Module 4 - Chemical Kinetics
Average Rate (change in concentration of NO)
- If we look at the change in concentration of NO between 5 and 10 μsec, we can calculate a rate by noting the change in concentration over this time interval, and obtain a rate of 1.4 M/s. - Similarly, if we define the rate in terms of the change in [N2] over the same time interval, we calculate a rate of 0.70 M/s. Note that this is half the rate of formation of product NO, as expected based on the reaction stoichiometry. - The reaction rates calculated in this manner represent average reaction rates and are only valid for the time interval in question. Any other time interval (e.g., between 25 and 30 μs) will yield a different value.
Molecular Orientation (ineffective collision)
- In the reaction sequence on the right, the reactants are oriented so that the O atom in NO is in line with the O3. Even if the collision energy is sufficient to break the O3 bonds, the free O atom is not in position to form a bond with the N atom in NO, and so the activated complex is not formed and no reaction occurs. - This dependence on collision orientation is reflected in the A constant and depends on the probability of a collision occurring in an appropriate orientation to yield products.
Average Rate
- Similarly, if we define the rate in terms of the change in [N2] over the same time interval, we calculate a rate of 0.70 M/s. Note that this is half the rate of formation of product NO, as expected based on the reaction stoichiometry. - The reaction rates calculated in this manner represent average reaction rates and are only valid for the time interval in question. Any other time interval (e.g., between 25 and 30 μs) will yield a different value.
Enzymes and Biocatalysis (plotted)
- The specificity of enzymes is due to the shape/structure of an active site, which is tailored to interact with a specific substrate using molecular recognition strategies discussed previously (e.g., chirality, complementary intermolecular forces). Enzymes are also used in many industrial applications because they convey the chiral specificity to produce enantiomerically pure products. This practice of using enzymes to catalyze reactions on a large scale is called biocatalysis. - The formation of an enzyme-substrate complex is analogous to the activated complex in a reaction mechanism and leads to the formation of product as shown in the two-step mechanism on the slide. - The rate of reaction is proportional to [S], and so can be considered as a first-order reaction. If the concentration of S is sufficiently large, however, all the active sites on the enzyme become filled and further increases in S do not change the rate of the reaction. Under these conditions, the reaction may appear to be zero-order in S.
Integrated Rate Law: First Order
Determining rate laws using initial rates has two disadvantages: 1. You must perform multiple experiments with varying initial concentrations. 2. You must determine the initial rate for each experiment accurately. It would be faster and easier if we could determine the rate law and rate constant by monitoring concentrations for one experiment. The integrated rate law approach allows this. - Consider the decomposition of ozone. Since there is only one reactant, we can write a generic rate law as rate = -Δ[O3]/Δt = k[O3]n. If we assume that the reaction is first order, then n = 1. - If we rearrange the rate law, isolating the concentration terms on one side and the kΔt term on the other side, we can apply integral calculus to obtain: ln[O3]t = -kt + ln[O3]o. - This integrated rate law takes the form of an equation for a straight line: y = mx + b, where y = ln[O3], m = slope (-k), x = time, and b (intercept) = the initial concentration of ozone, [O3]o. If the reaction is first order, then plotting [O3] versus time will yield a straight line with slope = -k.
The molecular models use the standard atomic color palette: C is black, N is blue, O is red, F is light green, and Cl is dark green. 2. Which representation is the overall reaction for the chlorine-catalyzed destruction of ozone?
E
Catalysts and the Ozone Layer
Ozone in photochemical smog is an irritant and contributes to the destruction of plants. In the stratosphere, however, it is useful because it absorbs UV radiation. In 1974, two chemists Sherwood Rowland and Mario Molina predicted that the presence of volatile organic compounds called chlorofluorocarbons (CFCs) would contribute to a significant depletion of the stratospheric ozone layer. This prediction was later supported by experimental evidence that demonstrated a thinning of the ozone layer and the appearance of an "ozone hole" over Antarctica. This phenomenon is illustrated in Figure 13.27. The ozone hole increased substantially from 1980 to 2000 (measured in millions of km2). This trend led to an international agreement called the Montreal Protocol, which limited the production of CFCs. Since then the ozone levels have started to increase and the ozone hole has decreased, but it will take a long time for the levels to recover to pre-1980 levels because of the persistence of these compounds in the environment.
Cars, Trucks, and Air Quality
Photochemical smog consists of various nitrogen oxides (NOx), ozone, and some small organic molecules. Smog is created by the interaction of sunlight with nitrogen oxides from vehicle exhaust and volatile organic compounds (VOCs) from manmade sources. -- The conditions that lead to the formation of smog, and the rate at which these compounds form, depend on chemical kinetics. We define the area of kinetics as the study of the rates of change of concentrations of substances involved in chemical reactions.
Ozone Layer
Photodecomposition of ozone is a naturally occurring process in the stratosphere, with a proposed two-step process as shown. The net reaction is the conversion of ozone to O2. The activation energy for the rate-determining step in this mechanism is 17.7 kJ/mol.
Graphical Determination of First-Order Reactions (calculating slope)
Typical data for the photodecomposition of ozone over time are presented in Table 13.6, along with a plot of the log of concentration versus time. Consistent with the expectations for a first order reaction, this plot produces a straight line. - Using the data, we can calculate the slope as Δln[O3]/Δt = -k = -1.1 × 10-3 s-1
Instantaneous Reaction Rates
We can also determine an instantaneous rate of a reaction as the tangential slope of the concentration plotted versus time. This is illustrated on the slide for the reaction between NO and O2 to form NO2. - The change in concentration of O2 over a specific time interval can be used to calculate the average rate of consumption of O2 during that time interval, but can also be used to determine the instantaneous rate at the midpoint of that time interval.
Effect of Temperature
We can also rationalize the effect of temperature on k by recalling how kinetic energy is related to temperature. - Consider the distribution of collision energies at two different temperatures as illustrated in Figure 13.19, in which T2 > T1. - On the graph, the dashed line represents Ea, the minimum collision energy required to form the activated complex and form products. As temperature increases, the fraction of collisions exceeding this minimum energy also increases, leading to an increase in k and an increase in the rate of the reaction.
Reaction Energy Profile
We can illustrate the role of activation energy by considering the relative internal energies in the reactants and products and the collision energy required to produce a reaction. An example of such an energy profile is provided in Figure 13.18. - Note the relative internal energy of the reactants on the left and the internal energy of the products on the right. - As the reaction proceeds, collisions occur between reactant molecules. If a collision has sufficient energy, it produces an intermediate state between reactants and products, called an activated complex. The energy difference between the original reactants and the activated complex is the activation energy. Collisions having sufficient energy to overcome this energy barrier and form the activated complex will yield products. - Note that the bigger the Ea, the fewer collisions you would expect to have sufficient energy and the slower the reaction rate (i.e., the smaller the value of k); hence, the negative exponential dependence of k on Ea in the Arrhenius equation.
Determination of Rate Laws (finding m & n)
- To find m (the reaction order with respect to [O2]), we can compare the relative change in reaction rate based on the change in [O2]. - The data show that a doubling of the [O2] produces a doubling of the rate. In other words, the rate is directly proportional to [O2]. The value of m = 1. Note that we could also verify this mathematically by setting up a ratio of the log of rates for trials 1 and 2 over the log of initial concentrations for trials 1 and 2, as illustrated in Equation 13.14 in text. - Similarly, we can determine the value of n by comparing the reaction rates for trials 1 and 3, with the initial [NO] for those trials. In this case, we see that doubling the initial [NO] results in an increase in reaction rate by a factor of four. In other words, rate is proportional to Δ[NO]2 or n = 2. - We can now write the rate law for this reaction as rate = k[O2][NO]2. Using the experimental data for any trial (rate, initial concentrations) we can insert the data and calculate the rate constant, k.
The image shows two graduated cylinders that contain aqueous solutions of copper(II) nitrate. Cylinder 1 has a concentration of 0.010 M. What is the concentration of the solution in cylinder 2? 0.010 M 0.005 M 0.020 M
0.020 M
The images represent three samples of nitrogen dioxide, a pollutant. Assuming the rate of the decomposition reaction depends on the frequency of the collisions between NO2 molecules, in which sample does the reaction proceed four times as rapidly as it does in sample (c)? (a) (b) Neither (a) nor (b)
A
The images represent three samples of nitrogen dioxide, a pollutant. At high temperatures, nitrogen dioxide decomposes to nitrogen monoxide and oxygen gas. In which of the samples above do the NO2 molecules collide most frequently—assuming the temperature of all three samples is the same? (a) (b) (c)
A
The molecular models use the standard atomic color palette: C is black, N is blue, O is red, F is light green, and Cl is dark green. 1. Which two images describe elementary steps which, when combined, depict the chlorine-catalyzed destruction of ozone?
A and H
Half-Life: First-Order Reactions
An important parameter in reaction kinetics is the half-life, which is the time needed for the initial concentration of a reactant to decrease by half. Half-life is inversely related to the rate constant; the larger the rate constant (i.e., the faster the reaction), the shorter the half-life. - To illustrate, consider the decomposition of N2O, nitrous oxide. It is another potent greenhouse gas and is produced by bacterial degradation of compounds in the soil. - The decomposition reaction is first order and is very rapid. A typical reaction curve of [N2O] versus time is presented in Figure 13.13. - Clearly, the reaction proceeds rapidly, with the initial concentration decreasing by half (from 16 → 8) in the first one-second interval. Note that in the next one-second interval, the concentration decreases by half again (from 8 → 4), and in the third one-second interval decreases by another half (from 4 → 2). - Each of these one-second intervals represents a half-life, designated as t1/2, for this reaction. Each successive half-life is the same time interval as the previous one. This is characteristic of a first-order reaction.
The image shows two graduated cylinders that contain aqueous solutions of copper(II) nitrate. Cylinder 1 has a concentration of 0.010 M. Which cylinder contains more dissolved solute? Cylinder 1 Cylinder 2 Cylinders 1 and 2 contain the same amount.
Cylinders 1 and 2 contain the same amount
Catalytic Converters
Another example of the action of catalysts is the catalytic converter that we mentioned at the beginning of this chapter. Catalytic converters are included in all automobiles manufactured in the United States in order to decrease the production of NOx compounds that contribute to photochemical smog. - Large surface areas of metal clusters in the honeycomb-like structure of the catalytic converter allows for rapid oxidation/reduction of the NOx species and speeds up their conversion to nonpolluting N2 and O2. - Because the catalyst (metal surface) is in a different phase than the reactants (gases), this is another example of a heterogeneous catalyst.
Graphical Determination of Ea
Another useful application of the Arrhenius equation is that it allows us to determine the activation energy of a reaction. - Consider again the log form of the Arrhenius equation: ln k = -Ea/RT + ln A - We have already examined how we can determine the rate constant, k, for a reaction once the rate law is known. If we run the reaction at different temperatures, we can determine the value of k at different temperatures. - Plotting ln (k) versus 1/T produces a straight line with a slope = -Ea/R and an intercept = ln A
Actual Reaction Rates (graphed)
As expected, the concentration of reactants decreases and the concentration of product increases as the reaction proceeds.
The molecular models use the standard atomic color palette: C is black, N is blue, O is red, F is light green, and Cl is dark green. 3. Which two images describe elementary steps that combine in an overall reaction in which NO3 is an intermediate?
B and D
Mechanisms: Zero Order
Before we leave the topic of mechanisms, let's consider an example in which a reaction mechanism can explain a zero-order reaction. - Reconsider the reaction: NO2 + CO→ NO + CO2 Rate = k[NO2]2 - How can we use reaction mechanisms to rationalize this result? - Consider the two-step mechanism presented on the slide. The sum of the elementary steps yields the observed balanced chemical equation, so the first requirement is fulfilled. If step 1 is the RDS; the rate law based on the molecularity of step 1 is consistent with the observed rate law, so the second requirement is fulfilled. The overall reaction is zero order with respect to [CO2] because it is not involved in the rate-determining step.
The images represent three samples of nitrogen dioxide, a pollutant. At high temperatures, nitrogen dioxide decomposes to nitrogen monoxide and oxygen gas. Which is the balanced equation describing the thermal decomposition of NO2? NO2 → NO + O2 NO2 → 2NO + O2 2NO2 → 2NO + O2 2NO2 → N2 + 2O2
C. 2NO2 → 2NO + O2
Reaction Mechanisms (reaction/steps)
Consider the reaction: 2 NO2(g) → 2 NO(g) + O2(g) cc- We may propose that this reaction takes place in two separate steps: - Step 1: Two molecules of NO2 react to produce NO and NO3. - Step 2: The NO3 molecule decomposes to produce a molecule NO and O2. There are several things to note about this mechanism. First, the NO3 produced in step 1 is an intermediate—it appears as a product of the first step and is consumed as a reactant in the second step. In addition, step 1 is bimolecular because it involves two reactant species; while step 2 is unimolecular (only one species is involved). Finally, if we add the two steps together and cancel the intermediate species, we obtain the balanced chemical equation of interest. - Another point of interest is that we can write rate laws based on the molecularity of each of the elementary steps simply by examination. For example: - Step 1: Rate = k1[NO2]2 - Step 2: Rate = k2[NO3] We can do this for elementary steps because we are stipulating which molecules (and how many) are involved in the reaction at this step. We cannot do this for the overall balanced reaction because we do not know exactly what is happening at the molecular level. But how do we know if this mechanism is feasible? To satisfy this requirement, the mechanism must be consistent with the experimentally observed rate law. To address this issue, we must identify the rate-determining step (RDS), or the slowest step. Since the reaction cannot proceed faster than the slowest step, the rate law obtained from the slowest elementary step must match the experimentally observed rate law.
Important Reactions in Smog
Consider the various reactions involved in the creation of photochemical smog. - The first reaction takes place within the internal combustion engine and involves the reaction between N2 and O2 to form NO. The reaction is highly endothermic, but the high temperatures within the combustion engine make it possible for this reaction to proceed. - In the second reaction, the NO produced within the engine is released to the atmosphere, where it combines with O2 to form NO2, the brown gas commonly associated with smog. Note that this reaction is exothermic and, therefore, energetically favorable. - In the third reaction, sunlight provides the energy to break bonds in NO2 to form NO and O atoms. These O atoms can then combine with O2 to form ozone or react with water vapor to form OH radicals. - The OH radicals are highly reactive species that can react with VOCs to form other substances, such as peroxyacetyl nitrate.
The image shows two graduated cylinders that contain aqueous solutions of copper(II) nitrate. Cylinder 1 has a concentration of 0.010 M. Which cylinder contains the more concentrated solution? Cylinder 1 Cylinder 2 Cylinders 1 and 2 are the same concentration.
Cylinder 1
Reaction Mechanisms (reaction energy profile)
Examine the reaction energy profile for the proposed reaction mechanism: - Step 1 has a large activation energy, and thus must be a very slow step. - Step 2 has a relatively small activation energy, and proceeds more rapidly. Once intermediates are produced in step 1 they are rapidly converted to products in step 2. Therefore, the overall reaction rate is determined by the rate of reaction in step 1, and the observed rate law is the same as the rate law obtained from the molecularity of step 1. - Rate = k[NO2]2
Reaction Order
Experimental observations and theoretical considerations tell us that the reaction rate depends on reactant concentrations, but they do not tell us the exact dependency. Based on experimental data we can determine the nature of this dependency and express it as reaction order, a parameter that tells us how a reaction rate depends on reactant concentration. - Knowing the reaction order also provides insight into how a reaction might proceed on the molecular level. For example, which molecules must collide and which bonds must break in order to form the observed products. - We can summarize the relationship between reaction rates and the concentrations of reactants using a rate law, which is a mathematical relationship that defines the experimentally observed relationship between rates and concentrations (i.e., reaction orders). - We can also determine the overall order of the reaction as the sum of the reaction orders of the individual reactants.
The molecular models use the standard atomic color palette: C is black, N is blue, O is red, F is light green, and Cl is dark green. 4. Which image shows the photodecomposition of a chlorofluorocarbon?
G
CFC Emissions and Ozone
How do CFCs contribute to the destruction of the ozone layer? These compounds are volatile and extremely stable and remain in the atmosphere for long periods of time so that they ultimately can disperse up into the stratosphere. - In the stratosphere, these compounds encounter UV radiation of sufficient energy to cause decomposition and produce Cl atoms. - These Cl atoms can then interact with ozone via the two-step mechanism shown. The activation energy of the RDS for this mechanism is only 2.2 kJ/mol, so this reaction mechanism is faster, resulting in an increased rated of ozone destruction compared to the natural process presented previously. - Note that the overall reaction is the same as that for the natural photodecomposition of ozone from the previous slide. Note also the intermediate species ClO, which appears in step 1 and is consumed in step 2. But what about Cl? It appears in step 1 and is regenerated in step 2 so it can be recycled over and over. It clearly affects the rate of the reaction, but does not appear in the balanced chemical equation and is not consumed during the reaction. - Chlorine is a catalyst, a substance that increases the rate of a reaction but is not consumed during the reaction.
Integrated Rate Law: Second-Order Reactions (Plotted)
In Figure 13.14, we have plotted the concentration data as ln[NO2] versus time and as 1/[NO2] versus time. - The ln[NO2] plot is curved, but the 1/[NO2] plot is linear, consistent with the expectation for a second-order reaction. Thus, the rate law for this reaction is rate = k[NO2]2. We can calculate the rate constant, k, as the slope of the plot.
Integrated Rate Law: Second-Order Reactions
In the introduction we described the photodecomposition of NO2 and its contributions to photochemical smog. NO2 can also undergo thermal decomposition via the reaction shown on the slide. - Since there is only one reactant we might expect this reaction to be first order with regard to NO2, but the concentration data versus time are not consistent with this observation. Can we use the integrated rate law approach to determine the order of the reaction? - Let's assume the reaction is second order, with a rate law of rate = Δ[NO2]/Δt = k[NO3]2. - Following the same logic as for the first-order rate law, we rearrange the rate law equation and integrate with respect to time (dt), which produces an integrated rate law of 1/[NO2]t = kt + 1/[NO2]o - Again, this equation takes the form of a straight line. Plotting 1/[NO2] versus time produces a linear plot with a slope = k and an intercept = 1/[NO2]o.
Initial Reaction Rates
In the previous section we examined how we could calculate both the average reaction rate over a specific time interval and the instantaneous rate as the tangential slope at a given point. - Consider the concentration data plotted in Figure 13.9. The tangential slope of the curve at points b and c represent instantaneous rates (tangential slopes) at specific times in the reaction. The slope at point a when t = 0 represents what is known as the initial rate. - Clearly the reaction rate (i.e., tangential slope of the concentration curve) decreases as the reaction proceeds. Eventually, we can imagine that the rate of change would approach zero, at which point concentrations of reactants and products no longer change. - This behavior is typical for most reactions. The most rapid reaction rates occur early in the reaction and reaction rates decrease over time. We can explain this in terms of molecular collisions. As concentrations decrease over time, the number of collisions between reacting species decreases and the reaction rate slows down accordingly. - Exactly how reaction rates depend on concentrations of reactants, however, is not always obvious. If we double the concentration of one reactant, the reaction rate may also double. If we double the concentration of another reactant, however, the reaction rate might quadruple. We need a way to determine the nature of the dependency of rates on concentrations of reactants.
Variations of Smog Components
In the reactions involved in smog formation, the products of some reactions are the reactants in others. Therefore, the relative rates of these reactions influence at what time during the day pollutants appear and how long they persist. - Note in Figure 13.4, for example, that the maximum concentration of NO occurs during morning rush hour. - As the NO is converted to NO2, the concentration of NO2 increases and NO decreases. - Later in the day, sunlight breaks down the NO2 to form free O atoms, which then react to form ozone. Hence, the O3 concentration begins to increase during the afternoon. - The catalytic converter, which is now a standard feature on automobiles in this country, has helped to minimize the problem of photochemical smog by removing NO from exhaust gases. Knowledge of chemical kinetics has made it possible to create catalytic converters and other devices that can be used to improve air quality.
The Rate Law and Rate Constant
Included in the rate law is a proportionality constant, k, called the rate constant that relates the concentration of reactants to the observed reaction rate. - Consider a generic reaction: A + B → C - We can write a generic rate law for this reaction as: Rate = k[A]m[B]n - Where the rate is defined in terms of the change in concentration of a given species over time (can be initial rate or instantaneous rate), k is the rate constant, [A] and [B] are the concentrations of reactants at a given time, and m and n are the reaction orders with respect to the reactants A and B. The values of m, n, and k must be determined experimentally.
Zero-Order Reaction
It is also possible that the rate of a reaction may be independent of the concentration of a given reactant. Such reactions are called zero-order reaction with regard to that reactant and the rate law can be written as rate = -Δ[X]/Δt = k[X]o = k. Applying the same logic as for previous examples, we can rearrange and integrate to obtain: [X]t = -kt, and t1/2 = [X]o/2k. We will discuss some examples of zero-order reactions in a later section.
Reaction Rates
Let's return to the reaction between N2 and O2 that we introduced at the beginning of the chapter. The amount of NO formed in a given period of time depends on how fast the reaction proceeds. - We can define the rate of the reaction in terms of the change in NO over time. The change in concentration can be calculated as the final concentration (at some time) minus the initial concentration. Since NO is a product, its concentration is increasing over time and so the rate is positive. - Alternatively, we can express the reaction rate in terms of the rate at which a given reactant is consumed. Alternative expressions for the rate of consumption of N2 and O2 are shown on the slide. In all cases the rate is expressed in terms of a change in concentration of a reactant or product (in units of moles/L) as a function of time.
Determination of Rate Laws
Let's revisit the reaction between N2 and O2 to form NO. - We can write a generic rate law for the reaction as: Rate = k[N2]m[O2]n - The values of m and n can be determined experimentally by observing how the initial rate of the reaction changes as the initial concentration of reactants is systematically changed. - Consider the reaction data in Table 13.3. Note that the initial concentration of NO is the same in trials 1 and 2, but the concentration of O2 changes by a factor of two. Any change in the rate of the reaction in this case must be due to the change in [O2]. - Similarly, the initial [O2] remains the same between trials 1 and 3, but the initial [NO] changes by a factor of 2. Any change in the reaction rate in this case must be due to the change in [NO]. We can use these data to mathematically determine the reaction order with respect to [NO] and [O2].
Note on Important Reactions in Smog
Note that these reactions are linked—products of one reaction also serve as reactants in another reaction. As a result, the system is very complex. In order to understand the phenomenon of photochemical smog, we need to understand the relationship between each of the individual reactions, as well as the factors that will affect how fast a given reaction may proceed.
Energy Profiles (High Ea)
Now consider the energy profile for the reverse reaction between NO2 and O2. - For the reverse reaction illustrated in Figure 13.21b, the Ea is now very large, which translates to a very small value for k and a very slow reaction rate for the reverse reaction. - The size of the activation energy depends on the direction in which the reaction is considered. In this example, the activation energy for the forward reaction is substantially smaller than the Ea for the reverse reaction, so the forward reaction will proceed much more rapidly than the reverse reaction (if concentrations of reactants and products are comparable).
Reaction Mechanisms
So far we have examined how the rate law defines the relationship between reaction rate and the concentration of reactants. We have also discussed how the rate law can be determined experimentally and how the value of the rate constant can provide insight into important parameters related to the reaction, such as the activation energy. But why are some reactions first order and others second order? To explain this we need to examine what happens at the molecular level during a reaction by proposing a reaction mechanism. - A reaction mechanism is a proposed series of steps that describe how a reaction occurs at the molecular level. - It consists of a series of elementary steps, which are chemical reactions that describe a single process in the overall reaction. - An elementary step may include reactants and products, but may also include intermediates, which are species that are produced in one elementary step and consumed in a subsequent step so that they do not appear in the overall balanced chemical equation. - We can define the molecularity of an elementary step based on the number of reactant species that are involved in that step. In order for a reaction mechanism to be feasible, the individual steps must add up to yield the observed overall balanced chemical equation and it must be consistent with the observed rate law.
Actual Reaction Rates (presented)
Suppose we ran a reaction in the laboratory to determine the rate of the reaction between N2 and O2. We can express the rate either in terms of the rate of consumption of N2 or O2, or in terms of the rate of formation of NO. - A typical set of concentration data for these species are presented in Table 13.1
Enzymes and Biocatalysis
The catalytic surfaces in catalytic converters are capable of catalyzing a variety of reactions. Catalysts are also very useful in biochemistry; in contrast, however, these biochemical catalysts are highly selective and typically only work for a very specific reaction. Such biochemical catalysts are called enzymes and the substances with which they act catalytically are called substrates. Enzymes are very important in metabolic pathways, sequences of reactions that are responsible for the synthesis of complex materials from simpler compounds.
Energy Profiles (Low Ea)
The energy profile shows the changes in internal energy that occur as the reaction progresses from reactants to the activated complex to products. It allows us to determine the Ea for the reaction and also allows us to determine the enthalpy of the reaction as the difference between the internal energies of reactants and products. - Consider the energy profile for the reaction between nitric oxide and ozone. As illustrated in Figure 13.21a, the Ea for the forward reaction is very small, which translates to a very large value for the rate constant k and a fast reaction rate.
Integrated Rate Law: Second-Order Reactions (General Form)
The general form of the integrated rate law is: 1/[X]t = kt + 1/[X]o. Plotting 1/[X] versus time produces a linear plot with a slope = k and an intercept = 1/[X]o. We can use the integrated rate law to determine the relationship between half-life and k for a second-order reaction. At t = t1/2, [X]t = 0.5[X]o. By substitution and rearrangement we obtain: t1/2 = 1/(k[X]). The half-life is still inversely related to k, as expected, and is now also dependent on the concentration of reactant. As the reaction proceeds and the [X] decreases, the half-life will increase! Half-life for the second order reaction depends on concentration whereas first order is independent of concentration.
Mathematical Determination of Ea
The graphical approach generally requires the determination of k at multiple temperatures in order to produce a reliable graph. Alternatively, we can calculate Ea mathematically if we know the value of k at two different temperatures. - Consider the Arrhenius equation for ln (k) at two different temperatures, as shown. - If we subtract one equation from the other and rearrange, we obtain the equation shown. If k is known at two different temperatures, and R is a constant, this relationship can be used to calculate the value of Ea for a reaction.
Instantaneous Reaction Rates (tangential slope)
The instantaneous rate can be seen more clearly as the tangential slope in this expanded section of the [O2] curve. - Using the change in O2 concentration between 1000 and 3000 sec, we can calculate the average rate of consumption of O2 as 6.0 × 10-7 M/s. This calculated rate represents the average rate over the time interval from 1000 to 3000 sec, but also represents the instantaneous rate at the midpoint of this time interval (at 2000 sec).
Other Reaction Orders
The integrated rate law for a second-order reaction only applies to a rate law that is second order in a particular reactant. It does not apply to a reaction that is second order overall, but is first order in two different reactants. - Consider a reaction, NO2 + O3 → NO3 + O2 with the observed rate law of rate = k[NO2][O3]. - The reaction is first order in both reactants and is second order overall. We can still use the integrated rate law to verify the reaction order if the concentration of one reactant is substantially greater than the other reactant so that its concentration does not change significantly over the course of the reaction. - Let's assume that the [O3] >> [NO2] in the above reaction. Under these conditions, the [O3] does not change significantly as the reaction proceeds and can be considered to be relatively constant. We can rewrite the rate law as Rate = (k[O3]) [NO2] = k'[NO2] - This rate law looks like a first-order rate law and can be dealt with in the same way using the integrated rate law. Such reactions are called pseudo-first-order reactions.
Relative Reaction Rates
The rate of formation of product is not necessarily the same as the rate of consumption of a reactant, but the two are related based on the stoichiometry of the reaction. - If we examine the balanced chemical reaction we see that one mole of N2 reacts with one mole of O2. Therefore, the rate of consumption of O2 and N2 must be the same. - For every mole of N2 and O2 in that reaction, however, two moles of NO are formed. Therefore, the rate of formation of NO is twice the rate of consumption of N2 or O2. - In order to set the rates equivalent to each other we must take stoichiometric factors into account. In this case, we divide the rate of formation of product [NO] by the stoichiometric factor of two. In general, this approach can be applied to define reaction rates for species in any balanced chemical equation. Although the balanced chemical equation will allow us to predict relative rates of formation and consumption of products and reactants, the actual reaction rates must be measured experimentally.
Arrhenius Equation
The relationship between the reaction rate constant, k, T, and Ea is known as the Arrhenius equation and is shown on the slide. - In addition to T (absolute temperature, Kelvin) and Ea, the equation includes the gas constant R [in units of J/(mol·K)] and a factor A, called the frequency factor, which is the product of collision frequency and an orientation factor, which we will discuss later. - The log form of the Arrhenius equation is: ln(k) = -Ea/RT + ln(A)
Energy Profiles for O3 Decomposition
The two reaction pathways for the decomposition of ozone are illustrated in the energy profiles in Figure 13.28. In this case, the Cl catalyst speeds up the decomposition by lowering the activation energy of the reaction pathway, leading to an increased reaction rate. - Chlorine is a catalyst, by definition, because it speeds up the reaction but is not consumed during the reaction. - Chlorine is called a homogeneous catalyst in this case because it is in the same phase (gas phase) as the reactants and products.
Graphical Determination of First-Order Reactions (plot)
This plot shows a curved line with changing rates over time. It is difficult to determine the rate constant for the reaction from this plot.
Molecular Orientation (formation of activated complex)
To understand the role of collision orientation and its contribution to the frequency factor, consider the reaction: O3 + NO → O2 + NO2. - In the reaction sequence on the left, the O3 and NO molecules collide in an orientation that allows one of the O atoms to form a bond with the N atom in NO, forming the activated complex and leading to products.
Half-life: First-Order Reactions - Integrated Rate Law
To understand why the half-life is constant, reconsider the integrated rate law for a first order reaction: ln([X]t1/2/[X]o) = -kt. - If we assume that t = t1/2, then the ratio of [X]t1/2/[X]o = 0.500, and the integrated rate law reduces to ln(0.500) = -kt1/2, which can be rearranged to produce t1/2 = 0.693/k. - Note that for a first-order reaction, the half-life depends inversely on k (as expected) and is independent of reactant concentration.
Graphical Determination of First-Order Reactions (table)
Typical data for the photodecomposition of ozone over time are presented in Table 13.6, along with a plot of concentration versus time.
Effects of Concentration on Reaction Rates
We can rationalize the dependence of reaction rates on concentrations by considering the illustration in Figure 13.10 for the reaction between NO and O3. The rate law indicates that the rate is first order with respect to NO and O3. - For a reaction to occur there must be a collision between reacting species (NO and O3). In the first flask, the rate depends on the collisions as indicated. - Moving to the second flask, the concentration of NO has doubled, so the probability of a collision between an NO and an O3 molecule also doubles. - Moving to the third flask, we now double the concentration of O3. The probability of collisions increases by another factor of two, or is now four times greater than in the first flask. - Similarly, increasing the concentration of NO and O3 by a factor of three each, compared to the first flask, increases the probability of a collision by 32 or by a factor of nine.
Chemical Kinetics
We define chemical kinetics as the study of the rate of change of concentrations of substances involved in a chemical reaction. We define the reaction rate as how rapidly a reaction occurs, which is related to the change in concentration of a reactant or product over time. Reaction rates are affected by several factors: - Physical state of reactants—in general, reaction rates are faster for substances in the liquid or gas phase than in the solid state. - Concentrations of reactants—the more collisions there are between particles, the more likely an interaction will lead to reaction. As concentrations increase, interactions between particles increase. - Temperature—the average kinetic energy of particles, and collisions between particles, increases with temperature. As the rate of collisions and collision energy increase, the likelihood of reaction increases. - Catalyst—the presence of a catalyst tends to increase the rate of a chemical reaction. We will explore each of these factors individually.
Factors Affecting Rate
We have already examined how concentrations of reacting species affect the rate of the reaction and how this is reflected in the rate law by reaction order. But what determines the value of k in the rate law? -As molecules collide, bonds in reactants are broken which allows new bonds to form to yield the products. The collisions between molecules must be of sufficient energy to break the bonds in reactants or no reaction occurs. The temperature of a reaction will affect the energy of the collisions because the kinetic energy of molecules is directly related to temperature. As T increases we would expect reaction rate to increase since kinetic energy of the molecules increases. - The minimum amount of collision energy required for reaction is called the activation energy (Ea). Every reaction has a characteristic activation energy, expressed in kJ/mol. - These factors (T and Ea) are reflected in the rate law by incorporation into the rate constant.
Reaction Mechanisms (rearrangement)
What about the reverse reaction? 2 NO(g) + O2(g) → 2 NO2(g) We previously determined that the experimentally observed rate law for this reaction was rate = k[NO]2[O2]. How do we rationalize this based on the proposed mechanism? Revisit the energy profile on the previous slide in terms of the activation energies of the two elementary steps: Step 1: NO(g) + O2(g) → NO3(g) Rate = k1[NO][O2] Step 2: NO3(g) + NO(g) → 2 NO2(g) Rate = k2[NO3][NO] In this case, the sum of the elementary steps yields the balanced chemical equation, but neither of the rate laws obtained from the molecularity of the elementary steps match the experimentally observed rate law. How do we resolve this? Consider the case in which step 2 is slow (i.e., the rate-determining step), while step 1 is fast and reversible. In other words, we can write expressions for the rate law for step 1 in both the forward and the reverse direction. The rate law for the RDS contains an intermediate, but we can take advantage of the reversibility of step 1 to resolve this issue. - If we set the two rate laws for the forward and reverse reactions for step 1equal to each other and rearrange, we obtain an equivalent expression for the [NO3], which we can then substitute into the rate law for the RDS. The resulting rate law agrees with the experimentally observed rate law. One final caveat—although a proposed mechanism is feasible and consistent with the observed rate law that does not prove that it is the correct mechanism. We might obtain additional information to support the proposed mechanism (i.e., we might be able to detect the presence of an intermediate that is consistent with the mechanism), but there might be other steps involved that we have not taken into consideration.