### Rates of Reaction

First of all, we should define what, exactly, a reaction rate is. Chemists define the rate of reaction as the amount of a particular reactant consumed in (moles per liter) per second, or mol/(L * sec), at a particular time. For example, reactant A may be being consumed at 0.27 moles per liter per second, 30 seconds after the reaction begins; or the rate of reaction can be expressed in an average manner, such as the rate of consumption over a time period. For instance, you can say that for 30 seconds, reactant A was being consumed at an average rate of .5 moles per liter per second. The first type of rate is called an instantaneous rate, since it measures the rate of consumption at one specific time. In particular, the instantaneous rate at the beginning of the reaction is called the initial rate. The second type of rate expression is called, appropriately, an average rate.

To begin our in-depth discussion how to analyze and calculate of rates of reaction, we need to consider the actual mechanism of a reaction. Except in the very simplest reactions, most chemical processes occur over a series of elementary steps. For instance, the reaction of Br2 gas and NO gas to produce BrNO, another gas, takes place in two steps:

Step 1:  Br2 (g) + NO (g) Br2NO (g)
Step 2:  Br2NO (g) + NO (g) 2 BrNO (g)

Overall:  Br2 (g) + 2 NO (g) 2 BrNO (g)

In step one, Br2NO is generated, only to be consumed in the second step. When the steps are added up to give the overall reaction, the Br2NO cancels and so is not included in the reaction equation. This substance, which has a very short life, is called a reaction intermediate. Most reactions have two or three steps, but some occur in a single step, while others take many more steps to continue.

In each step of a reaction, the reacting molecules must collide, with both sufficient energy and the proper orientation; if the molecules collide in an improper way, the reaction will not take place. So, we can write a provisional rate law for a single step:

Rate = N * fe * fo

where N is the number of collisions per second (more collisions means more opportunities for a reaction), fe is the fraction of molecules with sufficient energy, and fo is the fraction of molecules with the correct collision orientation. To increase the odds of a successful reaction, we can increase the number of collisions (by increasing temperature and/or the concentration of reactants) or increase the average energy of the reactants (again, by increasing temperature). When reactions involve solids, surface area takes the place of concentration; more surface area yields more possible reaction sites. This is why a pile of flour does not burn, but wheat dust in a grain elevator can cause an explosion: the dust has far more surface area. The term relating to collision geometry is fixed. This value, called the steric factor, depends on how difficult it is for the reacting molecules to collide properly.

Now, we see two things that affect reaction speed: the concentration of reactants and temperature. Let's discuss concentration first. There are three different types of reaction steps, in which one, two, or three molecules react (called unimolecular, bimolecular, and termolecular, respectively). The probability that four or more molecules will happen to collide at once and with the proper orientations is so low that we don't consider it.

Unimolecular: A good example of this type of reaction is the decomposition of hydrogen peroxide:

H2O2 (l) 1/2 O2 (g) + H2O (l)

In this case, the rate is proportional only to the concentration of peroxide (we're ignoring temperature until later). We can also say that the rate of reaction is equal to the concentration of hydrogen peroxide multiplied by some constant, which we call k (this is not the same as K, the reaction equilibrium constant). So, the rate equation is:

Rate = k[H2O2]

Since the reaction is dependent on just one factor, unimolecular reactions (or steps) are said to be first-order. First-order reactions have a general rate law of:

Rate = k[A],

Bimolecular: These types of reactions (or elementary steps) involve two molecules colliding; they can be two different molecules, or two of the same molecule. An appropriate example is the first step of the example reaction above:

Br2 (g) + NO (g) Br2NO (g)

Thus, the speed of the reaction is dependent on the concentration of both products, giving us this rate law:

Rate = k[BR2][NO]

This reaction type, since it is proportional to the concentrations of two reactants, is called second-order. The general format for second-order reactions is:

Rate = k[A][B] (or k[A]2, if two of the same reactant are required)

Termolecular: These reactions or steps involve the collision of three molecules; these can be three separate types of molecule, two of one type and one of another, or three of the same molecule. Termolecular reactions are rare, so we'll just give you the rate law:

Rate = k[A][B][C] (or k[A]2[B], or k[A][B]2, or k[A]3, depending on the reactant format)

Zero-order: Other reactions, particularly the catalyzed reactions we will discuss next, have a rate that does not depend on the concentrations of the reactants at all; only the rate constant sets the rate of reaction. These reactions are called zero-order, and their rate law is:

Rate = k

Now that you know how to write the rate law for an elementary step, you're probably wondering how this relates to the rate for an entire equation. As in many aspects of chemistry, an analogy is perhaps the best way to explain. Let's say you want to send a telegram to your cousin in Colorado. You call the telegram agency and give them her address, which takes about a minute. The agency forwards the request electronically to their Denver office, a process that lasts less than a second. Finally, a messenger drives from Denver, up a windy, snowy mountain road, to your cousin's house, which takes about two hours.

Of the three steps in the process, the final one mostly determines the speed of the entire event--speeding up the first two would make little difference, while sending the telegram via e-mail instead of hand-delivering it would shorten the delivery time dramatically.

This concept applies to chemical reactions as well. A typical reaction is composed of several elementary steps, of which one is often much slower than the rest. Since it is this step that determines in large part the speed of the whole reaction, it is called the rate-determining step, or "RDS." The rate equation for the RDS becomes the rate equation for the entire reaction. The other steps occur so quickly that their impact on the overall rate is negligible. So, if you know the rate law, you can find the slowest step (and vice versa).

The rate equation for a reaction cannot be predicted from theory--it must be determined by experiment. Therefore, you will often be given data relating the concentration of a substance (usually a reactant) with a time coordinate. Alternately, problems may provide data on multiple experiments, with varying initial concentrations and concentration changes. You will then discovery the order of the reaction, the rate law, the reaction mechanism, and the RDS of the reaction. While this may seem like a daunting task, we'll walk you through the process with some sample problems. And don't forget--we still have to deal with temperature and rate laws!

Let's start by analyzing a problem in which multiple initial rates of reaction are given, along with varying initial concentrations. We'll use the reaction for the desstruction of ozone by NO, producing oxygen and NO2.

This is the usual type of table given in a rate-law problem:

 Experiment # Initial [O3] Initial [NO] Initial Rate of Formation of NO2 1 .01 M .01 M x 2 .01 M .02 M 2x 3 .02 M .01 M 2x 4 .02 M .02 M 4x

Given this information, what is the rate law expression? Well, let's compare the first and second experiments: the concentration of NO was doubled, and the rate of formation doubled, so [NO] must appear once (no powers) in the rate law. By comparing the first and third experiments, the concentration of ozone doubled, again doubling the rate, so ozone must also appear once as [O3]. Therefore, the rate law must be:

Rate = k[NO][O3]

The steps below are a possible reaction mechanism. Using the rate expression, which step must be the slowest (which is the RDS)?

O3 + NO O + NO3

O + O3 2 O2

NO3 + NO 2 NO2

Well, because only ozone and nitrogen oxide appear in the rate law, the RDS must be the first step.

Other rate-law problems will present you with a declining concentration of a particular reactant. For zero-order reactions, graphing the concentration of the reactant versus time gives a line, whose slope is the negative rate constant (-k, so to find k, multiply the slope by -1). Mathematically, you can find the rate constant using the following equation:

[R]0 - [R]t = kt

For first-order reactions, graphing the natural log of the concentration (ln [R]) versus time will yield a straight line. The slope of this line is the opposite of the rate constant (or -k). To find the constant mathematically, use the equation:

ln([R]t/[R]0) = -kt

where t is the time in seconds, [R]t is the concentration at the measured time, [R]0 is the initial concentration, and k is the rate constant.

For second-order reactions, graphing 1/[R] versus time will give a line whose slope is the rate constant, k. Mathematically, the equation for the rate constant is:

1/[R]t - 1/[R]0 = kt

where the variables are the same as above.

Temperature changes also have very important effect on reaction rates. For instance, the conversion of a hydrocarbon (such as heptane, an ingredient in gasoline) into water and carbon dioxide can take hundreds of years at room temperature; when heptane and oxygen are lit, the mixture burns at very high speed as the heat of combustion continues to raise the temperature. As we indicated above, the reaction rate is dependent on a constant, k, which is in turn related to the fraction of collisions with correct orientation and the fraction of molecules with the required collision energy. In 1889, Svante Arrhenius of Sweden developed an equation relating k to these factors. The Arrhenius equation, named in his honor, is given below:

k = Ae-Ea/RT

where A is the frequency of successful collisions when the reactant concentrations are 1M, expressed in liters per (moles x seconds); e is the mathematical constant (about 2.71828); Ea is the activation energy of the reaction or step in kilojoules (kJ); R is the universal gas constant, 8.314 joules per (moles x kelvins), which we will explore further in the "States of Matter" chapter; and T is the temperature in kelvins. If we take the natural law of both sides, we get the equation:

ln k = -Ea/(RT) + ln A

To make the equation easier to use, we want to get rid of A. We can use two different versions of this equation, at different temperatures, to eliminate it:

1st Equation: ln k1 = -Ea/(RT1) + ln A
2nd Equation: ln k2 = -Ea/(RT2) + ln A

Subtracting the two equations will eliminate A and yield the following equation (ln A - ln B = ln (A/B), factoring out -Ea/R):

ln (k1/k2) = -Ea/R(1/T1 - 1/T2)

The actual rates of reaction (in moles per liter per second) may be substituted for the rate constant if desired. This powerful equation has three main uses: first, we can find the activation energy if two rate constants (or rates) and their temperatures are known; second, we can find a rate constant for a given temperature if we know another rate constant and its temperature and the activation energy; or third, we can find an appropriate temperature for a known rate constant, also given the activation energy and a rate constant/temperature pair.

Alternately, you can find the activation energy by plotting ln k versus 1/T on a graph. The slope of the line is -Ea/R, and since you know R (it is a constant), you can find the activation energy. You should know how to use both methods, although the first is probably more common.

Well, this wraps up reaction rates! This is a very long and detailed lesson, but it is also one of the most useful and important in this chapter. If our explanation is unclear, discuss it on the Message Board or e-mail us; we'll try to make things more understandable, to the best of our ability. Or, if you need more practice, AP exam review books or textbooks include numerous sample problems for your enjoyment.

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