Chemical Kinetics Chapter 12

Transcription

1 Chemical Kinetics Chapter 12 With the exception of Section 11.2, Chapter 11 is beyond the scope of the AP exam. These Tour de France cyclists generate a great amount of kinetic energy as they ride through the mountains.

2 Introduction! In thermo, we determined whether a reaction will occur spontaneously (sign of ΔG)." Spontaneous does NOT mean fast.! Chemical kinetics: The area of chemistry that concerns reaction rates."

3 Consider the decomposition of nitrogen dioxide." 2 NO 2 (g) 2 NO(g) + O 2 (g)" Experiment:! Start with a flask of nitrogen dioxide at 300 C and measure [all three substances] as decomposition occurs." Recall: [x] means molar concentration of x (mol L)."

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7 Reaction rate:! Δ[reactant or product] per unit time." Consider Rate avg at which [NO 2 ] changes from" t = 0 to t = 50 s." Change in [NO 2 ] Time elapsed = Δ[NO 2] Δt = [NO 2] t=50 [NO 2 ] t=0 (50. 0) s = mol L mol L 50. s = mol (L s)

8 Note:! Δ[NO 2 ] < 0 because it s decreasing. However, it s customary to work with positive reaction rates:" Rate = Δ[NO 2 ] Δt = ( mol (L s)) = mol (L s)

9 Note:! Rate is not constant, but decreases with time." Most reaction rates Δ with time since concentrations Δ with time. (Makes sense.)"

10 Rate can also be defined in terms of the products. Must take molar coefficients into account." 2 NO 2 (g) 2 NO(g) + O 2 (g)" 1. NO is produced at the same rate as NO 2 is consumed. (Both have coefficient of 2, also verified graphically.)" 2. Rate of NO production is twice the rate of O 2 production." and 2. Can be verified by calculation (textbook)."

11 Rate of consumption" of NO 2 " Rate of =" production ="2(rate of production of O 2 )" of NO"

12 Rate Law: Describes the dependence of the initial rate of a reaction on the [reactant(s)]. Must be determined experimentally." A + B C" General form: Rate = k [A] m [B] n! k rate constant" m order with respect to A" n order with respect to B" Note: m and n are most often integers (including zero), but can also be fractions."

13 Textbook Example, pages :! Consider the following reaction." NH 4+ (aq) + NO 2 (aq) N 2 (g) + 2 H 2 O(l) Table 12.4 gives initial rates obtained from three experiments involving different initial concentrations of reactants."

14 Textbook Example, pages :! Consider the following reaction." NH 4+ (aq) + NO 2 (aq) N 2 (g) + 2 H 2 O(l) Table 12.4 gives initial rates obtained from three experiments involving different initial concentrations of reactants." (a) Determine the order with respect to NH 4+." (b) Determine the order with respect to NO 2." (c) What is the overall order of the reaction?" (d) Determine the rate constant for the reaction, including units." (e) Write the rate law for the reaction."

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19 Example 12.1, page 562:! The reaction between bromate ions and bromide ions in acidic aqueous solution is given by the equation" BrO 3 (aq) + 5 Br (aq) + 6 H + (aq) 3 Br 2 (l) + 3 H 2 O(l) Table 12.5 gives the results from four experiments."

20 Example 12.1, page 562:! The reaction between bromate ions and bromide ions in acidic aqueous solution is given by the equation" BrO 3 (aq) + 5 Br (aq) + 6 H + (aq) 3 Br 2 (l) + 3 H 2 O(l) Table 12.5 gives the results from four experiments." (a) Using the information in the table, determine the order of the reaction with respect to each of the following. Justify your answers." i. BrO " 3 ii. Br " iii. H + " (b) Using your answers to part (a) above, complete the following:" i. Write the rate law for the reaction. " ii. Calculate the value of the rate constant, k, for the reaction, including appropriate units."

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25 The Integrated Rate Law: Rate law using concentration and time.! It is also useful to have rate laws that relate the rate constant k to the way that concentrations change over time." " For simplicity, consider reactions involving a single reactant:" aa products

26 Zero-Order Rate Law: Example, page " Consider the decomposition of gaseous dinitrogen monoxide." 2 N 2 O(g) 2 N 2 (g) + O 2 (g)" The reaction takes place on a platinum surface. Although [N 2 O] is 3 as great in (b) as in (a), the rate of decomposition of N 2 O is the same in both cases because the platinum surface can accommodate only a certain number of molecules. As a result, the reaction is zero-order."

27 First-Order Rate Law: Example 12.2, page 565" Consider the decomposition of gaseous dinitrogen pentoxide." 2 N 2 O 5 (g) 4 NO 2 (g) + O 2 (g)" Construct a plot of ln [N 2 O 5 ] versus time."

28 First-Order Rate Law: Example 12.2, page 565" Consider the decomposition of gaseous dinitrogen pentoxide." 2 N 2 O 5 (g) 4 NO 2 (g) + O 2 (g)" (a) How can one verify that the rate law is first order? Explain."

29 First-Order Rate Law: Example 12.2, page 565" Consider the decomposition of gaseous dinitrogen pentoxide." 2 N 2 O 5 (g) 4 NO 2 (g) + O 2 (g)" (b) Calculate the value of the rate constant, k. Be sure to include units."

30 First-Order Rate Law: Example 19.3, page 897" One of the most frequent examples of a first-order process on the AP exam is radioactive (nuclear) decay.! Technetium-99m (Tc, atomic number 43) is used to form pictures of internal organs in the body and is often used to assess heart damage. The m for this nuclide indicates an excited nuclear state that decays to the ground state by gamma emission. The rate constant for decay of 99m Tc is known to be h 1. What is the half-life of this nuclide?"

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32 First-Order Rate Law: Example 19.4, page 898 (reformatted )" One of the most frequent examples of a first-order process on the AP exam is radioactive (nuclear) decay.! The half-life of molybdenum-99 (Mo, atomic number 42) is 66.0 h. How much of a 1.00 mg sample of molybdenum-99 is left after 330. h?" (A) mg" (B) mg" (C) mg" (D) mg"

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34 Second-Order Rate Law: Example 12.5, page 569" Butadiene (C 4 H 6 ) reacts to form its dimer according to the equation" 2 C 4 H 6 (g) C 8 H 12 (g)"

35 Second-Order Rate Law: Example 12.5, page 569" Butadiene (C 4 H 6 ) reacts to form its dimer according to the equation" 2 C 4 H 6 (g) C 8 H 12 (g)"

36 Second-Order Rate Law: Example 12.5, page 569" Butadiene (C 4 H 6 ) reacts to form its dimer according to the equation" 2 C 4 H 6 (g) C 8 H 12 (g)" (a) How can one verify that the rate law is second order? Explain."

37 Second-Order Rate Law: Example 12.5, page 569" Butadiene (C 4 H 6 ) reacts to form its dimer according to the equation" 2 C 4 H 6 (g) C 8 H 12 (g)" (b) Determine the rate constant, k, for the reaction. Be sure to include units."

38 Second-Order Rate Law: Example 12.5, page 569" Butadiene (C 4 H 6 ) reacts to form its dimer according to the equation" 2 C 4 H 6 (g) C 8 H 12 (g)" (c) What is the half-life for the reaction under the initial conditions" of this experiment?"

39 The Integrated Rate Law: Rate law using concentration and time.! It is useful to have rate laws that relate the rate constant k to the way that concentrations change over time."

40 Collision Theory: A model for chemical kinetics." 1 Central idea: Molecules must collide to react. Two requirements must be satisfied for reactants to collide successfully (to rearrange to form products):" i. The collision must involve enough energy to produce the reaction; that is, the collision energy must equal or exceed the activation energy, E a.! ii. The relative orientation of the reactants must allow formation of any new bonds necessary to produce products."

41 Textbook Example, pages :! Consider the decomposition of BrNO in the gas phase." 2 BrNO(g) 2 NO(g) + Br 2 (g)"

42 Textbook Example, pages :! Consider the decomposition of BrNO in the gas phase." 2 BrNO(g) 2 NO(g) + Br 2 (g)"

43 Arrhenius Equation:! Page 580" k = Ae E a RT ln k = E a RT + ln A

44 Collision Theory: A model for chemical kinetics" 2 Reaction rate increases with increasing [reactants]." "If there are more reactant molecules moving around in a given volume, then more collisions will occur."

45 Collision Theory: A model for chemical kinetics" 3 Reaction rate increases with increasing temperature." "Higher T means the molecules are moving faster greater average kinetic energy. " "Higher T more reactant molecules colliding with each other with enough energy (E a ) to cause a reaction."

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47 Catalyst: Substance that speeds up a reaction without being consumed itself." Does not appear in balanced equation." Lowers E a."

48 Reaction Mechanisms" Most chemical reactions occur by a series of steps called the reaction mechanism."

49 Consider the reaction between nitrogen dioxide and carbon monoxide:" NO 2 (g) + CO(g) NO(g) + CO 2 (g)" The rate law for this reaction is known from experiment to be" Rate = k [NO 2 ] 2 " The mechanism is thought to involve the following steps:" NO 2 (g) + NO 2 (g) NO 3 (g) + CO(g) k " 1 NO 3 (g) + NO(g) k " 2 NO 2 (g) + CO 2 (g)

50 NO 2 (g) + NO 2 (g) NO 3 (g) + CO(g) k " 1 NO 3 (g) + NO(g) k " 2 NO 2 (g) + CO 2 (g) NO 3 (g) is an intermediate, a species that is neither a reactant nor a product but that is formed and consumed during the reaction sequence."

51 A reaction mechanism is a series of elementary steps that must satisfy two requirements:" 1 The sum of the elementary steps must give the overall balanced equation for the reaction." 2 The mechanism must agree with the experimentally determined rate law." NO 2 (g) + NO 2 (g) NO 3 (g) + CO(g) k " 1 NO 3 (g) + NO(g) k " 2 NO 2 (g) + CO 2 (g) slow! fast! The slow step in a reaction mechanism is called the rate determining step."

52 Chemical Equilibrium: An Introduction" Many chemical reactions stop far short of completion and some only occur to a slight extent." Chemical Equilibrium:! Concentrations of all reactants & products remain constant with time." Not static, but highly dynamic." Rate of forward reaction = Rate of reverse reaction"

53 Chemical Equilibrium: An Introduction" a A + b B c C + d D " Equilibrium expression:" K = [C]c [D] d [A] a [B] b K or K eq equilibrium constant" [ ] molar concentration" *Note: K is customarily unitless."

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