.1 Lesson Essential Question: How is the rate of a reaction determined?

9 Example Using stoichiometry allows rates to be set equal to each other! Thus the rate of disappearance of a reactant can be said to equal the rate of appearance of a product.

10 Kinetics l 2A + B  C + 3D (generic reaction) Rate =  [substance]  t Rate = -½  [A]/  t  t =  C  t =  D  t l Notice effect of coefficients! l Why are the first two negative? l Because reactant concentrations decrease! l This gives the initial reaction rate- how quickly the reaction occurs as soon as the reactants are mixed. Change in [ ] per unit time

11 Practice l N 2 + 3H 2  2NH 3 l What would the initial rate equal in terms of each substance in the reaction? Rate =  [N 2 ] = -1/3  [H 2 ] = ½  [NH 3 ]  t  t  t


13 l Rate = k[NO 2 ] n l This is called a rate law expression. l Only reactants are used! Why? l k is called the rate constant.  k does not change for a given reaction UNLESS T changes or a catalyst is added. l n is the order of the reactant(s) - usually a positive integer. l Value of exponent can ONLY be determined experimentally! Does NOT com from coefficients! 2 NO 2  2 NO + O 2 Rate Law Expression

14 l When n = 1: if reactant concentration is doubled, the rate doubles; if reactant concentration is tripled, the rate triples. First order with respect to this substance. l When n = 2: if reactant concentration is doubled, the rate increases four times (2 2 = 4). Second order with respect to this substance. l When n = 0: reactant concentration has no effect on the rate; it simply needs to be present in some amount. Zero order with respect to this substance. l Can add orders together for all reactants to get the overall order for the entire reaction. Understanding Rate Laws & Orders: Understanding Rate Laws & Orders: Rate = k[NO 2 ] n

15 Determining Rate Order & k 2NO (g) + O 2 (g)  2NO 2 (g) Initial [NO] Initial [O 2 ] Rate of NO 2 formation (M/s) Can determine rate orders by inspection or mathematically. Second order with respect to NO and first order with respect to O 2.

16 Determining Rate Order & k l Once you know the rate orders for each reactant you can solve for K. How? l Choose any experiment and the corresponding concentration values and rate. Plug them into the rate equation and solve for k. l k = 5 x 10 4 /M 2 s

17 Sections 1&2 Homework l Pg. 567 #10, 13, 26, 28

18 AP Practice Question What is the rate law for the following: CO (g) + Cl 2 (g)  COCl 2 (g) a) Rate = k[CO] c) Rate = k[CO][Cl 2 ] b) Rate = k[CO] 2 [Cl 2 ] d) Rate = k[CO][Cl 2 ] 2 ExperimentInitial [CO] MInitial [Cl 2 ] MInitial Rate x x x

19 AP Practice Question The reaction (CH 3 ) 3 CBr (aq) + H 2 O (l)  (CH 3 ) 3 COH (aq) + HBr (aq) has the following rate law: Rate = k[(CH 3 ) 3 CBr]. What will the effect of decreasing the concentration of (CH 3 ) 3 CBr? a) Reaction rate will increase. b) More HBr will form. c) Reaction rate will decrease. d) Reaction will shift to the left.

20 Another Example l For the reaction: BrO Br - + 6H +  3Br H 2 O l The general form of the rate law is: Rate = k[BrO 3 - ] a [Br - ] b [H + ] c l Use experimental data to determine the values of a, b, and c. l Then determine the reaction order for each reactant and the overall reaction.

21 Initial concentrations (M) Rate (M/s) BrO 3 - Br - H+H+H+H x x x x x x x x 10 -3

22 Initial concentrations (M) Rate (M/s) BrO 3 - Br - H+H+H+H x x x x x x x x Now we have to see how the rates change with concentration!

23 Answers to Example l The powers of the reactants are as follows: a = 1; b = 1; c = 2, so: Rate = k[BrO 3 - ] [Br - ] [H + ] 2 l The reaction is first order with respect to BrO 3 - and Br -, second order with respect to H +, and fourth order overall. l How would you calculate the value of the rate constant? k = 8.0 L 3 /mol 3 ∙s


25 Integrated Rate Law l Allows us to determine information about concentrations over time, not just at a particular instant during the reaction. l For example, with the integrated rate law we can determine how long it will take to use up a reactant, form a product, etc. l Integrated rate laws come from setting the rate law and rate expression equal to each other for each order, and integrating over time.

26 First Order Integrated Rate Law l ln[A] t – ln[A] 0 = -kt l [A] t = concentration of reactant at a given time, t l [A] 0 = initial concentration of reactant l Can be written as: ln[A] t = -kt + ln[A] 0 l Has the format of a line! l Note: slope = -k so adjust signs accordingly when solving for k! l Graphing ln[A] vs. time gives a straight line for first order!

27 Second Order Integrated Rate Law l 1/[A] t – 1/[A] 0 = kt l Can be written as: 1/[A] t = kt + 1/[A] 0 l This also has the format of a line! l Note: slope = k so no need to adjust signs! l Graphing 1/[A] vs. time gives a straight line for second order!

28 Practice HI decomposes in a second order reaction. The value of k = 2.40x /(Ms) at 25°C. How long will it take the concentration of HI to decrease from 0.200M to 0.190M at 25°C? 1.10 x s

29 Half Life Half life: time required for a reactant to reach half of its original concentration. Does not depend on concentrations for first order; only depends on k: t 1/2 = ln2/k = 0.693/k Does depend on concentrations for second order, and also k: t 1/2 = 1/k[A] 0 Note: radioactive decay is first order!

30 AP Practice Question The rate constant, k, for radioactive beryllium-11 is 0.049/s. What mass of a 0.500mg sample of beryllium-11 remains after 28 seconds? a) 0.250mg b) 0.125mg c) mg d) 0.375mg Find how many half lives make up the time given. Then halve the initial amount that many times. t 1/2 = 14s. 14 goes into 28 two times, so Be goes through two half lives; divide by two twice.

31 Practice l The rate constant for the radioactive decay of thorium-232 is 5.0x /year. What is the half life of thorium? x years

32 Practice HI decomposes in a second order process to its elements. The value of the rate constant is 2.40 x /Ms at 25°C. What is the half life for the decomposition of 0.200M HI at 25°C? l 2.08 x seconds

33 Determining Rate Order Example l If you are given data on a reaction (concentrations of reactant and times), plot the data to determine the order. l Plot ln[A] vs. time (first order) and 1/[A] vs. time (second order).  Whichever gives a straight line is the order of the reaction! l To find k, determine the slope of the line (remember, m = –k for first order and m = k for second order). l Use appropriate half life equation to calculate t 1/2.

34 Determining Order Example l The following data was collected for the reaction: 2C 4 H 6 (g)  C 8 H 12 (g) l What is: the reaction order, the rate constant value, and the half life of the reaction? [C 4 H 6 ] (mol/L)Time (±1s)

35 Determining Order Example l CAN’T determine order by inspection because the time at which [ ]’s are measured are not the same! MUST use integrated rate laws/graphing. [C 4 H 6 ] (mol/L)Time (±1s)

36 Zero Order Rate Law l Rate = k[A] 0 = k  Rate does not change with concentration. l Integrated: [A] = -kt + [A] 0 l Half life: t 1/2 = [A] 0 /2k

37 l Typically happen on a surface because the surface area stays constant.  Rate of contact to surface impacts rate of reaction, not the concentration of the substance on the surface. l Also applies to enzyme chemistry.  Rate depends on how quickly enzyme works. When Do Zero Order Rate Laws Occur?

38 AP Practice Question A reaction follows the rate law: Rate = k[A] 2. Which of the plots below will give a straight line? a) 1/[A] vs. 1/time b) [A] 2 vs. time c) 1/[A] vs. time d) ln[A] vs. time

39 AP Practice Question For the following reaction: NO 2 (g) + CO (g)  NO (g) + CO 2 (g) the rate law is: Rate = k[NO 2 ]. If a little CO is added to a mixture that had 0.10M NO 2 and 0.20M CO, which statement is true? a) Both k and the rxn. rate remain the same. b) Both k and the rxn. rate increase. c) Both k and the rxn. rate decrease. d) Only k increases; rxn. rate stays the same.

40 AP Practice Question The decomposition of ammonia to the elements is a first-order reaction with t 1/2 = 200s at a certain T. How long will it take the partial pressure of NH 3 to decrease from 0.100atm to atm? a) 200s b) 400s c) 800s d) 1,000s Remember- P is directly linked to # of moles for gases! Treat pressures like amounts. How much pressure would one half life give? Is this the final pressure being asked for?

41 Section 4 Homework/Examples l Pg. 569 # 31, 39, 43, 46

42 RATE LAWS: A SUMMARY Section 12.5

43 Important Points 1. Forward reaction is important- rate laws will only contain reactants. 2. Two types of rate laws: Rate Law Shows how rate depends on concentration. Integrated Rate Law Shows how concentration depends on time.

44 Important Points cont. 3. Type of law depends on type of data collected. Can write the other if we have one. 4. Initial rate method used if several experiments are done. Order of reactant can be determined 5. Can graph experimental data to determine rate law (order). K can obtained from the slope

45 Summary of Rate Laws

46 Further Look at Activation Energy l Activation energy (E a ): the minimum energy required to start a reaction.  Need E to break bonds!  KE when molecules collide can supply the energy needed to break bonds and start the reaction.  This is why higher T increases rates!  E is then released when new bonds form.

47 Activation Energy Graph (activated complex) The activated complex/transition state is the orientation of the reactants as bonds are breaking and forming. Note the difference!

48 Arrhenius Equation l Relationship between T and k can be used to find the activation energy: k = Ae -Ea/RT l k = rate constant; A = frequency factor; E a = activation E; R = ideal gas constant; T = temperature in K l Take ln of both sides: ln k = -(E a /R)(1/T) + ln A l ln k vs 1/T is a straight line: slope = -E a /R. Since R is known, E a can be calculated. l A could also be calculated: y-intercept.

49 AP Practice Question The energy difference between the reactants and the transition state is: a) the free energy. b) the heat of reaction. c) the activation energy. d) the kinetic energy.

50 AP Practice Question The purpose for striking a match against the side of a box is to: a) supply the free energy for the rxn. b) supply the activation energy for the rxn. c) supply the heat of reaction. d) supply the kinetic energy for the rxn.

51 REACTION MECHANISMS Section 12.6 Lesson Essential Question: 1)How do the steps of a reaction impact the rate of the reaction?

52 Reaction Mechanisms l The reaction mechanism is a series of steps that occur in a chemical reaction. l Reactions do not typically occur in one step. Why might this be?  Think about orientations and energy!  Particles need to collide in the correct location AND with enough energy.

53 Reaction Mechanisms l Kinetics can tell us about the mechanism; balanced equations can’t. l Ex: NO 2 (g) + CO (g)  NO (g) + CO 2 (g) l The proposed mechanism is: NO 2 (g) + NO 2 (g)  NO 3 (g) + NO (g) NO 3 (g) + CO (g)  NO 2 (g) + CO 2 (g) l How do people know that the overall reaction doesn’t happen in one step? l Experimental data suggests the following rate law: Rate = k[NO 2 ] 2. No CO is present!

54 Reaction Mechanisms l A mechanism should always end up giving the overall reaction being considered.  Think of this part like Hess’ Law- cancel out species that appear on opposite sides to arrive at the overall reaction. l Examine the previous mechanism: NO 2 (g) + NO 2 (g)  NO 3 (g) + NO (g) NO 3 (g) + CO (g)  NO 2 (g) + CO 2 (g) NO 2 (g) + CO (g)  NO (g) + CO 2 (g) +

55 Reaction Intermediates l Example continued: NO 2 (g) + NO 2 (g)  NO 3 (g) + NO (g) NO 3 (g) + CO (g)  NO 2 (g) + CO 2 (g) NO 2 (g) + CO (g)  NO (g) + CO 2 (g) l Reaction intermediates are substances that are produced and consumed. l What is/are the intermediate(s) above? l Only NO 3 because it is first produced, and later consumed. NO 2 is there to start with! +

56 Rate-Determining Step l Each step in a mechanism can be called an elementary step. l One elementary step is always slower than the rest; it’s called the rate-determining step (RDS). l This step is essentially the only one affecting the rate of the overall reaction since the other steps, in comparison, are much faster.

57 Rate-Determining Step l What does this mean? l Rate laws CAN be written directly from the RDS! l We CANNOT do this with the overall reaction being examined (as we saw in previous sections) because it is not the RDS!

58 Acceptable Reaction Mechanisms l Two requirements for acceptable mechanisms:  Using a process like Hess’ Law, the proposed mechanism must equal the overall reaction.  The RDS must equal the experimentally determined rate law.  Rate cannot be faster than the slowest step!

59 l Example continued: NO 2 (g) + NO 2 (g)  NO 3 (g) + NO (g) slow NO 3 (g) + CO (g)  NO 2 (g) + CO 2 (g) fast l Which is the RDS?  The first one! l The rate law for the overall reaction is: Rate = k[NO 2 ][NO 2 ]  Rate = k[NO 2 ] 2 Rate-Determining Step

60 l 2NO 2 + F 2  2NO 2 F l Proposed mechanism: NO 2 + F 2  NO 2 F + F slow F + NO 2  NO 2 Ffast l What is the RDS? l What is the rate law for this reaction? l What is/are the intermediate(s)? Reaction Mechanism Practice Rate = k[NO 2 ][F 2 ] first elementary step F

61 Molecularity l Molecularity tells you the number of molecules colliding. l Unimolecular- one molecule breaks down or rearranges. l Bimolecular- two molecules collide. l Termolecular- three molecules collide. l Which of these would you expect to be least likely to occur? l Termolecular; the odds of three molecules colliding in exactly the correct orientation is highly unlikely.


63 Example l 2NO 2 (g) + F 2 (g)  2NO 2 F(g) l Rate = k[NO 2 ][F 2 ] l Suggested mechanism: NO 2 + F 2 NO 2 F + F (slow) F + NO 2 NO 2 F (fast) l Is this an acceptable mechanism (does it satisfy the two requirements)? l Being acceptable does not mean it is the correct mechanism! k1k1 k2k2 YES

64 Example l Another possible mechanism: NO 2 + F 2  NOF 2 + O (slow) NO 2 + O  NO 3 (fast) NOF 2 + NO 2  NO 2 F + NOF (fast) NO 3 + NOF  NO 2 F + NO 2 (fast) l Perform additional reactions to decide on the most likely mechanism. l You can determine the rate law for a reaction given a proposed mechanism.  Use RDS information!

65 CATALYSTS Section 12.8

66 Catalysts l Speeds up a reaction without being consumed; the catalyst will still be there in the end although it may change along the way. l An iron ion will speed up the decomposition of hydrogen peroxide: 2Fe +3 + H 2 O 2  2Fe +2 + O 2 + 2H + 2Fe +2 + H 2 O 2 + 2H +  2Fe H 2 O l Notice the iron ends up exactly how it started!

67 Homogeneous Catalysts l Homogeneous catalysts (same state as reactants) allow reactions to proceed by a different mechanism - a new pathway. l New pathway has a lower activation energy. l More molecules will be able to overcome this activation energy. Does not change overall  E.  Starting and ending energies of reactants and products are not altered.


69 Homogeneous Catalysts l Two additional examples:  Enzymes are biological catalysts.  Chlorofluorocarbons (CFC’s) catalyze the decomposition of ozone.

70 Heterogeneous Catalysts l Heterogeneous catalysts (different state from reactants) allow more surface for the reaction to occur AND allow for better orientations of reactants. l Typically the catalyst is a solid for liquid or gaseous reactants. l Ex: making ammonia (Haber process) requires an iron catalyst for nitrogen and hydrogen gases to react.

71 Additional Info. l For reactions in equilibrium (when both forward and reverse reactions are considered), BOTH are affected by adding a catalyst.  The rates for both increase since they share the same E a.


73 Sections 7&8 Homework l Pg. 571 #49, 50, 51, 53, 55, 65

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