The first assumption we will put into our theory of kinetics is that two molecules must collide for a reaction to occur between them.

Chapter 18 Chemical Kinetics: Mechanisms In the last chapter we went through the mechanics of how you extract rate constants and order parameters from experimental data. In this chapter we will ge tthe theory of what these numbers mean 18-1 Mechanisms Up to now we have worked with balanced chemical equations. These equation show the stoichiometric relationships between products and reactants, but it says nothing about how fast the reaction goes, or how the reaction actually occurs. This is what we will try to do now. The first assumption we will put into our theory of kinetics is that two molecules must collide for a reaction to occur between them. For now let s define an elementary step as a chemical reaction that describes the simple collision between two molecules. I will come back with a better definition later We find that many chemical reactions appear to be more complicated than a simple single collision, and seem to involve one or more elementary steps, thus we will describe a chemical reaction as a series of elementary steps called a reaction mechanism Key concept: A reaction mechanism is a series of elementary steps that add up to the balanced stoichiometric equation The reaction mechanism ends up becoming a step-by-step description of what happens in a chemical reaction Let s start with the reaction NO (g) + CO(g) NO(g) + CO (g) The simplest assumption is that NO collides with CO, and one O is transferred from the NO to the CO in this reaction to make it complete. Reaction mechanism Step 1: NO (g) + CO(g) NO(g) + CO (g) If this were the case, then the reaction rate would depend on the concentrations of both NO and CO because both molecules are involved in the reaction. Thus rate = k[no ][CO]. Kinetic analysis of the experimental data tells us this is wrong. The experimental data says that rate = k[no ] and so is not consistent with the real data

Let s try another mechanism Reaction Mechanism Step 1: NO (g) + NO (g) NO 3(g) + NO(g) Step : NO (g) + CO(g) NO (g) + CO (g) 3 Key Concept: Test 1 of a reaction mechanism: Does the sum of the elementary steps = the stoichiometric equation for the reaction? Test1 for this reaction Sum of step 1 and step = NO 3(g) + CO(g) + NO (g) + NO (g) NO 3(g) + NO(g)+NO (g) + CO (g) =CO(g) + NO (g) NO(g) + + CO (g) Passes check 1! In this mechanism NO 3 was created in one step and destroyed in another, so it never appears in the final balanced reaction. Key Concept: An intermediate species, or intermediate, is a molecule that is produced in one elementary step and destroyed in subsequent step and therefore does not appear in the overall reaction. Intermediates are usually unstable, reactive compounds. While unstable, sometimes they can be detected in a reaction mixture, even it they exist only for a microsecond. Back in the last chapter I said that you cannot write a rate law equation from the stoichiometric equation. Why? Because stoichiometry only relates mole ratios of products and reactants to each other and has nothing to do with kinetics or the chemical mechanism. However since an elementary step describes a molecular collision and not a stoichiometeric relationship, you CAN write the rate law from the molecularity of an elementary step. An elementary step that involves the collision between two molecules is called a bimolecular step. An elementary step that involves a change in a single molecule is called a unimolecular step Let s go back to our definition of elementary step and try to improve it a bit

Key Concept: An elementary step is a single step in a reaction mechanism that describes how one molecule is transformed or how two molecules collide to produce one or more new molecules. The reaction rate law of an elementary step may be written from the molecularity of that step 3 Example problem Possible clicker question? The reaction NO + F NOF Is thought to undergo a step mechanism: Step 1: NO + F NOF + F Step : F + NO NO F Does the sum of the elementary steps = stoichiometric equation? (YES) Does this reaction have an intermediate? (YES F) Write the reaction rate laws for each step Step 1 rate = k[no ][F] Step rate = k[f][no ] 18- Activation Energy 18-3 Arrhenius Equation We haven t finished with our reaction mechanism yet, but I want to cover the next subject first. In this discussion I will mix parts of 18- and 18-3 together So now for our model of kinetics we say a reaction involved a collision between molecules or a transformation within a molecule, and for each elementary step we can write a reaction rate equation: rate = k [ ].[ ]... Where each concentration represents a molecule that takes part in that step But what does the k mean? The k has to do with how fast the reaction occurs. With that in mind, remember in the section on gases we could calculate how quickly one molecule collides with another molecule? o 31 In a 1L mixture of gases at 1 bar and 5 C a molecule has about 10 collisions/sec If every collision in a gas resulted in a reaction then 31 7 rate of reaction = 10 collision x 1 mol = 10 mol/l sec L sec 6.0x10 3 In fact almost no reactions occur at this high a rate, and most reaction occurs at rates 100' s or 1000's of time slower. Why?

Think demolition derby. Do all the collisions in a demolition derby end with one car dead and the other one driving away in fine shape? Does a driver care about where he hits the other car with his car? No and Yes. A slight bump between two cars won t put either one out of commission, and the drivers of demo derby cars are always trying to run their rear end into the other car. Bottom line, the energy of the collision and the orientation of the colliding objects makes a big difference in whether the collision is successful or not. The same is true in our collisional theory of chemical kinetics. Not only do we have collisions, but the collision has to have enough energy and be in the right orientation before a reaction will occur. Figure 18.1 if you want it rate of reaction = Collisional X Fraction of collisions X fraction of collisions Frequency With right orientation With enough energy Looking at individual terms Collisional frequency [A][B]...The higher the concentrations, the more collisions. This term will also depend on T, the higher the T, the faster the molecules move, and the more collisions you will have Energy of Collisions - The minimum energy required to cause a reaction between two molecules is called the activation energy (E a ). Figure 18. shows the energy distribution of molecules at two different temperatures. As you can see as the T increases, the number of molecules that exceed a fixed activation energy increases Fraction of collisions with right orientation - a factor that says waht fraction of the collisions are oriented correctly Putting these factors into a single equation k = zpe -Ea/RT where z is the collisional frequency p is the orientation factor, called the steric factor E a is the activation energy R is the gas constant in Joules T is the temperature in K -Ea/RT e puts in the temperature dependence of the # of molecules with E > Ea Since z and p do not vary significantly with T they are often lumped together into a single factor A, in a equation called the Arrehenius equation k =Ae -Ea/RT 4

5 k then, varies with temperature through an exponential function. Let s see if we can come up with an equation that shows how k varies with T that doesn t have -Ea/RT that nasty e term Taking the ln of both sides -Ea/RT ln(k) = ln(a) + ln(e ) ln(k) = ln(a) -Ea/RT At this point, if you have k at several temperatures you can determine Ea by plotting ln(k) vs 1/T. Why? Remember your line of best fit? Y = mx + b Y = ln(k) X =1/T m = slope = -Ea b = intercept = ln(a) On a test you won t have time to plot a bunch of points, so I want to show you a simpler two point equation Now let s get k 1 at T 1 and k at T ln(k ) = ln(a) -Ea/RT ln(k ) = ln(a) -Ea/RT 1 1 ; Now subtract the first equation from the second ln(k ) - ln(k ) = ln(a) -Ea/RT - ln(a) -Ea/RT 1 1 dropping out the ln(a) terms since they are equal ln(k ) - ln(k ) = -Ea/RT - (-Ea/RT ) 1 1 Grouping the -Ea/R terms together ln(k ) - ln(k ) = -Ea/R(1/T - 1/T ) 1 1 Using the properties of ln ln (k /k 1) = -Ea/R (1/T - 1/T 1) Now getting rid of the - sign on the right side by switching T 1 and T around ln (k /k ) = Ea/R (1/T - 1/T ) 1 1 Key Equations k = zpe -Ea/RT k =Ae -Ea/RT You probably won t do any calculations with these two equations, but you need to know what they mean ln(k) = ln(a) -Ea/RT ln (k /k ) = Ea/R (1/T - 1/T ) -or- ln (k /k ) = Ea/R[ (T -T )/(T T ) 1 1 1 1 1 These equations you will calculate with

6 Sample calculations: Find the activation energy E a for the reaction NO + O3 NO + O Given the data in the following table: T(K) k(l/mol s) 195 1.08x10 9 30.95 60 5.4 98 1.0 369 35.5x10 9 What we really want is a plot of ln(k) vs 1/T so: 1/T ln(k) -3 5.13x10 0.65 4.35 1.81 3.85.41 3.36 3.1-3.71x10 4.9 Plotting in excel -E a/r = -1489 3 3 E = 1.489x10 (R) = 1.489x10 (8.3145) = 1.4 kj/mol a slope = -1489 = -E a/r For shorter problems we can derive a simple equation for point determination of activation energy

7 Using the point equation ln(k /k ) = E /R (1/T -1/T ) 1 a 1 And making the first point T and the last point T 1 (it doesn t matter which is which) 9 9 ln(1.08x10 /35.5x10 ) = E a/8.3145(1/369-1/195) ln(1.08/35.5) = E a/8.3145(.0071 -.00513) -3.493=E a/8.3145 ( -.004 E a = 3.493(8.3145)/.004 = 1 kj about the same, just some roundoff error on the calculation The meaning of activation energy Now that you know how to find E a from experimental data, let s try to get a better understanding for what it means Earlier I said that the activation energy is the minimum energy required for a properly oriented collision to create a chemical reaction. It is useful to think of an activation energy in terms of an activation energy diagram Figure 18.3 or 18.4 on the board In this diagram we have the energy of the reactants on the left side, the energy of the products on the right side, and you can think of the activation energy as the height of a hill between the two that has to be overcome before a reaction will occur Example problems clicker type Determine if figure 18.3 is an exothermic or endothermic reaction Determine if figure 18.4 is exothermic or endo thermic If the reaction in 18.3 has a ÄH of -10 and an activation energy of +30, what is the activation energy for the REVERSE reaction If the reaction in 18.4 has a E a of 40 in the forward direction and an E a of 0 in the reverse direction, what is the ÄH for the reaction in the forward direction The top of the energy hump of an activation energy diagram has a useful physical interpretation. It is often interpreted as the energy of a transition state or activated complex For example the reaction - - CH3Cl(aq) + OH (aq) CH3OH(aq) + Cl (aq) has kinetics where rate = k [CH3Cl][OH - ] that says there is a key elementary step where the CH3Cl molecule collides with - an OH ion

8 One can theorize that this goes through an intermediate as shown in the mechanism shown on page 656 The square bracket with the funky superscript indicates a proposed intermediate or activated complex that occurs at the peak of the activation energy hill. Since this is at a higher energy than either the product or reactant, it is not stable, so you cannot isolate it, but sometimes you can gather data (usually spectroscopically) that implies it is there Key Definitions Be able to define these terms that you have seen in these two sections: Activated complex or transition state activation energy steric factor frequency factor activation energy diagram 18-4 Rate-Determining Step Now that we can interpret the kinetics of an elementary step, Let s return to putting the elementary steps together to come up with a reaction mechanism Returning to the reaction NO (g) + CO(g) NO(g) + CO (g) We had learned that it wasn t a simple 1 step reaction and had proposed the following mechanism: Step 1: NO (g) + NO (g) NO 3(g) + NO(g) Step : NO (g) + CO(g) NO (g) + CO (g) 3 The sum of the elementary steps = the stoichiometric equation so it passed check 1! Now on to check, does it pass experimental kinetic data? Well we have elementary steps The first would have rate = k[no ] The second would have rate = k[no ][CO] Which rate should we focus on? Key Concept: In general, in any reaction mechanism, the rate determining step is the slowest step in a reaction, and that single slow step determines the overall kinetic of the reaction. 3

If you have a hard time with this concept, think of driving to Rapid through a one lane construction zone; if you are stuck behind granny doing 40, you won t get to Rapid any faster than her, even if you do have a hot car! So if we propose that step 1 is the slow, rate determining step of our reaction mechanism. Our mechanism agrees with experimental data, and our proposed mechanism is consistent with experimental data 9 Example problem Possible clicker question? Earlier we proposed that the reaction NO + F NOFundergos a step mechanism: Step 1: NO + F NOF + F Step : F + NO NO F If the first reaction is the rate determining step, what should the experimental rate law be for the overall reaction? If the second reaction is the rate determining step, what should the experimental rate law be for the overall reaction? What we have been doing in bits and pieces is using experimental data to try to prove or disprove a proposed theoretical reaction mechanism. I would like to formalize what we have been doing Key Concept To show that a reaction mechanism is consistent with experimental data you must show two things: 1. The sum of the elementary steps equal the overall balanced reaction. The mechanism revealed in the elementary steps must agree with the experimental rate law. Does this prove that the mechanism is correct? NO it only shows it is consistent with the data. It is usually possible to come up with several different mechanisms, all of which are consistent with the data. One almost never proves a reaction mechanism to be correct 18-5 Reversible Reactions Up to this point, we have discussed reaction going in 1 direction only, however in many reactions both forward are reverse reactions are significant (That was one reason for doing the method of initial rates) Let s look at the reaction NO(g) + O (g) NO (g) One proposed mechanism is:

10 First, does the sum of the elementary steps = the stoichiometric equation? Yes, so mechanism might work Second does the mechanism agree with experimental kinetics? With the second step being slow you say, No the problem, I just learned that the slow step is the rate determining step so rate = k [N O ][O ] The problem is that you have used an intermediate NO in your kinetic analysis, and I have said several times that intermediates are very difficult to find, much less measure their concentration, so we can t confirm that step with experimental data! However, let s see what the fast equilibrium can do for us. To be at equilibrium things cannot be changing. That means that the rate of a forward reaction has to be equal to the rate of the reverse reaction. In this case the rate of the forward reaction = k 1[NO] The rate of the reverse reaction = k -1[NO ] and k [NO] =k [N O ] 1-1 Ah hah I see a cute trick [NO ] = k 1/k -1[NO] and we can substitute this into the previous rate equation: rate = k [NO ][O ] =k k 1/k -1[NO] [O ] =k [NO] [O ] where k =k k /k 1-1 This is indeed the form of the rate law we find experimentally, so this is a plausible reaction mechanism Note one could also propose a 1 step mechanism for this reaction in which 1 molecule of O and two molecules of NO collide in a termolecular step Step 1: NO + O NO rate = k[no] [O ]

This mechanism is also consistent with the experimental data. This illustrates what I said earlier about you can often find alternate reaction mechanisms that are consistent with the experimental data. In this case, most chemists do not like the termolecular step mechanism Can you think of why? Here is a hint. How often do you see a 3 car crash where all three cars hit the same place at exactly the same time? 11 18-6 Catalysis Key Concept A catalyst is a substance that increases the rate of a reaction, but is not consumed in a reaction How does it do this? It provides a different and faster reaction mechanism that is available when the catalyst is not present. How can the alternate pathway be faster? It must have a lower activation energy. Take example of hydrogenation fo a C=C double bond. When done in the gas phase you have to have a very high pressure of both gases to get them to collide. Then the collision has to be oriented correctly so the hydrogens line up with the double bond, and the collision has to be of a high enough energy to make the double bond break. When done with a solid metal catalyst (Shown in power point slide) The hydrogen first gets absorbed onto the metal surface and the metal reacts with the hydrogen to replace the H-H single bond with metal hydride bonds to the metal surface. Next the C=C double bond adheres to the surface. And while it is tethered to the surface it picks up the hydrogens and gets converted to it product. A totally different mechanism, but this one takes lots lower pressures of H and C=C double bond and much lower temperatures because the activation energy is so much lower You should notice that a catalyst does not change the overall ÄH rxn (as shown in figure 18.7) so it cannot make unfavorable reaction favorable. All it can do is speed up a reaction by lowering the activation energy of the reaction. The example I just gave you is called a heterogenous catalyst because the solid is a different phase than the gas phase of the reaction. You book gives you two examples of different homogeneous catalysts that are in the same phase as the reactants Key Definitions Homogeneous catalyst - is in the same phase as the reactants Heterogeneous catalyst - In a different phase than the reactants

18-7 Enzyme Kinetics Your body uses thousands of different enzymes to run all of the biological reaction needed for life. Most biological enzymes are proteins and we call these protein catalysts enzymes Key definition: An Enzyme is a biological catalyst made from protein. The reactant in a enzymes catalyzed reaction is called the substrate The place where the substrate reacts on the enzyme is called the active site. An example of a biological reaction that is catalyzed by an enzyme is the reaction of glucose with ATP to form glucose-6-phosphate and ADP REACTION page 668 This is the first step in the metabolism of glucose to CO and a very important reaction for virtually all aerobic organisms Figure 18.10 gives you an idea of the size of the enzyme compared to the size of the small molecules it reacts Molar mass of enzymes 10,000 to 50,000 to millions! 1 Experimental studies have shown that many enzymatic reactions have follow the kinetic shown in the next equation: rate of reaction = ÄP/Ät = k[s]/([s] + K M) Where [S] refers to the substrate (reactant) concentration Where k is a constant that is ties to the maximum rate of the reaction and K m is a constant that ties to how well the substrate binds to the enzyme This hyperbolic function looks like figure 18.11 Look at the equation and figure 18.11 at low concentrations of S (left hand side of figure) S < K m so it drops out of the denominator and rate ~ K/K [S] so rate goes up linearly with [S] M At high concentrations of S (right side of curve) [S]>Km, so K m drops out of the denominator and rate = k[s]/[s] and the rate hits a maximum rate of k The book goes over some of the details of a mechanism that has been proposed to explain the observed kinetics, but I think I will leave that for a Biochem class