L.S.T. Leung Chik Wai Memorial School F.6 Chemistry Chapter 15 Chpt. 15 : p.1 THE EFFECT OF TEMPERATURE CHANGE ON REACTION RATE REVIEW As the temperature of a reaction mixture increases, the reaction rate increases, it is generally found that a temperature rise of about 100C approximately doubles the rate of reaction. This can be explained in terms of 1. activation energy and 2. the collision theory. II. (A) EFFECT OF TEMPERATURE CHANGE ON REACTION RATE IN TERMS OF ACTIVATION ENERGY Activation Energy <1> During a reaction, bonds are first broken and others are then formed. Energy is required to break certain bonds and start the process, whether the overall reaction is exothermic or endothermic. <2> Particles will not always react when they collide because they may not have sufficient energy for appropriate bonds to break. If a reaction is to occur, the colliding reactant particles must process more than a certain minimum amount of energy to overcome an energy barrier. This “minimum” amount of energy is known as activation energy, Ea. Conversely, if the reactant particles do not possess the activation energy required for a reaction, this reaction will not occur. If reactant particles possess energy equal to or greater than the activation energy, they can cross the ‘energy barrier’. Reaction occur. (B) Effect of Temperature change on Reaction Rate <1> When temperature increases, the reactant particles will possess more energy. <2> More reactant particles possess energy equal to or greater than the activation energy. as a result, reaction rate increases. L.S.T. Leung Chik Wai Memorial School F.6 Chemistry Chapter 15 III. (A) Chpt. 15 : p.2 APPLICATION OF THE ARPHENIUS EQUATION TO DETERMINE THE ACTIVATION ENERGY OF A REACTION Arrhenius equation The relationship between the rate constant and the temperature for a given reaction is given by the Arrhenius equation, which is expressed as k A exp( Ea ) RT where k = rate constant of the reaction A = Arrhenius factor (of constant or coefficient) exp = exponential constant = 2.718 Ea = activation energy of the reaction T = temperature of the reaction mixture in the absolute scale (in K) i.e. ( 0C +273) R = universal gas constant = 8.314 JK1molt Note Ea ) suggest that RT increasing the temperature increases greatly the proportion of high kinetic energy molecules. <1> The exponential factor, exp( <2> It is obvious that Raising T makes Ea / RT smaller, k will increase and the reaction is faster. Lowering T makes Ea / RT larger, k will decrease and the reaction is slower. <3> Similarly, If Ea is low, Ea / RT will be small. K will be large and the reaction will be fast. If Ea is high, Ea / RT will be large. K will be small and the reaction will be slow. L.S.T. Leung Chik Wai Memorial School F.6 Chemistry Chapter 15 (B) Chpt. 15 : p.3 Natural Log of The Arrhenius Equation Taking natural log of the Arrhenius equation gives ln k ln A Ea RT log k log A or Ea 2.303RT Measuring the rate constant at two different temperature provides enough information to evaluate the activation energy. The Activation energy and the rates or rate constants (k1 and k2) at two temperatures (T1 and T2) can be related by the equation: ln( k1 Ea 1 1 ) x( ) k2 R T1 T2 Exercise 1 If the rate constant of a reaction at 310K is double that a 298K, calculate the activation energy of the reaction. (Gas constant = 8.314 JK-1mol-1) ANSWER Exercise 2 In the gas reaction 2NO(g) + 2H2(g) N2(g) + 2H2O(g) doubling the initial concentration of NO made the initial rate four times as fast, doubling the initial concentration of H2 made the reaction twice as rapid. (a) (b) (c) deduce the rate law. what are the units of rate constant. Calculate the activation energy of the reaction if the rate constant at 1115K is double the value at 1093K. ( Gas constant ,R = 8.314 JK-1mol-1) ANSWER L.S.T. Leung Chik Wai Memorial School F.6 Chemistry Chapter 15 Exercise 3 At 300°C, the rate constant for the reaction is 2.41x10-10 s-1. At 400°C. k equals to 1. 16x106s-1. Determine the value of (a) the activation energy (in kJ mol-1) and (b) the Arrhenius factor A for this reaction. (Gas constant, R = 8.314 J K-1 mol-1 ) ANSWER Exercise 4 For the hypothetical reaction A B + C, the rate constant is 3.91x104 mol1dm3s-1 at 370°C and 4.05x102 mol-1dm3s-1 at 470°C. calculate (a) the activation energy, and (b) the rate constant of the reaction at 450°C. ( Gas constant = 8.314 J K-1 mol-1 ) I ANSWER Chpt. 15 : p.4 L.S.T. Leung Chik Wai Memorial School F.6 Chemistry Chapter 15 Chpt. 15 : p.5 (C) Arrhenius Plot The activation energy is determined from the dependence of the natural logarithm of rate constant on the reciprocal of temperature. When the rate constant is known at more than two temperatures, the precision of the determination of the activation energy is increased by the Arrhenius plot, in k against 1/T. The equation Ea ln k ln A RT can be rewritten as Ea 1 ln k ln A ( ) R T to show the temperature dependence. This expression is of the type y = mx + c, a plot of the values of ln k at different temperatures against l/T will give a Straight-line graph with a slope (Ea/RT) and an intercept of ln A. Hence the activation energy and the Arrhenius Factor can be determined. Example This experiment is about the determination of the activation energy of the reaction between bromide ion and bromate(V) ion in acid solution: 5Br(aq) + BrO3(aq) + 6H+(aq) 3Br2 (aq) + 3H2O(l) The progress of the reaction is followed by adding a fixed amount of phenol together with some methyl red indicator. The bromine produced during the reaction reacts very quickly with phenol. Once all the phenol is consumed, any further bromine produced bleaches the indicator immediately. (a) Investigation of the activation energy of the reaction Experimental procedure <1> Place phenol solution, bromide solution and bromate solution and methyl red indicator into a boiling tube. <2> place sulphuric acid into another boiling tube. <3> Place both boiling tubes into a beaker of water (water bath) which is maintained at 30°C. Allow the contents of both tubes reach the temperature of the water bath. <4> Mix the contents of the two tubes. Start the stop watch and swirl the tube gently. Keep the tube in the water bath throughout the experiment. <5> Record the time (t) taken for the complete disappearance of the red colour. <6> Repeat the above steps, maintaining the reaction temperature at different temperatures, e.g. 35°C, 40°C. 45°C, 50°C. ( Temperature must be made the only variable in this experiment. ) L.S.T. Leung Chik Wai Memorial School F.6 Chemistry Chapter 15 Chpt. 15 : p.6 Principle of determination The time for the reaction to proceed to a certain extent is determined. (t denotes the time a definite quantity of bromine is produced at different temperatures.) Reaction rate = (concentration change/time) In general, the rate constant is related to the time of reaction by where t = the time for methyl red to be bleached k 1/t For the Arrhenius equation k = exp(-Ea/RT), the following expression is obtained. Ea 1 ln k ln A ( ) R T Putting k = concentration change (c)/t, plot the graph c Ea 1 ln ln A ( ) t R T using t at different temperatures Ea 1 ln t ln c ln A ( ) R T ln c and ln A are constants. The slope of the graph in t against 1/T is (Ea/R) Knowing the value of the gas constant R, Ea can be determined. (B) Necessary apparatus required for the determination Thermometer, stop watch, water bath, volume measuring device (e.g. dropper, burette, measuring cylinder) Exercise 5 The table below gives the rate constants obtained at different temperatures for the reaction 2N2O5(g) 2N2O4(g) + O2(g) Temperature/ °C Rate constant / s-1 10 3.83 x 10-6 20 1.71 x 10-5 30 6.94 x 10-5 40 2.57 x 10-4 50 8.78 x 10-4 Determine (a) the activation energy and (b) the Arrhenius factor for this reaction by plotting the graph for Ea 1 ln k ln A ( ) [Gas constant, R = 8.314 J K-1 mol-1] R T L.S.T. Leung Chik Wai Memorial School F.6 Chemistry Chapter 15 Chpt. 15 : p.7 L.S.T. Leung Chik Wai Memorial School F.6 Chemistry Chapter 15 Chpt. 15 : p.8 THE INTERPRETATION OF RATES OF GASEOUS RACTIONS AT MOLECULAR LEVEL I. DISTRIBUTION OF MOLECULAR SPEEDS IN A GAS • At any given temperature, all gases have the same average kinetic energy. • However, the motion of gas molecules is random. It follows that the collisions between molecules also occur randomly and involve a transfer of energy. • some collisions result in a gain of kinetic energy for one molecule and a loss of kinetic energy for the other. • If a molecule undergoes a series of collisions such that each collision adds to its kinetic energy, it will end up with a kinetic energy higher than the average. Conversely, if a molecule undergoes a series of collisions such that each collision results in a loss of kinetic energy, it will end up with a kinetic energy lower than the average. It can be concluded that Since molecules undergo continual random collisions, the molecules of a gas at constant temperature do not travel with the same speed. This results in a distribution of molecular speeds in a gas. II. GRAPHICAL REPRESENTATION OF THE MAXWELL-BOLTZMANN DISTRIBUTION AND ITS VARIATION WITH TEMPERATURE Maxwell—Boltzmann distribution curve is a plot of distribution of molecular kinetic energies (or speeds) different temperatures. Interpretation <1> The total area under the curve is proportional to the total number of molecules, the area under any potion of the curve is proportional to the number of molecules with the energies in that range. <2> The curve shows that some molecules have very low or very high speeds. However, most molecules have intermediate speeds. This results a normal distribution curve. L.S.T. Leung Chik Wai Memorial School F.6 Chemistry Chapter 15 (A) Chpt. 15 : p.9 Variation of Maxwell—Boltzmann curve with temperature The effect on the Maxwell—Boltzmann distribution curve of increasing the temperature is shown below: The spread of the Maxwell-Boltzmann distribution increases with increasing temperature It can be seen that at a higher temperature, the following changes occur: <1> The peak of the curve moves to the right, so that the mean energy of the molecules increases and the proportion of molecules having higher energy increases. <2> The curve flattens so that there is a wider distribution of energies and the proportion of molecules with the most probable speed decreases. <3> The area under the curve is still the same as that of the one at lower temperature, as the total number of molecules in the sample remains the same. III. SIMPLE COLLISION THEORY The collision theory explains chemical reactions at the molecular level. It is developed from the kinetic theory of gases to account for the effects of concentration and temperature on reaction rate. According to the collision theory <1> Chemical reactions in the gas phase are due to collision of reactant particles. <2> Not all collisions results in a reaction. For a collision to be effective such that a reaction can occur, the following conditions must be necessary (i) The reactant particles must collide with kinetic energy greater than the activation energy Ea (a certain threshold) to break the bonds that need to be broken, and (ii) The reactant particles must collide in the right direction (i.e. correct orientation or collision geometry) so that new bonds can form. As the number of effective collisions increases, the rate constant and the reaction rate increase. L.S.T. Leung Chik Wai Memorial School F.6 Chemistry Chapter 15 Chpt. 15 : p.10 collision theory In general, no of effective collisions = collision frequency x fraction of collision x with correct orientations fraction of gas molecules with K.E. greater than the Ea where fraction of gas molecules with kinetic energy greater than the activation energy is given by the Maxwell—Boltzmann energy distribution curve nEa Ea exp[ ] ntotal RT Since rate constant is proportional to the neither of effective collisions, the Arrhenius equation can be derived Ea ) Rate constant k A exp( RT (A) Temperature and the collision theory When the temperature is raised, there is a greater proportion of molecules with kinetic energy more than the activation energy than there is at the lower temperature. Interpretation <1> Ea represent the minimum collision energy necessary for the reaction to occur. <2> The area under the curve to the right of the activation energy represents the proportion of particles that collide with kinetic energies greater than the activation energy, Ea. L.S.T. Leung Chik Wai Memorial School F.6 Chemistry Chapter 15 Chpt. 15 : p.11 <3> The increase in the reaction rate with temperature corresponds closely to the ratio of the corresponding shaded areas. <4> Therefore, as temperature changes, the area under the distribution curve for E > Ea changes. This means that At higher temperatures, more molecules collide with energy greater than the activation energy and so the rate of reaction increase. <5> The fraction of effective collisions increases exponentially with temperature. Approximately. 10°C rise in temperature doubles the number of molecules enough to cross the activation energy barrier. Exercise 1 Discuss the effect of temperature on the rate of a reaction in terms of the Arrhenius equation. ANSWER (B) Concentration and the Collision Theory The increase in the rate of a reaction with the increase of concentration of one or more reactant particles can be explained by the Collision Theory. • As the concentration of the reactant increases, the frequency of collisions increases. • The probability of effective collisions (collisions with correct orientations and sufficient energy for the reaction to occur) also increases. • Reaction rate therefore increases. L.S.T. Leung Chik Wai Memorial School F.6 Chemistry Chapter 15 Chpt. 15 : p.12 ENERGY PROFILE I. ENERGY PROFILE Most reaction occur in more than one step. Each intermediate product has its specific potential energy. Energy profile is a representation of the changes in potential energy (against time) during a reaction. In an energy profile, the potential energies of the followings are shown: 1. reactants 2. products 3. intermediates 4. transition states. II TRANSITION STATE THEORY The transition state theory is developed to explain the rate of the reaction. It is mainly concerned with the events during collision. In the transition state theory, <1> As reactant particles collide and reaction takes place, they are temporarily in a less stable than the reactants or products. The atoms are rearranging themselves. <2> As the atoms are separated, the potential energy of the system increases. This results in an energy barrier between the reactants and products. <3> An activated complex in the transition state is formed. Note <1> The activated complex only exists for a fraction of a second. It is in equilibrium with the reactants and the products: [A.....B] C A+B ( Product ) activated complex in transition state <2> The activated complex in the transition state possesses the maximum potential energy in the energy profile. It may decompose to the product or reactant. <3> The products are formed only if the colliding particles have sufficient energy to overcome the energy barrier. The energy required is referred to as the activation energy. <4> The higher the activation energy, the lower the number of effective collisions. L.S.T. Leung Chik Wai Memorial School F.6 Chemistry Chapter 15 <5> Chpt. 15 : p.13 Distinguish between an intermediate and an activated complex with the aid of an energy profile An intermediate: It is an unstable species at a potential minimum which is relatively unstable compared to the reactants or products. (It can be detected or even isolated). An activated complex: It refers to a highly unstable species with the arrangement of nuclei and electrons at the maximum potential energy. It exists in a state between the reactants and products called the transition state. III. SINGLE STATE AND MULTIPLE—STAGE REACTION (A) Single stage reaction If a reaction between AB an C t es place in one—step process, A—B + C [A….B….C] A + B—C transition state the energy profile for the reaction can be represented as follows L.S.T. Leung Chik Wai Memorial School F.6 Chemistry Chapter 15 Chpt. 15 : p.14 Example : Alkaline hydrolysis of 1-bromobutane C4H9Br + OH- C4H9OH + BrRate of hydrolysis = k[C4H9Br][OH-] It can be deduced that • both 1-bromobutane and hydroxide ion are involved in the rate determining step, • the hydrolysis of 1-bromobutane is a single-stage reaction. The mechanism may be the following: C4HgBr + OH- [HO••••C4H9•••Br] C4H9OH + BrThe reaction is a bimolecular second order reaction. (B) Multi-stage reaction Multi-stage reactions are more common than single-stage reactions. For a reaction between DE and F, which takes place via a two-step or multi-step process, an intermediate is formed. D--E intermediate + F slow intermediate fast Overall Reaction : D—E + F D--F D—F + E + E The energy profile for this two-stage process can be represented as follows: Example : hydrolysis of 2-bromo-2-methylpropane (CH3)3CBr + OH (CH3)3COH + Brrate of hydrolysis = k[(CH3)3CBr] The reaction is independent of the alkali concentration, i.e. the reaction is zero order with respect to OH-. L.S.T. Leung Chik Wai Memorial School F.6 Chemistry Chapter 15 Chpt. 15 : p.15 It can be deduced that • only 2-bromo-2-methylpropane is involved in the rate determining step. • the hydrolysis is a two—step process, with the proposed mechanism as follows: This hydrolysis is a unimolecular first order reaction. IV. MECHANISM <1> Mechanism is a sequence of simple steps to account for the overall chemical reaction that takes place (in terms of bond breaking and bond formation). <2> Each of the steps in a mechanism is known as an elementary step or intermediate step. <3> Molecularity refers to the number of particles (molecules, ions or atoms) involved in each elementary step. Example : The rate determining step in the hydrolysis of 2-bromo-2-methyl propane slow (CH3)3CBr (CH3)3C+ + Bris unimolecular because only (CH3)3CBr is involved. The hydrolysis of 1-bromobutane (CH3)3CBr + OH- (CH3)3COH + Bris bimolecular because the rate-determining step involves 2 reactant particles, (CH3)3Br and OH-. V. RATE DETERMINING STEP The rate determining step in a multi-stage reaction is the slowest step in the reaction. The rate-determining step control the overall rate of reaction. Note: <1> <2> The activation energy for the rate-determining step greatest. In the hydrolysis of 2-bromo-2-methylpropane, the intermediate step slow (CH3)9CBr (CH3)3C+ + Br is the slow step in the reaction because it requires the greatest activation energy. The r.d.s. determines the overall rate of the hydrolysis because once this step has proceeded, the following intermediate step will proceed very quickly. fast (CH3)3C+ (CH3)3COH L.S.T. Leung Chik Wai Memorial School F.6 Chemistry Chapter 15 Chpt. 15 : p.16 <3> The molecularity of the rate-determining step is usually the order of the reaction. Exercise 2 Consider the energy profile shown below for a 3—step reaction: (a) (b) ANSWER Which is the rate-determining step in the reaction? Explain. Is the reaction exothermic or endothermic? Explain. Exercise 3 The following exothermic reaction: A2 + 2B 2AB proceeds in two steps: 1. A2 + B intermediate (slow) 2. intermediate + B 2AB (fast) (a) (b) ANSWER Sketch the energy profile against reaction coordinate for the two-step reaction. Indicate clearly on your diagram the transition state(s), the intermediate, and the enthalpy change of reaction for the overall reaction. If A2 is both a reactant. and a solvent for the reacting system, determine the order of reaction for the overall reaction. Give reasons for your answer. L.S.T. Leung Chik Wai Memorial School F.6 Chemistry Chapter 15 Chpt. 15 : p.17 Exercise 4 In acid solution, bromate ions slowly oxidize bromide ions to bromine: BrO3-(aq) + 5Br-(aq) + 6H+(aq) 3Br2 + 3H2O(l) To investigate the rate of this reaction, the following experimental data are obtained at 298K: Mixture A B C D (a) (b) (c) (d) Volume of molar bromate /cm3 50 50 100 50 Volume of Volume of molar molar H+(aq) bromide /cm3 /cm3 250 300 250 600 250 600 125 600 Volume of water /cm3 400 100 50 225 Relative rate of formation of bromine 1 4 8 2 Suggest a titrimetric method for determining the rate of reaction at various time of the experiment. Name one physical method which can be used to determine the rate of reaction. (Details are not required.) What is the rate equation for the reaction? Work out the units of the rate constant. A proposed mechanism for the reaction is Is this mechanism consistent with the rate equation for the reaction in (c)? Give your reasons and discuss each of the three reactants. ANSWER L.S.T. Leung Chik Wai Memorial School F.6 Chemistry Chapter 15 Chpt. 15 : p.18 CATALYSTS AND THEIR EFFECT ON REACTION RATES I. CATALYSTS 催化劑 A catalyst is a substance which alters改變 the rate of a particular chemical reaction, but itself is chemically unchanged at the end of the reaction. Note: <1> The catalyst may be changed physically or it may undergo temporary chemical change. <2> negative catalyst— a catalyst which slows down a reaction. positive catalyst — a catalyst which speeds up a reaction. <3> Catalytic action often continues until all the reactants have been converted into products. Reactants + catalyst Products + catalyst Catalyst goes back to catalyze more reactions. II. WORKING PRINCIPLE OF CATALYST AND THE EFFECT OF CATALYST ON REACTION RATE 催化劑的工作原理及對反應速率的影響 Since the presence of catalyst does not change the reactant concentration, the catalyst must cause the rate constant to increase or decrease. The working principles of a catalyst to increase the reaction rate are listed as follows: <1> A positive catalyst can increase the rate constant, and hence the reaction rate by providing an alternative pathway from reactants to products which has a lower activation energy. <2> In terms of collision theory, a catalyst increases reaction rate by (i) bringing the reactant molecules closer together, thereby increasing the collision frequency 使反應物聚在一起，增加它們的撞擊頻率。 (ii) orientating the reactant molecules so that they achieve the correct collision geometry. 將反應物分子導向，使它們達到正確撞擊位置。 Exercise 1 Discuss, in terms of the Arrhenius equation, the effect on the reaction rate of lowering the activation energy. ANSWER L.S.T. Leung Chik Wai Memorial School F.6 Chemistry Chapter 15 Chpt. 15 : p.19 (A) Provision of an alternative pathway for a reaction 提供另一反應途徑 In a catalyzed reaction, the catalyst participates in the reaction by providing an alternative reaction pathway (or mechanism) for the production of the products. • positive catalyst — provides an alternative pathway with lower activation energy. negative catalyst — provides an alternative pathway with higher activation energy. • Energy profiles showing the effect of a positive catalyst on the activation energy. • The smaller Ea means that in the reaction mixture, there is a greater total fraction of molecules possessing sufficient energy to react. • Thus, a positive catalyst can increase the chance of effective collisions, and hence greater reaction rate. L.S.T. Leung Chik Wai Memorial School F.6 Chemistry Chapter 15 (B) Chpt. 15 : p.20 Mechanism for the alternative pathway provided by the catalyst <1> In general, a catalyst is chemically involved in the catalysed reaction. It is consumed in one reaction step and regenerated in a subsequent step. A + Catalyst A—catalyst A—catalyst + B A—B + Catalyst ---------------------------------------------------------------------------Overall reaction: A + B A—B <2> The involvement of a catalyst causes the reaction to take place via an alternative pathway (the catalytic pathway). <3> The uncatalysed reaction is replaced by two other reactions, both of which have lower activation energies, as illustrated below: In conclusion, in a catalysed reaction, • The enthalpy of reaction, H, is unchanged. Only the activation energy is altered. • The reaction rate is increased as the activation energy is lowered in the catalytic pathway. Exercise 2 Does the use of a suitable catalyst in a reaction change (a) the order of the reaction, and (b) the enthalpy of reaction? Explain your answer. ANSWER L.S.T. Leung Chik Wai Memorial School F.6 Chemistry Chapter 15 Chpt. 15 : p.21 III. CHARACTERISTIC OF CATALYSTS 催化劑的特徵 Most important characteristics <1> It does not alter the equilibrium position of a reversible reaction because it increases both forward and backward rates. 不會改變可逆反應的平衡位置 of <2> It can increase the rate of production in many industrial processes (e.g. Haber process and contact process) but they do not increase the yield of product. It only increases the rate attainment of the equilibrium position, and hence the reaction rate. 能加快生成產物的速度，但不能提高產物的數量 Other characteristics of catalyst: <1> <2> <3> the <4> <5> <6> <7> <8> <9> A catalyst is chemically unchanged and are not used up at the end of a reaction. The action of a catalyst is specific. A certain catalyst can only catalyze some particular reactions, but have no effects on others. 一些催化劑只可對某反應起作用，其他的反 應並不適用 A catalyst must be able to bind結合to the reactant particles reversibly. If the bonding is too weak, the catalyst will be less powerful; if the binding is too strong, the reactant particles will be permanently bonded to the catalyst and cannot be converted into product. Its physical state may be altered after the reaction. In general, small amount of catalyst is enough to increase the rate of reaction. Adequate足夠amount of catalyst is enough. Further addition of catalyst cannot does not change the reaction rate. Many catalysts are transition metals and/or their compounds, e.g. palladium, nickel, vanadium (V) oxide, manganese(IV) oxide. This is because the transition metal can use low energy vacant d orbitals to form dative bond with the reactants. Impurities may bind to a catalyst irreversibly and “poison” the catalyst. The catalyst is destroyed and loses the catalytic function. (e.g. lead compounds in car exhaust would poison the platinum catalyst in catalytic converter.) Finely divided (solid) catalyst gives a larger catalytic surface and is therefore more effective than a large catalyst. IV. AUTOCATALYSIS 自動催化 For most catalytic reactions, the amount of catalyst remains constant with time. However, in autocatalytic reaction, the amount of catalyst increases as the reaction proceeds. In autocatalysis, the product of reaction becomes the catalyst for the reaction. Example :Effect of Mn2+ on the reaction between MnO4- and C2O42- in acidic medium The Mn2+ ion, formed in the oxidation of ethane-1,2-dioic acid (oxalic acid) by acidified potassium manganate (VII) catalyses the reaction: 2MnO4- (aq) + 16H+(aq) + 5C2O42- 2Mn2+ (aq) + 8H2O(l) + 10CO2(g) The following curve of plotted against time shows the variation of reaction rate in the course of the autocatalytic reaction. L.S.T. Leung Chik Wai Memorial School F.6 Chemistry Chapter 15 Chpt. 15 : p.22 Interpretation: <1> At the beginning of the reaction, the rate is slow because the collision of negatively charged MnO4- and C2O42- is difficult. <2> Once a certain amount of Mn2- has been formed, the reaction rate increases because Mn2+ acts as the catalyst for the reaction. <3> Reaction rate falls as the concentrations of reactants decreases. Exercise 3 Potassium manganate(VII) will oxidize ethanedioic acid (oxalic acid). (CO2H)2 to carbon dioxide and water in the presence of excess acid: 2MnO4-(aq) + 16H+ (aq) + 5C2O42- 2Mn2+(q) + 8H2O(l) + 10CO2(g) 2 experiments are devised to investigate the effect of Mn2+ ion on the reaction. Solution 0.2 M ethanedioic acid 0.2 M manganese (II) sulphate 2 M sulphuric acid water Experiment 1 100 cm3 0 5 cm3 95 cm3 Experiment 2 100 cm3 15 cm3 5 cm3 80 cm3 50 cm3 of 0.02 M potassium manganate(VII) are added to each reaction mixture. 10 cm 3 portion of the mixture is withdrawn and run up into a conical flask. This sample of reaction mixture is then added to about 10 cm3 of 0.1 M potassium iodide solution. This stops the reaction and releases iodine equivalent to the residual manganate(VII) ions. The iodine liberated is titrated with 0.01 M sodium thiosulphate. and a little starch solution is added near the end point. Further portions of the reaction mixture are removed and treated in the same way. The titration results are plotted for each experiment. (a) (b) Describe how the reactant concentration at various time intervals is derived from the titration results. The titration results are given in the curve shown below: (c) Interpret curve(2) briefly. Give an explanation for the abnormality observed in curve 1. Name a physical method which can be used to follow the reaction. L.S.T. Leung Chik Wai Memorial School F.6 Chemistry Chapter 15 Chpt. 15 : p.23 L.S.T. Leung Chik Wai Memorial School F.6 Chemistry Chapter 15 V. Chpt. 15 : p.24 HOMOGENEOUS 均相催化 AND HETEROGENEOUS CATALYSIS多相催化 There are two main types of catalysts, i.e. homogeneous and heterogeneous catalysis. Note: Enzymes, which are very important biological catalysts, do not fit into either of the above cataegories. (A) Homogeneous catalysis In homogeneous catalysis, the catalyst and the reactants are in the same phase, i.e. physical state. Note <1> Usually, the catalyst forms an intermediate compound which decomposes at a later stage. <2> Homogeneous catalysis in the liquid state may be illustrated by acid-catalyzed esterification: CH3COOH (l) + C2H5OH (l) CH3COOC2H5 (l) + H2O (l) This reaction between a carboxylic acid and an alcohol is catalysed by a little concentration sulphuric acid. Exercise 4 The reaction S2O82-aq + 2I-(aq) 2SO42-(aq) + I2(aq) is catalysed by Fe2+(aq), a transition metal ion. (a) Is this a homogeneous catalysis or heterogeneous catalysis? Explain. (b) Outline the essential steps to show how the catalytic activity of Fe2+ (aq) can be illustrated experimentally. ANSWER (B) Heterogeneous catalysis In heterogenous catalysis, the reactants and the catalyst are in two different phase. Note: <1> The behaviour of heterogeneous catalysts is explained by the Adsorption Theory (a) The reactants come together by adsorption on the surface of the catalyst. The bonds of the reactants are weakened. (b) Reaction then occurs on the catalytic surface. (c) After the product has been formed on the surface of the catalyst, the product molecules are desorbed from the catalyst. L.S.T. Leung Chik Wai Memorial School F.6 Chemistry Chapter 15 Chpt. 15 : p.25 <2> Adsorption Theory can be illustrated by the catalytic hydrogenation of ethene: <3> Generally, heterogeneous catalysis involves the reaction of two gases on the surface of a catalyst. Examples include: 1. the conversion of harmful pollutants from car exhausts into harmless substances in catalytic converters 2. the hydrogenation of unsaturated oils (or alkenes) using nickel as catalyst. 3. the conversion of sulphur dioxide into sulphur trioxide in the Contact process 4. the production of ammonia from hydrogen and nitrogen in the Haber process <4> In conclusion, A heterogeneous catalyst provides a surface onto which reactants are adsorbed and from which products of reaction are desorbed. <5> Catalytic decomposition of hydrogen peroxide solution with manganese(IV)oxide (MnO2) solid is an example of heterogeneous catalysis of a reaction in solution MnO2 2H2O2(aq) 2H2O(l) + O2(g) L.S.T. Leung Chik Wai Memorial School F.6 Chemistry Chapter 15 Chpt. 15 : p.26 VI. APPLICATIONS OF CATALYSTS Catalysts are widely used in many industrial processes, e.g. Haber process, contact process. They increase the rate of production but they do not increase the yield of products. (A) Contact process (Production of sulphuric acid) The major steps of the contact process are summarized in the table below: Step 1 Description Sulphur dioxide is produced by roasting sulphur or heating metal sulphides in air. Equation S(s) + O2(g) SO2(g) 2ZnS(s) + 3O2(g) 2ZnO(s) + 2SO2(g) 4FeS2 + 11O2(g) 2Fe2O3(s) + 8SO2(g) 2 Sulphur dioxide and air are purified to prevent poisoning the catalyst in step (3). 3 Sulphur dioxide is oxidized by air to sulphur trioxide. Conditions: 450 °C, atmospheric pressure,1 atm platinum or vanadium (V)oxide catalyst. 4 Sulphur trioxide is dissolved in concentrated sulphuric acid to form oleum (fuming sulphuric acid) in the absorber. Oleum is dissolved in water to SO3(g) + H2SO4(l) H2S2O7(l) 5 Oleum is dissolved in water to give sulphuric acid. H2S2O7(l) + H2O(l) 2H2SO4(l) 2SO2(g) + O2(g) 2SO3(g) Note <1> The conversion of sulphur dioxide (SO2) to sulphur trioxide (SO3) involves the catalytic action of platinum or vanadium (V) oxide. <2> The flow diagram for the manufacture of sulphuric acid is shown below: L.S.T. Leung Chik Wai Memorial School F.6 Chemistry Chapter 15 Chpt. 15 : p.27 (B) Haber process (Production of ammonia) Catalyst used: finely divided iron, platinum or copper In the Haber process. nitrogen and hydrogen combine directly to manufacture ammonia. Procedure <1> Initially, ammonia (1 part) and hydrogen (3 parts) are purified, dried and mixed. The mixture is compressed to 200 atmospheres at 500°C and then passed over a catalytic chamber containing the catalyst. The following reaction occurs: - Fe N2(g) + 3H2(g) 2NH3(g) <2> The ammonia produced is cooled and condensed to liquid and collected. <3> The unreacted nitrogen and hydrogen re—enter the catalyst chamber for recycling. (C) Hydrogenation of unsaturated oils Margarine and other fats are made by passing hydrogen through a heated vegetable oil in the presence of nickel catalyst. Under these conditions, hydrogen adds across some or all the double bonds in the oil. finely divided nickel, —CH=CH— unsaturated oil + H2 200 C 0 —CH2—CH2— saturated oil Note: <1> The saturated oils forms high-melting point soft solid fats on cooling. <2> Hydrogenation of unsaturated oils and alkenes using nickel, palladium or platinum catalyst is an example of heterogeneous catalysis. (D) Catalytic converters in car exhausts systems A catalytic converter is used in some exhaust systems to convert some pollutants from car exhausts (e.g. carbon dioxide, nitrogen monoxide and unburnt hydrocarbons) into relatively harmless substances (e.g. carbon dioxide, nitrogen and water). Working principles : <1> The exhaust gases are passed through a “honeycomb” of small beads coated with metal catalysts such as platinum (or palladium) and rhodium. The “honeycomb” increases the surface area of the catalyst for better catalytic action. <2> The converter is heated and extra air is pumped into it as shown. A catalytic coverter Lok Sin Tong Leung Chik Wai Memorial School F.6 Chemistry Chapter 15 Chpt.15:p28 <3> Reactions: (i) Nitrogen oxides react with carbon dioxide when they pass over the platinum catalyst to form carbon dioxide and nitrogen: 2CO(g) + 2NO(g) 2CO2(g) + N2(g) (ii) Unburat hydrocarbons and carbon monoxide are oxidized by the air pumped into the catalytic converter, forming water and carbon dioxide. <4> Cars which include catalytic converters in their exhaust systems must use lead—free petrol to prevent the catalyst from being coated with and poisoned by lead. Conclusion Many industrially important heterogeneous catalyst are d-block transition metals or their compounds, e.g. nickel, vanadium, manganese, manganese(IV) oxide. They use the 3d and 4s subshell to adsorb reactant molecules and lower the activation energies by weakening the bonds in the adsorbed molecules. VII. ENZYMES AS EXAMPLES OF BIOLOGICAL CATALYSTS Enzymes are proteins that catalyze specific biochemical reactions in living systems. They are often called biological catalysts. Without them, most biochemical reactions would be too slow to sustain life. For example, enzymes obtained from yeasts have long been used in the production of alcohol by fermentation. The fermentation. of glucose to form ethanol can be represented by the following equation: C6H12O6 • enzyme 2C2H5OH+ 2C02 Nowadays. enzymes have many uses, such as in the manufacture of biological washing powders. These washing powders contain enzymes which can break down stains caused by blood, egg, sweat and fats. They have the advantage of removing stains even at normal temperature.