16.1 What is chemical kinetics?

Chemical kinetics is the measurement and description of the rates of chemical reactions. It seeks to develop mathematical equations that describe and predict the course of chemical reactions in terms of the species involved in the reaction. The mathematical equations are called rate laws. There are seven types of species that can be involved in chemical reactions: reactants; products; intermediates; activated complexes; spectators (also called third bodies); catalysts; and inhibitors.

• Reactants and products are stable chemical entities (atoms and molecules) that appear on either the left- or right-hand side of a balanced chemical equation.
  
• Intermediates are also stable species but they only appear after the start of a reaction, and are consumed during the course of a reaction. They do not appear in the balanced stoichiometric equation. By stable, we generally mean something that can be isolated and put in a bottle. However, a better definition might be something with a sufficiently long lifetime. A unit of time of relevance to chemical lifetime is a vibrational period, which as we have seen is on the order of a picosecond. If a species can survive for many vibrational periods, it is essentially a stable chemical species that can be isolated, at least for the sake of spectroscopic investigation, if not for inspection in a bottle.
  
• An activated complex is not a stable chemical species. Its isolation and observation is a ‘Holy Grail’ in the study of chemical dynamics, and this has only been achieved in a select few examples. An activated complex is a species that may last for no more than the time it takes for a collision to occur, and certainly no more than a few vibrational periods. Nonetheless, it may be the most important species in a description of the dynamics of a reaction since it is the species found in the transition state, which we shall define below.
  
• A spectator is a species that is involved in collisions with reactants, intermediates and products but is not itself transformed chemically during the course of a reaction.
  
• Catalysts change the rate and selectivity of reactions but are not consumed during the reaction.
  
• An inhibitor is a species that decreases the rate of a reaction. In some cases a species may take on more than one role; for example, a product may also act as an inhibitor.

Chemical kinetics can be completely empirical. That is, we may only want to know the identities and concentrations of the species involved in a reaction at any given time. In this case, the descriptors of chemical kinetics are a series of rate equations in either differential or integrated form. The differential form tells us how the rate is related to concentrations, while the integrated form tells us how the concentrations change with time. The equations are determined by a fit to the experimental data. Since they are empirically derived with little interpretive knowledge of the chemical reaction dynamics, they are generally accurate for interpolation, but their use for extrapolation is limited to conditions only very closely related to the conditions used while acquiring the experimental data. For example, assume that we measured the concentrations of A, B, and C (all behaving as ideal gases) every minute for 10 min after 10 Pa of A and 10 Pa of B are mixed in a constant volume vessel held at 298 K. Curves of the concentrations of A, B, and C versus time are then plotted and fitted to mathematical equations. The derived equations would be highly accurate (assuming the data are good) if we used them to calculate the concentrations at t = 8.5 min. This is an interpolation – the calculation of concentrations at a time for which we have no data, but which lies within the range of times measured. The equations would probably also be accurate for extrapolation to t = 10.5 min, a time just outside the range measured. Use of the equations at t = 100 min would be dubious. Use of the equations at T = 500 K would not be justified without further information.

Empirical rate laws can have any arbitrary mathematical form. Empirical rate laws can be highly accurate and useful, especially if we always run a given reaction under the same conditions – that is, as long as we only need information for interpolation. However, if we want to understand how reactions change when conditions are changed, we either need to increase the range of empirical data, or else we need to understand how the empirical equations will change as we change conditions based on the chemistry that is occurring in our system. This understanding will allow us to make justifiable extrapolations and to judge the certainty with which these extrapolations will be accurate. In order to understand how rate laws change when reaction conditions change, we need to understand the dynamics of the reaction.

Figure 16.1 illustrates what we can learn immediately from kinetics data, and how we can use kinetics data to learn about the way in which chemical reactions occur. Ammonium carbamate was heated to 50 °C and allowed to develop its equilibrium vapor pressure. The pressure was monitored in a small tube held at room temperature, which is shown as the red curve. At a time near 3800 s on the time axis, a valve was opened to allow the warm vapor into the cool region. The pressure shoots up and then decays exponentially. The exponential nature of the decay is confirmed not only by the fitted blue curve to the pressure versus time data, but also by a plot of the natural logarithm of the pressure (normalized to the maximum value of pressure after subtracting the background value), in which a linear relationship is found.

What does this data mean? Empirically, it means that over the region of the blue curve, the pressure is quite accurately described by p(t) = p_bkg + p₀e^{− kt}. However, with further investigation of the relationship of kinetics to dynamics, and with some further experimental input, we will also be able to use this data to construct at least a partial description of the reaction dynamics. In this case, the rate-determining step of the reaction is the adsorption of CO₂. This is a simple first-order process; hence, the exponential decrease in pressure as a function of time as the surface reaction consumes NH₃ and CO₂ and turns them into solid NH₂CO₂NH₄. Now, we need to learn how we can relate mathematical descriptions of rates and concentrations versus time to the dynamics of reactions.

16.2 Rates of reaction and rate equations

Chemistry is all about molecules and compounds, so we start with a balanced chemical reaction

(16.1)

The reaction involves four species: two reactants and two products. There are four individual rates associated with this reaction, each of which we can express in terms of the rate of change of the molar concentration of each species: d[A]/dt, d[B]/dt, d[C]/dt, and d[D]/dt. However, these rates are not independent. The stoichiometry of this reaction tells us that when 4 mol of A are consumed, 3 mol of B are consumed while 2 and 1 mol of C and D, respectively, are formed. Therefore, if A were consumed at a rate of d[A]/dt = −4 mol l⁻¹ s⁻¹ (the negative sign indicating consumption), the rate of formation of D must be d[D]/dt = +1 mol l⁻¹ s⁻¹.

If we know the stoichiometry of a reaction (and there are no side reactions), then knowledge of the rate of change of any one species in the reaction lets us know the rate of change of all the other species. Thus, we can define a single parameter that we call the rate of reaction R. If we know the rate of reaction and the stoichiometry of the reaction, we can determine the rate of change of each and every species in the reaction.

The rate of reaction R is the rate of change in the number of moles of a hypothetical product that has a stoichiometric coefficient of 1.

According to this definition, the reaction rate is always positive. If we perform a reaction at constant volume, then the amount measured in moles is directly proportional to the concentration. If we perform a gas-phase reaction at constant temperature and constant volume, the partial pressure of a gas is directly proportional to the amount of that gas. Therefore, we can equivalently define the reaction rate in terms of rate of change of amount, concentration or pressure under the appropriate conditions.

For the specific example given above, R = d[D]/dt. But not all reactions have a product with a stoichiometric coefficient of 1. We introduce the general mathematical definition of the rate of reaction based on the stoichiometric number of the ith species ν_i,

(16.2)

where V is the volume and ξ is the extent of reaction. IUPAC recommends the use of R, r or v as symbols for reaction rate, and all three are commonly encountered. Always inspect the rates you determine to ensure that they pass basic tests of sensibility. The reaction rate is always positive by definition. The rate of change of products is also positive, but the rate of change of reactants must be negative. Furthermore, the stoichiometry determines how one rate is related to another. In the above example, the rate of production of C must be twice the rate of formation of D, and so forth.

16.2.1 Example

For the reaction 2NOBr(g) → 2NO(g) + Br₂(g) under a given set of conditions, the rate of appearance of NO is found to be 1.6 × 10⁻⁴ mol dm⁻³ s⁻¹. What are the stoichiometric numbers, the rate of reaction, and the rate of NOBr consumption?

Stoichiometric numbers are signed quantities. Their absolute values appear in the balanced chemical equations. The values for this reaction are ν_NO = 2, ν_Br = 1, and ν_NOBr = −2. The rate of reaction is

Since Br₂ is a product with stoichiometric coefficient of +1, the rate of reaction is the same as the rate of formation of Br₂. Equation (16.2) is then used to find to the rate of NOBr consumption,

Therefore

16.2.2 Directed practice

For the reaction BrO⁻₃ + 5Br⁻ + 6H⁺ → 3Br₂ + 3H₂O, calculate the rate of reaction and the rates of formation or consumption of all products and reactants if H⁺ is being consumed at a rate of 1.0 × 10⁻⁶ mol m⁻³ s⁻¹.¹

Having defined the rate of reaction, we now turn to the rate equations. In general, an empirical rate equation relates the rate to the concentration of each stable species involved in a reaction. In the above example, assuming that only the reactants and products are involved in the reaction then a general form of the empirical rate equation is

(16.3)

The reaction order is n = α + β + χ + δ, where α, β, χ, and δ are the partial orders or the orders with respect to A, B, C, and D, respectively. The rate constant k is defined as a positive number, and its units change depending on the reaction order, the units used for concentration, and the unit of time (though second is the preferred unit). In principle, the partial orders can be positive or negative, integer, or noninteger.

16.3 Elementary versus composite reactions

Having stated that empirical rate laws can have any arbitrary form, we are left with an infinite number of possible mathematical equations. At this point, we need to find some order among this chaos. Fortunately, chemical dynamics is not so random and it allows us to concentrate our mathematical treatment on a few simple cases to start. Here, the distinction between simple and complex reactions helps greatly to clarify the mathematics that we need to know. An elementary reaction is a one-step reaction that usually only involves the breaking and making of one or two bonds. An example is

This reaction occurs when NO strikes O₃. A bond is formed between the N of NO and one of the O atoms in O₃, while simultaneously the bond of that O atom to another O in O₃ is broken. The reaction has a molecularity of two. The molecularity of a reaction indicates the number of collision partners involved in the reaction. Common values of molecularity are 1 (a unimolecular reaction) and 2 (bimolecular reaction). A molecularity of 3 (termolecular reaction) or higher is uncommon.

Elementary reactions have a very interesting property regarding their kinetics: their reaction order is the same as their molecularity. The reaction above is a second-order reaction because it has a molecularity of 2. The reaction is second-order because it is first-order in NO and first-order with respect to O₃. As is commonly found for most reactions, the partial orders for the products are both zero and they do not appear in the rate equation. The fundamental principle of chemical kinetics for elementary reactions is this:

For an elementary reaction, the order of reaction is always the same as the molecularity, the partial order with respect to a specific reactant is the same as its stoichiometric coefficient, and products do not appear in the rate equation.

Since we have stated that most elementary reactions have a molecularity of 1 or 2, first-order and second-order rate equations are the most commonly encountered for elementary reactions. Under certain circumstances, when a reaction occurs from a nondepleting reservoir, such as evaporation from a bulk liquid, an elementary reaction may exhibit zeroth-order kinetics.

A composite reaction is a reaction that has a mechanism involving two or more elementary steps. This added complexity means that the fundamental principle of chemical kinetics for composite reactions is:

Neither the reaction order nor the partial order with respect to any reactant or product of a composite reaction can be read from the stoichiometric equation.

In principle, the rate equation of a composite reaction can take any arbitrary form. However, many composite reactions contain a rate-determining step (RDS).

A rate-determining step is the elementary step in the reaction mechanism that has the greatest influence on the reaction rate.

The presence of a rate-determining step often simplifies the kinetic behavior of the composite system such that the overall rate equation resembles the rate equation of the elementary step. Therefore, the most commonly encountered reaction orders are first and second, with occasional occurrence of zeroth-order and rare occurrence of third-order. When fractional orders arise, you can be sure that the reaction mechanism is complex and/or that the system does not possess a well-defined rate-determining step. Fractional orders are also commonly obtained for heterogeneous reactions. We will treat the kinetics of heterogeneous reactions in Chapter 18.

16.4 Kinetics and thermodynamics

Kinetics is perhaps the most important topic in all of chemistry. This may seem odd because it is thermodynamics that tells us whether a reaction is possible or not. Thermodynamics determines the concentrations of reactants and products at equilibrium. The laws of thermodynamics are essentially absolute and cannot be violated in an equilibrated system. Even for nonequilibrium systems, conservation of energy and evolution toward higher entropy are difficult to overcome. Therefore, thermodynamics has incredible predictive ability and must always be considered to determine limiting cases that cannot be exceeded. However, the state functions of thermodynamics, such as Gibbs energy change, are path-independent. The statements that all systems evolve to the lowest possible Gibbs energy or that all systems evolve until the change in Gibbs energy is zero have attached to them the caveat “… as long as a path is available.”

Kinetics is all about the path. Kinetics describes the rate of evolution of a system. It determines how long it takes for a system to achieve equilibrium, if at all. You may have heard the phrases that a reaction is run under thermodynamic control or under kinetic control. More accurately, the kinetics of some reactions is such that the thermodynamic limit is achieved on the time scale that the reaction was performed. A so-called ‘kinetically controlled reaction’ simply has not been performed for a sufficiently long period to reach thermodynamic equilibrium. Another way to phrase this is that while what is forbidden by thermodynamics is forbidden, that which is allowed by thermodynamics is only a limit that, while it cannot be exceeded, need not necessarily be approached.

Kinetics also determines in many ways whether a reaction is useful or not, and/or whether it is dangerous or not. Thermodynamics can tell us which reaction are candidates to create explosions, but not if they will explode. Combustion of glucose ignited by a spark in oxygen releases the same amount of energy per mole of glucose as combustion of glucose in the body. The overall thermodynamics are the same, but the utility and effect on the body are not at all the same.

16.5 Kinetics of specific orders

The molecularity of elementary reactions determines their kinetic order. Therefore, we expect first- and second-order kinetics to be commonly observed. Certain experimental conditions lead to zeroth-order kinetics. We now treat the mathematics of these three commonly encountered cases and the related topic of pseudo-order kinetics. In the examples below, we assume our reactants and products are acting ideally and we are free to use concentrations instead of activities.

16.5.1 First-order kinetics

Consider the simple unimolecular elementary reaction of 1 mol of A transforming to 1 mol of P performed at constant volume and temperature. Initially, that is at t = 0, A is present at concentration [A]₀ and there is no P present. When the concentration of A changes by −x, the stoichiometry demands that the concentration of P increases by x. Starting with a stoichiometric reaction, we can represent this in tabular form as

Since the reaction is first-order, we write the differential rate law for the consumption of A with effective rate constant k_A as

(16.4)

(16.5)

The use and meaning of an effective rate constant will be explained below when we consider reactions of different stoichiometry.

To obtain the integrated form of the rate equation, we use separation of variables to move all terms involving x to one side, and all terms involving t to the other side. Then, the boundary conditions in the table above are used to define the integrals to be evaluated.

(16.6)

(16.7)

The rate of appearance of P is also governed by first-order kinetics. Using the stoichiometry of the reaction (in other words, substituting [A] = a₀ – x or [P] = x) we can rewrite Eq. (16.7) for the rates of reagent consumption as well as the rate of product formation,

(16.8)

(16.9)

Equations (16.8) and (16.9) are plotted in Fig. 16.2. The plot shows that the concentration of A decreases as a result of first-order kinetics in a simple exponential fashion. Because of the 1:1 stoichiometry, note that the decay constant is the same as the true rate constant. That is, the effective rate constants k_A and k_P are equal to the true rate constant k defined by

(16.10)

In other words, the true rate constant is the rate constant that determines the rate of reaction. An effective rate constant defines the rate of disappearance of a specific reactant or the rate of appearance of a specific product.

The reason why we need to introduce both effective rate constants with respect to a specific reactant or product and the true rate constant – which is simply called the rate constant if it need not be differentiated from an effective rate constant – is because not all first-order reactions have 1:1 stoichiometry. For example, there are reactions such as 2A → P or A → 2P or A + B → P, and so on. With the definition of x used above, the procedure for obtaining Eqs (16.7) and (16.8) is valid for all stoichiometries. What changes is the relationship of x to [P] (and [B] if a second reagent with kinetic order zero is involved) and the relationship of k_A to k. For instance, the reaction A → 2P has [A] = a₀ – x but [P] = 2x. Therefore, and [P] = 2a₀(1 − exp ( − k t)).

16.5.1.1 Directed practice

Confirm substitution into Eq. (16.7) leads to Eqs (16.8) and (16.9).

The exponential form of Eq. (16.8) means that if we plot ln([A]/a₀) versus time, we should obtain a straight line with slope equal to −k_A. This is known as the method of integration. The stoichiometry of the reaction then determines the relationship between k_A and k.

16.5.1.2 Example

A first-order reaction A → P is 40% complete after 1 min. (a) What is the value of the rate constant? (b) How long does it take to reach 80% completion?

(a) That the reaction is 40% complete in 1 min means that the concentration at t = 60 s, [A]_{t = 60 s}, is (1 – 0.4) of the initial value. In mathematical terms [A]_{t = 60 s} = (1 – 0.4)[A]₀ = 0.6 [A]₀. Rearrange Eq. (16.8) and solve for the rate constant k.

(16.11)

(b) Then at 80% completion, [A] = (1 – 0.8) a₀ = 0.2 a₀. Solving Eq. (16.11) for t, we find

First-order kinetics is widely observed in any type of phenomenon involving a well-defined lifetime. The lifetime is defined by τ = 1/k. Radioactive decay, fluorescence and phosphorescence all follow first-order kinetics with well-defined lifetimes in the simplest cases. The same is often observed for the initial stages of growth of bacterial colonies.

16.5.2 Second-order kinetics

There are four types of second-order reactions to consider. The first type involves the reaction of two different species which both have the same stoichiometric coefficient of one, as in the reaction

In its most general case, the concentrations of A and B need not be the same. The second-order empirical rate equation takes the form

(16.12)

The differential rate equation is written most generally by stating it in terms of the effective rate constant k_A as well as x, a₀ and b₀,

(16.13)

Integrating the rate equation with the conditions a₀ ≠ b₀ and x = 0 at t = 0, we obtain

(16.14)

If we plot the left-hand side of Eq. (16.14) versus t, the slope is equal to k_A. Due to stoichiometry k = k_A = k_B.

The second type is a reaction of the same stoichiometry but in the special case of a₀ = b₀. Now, however, the boundary conditions of the system are different, which allows us to rewrite Eq. (16.13) as

(16.15)

Integration yields

(16.16)

A plot of x/a₀(a₀ − x) versus t yields a straight line of slope k_A and again k = k_A = k_B.

16.5.2.1 Directed practice

As an alternative to Eqs (16.15) and (16.16), state the rate in terms of d[A]/dt and show that a plot of 1/[A] versus t should be linear with slope k_A, as in Eq. (16.17).

If A and B are the same chemical species, then the two reactants must have the same concentration. We now have the reaction

The solution is the same as for A + B with equal concentrations,

(16.17)

but we must take account of the factor of two when evaluating concentrations and the rate constants. A plot of 1/[A] versus t, as shown in Fig. 16.3, yields a straight line with slope equal to k_A, but now k_A = 2k.

Both, Eqs (16.16) and (16.17) describe the change in concentration of a reactant if second-order kinetics are followed. The equations describe the concentration of A versus time for all of the following situations: (i) A + B → P and with partial order of 1 for both A and B; (ii) 2A → P and second-order with respect to A; and (iii) 2A + B + ∙∙∙ → P and second order with respect to A, as long as the concentrations of other reactants are constant. If both Eqs (16.16) and (16.17) hold, which one should be used to determine the effective rate constant k_A? The answer is that either one can be used but one or the other may be easier to use based on the system under investigation. If the concentration of A is directly measureable, then Eq. (16.17) is a sensible choice as [A] is directly related to the measured parameter. If, on the other hand, the concentration of a product is more easily measured, then Eq. (16.16) is a logical choice since x is directly related to the measured parameter. If both concentrations can be measured, then analysis with both forms allows for independent determination of the kinetic parameters.

16.5.2.2 Example

The rate constant at 25 °C and constant volume for the gas-phase reaction

is k = 4.0 × 10⁻² kPa⁻¹ s⁻¹. If the initial partial pressures of both NO and HNO₃ are 50 kPa, calculate the total pressure, the partial pressure of NO, and the partial pressure of NO₂ after 1 s, 5 s, 10 s, and 50 s.

As long as the reaction is performed at constant volume and temperature, we may substitute pressure for concentration. Thus,

where p₀ = 50 kPa. The results are shown in Fig. 16.4 and the table below.

t/(s)	0	1	5	10	50
ptotal/(kPa)	100	100	100	100	100
p(NO)/(kPa)	50	17	4.5	2.4	0.50
p(NO2)/(kPa)	0	33	45.5	47.6	49.5

16.5.3 Pseudo-order kinetics

For second- and higher-order reactions in which more than one reactant contributes to the overall order of reaction, the kinetic order is sensitive to the relative concentration of the reactants. If all but one of the reactants is present in great excess, then the concentrations of the species in excess will not change appreciably during the course of the reaction. The reaction kinetics will only depend on the kinetic dependence of the one low-concentration species. This situation is known as pseudo-order kinetics.

Consider a reaction of the type

with the second-order empirical rate equation

If [B] » [A], then [B] does not change significantly (say <5%) during the course of the reaction. Therefore, [B] can be taken as a constant and we can define an effective rate constant according to

(16.18)

The rate equation then becomes

(16.19)

To determine k, we first determine k^* at constant [B] as we would for any first-order reaction. Then, after determining the rate for several different values of [B], a plot of k^* versus [B] should yield a straight line with slope equal to k.

This is commonly observed for solvolysis reactions. For example, the reaction of an alkyl halide with water to form an alcohol as in

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