1 Chemical Kinetics Chapter 14 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 動力 : 移動與變化

2 Chemical Kinetics Thermodynamics – does a reaction take place? Kinetics – how fast does a reaction proceed? When the reaction process will be terminated? Reaction rate is the change in the concentration of a reactant or a product with time (M/s). Ex: 光合作用, 核反應 : ~10 -6 s 但石墨轉化成鑽石結構卻要上百年 1. 化學工程師花很多時間再改善反應速率而非僅是產率 2. 實際的應用可以在 : 藥物設計、污染防治、食品處理。

3 A B rate = -  [A] tt rate =  [B] tt  [A] = change in concentration of A over time period  t  [B] = change in concentration of B over time period  t Because [A] decreases with time,  [A] is negative. Average rate

4 A B rate = -  [A] tt rate = [B][B] tt 單位時間的濃度變化 M/s

5 Br 2 (aq) + HCOOH (aq) 2Br - (aq) + 2H + (aq) + CO 2 (g) time 393 nm light Detector  [Br 2 ]   Absorption red-brown t 1 < t 2 < t 3

6 Br 2 (aq) + HCOOH (aq) 2Br - (aq) + 2H + (aq) + CO 2 (g) average rate = -  [Br 2 ] tt = - [Br 2 ] final – [Br 2 ] initial t final - t initial slope of tangent slope of tangent slope of tangent instantaneous rate = rate for specific instance in time

7 比例常數 濃度與反應速率成正比

8 2H 2 O 2 (aq) 2H 2 O (l) + O 2 (g) PV = nRT P = RT = [O 2 ]RT n V [O 2 ] = P RT 1 rate =  [O 2 ] tt RT 1 PP tt = measure  P over time Decomposition of Hydrogen Peroxide: 藉由量測壓力來得知反應速率 M

9

10 Reaction Rates and Stoichiometry 2A B Two moles of A disappear for each mole of B that is formed. rate =  [B] tt rate = -  [A] tt 1 2 aA + bB cC + dD rate = -  [A] tt 1 a = -  [B] tt 1 b =  [C] tt 1 c =  [D] tt 1 d

11 Write the rate expression for the following reaction: CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (g) rate = -  [CH 4 ] tt = -  [O 2 ] tt 1 2 =  [H 2 O] tt 1 2 =  [CO 2 ] tt

12 The Rate Law The rate law expresses the relationship of the rate of a reaction to the rate constant and the concentrations of the reactants raised to some powers. aA + bB cC + dD Rate = k [A] x [B] y Reaction is xth order in A Reaction is yth order in B Reaction is (x + y)th order overall Reaction Order: ( 反應級數 )

13 F 2 (g) + 2ClO 2 (g) 2FClO 2 (g) rate = k [F 2 ] x [ClO 2 ] y Double [F 2 ] with [ClO 2 ] constant Rate doubles x = 1 Quadruple [ClO 2 ] with [F 2 ] constant Rate quadruples y = 1 rate = k [F 2 ][ClO 2 ] [ClO 2 ] 的次方為 1 但是其化學劑量係數是 2 。

14 F 2 (g) + 2ClO 2 (g) 2FClO 2 (g) rate = k [F 2 ][ClO 2 ] Rate Laws Rate laws are always determined experimentally. Reaction order is always defined in terms of reactant (not product) concentrations. The order of a reactant is not related to the stoichiometric coefficient of the reactant in the balanced chemical equation. 1 速率定理

15 Determine the rate law and calculate the rate constant for the following reaction from the following data: S 2 O 8 2- (aq) + 3I - (aq) 2SO 4 2- (aq) + I 3 - (aq) Experiment [S 2 O 8 2- ][I - ] Initial Rate (M/s) x x x rate = k [S 2 O 8 2- ] x [I - ] y Double [I - ], rate doubles (experiment 1 & 2) y = 1 Double [S 2 O 8 2- ], rate doubles (experiment 2 & 3) x = 1 k = rate [S 2 O 8 2- ][I - ] = 2.2 x M/s (0.08 M)(0.034 M) = 0.08/M s rate = k [S 2 O 8 2- ][I - ]

16 First-Order Reactions A product rate = -  [A] tt rate = k [A] k = rate [A] = 1/s or s -1 M/sM/s M =  [A] tt = k [A] - [A] is the concentration of A at any time t [A] 0 is the concentration of A at time t=0 [A] = [A] 0 e −kt ln[A] = ln[A] 0 - kt 分別對時間與濃度積分

17 EX:

18 2N 2 O 5 4NO 2 (g) + O 2 (g) Graphical Determination of k

19 For the gas-Phase reactions:

20 EX:

21 First-Order Reactions The half-life, t ½, is the time required for the concentration of a reactant to decrease to half of its initial concentration. t ½ = t when [A] = [A] 0 /2 ln [A] 0 [A] 0 /2 k = t½t½ ln 2 k = k = 對一級反應來說, 半生期與反應物的初始濃度無關。舉例來說 當濃度由 1.0M 降到 0.50M 或從 0.10M 到 0.050M, 所需要的時間 都一樣。此外, 半生期越短， K 值越高, 反應速率越快 : EX: 24 Na(14.7 h) 比 60 Co(t ½ =5.3 year) 的蛻變速率快。

22 What is the half-life of N 2 O 5 if it decomposes with a rate constant of 5.7 x s -1 ? t½t½ ln 2 k = x s -1 = = 1200 s = 20 minutes How do you know decomposition is first order? units of k (s -1 )

23 EX

24 A product First-order reaction # of half-lives [A] = [A] 0 /n

25 Second-Order Reactions A product rate = -  [A] tt rate = k [A] 2 k = rate [A] 2 = 1/M s M/sM/s M2M2 =  [A] tt = k [A] 2 - [A] is the concentration of A at any time t [A] 0 is the concentration of A at time t=0 1 [A] = 1 [A] 0 + kt t ½ = t when [A] = [A] 0 /2 t ½ = 1 k[A] 0 另一種二級反應為 A+B  product 二級反應的半生期與反應物濃度成反比, 反應初期濃度高, 碰撞頻率高, 所以半生期短

26 Zero-Order Reactions A product rate = -  [A] tt rate = k [A] 0 = k k = rate [A] 0 = M/s  [A] tt = k - [A] is the concentration of A at any time t [A] 0 is the concentration of A at time t = 0 t ½ = t when [A] = [A] 0 /2 t ½ = [A] 0 2k2k [A] = [A] 0 - kt 零級反應很少見，許多零級反應發生在金屬表面， 例如 : N 2 O 在 Pt 表面分解成 N 2 跟 O 2 ，當 Pt 的結合 部位都被占滿後，反應速率與 N 2 O 濃度無關。 催化反應

27 Summary of the Kinetics of Zero-Order, First-Order and Second-Order Reactions OrderRate Law Concentration-Time Equation Half-Life rate = k rate = k [A] rate = k [A] 2 ln[A] = ln[A] 0 - kt 1 [A] = 1 [A] 0 + kt [A] = [A] 0 - kt t½t½ ln 2 k = t ½ = [A] 0 2k2k t ½ = 1 k[A] 0

28

29 Activation Energy and Temperature Dependence of Rate Constants

30 Temperature Dependence of the Rate Constant E a is the activation energy (J/mol) R is the gas constant (8.314 J/Kmol) T is the absolute temperature A is the frequency factor ln k = - EaEa R 1 T + lnA (Arrhenius equation) Alternate format:

31 The Collision Theory of Chemical Kinetics 分子動能大, 與另一個分子碰撞後, 部分動能轉成為振動能 ; 因此若初 始的動能夠大，碰撞造成的振動 足以破壞化學鍵結，即可以反應為 生成物。

32 Exothermic ReactionEndothermic Reaction The activation energy (E a ) is the minimum amount of energy required to initiate a chemical reaction. A + B AB C + D + + 活化能 : 克服不穩定的過度態 (transition state) 放熱 吸熱

33 Arrhenius Equation The relation between K and T and be described by the Equation: k= A e -Ea/RT 碰撞頻率 速率常數與碰撞頻率成正比，活化能減少或是溫度增加，都將使 速率常數增加

34 Alternate Form of the Arrhenius Equation At two temperatures, T 1 and T 2 or 在已知兩溫度的速率常數, 就可以求得活化能。

35

36 The orientation of the molecules Arrhenius Equation 僅反映出碰撞頻率的因子，但是對於複雜的反應，化學分子間在碰撞時的位向也會影響反應 因此需要進行修正 ~

37 The modified Arrhenius Equation equation:

38 Reaction Mechanisms The overall progress of a chemical reaction can be represented at the molecular level by a series of simple elementary steps or elementary reactions. The sequence of elementary steps that leads to product formation is the reaction mechanism. 2NO (g) + O 2 (g) 2NO 2 (g) N 2 O 2 is detected during the reaction! Elementary step:NO + NO N 2 O 2 Elementary step:N 2 O 2 + O 2 2NO 2 Overall reaction:2NO + O 2 2NO 2 + 反應機構 基本反應 intermediate 雙分子, 單分子，三分子反應 ( 一般要三個分子同時碰撞機率低 )

39 2NO (g) + O 2 (g) 2NO 2 (g) Mechanism:

40 Elementary step:NO + NO N 2 O 2 Elementary step:N 2 O 2 + O 2 2NO 2 Overall reaction:2NO + O 2 2NO 2 + Intermediates are species that appear in a reaction mechanism but not in the overall balanced equation. An intermediate is always formed in an early elementary step and consumed in a later elementary step. The molecularity of a reaction is the number of molecules reacting in an elementary step. Unimolecular reaction – elementary step with 1 molecule Bimolecular reaction – elementary step with 2 molecules Termolecular reaction – elementary step with 3 molecules

41 Unimolecular reactionA productsrate = k [A] Bimolecular reactionA + B productsrate = k [A][B] Bimolecular reactionA + A productsrate = k [A] 2 Rate Laws and Elementary Steps Writing plausible reaction mechanisms: The sum of the elementary steps must give the overall balanced equation for the reaction. The rate-determining step should predict the same rate law that is determined experimentally. The rate-determining step is the slowest step in the sequence of steps leading to product formation. 基本反應的反應級數等於其化學劑量比

42 Sequence of Steps in Studying a Reaction Mechanism

43 The experimental rate law for the reaction between NO 2 and CO to produce NO and CO 2 is rate = k[NO 2 ] 2. The reaction is believed to occur via two steps: Step 1:NO 2 + NO 2 NO + NO 3 Step 2:NO 3 + CO NO 2 + CO 2 What is the equation for the overall reaction? NO 2 + CO NO + CO 2 What is the intermediate? NO 3 What can you say about the relative rates of steps 1 and 2? rate = k[NO 2 ] 2 is the rate law for step 1 so step 1 must be slower than step 2

44 EX: 過氧化氫的分解反應 與碘化氫的生成反應

45

46 Experimental Support for Reaction Mechanisms 利用含有氧 18 同位素的水， 來取代純水，分析水解與 光合作用的反應 乙酸甲酯 乙酸

47 善用分析工具來檢測化學反應機制

48 Carbon atoms migration on metal surface: 1.Carbons overcoming the energy barrier and moving to the metal surface 2. surface-diffusing carbon atoms would be trapped by defects or step edges on metal surface Nano Lett. 2011, 11, 297-–303

Catalysis reaction 提升溫度可以加速反應進行, 但生成物也有可能進行其他反應, 因而導致產率的下降. 催化劑是一個有效的解決辦法。

50 Catalyst: 反應過程中沒有增減 氯化鉀的熱解反應 改變反應的路徑, 降低反應的活化能

51 A catalyst is a substance that increases the rate of a chemical reaction without itself being consumed. EaEa k rate catalyzed > rate uncatalyzed E a < E a ′ UncatalyzedCatalyzed 反應位能不變, 僅改變活化能 但注意 : 逆反應的速率也會增加

52 In heterogeneous catalysis, the reactants and the catalysts are in different phases. In homogeneous catalysis, the reactants and the catalysts are dispersed in a single phase, usually liquid. Haber synthesis of ammonia Ostwald process for the production of nitric acid Catalytic converters Acid catalysis Base catalysis 均相與非均相催化反應

53 N 2 (g) + 3H 2 (g) 2NH 3 (g) Fe/Al 2 O 3 /K 2 O catalyst Haber Process ( 哈伯法製氨 ) 1.NH3 是重要的無機物質，工業需求大。 2. 從 N 2 和 H 2 要直接合成 NH 3 ，反應極慢，而升溫加速反應 又會使 NH 3 分解。 年發現金屬催化製氨的反應。 共價鍵變弱導致分解，反應後結合成 NH3 後脫附金屬表面

54 催化劑對於材料合成的重要性 : 從奈米材料的合成觀點來說 Carbon nanotube Si Nanowire

55 Ostwald Process( 硝酸的製備 ) Pt-Rh catalysts used in Ostwald process 4NH 3 (g) + 5O 2 (g) 4NO (g) + 6H 2 O (g) Pt catalyst 2NO (g) + O 2 (g) 2NO 2 (g) 2NO 2 (g) + H 2 O (l) HNO 2 (aq) + HNO 3 (aq) 亞硝酸加熱後 又可以形成 NO

56 Catalytic Converters CO + Unburned Hydrocarbons + O 2 CO 2 + H 2 O catalytic converter 2NO + 2NO 2 2N 2 + 3O 2 catalytic converter

57 Homogenous catalysis ( 均相催化 )

58 Enzyme Catalysis 酵素可以使生化的反應提升到 倍 具有高度選擇性, 每個酵素只對特定分子反應 活性部位 : lock-and-key model

59 Binding of Glucose to Hexokinase 己糖激酶

60 rate =  [P] tt rate = k [ES] Enzyme Kinetics 所有活性部位都被占滿後

How does a catalyst increase the rate of a reaction? For the reactionthe frequency factor A is 8.7 × s −1 and the activation energy is 63 kJ/mol. What is the rate constant for the reaction at 75°C? Sketch a potential-energy-versus-reaction-progress plot for the following reactions: S(s) + O 2 (g) → SO 2 (g) ΔH° = − kJ/mol Cl 2 (g) → Cl(g) + Cl(g) ΔH° = kJ/mol