Chemical Kinetics Reaction Rates Up to this point in the course our concern with chemical equations has focused upon understanding, reactants, products, stoichiometry, and states. We have also looked at the energy associated with various reactions and physical processes, and we have also had a brief introduction to thinking about entropy associated with various physical processes. In the section on the solution process, we also began to think about the rates of the physical processes of dissolving and crystallization. The relative rates of these two processes determine whether a crystal dissolves, grows, or is in dynamic equilibrium We are now going to begin to consider the reaction rates of the chemical reactions that we have been studying. This area of chemistry is referred to as chemical kinetics. The rates of chemical reactions can be relatively fast, or slow, and can also be influenced by various factors, including: Concentrations of reactants. Generally speaking, the higher the concentration of reactants, the faster the rate of the reaction. The temperature. The higher the temperature, the faster the rate of the reaction. The presence of a catalyst. A catalyst is able to increase the rate of a reaction, although the catalyst itself is neither created nor destroyed when performing this function. Catalysts don't "cause" a reaction, they just speed it up. The surface area of the reactants or catalyst. Reactions that involve solids often proceed faster is the solid is a fine powder instead of big chunky bits. The physical difference between a fine powder, and big chunky bits, will be the surface area. (You typically will try to start a fire using kindling, and will not try to put a match to a large log) "Speed" The speed of any activity (e.g. running, reading, cooking hamburgers, etc) involves quantifying how much you accomplish in a specific amount of time We can also quantify, or measure, the speed of a chemical reaction (also known as its reaction rate) An example of a simple chemical reaction: Let's assume that this reaction does not occur instantaneously, and therefore, it takes some time At the beginning of the experiment, our sample is composed 100% of the A molecules (Note: the start of experiments that measure reaction rates is usually referred to a "T 0 ") As time goes by, the A molecules are chemically converted into the B molecules Thus, over time, the number of A molecules in the sample decreases and the number of B molecules increases 1
The reaction rate is a measure of how quickly the A molecules (not the mass, for this is not a measure of stoichiometry) are consumed, or how quickly the B molecules are produced The reaction rate can be expressed by measuring the change in the number of A or B molecules per unit of time o Numbers of molecules are quantified using moles (a convenient way to keep track of the large number of molecules in such experiments) o Time is often seconds or minutes In the reaction of A B, the reaction rate can be determined by measuring the increase in the amount of the B molecules over time. (Note: conversely we could monitor the decrease in the levels of the A molecules). This is the average reaction rate, or "rate of reaction" ΔT here refers to some period or unit of time. The change in the number of moles of product B would be determined by measuring the number of moles of B at the start of the time interval and comparing it to the number of moles of B at the end of the time interval Δ(moles of B) = (#moles of B at end of time interval - #moles of B at start of time interval) and Δ(moles of A) = (#moles of A at end of time interval - #moles of A at start of time interval) The following is a plot of some experimental data for this type of reaction. The plot displays time (in minutes) along the x-axis, and the number of moles of the A reactant and B product along the y-axis: Notice a couple of things: As the amount of reactant A decreases, the amount of product B increases (just like the equation says!) The stoichiometry of the balanced equation is reflected in the concentrations of A and B: at each time point the number of moles of A plus B equals 1.0 2
o We started with 1.0 mole of A. According to the stoichiometry, if 0.70 moles of A have been used up in the reaction, it must mean that 0.30 moles of A remain and 0.70 moles of B have been produced (see the 40 minute time point) The rate of disappearance of A (or appearance of B) does not appear to be linear with time o The reaction appears to slow down with time The reaction rate is not constant but changes with time We can determine the average reaction rate for each of the 10 minute intervals that data was collected: For the first 10 minute period, Δ(moles of B) = (moles of B at t = 10 min) - (moles of B at t = 0 min), and the time interval is 10 min The average reaction rate for the other 10 minute intervals in calculated similarly (Δt always 10 min) Results of the calculation of the average reaction rate for the appearance of the B product: Time Interval (min) Average Reaction Rate (moles/min) 0-10 0.026 10-20 0.02 20-30 0.014 30-40 0.01 40-50 0.008 50-60 0.006 Note that the reaction rates have units of moles/min in these calculations (they describe the amount of the reaction in a unit of time) How do the reaction rates for the formation of B relate to the reaction rates for the disappearance A? In the first 10 minutes, 0.026 moles of B were formed Thus, in the first 10 minutes, 0.026 moles of A disappeared, i.e. a negative reaction rate for the A component For the stoichiometry of this reaction (1:1), the reaction rate of A is equal in magnitude to the reaction rate of B, but opposite in sign Rates in terms of concentrations The volume of reactions is typically held constant. Instead of monitoring the absolute number of moles during a reaction it is more common to monitor concentration during a reaction 3
Reaction rates are therefore typically in units of M/sec, or M/min (i.e moles/l*sec, or moles/l*min) The reaction of butyl chloride with H 2 O to produce butanol and hydrochloric acid: C 4 H 9 Cl(aq) + H 2 O(l) C 4 H 9 OH(aq) + HCl(aq) (A + B C + D) The experiment is setup in the following way: We start by preparing a 0.10 M solution of C 4 H 9 Cl in H 2 O As soon as we mix the C 4 H 9 Cl in H 2 O to make a 0.10 M solution we call that the T 0 timepoint and we start the clock running Then we measure the concentration of C 4 H 9 Cl at various times Here's the raw data: Time (sec) [C 4 H 9 Cl] (M) 0 0.100 50 0.0905 100 0.0820 150 0.0741 200 0.0671 300 0.0549 400 0.0448 500 0.0368 800 0.0200 10,000 (undetectable) 4
In the above graph we have determined the average reaction rate over the time period 300-400 sec. The general trend of the plot of [C 4 H 9 Cl] vs. time is not a straight line: the reaction slows down over time Therefore, when considering the period of time from 300 to 400 seconds, the reaction rate is faster at 300 sec and slightly slower at 400 sec The average reaction rate is the average over that time period As we shorten the time period, there is less difference between the starting and ending reaction rates and the average reaction rate If the time period was infinitely small, the average reaction rate would be the instantaneous reaction rate for that time The instantaneous reaction rate is equal to the slope of a line tangent to the curve of [C 4 H 9 Cl] vs. time The slope of the tangent line is found by determining the rise/run for the line. Pick any two convenient time points and figure out Δt and Δ[C 4 H 9 Cl] 5
In this case, at 350 sec the slope of the tangent line is approximately Since we are concerned with the disappearance of C 4 H 9 Cl, the reaction rate is the negative of the slope (i.e. -Δ[A]/Δt) Therefore, the instantaneous reaction rate at 350 sec for the disappearance of C 4 H 9 Cl is 1.6 x 10-4 M/sec Reaction Rates and Stoichiometry In the above reaction the stoichiometry of the reaction is such that for every 1.0 mole of C 4 H 9 Cl reactant consumed we produce 1.0 mole of C 4 H 9 OH product. What about for other reactions where the stoichiometry is not 1:1? The reaction can be monitored by measuring the disappearance of HI(g) or by measuring the appearance of either the H 2 (g) or I 2 (g) However, one mole of either H 2 (g) or I 2 (g) will be produced for every two moles of HI(g) consumed Therefore, the rate of reaction of the consumption of HI(g) will be twice as rapid as the rate of production of H 2 (g) or I 2 (g) In other words, to correct the reaction rates for the stoichiometry we divide the rate for each component by their coefficient in the balanced equation 6
This general equation that describes the reaction rates normalized for the stoichiometry is termed the General Rate of Reaction or the General Rate Law During a chemical reaction: The Dependence of Rate on Concentration The reaction often starts out at a fast rate and then slows down as time goes on Also, the concentration of reactants also decreases over time (as they are consumed in the reaction) We can speed the reaction back up by adding more reactant(s), i.e. increasing the concentration of reactant(s) How does the starting concentration of a reactant affect the initial reaction rate? Consider the following reaction of ammonium ion (NH 4 + ) with Nitrite ion (NO 2 - ): NH 4 + (aq) + NO 2 - (aq) N 2 (g) + 2H 2 O(l) + - As far as the stoichiometry is concerned, NH 4 NO 2 N 2. Therefore, we could monitor the reaction rate by monitoring the changes in concentration, over time, of NH + 4 or NO - 2, or by the volume of the released N 2 gas; all of these rates will be equal The key point is to determine the initial instantaneous reaction rate for a variety of different starting concentrations of reactants How does this data look? Situation #1: let's keep the concentration of nitrite ion constant, and vary the concentration of ammonium ion (You might be asking how could we do this if we have an ionic compound of ammonium nitrite. Surely we would end up adding equal amounts of each ion. However, we could add a different ammonium salt, and it is counter ion could be unreactive and is thus a spectator ion in the reaction. The net result would be the addition of ammonium ion) [NH + 4 ] (M) [NO - 2 ] (M) Initial reaction rate (M/s) 0.01 0.20 5.4 x 10-7 0.02 0.20 10.8 x 10-7 0.04 0.20 21.5 x 10-7 0.06 0.20 32.3 x 10-7 7
The general relationship between concentration and initial reaction rate for the NH + 4 ion, is that the reaction rate is directly proportional; if we double the concentration of NH + 4 ion, then the reaction rate doubles Situation #2: let's keep the concentration of ammonium ion constant, and vary the nitrite ion concentration [NH + 4 ] (M) [NO - 2 ] (M) Initial reaction rate (M/s) 0.20 0.02 10.8 x 10-7 0.20 0.04 21.5 x 10-7 0.20 0.06 32.3 x 10-7 0.20 0.08 43.3 x 10-7 The general relationship is pretty much the same thing as seen when varying the ammonium ion. In other words, the reaction rate is directly proportional to the concentration of nitrite ion. Conclusion: For example, if the concentration of ammonium ion were doubled, and the concentration of nitrite ion were doubled, we would expect the initial reaction rate to increase by a factor of 4 We can write an exact equation by introducing a proportionality constant, k: This overall equation is called a rate law The constant, k, in the rate law is called the rate constant How is the rate constant, k, determined for a particular reaction? We need to know the particular rate law for a reaction We need a set of data that provides us with reaction rate information for different concentrations of reactant(s) Rearranging the rate law for the above reaction to solve for k: Let's try one of the data points from above: [NH 4 + ] (M) [NO 2 - ] (M) Initial reaction rate (M/s) 0.20 0.02 10.8 x 10-7 8
k is a constant, and the same value for k is found when using any data point from above If we know the rate law for a particular reaction, and the rate constant, k, then we can determine the instantaneous initial reaction rate for any given concentration of reactants What is the initial reaction rate for the above reaction if we combine 0.5M NH 4 + with 1.0M NO 2 -? Reaction Order Rate laws have the general form of: Rate = k [reactant 1] n [reactant 2] m The exponents m, and n, are called reaction orders The sum of the reaction orders (m + n + ) is the overall reaction order In the above reaction of ammonium ion with nitrite ion, the reaction order with respect to ammonium is 1 (or first order), and the reaction order with respect to nitrite is 1 (or first order). The overall reaction order is 2 (or second order overall) Reaction orders can only be determined EXPERIMENTALLY Reaction orders are commonly 0, 1 or 2. However, they can also be fractional, or negative A reaction order of 1 mans that the reaction rate is directly proportional to the concentration of that reactant. A reaction order of 2 means that the reaction rate is directly proportional to the square of the reactant. An order of zero means that the rate does not depend on the concentration of the reactant as long as some of the reactant is present. The sum of all the reactant orders is called the order of the reaction, or overall order. It is important to understand that the orders in the rate law may or may not match the coefficients in the balanced equation. These orders must be determined experimentally. CHCl 3 (g) + Cl 2 (g) CCl 4 (g) + HCl(g) Rate Law: Rate = k [CHCl 3 ] [Cl 2 ] 1/2 Units of Rate Constants k The units of the rate constant, k, depends upon the overall reaction order of the rate law In the above reaction of ammonium ion with nitrite ion, the overall reaction order was second order: 9
k will therefore have units of: Units of k for a first order overall reaction order would be: Using initial rates to determine Rate Laws Rate laws (Reaction Orders) must be determined experimentally The key is to characterize the initial rates of reaction for different concentrations of reactants o If a reaction is zero order for a particular reactant, then changing it's concentration will have no effect upon the reaction rate o o If a reaction is first order for a particular reactant, then changing it's concentration will cause a direct, proportional change in the reaction rate. In other words, doubling the concentration will double the reaction rate, etc. If a reaction is second order for a particular reactant, then changing its concentration will cause an exponential change in the reaction rate. In other words, doubling the concentration will result in a four-fold increase (2 2 ) in reaction rate; tripling the concentration will result in a nine-fold increase (3 2 ) in reaction rate. The Change of Concentration with Time Rate laws tell us what the rate of a reaction is in terms of the concentration(s) of reactant(s) Rate laws can be converted into equations that tell us what the concentrations of the reactants (or products) are at any time during the course of the reaction. First Order Reactions A reaction with a single reactant, where the reaction rate is linearly proportional to the concentration of the reactant (i.e. the 1 st power of the reactant) is a first order reaction: 10
What this says is that the rate at which the concentration of reactant A decreases over time (i.e. the rate at which it is used up in the reaction) is proportional to the concentration of reactant A. This makes sense, in that we expect the rate to slow down as A is used up (and its concentration decreases) The proportionality constant, k, will have units of inverse time, like sec -1 or min -1, so that k[a] will have units of M/sec or M/min (which are appropriate units for rates of change of concentration) This can be rearranged to relate the effects of a change in time to a change in the concentration of reactant A: Using calculus, this equation can be transformed (i.e. integrated) to yield an equation that relates the concentration of A at the start of the reaction [A] 0, to its concentration at any other time t, [A] t : Rearranging to solve for [A] t : y = mx + b This equation relates the concentration of reactant A after some time t, if given the initial concentration ([A] 0 ) and rate constant k. This equation actually has the form of a linear equation, y = mx + b o y = ln[a] t o slope (m) of the line = -k o x = t o y-intercept of the line = ln[a] 0 Therefore, for a first-order reaction, the plot of ln[a] t (y values) versus time, t, (x values) yields a straight line with a slope of -k and a y-intercept of ln[a] 0 For a first order reaction the equation: can be used to determine: 1. The concentration of a reactant remaining at any time after reaction has started (i.e. [A] t ), if you know k, A 0, and t 2. The time required for a given fraction of a sample to react (i.e. solving for t if you know the ratio of [A] t /[A] 0 ) and k 3. The time required for a reactant concentration to reach a certain level - as in "half-life" calculations (see below) 11
Half-life The half-life of a reaction, also known as t 1/2, is the amount of time it takes for the concentration to drop to one-half of its initial level In other words, the point in time where [A] t = 1/2[A] 0 We can determine t 1/2 in a first order reaction by substituting in 1/2[A] 0 for [A] t : Note that the half-life is independent of the concentration. This means that if you randomly choose some time to calculate the concentration of reactant, exactly 0.693/k seconds later, the concentration will be 1/2 of what it was Radioactive decay behaves this way also Second-Order Reactions A second order reaction, by definition, can be the result of: A reaction involving a single reactant whose rate depends upon the reactant concentration raised to the second power: Rate of reaction = k[a] 2 A reaction involving two reactants, whose rate depends on the first power of each reactant: Rate of reaction = k[a][b] For a reaction that is second order in just one reactant (A), the rate law is given by: Rate = -Δ[A]/Δt = k [A] 2 y = mx + b This is another linear function (y = mx +b) with o Slope (m) = rate constant, k o x values are time o y values are corresponding inverse concentrations of reactant 12
o the y-intercept is the inverse of the value of the starting concentration Note: one way to distinguish between first- and second-order reaction laws is to graph both ln[a] t and 1/[A] t versus time. If the plot is a straight line with ln[a] t, then it is first order; if it is linear with the 1/[A] t values, then it is second order. What is the half-life (t 1/2 ) of a second order reaction? For second-order reactions, the half-life (t 1/2 ) is dependent upon the initial concentration Temperature and Rate The rates of most chemical reactions increase at the temperature rises Rate expressions describe reaction rates in terms of concentrations and rate constants (k) Rate = k[a], or Rate = k[a] 2, or Rate = k[a][b], etc. Temperature changes do not affect concentrations Therefore, the observed effects of temperature upon reaction rates must be due to changes in the value of the rate constant k. In particular, k increases with increasing temperature Beginning near room temperature, the reaction rates of many common reactions roughly double with each 10 o C rise in temperature (the actual rate of change must still be determined experimentally). The Collision Model Overall reaction rates can be increased by: Increasing the concentration of reactants Increasing the temperature The underlying physical interpretation for these two observations is the Collision Model of chemical kinetics The main point of the Collision Model is that 13
1. molecules must physically collide in order to react 2. The more collisions that occur over a given period of time, the faster the reaction rate Increasing concentration (i.e. the number of molecules in a given volume) will increase the number of collisions and result in a faster reaction rate What about increasing temperature? The Kinetic Molecular Theory of Gases and the effect of temperature upon rates: Increasing temperature results in an increase in the velocity of molecules As molecules mover faster, there are more collisions per unit of time Not only are there 1) more collisions, but 2) the collisions are harder (i.e. the impacts involve greater energy levels) Does every collision result in a reaction? Only a small fraction of collisions (~1 in every 10 13 collisions!) results in a reaction. Why doesn't every molecular collision result in a reaction? Activation Energy 1888 Swedish chemist Svante Arrhenius proposed: Molecules must possess a certain minimum amount of energy in order to react The available energy is related to the kinetic energy of the molecular collision The kinetic energy of impact can be used to stretch, bend and break covalent bonds, resulting in chemical reactions (recall that in a typical chemical reaction some bond is broken and a new bond is formed) If molecules are moving too slowly, they collide with insufficient energy, and just bounce off each other instead of reacting The minimum energy needed to cause a particular chemical reaction is called the Activation Energy Activation Energy is symbolized by E a The value of E a is dependent upon the particular reaction in question (it is a different value for different reactions) Although a reaction may be energetically favorable overall (i.e. ΔEnergy = ΔH rxn = a negative value) the rate of reaction depends upon the magnitude of the activation energy; the higher the activation energy, the slower the reaction In the above energy diagram for the reaction A B we have the following features: 14
1. Overall, the reaction is energetically favorable. In other words, the product, B, is at a lower energy level than the reactant, A. Energetically, the reaction will proceed with a net release of energy (i.e. goes downhill energetically as it goes from A B) 2. However, for the reaction to proceed, there is an activation energy barrier that molecule A will have to overcome Molecules of A will have to acquire enough energy to overcome E a in order for the reaction to proceed. This energy will come from the kinetic energy associated with molecular collisions The conversion of methyl isonitrile (H 3 CNC) to acetonitrile (H 3 CCN): The conversion appears to involve a swapping of the triple bond N-C group Conceptually, the reaction may proceed through an intermediate state in which the triple-bond N- C portion of the molecule is sitting sideways (denoted in brackets above) In order for this group to rotate, the CH 3 -N bond must stretch and break. This will require the input of energy. This required input of energy is reflected in the activation energy barrier being higher than the energy level of methyl isonitrile The intermediate structure (in brackets above) is a high-energy intermediate called the transition state, or activated complex After the CH 3 -N bond is broken, the new C-C bond forms. This results in the release of energy. The formation of the C-C bond leading to the acetonitrile structure releases energy and thus the energy diagram decreases after the activation energy. Acetonitrile is a lower energy structure than methyl isonitrile (the C-C bond is a lower energy bond than C-N). The reaction is exothermic (energy is released). What are the key properties of the above energy landscape for the conversion of methyl isonitrile to acetonitrile that determines the rate of the reaction? The change in energy, ΔE, has no effect upon the rate of the reaction The rate depends upon the magnitude of the activation energy E a Why doesn't B convert back into A? Note that for the backwards reaction, there are two issues: 1. The reaction of B A is energetically unfavorable (i.e. is endothermic, and requires the input of energy. However, we have seen that entropic contributions can drive endothermic reactions in some cases. 15
2. Note that the activation energy for the reverse reaction is equal to E a in magnitude than E a alone. Thus, not only is the reverse reaction energetically unfavored, the rate of the reverse reaction is much slower due to the larger activation energy "barrier". What fraction of molecules has enough kinetic energy to overcome the activation energy barrier, and how does temperature affect this? The kinetic energy (speed) distribution of gas molecules at two different temperatures: Increasing the temperature increases the fraction of molecules with higher speeds If the activation energy is the minimum needed for the reaction to proceed (i.e. to overcome the activation energy barrier) then at higher temperatures more molecules will have that amount of energy. The reaction rate will be proportional to the number of molecules that have the minimum required activation energy (i.e. the area of the curves to the right of the minimum activation energy line). Although collisions with enough energy occur, why do only a small fraction of collisions result in a productive reaction? In addition to the activation energy requirement collisions between molecules may need a correct orientation for the chemical reaction to occur 16
The particular reactive atoms may need to collide in the appropriate geometry or juxtaposition For example in the reaction Cl + NOCl NO + Cl 2 the O-Cl bond is broken and a Cl-Cl bond is formed. In this case, it may be essential for the incoming Cl atom to collide in the correct orientation with the Cl atom of NOCl (all other collisions being unproductive despite having appropriate kinetic energy) The Arrhenius Equation Arrhenius studied the relationship between the increase in reaction rate and increasing temperature: The increase in reaction rate (k) is not linear with temperature The relationship between the reaction rate and temperature was found by Arrhenius to be: k is the rate constant, E a is the activation energy, R is the gas constant (8.314 J/mol K), and T is the temperature (Kelvin) The term A is the frequency factor. o It is related to the frequency of collisions and the probability that the collisions are productive (i.e. correctly oriented) o It is specific for the particular reaction Plotting the Arrhenius Equation If we take the natural log of both sides of the Arrhenius equation we get: 17
Thus, this will have the form of a linear equation if we plot lnk vs. 1/T. The slope of this line will be equal to (-E a /R) and the y intercept will be lna Thus, for this data, -E a /R = -6014, and E a therefore equals 50,000 J/mol. lna = 9.210, therefore, A = 10,000 s -1 Another form of the Arrhenius Equation If we know the reaction rate at two different temperatures, T 1 and T 2 we can calculate the activation energy, E a, without knowing the value for the frequency factor, A Subtracting the equation for lnk 2 from lnk 1 gives: This relationship allows us to determine the rate constant, k 1, at some temperature, T 1, when we know the activation energy, E a, and the rate constant, k 2, at some other temperature, T 2 Reaction Mechanisms 18
The balanced chemical equation provides no information about how a particular reaction occurs (it provides information about the net overall reaction) The process by which a reaction occurs is called the reaction mechanism The reaction mechanism can provide details about the order in which bonds are broken and reformed, and the changes in the relative positions of atoms, during the course of the reaction (i.e. the formation of intermediate compounds during the reaction) Elementary steps Reactions take place as a result of collisions between reacting molecules The reaction of NO and O 3 to form NO 2 and O 2 occurs as the result of a single collision (with sufficient energy) of correctly oriented molecules of NO and O 3 : This single collision event is called an elementary reaction step (or elementary process) - there are no other "hidden" reactions. The number of molecules that participate as reactants in an elementary reaction step defines the molecularity of the step If a single molecule is involved, the reaction is said to be unimolecular o The reaction involving the conversion of methyl isonitrile to acetonitrile (CH 3 NC CH 3 CN) is an example of a unimolecular reaction (only one reactant, methyl isonitrile, is involved in collisions to produce acetonitrile) If two reactant molecules are involved (as with the NO and O 3 reaction above), the reaction is said to be bimolecular Elementary reactions involving the simultaneous collision of three different reactant molecules is called a termolecular reaction Note: It is rare to find an elementary reaction step involving three reactants; and elementary reactions involving four reactants is considered so unlikely that they are not proposed as likely elementary reactions Balanced equations represent the net change of reactants to products in a chemical reaction Chemical reactions, however, often occur by a multi-step mechanism, that consists of a sequence of elementary reaction steps: NO 2 (g) + CO(g) NO(g) + CO 2 (g) The above reaction actually proceeds in two elementary reaction steps: o Two NO 2 (g) molecules collide to produce NO 3 (g) and NO(g) (i.e. an oxygen atom is transferred to one of the NO 2 molecules during this collision): NO 2 (g) + NO 2 (g) NO 3 (g) + NO(g) o A molecule of NO 3 (g) collides with a molecule of CO(g) to produce NO 2 (g) and CO 2 (g) (i.e. an oxygen is transferred from NO 3 to CO during this collision) NO 3 (g) + CO(g) NO 2 (g) + CO 2 (g) 19
o Each of these above elementary reaction steps is bimolecular The sum of the individual elementary reaction steps yields the balanced chemical equation of the overall process: NO 2 (g) + NO 2 (g) NO 3 (g) + NO(g) + NO 3 (g) + CO(g) NO 2 (g) + CO 2 (g) NO 2 (g) + NO 2 (g) + NO 3 (g) + CO(g) NO 3 (g) + NO(g) + NO 2 (g) + CO 2 (g) NO 2 (g) + NO 2 (g) + NO 3 (g) + CO(g) NO 3 (g) + NO(g) + NO 2 (g) + CO 2 (g) NO 2 (g) + CO(g) NO(g) + CO 2 (g) In the balanced chemical equation, the NO 3 (g) component does not appear anywhere o NO 3 (g) is formed in the first elementary reaction step o This NO 3 (g) is consumed in the second elementary reaction step The NO 3 (g) is termed a reaction intermediate Multi-step reaction mechanisms involve one or more reaction intermediates Rate Laws of Elementary Steps Every reaction is made up of one or more elementary reaction steps The combined effects of the individual rate laws and relative reaction rates of these individual elementary steps will determine the overall rate law and reaction rate Each elementary reaction step will have: o It's own rate law o It's own reaction rate Since the balanced overall equation does not provide details of the underlying elementary reaction steps, the overall rate law must be determined experimentally (i.e. the individual elementary reaction steps must be identified experimentally) If we know that a reaction is an elementary step, then we know its rate law The rate law of any elementary step is based directly on its molecularity Unimolecular elementary reaction step A unimolecular process is one like: A product(s) (i.e. decomposition reaction) As the concentration of A increases, the reaction rate increases The rate of a unimolecular process will be first order: Bimolecular elementary reaction step Rate = k[a] A bimolecular process is one like: A + B product(s) The reaction rate increases with increases in both the concentration of A and B The rate of a bimolecular process will be second order: Rate = k[a][b] The second order rate law can be explained by the molecular collision theory: the more collisions of A with B, the more likely the reaction will occur, and the faster the rate. Increasing collisions of A with B can be achieved by increasing the concentration of either A or B 20
Elementary Steps and Their Rate Laws The rate laws for possible different types of elementary reaction steps are listed below: Molecularity Elementary reaction step Rate law Unimolecular A product(s) Rate = k[a] Bimolecular A + A product(s) Rate = k[a] 2 Bimolecular A + B product(s) Rate = k[a][b] Termolecular A + A + A product(s) Rate = k[a] 3 Termolecular A + A + B product(s) Rate = k[a] 2 [B] Termolecular A + B + C product(s) Rate = k[a][b][c] The rate laws reflect the molecularities of the elementary reaction steps You cannot tell, just by looking at a balanced chemical equation, what the underlying elementary reaction steps are, particulary since the balanced equations do not list reaction intermediates that might exist. Rate Laws of Multistep Mechanisms Most chemical reactions involve several elementary steps The different elementary reaction steps will have different intrinsic rates The slowest of the elementary reaction steps will determine the overall rate of the reaction, and is called the rate determining step The reaction of nitrogen dioxide with carbon monoxide to produce nitric oxide and carbon dioxide: NO 2 (g) + CO(g) NO(g) + CO 2 (g) Experimental results indicate that this reaction is second order in NO 2 and zero order in CO Rate = k[no 2 ] 2 (and rate is independent of the concentration of CO) What might be the underlying elementary steps that would account for the observed reaction orders? The proposed underlying elementary steps include an initial step that is bimolecular for NO 2 (g) (Rate = k 1 [NO 2 ] 2 ) There is a second step that is also bimolecular (Rate = k 2 [NO 3 ][CO]) The rate constants for the two reactions are quite different; k 1 is very slow and k 2 is fast 21
Step #1 is the rate limiting step. The intermediate NO 3 is slowly produced by step 1 and consumed immediately by step 2. NO 3 does not have a chance to build up, and therefore, CO is essentially always present in vast excess to NO 3. Therefore CO is never a limiting reagent, and the reaction does not depend upon the CO concentration Since step 1 is the rate limiting step, the overall reaction rate depends upon this step, and the observed reaction kinetics are second order in NO 2 and zero order in CO An alternative hypothesis for the reaction mechanism We might have hypothesized that the NO 2 (g) and CO(g) could combine in an elemental bimolecular reaction: Rate = k[no 2 ][CO] However, this would not fit the experimentally determined reaction orders Mechanisms with an initial fast step The gas phase reaction of nitric oxide, NO, with bromine, Br 2 : 2NO(g) + Br 2 (g) 2NOBr(g) The experimentally determined rate law for this reaction is second order in NO and first order in Br 2 : Rate = k[no] 2 [Br 2 ] We would like to be able to find a reaction mechanism that is consistent with the observation of the reaction orders. What about a single termolecular reaction? NO(g) + NO(g) + Br 2 (g) 2NOBr(g) (Rate = k[no] 2 [Br 2 ]) Well, this would do it. But, temolecular reactions are quite rare, so (unfortunately) it is unlikely that this is actually what is going on at the molecular level. We need another (more likely) solution How about this: The initial step is very fast, and the second step is slow (and is the rate limiting step) The rate limiting step reaction kinetics would be: Rate = k 2 [NOBr 2 ][NO] The reaction is first order in both NO and NOBr 2 NOBr 2 is an intermediate in the reaction and we aren't able to measure it accurately during the reaction. However, there are some assumptions we can make about it that can help us come up with an equation for its concentration. Here are the assumptions: o Since NOBr 2 is made quite quickly (in the first step) and the second step is slow, the concentration of NOBr 2 builds up. 22
o As NOBr 2 builds up, its fate would seem to have two possibilities: a) it can be collide with NO(g) to produce NOBr (as discussed above) - but that is a slow reaction, b) it can decompose back to NO(g) and Br2(g): Since step 2 (above) is the rate limiting step (i.e. a bottleneck) we assume that the rate of formation and rate of decomposition of NOBr 2 (g) come to equilibrium: Rate of formation = Rate of decomposition k 1 [NO][Br2] = k -1 [NOBr 2 ] From this equilibrium assumption, we can solve for the concentration of [NOBr 2 ]: [NOBr 2 ] = (k 1 /k -1 )[NO][Br 2 ] We can now substitute the value for [NOBr2] back into our original rate description for the rate limiting step of the reaction: Rate = k 2 [NOBr 2 ][NO] Rate = k 2 (k 1 /k -1 )[NO][Br 2 ][NO] Rate = k 2 (k 1 /k -1 )[NO] 2 [Br 2 ] Thus, the reaction is second order in NO and first order in Br 2. And the observed rate constant is actually equal to k 2 (k 1 /k -1 ) When we have a fast initial step, followed by a slow second step, we can determine the concentration of an intermediate by assuming that it achieves an equilibrium concentration Catalysis A catalyst is a substance that changes the speed of a chemical reaction (i.e. increasing the reaction rate) without undergoing a permanent chemical change itself during the process A chemical reaction may be energetically favorable (i.e. may be exothermic), but if the activation barrier is high (i.e. the activation energy is high) the reaction rate may be extremely slow A lot of research is performed to identify new catalysts for chemical reactions of commercial interest (time is money, after all) Also, a lot of research is performed to develop methods to inhibit the action of certain catalysts, so that some reactions will not occur (e.g. certain biochemical reactions that rely upon the action of catalytic molecules) It turns out that while may reactions and physical processes required for living systems are thermodynamically favored (i.e. spontaneous) their rate is so slow that a living system will die just waiting for such processes to occur. Thus, catalysts are essential to living systems - they are necessary in order to get required reactions to occur on a timescale that is useful for living systems. Proteins form the basis of almost all catalysts and are known as enzymes. For example, viruses also rely on certain enzymes for 23
Homogenous Catalysis function. If such enzymes can be selectively inhibited, then a virus will not be able to reproduce or will not infect a host cell (i.e. you and I). A homogenous catalyst is present in the same phase as the reactants The Arrhenius equation states that the rate constant, k, of a reaction is directly proportional to the frequency factor (A), inversely proportional to the activation energy E a, and directly proportional to the temperature: note: the negative exponent means that k is large when the value of E a /RT is small (and vice versa). E a /RT will be small when E a is small or T is large (R is a constant). Thus, k is large when T is large or E a is small. As a biological system utilizing chemical reactions you can prove the relationship of k to T. Next time you are in the snow (or in a walk-in freezer) see how rapidly you can open and close your fist. Try this again in a warm environment. You will find that it is much slower in the cold. If a catalyst is to increase the reaction rate, k, it would appear to be able to do so by one of two ways: 1. Increase the frequency factor, A (i.e. in some way increase the rate of successful molecular collisions) 2. Decrease the activation energy, E a Generally speaking, a catalyst typically increases reaction rates by lowering the activation energy, E a. This is related to the energy required to stretch and break a bond, thus, catalysts must facilitate this process in some way. Also, a catalyst often lowers the overall activation energy for a reaction by providing a completely different reaction mechanism for the reaction. In other words, a different set of underlying elementary reaction steps. Heterogeneous Catalysis A heterogeneous catalyst exists in a different phase from the reactant molecules - often as a solid in contact with either a gaseous or liquid reactants Many important industrial reactions are catalyzed by the surfaces of special solid materials Heterogeneous catalysts are often composed of metals or metal oxides 24
The greater the surface area of a heterogeneous catalyst, the more reactions can take place. Thus, in manufacturing heterogeneous catalysts techniques are used to maximize the surface area (e.g. using highly porous structures) Adsorption: the binding of molecules to a surface Absorption: refers to the movement of molecules into the interior of a material The initial step in heterogeneous catalysis is the adsorption of reactants onto the surface of a catalyst The surface of metal catalysts are highly reactive in comparison to interior atoms o Interior atoms have fully satisfied bonding interactions with neighbor atoms o Atoms on the surface lack a complete set of bonding partners, and thus can bind and react with other molecules in the environment The reaction of ethylene and hydrogen gas to form ethane, and the catalysis of this reaction by platinum metal The reaction of ethylene gas (C 2 H 4 ) and hydrogen gas to form ethane gas (C 2 H 6 ) is exothermic, but occurs very slowly (due to a large activation energy) C 2 H 4 (g) + H 2 (g) C 2 H 6 (g) (kinetically slow) In the presence of a metal surface (e.g. in the presence of platinum powder) the reaction proceeds rapidly even at room temperature Binding and interaction with the metal atoms promotes dissociation of the hydrogen atoms (lowering the activation energy associated with this process Furthermore, binding of both reactants can increase the probability of collision (increasing the frequency factor A). In other words, the diffusion of the molecules on the surface is diffusion in 2-25
dimensions, rather than 3. This "contains" the reacting molecules in a smaller area and increases the likelihood of collision and therefore, reaction Some commercially important metal catalysts: Platinum can catalyze the oxidation of CO to CO 2 CO(g) + 1/2O 2 (g) CO 2 (g) reaction catalyzed in presence of Pt(s) Rhodium can catalyze the decomposition of NO 2NO(g) N 2 (g) + O 2 (g) reaction catalyzed in presence of Rh(s) CO(g) and NO(g) are environmentally toxic gasses produced by the incomplete combustion of hydrocarbons in a nitrogen-containing atmosphere. Rhodium and platinum catalysts in an automobile exhaust can perform chemistry, at the necessary kinetic rate, to render these as less harmful gasses The interaction of the metal catalyzes the reaction. There are several things to consider: The reaction rate is increased, by lowering of the overall activation energy, but the energy difference between the reactants and products has not been altered The reaction requires specific bonds to be broken (H 2 ) and this requires the input of energy. The formation of specific bonds (C-H) is exothermic and releases energy. o The difference between the two processes is the overall reaction enthalpy (which a catalyst does not change) o If the catalyst makes it easier to break a specific bond, then to maintain the overall reaction enthalpy, the formation of the appropriate new bonds must not result in the release of as much energy (as in the absence of the catalyst). In other words, if it takes less energy to break a bond, then conversely, not as much energy is released when forming a new bond. Enzymes Many chemical reactions associated with life are energetically favorable, but kinetically slow (i.e. exothermic, but high activation energy and therefore a slow reaction rate) Biological catalysts are needed Proteins have evolved to serve as specific and effective catalysts for many biological reactions (reactions to break down food to produce energy and building products, and reactions to use this energy and building materials to build a new you; reactions to replicate your genetic material; reactions that let you contemplate the words you are now reading). These protein catalysts are called enzymes. o Proteins are polymers of amino acids connected by a peptide bond (NH-CO) o Each amino acid is about 10 atoms or so in size, consisting of combinations of N, C, O and S atoms. Each amino acid is about 110 amu in mass. The average protein is a linear polymer of about 150 amino acids (some smaller, some much larger) o The protein polymer typically folding up into a unique 3-D structure. For enzymes, the "active" site (where the catalysis occurs) is a very small region of the overall protein structure The "Lock and key" model for enzyme based catalysis describes the protein active site as being complementary (i.e. the "lock") to the reactant (also called a substrate) that it will catalyze (i.e. the "key") o The reactive groups in the enzyme important for catalysis are located in or near the substrate binding site. The region of catalysis is called the active site 26
The number of catalyzed reaction events, called the turnover number, is on the order of 1,000 to 10 million per second. These high rates indicate that enzymes have reduced the energy barrier to a very low activation energy The proteins in your body will help catalyze the oxidation of hydrocarbons (fats) or carbohydrates (sugars), and both a form of reduced carbon, to produce CO 2 (g) and H 2 O(g) and this will occur in an aqueous environment 2000 Dr. Michael Blaber C 6 H 12 O 6 (aq) + 6O 2 (aq) 6CO 2 + 6H 2 O(aq) (release of much energy) 27