The Rates Of Chemical

Foundations to Chemistry - adapted from "Chemistry, Matter and the Universe"

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15. The Rates of Chemical Reactions Introduction

The following web pages consist of a two frame layout. The left handside frame is the "text" frame which contains references to the illustrations in the right handside "illustration" frame. The back and next arrow keys shown in the navigation bar enable movement through the text. Arrow keys also appear in the "illustration" frame in many instances, but these only control the "illustration" frame's content. In addition to the next and back buttons, navigation is possible by clicking on the menu bar across the top of the pages. Clicking on an icon once will open a block of relevant links to chose from. Clicking on these links enables navigation through the site. Clicking on the aforementioned icon once more will close the block of links. Finally I hope that you enjoy viewing these web pages and that it adds to your understanding of the rates of chemical reactions.

A fanciful analogy symbolizing activatio catalytic pathways

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Foundations to Chemistry - adapted from "Chemistry, Matter and the Universe"

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15. The Rates of Chemical Reactions Introduction

The following web pages consist of a two frame layout. The left handside frame is the "text" frame which contains references to the illustrations in the right handside "illustration" frame. The back and next arrow keys shown in the navigation bar enable movement through the text. Arrow keys also appear in the "illustration" frame in many instances, but these only control the "illustration" frame's content. In addition to the next and back buttons, navigation is possible by clicking on the menu bar across the top of the pages. Clicking on an icon once will open a block of relevant links to chose from. Clicking on these links enables navigation through the site. Clicking on the aforementioned icon once more will close the block of links. Finally I hope that you enjoy viewing these web pages and that it adds to your understanding of the rates of chemical reactions.

A fanciful analogy symbolizing activatio catalytic pathways

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Foundations to Chemistry - adapted from "Chemistry, Matter and the Universe"

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15. The Rates of Chemical Reactions Introduction

Some reactions are not 100% complete, even after an infinite period of time they remain as a mixture of reactants and products after all visible reaction has ceased. A competition exists between forward and reverse reactions, and equilibrium is a condition of balance between these opposing processes. We now come to the question: Why do some reactions, which by their free energy values should be spontaneous and far from equilibrium, sit inert and unreactive for years, whereas other reactions go with explosive rapidity? The decomposition of NO to nitrogen and oxygen is thermodynamically spontaneous, so why do we have photochemical smog from oxides of nitrogen?

The activation energy is symboliz hill that the boulder must surmou it can roll off the mountain.

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15. The Rates of Chemical Reactions Introduction

If all combustions with oxygen liberate free energy, and the atmosphere is full of oxygen, then why doesn't everything that is potentially flammable burn at once, including ourselves? The answer is that these decompositions and combustions, although thermodynamically spontaneous, occur at miniscule rates at room temperature. The rates of chemical reactions and the factors that affect them are the subjects of this chapter. The central theme to be developed in this chapter is that the rate of a chemical reaction depends on its reaction mechanism . Two molecules coming together must collide and rearrange their atoms to make product molecules. The intermediate arrangements of atoms may have a high energy, and if so, the reaction will be slow because not all colliding molecules will have enough energy to rearrange properly. The concept of an "activation-energy " barrier to reaction is illustrated with the mountain analogy on the previous page. The boulder cannot roll off the edge of the mountain without first surmounting the activation barrier crowned by a double dagger symbol (the conventional indication of an activated intermediate state). A catalyst makes a chemical reaction go faster by providing an alternate path with a lower activation-energy barrier. This is symbolized by the winding path down the side of the mountain.

A catalyst provides an alternate pathway with a lower activaio barrier

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Foundations to Chemistry - adapted from "Chemistry, Matter and the Universe"

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15. The Rates of Chemical Reactions Experimental Rate Laws

A rate law is an equation that relates the rate of disappearance of reactants, or appearance of products, to the reactant concentrations. The simplest type of reaction is the spontaneous decomposition of isolated molecules or atoms, and it is encountered most commonly in radioactive decay of unstable nuclei. The rate law for the breakdown of carbon-14 nuclei is

The expression

should be read as "the rate of change of carbon-14 concentration with time." (For any quantity, x, whose value changes with time, the expression dx/dt means "the rate of change of x with time.") The rate law just given can be translated as: "The rate of disappearance of carbon-14 atoms is proportional to the number of carbon-14 atoms that are present per liter and available for decay." Since each atom has the same inherent probability of decaying during a specified time interval, and since the probability of one atom's decaying is independent of the presence or absence of other atoms, this is the rate law that would be expected intuitively. [C-14] represents the concentration of carbon-14, and k is the rate constant.

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Foundations to Chemistry - adapted from "Chemistry, Matter and the Universe"

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15. The Rates of Chemical Reactions Experimental Rate Laws

A rate law is an equation that relates the rate of disappearance of reactants, or appearance of products, to the reactant concentrations. The simplest type of reaction is the spontaneous decomposition of isolated molecules or atoms, and it is encountered most commonly in radioactive decay of unstable nuclei. The rate law for the breakdown of carbon-14 nuclei is

The expression

should be read as "the rate of change of carbon-14 concentration with time." (For any quantity, x, whose value changes with time, the expression dx/dt means "the rate of change of x with time.") The rate law just given can be translated as: "The rate of disappearance of carbon-14 atoms is proportional to the number of carbon-14 atoms that are present per liter and available for decay." Since each atom has the same inherent probability of decaying during a specified time interval, and since the probability of one atom's decaying is independent of the presence or absence of other atoms, this is the rate law that would be expected intuitively. [C-14] represents the concentration of carbon-14, and k is the rate constant.

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Foundations to Chemistry - adapted from "Chemistry, Matter and the Universe"

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15. The Rates of Chemical Reactions Experimental Rate Laws

A rate law is an equation that relates the rate of disappearance of reactants, or appearance of products, to the reactant concentrations. The simplest type of reaction is the spontaneous decomposition of isolated molecules or atoms, and it is encountered most commonly in radioactive decay of unstable nuclei. The rate law for the breakdown of carbon-14 nuclei is

The expression

should be read as "the rate of change of carbon-14 concentration with time." (For any quantity, x, whose value changes with time, the expression dx/dt means "the rate of change of x with time.") The rate law just given can be translated as: "The rate of disappearance of carbon-14 atoms is proportional to the number of carbon-14 atoms that are present per liter and available for decay." Since each atom has the same inherent probability of decaying during a specified time interval, and since the probability of one atom's decaying is independent of the presence or absence of other atoms, this is the rate law that would be expected intuitively. [C-14] represents the concentration of carbon-14, and k is the rate constant.

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15. The Rates of Chemical Reactions Experimental Rate Laws

Since the carbon-14 concentration decreases with time, d[C-14]/dt is a negative number;thus a minus sign is required on the right side of the equation. It is common to bring the minus sign to the left side, beside the rate term, and to write the rate equation as

One reaction for which the rate law has been established is the hydrogen-iodine reaction. It depends on the concentrations of H2 and I2

The only unfamiliar aspect about this rate equation is the dx/dt style of writing rates. This equation should be translated as: "The rate of disappearance of H2 and the rate of disappearance of I2, both are half the rate of appearance of HI (since two HI molecules are produced), and each of these is proportional to the product of H2 and I2 concentrations.

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15. The Rates of Chemical Reactions Experimental Rate Laws

Since the carbon-14 concentration decreases with time, d[C-14]/dt is a negative number;thus a minus sign is required on the right side of the equation. It is common to bring the minus sign to the left side, beside the rate term, and to write the rate equation as

One reaction for which the rate law has been established is the hydrogen-iodine reaction. It depends on the concentrations of H2 and I2

The only unfamiliar aspect about this rate equation is the dx/dt style of writing rates. This equation should be translated as: "The rate of disappearance of H2 and the rate of disappearance of I2, both are half the rate of appearance of HI (since two HI molecules are produced), and each of these is proportional to the product of H2 and I2 concentrations.

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15. The Rates of Chemical Reactions Order of a Reaction

The order of an experimental rate equation describes how the rate depends on the power of the concentration terms. The rate of breakdown of carbon-14 is proportional to the first power of C-14 concentration, so the reaction is first order. The hydrogen-iodine reaction rate depends on the product of two concentrations, so the reaction is second order in overall concentration. One also can focus on the individual reactants, and say that the HI reaction is separately first order in H2 and I2 concentrations, since these each occur in the rate equation with a firstpower dependence. Example. The decomposition of N205 vapor,

has been observed experimentally to follow the rate law

What is the order of this reaction?

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15. The Rates of Chemical Reactions Order of a Reaction

Solution. The reaction is a first-order decomposition, because the rate depends on the first power of N205 concentration. Notice that the order of a reaction is an experimentally measured quantity, which does not depend on how you write the equation for the reaction. Writing the reaction with a coefficient 2 in front of the N205 term in the above example does not make the reaction second order, any more than writing the same reaction as

would make it first order. To determine the order of a chemical reaction, you must carry out real experiments. Example.The reduction of NO with H2,

is observed to depend on the NO and H2 concentrations in the following way:

with a dependence on the first power of the H2 concentration rather than [H2]2 as the balanced equation might suggest. What is the order of the overall reaction, and what is its order with respect to each of the reacting substances?

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15. The Rates of Chemical Reactions Order of a Reaction

Solution. The reaction is third order overall, is second order in NO concentration and is first order in H2 concentration. Observe once again that the simple answer that you might expect from looking at the coefficients of NO and H2 in the balanced equation does not correspond with the experimental facts. Example. The reaction of chlorine with chloroform to yield carbon tetrachloride and hydrogen chloride is

and the observed rate expression for production of HCl is

What is the overall order of reaction, and what is the order with respect to each reactant? Solution. The reaction is first order in chloroform concentration, half order in chlorine concentration, and one-and-a-half order overall. Note that there is no reason that the order cannot be fractional in one or more concentrations. The reason that the coefficients in the balanced equation and in the rate law generally do not agree is that the actual chemical reaction mechanism usually is not a simultaneous collision of as many molecules as the coefficients indicate. Instead, the overall mechanism is built up from a series of smaller steps.

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15. The Rates of Chemical Reactions Order of a Reaction

Four- and five-body simultaneous collisions are very improbable events. A reaction in which four or five molecules interact is much more likely to occur in a series of steps, in which two reactant molecules first collide and form an intermediate substance, this substance collides with the next reactant molecule, and so on. The observed rate law is a summary of all of these steps, and may depend on reactant concentrations in a complicated way. The form of the experimental rate law is the first step on the way toward unscrambling the actual mechanism of reaction, but it does not give us the entire story. Only for simple, one-step reactions will the order of the rate law necessarily agree with the coefficients of the balanced equation. This occurs almost exclusively with first-order decompositions and with those relatively rare second-order collisions that are uncomplicated by further reactions. What we have just said applies to rate expressions, but not to equilibriumconstant expressions. The exponential coefficients in the equilibrium-constant expression do match the coefficients in the balanced chemical equation. The equilibrium-constant expression for the HI reaction is

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15. The Rates of Chemical Reactions Order of a Reaction

The equilibrium constant expression for the reduction of NO with H2 is

with a second-power dependence on H2 concentration in the denominator. The reason for the apparent contradiction between this equilibrium-constant expression and the rate law is that any complications of mechanism that affect the forward reaction and its rate law also affect the reverse reaction in the same way. They both cancel out in the overall equilibrium-constant expression. Although the correct rate law for the forward or reverse reaction cannot be deduced from the balanced chemical reaction alone, the equilibrium-constant expression can be. A classical example of complications in a rate expression is the reaction analogous to HI production, but with bromine instead of iodine:

The equilibrium-constant expression is:

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15. The Rates of Chemical Reactions Order of a Reaction

However, the rate of the forward reaction when little or no HBr is present is observed to depend on H2 and Br2 concentrations in the following way:

The rate of production of HBr with time is proportional to the product of the H2 concentration and the square root of the Br2 concentration. Under these conditions, the reaction has an overall order of 3/2. After appreciable amounts of HBr have accumulated, the overall rate law is

(Notice that this expression reduces to the simpler form when the ratio [HBr]/ [Br2] is close to zero.) When the rate law becomes this complex, the concept of reaction order begins to lose meaning. Such a complex rate behavior occurs because the actual reaction mechanism is a series of steps, one after the other. We will examine this chain reaction later on. If the mechanism were simply a collision of H2 and Br2 molecules, the rate law would be

The fact that this is observed by experiment to be the wrong rate law tells us that the simple reaction mechanism also is wrong.

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15. The Rates of Chemical Reactions

Exponential First Order Decay

The first-order rate law for the decay of carbon-14 nuclei,

is a differential equation. It tells us how the change in with time depends on . It would be desirable to have another expression that simply told how the varies with time in a first-order decay process:

[C-14] = some function of t Elementary calculus shows us how we can derive an expression for concentration versus time from a rate equation, using the process of integration. The general method is beyond the scope of this chapter, but we can give the result for a first-order decay:

[C-14] = [C-14]oe-kt Starting from an initial concentration at time t = O of [C-14]o the concentration of carbon-14 at some later time, t, decreases exponentially, as shown above. One property of exponential decay is that, if after a certain time interval the has fallen by half, then after another interval of equal length the concentration will have fallen by half again, or to one quarter its original value.

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15. The Rates of Chemical Reactions

Exponential First Order Decay

After another equal time interval the concentration will be at an eighth its starting value, then a sixteenth, and so on. The time required for any beginning quantity of material to decay by a first-order process to half its starting is known as the "half-life" for the decay. The faster the decay the shorter the half-life. For carbon-14 the half-life is 5570 years, which means that if an experiment is begun with one gram of pure carbon-14, only a half gram will be left after 5570 years. In 11,140 years only a quarter gram will remain, and after 16,710 years, one eighth gram will remain. Unstable nuclei vary widely in their decay rates or half-lives: uranium-238 has a half-life of 4,510,000,000 years, whereas the elusive polonium-213 nucleus has a half-life of only 4.2 millionths of a second. Since the half-life, t½ is the time required for the ratio [C-14]/[C-14]o to decrease to 0.5 , the half-life and the decay rate constant, k, are related by the expression

0.5 = e-kt½

ln2 = +kt1/2

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15. The Rates of Chemical Reactions

Exponential First Order Decay

If either the rate constant or the half-life is known, the other can be calculated. Half-lives usually are used because they have an immediate physical meaning. The exponential decay curve (fig 4 opposite) can be used to give the rate law a physical meaning. The rate of change of concentration with time, d[C-14]/dt, is simply the slope of the first-order decay curve at any time, t. Because carbon-14 is disappearing, the slope is negative. By drawing a tangent line to the decay curve at several points and examining the slope, you should be able to verify that the slope of the curve at any time t is proportional to the remaining concentration of carbon-14, measured in the vertical direction. This is what the rate law for a first-order process means:

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15. The Rates of Chemical Reactions Reaction Mechanisms

Thus far we have established that the rates of chemical reactions depend on the concentrations of reactant species, and that this dependence often has a complicated form that cannot be predicted directly from the coefficients of the balanced chemical equation (as the equilibriumconstant expression can). The reason for these complications is that the actual mechanism of reaction may involve a series of small, one or two-molecule reactions, with intermediate complexes of atoms that are used up in subsequent steps. The rate equation only summarizes the overall process and does not tell us what is happening in the individual steps. Nevertheless, if we can come up with a series of hypothetical reaction steps that faithfully reproduce the observed rate expression, then we feel confident that our hypothetical mechanism must be close to the actual mechanism. Molecules react when they collide, provided that the collision generates enough energy to tear the atoms apart from one another and rearrange them into new molecules. Bimolecular collisions (between two molecules) are common in gases, but simultaneous trimolecular collisions are a thousand times rarer, and four-molecule simultaneous impacts are so infrequent as to be eliminated from consideration. Then how does the following multimolecular smog reaction take place?

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15. The Rates of Chemical Reactions Reaction Mechanisms

One possible mechanism would be the collision of two N0 molecules to form the unstable but perfectly legitimate N202 molecule, followed either by its decomposition to N0 again or its collision with 02 to form N02:

This mechanism is shown in fig 5 opposite. If we assume that the buildup and breakdown of N202 are very fast reactions in comparison with the collision of N202 and 02, and that the first two reactions are in equilibrium, then we can describe the association and dissociation reactions involving constants k1 and k2 in terms of an equilibrium constant:

k1[NO]2 = k2[N2O2]

The slower collision of N202 with 02 is called the rate-determining step, since the overall reaction rate depends on how fast the slowest step occurs. If the reaction takes place by collision of N202 and 02, then the rate of production of N02 will be proportional to the concentrations of these two molecules:

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15. The Rates of Chemical Reactions Reaction Mechanisms

It would be difficult or impossible to measure the concentration of N202 because the molecule is a short-lived reaction intermediate, but fortunately this is unnecessary. The equilibrium-constant expression relates the N202 concentration with that of the reactant N0 concentration:

Substituting in the rate equation, we get

in which k’ = k3Keq. This is the same rate expression that would have resulted if the reaction had occurred by the simultaneous three-molecule collision of N0, N0, and 02, but the mechanism just proposed assumes a series of two-molecule collisions instead. One cannot decide the actual mechanism of a reaction from the rate equation alone. For this example, it would be necessary to carry out chemical experiments to determine the presence or absence of reaction intermediates such as N202. Finding them would support the proposed mechanism; but not finding them might only mean that the chemical detection methods were not sensitive enough. This is why the number of proposed reaction mechanisms in the chemical literature is much mechanisms.

greater

than

the

number of

well-established

We will look at three other examples of reaction mechanisms, and see how they account for the observed rate expressions.

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15. The Rates of Chemical Reactions

The Hydrogen-Iodine Reaction

An example of a long-standing error in reaction mechanisms, which we and others have used so often because it apparently is so well-behaved. The reaction is

and the observed rate equation is

For nearly seventy-five years, everyone assumed that the process occurred by the collision of one hydrogen molecule and one iodine molecule, with enough energy to bring about reaction. Only as recently as 1967 was it demonstrated that the actual reaction involves the reversible dissociation of I2 molecules into atoms, followed by the reaction of these I atoms with H2 :

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15. The Rates of Chemical Reactions

The Hydrogen-Iodine Reaction

As with the NO reaction just discussed, one can write an equilibrium constant for the dissociation and reassociation of I2:

The rate of reaction is determined by the slowest step, and hence is

Using the equilibrium expression to eliminate the concentration of the short-lived I atom intermediate, produces a rate expression that is identical with that predicted from simple collision theory: [I]2 = Keq[I2]

How then can one decide which mechanism is right, bimolecular collision, or dissociation of I2 and subsequent reaction of I atoms? In 1967, J. H. Sullivan found an ingenious way to decide. At equilibrium at any given temperature, the iodine molecule and atom concentrations always will be linked by the equilibrium expression

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15. The Rates of Chemical Reactions

The Hydrogen-Iodine Reaction

Sullivan found a way to change the relative concentrations of I2 and I by using ultraviolet light from a mercury vapor lamp to cause more iodine molecules to dissociate:

In effect, the photons from the lamp gave this reaction a vigorous kick to the right. Sullivan then could alter the relative amounts of I2 and I at will, by controlling the amount of ultraviolet light from the lamp. The question became: Is the rate of reaction dependent on the concentration of I2 molecules, or of I atoms? The answer from the experiments was clear; the rate depends on the concentration of I atoms. The actual rate equation under all conditions is

The simpler form, which makes it look as if the rate depends on the first power of the concentration of iodine molecules, is valid only because in the absence of UV disturbance, I2 and I always are in rapid equilibrium. The Sullivan mechanism for HI was startling because it destroyed what was long believed to be a classical example of true bimolecular collision.

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15. The Rates of Chemical Reactions

The Hydrogen-Iodine Reaction

The mechanism given above seems to demand a three-body collision between one H2 molecule and two I atoms, but this, too, can be avoided by assuming a rapid equilibrium between H2 and another intermediate molecular species, H2I:

This is the mechanism outlined in fig 6 opposite. The rate then is given by

Once again, this is the same expression that would be expected from a simple bimolecular reaction. It emphasizes the truth of the statement that neither the overall chemical equation nor even the correct rate equation is enough by itself to tell you the detailed mechanism of a reaction.

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15. The Rates of Chemical Reactions Reduction of NO by H2

In an example discussed earlier we saw that the reaction of NO with H2 leads to a rate expression that is different from what might be expected from the equation of the reaction alone:

The reaction is first order in H2 concentration, not second order. One mechanism that will account for this rate behavior is

This mechanism is illustrated in fig 7 opposite. If all reactions except the k3 process are quite fast, then the rate equation is

You should be able to use the first equilibrium condition to eliminate N202 concentration, and show that the rate equation given previously is the result. The last reaction (k4) occurs so fast that it scavenges any N20 as rapidly as it is formed, and has no effect on the overall rate of the process. In effect, N2 is produced as fast as N20 appears, so

In a series of reactions, the slowest one has the greatest influence on reaction rates. To invoke a painful analogy, if it takes ten days to to get a certified letter from Los Angeles to the White House, then rushing to get it into the one o'clock mail instead of the three o'clock will make little difference in the long run.

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15. The Rates of Chemical Reactions Formation of HBr

The rather horrendous rate law for the HBr reaction,

arises because the true process is a chain reaction that involves first the dissociation of Br2 molecules into atoms, then the reaction of atoms with other H2 and Br2 molecules:

The latter two equations constitute a chain reaction, each one yielding a molecule of HBr and producing the reactant atom for the other chain step. There also are reactions that either damp down or reverse the chain process:

With these reactions and a certain amount of algebra, one can arrive at the observed rate expression in a straightforward though tedious manner. Although the rate equation looks complicated, we can understand it in terms of the HBr mechanism. For example, as [HBr] increases, its presence in the denominator decreases the rate of reaction. This happens because the chain-reversal reaction sends more HBr back to H2 molecules and Br atoms. At low HBr concentration, for which the ratio [HBr]/[Br2] is small in comparison with the rate constant k, the rate law simplifies to the 1½-order expression that we saw previously:

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15. The Rates of Chemical Reactions

Molecular Collisions and Reaction

All reactions other than spontaneous decompositions of single particles involve collisions between molecules. As we have just seen, some reactions that look simple from their overall stoichiometric equations actually involve elaborate chain mechanisms, and it is not possible to predict the true mechanism just by looking at the equation for the chemical reaction. Even these more complex processes, however, involve separate steps that usually are simple two-body collisions, and what we derive from bimolecular reactions below can be applied to each of the individual steps. The principles are the same for both simple and complex mechanisms. A reaction that does appear to proceed by a simple two-body collision is the following:

The rate of the forward reaction, by which NO2 and CO molecules collide and react, can be written as

rate = k[NO2] [CO] We know that we can increase the rate of the forward reaction by increasing the concentrations of NO2 and CO. Can we also find some way to increase the rate constant, k?

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15. The Rates of Chemical Reactions

Molecular Collisions and Reaction

The most obvious factor that affects k is temperature; k increases as T increases. All reactions go faster at higher temperatures, because the molecules move faster, and collide more often and more effectively. We could calculate k from basic principles if we knew the answers to two questions:

How frequently do two molecules in a reaction 1. collide? 2. What is the probability that when they do collide, they will react instead of rebounding and going harmlessly on their way? The collision frequency of gas molecules depends on the number of molecules in a given volume, how large they are, and how fast they are moving. The measurable variables that control these are concentration, molecular weight, and temperature. Molecules move faster as the temperature increases, and for a given temperature (or kinetic energy), lighter molecules move faster than heavier ones. The kinetic theory of gases can give us an exact expression for the collision frequency. What is the probability that any given collision will lead to chemical reaction rather than to recoil? The simplest possible model of chemical reactivity proposes a threshold energy, Ea, such that if the kinetic energy of two colliding molecules along their line of approach is greater than Ea they will react, but if the collision energy is less than Ea the molecules will rebound without reacting (see opposite).

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15. The Rates of Chemical Reactions

Molecular Collisions and Reaction

This threshold energy, Ea is called the energy of activation. It can be thought of as a barrier that the reaction must surmount before it can go to completion (see fig 10 opposite). From the usual random distribution of energy among molecules, the fraction of molecules that have enough kinetic energy to collide with energy Ea or greater is f= e-Ea/RT This fraction increases with temperature. At absolute zero, it has the value: f = e-Ea/0 = e- =0 This is reasonable, since if the molecules are motionless at absolute zero, none of them have enough energy to react. As the temperature approaches infinity, the fraction of molecules capable of reacting approaches unity, no matter how large the activation energy, Ea, might be: f=e-Ea/

=e-0= 1

At any finite temperature, the larger the activation energy, the smaller the fraction of molecules that have enough energy to surmount this barrier and react. The simple collision theory of chemical reaction states that the rate constant can be represented by k= Ae-Ea/RT in which A is a constant derived from the Collision frequency. It depends on the molecular weight, the molecular diameter, and the square root of the temperature.

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15. The Rates of Chemical Reactions

Molecular Collisions and Reaction

This temperature dependence of collision frequency is almost negligible in comparison with the exponential temperature dependence of the eEa/RT term, which gives the fraction of the colliding molecules with energy greater than Ea. The collision theory tells us how the experimental Rate constant should vary with temperature: There should be a straight-line relationship between the logarithm of k and the reciprocal temperature, 1/T, with a negative slope proportional to the activation energy, Ea

If we measure k at several temperatures and plot ln k against 1/T, we obtain an Arrhenius plot (fig 11 opposite). (This kind of plot is named after the Swedish chemist who first developed the theory. ) This is the most direct means of obtaining the activation energy of a reaction, and is convincing evidence for the entire picture of activationenergy barriers. In order that the reverse reaction occurs, the same energy level for the reaction intermediates must be achieved by the collision of product molecules. If product molecules are more stable than reactants (∆H0 negative), then the energy of activation will be greater for the reverse reaction than the forward.

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15. The Rates of Chemical Reactions

Molecular Collisions and Reaction

The relationships between energies of reactants, the activated state, and products are in the drawing for the NO2 + CO reaction shown opposite(fig 12). The activated state, which the molecules must reach before either the forward or the reverse reaction is possible, is 31.6 kcal mole-1 higher in energy (enthalpy) than the NO2 and CO reactants, and 85.7 kcal mole-1 higher in energy than the NO and CO2 products. The

difference

between

the

forward

and

reverse

Ea’s

is

the

thermodynamic enthalpy of reaction (∆H0= -54.1 kcal mole-1). In the collision theory of reaction, the activation energy is simply a barrier that the colliding molecules must surmount before they can react. The transition-state theory proposes that this activated state is an intermediate arrangement that the molecules go through, which has a real physical existence although it cannot be isolated and studied at leisure. In the NO2 + CO reaction, this transition state, or activated complex, would be one in which the O to be transferred has not yet been completely lost by NO, nor completely acquired by CO. If the two molecules approached end-on, the activated complex would be something like

The dots indicate attractions between atoms that are longer than normal covalent bonds, and quantum mechanical calculations suggest that they might be some 30% longer.

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15. The Rates of Chemical Reactions

Molecular Collisions and Reaction

The activated complex will decompose when molecular vibrations rupture either one of these bonds, and it could just as well result in reactant molecules as product molecules. In the transition-state theory, the rate of the forward reaction depends on the concentration of activated complex, and the probability that it will break down to form products and not reactants. We will find the picture of an activated complex as being a reaction intermediate useful, even though we will not go into the mathematics of the transtion state theory. The activated complex is the state at the top of the "mountain" in the drawing opposite. In the N02 + CO reaction, which we have been using as an illustration, the energy of activation of the reverse reaction (Ea = 85.7 kcal) is greater than that of the forward reaction (Ea = 31.6 kcal), as shown. By the simple collision theory, the rates of the forward and reverse reactions are ratef = 1 Afe-3.61/RT[NO2] [CO] rater = Are-85.7/RT[NO] [C02] The factors Af and Ar are approximately equal, thus most of the difference in the two rate constants lies in the exponentials that contain Ea.

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15. The Rates of Chemical Reactions

Molecular Collisions and Reaction

If we begin with equal of reactants and products, the forward reaction will be faster than the reverse because e-3.61/RT is greater than e-85.7/RT. More products will accumulate. equilibrium, defined by ratef = rater, will not be reached until the excess of products is great enough to compensate for the larger forward rate constant, which is a consequence of the smalleractivation energy of the forward reaction. The equilibrium constant, which is the ratio of products to reactants at equilibrium, therefore will be larger than 1.00. If the products are thermodynamically less stable than reactants (positive ∆H0 of reaction), then the reverse rate constant will be greater than the forward, equilibrium will be attained with an excess of reactants, and Keq will be less than 1.00. Both situations are diagrammed opposite.

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15. The Rates of Chemical Reactions

Speeding a Reaction by Catalysts

It is always possible to increase a rate constant and accelerate a reaction by increasing the temperature. For reactions with an activation energy of 12 to 13 kcal, at temperatures around 2980K the rate constant doubles with every 100 rise in temperature. (Can you prove this?) But as we have seen with NH3, there can be difficulties: The reverse reaction may be accelerated faster than the forward reaction, so that fewer products are obtained. The products or reactants may be unstable at elevated temperatures, or in special applications the surroundings may preclude the use of higher temperatures. For example, one cannot light a match to burn glucose in the human body; this reaction must be carried out at approximately 98.60F. It is for such reactions that catalysis becomes useful. In general, catalysts lower the activation barrier for a reaction (Ea) thereby making the rate constants larger and the reactions faster. This is represented schematically by the drawing opposite. Lowering Ea means finding an alternative pathway or mechanism for the reaction, in which the intermediate states (activated complexes) at all times are at a lower energy. Both the forward and the reverse reactions are speeded up by a catalyst, since lowering the forward Ea necessitates lowering the reverse Ea by the same amount. A catalyst has no effect on Keq or on the ultimate equilibrium conditions for a reaction; it only provides a way in which a spontaneous but slow reaction can arrive at equilibrium faster. If the reaction is not already thermodynamically spontaneous, a cataIyst will be of no use. Thermodynamics does not tell a chemist how to find a catalyst for a given reaction, but it does tell him when it is, or is not, worth his time to look for one.

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15. The Rates of Chemical Reactions

A Simple Catalytic Mechanism

How does a catalyst provide a reaction mechanism with a lower activation energy? Although inorganic and metallic catalysts have been used for decades in the chemical and petroleum industries, we are in the rather odd position of knowing more of the details of catalytic mechanisms for enzymes than for these simpler catalysts, mainly because of recent x-ray crystallographic structure analyses of enzyme molecules. Nevertheless, we can find a simple explanation of why platinum, nickel, or other clean metal surfaces are effective accelerators for reactions involving hydrogenation. Many hydrogenation reactions, such as the following, are catalyzed by metal surfaces:

It would be difficult to give these molecules enough kinetic energy in the gas phase to cause them to react upon colliding. The metal surface assists by adsorbing H2 molecules and pulling them apart into hydrogen atoms, which bind to metal atoms at the surface.(opposite)

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15. The Rates of Chemical Reactions

A Simple Catalytic Mechanism

These reactive H atoms then combine more rapidly with other molecules that collide with the surface. A particularly simple reaction involving H2 is isotope exchange with D2. The probable series of events for the reaction

is shown in the panels opposite(fig 15). In the first panel, an H2 molecule is pulled apart as it binds to two metal atoms on the catalytic surface. The energy that is required to dissociate the H2 molecule is gained from the energy of the two H-metal bonds that are formed. In panel two, a D2 molecule approaches the catalytic surface, and in panel three, one end of the molecule binds to another site on the surface. By forming a weak D-metal bond, the molecule weakens its own internal D-D bond, thereby making it more susceptible to attack by the nearby bound H atom. In panel four the central D atom is shared equally between H and D in a configuration that is analogous to the activated complex in the uncatalyzed reaction. This activated complex could decompose equally well in two ways: on to panels five and six, or back through panels three and two. In about half the cases it will continue on as in panel five, breaking the D-D bond entirely and leading in panel six to a released H-D molecule and a deuterium atom bound to the surface.

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15. The Rates of Chemical Reactions

A Simple Catalytic Mechanism

The energy released when two H atoms and a D2 molecule are bound to the catalytic surface is approximately equal to the energy required to dissociate the H2 molecule; thus panels one and three represent nearly

{graphics}

the same energy states. The contribution of the catalyst arises because the energy of the activated complex in panel four is not nearly as high as that of the intermediate complex in the gas phase:

The metal atoms help to hold the complex in place and stabilize it. The activation barrier therefore is lower, and the reaction takes place faster. Reaction energy profiles for uncatalyzed and catalyzed reactions are shown opposite. This particular kind of assistance by a catalyst is known as a "rack" mechanism, because it literally pulls molecules apart and weakens bonds, thereby making them more susceptible to attack. The molecules that bind to a catalytic surface and are acted upon are called "substrate" molecules. A catalytic surface, whether it be a clean surface of a finely divided metal or metal oxide, or the active site of an enzyme molecule, must be structured in such a way that it can bind substrate molecules from the reaction that it catalyzes, enable them to react, and release the products afterward. In the words of Emil Fischer, a turn-of-the-century enzyme chemist, a catalyst and its substrate molecules must fit one another like a lock and key.

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15. The Rates of Chemical Reactions Postscript

Catalysis And The Environment One aspect of the current pollution problem is that substances that should be degraded spontaneously by the atmosphere or by microorganisms are not, so they remain in the environment to cause trouble. Much of this trouble is our own fault. In the presynthetic era, most raw materials were natural substances taken from the environment. They blended back into the environment when we were through with them. They decayed, decomposed, and rotted, oxidized by the atmosphere and eaten by bacteria that had evolved with these substances as their natural foods. The problem is not quite so simple. A general oxidation catalyst would be a disaster. In a simpler era of chemistry, an intellectual nonproblem in the tradition of the Philosopher's Stone and the Perpetual Motion Machine was the Universal Solvent. (What would you keep it in?)

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15. The Rates of Chemical Reactions Postscript

The Universal Catalyst is in the same league. If some way could be found to make all spontaneous reactions rapid, then all organic matter on our planet, including ourselves, would be converted quickly into carbon dioxide, water, and nitrogen. We can exist in an oxygen atmosphere only because of the existence of activation-energy barriers to reaction. Catalysts do have a limited use in cleaning up the environment. One line of approach to automotive pollution is the use of afterburners for hydrocarbons. To get rid of oxides of nitrogen, catalytic beds are used to accelerate reactions of the type

Some of the practical difficulties include finding a catalyst that has a long lifetime and is not poisoned easily by lead and other components of commercial gasolines. While we search for the perfect catalyst for nitrogen oxide degradation, it might be well to keep in mind another reaction involving nitrogen:

While this reaction under standard conditions (including 1-molar nitric acid as product) is not quite spontaneous, the nitric acid concentration only has to fall to 0.44 mole liter-1 to make it spontaneous. Were it not for the high activation energy of this reaction, all of the water vapor, all of the oxygen, and a good part of the nitrogen, would be swept from our atmosphere, and the oceans would become a solution of dilute nitric acid.1 We had better be sure that the "perfect catalyst" for the smog reaction does not also catalyze nitrogen reactions in general!

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