Chemical Kinetics. Kinetics. Factors That Affect Reaction Rates. Factors That Affect Reaction Rates. Factors That Affect Reaction Rates

Transcription

1 Kinetics hemical Kinetics In kinetics we study the rate at which a chemical process occurs. esides information about the speed at which reactions occur, kinetics also sheds light on the reaction mechanism (exactly how the reaction occurs). Factors That ffect Reaction Rates Factors That ffect Reaction Rates The Physical State of the Reactants The oncentrations of the Reactants Temperature The Presence of a atalyst Surface area Physical State of the Reactants In order to react, molecules must come in contact with each other. The more homogeneous the mixture of reactants, the faster the molecules can react. Factors That ffect Reaction Rates The oncentrations of the Reactants s the concentration of reactants increases, so does the likelihood that reactant molecules will collide. ir hole Factors That ffect Reaction Rates Temperature t higher temperatures, reactant molecules have more kinetic energy move faster collide more often and with greater energy. Why do we keep food in fridge? The flame from the unsen burner shows the variation in color as more air is mixed with the natural gas. s more air is mixed with the natural gas, the rate of combustion reaction will increase and the flame gets more blue in color. 1

2 Factors That ffect Reaction Rates Presence of a atalyst atalysts speed up reactions by changing the mechanism of the reaction. atalysts are not consumed during the course of the reaction. Factors That ffect Reaction Rates Surface rea When more surface area of a reactant is exposed, a reaction will occur faster. steel bar will not ignite if exposed to a flame. In contrast, a steel wool pad will burn and ignite quickly. 1 Which of the following does not play a part in determining the rate of a reaction? Temperature oncentration of the reactants atalyst Surface area of the solid quilibrium constant 2 Food spoils at a faster rate at 25 degrees F compared to 5 degrees. Yes No 3 Why does a higher temperature cause a reaction to go faster? 4 Why does a higher reactant concentration make a reaction faster? Only because there are more collisions per second. Only because there are more collisions per second. Only because collisions occur with greater energy. Only because collisions occur with greater energy. There are more frequent collisions and they are of greater energy. There are more frequent collisions and they are of greater energy. 2

3 5 Why does a greater surface area cause a reaction to proceed faster? 6 What happens to a catalyst in a reaction? Only because there are more collisions per second. Only because collisions occur with greater energy. It is unchanged. It is incorporated into the products. It is incorporated into the reactants. There are more frequent collisions and they are of greater energy. It evaporates away. 7 If a catalyst is used in a reaction. Reaction Rates the energy of activation increases. different reaction products are obtained. 0 s 20 s 40 s the reaction rate increases. it evaporates away. 1 mol 0.5 mol 0.5 mol 0.25 mol 0.75 mol Rates of reactions can be determined by monitoring the change in concentration of either reactants or products as a function of time. Reaction Rates 4H9l(aq) + H2O(l) 4H9OH(aq) + Hl(aq) Time, t(s) [4H9l](M) In this reaction, the concentration of butyl chloride, 4H9l, was measured at various times. Reaction Rates 4 H 9 l(aq) + H 2 O(l) 4 H 9 OH(aq) + Hl(aq) Time, t(s) [ 4H 9l](M) verage Rate (M/s) x x x x x x x x verage Rate = [ 4 H 9 l] t The average rate of the reaction over each interval is the change in concentration divided by the change in time: 3

4 Reaction Rates 4 H 9 l(aq) + H 2 O(l) 4 H 9 OH(aq) + Hl(aq) Time, t(s) [ 4H 9l](M) verage Rate (M/s) Reaction Rates 4 H 9 l(aq) + H 2 O(l) 4 H 9 OH(aq) + Hl(aq) x x x x x x x x Note that the average rate decreases as the reaction proceeds WHY? plot of [ 4 H 9 l] vs. time for this reaction yields a curve like this. The slope of a line tangent to the curve at any point is the instantaneous rate at that time. [4H9l] (M) x 10 2 Instantaneous rate at t=0s Instantaneous rate at t=600s Time (s) ll reactions slow down over time. Therefore, the best indicator of the rate of a reaction is the instantaneous rate near the beginning of the reaction. Reaction Rates 4 H 9 l(aq) + H 2 O(l) 4 H 9 OH(aq) + Hl(aq) [4H9l] (M) x 10 2 Instantaneous rate at t=0s Time (s) Instantaneous rate at t=600s Reaction Rates and Stoichiometry 4 H 9 l(aq) + H 2 O(l) 4 H 9 OH(aq) + Hl(aq) In this reaction, the ratio of 4 H 9 l to 4 H 9 OH is 1:1. Thus, the rate of disappearance of 4 H 9 l is the same as the rate of appearance of 4 H 9 OH. Rate = [ 4 H 9 l] t = [4H9l] (M) x 10 2 [ 4 H 9 OH] t Instantaneous rate at t=0s Time (s) Instantaneous rate at t=600s 8 Which substance in the reaction below either appears or disappears the slowest? 4NH 3 + 7O 2 4NO 2 + 6H 2 O NH 3 O 2 H 2 O 9 Which of the following expressions could represent a reaction rate? time/mass concentration/time energy/time time/energy NO 2 The rates of appearance are the same as the rates of disappearance 4

5 10 Which of the following set of units represents a reaction rate? g/s mol/s s j/s s/j 11 flask is charged with 0.124mol of and allowed to react to form according to the reaction (g) (g). the following data are obtained for [] as the reaction proceeds. The average rate of disappearance of between 10 s and 20 s is mol/s flask is charged with mol of and allowed to react to form according to the reaction (g) (g). The following data are obtained for [] as the reaction proceeds: 13 The peroxydisulfate ion, S 2 O 2 8, reacts with the iodide ion in aqueous solution via the reaction: S 2 O 2 8 (aq) + 3I > 2SO 4 (aq) + I 3 (aq) The following data are collected for the concentration of iodide ion: 590 The average rate of disappearance of between s and s is M/s. 2.8 x x x x x 10 4 Reaction Rates and Stoichiometry Reaction Rates and Stoichiometry What if the ratio is not 1:1? 2HI(g) H 2 (g) + I 2 (g) Rate = 1 2 [HI] t [I 2 ] t For every 1 mole of I 2 that is produced, 2 moles of HI must decompose. So the rate of disappearance of HI is twice as fast as the appearance of I 2 or the appearance of H 2. = To generalize, then, for the reaction a + b c + d Rate = 1 [] 1 [] 1 [] 1 a t = = b t c t = d [] t 5

6 14 t elevated temperatures, dinitrogen pentoxide decomposes to nitrogen dioxide and oxygen: 2 N 2 O 5 (g) 4 NO 2 (g) + O 2 (g) When the rate of formation of O 2 (g) is M/s, the rate of decomposition of N 2 O 5 is M/s. 2.2 x x x x x t elevated temperatures, dinitrogen pentoxide decomposes to nitrogen dioxide and oxygen: 2 N 2 O 5 (g) 4 NO 2 (g) + O 2 (g) When the rate of formation of NO 2 (g) is M/s, the rate of decomposition of N 2 O 5 is M/s. 2.2 x x x x x 10 4 oncentration and Rate oncentration and Rate xperiment Initial[ NH4 + ] Initial [NO2 ] Observed initial number (M) [M] Rate (M/s) One can gain information about the rate of a reaction by seeing how the rate changes with changes in concentration x x x x x x10 7 NH 4 + (aq) + NO 2 (aq) N 2 (g) + 2 H 2 O(l) If we compare xperiments 1 and 2, we see that when [NH 4 + ] doubles, the initial rate doubles. oncentration and Rate xperiment Initial[ NH4 + ] Initial [NO2 ] Observed initial number (M) [M] Rate (M/s) x x x x x x10 7 NH 4 + (aq) + NO 2 (aq) N 2 (g) + 2 H 2 O(l) Likewise, when we compare xperiments 5 and 6, we see that when [NO 2 ] doubles, the initial rate doubles. oncentration and Rate This means Rate [NH + 4 ] Rate [NO 2 ] Rate [NH + 4 ] [NO 2 ] which, when written as an equation, becomes Therefore, Rate = k [NH + 4 ] [NO 2 ] This equation is called the rate law, and k is the rate constant. 6

7 Rate Laws rate law shows the relationship between the reaction rate and the concentrations of reactants. The exponents tell the order of the reaction with respect to each reactant. Since the rate law is Rate = k [NH 4 + ] [NO 2 ] the reaction is First order in [NH 4 + ] and First order in [NO 2 ]. Rate Laws Rate = k [NH 4 + ] [NO 2 ] The overall reaction order can be found by adding the exponents on the reactants in the rate law. This reaction is second order overall. Order of a reaction Order of the reaction can be determined only from experimental data. oncentration changes Rate independent zero order oncentration doubles Rate doubles first order 16 If the rate law for the reaction products is first order in and second order in, then the rate law is: Rate = k [][] Rate = k [] 2 [] 3 Rate = k [][] 2 Rate = k [] 2 [] Rate = k [] 2 [] 2 oncentration doubles Rate quadruples second order 17 What will happen to the rate of a reaction if the molarity of a firstorder reactant is doubled? rate will double 18 What will happen to the rate of a reaction if the molarity of a second order reactant is doubled? rate will triple rate will double rate will quadruple rate will triple rate remains unchanged rate will quadruple rate remains unchanged 7

8 Practice Question [NO] [O2] Initial rate M/s x x x10 2 R3 = K[NO] x [O2] y 5.64x10 2 R1 K[NO] x [O2] y 1.41x10 2 = k[0.0256] x [0.0125] y = 4 k[0.0126] x [0.0125] y 2 x = 4; x = 2 19 This reaction is order with respect to [OH ]. 1st 2nd 3rd 4th zero R2/R3 = K[0.0252] x [0.025] y =1.13x10 1 /5.64x10 2 k[0.0252] x [0.0125] y 2 y = 2 y = 1 R = K[NO] 2 [O 2 ] 20 This reaction is order with respect to [lo 2 ]. 1st 2nd 3rd 4th zero 21 The overall order of a reaction is 2. The units of the rate constant for the reaction are. M/s M 1 s 1 s 1 M 1 s / M 2 Integrated Rate Laws Integrated rate law equations can be used to determine the concentration of either a reactant or a product at any time. In a typical problem, you may be asked to find either: 1) concentration of a reactant at a given time, t, after the start of a rxn 2) the time it takes for a certain amount of reactant to be consumed 3) the time it takes for the concentration of a reactant or product to reach a certain level Integrated Rate Laws Using calculus to integrate the rate law for a first order process gives us R= k[] ln [] t [] = kt time 0 to t 0 Where [] 0 is the initial concentration of, and [] t is the concentration of at some time, t, during the course of the reaction. 8

9 Integrated Rate Laws Manipulating this equation produces ln [] t [] 0 = kt ln[] t ln[] 0 = kt First Order Processes ln [] t = kt + ln [] 0 Therefore, if a reaction is first order, a plot of ln [] vs. t will yield a straight line, and the slope of the line will be k. ln[] t = kt + ln[] 0 which is in the form y = mx + b First Order Processes First Order Processes H 3 N onsider the process in which methyl isonitrile is converted to acetonitrile. H 3 N H 3 N H 3 N H 3 N H 3 N This data was collected for this reaction at First Order Processes Second Order Processes Similarly, integrating the rate law for a process that is second order in reactant, R= k[] 2, we get 1 [] t = kt + 1 [] 0 also in the form y = mx + b When ln P is plotted as a function of time, a straight line results. Therefore, The process is first order. k is the negative of the slope: s 1. 9

10 Second Order Processes 1 []t = kt + 1 []0 So if a process is second order in, a plot of 1/[] vs. t will yield a straight line, and the slope of that line is k. Second Order Processes The decomposition of NO 2 at 300 is described by the equation and yields data comparable to this: 1 NO 2 (g) NO (g) + O 2 (g) 2 Time (s) [NO 2 ], M Second Order Processes Plotting ln [NO 2 ] vs. t yields the graph below The plot is not a straight line, so the process is not first order in []. Second Order Processes 1 [NO2] Graphing vs. t, however, gives this plot. ecause this is a straight line, the process is second order in [] Time (s) [NO 2 ], M ln [NO 2 ] Time (s) [NO 2 ], M 1 [NO2] Half Life Half Life For a first order process, this becomes ln 0.5[] 0 = kt 1/2 [] 0 ln 0.5 = kt 1/ = kt 1/ = t k 1/2 Half life is defined as the time required for one half of a reactant to react. ecause [] at t 1/2 is one half of the original [], [] t = 0.5 [] 0. NOT: For a first order process, then, the halflife does not depend on [] 0. 10

11 Half Life 22 Which of the following will yield a straight line for a first order reaction? For a second order process, [] 0 kt 1/2 = + 2 = kt 1/2 + [] 0 1 [] 0 1 [] 0 [] vs. time log [] vs. time ln [] vs. time ln [] vs. k 1/[] vs. k = = kt [] 0 [] 1/2 0 1 = t 1/2 k[] 0 23 Which of the following will yield a straight line for a second order reaction? 24 Which of the following will yield a straight line for a second order reaction? [] vs. time log [] vs. time ln [] vs. time 1/[] vs. time 1/[] vs. k [] vs. time Pressure, P vs. time 1/P vs. k ln P vs. time 1/P vs. time 25 The half life for Substance X is 22 minutes. How much of a 10.0 gram sample will remain after 66 minutes? 1.25 grams 1.33 grams 2.5 grams 5.00 grams 7.00 grams 26 medical courier must deliver 8.0 mg of a substance in four hours. If the half life of this substance is 2.0 hours, the original sample should be at least mg. 11

12 Temperature and Rate Maxwell oltzmann istributions Low temperature 4 High Temperature y Fraction of molecules a Temperature is defined as a measure of the average kinetic energy of the molecules in a sample. x 10 3 k(s 1 ) x Generally, as temperature increases, so does the reaction rate. This is because k is temperature dependent. Speed / K t any temperature there is a wide distribution of kinetic energies temperature ( 0) Fraction of molecules Maxwell oltzmann istributions a s the temperature increases, the curve flattens and broadens. Thus at higher temperatures, a larger population of molecules has higher energy. Fraction of molecules Maxwell oltzmann istributions If the dotted line represents the activation energy, then as the temperature increases, so does the fraction of molecules that can overcome the activation energy barrier. a s a result, the reaction rate increases. Speed / K Speed / K Maxwell oltzmann istributions This fraction of molecules can be found through the expression f = e a RT where a is the activation energy, R is the gas constan and T is the Kelvin temperature. Temperature and Rate The ollision Model In a chemical reaction, bonds are broken and new bonds are formed. Molecules can only react if they collide with each other. Fraction of molecules a Speed / K 12

13 Temperature and Rate The ollision Model In a chemical reaction, bonds are broken and new bonds are formed. Molecules can only react if they collide with each other. Furthermore, molecules must collide with the correct orientation and with enough energy to cause bond breakage and formation. l + lon l 2 + NO Temperature and Rate The ollision Model Svante rrhenius developed a mathematical relationship between the rate constant, k and a : k = e a RT where is the frequency factor, a number that represents the likelihood that collisions would occur with the proper orientation for reaction. efore collision ollision fter collision rrhenius quation Reaction oordinate iagrams Taking the natural logarithm of both sides, the rrhenius equation becomes: a which is also in the form y = m x + b Therefore, if k is determined experimentally at several temperatures, a can be calculated from the slope of a plot of ln k vs. 1/T. Potential energy Reaction pathway It is helpful to visualize energy changes throughout a process on a reaction coordinate diagram like this one for the rearrangement of methyl isonitrile. H 3N H 3N Methyl isonitrile to Methyl nitrile Δ Reaction oordinate iagrams Reaction oordinate iagrams Potential energy a Potential energy a Δ Reaction pathway Δ Reaction pathway 13

14 Reaction oordinate iagrams Reaction oordinate iagrams Potential energy a Potential energy a Δ Δ Reaction pathway Reaction pathway Reaction oordinate iagrams Reaction oordinate iagrams The diagram shows the energy of the reactants and products (and, therefore, ). Potential energy Potential energy The high point on the diagram is the transition state. a Reaction pathway Δ The species present at the transition state is called the activated complex. Reaction pathway The energy gap between the reactants and the activated complex is the activation energy barrier. Reaction oordinate iagrams Reaction oordinate iagrams In order for a reaction to occur, a minimum amount of energy must first be applied to the reactants. This minimum energy input is usually needed to break bonds. This minimum energy input is called the ctivation nergy, or a, energy of activation. ctivation energy is the energy gap between the reactants and the activated complex. 14

15 Reaction oordinate iagrams 27 This potential energy diagram represents a reaction that is with a H of. In this example, the P of the products are lower than the P of the reactants. Therefore, H is negative and the reaction is exothermic. exothermic, 10 kj exothermic, 25 kj endothermic, +15 kj endothermic, +25 kj 28 This potential energy diagram represents a reaction that is with a H of. 29 The activation energy for this reaction is exothermic, 5 kj exothermic, 20 kj endothermic, +5 kj endothermic, +25 kj +15 kj 15 kj +25 kj 25 kj 30 The activation energy for the reverse reaction is 31 What is another name for the activated complex? energy barrier +25 kj 25 kj transition state rate limiter +40 kj 40 kj collision group reactant or product 15

16 32 t what stage of a reaction do atoms have the highest energy? reactant stage product stage transition state stage 33 ctivation energy is. the heat released in a reaction the energy given off when reactants collide generally very high for a reaction that takes place rapidly an energy barrier between reactants and products atalysts atalysts increase the rate of a reaction by decreasing the activation energy of the reaction. Reaction Mechanisms This graph shows the decomposition of hydrogen peroxide both with and without a catalyst. Notice that the energies of reactants and products are unchanged by the catalyst. The sequence of events that describes the actual process by which reactants become products is called the reaction mechanism. Reaction Mechanisms Reaction Mechanisms Molecularity lementary Reaction Rate Law Reactions may occur all at once or through several discrete steps. ach of these processes is known as an elementary reaction or elementary process. Unimolecular > products Rate =k[] imolecular + > products Rate =k[] 2 imolecular + > products Rate = k[][] Termolecular ++ > products Rate = k[] 3 Termolecular ++ > products Rate =k[] 2 [] Termolecular ++ > products Rate = k[][][] The molecularity of a process tells how many molecules are involved in the process. 16

17 Multistep Mechanisms In a multistep process, one of the steps will be slower than all others. The overall reaction cannot occur faster than this slowest, rate determining step. Toll Toll Toll Toll Multistep Mechanisms In a multistep mechanism, the main idea about identifying the rate determining step is this: The rate law for the overall reaction is equivalent to the rate law for the slowest, or rate determining step slow step 1 step slow Slow Initial Step NO 2 (g) + O (g) NO (g) + O 2 (g) The rate law for this reaction is found experimentally to be Rate = k [NO 2 ] 2 O is necessary for this reaction to occur, but the rate of the reaction does not depend on its concentration. This suggests the reaction occurs in two steps. Slow Initial Step proposed mechanism for this reaction is Step 1: NO 2 + NO 2 NO 3 + NO (slow) Step 2: NO 3 + O NO 2 + O 2 (fast) The NO 3 intermediate is consumed in the second step. s O is not involved in the slow, rate determining step, it does not appear in the rate law. Toll Toll slow Fast Initial Step 2 NO (g) + r 2 (g) 2 NOr (g) The rate law for this reaction is found to be Rate = k [NO] 2 [r 2 ] This reaction is termolecular ecause termolecular processes are rare, this rate law suggests a two step mechanism. collision of three molecules in the correct orientation is very difficult Fast Initial Step proposed mechanism is K1 Step 1: NO + r 2 NOr 2 (fast) K2 Step 2: NOr 2 + NO 2NOr (slow) Step 1 includes the forward and reverse reactions. K 1 17

18 Fast Initial Step Fast Initial Step The rate of the overall reaction depends upon the rate of the slow step. The rate law for that step would be Rate = K 2 [NOr 2 ] [NO] ut how can we find [NOr 2 ]? NOr 2 can react two ways: 1. With NO to form NOr 2. y decomposition to reform NO and r 2 The reactants and products of the first step are in equilibrium with each other. Therefore, Rate f = Rate r Fast Initial Step Fast Initial Step ecause Rate f = Rate r, k 1 [NO] [r 2 ] = k 1 [NOr 2 ] Solving for [NOr 2 ] gives us K 1 K 1 [NO] [r 2 ]= [NOr 2 ] Substituting this expression for [NOr 2 ] in the rate law for the rate determining step gives Rate = K 2 K 1 [NO][r 2 ][NO] ` K 1 = k [NO] 2 [r 2 ] 34 Which step is the rate determining step, in a multistep reaction mechanism? the first step 35 What is the rate law for the overall reaction, given the following mechanism: Step 1: + (slow) Step 2: + 2 (fast) the last step the slowest step the fastest step not enough information given. Rate = [] 2 [][] Rate = [][] Rate = [][] Rate = [] 2 [] / [] 2 18

19 36 What is the rate law for the overall reaction, given the following mechanism: Step 1: + (fast) Step 2: + 2 (slow) Rate = [] 2 [][] Rate = [][] Rate = [][] Rate = [] 2 [] atalysts atalysts increase the rate of a reaction by decreasing the activation energy of the reaction. atalysts change the mechanism by which the process occurs. atalysts One way a catalyst can speed up a reaction is by holding the reactants together and helping bonds to break. utomobiles are equipped with catalytic converters, which are part of their exhaust systems. atalytic onverters The exhaust gases contain O, NO, NO 2, and unburned hydrocarbons that pass over surfaces impregnated with catalysts. The catalysts promote the conversion of the exhaust gases into O 2, H 2 O, and N 2. Shown here is hydrogen reacting with ethylene. oth molecules are adsorbed onto the metal (blue) surface which acts as the catalyst. 37 Why does a catalyst cause a reaction to proceed faster? Only because there are more collisions per second. Only because collisions occur with greater energy. Only because the activation energy is lowered. There are more frequent collisions and they are of greater energy. 38 catalyst can increase the rate of a reaction. y changing the value of the frequency factor () y increasing the overall activation energy (a) of the reaction y lowering the activation energy of the reverse reaction y providing an alternative pathway with a lower activation energy ll of these are ways that a catalyst might act to increase the rate of reaction. 19

20 nzymes nzymes are catalysts in biological systems. The substrate fits into the active site of the enzyme much like a key fits into a lock. substrate ctive site 39 Which of the following acts as catalysts in the body? arbohydrates Lipids Nucleic cids nzymes Water Substrate entering active site of enzyme nzyme/substrate complex nzyme/products complex Products leaving active site of enzyme Liver contains an enzyme called catalase which speeds up the decomposition of H2O2. 20