# Rates of Reactions and Rate Laws

#### Thermodynamics points the way and makes it possible

Chemical change is driven by the tendency of atoms and molecules to rearrange themselves in a way that results in the maximum possible dispersion of thermal energy into the world. The observable quantity that measures this spreading and sharing of energy is the free energy of the system. As a chemical change takes place, the quantities of reactants and products change in a way that leads to a more negative free energy. When the free energy reaches its minimum possible value, there is no more net change and the system is said to be in equilibrium.

The beauty of thermodynamics is that it enables us to unfailingly predict the net direction of a reaction and the composition of the equilibrium state even without conducting the experiment; the standard free energies of the reactants and products, which can be independently measured or obtained from tables, are all we need.

#### ...but it says nothing about how long it will take to get there!

It is worth noting that the concept of "time" plays no role whatsoever in thermodynamics. But kinetics is all about time.

The "speed" of a reaction — how long it takes to reach equilibrium — bears no relation at all to how spontaneous it is (as given by the sign and value of ΔG°) or whether it is exothermic or endothermic (given by the sign of ΔH°). Moreover, there is no way that reaction rates can be predicted in advance; each reaction must be studied individually. One reason for this is that

#### the stoichiometric equation for the reaction says nothing about its mechanism

By mechanism, we mean, basically, "who does what to whom". Think of a reaction mechanism as something that goes on in a "black box" that joins reactants to products:

The inner workings of the black box are ordinarily hidden from us, are highly unpredictable and can only be inferred by indirect means. Consider, for example, the gas-phase formation reactions of the hydrogen halides from the elements.

#### Three reactions that look alike, but are not

The thermodynamics of these reactions are all similar (they are all highly exothermic), but their dynamics (their kinetics and mechanisms) could not be more different.

 $H_2(g) + I_2(g) \rightarrow 2 HI(g)$ $H_2(g) + Br_2(g) \rightarrow 2 HBr(g)$ $H_2(g) + Cl_2(g) \rightarrow 2 HCl(g)$ Careful experiments, carried out over many years, are consistent with the simplest imaginable mechanism: a collision between the two reactant molecules results in a rearrangement of the bonds: One might be tempted to suppose that this would proceed in a similar way, . but experiments reveal that the mechanism of this reaction is far more complex. The reaction takes place in a succession of steps, some of which involve atomic H and Br. The mechanism of this reaction is different again. Although the first two reactions reach equilibrium in minutes to an hour or so at temperatures of 300 to 600 K, a mixture of hydrogen and chlorine will not react at all in the dark, but if you shine a light on the mixture, it goes off with a bang as the instantaneous reaction releases heat and expands the gas explosively.

... and what is particularly noteworthy is that these striking differences cannot be reliably predicted from theory; they were revealed only by experimentation.

### 2 The rates of chemical reactions

Chemical reactions vary greatly in the speed at which they occur. Some are essentially instantaneous, while others may take years to reach equilibrium.

#### Definition of reaction rate

The speed of a chemical reaction may be defined as the change in concentration of a substance divided by the time interval during which this change is observed:

$\text{rate}=\dfrac{\Delta \text{concentration}}{\Delta \text{tine}} \tag{2-1}$

For a reaction of the form $$A + B \rightarrow C$$, the rate can be expressed in terms of the change in concentration of any of its components

 $\text{rate}=-\dfrac{\Delta [A]}{\Delta t}$ $\text{rate}=-\dfrac{\Delta [B]}{\Delta t}$ $\text{rate}=\dfrac{\Delta [C]}{\Delta t}$

in which Δ[A] is the difference between the concentration of A over the time interval t2 – t1:

$\Delta [A] = [A]_2 – [A]_1 \tag{2-2}$

Notice the minus signs in the first two examples above. The concentration of a reactant always decreases with time, so $$\Delta [A]$$ and $$\Delta [A]$$ are both negative. Since negative rates don't make much sense, rates expressed in terms of a reactant concentration are always preceded by a minus sign in order to make the rate come out positive.

Consider now a reaction in which the coefficients are different:

$A + 3B \rightarrow 2D$

It is clear that $$[B]$$ decreases three times as rapidly as $$[A]$$, so in order to avoid ambiguity when expressing the rate in terms of different components, it is customary to divide each change in concentration by the appropriate coefficient:

$\text{rate}= -\dfrac{\Delta [A]}{\Delta t} = -\dfrac{\Delta [B]}{3\Delta t} = \dfrac{\Delta [D]}{2\Delta t} \tag{2-3}$

Example 1

For the oxidation of ammonia

$4 NH_3 + 3O_2 \rightarrow 2 N_2 + 6 H_2O$

it was found that the rate of formation of N2 was 0.27 mol L–1 s–1.

1. At what rate was water being formed?
2. At what rate was ammonia being consumed?

SOLUTION

a) From the equation stoichiometry, Δ[H2O] = 6/2 Δ[N2], so the rate of formation of H2O is

3 × (0.27 mol L–1 s–1) = 0.81 mol L–1 s–1.

b) 4 moles of NH3 are consumed for every 2 moles of N2 formed, so the rate of disappearance of ammonia is

2 × (0.27 mol L–1 s–1) = 0.54 mol L–1 s–1.

Comment: Because of the way this question is formulated, it would be acceptable to express this last value as a negative number.

#### Instantaneous rates

Most reactions slow down as the reactants are consumed. Consequently, the rates given by the expressions shown above tend to lose their meaning when measured over longer time intervals Δt. Note: Instantaneous rates are also known as differential rates.

Thus for the reaction whose progress is plotted here, the actual rate (as measured by the increasing concentration of product) varies continuously, being greatest at time zero. The instantaneous rate of a reaction is given by the slope of a tangent to the concentration-vs.-time curve. Three such rates have been identified in this plot.

An instantaneous rate taken near the beginning of the reaction (t = 0) is known as an initial rate (label (1)here). As we shall soon see, initial rates play an important role in the study of reaction kinetics. If you have studied differential calculus, you will know that these tangent slopes are derivatives whose values can very at each point on the curve, so that these instantaneous rates are really limiting rates defined as

... but if you don't know calculus, just bear in mind that the larger the time interval Δt, the smaller will be the precision of the instantaneous rate.

### 3  Rate Laws and Reaction Order

The relation between the rate of a reaction and the concentrations of reactants is expressed by its rate law. For example, the rate of the gas-phase decomposition of dinitrogen pentoxide

2N2O5 → 4NO2 + O2

has been found to be directly proportional to the concentration of N2O5:

rate = k [N2O5]

Be very careful about confusing equilbrium constant expressions with those for rate laws. The expression for Keq can always be written by inspecting the reaction equation, and it contains a term for each component (raised to the appropriate power) whose concentration changes during the reaction. For this reaction it is given by

In contrast, the expression for the rate law generally bears no necessary relation to the reaction equation, and must be determined experimentally.

More generally, for a reaction of the form

nA A + nB B + ... → products

the rate law will be

in which the exponents a and b are usually (but not always) integers and, we must emphasize once again, bear no relation to the coefficients nA ,nB.

Since the rate of a reaction has the dimensions of (concentration/time), the dimensions of the rate constant k will depend on the exponents of the concentration terms in the rate law. To make this work out properly, if we let p be the sum of the exponents of the concentration terms in the rate law

p = a + b + ...

then k will have the dimensions (concentration1–p/time).

#### Reaction order

The order of a rate law is the sum of the exponents in its concentration terms. For the N2O5 decomposition with the rate law k[N2O5], this exponent is 1 (and thus is not explicitly shown); this reaction is therefore a first order reaction. We can also say that the reaction is "first order in N2O5".

For more complicated rate laws, we can speak of the overall reaction order and also the orders with respect to each component. As an example, consider a reaction

A + 3B + 2C → products

whose experimental rate law is

rate = k[A] [B]2

We would describe this reaction as third-order overall, first-order in A, second-order in B, and zero-order in C.

Zero-order means that the rate is independent of the concentration of a particular reactant. But of course enough C must be present to allow the equilibrium mixture to form.

Example 2

The rate of oxidation of bromide ions by bromate in an acidic aqueous solution

$6H^+ + BrO_3^– + 5Br^– \rightarrow 3 Br_2 + 3 H_2O$

is found to follow the rate law

rate = k[Br][BrO3][H+]2

What happens to the rate if, in separate experiments, (a) [BrO3] is doubled;(b) the pH is increased by one unit; (c) the solution is diluted to twice its volume, with the pH kept constant by use of a buffer?

SOLUTION

1. Since the rate is first-order in bromate, doubling its concentration will double the reaction rate.
2. Increasing the pH by one unit will decrease the [H+] by a factor of 10. Since the reaction is second-order in [H+], this will decrease the rate by a factor of 100.
3. Dilution reduces the concentrations of both Br2 and BrO3 to half their original values. Doing this to each concentration alone would reduce the rate by a factor of 2, so reducing both concentration will reduce the rate by a factor of 4, to (½)×(½) = ¼ of its initial value.

#### Observing rate-vs.-concentration proportionality

In order to determine the value of the exponent in a rate equation term, we need to see how the rate varies with the concentration of the substance. For a single-reactant decomposition reaction of the form

A → products

in which the rate is –d[A]/dt, we simply plot [A] as function of time, draw tangents at various intervals, and see how the slopes of these tangents (the instantaneous rates) depend on [A].

• If doubling the concentration of A doubles the rate, then the reaction is first-order in A.
• If doubling the concentration results in a four-fold rate increase, the reaction is second-order in A.

Example 3

Use the tabulated experimental data to determine the order of the reaction

2 N2O5 → 4 NO2 + O2

Time (min) p(N2O5) [N2O5] mol L-1

Rate

mol L-1 min-1

0 301.6 0.0152
10 224.8 0.0113 3.4 × 10–4
20 166.7 0.0084 2.5
30 123.2 0.0062 1.8
40 92.2 0.0046 1.3
69.1 69.1 0.0035 1.0
SOLUTION

The ideal gas law can be used to convert the partial pressures of $$N_2O_5$$ to molar concentrations. These are then plotted (left) to follow their decrease with time. The rates are computed from the slopes of the tangents (blue lines) and their values plotted as a function of $$[N_2O_5]$$ and $$[N_2O_5]^2$$. It is apparent that the rates are directly proportional to $$[N_2O_5]^1$$, indicating that this is a first-order reaction.

### Initial rate method

When there is more than one reactant, the method described above is rarely practical, since the concentrations of the different reactants will generally fall at different rates, depending on the stoichiometry. Instead, we measure only the rate near the beginning of the reaction, before the concentrations have had time to change significantly. The experiment is then repeated with a different starting concentration of the reactant in question, but keeping the concentrations of any others the same. After the order with respect to one component is found, another series of trials is conducted in which the order of another component is found.

 Using the same dinitrogen pentoxide decomposition as in the previous example, we conduct a series of runs using five different initial concentrations of N2O5. Note that we use only the tangents at the beginning of each plot — that is, at timest=0. We then plot the five initial rates of consumption of  N2O5 as a function of its molar concentration. As before, we see that these rates are directly proportional to [N2O5]. The slope of this plot gives the value of the rate constant.

rate = (5.2 × 10–3) [N2O5] mol L–1 s–1

Example 4

A study of the gas-phase reduction of nitric oxide by hydrogen

2 NO + 2 H2 → N2 + 2 H2O

yielded the following initial-rate data (all pressures in torr):

experiment P(NO) P(H2)

initial rate
(torr s–1)

1
359 300 1.50
2
300 300 1.03
3
152 300 0.25
4
300 289 1.00
5
300 205 0.71
6
200 147 0.51

Find the order of the reaction with respect to each component.

SOLUTION

In looking over this data, take note of the following:

• The six runs recorded here fall into two groups, in which the initial pressures of H2 and of NO, respectively, are held constant.
• All the data are expressed in pressures, rather than in concentrations. We can do this because the reactants are gases, whose concentrations are directly proportional to their partial pressures when T and V are held constant. And since we are only interested in comparing the ratios of pressures and rates, the units cancel out and don't matter. It is far easier experimentally to adjust and measure pressures than concentrations.

Experiments 2 and 3: Reduction of the initial partial pressure of NO by a factor of about 2 (300/152) results in a reduction of the initial rate by a factor of about 4, so the reaction is second-order in nitric oxide.

Experiments 4 and 6: Reducing the initial partial pressure of hydrogen by a factor of approximately 2 (289/147) causes a similar reduction in the initial rate, so the reaction is first-order in hydrogen.

The rate law is thus

$\text{rate} = k[NO]^2[H_2].$

### Dealing with multiple reactants: the isolation method

It is not always practical to determine orders of two or more reactants by the method illustrated in the preceding example. Fortunately, there is another way to accomplish the same task: we can use excess concentrations of all the reactants except the one we wish to investigate. For example, suppose the reaction is

A + B + C → products

and we need to find the order with respect to [B] in the rate law. If we set [B]o to 0.020 M and let [A]o = [C]o = 2.00M, then if the reaction goes to completion, the change in [A] and [C] will also be 0.020 M which is only 1 percent of their original values. This will often be smaller than the experimental error in determining the rates, so it can be neglected. By "flooding" the reaction mixture with one or more reactants, we are effectively isolating the one in which we are interested.

#### Differential and integral rate laws

Measuring instantaneous rates as we have described on the previous page of this unit is the most direct way of determining the rate law of a reaction, but is not always convenient, and it may not even be possible to do so with any precision.

• If the reaction is very fast, its rate may change more rapidly than the time required to measure it; the reaction may be finished before even an initial rate can be observed.
• In the case of very slow reactions, observable changes in concentrations occur so slowly that the observation of a truly "instantaneous" rate becomes impractical.

The ordinary rate law (more precisely known as the instantaneous or differential rate law) tells us how the rate of a reaction depends on the concentrations of the reactants. But for many practical purposes, it is more important to know how the concentrations of reactants (and of products) change with time.

For example, if you are carrying out a reaction on an industrial scale, you would want to know how long it will take for, say, 95% of the reactants to be converted into products.

This is the purpose of an integrated rate law.

#### Expressing the "speed" of a reaction: the half-life

How long does it take for a chemical reaction to occur under a given set of conditions? As with many "simple" questions, no meaningful answer can be given without being more precise. In this case,

#### How do we define the point at which the reaction is "completed"?

A reaction is "completed" when it has reached equilibrium — that is, when concentrations of the reactants and products are no longer changing.

If the equilibrium constant is quite large, then the answer reduces to a simpler form: the reaction is completed when the concentration of a reactant falls to zero. In the interest of simplicity, we will assume that this is the case in the remainder of this discussion.

#### "How long?" may be too long

If the reaction takes place very slowly, the time it takes for every last reactant molecule to disappear may be too long for the answer to be practical. In this case, it might make more sense to define "completed" when a reactant concentration has fallen to some arbitrary fraction of its initial value — 90%, 70%, or even only 20%.

The particular fraction one selects depends on the cost of the reactants in relation to the value of the products, balanced against the cost of operating the process for a longer time or the inconvenience of waiting for more product. This kind of consideration is especially important in industrial processes in which the balances of these costs affect the profitability of the operation.

#### The half-life of a reaction

Instead of trying to identify the time required for the reaction to become completed, it is far more practical to specify the the time required for the concentration of a reactant to fall to half of its initial value. This is known as the half-life (or half-time) of the reaction.

#### The law of exponential change

The rate at which a reactant is consumed in a first-order process is proportional to its concentration at that time. This general relationship, in which a quantity changes at a rate that depends on its instantaneous value, is said to follow an exponential law.

Exponential relations are widespread in science and in many other fields. Consumption of a chemical reactant or the decay of a radioactive isotope follow the exponential decay law. Its inverse, the law of exponential growth, describes the manner in which the money in a continuously-compounding bank account grows with time, or the population growth of a colony of reproducing organisms.

The reason that the exponential function y = ex so efficiently describes such changes stems from the remarkable property that dy/dx = ex; that is, ex is its own derivative, so the rate of change of y is identical to its value at any point.

#### The first-order integrated rate law

The integrated rate law for a first-order reaction

A → products

is a common example of the law of exponential change. For a reactant A, its concentration [A]at time t is given by

$[A]_t = [A]_o e^{–kt}$

in which [A]is its initial concentration and k is the first-order rate constant.

The "e" in the exponential term is of course the base of the natural logarithms, and the negative sign in its exponent means that the value of this term diminishes as $$t$$ increases, as we would expect for any kind of a decay process.

A more convenient form of the integrated rate law is obtained by taking the natural logarithm of both sides:

$\ln [A] = –kt + \ln [A]_o\tag{4-1}$

This has the form of an equation for a straight line

y = mx + b

in which the slope $$m$$ corresponds to the rate constant $$k$$This means that, for a first-order reaction, a plot of $$\ln [A]$$ as a function of time gives a straight line with a slope of $$–k$$.

#### Half-life

After a period of one half-life, t = t½ and we can write

(in which we express the exponential as a function in order to make it stand out more prominently.) Taking logarithms of both sides (remember that ln ex = x) yields

Solving for the half-life, we obtain the simple relation

which tells us that the half-life of a first-order reaction is a constant. This means that 100,000 molecules of a reactant will be reduced to 50,000 in the same time interval needed for ten molecules to be reduced to five.

It should be clear that the rate constant and the half life of a first-order process are inversely related.

Example 5

The half-life of a first-order reaction was found to be 10 min at a certain temperature. What is its rate constant in reciprocal seconds?

SOLUTION

From the above equation, k = –0.693/(600 s) = 0.00115 s–1

The decay of radioactive nuclei is always a first-order process.

#### Second-order reactions

Integration of the second-order rate law

$-\dfrac{d[A]}{dt}=-k[A]^2$

yields

$-\dfrac{1}{[A]} = \dfrac{1}{[A]_0}+kt$

which is easily rearranged into a form of the equation for a straight line (try showing this yourself!) and yields plots similar to the one shown on the left below.

The half-life is given by

Notice that the half-life of a second-order reaction depends on the initial concentration, in contrast to its constancy for a first-order reaction. For this reason, the concept of half-life for a second-order reaction is far less useful.

#### Zero-order processes

In some reactions, the rate is apparently independent of the reactant concentration, in which case integration of the differential rate law yields

[A]t = [A]0 – k t

rate = k [A]0 = k

Note the word "apparently" in the preceding sentence; zero-order kinetics is always an artifact of the conditions under which the reaction is carried out. For this reason, reactions that follow zero-order kinetics are often referred to as pseudo-zero-order reactions.

Clearly, a zero-order process cannot continue after a reactant has been exhausted. Just before this point is reached, the reaction will revert to another rate law instead of falling directly to zero as depicted at the upper left.

There are two general conditions that can give rise to zero-order rates:

1. Only a small fraction of the reactant molecules are in a location or state in which they are able to react, and this fraction is continually replenished from the larger pool.
2. When two or more reactants are involved, the concentrations of some are much greater than those of others

This situation commonly occurs when a reaction is catalyzed by attachment to a solid surface (heterogeneous catalysis) or to an enzyme. For example, the decomposition of nitrous oxide

$N_2O \rightarrow N_2(g) + \dfrac{1}{2} O_2(g)$

in the presence of a hot platinum wire (which acts as a catalyst) is zero-order, but it follows more conventional kinetics when carried out entirely in the gas phase.

In this case, the $$N_2O$$ molecules that react are limited to those that have attached themselves to the surface of the solid catalyst. Once all of the sites on the limited surface of the catalyst have been occupied, additional gas-phase molecules must wait until the decomposition of one of the adsorbed molecules frees up a surface site.

Enzyme-catalyzed reactions in organisms begin with the attachment of the substrate to the active site on the enzyme, leading to the formation of an enzyme-substrate complex. If the number of enzyme molecules is limited in relation to substrate molecules, then the reaction may appear to be zero-order.

This is most often seen when two or more reactants are involved. Thus if the reaction

A + B → products

is first-order in both reactants so that

rate = k [A][B]

then if $$B$$ is present in great excess, the reaction will appear to be zero order in $$B$$ (and first order overall). This commonly happens when $$B$$ is H2O and the reaction is carried out in aqueous solution.

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23:29, 1 Jan 2014

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