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(1)UNIVERSITY OF CAMBRIDGE DEPARTMENT OF CHEMISTRY IA CHEMISTRY 2002/3. Kinetics of Chemical Reactions. potential energy. alternative products. reagents products. Lecturer: Dr James Keeler EMAIL: [email protected].

(2) 1. Chemical kinetics Kinetics is the study of the rates of chemical processes in an effort to understand what it is that influences these rates and to develop theories which can be used to predict them. A knowledge of reaction rates has many practical applications, for example in designing an industrial process, in understanding the complex dynamics of the atmosphere and in understanding the intricate interplay of the chemical reactions that are the basis of life. At a more fundamental level we want to understand what happens to the molecules in a chemical reaction – that is what happens in a single reactive encounter between two reagent molecules. By understanding this we may be able to develop theories that can be used to predict the outcome and rate of reactions. 1.1 Books Any general physical chemistry text (such as P W Atkins Physical Chemistry, OUP, any edition) will have several chapters on chemical kinetics. There is a nice small book about kinetics in the Oxford Chemistry Primers series: B G Cox Modern Liquid Phase Kinetics, OUP, 1994. For more detail (beyond the scope of this course), Reaction Kinetics by M J Pilling and P W Seakins (OUP 1995) is a good source.. 2. Rates, rate laws and rate constants. change in concentration, ∆c,in time ∆t ∆t We can talk about the rate of formation or loss of any species – reactant, intermediate or product. It is, however, important to specify which species we are talking about. The rate can be positive or negative: a positive rate means that the concentration is increasing with time e.g. a product; a negative rate means that the concentration is falling with time e.g. a reactant. The rate may vary with time (and concentration), so it is usual to define the rate over a very small time, ∆t. We think of the rate as the derivative of concentration with respect to time d[concentration] dt This derivative is the slope of a graph of concentration against time, taken at a particular time. From its definition it is clear that the units of a rate are concentration per rate =. 1. ∆c ∆t time, t. concentration. 2.1 Rate of reaction The rate is defined as. concentration, c. In this Section we will introduce the language and terms used to describe the rates of chemical reactions. At this stage we will not be concerned with the theory of reactions or mechanisms, but just stand back and describe the overall rates.. time The rate is the instantaneous slope, and this varies with time..

(3) Example 1 Exercise 1. unit time, for example mol dm–3 s–1 . There are other measures of concentration, for example in the gas phase pressure is proportional to concentration, so a rate can be expressed in torr min–1 (1 torr = 1 mm Hg, a measure of pressure). It is also common to express concentration not in moles per unit volume but in molecules per unit volume, so the rate would be expressed in molecules dm–3 s–1 or molecules cm–3 s–1 . Symbolically, concentration is often indicated by square brackets. So [A] means the concentration of species A, and [Br2] means the concentration of bromine, and so on. 2.1.1 Relation to stoichiometric equations Consider the reaction whose stoichiometric equation is → 2H 2 O O 2 + 2H 2  The stoichiometric equation shows how the number of moles of reactants and products are related; it must be balanced. This equation says that to form two moles of water, one mole of oxygen and two moles of hydrogen must react. It follows that the rate of consumption of hydrogen is twice that of oxygen. d[H 2 ] d[O 2 ] =2 dt dt The rate of formation of water is twice the rate of loss of oxygen, as two moles of water are formed from one mole of oxygen. As water is the product, the rate of change of its concentration is positive i.e. the concentration is increasing with time; however, the rates of change of the concentrations of both hydrogen and oxygen are negative as these concentrations are decreasing with time. In derivative form:. A warning: not everyone or every book is careful about using this convention, and you will find that the "rate" is often used in a loose sense, rather than being defined in the precise way given here. Be prepared to be flexible.. d[H 2 O] d[O 2 ] 1 d[H 2 O] d[O 2 ] or, equivalently – = −2 = dt dt 2 dt dt With all these different "rates", different for each species, how can we define the "rate of the reaction"? The usual procedure is to include the stoichiometric coefficients in the definition of the rate. These coefficients are the numbers in front of the species in the balanced chemical equation. So the stoichiometric coefficient is 1 for oxygen and 2 for both hydrogen and water. The rate of reaction, r, is defined for any species A as r=. 1 d[A ] ν A dt. [1]. where νA is the stoichiometric coefficient of species A in the balanced chemical equation. In addition the stoichiometric coefficients are given negative signs for reagents and positive signs for products. This definition ensures that the rate is always positive and the same for a given reaction no matter which species is considered.. Example 2 Exercise 2. 2.

(4) 2.2 Rate laws and rate constants Experimentally it is found that rates depend on the concentrations of the species involved in the reaction equation (and sometimes on the concentrations of species which do not appear at first sight to be involved!). The relation between the rate and these concentrations can often be expressed mathematically in the form of an equation called a rate law. Some rate laws are very simple and some are very complicated. A rate law may be determined experimentally (Section 4) or may be the result of a theoretical prediction, or both. Often reactions are found to have rate laws of the form r = k[A ] [B] K a. b. where k is constant, characteristic of a particular reaction, called the rate constant or the rate coefficient. The powers a, b ... are also constants: a is the order with respect to A, b is the order with respect to B. The overall order is (a + b + ...). Orders are commonly integers, but they do not have to be. For example, the solution reaction N. N Ph. Br. N+. N. Br–. Ph PhCH2Br. DABCO. is found to have the rate law r = k[ PhCH 2 Br ][DABCO] The reaction is second order overall, and first order with respect to both PhCH2 Br and DABCO. Some reactions have very much more complex rate laws. For example, in the gas phase, the reaction H 2 + Br2  → 2HBr has the rate law (Section 6.2) ka [H 2 ][Br2 ] 2 r= [Br2 ] + kb [HBr ] 3. For this reaction, no order with respect to Br2, and no overall order, can be defined. An enzyme, E, catalysing the conversion of a substrate, S, to products is often found to obey the rate law (Section 5.5.4) r=. kcat [E]0 [S] [S] + KM. Again, no order with respect to S can be defined. 2.2.1 Simple rate laws A first order rate law is one in which the rate is proportional to the 3.

(5) concentration raised to the power 1 (hence "first") r = k1st [A ]. 1. k1st is called a first order rate constant. The rate equation, and the reaction it describes, is said to be first order in A or it is said that the order with respect to A is one. As the units of r are concentration time–1 and the units of [A] are concentration, the units of k1st are found as r = k1st [A ]. 1. so k1st =. r conc. time –1 = = time –1 conc. [A]. –1 Hence first order rate constants have units of time . A second order reaction has the concentration raised to the power of 2. r = k2 nd [A ]. 2. k2nd is called a second order rate constant. The rate equation, and the reaction it describes, is said to be second order in A, or the order with respect to A is two. The units of k2nd are found to be conc.–1 time–1 . Example 3 Exercises 3 & 4. 2.2.2 Importance of rate laws 1. If we know the rate law and the constants in it we can use this to predict the rate for any set of conditions (concentrations). The rate law is thus a very succinct and practical way of expressing the rate. You might use this, for example, in a model of the atmosphere or in predicting the rate of an enzyme catalysed reaction. 2. The form of the rate law can tell us something about the mechanism of the reaction. This is a point which we will consider in more detail below. 3. Knowing the rate law enables us to separate the concentration dependence from the underlying, fundamental effect which is the size of the rate constant. 2.3 Effect of temperature Rate constants, and hence reaction rates, and are often found to depend strongly on temperature. It is therefore important to quote the temperature at which any rate constant is determined. Most commonly the rate goes up with temperature, but this is not always the case. For example the rate constant for the reaction 2NO + O2 → 2NO2 decreases with increasing temperature. Experimentally, a very large number of rate constants are found, to some level of approximation, to vary with temperature according to the Arrhenius equation: k (T ) = A exp. E a/ RT must be dimensionless for us to be able to take the exponential of it..  − Ea   RT . [2]. where we have written the rate constant as k(T) to emphasize that it depends on temperature. R is the gas constant, which has dimensions energy temperature–1 mol–1, so the product RT has dimensions of energy mol–1. Ea has units of energy mol–1 and is called the activation energy; it is usually expressed in kJ mol–1. Typical values for Ea are between 10 and 200 kJ mol–1 , for example: 4.

(6) 2N2 O5 (g) → 4NO2 (g) + O2 (g) EtO– + MeI → EtOMe + I– (EtOH solution) Cl(g) + O3 (g) → ClO(g) + O2 (g) CH(g) + CH4 (g) → C2 H4 (g) + H(g). Ea Ea Ea Ea. = 160 kJ mol–1 = 82 kJ mol–1 = 2.1 kJ mol–1 = –1.7 kJ mol–1. A is called the A factor or the pre-exponential factor. It is often found to be independent of temperature, or at least only weakly dependent on temperature. Since the exponential term is dimensionless, A must have the same dimensions and units as k. The Arrhenius equation predicts that, for a positive activation energy, the rate constant increases with temperature. The equation can be manipulated into a straight line form by taking natural logarithms of both sides:. increasing Ea k.  − Ea   RT . 0 200. E  1 ln k (T ) = ln A − a R T. of the form y = mx + c. So a plot of ln k against 1/T should be a straight line; such a plot is called an Arrhenius plot. The slope of the plot is –Ea /R (dimensions temperature, e.g. K), and the intercept with the vertical axis, when 1/T goes to zero, is ln A. We see that A can be identified as the rate constant at infinite temperature (or, more precisely, when RT >> Ea ) i.e. the maximum possible rate constant. In Section 3 we will see what the significance of A and E a are at the molecular level. It should be noted that many physical and chemical processes show approximate Arrhenius behaviour, even if the processes themselves turn out to be rather complex. This is partly because experimental data is often not of sufficient quality to detect minor deviations from the Arrhenius equation. 2.4 Relation to equilibrium constants Chemical equilibrium is a dynamic state, not one in which all reaction has stopped. The forward and back reactions continue just as they do when a system is away from equilibrium, but when equilibrium has been reached these forward and backward rates are equal. As a result, there is no change in the concentration of any of the species, even though the reactions are still going on. Suppose we have a reaction of the form C+D. A+B. Also suppose that rate of the forward reaction is kf [A][B] and that the rate of the back reaction is kr [C][D] (these are assumptions, we cannot say from the stoichiometric equation that these will be the rate laws). At equilibrium these two rates are equal kf [A ]eq [B]eq = kr [C]eq [D]eq where the "eq" subscripts have been added to emphasize that this relation is 5. T. The rate constant, k , plotted as a function of temperature for three values of the activation energy in the ratio 1:5:10. Note that the temperature scale does not start at zero.. ln A slope = –Ea/R ln k. k (T ) = A exp. A1. 0 An Arrhenius plot Exercise 5. 1/T.

(7) only true at equilibrium. Rearranging gives kf [C]eq [D]eq = kr [A ]eq [B]eq The quantity of the right is the equilibrium constant, Keq, so it follows that Keq =. kf kr. The equilibrium constant is therefore the ratio of the rate constants of the forward and backward reactions.. 3. Theories about reaction rates We would like to be able to develop a theory which will enable us to calculate – almost certainly with the aid of a computer – the rate constant of any reaction we care to specify. To develop such a theory will be first and foremost a test of our understanding of the fundamental processes involved. If the theory produces predictions which are in accord with experiment, then we may have some faith in the theory. Then, if the theory proves to be successful and our calculations reliable, we could go on to use the theory to predict the rates of unknown or littlestudied reactions. We might want to do this to avoid experiments, which are not always easy or possible. It turns out that although the theory is now quite well understood the calculations needed to predict values of rate constants are very challenging. With the currently available super-computers it is feasible to calculate rates for simple gas phase reactions (e.g. F + H2 → HF + H and H2 + OH → H2 O + H). Reactions in solution present an even greater challenge as we have to consider the role of a large number of solvent molecules. Even though we will not be able to actually make any calculations of rate constants ourselves, a great deal of insight into chemical reactions is obtained by studying the model on which the theory is based and looking at the general features we expect from such a model. This will lead to an interpretation of why rate laws have the form they do, the factors that influence the size of the rate constant, and how the Arrhenius equation arises. 3.1 What happens in a reaction? Suppose that two molecules A and B are going to react. When they are a large distance apart we imagine that each molecule is largely independent of the other. In A the atoms are held together in a certain geometry by electrons which we imagine as occupying MOs. As A and B approach one another, the electrons and nuclei from one molecule start to interact with those from the other. We usually describe this by saying that the MOs begin to interact and, as we saw in Reactions and Mechanisms in Organic Chemistry course, often the strongest interaction is that between the HOMO of one molecule and the LUMO of another. As a result of this interaction, bonds begin to be broken and new bonds 6.

(8) begin to be formed; new MOs are generated. As the rearrangement progresses further, the new molecules (the products) begin to gain their identity and move apart. The atoms and electrons (and MOs) can now be identified as belonging to one molecule or another; the reaction is complete. An example of this process is the familiar SN2 reaction: Nu. X. Nu. X. Nu. X products. HOMO/LUMO interaction. Chemical reactions are therefore seen at a fundamental level to involve electronic rearrangements and orbital interactions. The theory needed to understand them is the same theory needed to study the electronic structure and bonding in molecules i.e. quantum mechanics, in the form of MO theory. 3.1.1 Potential energy surfaces Using MO theory, we can calculate (at least in principle) the energy of a molecule, or come to that any arbitrary arrangement of nuclei. All that we need to do is to find the energies of the MOs, assign the electrons to these, and hence compute the overall electronic energy the system. As two molecules A and B come together and start to interact, we can imagine computing the total energy of the system. When A and B are far apart, the energy will simply be the sum of the energies of A and B. However, as they approach the energy will depend on the details of how the atoms and orbitals in A and B are interacting; eventually we will reach the transition state whose energy we can also calculate. Finally, as the product molecules move apart, the energy will tend to the sum of the energies of the products. So, we can imagine that the total energy of the system varies continuously throughout the reaction. The problem is that the energy will depend on the precise arrangement of atoms, and if A and B have any complexity at all there will be an almost mind-boggling different number of ways in which the two molecules can approach one another and interact. In principle we can calculate the energy of each of these arrangements of the atoms in A and B as they interact. What we end up with is the idea of a potential energy surface (PE surface or PES) which gives the energy (the potential energy) as a function of the positions of all the atoms in the system. For all but the simplest molecules, such a surface is a function of a very large number of distances and angles. It cannot simply be visualized as energy as a function of the x- and y-coordinates; rather it is multidimensional. Difficult though it may be a visualize or draw, the PE surface is key to our understanding of reactions. We imagine the atoms as "moving" on this potential energy surface. They start out at one position which corresponds to reactants, move along some path over the surface as they rearrange 7.

(9) themselves, and then end up at a position which corresponds to products.. potential energy. alternative products. reagents products. Note that the stable molecules, the products and reactants, exist in potential energy minima. There may be several such minima on the surface. For example, the reaction between H and ClBr may give HCl + Br or HBr + Cl; the alternative products will correspond to different minima on the surface. 3.1.2 The lowest energy pathway and the transition state As the reactants and products are in potential energy minima, it will always be the case that extra energy has to be put in order to traverse the path between them. The amount of extra energy will depend on the details of the potential energy surface and the path taken. Of the many possible paths which lead from reactants to products, the one which involves the least expenditure of energy turns out to be the one which is most favoured. The point of highest energy on this pathway is called the transition state; it corresponds to an arrangement of the atoms of the reacting molecules in which there are partially made and partially broken bonds. Just exactly where the transition state is depends on the details of the PE surface. Why is the lowest energy pathway the one which the reaction takes? The answer lies in understanding the way in which energy is distributed amongst the molecules. To move from reactants to products by the minimum energy pathway still needs a certain amount of energy; to be precise the extra energy, E, needed is the difference between the energy of the transition state, and the energy of the reactants. Only a fraction of the molecules will have this extra energy, and this fraction is, according to the Boltzmann distribution, exp.  −E   RT . As this fraction is an exponential function of the energy, increasing E by just a small amount (1 or 2 kJ mol–1) reduces greatly the number of molecules which have the required extra energy. Thus, very few molecules are able to take pathways which are higher in energy than the lowest energy pathway and so, for all practical purposes, we can safely assume that all the reactants 8.

(10) which go to products do so via this minimum energy route. 3.1.3 The transition state The transition state is not a molecule in the conventional sense. Molecules exist in potential energy minima; this means then when they distort there is a force which pushes them back to their equilibrium geometry. The PE surface at the transition state is a minima in some directions, but along the path that leads from reactants to products it is at a maximum. This means that when the molecule distorts along this path there is no restoring force and the distortion continues; the transition state then "falls apart" either to products or back to reactants. As it exists at a PE maximum the transition state only has a fleeting existence, perhaps of the order of a the period of a molecular vibration, ~ 10–13 s. The recent advent of lasers capable of producing femto-second pulses has made it possible – just – to study something as short-lived as a transition state. The calculations which are needed to map out the PES are time consuming, even for the most powerful computers. A number of surfaces have been mapped out in great detail, including those for the reaction between H + H2 , F + H2 and H2 + OH → H2 O + H. It is essential to distinguish the transition state from an intermediate. An intermediate is a molecule like any other, existing in a PE minimum, and detectable by conventional means; it may be highly energetic compared to reactants and products, but it is not unstable in the sense that a transition state is.. potential energy. 3.1.4 The reaction coordinate and the reaction profile The minimum energy pathway from reactants to products is called the reaction coordinate. Moving along this pathway involves changing the positions of all of the atoms involved in the reaction in a complex way, so the "coordinate" is in fact a composite of many motions. We can plot the energy against the reaction coordinate to give a plot which is know as the reaction profile, which you have encountered before. transition state. Ea. products. reactants. reaction coordinate. The transition state exists at the potential energy maxima, and the difference in energy between this maxima and the reactants is called the activation 9. stable molecule. transition state.

(11) energy, Ea . The fraction of molecules coming together with sufficient energy to reach the transition state is exp(–Ea /RT). For a typical value of the activation energy of 50 kJ mol–1 , this means that only about 1 in 109 of the molecules which attempt to reach the transition state actually have enough energy to make it. We can imagine the molecules all queuing up like eager climbers at Everest base camp (the reactants); many set off towards the summit (the transition state), but few have the energy to make it, and so slide back down the slope. As we commented before, actually calculating the form of the potential energy surface is a challenging task. However, this discussion has allowed us to develop an understanding of the origin of the energy barrier to reaction, and to introduce the important concepts of the transition state and the reaction profile. 3.2 Collision theory Although we have not said so explicitly, we have assumed that reactions take place when molecules come together in a collision, and we have also assumed that the energy needed to overcome the activation barrier comes from the energy of the collision. This leads to the idea that we might be able to understand something about the rates of reactions by analysing molecular collisions. In its simplest form collision theory is rather straightforward to develop and apply but, as we shall see, the values of the rate constants that it predicts do not compare well with experiment. Nevertheless, the simple theory does aid our understanding of the nature of reactions at a microscopic level. 3.2.1 Gas kinetic theory This is a simple theory about the motions and energies of the particles (molecules and atoms) which make up gases. The theory is based entirely on classical mechanics (i.e. Newton's Laws). The theory treats the particles as objects whose size is much less than the typical distance between them. These particles have kinetic energy which causes them to move around in random directions, colliding with one another and with the walls of the container. All collisions are assumed to be elastic, which means that energy is conserved. With these constraints, it can be shown that for a macroscopic sample the energies and speeds of the particles are distributed according to the Maxwell distribution. All speeds are possible, but very high and very low speeds are very improbable. The mean speed, c , is given by 1. 8k T 2 c = B   πm  where kB is Boltzmann's constant, T is the temperature and m is the mass of the particle (in SI units, the mass must be in kg). When considering molecular collisions it turns out that we need to know the mean relative speed, crel , of two molecules, A and B; this makes sense 10.

(12) as collisions just depend on the relative motion of the two molecules. This mean relative speed is 1.  8kBT  2 crel =    πµ  µ is the reduced mass of A and B calculated using this formula; do not confuse this with the reduced mass of a diatomic molecule.. where µ is the reduced mass, given by  m m  µ= A B   mA + mB . Example 4. 3.2.2 Calculation of the collision rate If we want to calculate the rate of a reaction we need to calculate the rate at which the reactant molecules collide. However, we know that the orientation of molecules is important for reaction, so just calculating the number of collisions will not be sufficient. For example an SN2 reaction requires that the nucleophile attacks from the rear; an attack from the front is much less likely to be successful. Nu–. X–. Nu. X. In its simplest form, gas kinetic theory cannot cope with this subtlety; we simply have to assume that the molecules are structureless spheres. We will see that this is a major defect of the theory. It is relatively easy to compute the collision rate, ZAB, between a molecule A, which is represented by a sphere of radius rA, and a molecule B, represented by a sphere of radius rB. The details are given in the appendix; here, we will simply quote the result† ZAB = cA cBπ (rA + rB ) crel 2. 1. = cA cBπ (rA + rB ). 2.  8kBT  2    πµ . where cA and cB are the concentrations of A and B, respectively, in molecules per unit volume. In SI units the radii would be in m, the reduced mass in kg (not mass units) and the concentrations in molecules m –3 , hence the units of ZAB will be collisions per m3 per second (m–3 s–1 ). 2 The quantity π (rA + rB ) is an area and is often called the collision crosssection, σ. Using this we can re-write the collision rate as 1. ZAB.  8kBT  2 = cA cB σ    πµ . 3.2.3 Calculation of rate constants We will now assume that a collision between A and B will result in a reaction provided that the energy of the collision is sufficient to overcome the †. You are not required to know how to prove this expression nor remember its precise form.. 11. Typical values for σ (in nm 2) are 0.24 for collisions between H2 molecules, 0.43 for N2, 0.30 for CO, 0.21 for He, 0.91 for Cl2 and 0.40 for O2.. Example 5 Exercise 6.

(13) energy barrier for reaction which was described in section 3.1.4. The fraction of collisions with energy in excess of E a , the activation energy, is given by exp(–Ea /RT) (see section 3.1.4). So the number of successful collisions per unit volume per unit time is ZABexp(–Ea /RT). Each of these collisions leads to a molecule of product, so the number of moles of product formed per unit time per unit volume is ZABexp(–Ea /RT)/L, where L is Avogadro's number. This is the rate expressed in moles per unit volume per unit time. So: r=. 1 ZAB exp( − Ea RT ) L 1.  8kBT  2 1 = cA cB σ   exp( − Ea RT ) L  πµ . [3]. If the reaction takes place when A and B collide, as we have assumed, we expect it to be first order in A and in B; the rate law is cA cB [4] L L where we have divided the concentrations, ci, by Avogadro's number so that they are in the more usual units of moles per unit volume. Thus the rate will be in moles per unit volume per unit time. Then, comparing Eq. [3] and Eq. [4] we can find an expression for the second order rate constant as r = k2nd. 1. k2nd.  8k T  2 = σ  B  L exp( − Ea RT )  πµ . [5]. k2nd will be in m3 mol–1 s–1. 3.2.4 Comparison with the Arrhenius equation The expression for the rate constant, Eq. [5], has a similar form to the Arrhenius equation (Section 2.3) 1. collision theory k2nd Arrhenius.  8k T  2 = σ  B  L exp( − Ea RT )  πµ . k = A exp( − Ea RT ). Comparing these two shows that the A-factor is related to the collision rate –to be precise it is the collision rate per unit concentration of A and B 1. Acoll theor = π (rA + rB ). Example 6 Exercise 7. 2.  8kBT  2   L  πµ . [6]. The activation energy in the Arrhenius expression can be identified with the extra energy needed to reach the transition state. Collision theory predicts that the A-factor depends on temperature, 1 varying as T 2 . However, the exponential factor has a much stronger temperature dependence, and this tends to swamp the weak temperature dependence of the A-factor.. 12.

(14) 3.2.5 Comparison with experiment In order to test our expression, Eq. [6], we need to find values for the radii and the activation energy. The radii can be obtained from measurements of properties of gases, such as viscosity and effusion. Using gas kinetic theory, these measurements of bulk properties can be related back to the size of the atoms or molecules involved. As collision theory does not predict the value of the activation energy, we will simply compare experimental and predicted values of the preexponential factor, A. +. reaction (gas phase). T/K. Aexp Acoll theor 3 / dm mol–1 s–1 / dm3 mol–1 s–1. CH3+CH3→C2Η6 2NOCl→2NO + Cl2 H2+C2H4→C2Η6 Diels-Alder reaction* K+Br2→KBr+Br. 300 470 800 500 600. 2.4 × 1010 9.4 × 109 1.24 × 106 1.5 × 106 1.0 × 1012. 1.1 × 1011 5.9 × 1010 7.3 × 1011 3.0 × 1011 2.1 × 1011. Aexp /Acoll theor 0.22 0.16 1.7 × 10–6 5 × 10–6 4.8. The general picture is clear: with one exception collision theory overestimates the value of the A-factor. The overestimate varies from a factor of 5 for a rather simple reaction such as CH3+CH3→C2Η6, to a factor of more that 106 for the Diels-Alder reaction. 3.2.6 Steric factors It is not surprising that collision theory predicts A-factors to be too large as in the derivation we have assumed that all collisions with sufficient energy lead to reaction. Clearly this assumption is not valid as we know that orientation of the reactants is important. By assuming that the molecules are hard spheres, simple collision theory ignores orientational effects entirely. The fraction of sufficiently energetic collisions which lead to reaction is called the steric factor, p. It can be given a value by comparing the experimental and predicted A-factors p=. Aexperiment Acoll theor. Typically p is much less than 1, but there are a few reactions which have p greater than 1. These are reactions in which it appears that the molecules interact over greater distances than the radii found from gas kinetic theory would indicate.. 4. Experimental determination of rate laws and rate constants We need to determine, by experiment, the form of the rate law, and hence the associated rate constants, for a number of reasons: 1. To extend our knowledge of the rates of chemical processes, so that 13. *Diels-Alder reaction between two cyclopentadiene molecules..

(15) predictions can be made. 2. To compare the measured rate constants with those predicted by theory. 3. To study the mechanisms of chemical reactions. The last point is one we will consider in more detail in Sections 5 and 6 where we will find that a knowledge of the form of the rate law can be used to find out something about the mechanism of a complex reaction. The basic measurements we can make are concentration as a function of time. We then use various methods to determine the rate law from this raw data. Section 4.1 first considers how this data can be analysed and the Section 4.2 considers how these data are obtained. 4.1 Fitting data to rate laws There are a very large number of possible rate laws, so there is really no way that we can look at some data of concentration against time, and determine the rate law directly. The way we proceed is to propose a rate law and then see if the data can be made to fit it. If the fit is acceptable to within the errors of the experimental data we say that the proposed rate law is consistent with the data. If the fit is not good enough, another law will have to be proposed and tested against the data. Rate laws are essentially differential equations, and so need to be integrated (solved) in order to see if the data fits the law. A few simple rate laws can be solved "by hand", but most can only be solved numerically using a computer program. There are many computer algorithms available for tackling this problem. We will investigate just the simplest rate laws which can be integrated by hand. 4.1.1 First order rate law For a reaction of the form A → products, a first order rate law takes the form r = k1st [A ]. 1. writing the rate in the differential form this becomes d[A ] = − k1st [A ] [7] dt We need the minus sign as A is a reactant and so its concentration decreases with time i.e. the slope of a graph of [A] against time will be negative. The variables [A] and t in Eq. [7] can be separated to give 1 d[A ] = −k1st dt [A] which can then be integrated 1. ∫ [A] d[A] = ∫ − k. 1st. dt. ln[A ] = − k1st t + const.. 14.

(16) The constant of integration can be removed by supposing that at t = 0 [A] = [A]0, the initial concentration of A; so. [A] [A]0. ln[A ]0 = const. Substituting this gives the final form This can also be expressed as. [A] = [A]0 exp(− k1st t ). t. [8] [9]. 4.1.1.1 Straight line graph for first order kinetics Equation [8] is of the form y = mx + c, so a graph of ln[A] against time will be a straight line, with slope –k1st and intercept ln[A]0. We do not need to know the absolute concentration in order to determine the first order rate constant, just some quantity which is proportional to concentration. The argument goes like this. Suppose that the measured quantity, I, is proportional to concentration: I = γ [A], where γ is the constant of proportion whose value we do not know. [A] and [A]0 can be written in terms of I: [A] = I/γ and [A]0 = I0/γ. Equation [8] becomes. Plot of [A] against time for a first order reaction. [A] falls exponentially from [A] 0 towards zero.. ln [A]0 ln [A]. ln[A ] = − k1st t + ln[A ]0. t. ln( I γ ) = − k1st t + ln( I0 γ ) – ln γ + ln I = − k1st t + − ln γ + ln I0 ln I = − k1st t + ln I0 The unknown constant γ cancels, and so a plot of ln I against t has slope –k1st , just as before. The need for only a relative measure of concentration is a very useful and unique feature of analysing first order data. 4.1.1.2 More complex first order rate laws Equation [7] expresses the rate in terms of the reactant. What happens if we want to express the rate in terms of the appearance of a product? For example, if the reaction has the stoichiometric equation A→B we can express the rate law as d[B] = k1st [A ] [10] dt Note that the derivative is positive as the concentration of B grows as the reaction proceeds. We cannot solve this differential equation as it stands since both [A] and [B] vary with time. However, we can find [B] at any time by expressing it in terms of [A]. This can be done by recognizing that, due to the stoichiometry of the reaction: [B] = [A]0 – [A]. [11]. where [A]0 is the initial concentration of A. In words, each mole of [B] that is formed must be as a result of the loss of a mole of [A]. Equation [11] can be rearranged to 15. slope = –k1st. Exercise 8.

(17) [A] = [A]0 – [B]. [12]. and this can be substituted into Eq. [9]. [A]0. [A] = [A]0 exp(− k1st t ). conc.. B. [9]. to give A t. Growth of the product [B] and decay of the reactant [A] as a function of time. At all times [A] + [B] = [A] 0.. [A]0 − [B] = [A]0 exp(− k1st t ). [B] = [A]0 (1 − exp(− k1st t )). [13]. As expected, this predicts that [B] rises from zero and tends towards [A]0 at long times. If the stoichiometric equation were different, for example A → 2B the relationship between the concentrations is different as two moles of B are produced from each mole of A consumed. Then. Exercises 9, 10 & 11. [A] = [A]0 –. 1 2. [B] or [A] +. 1 2. [B] = [A]0. 4.1.2 Second order rate law For a reaction of the form A → products, a second order rate law takes the form d[A ] 2 [14] = − k2nd [A ] dt As before, the variables [A] and t in Eq. [14] can be separated to give 1 d[A ] = −k2 nd dt [A]2 which can then be integrated 1 ∫ [A] d[A] = ∫ − k 2. 2nd. −. dt. 1 = − k2nd t + const. [A]. The constant of integration can be found by supposing that at t = 0 [A] = [A]0; so −. 1 = const. [A]0. Which gives the final form 1 1 = k2nd t + [A] [A]0. 16. [15].

(18) 4.1.2.2 More complex second order rate laws Consider a reaction with the stoichiometry A+B→C and rate law d[A ] = − k2nd [A ][B] dt As [A] and [B] are related by the stoichiometry of the reaction it is possible to solve this differential equation, but it is not that easy. Any more complex rate laws really are impossible to solve "by hand". If at time zero the concentrations of A and B are equal, then at all subsequent times they will remain equal because of the 1:1 stoichiometry of the reaction: [A] = [B]. Then the rate law becomes d[A ] 2 = − k2nd [A ] dt which is one we have already solved. 4.1.3 Isolation method As rate laws become more complex, it rapidly becomes difficult to integrate them by hand. A way out of this is to use the isolation method as way of simplifying the rate laws. Suppose we have a reaction A + B → products and we suspect that the rate law is d[B] 2 = − k3rd [A ] [B] dt The idea is to make measurements with one reagent in great excess, typically more than 50 times the concentration of the other. If A is in excess, we can say that its concentration will not change very much during the course of the reaction, i.e. as [B] drops from its initial value to zero. The rate law becomes d[B] 2 = − k3rd [A ]0 [B] dt where [A]0 is the initial excess concentration of [A]. In this situation, 2 k3rd [A ]0 is constant and can be written as keff ; the rate law becomes d[B] = − keff [B] dt. where keff = k3rd [A ]0 2. 17. 1/[A]. 4.1.2.1 Graphs for second order rate laws A plot of [A] against time shows a hyperbolic decrease starting from [A]0 and falling asymptotically towards zero at very long times. Equation [15] is of the form y = mx + c, so a graph of 1/[A] against time will be a straight line, with slope k2nd and intercept 1/[A]0. In contrast to the case of a first order process, we need to know absolute concentrations in order to find a value for the rate constant.. slope = k2nd. 1/[A]0 0. t.

(19) Exercise 12. The rate law is now looks like that of a first order reaction – we usually call this pseudo first order because it only appears to be so; underneath lurks greater complexity. Likewise keff is called a pseudo first order rate constant as it is not really a rate constant (it depends on concentration). Having found keff we can find k3rd , which is what we really want to know, from keff = k3rd [A]0 2 . If several measurement have been made at different excess concentrations of A, the value of k3rd can be found from a plot of keff against [A]0 2 . Equally well we could have chosen to put B in excess, in which case the rate law would become. which, as k3rd [B]0. d[A ] 2 = − k3rd [A ] [B]0 dt is effectively constant, can be written. d[A ] 2 = − keff,2 [A ] where keff,2 = k3rd [B]0 dt Now keff,2 is a pseudo second order rate constant. There are problems with the isolation method. Firstly, putting one reagent in such large excess may make the rate inconveniently fast. Secondly, such conditions may alter the mechanism of a complex reaction. Nevertheless, the isolation method is a popular approach to analysing complex rate laws. In practice one would put all suspected reagents but one in excess so that the order and the pseudo-rate constant for that species could be determined. The same procedure would then be repeated for all the other reagents. 4.1.4 Differential method If the rate law is of the form r = k[ A ]. n. [16]. then an appealing method of finding the value of the order, n, is to plot the log of the rate against the log of [A] ln r = ln k + n ln[A ] Such a graph will have slope n. This method, called the differential method, is convenient as, in contrast to the method of Sections 4.1.1and 4.1.2 we do not have to make any assumptions about the value of n. The drawback is that rather than plotting a function of concentration we have to plot rates, and rates are much harder to measure than concentrations. The rate is the slope of a graph of concentration against time, and as such a graph is usually curved taking an accurate slope is not at all easy. More complex laws can, of course, be manipulated into the form of Eq. [16] by using the isolation method described in Section 4.1.3 4.1.5 Half lives The half life of a reaction is defined as the time it takes for the concentration of a specified reagent to fall to half of its initial value. For a first order reaction it turns out that the half life is independent of the initial concentration: 18.

(20) To prove this we start with the integrated rate law, Eq. [8] ln[A ] = − k1st t + ln[A ]0. [8]. At time t = 0 the concentration is [A]0; after a half life, t = t 1 , the 2 concentration is [A]0/2. Substituting these values into Eq. [8] we find. [A] [A]0. ln([A ]0 / 2) = − k1st t 1 + ln[A ]0. [A]0/2. 2. [. ]. t1 =. 1 ln[A ]0 − ln([A ]0 / 2) k1st. t1 =. 1 [A]0 ln k1st [A ]0 / 2. t1 =. 1 ln 2 k1st. 2. 2. 2. 0. t1/2. We have shown that the half life is independent of the initial concentration. This means that for a first order reaction the time taken for the concentration to halve from a particular value is constant throughout the reaction. In reality, measuring the half life is not a very accurate way of finding the rate constant; we would do much better to plot the data as suggested in Section 4.1.1.1. However, it may be useful when appraising kinetic data roughly to keep in mind this property of the half life and its relation to the rate constant. We can use a similar method to find the half life of a second order reaction, starting from Eq. [15]. The result is t1 = 2. 1 k2nd. Exercise 13. 1 [A]0. In contrast to the first order case, this half life does depend on the initial concentration. The half life is a useful way of specifying the extent of a reaction. If a reaction has been taking place for several half lives we can say that a significant amount of reaction has taken place and the process is well on the way to completion or equilibrium. On the other hand, a process which has been observed for less than a half life since it started has not taken place for long enough to be characterised very well. 4.1.6 Comparing rate constants As the half life of a first order reaction is independent of concentration it is easy simply to look at the rate constant and get an immediate feel for how fast it is. For example, the reaction 2N2 O5 → 4NO2 + O2 has a first order rate constant of about 2000 s–1 at 400 K; the half life is therefore 0.3 ms, which makes the reaction quite fast. On the other hand the second order reaction 2NOBr → 2NO + Br2 -1. (gas phase) 3. –1. has a rate constant at 280 K of 0.8 mol dm s . Is this "faster" or "slower" 19. t.

(21) than the first order reaction above? To answer this we need to know the concentration of the reactants. Suppose that the pressure of NOBr is 0.1 atmospheres, corresponding to a concentration of 4.4 × 10– 3 mol dm–3; using the above expression we find that the half life is 280 s – so, under these conditions, the reaction is much slower than the decomposition of N2 O5 . The key point to understand here is that we cannot compare directly the rates of reactions of different orders by comparing their rate constants; concentration has to be taken into account. 4.2 Obtaining kinetic data The raw data needed for the analysis described in Section 4.1 is concentration as a function of time. In this Section we will investigate how we go about measuring such data. This is the raw material of kinetic studies. Kinetic measurements represent quite a challenge to the experimentalist. Firstly, reactions proceed on a vast range of different timescales – varying from the almost geological to sub nano-second. We need all sorts of different strategies for making measurements over this range. Secondly, many reactions involve complex mixtures, perhaps with the species in vastly different concentrations; we want to be able to measure the concentrations of all these species individually. Thirdly, we want to be able to do all this without interfering with the reaction mixture – this points to the use of physical methods of measuring concentration, which are non-invasive. Finally, it would be nice to be able to automate taking concentration readings. First we will look at methods of measuring concentration, and then consider the special difficulties of making measurements on fast reactions. 4.2.1 Light absorbance As you have seen in earlier lectures, molecules absorb light at characteristic frequencies. These frequencies are associated with transitions between energy levels, for example, those associated with vibrations or electrons. For kinetic studies, the commonest kinds of transitions to use are those in the visible or ultra-violet (UV) part of the spectrum (1000 – 200 nm). These transitions come about when the absorbed photon causes the molecule to move to an excited electronic energy level. All molecules absorb in the UV to some extent, but it takes the presence of special groups to cause strong absorptions in the visible region. Here are some examples of species that have been monitored by UV/vis absorption in kinetic experiments: COO– Br2. H2 O H2 O. NO2 400 nm. 495 nm. SCN OH2 Fe OH2 OH2 490 nm. CH3. N MeO. 216 nm. O. 278 nm. The extent to which light is absorbed is related directly to the concentration of the absorbing species by the Beer-Lambert Law. 20.

(22) I = I0 exp( −εcl ). l. [17]. where I0 is the intensity of the light entering the medium, I is the intensity of light exiting the medium, l is the path length through which the light passes, c is the concentration and ε is the extinction coefficient. The value of the extinction coefficient depends on the species absorbing the light and the wavelength. The path length is related to the physical dimensions of the container holding the sample. Typically, a glass or quartz "cuvette" of known dimensions (a 1 cm path length is common) is used for solutions. Equation [17] can be rearranged to give the concentration. I0. I. ln I = ln I0 − εcl I0 = εcl I The quantity ln(I0/I) is called the absorbance, A, and it is common for spectrophotometers to read this out directly. Absorbance is directly proportional to concentration. ln. A = εcl To turn measured absorbances into absolute concentrations, we need to know εl. Usually this is done by measuring the absorbance of a series of solutions of known concentration and then simply plotting A against c; the slope is εl. If we are studying first order processes then only a relative measure of concentration is needed, so it is sufficient just to know the absorbance (Section 4.1.1.1). An example of the use of light absorption to follow a reaction is the bromination of acetone (under acid conditions) O. O Br2. Br. Br–. Of all the species involved, only Br2 absorbs significantly at 495 nm and so measuring absorbance at this wavelength enables us to measure its concentration. Absorbance measurements are very convenient; they are non-invasive, rapid and can easily be automated. As different species absorb at different wavelengths it is possible to study different molecules in a reaction mixture. However, UV/vis absorptions tend to be rather broad (especially in solution), so it is possible that more than one species will absorb at a given wavelength. The same principle applies to IR measurements. Proton NMR can also be used to monitor concentrations, as under the right conditions the peak height is proportional to concentration. It has the advantage that individual chemical species can easily be followed, but NMR is not very sensitive so as an analysis method it is not very fast. 4.2.2 Conductivity measurements Ions in solution conduct electricity and the resistance of such a solution depends on the concentration of the ions and their identity. In fact the conductance, which is the reciprocal of resistance, is directly proportional to the concentration of the ions. 21. A typical value for the extinction coefficient for a transition which absorbs "strongly" is 10,000 mol–1 dm 3 –1 cm ; a "weak" absorption might have an extinction coefficient three orders of magnitude lower. Example 7.

(23) If a reaction involves a change in the number of charged species it is possible to monitor the progress in the reaction by measuring the change in conductance. An example is the following reaction, in which neutral species react to give charged ones; the reaction can therefore be monitored by the increase in conductance. N. N Ph. Br. N+. N. Br. –. Ph. The contribution that an ion makes to the conductance depends on the type of ion. In water, H3O+ and OH – ions are very mobile and are thus good conductors.. The following reaction (in aqueous solution) involves no change in the number of ions, but as the identity of the ions change there is still a measurable change in conductance O. O OEt. Exercise 14. + OH –. O. –. + EtOH. The relationship between conductance and concentration can be rather involved, so it is really only convenient to use this method for relative measurements of concentration i.e. appropriate for first order kinetics, Section 4.1.1.1 It is relatively easy to measure conductance using some kind of bridge or more modern electronic device; the measurement is noninvasive and rapid. 4.2.3 Gas pressure The pressure, p, that a gas exerts is related to the volume, V, the number of moles n, and the temperature, T, by the ideal gas equation pV = nRT This can be rearranged to give n p = V RT the fraction n/V is a measure of concentration; in SI it would be in moles m–3. The relationship tells us that at constant temperature pressure is directly proportional to concentration. In a reaction we will generally have a mixture of gases, and so we can talk about the partial pressure, pi, of each gas i. The partial pressure of a gas is the pressure that the gas would exert if it occupied the volume on its own. Dalton's Law states that the total pressure, pT, is the sum of the partial pressures: pT = p1 + p2 + p3 + K Each partial pressure follows an ideal gas law ni p = i V RT where ni is the number of moles of gas i. Therefore the partial pressure of species i is a measure of its concentration. In a reaction involving gases the only measurable pressure is the total pressure exerted by the system. However, to make kinetic measurements we need to know the concentration, that is the partial pressure, of the 22.

(24) individual species. We need, therefore, to find some way of relating these two. An example of how this is done is considered here. Consider the reaction between NO and O2 in the gas phase 2NO + O2 → 2NO2 Overall three moles of gas go to two, so as the reaction proceeds there is a measurable decrease in the total pressure. The total pressure is the sum of the partial pressures pT = pNO + pNO 2 + pO 2 There are three unknown quantities on the right, and only one measurable quantity, the total pressure; to solve this we are going to have to introduce some extra restrictions. Suppose that at time zero there is no NO2 present, and that the NO and O2 are mixed in a 2:1 ratio, just as the stoichiometry of the reaction. If the initial partial pressure of O2 is pO 2 ,0 , then the initial partial pressure of NO is 2 pO 2 , 0 , and so the initial total pressure, p0 , is 3 pO 2 , 0 . This is the situation set out on line 1 of the Table pO 2. pNO. pNO 2. pT. 1 2 3. pO 2 ,0 pO 2 ,0 − ∆pO 2 1 3 p0 − ∆pO 2. 2 pO 2 , 0 2 pO 2 , 0 − 2∆pO 2. 0 2∆pO 2. 3 pO 2 , 0 ≡ p0 3 pO 2 , 0 − ∆pO 2. 2∆pO 2. p0 − ∆pO 2. 4. pT − p0. 2 3. p0 − 2∆pO 2. 2 3. Suppose after time t the partial pressure of O2 has fallen by ∆pO 2 ; due to the stoichiometry of the reaction the partial pressure of NO will fall by 2∆pO 2 and the partial pressure of NO2 will increase by 2 ∆pO 2 . This is the situation set out on line 2 of the Table. Our aim now is to find pO 2 in terms of measurable quantities. There are two steps. In going from line 2 to line 3 in the Table we have used the fact that p0 = 3 pO 2 , 0 to substitute for pO 2 ,0 . From line 3 we then see that pT = p0 − ∆pO 2 from which it follows that ∆pO 2 = p0 − pT . This value of ∆pO 2 is used in going from line 3 to 4. So, finally, on line 4 we have an expression for pO 2 in terms of measurable quantities: pO 2 = pT − 23 p0 . There are many devices capable of measuring pressure, from simple manometers (columns of mercury) to various electronic devices, some of which read absolute pressures. If concentration is measured in pressure units, it is common to express rate constants in pressure units too. So, for example, a second order rate constant might be expressed in mmHg–1 s–1. 4.2.4 Electrochemistry As we have seen, the EMF (voltage) generated by an electrochemical cell depends on the concentration of the species in the cell. By constructing a cell with an appropriate electrode it is possible to measure the concentration of a 23. Exercises 15 & 16.

(25) wide variety of ions in solution. Such electrodes can give both absolute and relative measurements of concentration; they are non-invasive, sensitive to particular ions (selective) and have a rapid response. They are widely used when studying reactions of ions in solution. An example of the use of electrochemistry is in studying the oxidation of formic acid by bromine in aqueous solution: Br2 + HCOOH → 2Br– + 2H+ + CO2 The calomel electrode is there to complete the cell (we need two electrodes); it plays no role in the reaction.. Two electrodes are dipped into the solution: a calomel reference electrode and a platinum electrode. At the platinum electrode the redox process is Br2 + 2e– → 2Br– and so the voltage produced by the Pt electrode depends on the concentration of the bromine and the bromide. It can be shown that the voltage produced by the cell, E, is RT [Br – ] E=E − ln 2F [Br2 ]. 2. 0. where E0 is the standard EMF (voltage) produced by the cell (a known quantity), R is the gas constant and F is the Faraday, 96485 C mol–1. If the reaction is run with an excess of Br– then the voltage produced by the cell depends solely on the concentration of Br2 ; the progress of the reaction can therefore be monitored. 4.2.5 Chemical methods Concentration can be measured by "classical" methods of chemical analysis, such as titrations. Such methods are rather slow, and so it is usually necessary to extract some of the reaction mixture (an "aliquot") and then stop (quench) the reaction so that no further reaction takes place during the analysis. Examples of such quenching procedures might be sudden cooling or chemical removal of one of the reagents. For example, the alkaline hydrolysis of an ester can be followed by taking samples of the reaction mixture, quenching them in a know excess of acid to stop the reaction and then titrating the remaining acid against standard alkali. In such a way the concentration of alkali in the reaction mixture can be determined. A somewhat more sophisticated chemical analysis uses a gas chromatogram (GC), possibly in conjunction with a mass spectrometer. A gas chromatogram separates the components of a mixture by using an inert carrier gas to sweep them down a column (typically about a metre long) which is packed with some sort of porous solid material. Different components travel at different speeds down the tube and so are detected at different times after injecting the sample; their concentrations can thus be measured separately. The detector may be a simple one which detects a change in the thermal conductivity of the carrier gas resulting from the presence of other molecules, or it may be a mass spectrometer which can not only detect the presence of molecules but also identify them. A mass spectrometer can also 24.

(26) be used on its own to measure and identify species directly – that is without the use of a GC. An example of using this method of analysis is in studying the photolysis of acetaldehyde (photolysis means a reaction, usually destructive, brought about by irradiation with light). There are a number of possible products resulting from what turns out to be rather a complex reaction CH3 CHO + hυ → CH4 + CO + C2 H6. {not a balanced equation}. The rate of production of all three possible products can be monitored by passing small amounts of the reaction mixture into a GC at known times and measuring the heights of the three separate peaks which appear. Another example of analysis using a GC is the pyrolysis of ethane (pyrolysis means destruction by heating) C2 H6 → CH2 =CH2 + H2 + other products {not a balanced equation} As a further example, a mass spectrometer was used to monitor the rates of the following two reactions which are both important in the atmospheric oxidation of ethane C2 H5 + O2 → C2 H5 O2 C2 H5 + O2 → C2 H4 + HO2 The two alternative products, C2 H5 O2 and C2 H4 have different masses are so can be detected separately by the mass spectrometer. 4.2.6 Studying fast reactions Whatever method we choose for monitoring the concentration of the reacting species we also have to consider how the reagents are going to be mixed and the reaction initiated. This is a serious problem for fast reactions. It is no good if it takes 1 second to mix the reagents if the reaction is over in 100 ms! Further, our method of measuring concentration has to be fast enough to make measurements over the time scale of the reaction. A great deal of effort has been put into developing methods for measuring such fast reactions; we will consider some of these here. Continuous flow reagent from reservoir observe. mixing zone. waste d reagent from reservoir. The idea here is to flow the reagents together (they can be gases or solutions), and after mixing let them continue to flow down the reaction tube. Different distances down the tube correspond to different times after mixing and initiating the reaction. As the reagents are constantly replenished, we can take our time in making observations at any distance (and hence time). Essentially the method is a way of achieving rapid mixing and then being 25.

(27) The green glow is called the air afterglow and was first observed in the early part of the 20th century.. able to monitor the reaction at leisure, rather than in "real time". Mixing can be complete in under 1 ms, so reactions with half-lives on the ms timescale can studied. A typical flow velocity is 10 m s–1, so a distance of 1 cm corresponds to 1 ms of reaction. Usually the reaction is monitor by spectrophotometry, which is a very rapid technique. The main disadvantage of this method is that you needs lots of reagents, particularly if you are trying to observe a fast reaction where the flow rate needs to be high. An example of using this method in solutions is studying the complexation reaction between Fe2+ and thiocyanate, SCN– , in aqueous solution. The resulting complex absorbs strongly in the visible part of the spectrum and is thus easily monitored by spectrophotometry. The method is perhaps used more in the gas phase; in this case the flow is created by pumping on the "waste" end of the tube and supplying gases at the other. For example, the reactions between oxygen atoms and other species can be studied by generating O atoms in an electric discharge and then flowing them down a tube. Reagents can be introduced into the tube and the rate of their reaction with O atoms studied. The problem is how to measure the O atom concentration in the flow tube. One way to do this is as follows. A small amount of NO is introduced into the flow tube. The NO reacts very quickly with O to produce electronically excited NO2 , denoted NO2 * ; this then rapidly emits a photon (in the green part of the spectrum) which is detected using a photomultiplier. Finally, the NO2 the reacts with another O atom regenerating the NO. O + NO → NO2 * NO2 * → NO2 + hυ (detected) O + NO2 → NO + O2. Exercise 17. As all of these reactions are fast and the NO is regenerated, the intensity of the green emission is directly proportional to the O atom concentration. Stopped flow reagent. observe. trigger initiates observation when plunger hits stop. plunger moves back until hits stop reagent. This apparatus is similar in conception to the continuous flow method, in that the reagents are flowed together to ensure rapid mixing. However, the difference is that the reacting mixture pushes a plunger back until it hits a stop; then the flow is halted and measurement, at a fixed point, is initiated (again, usually using spectrophotometry). Observations are made at a point at which mixing is complete. The stopped flow method is only used for 26.

(28) solutions. In contrast to continuous flow, stopped-flow requires measurements in real time. However, it has the advantage of not requiring large amounts of reagents (as little as 1 cm3 may be sufficient). Generally the time resolution is a little faster then for continuous flow. Two examples, both taken from studies on enzymes: (1) The enzyme myeloperoxidase (found in leukocytes) catalyses the formation of HOCl from H2 O2 and Cl– ; the HOCl is though to play a part in the antimicrobial activity of these cells. The reaction was studied using a stopped flow system not only to find out about the overall rate of the reaction but also to monitor two transient intermediates which are formed along the reaction pathway. These absorb at different wavelengths, so their build-up and decay could be monitored separately. (2) The enzyme lactate dehydogenase catalyses the oxidation of lactate to pyruvate; the oxidizing agent is to the cofactor NAD+ OH. O CO2–. NAD. +. CO 2 pyruvate. lactate. –. NADH. The enzyme binds NADH rather strongly, and in elucidating the details of the mechanism it was necessary to find out the rate constant for the dissociation of the NADH-enzyme complex. This was studied using the stopped flow technique, and the reaction was monitored by a change in fluorescence from the NADH when it binds to the enzyme. Flash photolysis The idea of this method is to use a short, intense flash of light (now often from a laser) to generate some reactive species, typically free radicals. Then, the reaction of these species with other reagents present in the system is monitored, usually spectrophotometrically. This is a useful method for studying the fast reactions of radicals; there is no mixing problem as the reagents are generated in situ. However, measurements do have to be made in "real time", so rapid detection is needed. Two examples: (1) The recombination of two CH3 radicals, to give ethane, was studied by generating the methyl radicals by photolysis of acetone (a pulse a laser light of duration tens of ns was used). The concentration of the CH3 radicals was followed spectrophotometrically. (2) The reaction, in solution, of a carbocation was followed after it has been generated using a laser pulse laser pulse 20 ns, 248 nm. Ph. + nucleophiles, Nu–. Ph. Ph Nu. Cl Ph. Ph. Ph. Second order rate constants of up to 109 mol–1 dm3 s–1 have been measured in this way. 27. Fluorescence is when light of one wavelength is emitted by a molecule when it is irradiated by light of another (usually shorter) wavelength. The irradiation causes electrons to move to higher energy levels and these then emit light (fluoresce) when they drop back down to lower levels. The technique of flash photolysis was pioneered by Norrish and Porter, working in this Department in the 1950s; in 1967 they received a Nobel Prize for their work.. If two radicals come together by definition they have sufficient energy to fall apart again. How then, do radicals recombine? Somehow they have to lose the excess energy before they fall apart. Kinetic studies, on such things as methyl radical combination, are used to unravel how this excess energy is disposed of..

(29) 5. Analysing the kinetics of complex reactions We have already come across the idea that many reactions take place in a series of steps, often involving the formation of intermediates. This set of steps is called the mechanism. For example, the alkaline hydrolysis of some esters is thought to proceed by the following mechanism –. HO O–. O. O. OH. EtO– OEt. OEt. OH. Another example is the set of steps believed to be responsible for the destruction of ozone in the upper atmosphere Step 1: Cl + O3 → ClO + O2 Step 2: ClO + O → Cl + O2 The Cl used up in step 1 is regenerated in step 2, and so can go on to destroy more ozone. Each of the reactions which comprise the mechanism is called an elementary step; it is elementary in the sense that we believe that it takes place in a single reactive encounter between the species involved. The elementary steps are thus the basic building blocks of a complex reaction and cannot be broken down further. A mechanism can be defined as a proposed set of elementary steps which account for the overall features of the reaction. One of the aims of studying chemical kinetics is to be able to determine the mechanisms of complex reactions, and this Section is concerned with just how kinetics can help in such a study. However, a study of chemical kinetics alone is unlikely to be able to determine a mechanism, especially a complex one. We need to use other information, such as the direct detection of intermediates (for example, by spectroscopy), trapping and labelling experiments, and above all our chemical intuition. However, chemical kinetics does play an important role in such studies. The way we proceed with a complex reaction is first to propose a mechanism and then proceed to analyse its chemical kinetics. We can then compare this with experiment and see if the two agree. If they do, the proposed mechanism can be said to be consistent with the kinetic evidence; if they do not agree, we have to go back and modify the proposed mechanism. There is no way that we can "determine" the mechanism just from kinetic data, all we can do is propose a mechanism and see if it fits the data. The details of the mechanism we propose will depend on other information available to us and our chemical intuition. 5.1 Rate laws for elementary reactions We can always write down the rate law for an elementary reaction just by inspecting the stoichiometric equation. This is because we believe the reaction to take place in a single encounter between the species involved. For example, for the reaction Cl + O2  k → ClO + O 28.

(30) we can write down straight away the rate of loss of reagents or gain of products: d[O 2 ] d[Cl] = − k[Cl][O 2 ] = − k[Cl][O 2 ] dt dt d[O] d[ClO] = k[Cl][O 2 ] = k[Cl][O 2 ] dt dt As before, the rate has to be negative for a reagent, as it is lost in the reaction, but it is positive for a product as it is gained. 5.2 Rate laws for complex reactions Once we specify a reaction mechanism we can easily write down the rate laws for all the steps. As an example we will consider the reaction of HBr with an alkene (in solution). The overall reaction is Br HBr R. R. but it is not believed that this takes place in a single encounter between the reactant molecules. Rather, it is thought to take place by two sequential elementary steps, each of which is a single reactive encounter. The mechanism is thought to be: Br–. H Br k1. k2. R. R. R A. Br. C+. P. The rate of formation of the product, P, is d[ P ] = k2 [C + ][Br – ] dt This depends on the concentration of the intermediate carbocation, C+. Now this species is created in step 1, and destroyed in step 2, so its overall rate of change depends on the rates of both steps d[C + ] = + k1[A ][HBr ] − k2 [C + ][Br – ] dt The first term, which is positive, is the rate of step 1 in which C+ is formed; the second term, which is negative, is the rate of step 2 in which C+ is destroyed. Using this approach we can write down expressions for the rate of change of all species involved in a given mechanism. Even for this simple reaction the mathematics has become a bit involved. To find out exactly how [P] depends on time – which we would need to do to compare the predictions of the mechanism with experiment – we need to integrate these differential equations, just as we did in Section 4.1. This is not easy as the variables from one equation appear in the other; that is the equations are coupled. In fact this pair can be solved relatively easily, but if we add a few more steps to the mechanism we rapidly end up in a situation where the solution becomes impossible by hand. 29. Example 8 Exercise 18.

(31) concentration. It is possible to solve such coupled differential equations numerically using a computer algorithm. Indeed, a huge amount of effort has gone into this approach as it is essentially the only way of handling complex kinetic systems such as the atmosphere, combustion and so on. For our purposes, however, we need to find some simpler approaches which, although they may not be universally applicable, will enable us to get a handle on the kinetics of complex reaction schemes without resorting to computers. All of these methods remove some of the complex time dependence by making assumptions about the relative rates of the processes involved. 5.3 Sequential reactions Suppose that we have two sequential reactions which for simplicity we will assume are both first order. 1.0. C 0.8. 0.6. 0.4. B. 0.2. A 0.0 0. 1. 2. 3. 4. 5. t/s Concentration against time for two sequential reactions. Here k 1 = 4 s–1 and k 2 = .5 s–1.. concentration. 1.0. C. 0.8. k1 →B A . [1]. k2 →C B . [2]. We can imagine two extreme (limiting) cases: (1) Rate constant of process [1] is very much greater than that of process [2]: k1 >> k2 In this case A is converted rapidly into B, leading to an accumulation of B which is more slowly converted into C (see the graph opposite); the production of C mirrors the fall in B. The rate at which the product C is formed depends on the rate of process [2] – it is said that step [2] is the rate determining step. (2) Rate constant of process [2] is very much greater than that of process [1]: k2 >> k1 . In this case A is converted slowly into B, and then B reacts to form C almost as soon as B is formed (see graph below); the production of C mirrors the fall in A. The rate at which the product C is formed depends on the rate of process [1] – step [1] is now the rate determining step. In this limit there is very little B present at any time as the moment it is produced it reacts. We will refer again to this Section 5.5. If either of these limiting cases applies then the kinetics is greatly simplified, as the overall rate of production of the product depends only on the step with the smallest rate constant. The same is true for more complex schemes of sequential reactions. 0.6. 0.4. 0.2. A B. 0.0 0. 1. 2. 3. 4. 5. t/s Concentration against time for two sequential reactions. Here k 1 = .5 s–1 and k 2 = 4 s–1. k3 k1 k2 k4 A  → B  → C  → D  →E If the step from B to C has the smallest rate constant, then the rate of formation of the product E depends only on k2 . If the reactions are not first order things become more complex as the relative rates of the steps depend not only on the rate constants but also on the concentration of the species involved. Consider the following sequential reactions k1 k2 A + B  → C  →D The first step is second order, and so its rate relative to the second step can be. 30.

(32) altered by varying the concentrations of A or B. Things get more complex if the reactions are not simply sequential; to handle these we need some further tools introduced in the next two sections. 5.4 The pre-equilibrium hypothesis In many mechanisms the species directly involved in the rate limiting step are in equilibrium with the reagents. A typical example is the acid catalysed hydrolysis of esters: H OH2. O. OEt. k–1. E HO H2 O. k1. k2 OEt. H2 O. OH. OEt EH+. H2 O. RDS. OH OEt. EH+. The carbonyl group of the ester (E) is protonated on the oxygen and it is this species (EH+) which is subject to nucleophilic attack by water in the rate determining step. As the attack by water is slow, it is quite likely that EH+ will lose its proton and return to ester before it is attacked by H2 O – the first reaction is reversible. In the pre-equilibrium hypothesis we assume that (1) E, H3O+ and EH+ are always in equilibrium which implies that (2) the reaction of EH+ with H2O is slow enough that the equilibrium is not perturbed. The latter reaction is therefore rate determining. The reaction scheme is k1 E + H 3O +  → EH + + H 2 O. [1]. k −1 EH + + H 2 O  → E + H 3O +. [ −1]. EH + + H 2 O → rest of reaction [2] k2. For the equilibrium hypothesis to apply the rate of process [–1] must be much faster that the rate of process [2], so that equilibrium is established between E, H3O+ and EH+. If this is so, then the rates of [1] and [–1] are equal, as this is the definition of equilibrium (see Section 2.4):. 31. Exercises 19 and 20.

(33) Ea,–1 Ea,2 Ea,1. EH++H2O. E+H3O+. at equilibrium : rate of [1] = rate of [–1]. [. ]. [. ]. k1[ E ] H 3O + = k–1 EH + [H 2 O]. [18]. Alternatively, we could write an equilibrium constant for the equilibrium EH + + H 2 O. E + H 3O + products. Energy profile for the case where pre-equilibrium applies. The intermediate has a lower energy barrier (E a,–1 ) to returning to reactants than it does to go on to products (E a,2 ). Note that there is no particular requirement on the size of E a,1 .. Keq =. [EH + ][H 2O]. [19]. [E][H 3O+ ]. By comparing Eq. [18] and Eq. [19] we find, as expected from Section 2.4, that k1 k−1. Keq =. The rate of formation of products is assumed to be controlled by the rate of process [2] i.e. this is the rate limiting step. rate = k2 [ EH + ][H 2 O]. [20]. An expression for [EH+] can be found from Eq. [19]. [EH + ] = Keq. [E][H 3O+ ]. [ H 2 O]. Substituting this into Eq. [20] we find rate = k2 [H 2 O] × Keq. [E][H 3O+ ]. [ H 2 O]. = k2 Keq [ E ][H 3O + ]. or. k2. k1 [E][H 3O+ ] k−1. [21]. The final form of the rate equation is easy to handle as it only involves the reacting species and no intermediates; the differential equation is simple enough that it can be integrated in the ways described in Section 4.1. Let us assume that in an experiment it is found that the rate law is of the form rate = kexp [ E ][H 3O + ] Comparison of this with the theoretical rate law of Eq. [21] enables us to identify kexp as k1k2/k–1 . In other words the "rate constant" kexp is not really a rate constant but a composite of rate constants of three elementary processes. For different esters we could then interpret changes in kexp in terms of changes in either Keq (i.e. the basicity of the carbonyl), or k2 (i.e. the reactivity of the protonated carbonyl), or both. This pre-equilibrium approach is often successful for intermediates which involve simple protonation or de-protonation, as such processes are generally fast compared to breaking or making bonds to heavier atoms. Example 9. 32.