Chapter 12 Chemical Kinetics

12.1 Reaction Rates 12.2 Rate Laws: An Introduction 12.3 Determining the Form of the Rate Law 12.4 The Integrated Rate Law 12.5 Reaction Mechanisms 12.6 A Model for Chemical Kinetics 12.7 Catalysis

Energy required for Wheelchair Bicyclists Energy required for Runner Energy required for Wheelchair Bicyclists Your body coverts the carbohydrates you eat into glucose. Glucose is then immediately used by your body for energy or is stored in the muscles as glycogen

The combustion of fuel in a race car The breaching of whale The combustion of fuel in a race car

Change in concentration of a reactant or product per unit time. Reaction Rate Change in concentration of a reactant or product per unit time. [A] means concentration of A in mol/L; A is the reactant or product being considered.

There are several factors that determine the rate of a specific reaction and those are expressed in the "collision theory" that states that for molecules to react, they must: collide have the right energy have the right geometry

To increase the rate, you must make the above more likely to occur To increase the rate, you must make the above more likely to occur. This is possible by changing other factors such as: increasing the surface area (of solids)-this allows for more collisions and gives more molecules the right geometry. increasing the temperature-this gives more molecules the right energy (also called the activation energy, Ea). increasing the concentration (of gases and solutions)-this allows for more collisions and more correct geometry using a catalyst-helps molecules achieve the correct geometry by providing a different way to react.

Increase temperature

using a catalyst

影響反應速率的因素 碰撞 反應物濃度 溫度 反應物的本質 催化劑

碰撞 一個反應能否進行?首先反應物的粒子間必須能相互碰撞或彼此相當接近，例如鈉和水能起激烈反應，若將鈉貯存於煤油中和水隔絕，則無法和水產生氫氣。 但是反應物的分子相互碰撞，是否能保證反應一定會發生?例如氮氣和氧氣為大氣中的主要成分，但是正常情況下並不能生成氮的氧化物。 反應發生時，反應物的粒子之間必須相互碰撞，且碰撞時的能量要超過反應的活化能(Ea) 方為有效的碰撞，否則粒子僅做彈性碰撞而無法反應，此一能量常因不同的反應而不同。此外反應是否能進行，碰撞時的角度方位也佔很重要的因素。

反應物濃度 按照碰撞學說，若反應物分子碰撞的頻率愈高，反應速率就可能愈快。實驗顯示，在均相反應(homogeneous reaction)中，反應速率和反應物濃度某次方成正比，而在非均相反應(heterogeneous reaction)中，則和反應物的濃度關係不大，而和接觸面積關係較大。在日常生活中反應速率隨碰撞頻率加大而增快的例子很多，例如木塊不易點燃，削成木屑則易燃，主要是氧氣分子和木屑因為接觸的面積增大而使碰撞頻率加大；實驗室的藥品大部份先配製成溶液再進行實驗，因為溶液中反應物的粒子能均勻混合，因此碰撞機率大增。增大反應物的濃度，雖能增大碰撞頻率，卻不一定能增快反應，必須由實驗數據加以判斷。

溫度 夏天的食物特別容易腐壞，放在冰箱中便能保存較長的時間。實驗室常用酒精燈加熱反應物，以縮短反應所需的時間，依經驗法則粗略的估計，溫度每提高10°C反應速率便大約增快一倍。為什麼溫度升高時，會使反應速率加快？前曾提到分子的平均動能和絕對溫度成正比。溫度提高時，分子的碰撞頻率也隨之增加，但真正加快反應速率的主要原因是因溫度增加，高能分子的數目大幅增加，使有效的碰撞次數加多，反應速率因而增快。由反應速率定律式 R = k〔A]ｍ[B]ｎ…中可以看出，升高溫度使反應速率加快的最主要因素是因為 k 值變大。在定溫下 k 值雖為常數，但在不同的溫度下，會有不同的數值。

反應物的本質 同一反應的速率會受濃度及溫度的影響，但是不同的反應即使濃度、溫度完全相同，其反應速率依然有快有慢，例如下列二反應： [式7-23] 反應物的本質 同一反應的速率會受濃度及溫度的影響，但是不同的反應即使濃度、溫度完全相同，其反應速率依然有快有慢，例如下列二反應： 酸性溶液中，鐵(II)離子加入過錳酸根溶液，過錳酸根離子的紫色立即消失，而在另式中，草酸根離子雖然和過錳酸根溶液充分混合，紫色仍能維持相當久而不褪色。因為鐵(II)離子變成鐵(III)離子時不涉及鍵的破壞，而C2O42-變成CO2時必須先打斷一些鍵，才能形成新的鍵結(Ea高)，所以反應速率和鍵結的情形有關，而鍵結的情形則取決於物質的本質。若反應不涉及鍵的破壞如離子間的反應，則反應速率較快。若反應物分子間的原子必須重新分解再組合，則反應較難進行。

催化劑 增溫雖然可以加快反應速率，但是加熱卻耗能、造成污染，很多反應也不允許在高溫下進行(例身體內的化學變化、酒的釀造等)。因此有另一種改變反應速率的方法，即在反應物中加入另一物質參與反應，使反應速率改變，此種能改變反應速率但反應前後質量並未減少的物質稱為催化劑(catalyst)，亦稱為觸媒，例如實驗室常用來製備造氧氣的反應： KClO3 (s) → 2KCl (s) + 3O2 (g) 既使強熱的情況下，產生氧氣的速率也非常緩慢，但是氯酸鉀若和黑色的二氧化錳混合加熱時，氧氣則迅速產生，而反應結束後二氧化錳的總量並未改變，MnO2在整個反應過程中即做為催化KClO3分解的角色。

另外將葡萄糖放入37℃的水中，經過數天仍保持原狀無法氧化，但是一旦進入相同 37℃的人體，則迅速氧化生成二氧化碳及水： C6H12O6 (aq) + 12O2 (g) → 12CO2 (g) + 11H2O(l) 因為葡萄糖在人體內，受到數種催化劑的影響而加快氧化，此種在生化反應的催化劑稱為酵素或酶，多由蛋白質構成。不同的生化反應必須由特定的媒加以催化，例如在唾液中有消化澱粉的澱粉酶，在胃液中有消化蛋白質的胃蛋白酶。酶在生物體中的含量雖少，卻不能少於一定含量，否則會造成各項生理機能失調。

酶的另一個特質，必須在適宜的溫度下進行，高溫時會減低效率或完全失去功用。 催化劑無法使反應物粒子的碰撞頻率增加，也不能使粒子的平均動能增加，為何能使反應加快？催化劑可與反應物形成位能較低的活化複體，提供不同的反應途徑以降低化學反應的活化能，因此超過發生反應所需最低能量的分子數目大幅增加，致使有效碰撞頻率增多，反應速率加快。另外催化劑雖能縮短反應進行的時間，卻無法改變生成物的產率。 Copyright © Cengage Learning. All rights reserved

The coefficients in the balanced reaction relate the rate of disappearance of reactants to the rate of production of products. From the balanced reaction, the rate of production of P4 will be 1/4 the rate of disappearance of PH3, and the rate of production of H2 will be 6/4 the rate of disappearance of PH3. By convention, all rates are given as positive values.

The Decomposition of Nitrogen Dioxide Copyright © Cengage Learning. All rights reserved

The Decomposition of Nitrogen Dioxide Copyright © Cengage Learning. All rights reserved

Figure 12.2: (a) (b) (c) Representation of the Reaction 2NO2(g) → 2NO(g) + O2(g)

Average Rate of Decompositionof Nitrogen Dioxide as a Function of Time

Value of the rate at a particular time. Instantaneous Rate Value of the rate at a particular time. Can be obtained by computing the slope of a line tangent to the curve at that point. Copyright © Cengage Learning. All rights reserved

Rate of consumption of NO2 Rate of production of NO Instantaneous Rate Rate of consumption of NO2 Rate of production of NO 2Rate of production of O2 ＝ ＝ Copyright © Cengage Learning. All rights reserved

Shows how the rate depends on the concentrations of reactants. Rate Law Shows how the rate depends on the concentrations of reactants. For the decomposition of nitrogen dioxide: 2NO2(g) → 2NO(g) + O2(g) Rate = k[NO2]n k = rate constant n = order of the reactant Copyright © Cengage Learning. All rights reserved

Rate Law Rate = k[NO2]n The concentrations of the products do not appear in the rate law because the reaction rate is being studied under conditions where the reverse reaction does not contribute to the overall rate. Copyright © Cengage Learning. All rights reserved

Rate Law Rate = k[NO2]n The value of the exponent n must be determined by experiment; it cannot be written from the balanced equation. Copyright © Cengage Learning. All rights reserved

Types of Rate Laws Differential Rate Law (rate law) – shows how the rate of a reaction depends on concentrations. Integrated Rate Law – shows how the concentrations of species in the reaction depend on time. Copyright © Cengage Learning. All rights reserved

Rate Laws: A Summary Because we typically consider reactions only under conditions where the reverse reaction is unimportant, our rate laws will involve only concentrations of reactants. Because the differential and integrated rate laws for a given reaction are related in a well–defined way, the experimental determination of either of the rate laws is sufficient. Copyright © Cengage Learning. All rights reserved

Rate Laws: A Summary Experimental convenience usually dictates which type of rate law is determined experimentally. Knowing the rate law for a reaction is important mainly because we can usually infer the individual steps involved in the reaction from the specific form of the rate law. Copyright © Cengage Learning. All rights reserved

Determine experimentally the power (exponent n) to which each reactant concentration must be raised in the rate law. 反應速率的定義雖如上式所列，可由任一反應物或生 成物的濃度來計算，但實際在測量時應選擇何者較為 方便?一般會以巨觀上有明顯變化的物理量來做觀察 較為理想 。例如前式的反應物NO2為紅棕色氣體，可 以由顏色的深淺求出濃度的大小變化(比色法) Copyright © Cengage Learning. All rights reserved

2N2O5 (soln)→ 4NO2 (soln) + O2(g) (at 45 ºC)

2N2O2 (soln)→ 4NO2 (soln) + O2(g) (at 45 ºC) [N2O5] Rate(mole/L·s) 0.9M 5.4 ×10-4 0.45M 2.7 ×10-4

Method of Initial Rates The value of the initial rate is determined for each experiment just after the reaction begins( t = 0 ). Several experiments are carried out using different initial concentrations of each of the reactants, and the initial rate is determined for each run. The results are then compared to see how the initial rate depends on the initial concentrations of each of the reactants. Copyright © Cengage Learning. All rights reserved

Overall Reaction Order The sum of the exponents in the reaction rate equation. Rate = k[A]n[B]m Overall reaction order = n + m k = rate constant [A] = concentration of reactant A [B] = concentration of reactant B Copyright © Cengage Learning. All rights reserved

Concept Check How do exponents (orders) in rate laws compare to coefficients in balanced equations? Why? The exponents do not have any relation to the coefficients (necessarily). The coefficients tell us the mole ratio of the overall reaction. They give us no clue to how the reaction works (its mechanism). The exponents do not have any relation to the coefficients (necessarily). The coefficients tell us the mole ratio of the overall reaction. They give us no clue to how the reaction works (its mechanism). Copyright © Cengage Learning. All rights reserved

Rate = k[A] Integrated: ln[A] = –kt + ln[A]o First-Order [A] = concentration of A at time t k = rate constant t = time [A]o = initial concentration of A Copyright © Cengage Learning. All rights reserved

n = 1 first order n = 2 second order n = 0 zero order Integrated Rate Law – shows how the concentrations of species in the reaction depend on time n = 1 first order n = 2 second order n = 0 zero order Copyright © Cengage Learning. All rights reserved

上式是一級反應速率方程式的直線型式，實驗數據滿足上式的反應皆為一級反應。由於一級反應是自然界最常見的狀況 First-Order Integrated: ln[A] = –kt + ln[A]o 上式是一級反應速率方程式的直線型式，實驗數據滿足上式的反應皆為一級反應。由於一級反應是自然界最常見的狀況 y = -kt + b → t versus ln[A] is a straight line [A] = concentration of A at time t k = rate constant t = time [A]o = initial concentration of A Copyright © Cengage Learning. All rights reserved

一級反應速率方程式也有其他的型式，例如： 或者… 等，但屬罕見。 Copyright © Cengage Learning. All rights reserved

Plot of ln[N2O5] vs Time Copyright © Cengage Learning. All rights reserved

Half–life does not depend on the concentration of reactants. First-Order Half–Life: k = rate constant Half–life does not depend on the concentration of reactants. Copyright © Cengage Learning. All rights reserved

Half-Life of Reactions Copyright © Cengage Learning. All rights reserved

遵循一級動力學的化學反應: H2O2 有強的氧化能力，市面上商品都是水溶液的形式，在常溫下不穩定，易分解為水和氧氣(添加二氧化錳催化劑則更容易），需冷藏。具殺菌力，稀薄的 H2O2水溶液可作傷口清潔消毒（雙氧水）。 口服會對人體造成很大的傷害，但其能將有機物中的暗色成份氧化，有漂白的功能，有不肖商人用 H2O2 漂白不新鮮的肉品、筍乾、回收的竹筷等。H2O2 的分解反應如下： 平衡顯示 2 莫耳的 H2O2 參與反應，但這不是一個基本反應，而是具有複雜反應機制的連續反應，實際上 H2O2 的分解是一級動力學反應，其速率方程式為： Copyright © Cengage Learning. All rights reserved

有許多二級、三級甚至更高級次的化學反應，卻表現出一級反應動力學模式。例如 COS 在水中的分解反應： 一般環境工程領域，常用 H2O2 來作水質的高級處理，用其強氧化力以去除水中微量的殘留物。其分解的產物只有水和氧氣，不會造成二次污染，但如分解不完全，飲用水中殘留量太高，就可能造成傷害，所以必須精確的計算其反應速率，控制安全的劑量。 有許多二級、三級甚至更高級次的化學反應，卻表現出一級反應動力學模式。例如 COS 在水中的分解反應： 事實上這是一個二級反應，其速率方程式為： Copyright © Cengage Learning. All rights reserved

但實際測量的結果卻顯示，反應速率與水無關，只與 COS 的 濃度成正比。實驗所得的速率方程式為一級反應(如下所示） 這種現象稱作假分子 pseudo-molecular效應，水分子好像不 參與反應。這是因為反應在水溶液中進行，系統中水的 量與 COS 比起來多得太多了，即使 COS 完全消耗掉了，水 的量仍然測不出有甚麼變化。因變化不大，所以併入常數項 ，使得整個速率方程式呈現一級反應的型式。 （k’=k·[H2O]） Copyright © Cengage Learning. All rights reserved

自然環境中有許多複雜的化學反應，因為假分子現象而表 現出一級反應的型式，稱作假一級反應pseudo-first order reaction。水中 BOD 的分解，有機氯農藥以及大多數微量 有機毒物在土壤中的自然分解都是假一級反應。 一般常將複雜的高級次反應簡化成假一級反應。通常的作法 是只觀測一種目標反應物的濃度，而令其他反應物的濃度一 直維持在充分過量的情況，使其他反應物在整個反應過程中 的濃度幾乎固定，同時盡量只測量極短時間內的目標反應物 濃度變化，如此可使逆反應的比例很小。（目的是要解決實 際的問題，簡化複雜的現象，雖然不符合真實的情況，但只 要經驗上行得通，亦是無妨。） Copyright © Cengage Learning. All rights reserved

放射性同位素的衰變反應都是一級反應，而且速率常數與溫 度無關，這點與化學反應不同。放射性同位素都具有恆定的 半衰期，可以用來作礦物的年代鑑定dating，而礦物年代鑑 定則可用來作污染源的判斷。 Copyright © Cengage Learning. All rights reserved

Exercise A first order reaction is 35% complete at the end of 55 minutes. What is the value of k? ln(0.65) = –k(55) + ln(1) k = 7.8 x 10-3 min-1. If students use [A] = 35 in the integrated rate law (instead of 65), they will get k = 1.9 x 10-2 min-1. Note: Use the red box animation to assist in explaining how to solve the problem. Copyright © Cengage Learning. All rights reserved

常見的二級反應動力學模式有兩種型式，第一種型式的二級反應為兩個相同的分子進行反應： 2A → Products [A] = concentration of A at time t k = rate constant t = time [A]o = initial concentration of A Copyright © Cengage Learning. All rights reserved

煙道中二氧化氮 NO2 的熱解反應，以及在大氣上層臭氧O3的熱解反應，都是屬於第一種型式的二級反應： Copyright © Cengage Learning. All rights reserved

第二種型式的二級反應為兩個不同的分子進行反應： A + B → Products 其速率方程式為： 在大氣中，一氧化氮 NO 與臭氧O3 的反應，以及煙道中二氧化氮 NO2 與一氧化碳 CO 的反應，都是第二種類型的二級反應： Copyright © Cengage Learning. All rights reserved

Plot of ln[C4H6] vs Time and Plot of 1/[C4H6] vs Time Slope k 截距1/〔C4H6〕0 Copyright © Cengage Learning. All rights reserved

Each successive half–life is double the preceding one. Second-Order Half–Life: k = rate constant [A]o = initial concentration of A Half–life gets longer as the reaction progresses and the concentration of reactants decrease. Each successive half–life is double the preceding one. Copyright © Cengage Learning. All rights reserved

Write the rate law for this reaction. rate = k[A]2 b) Calculate k. Exercise For a reaction aA  Products, [A]0 = 5.0 M, and the first two half-lives are 25 and 50 minutes, respectively. Write the rate law for this reaction. rate = k[A]2 b) Calculate k. k = 8.0 x 10-3 M–1min–1 Calculate [A] at t = 525 minutes. [A] = 0.23 M a) rate = k[A]2 We know this is second order because the second half–life is double the preceding one. b) k = 8.0 x 10-3 M–1min–1 25 min = 1 / k(5.0 M) c) [A] = 0.23 M (1 / [A]) = (8.0 x 10-3 M–1min–1)(525 min) + (1 / 5.0 M) Copyright © Cengage Learning. All rights reserved

三級反應third order kinetics： 一般很少面臨三級反應的問題，即使有也是想辦法將它控制成假一級反應來處理。不過為了萬全之計，把常見的三級反應的速率方程式加以列出。 常見的三級反應動力學模式有三種型式，第一種型式的三級反應為三個相同的分子進行反應： integrated Copyright © Cengage Learning. All rights reserved

第二種型式的三級反應為兩種不同的分子以1：2的比例進行反應： 第三種型式的三級反應為三種不同的分子以1：1：1的比例進行反應： 煙道或空氣中的一氧化氮被氧氣氧化的反應，是三級反應的例子之一： Copyright © Cengage Learning. All rights reserved

Rate = k[A]0 = k Integrated: [A] = –kt + [A]o Zero-Order [A] = concentration of A at time t k = rate constant t = time [A]o = initial concentration of A Copyright © Cengage Learning. All rights reserved

Plot of [A] vs Time Copyright © Cengage Learning. All rights reserved

Figure 12.7: (a) (b) Decomposition Reaction 2N2O(g) → 2N2(g) + O2(g) Zero order reaction are often encountered when a substance such as a metal surface or an enzyme is required for the reaction to occur. Such as 2N2O(g) → 2N2(g) + O2(g) On a hot Pt surface k is a constant no matter how high the 〔N2O〕is.

Zero-Order Half–Life: k = rate constant [A]o = initial concentration of A Half–life gets shorter as the reaction progresses and the concentration of reactants decrease. Copyright © Cengage Learning. All rights reserved

不同反應級次的速率常數的單位是不相同的，這點可經由單位分析dimensional analysis的方法證明。單位分析是根據，等式的兩邊除了函數及數值必須相等之外單位也必須相等的原理來運作的。 Copyright © Cengage Learning. All rights reserved

Copyright © Cengage Learning. All rights reserved

Rate Laws Copyright © Cengage Learning. All rights reserved

Summary of the Rate Laws Copyright © Cengage Learning. All rights reserved

Zero order First order Second order 4.7 M 3.7 M 2.0 M Exercise Consider the reaction aA  Products. [A]0 = 5.0 M and k = 1.0 x 10–2 (assume the units are appropriate for each case). Calculate [A] after 30.0 seconds have passed, assuming the reaction is: Zero order First order Second order a) 4.7 M [A] = –(1.0×10–2)(30.0) + 5.0 b) 3.7 M ln[A] = –(1.0×10–2)(30.0) + ln(5.0) c) 2.0 M (1 / [A]) = (1.0×10–2)(30.0) + (1 / 5.0) 4.7 M 3.7 M 2.0 M [A] = –(1.0×10–2)(30.0) + 5.0 ln[A] = –(1.0×10–2)(30.0) + ln(5.0) (1 / [A]) = (1.0×10–2)(30.0) + (1 / 5.0) Copyright © Cengage Learning. All rights reserved

Most chemical reactions occur by a series of elementary steps. Reaction Mechanism Most chemical reactions occur by a series of elementary steps. An intermediate is formed in one step and used up in a subsequent step and thus is never seen as a product in the overall balanced reaction. Copyright © Cengage Learning. All rights reserved

NO2(g) + NO2(g) → NO3(g) + NO(g) NO3(g) + CO(g) → NO2(g) + CO2(g) A Molecular Representation of the Elementary Steps in the Reaction of NO2 and CO NO2(g) + CO(g) → NO(g) + CO2(g) NO2(g) + NO2(g) → NO3(g) + NO(g) NO3(g) + CO(g) → NO2(g) + CO2(g) Copyright © Cengage Learning. All rights reserved

Elementary Steps (Molecularity) Unimolecular – reaction involving one molecule; first order. Bimolecular – reaction involving the collision of two species; second order. Termolecular – reaction involving the collision of three species; third order. Copyright © Cengage Learning. All rights reserved

Rate-Determining Step A reaction is only as fast as its slowest step. The rate-determining step (slowest step) determines the rate law and the molecularity of the overall reaction. Copyright © Cengage Learning. All rights reserved

Reaction Mechanism Requirements The sum of the elementary steps must give the overall balanced equation for the reaction. The mechanism must agree with the experimentally determined rate law. Copyright © Cengage Learning. All rights reserved

Decomposition of N2O5 Copyright © Cengage Learning. All rights reserved

Step 2: NO2 + NO3 → NO + O2 + NO2 (slow) Decomposition of N2O5 2N2O5(g)  4NO2(g) + O2(g) Step 1: N2O5 NO2 + NO3 (fast) Step 2: NO2 + NO3 → NO + O2 + NO2 (slow) Step 3: NO3 + NO → 2NO2 (fast) 2( ) Copyright © Cengage Learning. All rights reserved

Molecules must collide to react. Main Factors: Activation energy, Ea Collision Model Molecules must collide to react. Main Factors: Activation energy, Ea Temperature Molecular orientations Copyright © Cengage Learning. All rights reserved

Energy that must be overcome to produce a chemical reaction. Activation Energy, Ea Energy that must be overcome to produce a chemical reaction. Copyright © Cengage Learning. All rights reserved

Transition States and Activation Energy Copyright © Cengage Learning. All rights reserved

2BrNO → 2NO + Br2 2Br-N bonds be broken(243kj/mole) , the energy comes from the reacting molecules before the collision. 1Br-Br bond be formed Copyright © Cengage Learning. All rights reserved

Figure 12.11: Plot of Number of Collisions with a Particular Energy at T1 and T2

Figure 12.12: (a) (b) (c) Possible Orientations for Collisions Between Two BrNO Molecules

For Reactants to Form Products Collision must involve enough energy to produce the reaction (must equal or exceed the activation energy). Relative orientation of the reactants must allow formation of any new bonds necessary to produce products. Copyright © Cengage Learning. All rights reserved

The Gas Phase Reaction of NO and Cl2 Copyright © Cengage Learning. All rights reserved

R = gas constant (8.3145 J/K·mol) T = temperature (in K) Arrhenius Equation A = frequency factor Ea = activation energy R = gas constant (8.3145 J/K·mol) T = temperature (in K) Copyright © Cengage Learning. All rights reserved

Linear Form of Arrhenius Equation y = mx + b y = ln(k) m = -Ea/R = slope x = 1/T b = ln(A) =intercept Copyright © Cengage Learning. All rights reserved

Linear Form of Arrhenius Equation Copyright © Cengage Learning. All rights reserved

Exercise Chemists commonly use a rule of thumb that an increase of 10 K in temperature doubles the rate of a reaction. What must the activation energy be for this statement to be true for a temperature increase from 25°C to 35°C? Ea = 53 kJ Ea = 53 kJ ln(2) = (Ea / 8.3145 J/K·mol)[(1/298 K) – (1/308 K)] ln(2) = (Ea / 8.3145 J/K·mol)[(1/298 K) – (1/308 K)] Copyright © Cengage Learning. All rights reserved

A substance that speeds up a reaction without being consumed itself. Catalyst A substance that speeds up a reaction without being consumed itself. Provides a new pathway for the reaction with a lower activation energy. Copyright © Cengage Learning. All rights reserved

Energy Plots for a Catalyzed and an Uncatalyzed Pathway for a Given Reaction Copyright © Cengage Learning. All rights reserved

Effect of a Catalyst on the Number of Reaction-Producing Collisions Copyright © Cengage Learning. All rights reserved

Catalysts are classified as homogeneous or heterogeneous A homogeneous catalyst is one that is present in the same phase as the reacting molecules. A heterogeneous catalyst exist in a different phase, usually as a solid. Copyright © Cengage Learning. All rights reserved

Heterogeneous Catalyst Most often involves gaseous reactants being adsorbed on the surface of a solid catalyst. Adsorption – collection of one substance on the surface of another substance. Copyright © Cengage Learning. All rights reserved

Heterogeneous Catalyst Adsorption and activation of the reactants. Migration of the adsorbed reactants on the surface. Reaction of the adsorbed substances. Escape, or desorption, of the products. Copyright © Cengage Learning. All rights reserved

Heterogeneous Catalysis of the Hydrogenation of Ethylene 1.reactants above the metal surface. 2.H2 is adsorbed onto the surface, forming metal-H bonds. The π bonds in C2H4 is broken. 3.The adsorbed molecules and atoms migrate toward each other, form C-H bonds. 4.C2H6 molecules escspe from surface.

Figure 12.16: Schematic of Exhaust Gases from an Automobile

Exists in the same phase as the reacting molecules. Homogeneous Catalyst Exists in the same phase as the reacting molecules. Enzymes are nature’s catalysts. Copyright © Cengage Learning. All rights reserved

Homogeneous Catalysis Copyright © Cengage Learning. All rights reserved

In the lower atmosphere Copyright © Cengage Learning. All rights reserved

In the upper atmosphere Copyright © Cengage Learning. All rights reserved

Data from 2006 TOMS Earth Probe NASA

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Freon-12 Molecule