Chemical Reactions (Differential Equations)

S. F. Ellermeyer and L. L. Combs

This module was developed through the support of a grant from the National Science Foundation (grant number DUE-9752555)

Contents

1  Introduction
    1.1  Units of Measurement and Notation
2  Rates of Reactions
    2.1  The Rate Law
    2.2  Example
    2.3  Exercises

1  Introduction

The net chemical reaction that takes place when water (H2O) is formed from hydrogen (H2) and oxygen (O2) is

2H2+O2®2H2O.
(1)
The notation used to describe this reaction indicates that two molecules of hydrogen combine with one molecule of oxygen to form two molecules of water. Chemists use this type of notation in describing any chemical reaction. For example, the net reaction for the formation of nitrogen dioxide (N02) from nitrous oxide (NO) and oxygen (O2) is
2NO+O2®2NO2,
(2)
which indicates that two molecules of nitrous oxide combine with one molecule of oxygen to form two molecules of nitrogen dioxide. The reaction (2) is one of great practical interest in our modern world because it is the first of a series of reactions that produce ``photochemical smog'' from nitrous oxide that is emitted from the exhaust gasses of automobiles.

Another example of a chemical reaction is the decomposition of hydrogen peroxide (H2O2) into water and oxygen. The net reaction for this is

2H2O2®2H2O+O2.
(3)
Since reaction (3) is catalyzed by light, hydrogen peroxide left to sit in a clear bottle would rather quickly turn into water and oxygen. It is because of this reaction that hydrogen peroxide is usually sold in brown (non-transparent) plastic bottles.

In general, any chemical reaction can be described by using a notation of the form

aA+bB® cC+dD
(4)
where the symbols A and B stand for the reactants of the reaction and the symbols C and D stand for the products of the reaction. The constants a, b, c, and d, which indicate the proportions in which the reactants combine and the products are formed, are called stoichiometric coefficients. Note that reactions (1) and (2) both have the form
2A+B®2C;
whereas, reaction (3) has the form
2A®2C+D.

Unfortunately, stoichiometric descriptions of reactions such as (4) do not tell the whole story. They only tell us the net result of a reaction, without telling us how the reaction takes place. Most chemical reactions actually have a mechanism involving the formation of intermediate products. For example, the net reaction (2) for the formation of nitrogen dioxide actually consists of three subreactions-all of which occur simultaneously. A proposed mechanism for this reaction is

2NO
® N2O2   (NO reacts with itself toform N2O2.)
N2O2
®2NO   (NO is reformed from someof the N2O2.)
N2O2+O2
®2NO2   (Some of the N2O2 reacts with O2 to form NO2.)
The intermediate product of dinitrogen dioxide (N2O2) does not appear in the net reaction (2), but it is involved in the mechanism of the reaction and, hence, a mathematical model of the reaction must take the presence of N2O2 into account.

Because most chemical reactions that are important for a scientific understanding of our world involve complex mechanisms, the development of mathematical models of such reactions must be preceded by a substantial amount of theoretical and experimental work in Chemistry aimed at gathering an understanding of the mechanisms. In this module, we will restrict our attention to the study of simple chemical reactions. Simple reactions are reactions that do not involve complex mechanisms. The study of simple reactions is a good starting point for learning some of the mathematics that also pertains to the study of more complex reactions.

1.1  Units of Measurement and Notation

Since molecules are very small, quantities of molecules are measured in units of moles. One mole of molecules is an Avogadro's number of molecules. Avogadro's number is approximately 6.022×1023 . Hence, for example, two moles is the same as 1.2044×1024 molecules. Concentrations of molecules in a solution are measured in units of molarities (M). One molarity is one mole of solute per liter of solution. For example, a 2 M aqueous solution of sodium chloride (NaCl) is a solution consisting of two moles of NaCl per each liter of solution. The notation [A] denotes the concentration (in molarities) of a molecule A in solution. Thus, if we write [NaCl] = 2 M, we mean that we have a solution with a 2 M concentration of sodium chloride.

2  Rates of Reactions

Consider a simple chemical reaction of the form

2A+B® C.
This reaction involves the combination of two molecules of A and one molecule of B to form one molecule of C. Because we are assuming that the reaction is simple (that is, there are no intermediate steps), the concentration of A decreases at twice the rate that the concentration of B decreases. Thus, we have
d[ A]
dt
 = 2 d[ B]
dt
 .
Also, the concentration of C increases at the same rate that the concentration of B decreases, so
d[ C]
dt
 = - d[ B]
dt
 .
We can summarize the above two equations by writing
- 1
2
 d[ A]
dt
 = - d[ B]
dt
 = d[ C]
dt
 .
In general, for a simple chemical reaction of the form
aA+bB® cC+dD,
we have
- 1
a
 d[ A]
dt
 = - 1
b
 d[B]
dt
 = 1
c
 d[ C]
dt
 = 1
d
 d[ D]
dt
and we define the rate of the reaction, v( t) , as this common value. That is,
v = - 1
a
 d[ A]
dt
 = - 1
b
 d[B]
dt
 = 1
c
 d[ C]
dt
 = 1
d
 d[ D]
dt
 .
Since the stoichiometric coefficients a, b, c, and d have no units, v has units of M/time. We stress that the reaction rate v is a function of time (t) because the reaction slows as the reactants are used up during the course of the reaction. The units of measurement of time for a particular reaction depend on the speed of the reaction. For a reaction that proceeds very quickly, it might be appropriate to measure time in seconds or milliseconds, whereas; for a very slow reaction such as the decay of some radioactive compounds, time is measured in years.

2.1  The Rate Law

An important concept of chemistry that is crucial to the development of mathematical models of chemical reactions is the Rate Law. For the homogeneous reaction

aA+bB® cC+dD
with reaction rate v (as defined in the previous section), the Rate Law gives the equation
v( t) = k[ A] a[ B] b.
(5)
Equation (5) is the framework on which mathematical models of chemical reactions are built. In this equation, the constant of proportionality, k, is called the rate constant of the reaction, and the constants a and b are called the order of the reaction with respect to the reactants A and B respectively. The constants k, a, and b can be determined only by actual chemical experiments. In general, except possibly by coincidence, a and b are not related to the stoichiometric coefficients a and b. Also, a and b have no units of measurement and the units of k are determined by the values of a and b in the following way: Recalling that v has units of M/time and [ A] and [ B] both have units of M, writing equation (5) in terms of units, we obtain
M
time
 = ( units of k) ·(M) a·( M) b = (units of k) ·Ma+b.
From this we obtain,
units of k = M
time·Ma+b
 = M1-a-b·time-1.

2.2  Example

As an example of the Rate Law, suppose that we have a reaction

2A+B® C+2D
which is first order with respect to A and B (that is, a = b = 1), and has rate constant k = 0.02 M-1· min-1. Suppose also that this reaction takes place in a container of fixed volume. The Rate Law then gives us that the rate of this reaction is
v = 0.02[ A] 1[ B] 1.
(6)

Note that the stoichiometric coefficients play no role in the rate equation (6). However, if we wish to rewrite the rate equation in terms of the rate of consumption of one of the reactants (or the rate of formation of one of the products), then the stoichiometric coefficients do play a role. For example, using the fact that

v = - 1
2
 d[ A]
dt
 ,
we obtain the rate equation
- 1
2
 d[ A]
dt
 = 0.02[ A] [B]
or equivalently
d[ A]
dt
 = -0.04[ A] [ B]
(7)
for the rate of consumption of reactant A.

Likewise, we also obtain the following differential equations for the concentrations of B, C, and D:

d[ B]
dt
= -0.02[ A] [ B]
d[ C]
dt
= 0.02[ A] [ B]
d[ D]
dt
= 0.04[ A] [ B].

2.3  Exercises

Consider the simple bimolecular reaction

A+B® C
(8)

  1. Given that reaction (8) is first order with respect to A and B, and given that the rate constant for this reaction is k = 0.025 M-1· sec-1, use the Rate Law to write down a system of differential equations for [ A] and [ B] :
    d[ A]
    dt
    = ________
    		(9)
    d[ B]
    dt
    = ________

  2. System (9) that you derived in part 1 should have infinitely many equilibrium points. Identify these equilibrium points and draw a picture of the ( [ A] ,[ B] ) plane indicating where the equilibrium points are. Are all of these equilibrium points relevant to the study of the chemical reaction? If not, then which are the relevant ones? In terms of the chemical reaction, what is the significance of each equilibrium point?

  3. Find the direction vector of system (9) at the point ( 6,4) . Draw this direction vector in the ( [A] ,[ B] ) plane. What does this direction vector tell you about the chemical reaction? Find the direction vector at (4,2) and draw it. What does this vector tell you about the reaction? What is similar about these two direction vectors and what is different about them?

  4. Produce a rough sketch of the direction field for system (9). (You can do this with a computer, but you may realize that you don't really need a computer to do it.) What special feature does the direction field have that makes it particularly easy to see what solution curves of system (9) look like in the ( [ A] ,[ B]) plane?

  5. Using your direction field, do a hand (or computer) sketch of the solution of system (9) with initial condition ([ A] 0,[ B] 0) = ( 6,4) . Which equilibrium point does this solution converge to as t®¥.

  6. Sketch (by hand or computer) the [ A] vs. t and [ B] vs. t graphs of the solution from exercise 5.

  7. For the solution from exercise 5, find an equation relating [ A] and [ B] .

  8. Even though the model system (9) is nonlinear, it is possible to find its exact solution analytically (a rarity for nonlinear systems). Find the solution of system (9) with initial conditions [ A] 0 = 6 M and [ B] 0 = 4 M. HINT: The relation that you found between [ A] and [B] in exercise 7 can be used to decouple system (9).

  9. Discuss whether or not the exact solution you found in part 8 is consistent with your earlier findings in parts 1-6. For example, what do you get when you use your exact solution to compute limt®¥[ A] and limt®¥[ B] ?

  10. Time to generalize! Write an initial value problem that models the reaction A+B® C which is assumed to be first order with respect to both A and B and which has rate constant k > 0 and initial reactant concentrations [ A] 0 and [ B] 0. (You may assume that [ A] 0 ¹ [ B] 0. As you proceed, you should see why the case [ A] 0 = [B] 0 must be handled separately.)

    Solve the initial value problem you have written and discuss whether or not your solution tells you what you expect it to about the chemical reaction. In particular, discuss the long term behavior (as t®¥) of your solution and discuss what role is played by the rate constant, k, and by the initial concentrations, [ A] 0 and [ B] 0.

  11. Repeat as many as possible of the exercises in parts 1-9 using the reaction
    A+B® C
    with rate constant k > 0, but assuming that this reaction is first order with respect to A (a = 1) and second order with respect to B (b = 2). Discuss any difficulties that are encountered by assuming that b = 2 (as opposed to b = 1).

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On 15 Jan 2000, 14:11.