Reaction Rates and Chemical Kinetics. Factors Affecting Reaction Rate [O 2. CHAPTER 13 Page 1

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1 CHAPTER 13 Page 1 Reaction Rates and Chemical Kinetics Several factors affect the rate at which a reaction occurs. Some reactions are instantaneous while others are extremely slow. Whether a commercial process is economically feasible may depend on its kinetics. Or, for example the rate at which certain compounds can destroy ozone could be of significant environmental significance. Reaction rate is defined as the rate of increase in the molar concentration of the products of a reaction per unit time or the rate of decrease in the molar concentrations of the reactants per unit time. Units of reaction rates are generally given as mol/(l. sec) Factors Affecting Reaction Rate 1. Concentration of reactants. As the population of reactants increase, the likelihood that they will collide increases.. Concentration of a catalyst. A catalyst can provide a good environment for a reaction to occur without actually being consumed by the reaction. 3. Temperature. As temperature increases, molecules can collide with greater energy and frequency. 4. Surface area of a solid reactant or catalyst. Increased surface area of a solid provides the exposure of more reactant molecules. An example of how reaction rate is expressed can be seen in the reaction: N (s) 4 NO (g) + O (g) Even though the "overall" rate of the reaction can be averaged over its entire duration, the rate changes with time and at any given instant. rate of formation of O = Change in Molarity of O [O t instantaneous rate of change Change in time [O [Ο τ time

2 Experimental Determination of Reaction Rate To follow reaction rate, the molar concentration of specific reactants or products must be monitored. This can be accomplished by: 1. titrating a small portion of the reaction mixture at different times during the reaction where a volumetric analysis is appropriate.. monitoring the partial pressures of the participants when reactions occur in the gas phase. 3. measuring the concentrations of reactants or products colorimetrically when appropriate, Dependence of Concentration on Reaction Rate In general, the "Rate Law" for a reaction is of the form:. CHAPTER 13 Page Rate = k [Molarity of Reactants m rate constant... each reaction has its own characteristic rate constant for example, the rate law for the reaction NO (g) + F (g) NO F (g) is Rate = k [NO [F note that both reactants are 1st order...see next section. In this example, doubling the concentration of either reactant will double the rate of the reaction. Note that doubling the concentration of both reactants will cause the reaction to go four times as fast. Reaction Order Sometimes a particular reactant will have a greater effect on the reaction rate..for example, the following reaction NO (g) + H (g) N (g) + H O (g) follows the rate law Rate = k [NO [H Note that doubling the concentration of H will only double the rate (since its concentration is only taken to the 1st power), however, doubling the concentration of NO will cause a fourfold increase in the rate (since its concentration in the rate law is squared). Consequently, this reaction is said to have a "second-order dependence" with respect to NO...but only a "first order dependence" with respect to H. The overall order of the reaction is the sum of the orders of all reactants.. in this case, the overall order is 3. So the power to which a reactant appears in its rate law determines its "order". Fractional, zero, and negative orders are also possible. the power to which the molarity of a particular reactant is raised is referred to as the "order" of the reaction with respect to that particular reactant...this will be discussed in the next section. The practical advantage of knowing the rate law for a given reaction would allow the chemist to predict which reactant to increase in concentration in order to have the greatest effect on its rate.

3 Experimental Determination of a Rate Law In order to determine the rate law of a reaction, it must be carried out several times, varying the initial concentration of only one reactant each time. The change in the initial reaction rate as each reactant's concentration is changed allows you to deduce the "power" to which that reactant is raised (that is..its "order") in the rate law. for example, CHAPTER 13 Page 3 Iodide ion is oxidized in acidic solution to triiodide, I 3-1, by hydrogen peroxide. H O (aq) + 3 I -1 (aq) + H+ (aq) I 3-1 (aq) + H O (l) A series of four experiments was run at different concentrations, and the initial rates of I 3-1 formation were determined (see the table on the next page). From these data we can obtain the reaction orders with respect to H O, I -1, and H +.This will lead us to the rate law for the reaction and allow us to calculate the rate constant. Remember that we can assume the rate law for the reaction above to be of the form Rate = k [H O m [I -1 n [H + p and you determine the reaction orders (exponents m,n, and p) by comparing two rate experiments in which all concentrations of reactants but one are held constant. Initial Concentration (mol/l) Initial Rate H O I -1 H + [mol/(l. s) Exp x 10-6 Exp x 10-6 Exp x 10-6 Exp x 10-6 Comparing experiment 1 and, we see that doubling the concentration of H O (with the other concentrations constant) simply doubled the rate of the reaction. This would imply the the H O term in the rate law must only be to the first power...that is, the reaction is first order with respect to H O. Comparing experiments 1 and 3, we see that doubling the concentration of I -1 (with the other concentrations constant) likewise doubles the rate of the reaction. Therefore the reaction must also be first order with respect to I -1. Comparing experiments 1 and 4 shows that doubling the concentration of H + (with the other concentrations constant) has no effect on the rate...this would imply a zero order dependence of the reaction with respect to H +. From these comparisons we can deduce the rate law to be Rate = k [H O 1 [I -1 1 [H + 0 or Rate = k [H O 1 [I -1 1

4 CHAPTER 13 Page 4 We can now calculate the rate constant of the reaction by substituting values from any of the experiments into the rate law. Using Experiment 1, you obtain 1.15 x 10-6 mol/l. s = k x mol/l x mol/l k = 1.15x10 6 mol L s mol L x 0.010mol L = 1.x10 L/(mol s) Using the Rate Law of an equation, we can derive a formula describing the change of concentration of the reactants with time. This is a useful tool in predicting, say, how long it would take for a reaction to be 50% complete. The derivations in the following sections will show you how to transform a rate law into a mathematical relationship between concentration and time. First Order Rate Law N 4 NO + O The decomposition of N to form NO and O in the reaction above is 1st order with respect to N. That is, its rate law is simply: Rate = k [N But the rate of disappearance of the N can be more simply expressed as: Change in N concentration Rate = = Change in time [N t Combining these two expressions in Calculus terms, we get: d[n dt = k[n To simplify the following derivation, let [N = [A giving us the more generic form: Rearranging this expression, we get d[a dt d[a [A = k[a = k dt If we integrate this expression from time = 0 (where [A = [A ) to time = t (where [A = [A 0 t ), we obtain: [A t t d[a = k [A dt [A o 0 ln[a t ln[a 0 = k(t 0) Since dx x = ln(x) or, we get ln [A t [A 0 = kt

5 CHAPTER 13 Page 5 Since ln (x) =.303 log(x), we can substitute to obtain Moving the constants to one side of the equation gives us the equation we're looking for. log [A t [A 0 = kt log [A t [A 0 = kt This equation describes the concentration of "A" at time, t, given the initial concentration of "A" ([A 0 ) Example...First-order concentration-time equation The rate constant for a given first-order reaction is 6. x 10-4 min -1. Calculate the time requried for the reaction to be 75% complete. Solution When the reaction is 75% complete (which implies that only 5% of "A" remains), then the ratio Substituting this ratio into the first-order concentration-time equation, we get Solving the equation for t Second Order Rate Law k, we get You'll be pleased to know that I'm not going to integrate the following equation for you...just trust me. Consider the reaction a A products And suppose it has the second-order rate law log(0.5) = kt.303 t = (.303)log(0.5) 6.x10 4 An example is the decomposition of nitrogen dioxide. = 36 min Rate = d[a dt = k[a NO (g) NO(g) + O (g) Using calculus you can obtain the following relationship between the concentration of A and the time. 1 [A t = kt + 1 [A 0 [A t [A 0 = 0.5 This the second-order equation relating the concentration of A with respect to time

6 Half-life CHAPTER 13 Page 6 As a reaction proceeds, the concentration of reactants decreases, because it is being consumed. The half-life, t 1/, of a reaction is the time it takes for the reactant concentration to decrease to one-half of its initial value. For a first-order reaction, the half-life is independent of the initial concentration. To illustrate this, we can look at the equation derived in the preceding section for a first -order reaction. log [A t [A 0 = kt.303 In one half-life, the concentration of "A" would be [A t 1 and would equal one-half of the initial concentration, [A 0. This means that the ratio of the concentration of "A" after one half-life to the initial concentration of "A" [A would be 1:. That is, t 1 So, substituting this ratio into the equation above gives us [A 0 = 1 log 1 kt = Solving this equation for t 1 we get: Since the log(1/) = we obtain our final relationship: t 1 = log( 1 ) x.303 k t 1 = k The most important thing to notice here is that, as mentioned before, the half-life of a first-order reaction has nothing to do with the initial concentration of the reactants. It depends only on the rate constant, k, of the reaction and therefore is a constant value. This implies that for a first-order reaction, after one half-life has passed, 1/ of the original reactants remain...after two half-lives have passed, 1/4 of the original remains...and after three half-lives have passed, 1/8 of the original reactants remain...all of this regardless of how much reactant you initially had. So, if you know the rate constant for a given first-order reaction, its a simple matter of using the equation above to determine its half-life. Example...Half-life Suppose the rate constant for a given first-order reaction is 1. x 10-3 sec -1. Calculate the half-life of the reaction. Solution Since t 1 = k we can simply substitute the given value for k and obtain t 1 = x10 3 sec 1 = sec

7 CHAPTER 13 Page 7 Half-Life of a Second-Order Reaction Using similar reasoning, we can show that the half-life of a second-order rate law, Rate = k [A,is described by the following expression. t 1 = 1 k [A 0 Note, that in this case, the half-life at any given point of a second order reaction depends on the concentration of "A" at that initial starting point. This means that the half-life of a second order reaction increases as the reaction proceeds. From the chemist's standpoint, the realization that the half-life of a given reaction changes with time is evidence that it is not first-order. Graphing Kinetic Data An alternative method for determining the order of a reaction (rather than running several experiments at various initial concentrations of reactants) is to graphically plot the data for a particular reaction. If we rearrange our equation derived for a first-order reaction, we obtain: log[a t = ( k.303 ) t + log[a 0 y = m x + b (General equation for a straight line) A comparison of these two equations illustrates that if we plot log[a t vs. t for a first-order reaction, we should obtain a straight line. (If the line is not straight, then this is evidence that your reaction is not first-order) You could then obtain the rate constant for the reaction from the slope of the line, since m = (-k/.303). Using the second-order rate law, we can also derive a linear relationship between concentration and time. (See the section on second-order rate law) In this case, you get a straight line if you plot 1 = k t + 1 [A t [A 0 y = m x + b 1 [A t vs. time. So, in summary, one can determine if a reaction is first or second order by simply plotting the experimental data as log[a t vs. time or 1/[A t vs. time and see which one gives you a linear relationship. the table below summarizes the relationships discussed for first-order and second-order reactions. Order Rate Law Concentration-time Equation Half-Life Graphical Plot 1 Rate = k [A log[a t = ( k ) t + log[a /k log[a vs. t Rate = k[a 1 = k t /(k[A [A t [A 0 ) 0 [A t vs. t

8 CHAPTER 13 Page 8 Temperature and Rate: Collision and Transition State Theories Rate constants vary with temperature and consequently the actually rate of a reaction is temperature dependent. Why the rate depends on temperature can be explained by collision theory. Collision Theory This theory assumes that for a reaction to occur, reactant molecules must collide with sufficient energy to react and with the proper orientation. The minimum energy required for a successful collision is called the Activation Energy, E a. Collision theory maintains that the rate constant for a reaction is the product of three factors: (1) Z, the collision frequency, () f, the fraction of collisions having sufficient energy to react, and (3) p, the fraction of collisions with the proper orientation to react. k = Z f p The factor, Z, is slightly temperature dependent and can be shown using the Kinetic Theory of gases which shows the relationship between the velocity of gas molecules and their absolute temperature. velocity = 3RT abs M m so Velocity is proportional to T abs This alone, however, does not explain the observed increases in the rates of reactions with only small increases in temperature. From Kinetic Theory, it can be shown that a 10 o C rise in temperature will only produce a % increase in the collision frequency. This does not account for the much greater increase in reaction rate that is observed with this temperature increase. So the collision frequency varies only slowly with temperature, however, f, the fraction of molecules with sufficient activation energy turns out to be very temperature dependent. It can be shown that f is related to E a by the following expression. f = e E a/rt Here e = , and R is the gas constant, which equals 8.31 J/(mol K) From the relationship above, one can see that as Temperature increases, the exponent in the above expression becomes less negative resulting in a much larger fraction of successful collisions. One can also see that if E a decreases, this also causes an increase in successful collisions. The reaction rate also depends on p, the proper orientation of the reactants when they collide. This factor is independent of temperature changes.

9 CHAPTER 13 Page 9 Transition State Theory Transition-State Theory explains the reaction resulting from the collision of two molecules in terms of an activated complex. An Activated complex (transition state) is an unstable grouping of atoms that can break up to form products. A simple analogy of transition state theory would be the collision of three billiard balls on a pool table. Suppose two of the spheres are coated with a slightly sticky adhesive and are stuck together. If we take a third sphere covered with an extremely sticky adhesive and collide it with our joined pair, it would likely stick to one of the joined spheres and provide sufficient energy to dislodge the other resulting in a new "pairing". At the instant of impact, when all three spheres are joined, we have an unstable transition state complex. (Unstable in the respect that energy is being transferred to the sphere that is leaving.) If we repeated this scenario several times we would find that some collisions were successful resulting a new "product" whereas some of the collisions would be unsuccessful (because of either insufficient energy transmission or improper orientation) returning us to our original "reactants". We could liken the energy that we provided to produce the collision as the activation energy, E a. Potential Energy Diagrams for Reactions REFER TO FIGURES AND IN YOUR TEXT FOR THE FOLLOWING DISCUSSION. To illustrate graphically the energy involved in the formation of a transition state,we can plot the potential energy of the reaction vs. time. Let's first look at a graph where the products are less stable than the reactants, that is, an reaction. Note that the forward activation energy is the energy necessary to form the activated complex (transition state). The H of the reaction is the net change in energy between reactants and products. In the case of exothermic reactions, where the products are more stable than the reactants, we see again that the activation energy is the energy required to form the activated complex (transition state). In this case the H has a negative value as the change in energy from reactants to products is negative. Note that in both of these graphs, the reverse reaction still must provide enough activation energy to form the activated complex (transition state). The Arrhenius Equation A graphical method for the experimental determination of Activation Energy We saw in the section on collision theory that the rate constant, k, for a reaction was the product of three factors: k = Z f p In that same section, it was shown that f, the fraction of molecules colliding with sufficient Energy of Activation could be expressed as: f = e E a/rt We also noted in that section that Z, the frequency of successful collisions was only slightly dependent on temperature, while p, the orientation factor was not temperature dependent at all. So, for all practical purposes, Z and p are constant.

10 CHAPTER 13 Page 10 If we were to combine the constant terms Z and p into one constant..let's call it A, we obtain the Arrhenius Equation. k=ae E a/rt The Arrhenius equation expresses the dependence of the rate constant on absolute temperature and Activation Energy. It is useful to recast the Arrhenius Equation in logarithmic form. Taking the natural logarithm of both sides of the equation we get: Or, in terms of logarithms to the base 10: If we relate this equation to the (somewhat rearranged) general formula for a straight line: logk=loga E a.303r (1) T y = b + m x This shows that a plot of log k vs. (1/T) should yield a straight line. The slope of this line would be -E a /(.303R), from which one could obtain the activation energy, E a. The intercept is log A. A more useful way to use this equation would be to look at two points on the line it describes, that is (k 1, (1/T 1 )) and (k, (1/T )). The two equations describing the relationships at each coordinate would be: logk 1 =loga and We can eliminate log A by subtracting the two equations to obtain: With this form of the equation, given the activation energy and the rate constant k 1 at a given temperature T 1, we can calculate the rate constant k at any other temperature T. Reaction Mechanisms Even though a balanced chemical equation may give the ultimate result of a reaction, what actually happens in the reaction may take place in several steps. These steps describe what is called the reaction mechanism. These individual steps in a larger reaction are referred to as elementary reactions. Elementary Reactions E a.303r ( 1 T 1 ) lnk=lna E a RT logk=loga Consider the reaction of nitrogen dioxide with carbon monoxide. E a.303rt logk =loga log k k 1 = E a.303r ( 1 T ) E a.303r ( 1 T 1 1 T ) NO (g) + CO (g) NO (g) + CO (g) This reaction is believed to take place in two steps: NO + NO NO 3 + NO (elementary reaction) NO 3 + CO NO + CO (elementary reaction)

11 CHAPTER 13 Page 11 Each step, called an elementary reaction, is a singular molecular event resulting ion the formation of products. The set of elementary reactions whose overall effect is given by the net chemical equation is called the reaction mechanism. Note that NO 3 does not appear in the overall net equation, but is formed as a "temporary" reaction intermediate. The overall chemical equation, which represents the net results of these two elementary reactions, is obtained by adding the two steps together and canceling any species common to both sides of the equation. NO + NO NO 3 + NO NO 3 + CO NO + CO Molecularity NO + NO + NO 3 + CO NO 3 + NO + NO + CO We can classify elementary reactions according to what is called their molecularity, that is the number of molecules on that must collide for the elementary reaction to occur. For example, a unimolecular reaction involves only one reactant molecule. A bimolecular reaction involves the collision of two reactant molecules and a termolecular reaction involves the collision of three reactant molecules. Higher molecularities are rare because of the small statistical probability that four or more molecules would all collide at the same instant. Rate Equations for Elementary Reactions. Since an chemical reaction may occur in multiple steps, there is no easy relationship between its overall equation and its rate law. However, for elementary reactions there is a somewhat simple relationship. For an elementary reaction, the rate is proportional to the concentrations of all reactant molecules involved in the reaction. For example, consider the generic equation A B + C The rate is dependent only on the concentration of A, that is, Rate = k[a. However, for the reaction A + B C + D the rate is dependent on both the concentration of A and B... Rate = k[a[b For a termolecular equation such as A + B + C D + E the rate is dependent on the populations of all three participants.. Rate = k[a[b[c

12 CHAPTER 13 Page 1 Note that if two molecules of a given reactant are required, it appear twice in the rate law. For example, the reaction A + B C would have the rate law Rate = k[a[a[b or Rate = [A [B So, in essence, for an elementary reaction, the coefficient of each reactant becomes the power to which it is raised in the rate law for that reaction. But since any overall reaction you look at is likely to have several elementary steps, the overall rate law would be determined by the combination of these steps making it impossible to predict the overall rate law by simply looking at the overall reaction. Rate Laws and Mechanisms Consider the reaction NO (g) + F (g) NO F (g) Experiments performed with this reaction show that its rate law is Rate = k[no [F that is, it appears to be only first-order with respect to each of the reactants even though the coefficient for NO in the overall reaction is two. This implies that the above reaction is not elementary, but instead must occur in more than one step in order result in the observed rate law. Rate Determining Step. In multiple step reactions, it is often the case that one of the elementary reactions in the sequence of steps is slower than the others. The overall reaction then cannot proceed any faster than this slowest, rate-determining step. Our previous example NO (g) + F (g) NO F (g) can be shown to occur in two elementary steps where the first step is much slower than the second. By adding the two steps together, you can see that this is the case. NO + F NO F + F (slow) NO + F NO F (fast) NO + F NO F Since the overall rate of this reaction is determined by the first elementary reaction above it seems logical then why the observed rate law is Rate = k [NO [F So in a mechanism where the first elementary step is the slow one (the rate determining step), the overall rate law is simply written as the elementary rate law for that slow step. A more complicated scenario occurs when the rate determining step is in the middle of the series as will be outlined in the next section.

13 CHAPTER 13 Page 13 Mechanisms with an Initial Fast Step There are cases where the rate determining step of a mechanism contains an intermediate that does not appear in the overall equation. The experimental rate law can only be written, however, in terms of substances that appear in the overall equation. Consider the decomposition of dinitrogen pentoxide. N (g) 4 NO (g) + O (g) This reaction is believed to follow this type of mechanism. N NO + NO 3 (fast, equilibrium) NO + NO 3 NO + NO + O (slow) NO 3 + NO NO (fast) It has been determined experimentally that the rate law is: k 1 k -1k k 3 Rate = k[n Let's show that the mechanism outlined above is consistent with the observed rate law. From the last section, we saw that the rate determining step (Step in this case) generally outlines the rate law of the overall equation, which in this case would be: Rate = k [NO [NO 3 (Rate law for the rate determining step) But, as mentioned before, the experimental rate law can only be written in terms of substances contained in the overall equation and cannot contain reaction intermediates. So, it is necessary to re-express the equation eliminating [NO 3. Looking at the first step, we see that it is fast and reversible. The rate of the forward decomposition is defined by the rate law. Rate = k 1 [N and the rate of the reverse reaction (the formation of N from NO and NO 3 ) is defined as Rate = k -1 [NO [NO 3 at the beginning of the reaction there is no NO or NO 3, but as they begin to form, the reverse reaction begins to pick up speed until finally, an equilibrium is established where the forward and reverse reactions occur at the same rate. Because this first elementary reaction is much faster than the second step, this equilibrium is reached before any significant reaction by the second step occurs. Since at equilibrium, the forward and reverse rates are equal for the first step, we can write: k 1 [N = k -1 [NO 3 [NO or, [NO 3 = k 1 [N k -1 [NO Substituting this expression into the rate law for the rate determining step, we get k 1 [N Rate = k [NO [NO 3 = k [NO x or, Rate = k [N = "k"[n k -1 [NO So, if we identify k k 1 /k -1 as "k", we've reproduced the experimental rate law. k 1 k -1

14 CHAPTER 13 Page 14 Catalysis A catalyst is a substance that provides a good "environment" for a reaction to occur thereby increasing the reaction rate without being consumed by the reaction. The reason a catalyst can affect the rate of a reaction and yet not be consumed is that the catalyst must participate in at least one step of the reaction and be regenerated in a later step. Its presence increases the rate of the reaction by either increasing the frequency factor, A (from the Arrhenius Equation) or more often by lowering the activation energy, E a, required for a successful reaction. Homogeneous Catalysis Homogeneous catalysis is the use of a catalyst in the same phase as the reacting species. Heterogeneous Catalysis Heterogeneous catalysis is the use of a catalyst that exists in a different phase from the reacting species, usually a solid catalyst in contact with a gaseous or liquid solution of reactants. Chapter 13 Assignments The following problems will be collected at the first exam. Review Questions: 1,6,7,8,11,16,0 Problems: 5,37,39,43,45,51,53,57,63,65,69,75,77,79,85,87,93,97,99,109 Exam Review Topics terms Catalyst Reaction rate Rate constant Reaction order Half-life Activation energy Activated complex Reaction mechanism Molecularity Rate determining step skills/operations Calculating average reaction rate Determining reaction order from Rate Law Determining Rate Law from initial rates Concentration-time equation for first-order reactions Relating t1/ to rate constant Find overall equation from a mechanism Write rate equation for an elementary reaction Determining Rate Law from a mechanism

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