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Chapter 9 Chemical Kinetics I. The Basic Ideas. ~Thermodynamics is concerned with the direction in which a process will occur , but in itself can tell us nothing about its rate . ~Chemical kinetics is the branch of physical chemistry that deals with the rates of
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Chapter 9 Chemical Kinetics I. The Basic Ideas ~Thermodynamics is concerned with the direction in which a process will occur, but in itself can tell us nothing about its rate. ~Chemical kinetics is the branch of physical chemistry that deals with the rates of chemical reactions and with the factors on which the rates depend. One reason for the importance of this subject is that it provides some of the information needed to arrive at the mechanism of a reaction. ~A vast amount of work has been done on reaction rates. It is now realized that some reactions go in a single step; they are known as elementary reactions and are dealt with in this chapter. ~Other reactions go in more than one step and are said to be composite, stepwise, or complex; such reactions are treated in Chapter 10.
9.1 Rates of Consumption and Formation Kinetic investigations are concerned with rates of change of concentrations of reactants and products. Consider, for example, a reaction Figure 9.1 shows schematically the variations in concentrations of A, B, and Y if we start the kinetic experiment with a mixture of A and B but no Y. At any time t we can draw a tangent to the curve representing the consumption of A, and the rate of consumption of A at that time is As a special case, we may draw a tangent at t = 0, corresponding to the beginning of the reaction; the negative of the slope is the initial rate of consumption of A. Figure 9.1 The rate of formation of Y, vY, is given by In this particular reaction the stoichiometric coefficients are different for the three species, A, B, and Y, and the rates of change of their concentrations are correspondingly different. Thus, the rate of consumption of B, vB, is three times the rate of consumption of A, vA, and the rate of formation of Y, vY, is twice the rate of consumption of A: Individually vA, vB, or vY would be ambiguous as the rate of reaction.
9.2 Rate of Reaction In Section 2.5 we introduced the concept of extent of reaction , and this forms the basis of the definition of rate of reaction. The rate of reaction is now defined as the time derivative of the extent of reaction divided by the volume: If the volume is constant, dni/V can be replaced by the concentration dci This quantity is independent of which reactant or product species is chosen. For a reaction occurring at constant volume the rate of reaction is ~A distinction must be made between v without lettered subscript, meaning reaction, and v with lettered subscript (e.g., vA), meaning rate of consumption or formation. ~Since the stoichiometric coefficients and therefore extents of reaction depend on the way in which the reaction is written (e.g., H2 + Br22HBror1/2H2+1/2Br2HBr), whenever rates of reaction are given the stoichiometricequation must be stated.
9.3 Empirical Rate Equations For some reactions the rate of consumption or formation can be expressed empirically by an equation of the form where kA, , and are independent of concentration and of time Similarly, for a product Z, where kZ is not necessarily the same as kA, When these equations apply, the rate of reaction must also be given by an equation of the same form: In these equations, kA, kZ, and k are not necessarily the same, being related stoichiometric coefficients; thus if the stoichiometric equation is
CH2 CH2 CH CH3 H2C CH2 Order of Reaction The exponent in the previous equations is known as the order of reaction with respect to A and can be referred to as a partial order. Similarly, the partial order is the order with respect to B. These orders are purely experimental quantities and are not necessarily integral. The sum of all the partial orders, + + …, is referred to as the overall order and is usually given the symbol n. A very simple case is when the rate is proportional to the first power of the concentration of a single reactant: Such a reaction is said to be of the first order. An example is the conversion of cyclopropane into propylene: Over a wide range of pressure the rate of this reaction is proportional to the first power of the cyclopropane concentration.
There are many examples of second-order reactions. The reaction is second order in both directions. For this reaction the rate from left to right is proportional to the product of the concentrations of H2 and I2: where k1 is a constant at a given temperature The reaction from left to right is said to be first order in H2 and first order in I2, and its overall order is two. The reverse reaction is also second order; the rate from right to left is proportional to the square of the concentration of hydrogen iodide:
~The rate of a reaction must be proportional to the product of two concentrations [A] and [B] if the reaction simply involves collisions between A and B molecules. ~Similarly, the kinetics must be third order if a reaction proceeds in one stage and involves collisions between three molecules, A, B, and C. ~There are a few reactions of the third order, but reactions of higher order (>3) are unknown. Collisions in which three or more molecules all come together at the same time are very unlikely; reaction will instead proceed more rapidly by a composite mechanism involving two or more elementary processes, each of which is only first or second order. single-step; elementary single-step; elementary
There is no simple connection between the stoichiometric equation for a reaction and the order of the reaction. An example that illustrates this is the decomposition of gaseous ethanal (acetaldehyde), for which the equation is We might think that because there is one molecule on the left-hand side of this equation, the reaction should be first order. In fact, the order isthree-halves: We will see in Section 10.5 that this reaction occurs by a composite free-radical mechanism of a particular type that leads to three-halves-order behavior.
Reactions Having No Order Not all reactions behave in the manner described by the following equation, and the term order should not be used for those that do not. For example, as we will discuss in Section 10.9, reactions catalyzed by enzymes frequently follow a law of the form where V and Km are constants; and [S] is the concentration of the substance as the substrate, that is undergoing catalyzed reaction. This equation does not correspond to a simple order, but under two limiting conditions an order may be signed. first order with respect to S zero order with respect to S
Rate Constants and Rate Coefficients The constant k that appears in rate equations that are special cases of following equation is known as the rate constant or the rate coefficient. reaction is believed to be elementary rate constant reaction is known to occur in more than one stage rate coefficient The units of the rate constant or coefficient vary with the order of the reaction. units of v are mol dm-3 s-1 unit of rate constant k is mol dm-3 s-1 units of v are mol dm-3 s-1 unitsof [A] are mol dm-3 unit of rate constant k is s-1 units of v are mol dm-3 s-1 unitsof [A] are mol dm-3 unit of rate constant k is dm3 mol-1 s-1
Table 9.1 n=0 n=1 n=2 n=2
We have seen that the rate of change of a concentration in general depends on the reactant or product with which we are concerned. The rate constant also reflects this dependence. For example, for the dissociation of ethane into methyl radicals, The rate of formation of methyl radicals, vCH3, is twice the rate of consumption of ethane, vC2H6: The reaction is first order in ethane under certain conditions The rate coefficient for the consumption of ethane, kC2H6, is one-half of that for the appearance of methyl radicals. In cases such as this it is obviously important to specify the species to which a rate constant applies.
Problems 9.1 The stoichiometric equation for the oxidation of bromine ions by hydrogen peroxide in acid solution is Since reaction does not occur in one stage, the rate equation does not correspond to this stoichiometric equation but is a. If the concentration of H2O2 is increased by a factor of 3 by what factor is the rate of consumption of Br- ions increased? b. If the rate of consumption of Br- ions is 7.2 10-3 mol dm-3 s-1, what is the rate of consumption of hydrogen peroxide? What is the rate of formation of bromine? c. What is the effect on the rate constant k of increasing the concentration of bromide ions? d. If by the addition of water to the reaction mixture the total volume were doubled, what would be the effect on the rate of change of the concentration of Br-? What would be the effect on the rate constant k? Solution a. 3 b. Both rates are 3.6 10-3 mol dm-3 s-1 c. None d. Rate of Br- disappearance decreased by a factor of 8; no effect on rate constant
Problems 9.2 A reaction obeys the stoichiometric equation Rates of formation of Z at various concentrations of A and B are as follows: What are and in the rate equation and what is the rate constant k? Solution
Problems 9.9 The reaction is second order in NO and first order in Cl2. In a volume of 2 dm3, 5 mol of nitric oxide and 2 mol of Cl2 were brought together, and the initial rate was 2.410-3 mol dm-3 s-1. What will be the rate when one-half of the chlorine has reacted? Solution
9.4 Analysis of Kinetic Results ~The first task in the kinetic investigation of a chemical reaction is to measure rates under a variety of experimental conditions and to determine how the rate are affected by the concentrations of reactants, products of reaction, and other substances (e.g., inhibitors) that may affect the rate. ~There are two main methods for dealing with such problems; they are the method of integration and the differential method. method of integration good fit compare this with the experimental variation of c with t poor fit differential method This method does not lead to any particular difficulties when there are complexities in the kinetic behavior experimental variation of c with t slopes cannot always be obtained very accurately
Method of Integration A first-order reaction may be of the type Suppose that at the beginning of the reaction (t = 0) the concentration of A is a0 and that of Z is zero. If after time t the concentration of Z is x, that of A is a0-x. The rate of change of the concentration of Z is dx/dt, and thus for a first-order reaction separation of the variables integration where I is the constant of integration x = 0 when t = 0 Figure 9.2
Second-order reactions can be treated in a similar fashion. There are now two possibilities: The rate may be proportional to the product of two equal concentrations. This case must occur when a single reactant is involved, as in the process The rate may be proportional to the product of two different ones.This case may also be found in reactions between two different substances, provided that their initial concentrations are the same. The rate may be expressed as separation of the variables integration where I is the constant of integration x = 0 when t = 0
Graphical methods can again be employed to test this equation and to obtain the rate constant k. Three possible plots are shown in Figure 9.3. Figure 9.3
If the rate is proportional to the concentrations of two different reactants, these concentrations are not initially the same, the integration proceeds differently. Suppose that the initial concentrations are a0 and b0, integration rate after an amount x (per unit volume) has reacted is x = 0 when t = 0 This equation can be tested by plotting the left-hand side against t; if a straight line is obtained, its slope is k. The main disadvantage of the method of integration is that the integrated expressions, giving the variation of x with t, are often quite similar for different types of reactions. For example, the time course of a simple second-order reaction is closely similar to that of a first-order reaction inhibited by products; unless the experiments are done very accurately, there is danger of confusion. We can test for inhibition by-products by measuring the rate after the deliberate introduction of products of reaction.
Problems 9.17 Vaughan reported the following pressure measurements as a function of time for the dimerization of 1,3-butadiene (C4H6) constant volume conditions at 326 C: t/min 3.25 12.18 24.55 42.50 68.0 P/Torr 618.5 584.2 546.8 509.3 474 The initial amount of butadiene taken would have a pressure of 632.0 Torr. Find whether the reaction follows first- or second-order kinetics and evaluate the rate constant. Solution total number of moles at any time t = n0-x t/min 3.25 12.18 24.55 42.50 68.0 PC4H6/Torr 605.0 536.4 461.6 386.6 317.2
Problems 9.19 Equation 9.45 applies to a second-order reaction of stoichiometry A + B Z. Derive the corresponding equation for a second-order reaction of stoichiometry 2A + B Z. Solution x = 0 when t = 0
Problems 9.22 Prove that for two consecutive first-order reactions the rate of formation of C is given by [A]0 is the initial concentration of A. Solution The rate of consumption of A is given by The rate of formation of B and C is given by
integrating factor integrating factor [B] = 0 when t = 0
Half-Life ~For a given reaction the half-life t1/2 of a particular reactant is the time required for its concentration to reach a value that is halfway between its initial and final values. ~The value of the half-life is always inversely proportional to the rate constant and in general depends on reactant concentrations. For a first-order reaction the rate equation is the following equation, and the half-life is obtained by putting x equal to a0/2: In this case, the half-life is independent of the initial concentration. Since, for a first-order reaction, there is only one reactant, the half-life of the reactant can also be called the half-life of the reaction. For a second-order reaction involving a single reactant or two reactants of equal initial concentrations, the rate equation is the following equation, and setting x = a0/2 leads to The half-life is now inversely proportional to the concentration of the reactant.
~In the general case of a reactant of the nth order involving equal initial reactant concentrations a0, the half-life, as seen in Table 9.1, is inversely proportional to a0n-1. ~For reactions of order other than unity the half-life is the same for different reactants only if they are initially present in their stoichiometric ratios; only in that case is it permissible to speak of the half-life of the reaction. ~The order of a reaction can be determined by determining half-lives at two different initial concentrations a1 and a2. The half-lives are related by and n can readily be calculated. This method can give misleading results if the reaction is not of simple order or if there are complications such as inhibition by products.
Table 9.1 n=0 n=1 n=2 n=2 Since the half-lives of all reactions have the same units, they provide a useful way of comparing the rates of reactions of different orders. Rate constants, as we have seen, have different units for different reaction orders. Thus, if two reactions have different orders, we cannot draw conclusions about their relative rates from their rate constants.
Problems 9.6 The rate constant for the reaction H+ + OH- H2O is 1.3 1011 dm3 mol-1 s-1. Calculate the half-life for the neutralization process if (a) [H+] = [OH-] = 10-1 M and (b) [H+] = [OH-] = 10-4 M. Solution For a second-order reaction,
Problems 9.4 A substance decomposes at 600 K with a rate constant of 3.7210-5 s-1. a. Calculate the half-life of the reaction. b. What fraction will remain undecomposed if the substance is heated for 3 h at 600 K? Solution a. b.
Problems 9.20 Derive the integrated rate equation for an irreversible reaction of stoichiometry 2A + B Z. the rate being proportional to [A]2[B] and the reactants present in stoichiometric proportions; take the initial concentration of A as 2a0 and that of B as a0. Obtain an expression for the half-life of the reaction. Solution x = 0 when t = 0
Example 9.1 The half-life of radium, , is 1600 years. How many disintegrations per second would be undergone by 1 g of radium? Solution The half-life in seconds is The decay constant is thus Radioactive disintegrations follow first-order kinetics and therefore have half-lives that are independent of the amount of radioactive substance present. The number of nuclei present in 1 g of radium is The number of disintegrations is therefore
Differential Method ~In the differential method, which was first suggested in 1884 by van't Hoff, the procedure is to determine rates directly by measuring tangents to the experimental concentration- time curves and to introduce these into the equations in their differential forms. ~The theory of the method is as follows. The instantaneous rate of a reaction of the nth order involving only one reacting substance is proportional to the nth power of its concentration: ~A plot of lnv against lna therefore will give a straight line if the reaction is of simple order; the slope is the order n, and the intercept is lnk. However, an accurate value of k cannot be obtained from such a plot.
Problems 9.11 The following results were obtained for the rate of decomposition of acetaldehyde: % decomposed 0 5 10 15 20 25 30 35 40 45 50 Rate/Torr min-1 8.53 7.49 6.74 5.90 5.14 4.69 4.31 3.75 3.11 2.67 2.29 Employ van't Hoff's differential method to obtain the order of reaction. Solution slope=2
Reactions Having No Simple Order Many reactions do not admit to the assignment of an order. Although such reactions can be treated by the method of integration, this procedure is seldom entirely satisfactory, owing to the difficulty of distinguishing between various possibilities. In such cases the best method is usually the differential method; rates of change of concentrations are measured accurately in the initial stages of the reaction, and runs are carried out at a series of initial concentrations. In this way, a plot of rate against concentration can be prepared and the dependence of rate on concentration can then be determined by various methods; one that is frequently used for enzyme reactions will be considered in Section 10.9.
Opposing Reactions One complication is that a reaction may proceed to a state of equilibrium that, differs appreciably from completion. The simplest case is when both forward and reverse reactions are of the first order: If the experiment is started using pure A, of concentration a0, and if after time t the concentration of Z is x, that of A is a0 - x. The rate of reaction 1 if it occurred in isolation is then equal to k1(a0 - x), while the rate of the reverse reaction is k-1x; the net rate of change of concentration of Z is thus If xe is the concentration of Z at equilibrium, when the net rate is zero, The equilibrium constant Kc, equal to xe/(a0 - xe), is equal to the ratio of the rate constants, k1/k-1.
integration x = 0 when t = 0 Figure 9.4 The amount of x present at equilibrium, xe, can be measured directly. If values of x are determined at various values of t, the rate constant k1 can be obtained by graphical means, as shown in Figure 9.4. Since the equilibrium constant Kc = k1/k-1, the rate constant k-1 is easily calculated. Problems 9.23
Problems 9.14 The reaction is first order in both directions. At 25 C the equilibrium constant is 0.16 and the rate constant k1 is 3.3 10-4 s-1. In an experiment starting with the pure cis form, how would it take for half the equilibrium amount of the trans isomer to be formed? Solution
9.5 Techniques for Very Fast Reactions The essential feature of all methods of studying the kinetics of a reaction is to determine the time dependence of concentrations of reactants or products. Some reactions are so fast that special techniques have to be employed in order for this to be possible. There are two reasons why conventional techniques are not suitable for very rapid reactions: 1. The time that it usually takes to mix reactants, or to bring them to a specified temperature, may be significant in comparison with the half-life of the reaction. An appreciable error therefore will be made, since the initial time cannot be determined accurately. 2. The time that it takes to make a measurement of concentration is significant compared with the half-life of the reaction.
Figure 9.5 ~The methods that are used in the study of rapid reactions fall into two classes: flow methods pulse methods ~Some of the more important technique for studying fast reactions are indicated in the lower part of Figure 9.5. ~The upper part of the diagram shows for comparison the half-lives or relaxation times for various chemical and physical processes.
Flow Methods ~The difficulty regarding the mixing of reactants can sometimes be overcome by using special techniques for bringing the reactants very rapidly into the reaction vessel and for mixing them very rapidly. ~With the use of conventional techniques, it takes from several seconds to a minute to bring solutions into a reaction vessel and to have them completely mixed and at the temperature of the surroundings. This time can be greatly reduced by using a flow technique. ~Two gases or two solutions can be introduced into a mixing chamber, and the resulting mixture caused to pass rapidly along a tube. Concentrations of reactants or products can be determined at various positions along the tube, corresponding to various times. ~Reactions having half-lives down to about 10-2 seconds can be studied by such techniques.
Stopped-flow Technique One useful modification is the stopped-flow technique, shown schematically in Figure 9.6. This particular apparatus is designed for the study of a reaction between two substances in solution. Figure 9.6 A solution of one of the substances is maintained initially in the syringe A, and a solution of the other is in syringe B. The plungers of the syringes can be forced down rapidly and a rapid stream of two solutions passes into themixing system. This is designed in such a way that jets of the two solutions impinge on one another and give very rapid mixing; with a suitable design of the mixing chamber it is possible for mixing to be essentially complete in 0.001 s. From this mixing chamber the solution passes at once into the reaction cuvette; alternatively, the two may be combined in the reaction chamber. If a reaction is rapid, it is not possible to carry out chemical analyses at various stages. This difficulty must be resolved by using techniques that allow properties to be determined instantaneously. For reactions in solution, spectrophotometric methods are commonly employed. Fluorescence, electrical conductivity, and optical rotation are also convenient properties to measure in such high-speed studies.
Pulse Methods The flow techniques just described are limited by the speed with which it is possible to mix solutions. There is no difficulty, using optical or other technique, in following the course of a very rapid reaction, but for hydrodynamic reasons it is impossible to mix two solutions in less than about 10-3 s. If the half-life is less than this, the reaction will be largely completed by the time that it takes for mixing to be achieved; any rate measurement made will be of the rate of mixing, not of the of reaction. The neutralization of an acid by a base, that is, the reaction under ordinary conditions has a half-life of 10-6 s or less (see Problem 9.6), and its rate, therefore cannot be measured by any technique involving the mixing of solutions.
~These technical problems were overcome by the development of a group of methods known as pulse methods. ~The first of these to be developed, in 1949, the flash-photolysis methods, are more conveniently dealt with later in Section 10.6. Another class of pulse methods comprises the relaxation methods, developed in the 1950s. The relaxation methods differ fundamentally from conventional kinetic methods in that we start with the system at equilibrium under a given set of conditions. We then change these conditions very rapidly; the system is no longer at equilibrium, and it relaxes to a new state of equilibrium. The speed with which it relaxes can be measured, usually by spectrophotometry, and we can then calculate the rate constants. There are various ways in which the conditions are disturbed. One is by changing the hydrostatic pressure. Another, the most common technique, is to increase the temperature suddenly, usually by the rapid discharge of a capacitor; this method is called the temperature-jump or T-jump method. It is possible to raise the temperature of a tiny cell containing a reaction mixture by a few degrees in less than l0-7 s, which is sufficiently rapid to permit the investigation of many fast processes. Some are too fast for this technique, and then flash photolysis (Section 10.6) must be employed.
The principle of the method is illustrated in Figure 9.7. Suppose that the reaction is of the simple type Figure 9.7 At the initial state of equilibrium, the product Z is at a certain concentration, and it stays at this concentration until the temperature jump occurs, when the concentration changes to another value that will be higher or lower than the initial value according to the sign of Ho for the reaction. Ho<0 From the shape of the curve during the relaxation phase, we can obtain the sum of the rate constants, k1 + k-1, as shown by the following treatment.
Let a0 be the sum of the concentrations of A and Z, and let x be the concentration of Z at any time; the concentration of A is a0-x. The kinetic equation is xe is the concentration of Z at equilibrium, integration x =(x)0 when t = 0 The quantity x thus varies with time in the same manner as does the concentration of a reactant in a first-order reaction.
Figure 9.7 We can define a relaxation timet* as the time corresponding to The relaxation time is thus the time at which the distance from equilibrium is 1/e of the initial distance (see Figure 9.7). If, therefore, we determine t* experimentally for such a system, we can calculate k1 + k-1. However, the ratio k1/k-1 is the equilibrium constant and it can be determined directly; hence, the individual constants k1 and k-1 can be calculated.
Problem 9.24 The dissociation of a weak acid can be represented as The rate constants k1 and k-1 cannot be measured by conventional methods but can be measured by the T-jump technique (Section 9.5). Prove that the relaxation time is given by where the concentration of the ions (Y and Z) is at equilibrium. Solution If a0 is the initial concentration of A, and x is the concentration of ions, At equilibrium
integration x =(x)0 when t = 0 Problem 9.35 and 9.36
Problem 9.37 The equilibrium H2OH++OH- has a relaxation time of about 40 s at 25 C. Find the values of the forward and reverse rate constants. Kw=[H+][OH-]=10-14 M2. Solution
