Research: Science and Education
What Is a Reaction Rate? Guy Schmitz Faculté de Sciences Appliquées, Université Libre de Bruxelles, CP 165/63, Av.F.Roosevelt 50, 1050 Brussels, Belgium;
[email protected] Le Vent (1) has discussed the definition of a reaction rate in this Journal and demonstrated the need for care when using this term. He pointed out correctly some pitfalls in the literature but ignored other ones. In the present article we start again at the beginning of the problem, progress stepby-step, and try to reach clear conclusions helpful to instructors in physical chemistry. We begin by considering closed systems and show that a reaction rate cannot be defined in terms of chemical species properties—amount of substance or concentration—but can be defined in terms of a reaction property—the extent of reaction. We state precisely the relationship with stoichiometry. Next we consider open systems and show that they require a new approach to the concept of reaction rates. It is however possible to give a definition applicable in open as well as in closed systems.
and will have to note that one mole of iodate gives 3 moles of iodine. Then r3 = (dnI2dt)3V and eq 2 is too restrictive. For a general reaction ν1 S1 + ν2 S2 + …
νp Sp + νp + 1 S p + 1 + …
we obtain
r =
(d n1 d t ) ν1V
= … =
r = ±
d cS
(1)
dt
where cS is the concentration of a substance S and r is the reaction rate. The plus or minus sign depends on whether S is a product or a reactant. Introducing the relationship between the concentration and the amount of substance nS for a product, we get: r =
d (n S V ) dt
=
n dV 1 d nS − S2 V dt V dt
The last term is not related to a reaction rate. Thus, if the volume is not constant, we have to replace eq 1 with
r = ±
(2)
V
Le Vent (1) showed the superiority of eq 2 over eq 1 on the basis of both the collision theory and the transition-state theory. The quantity dcS/dt is just an experimental quantity related to a reaction rate in a way we have to state precisely. Rate and Stoichiometry Let us consider the classical Dushman reaction (2): 3I2 + 3H2O
dξξ =
(4)
d nS
(5)
ν
where dnS is the change of the amount of substance S resulting from the considered reaction. Introducing this definition into eq 4 we obtain
r =
(d ξ d t )
(6)
V
The International Union of Pure and Applied Chemistry (IUPAC) proposed this definition, but with some ambiguities discussed by Le Vent (1). Actually, eq 6 is always correct as long as we consider a closed reactor. It must be clear that a reaction rate is a property of a reaction, not of chemical species. As stated by Cvitas (3), “When the reaction cannot be specified, neither can its advancement or rate be. In such a case only rates of consumption and rates of formation … are meaningful quantities.”
(3)
The stoichiometry of the Dushman reaction is not exactly described by eq 3 because iodine can react further. I2 + I −
•
I3−
(7)
At the time scale of the Dushman reaction, reaction 7 is nearly at equilibrium and we get a mixture of I3− and I2. If the iodide concentration is low the final product is mainly I2 but if the iodide concentration is high it is mainly I3−. Thus the rate of reaction 3 is not equal to (dnI2dt)3V. The correct equation is
According to eq 2 the rate is r3 = ᎑(dnIO3−dt)V. However, experimentally we will probably follow the I2 concentration www.JCE.DivCHED.org
= …
More than One Reaction
(d n S d t )
IO3− + 5 I − + 6H+
νp V
where the stoichiometric numbers νi are negative for the reactants and positive for the products.1 We can replace this set of equations with only one equation if we use the extent of reaction ξ (3, 4). The usual definition of ξ is
Amount of Substance and Concentration To measure a reaction rate, the concentration change of a substance involved in the reaction is usually measured. Thus, most textbooks define the rate as
(d np d t )
Vol. 82 No. 7 July 2005
r3 =
•
d n I + nI 2
3
−
dt
3V
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Equation 4 is not a good definition of a reaction rate when we have more than one reaction. Let us now consider eq 6. We have two extents of reaction and two rates, r 3 = (dξ3dt)V and r7 = (dξ7dt)V. The relationships with the amounts of substances are dnI2 = 3dξ3 − dξ7 and dnI3− = dξ7 giving d(nI2 + nI3−)dt = 3dξ3dt = 3Vr3. Equation 6 gives the correct expression even when we have more than one reaction. However, the relationship dnI2 = 3dξ3 − dξ7 shows that we have now a difficulty with the definition 5 of the extent of reaction. This brings us into the core of the problem, the stoichiometry of the studied system. A stoichiometric equation is a relation between chemical amounts of reactants and chemical amounts of products. It is formally correct if and only if it is balanced, if the numbers of atoms and charges is preserved. Equation 3 says simply that one mole of iodate reacts with five moles of iodide to give three moles of iodine. When the extent of reaction is equal to 0.1, 0.1 mole of iodate has reacted with 0.5 moles of iodide to give 0.3 moles of iodine. This is absolutely true, even when there are other reactions in the same system. However, in this case the amount of a substance produced by a reaction can be different from the change of its amount in the solution and the meaning of dnS in eq 5 can be misleading. For clarity, we should write explicitly for a reaction i: dξ i =
(d n S )i ν S, i
The subscript i emphasizes the difference between [dnS]i the amount of substance S produced or consumed by reaction i, and the change of the total amount dnS. The amount of I2 produced by reaction 3 is 3ξ3, the amount consumed by reaction 7 is ξ7 and the net production is dnI2 = 3dξ3 − dξ7. Complex Reactions Toby and Tobias (5) have shown that there are two approaches to the stoichiometry of a complex reaction, the kinetic approach and the material-balance approach. They are appropriate in different contexts. At first, we study a reaction experimentally and obtain its rate law. After that, we try to explain our observations and try to build a mechanism. Different mechanisms can give different stoichiometric relationships but the stoichiometry must be known first and a mechanism that does not give the correct stoichiometry must be rejected. The stoichiometry is an experimental result and “The decoupling between stoichiometry and kinetics is particularly important for dealing with non-elementary reactions” (6). Reaction 3 is not an elementary reaction. However, if we can make the steady-state approximation for the intermediates, its contribution to the evolution of the system can be described using only one reaction rate (7). The definition of r3 is independent of the mechanism, it is based purely on stoichiometric considerations. Open Systems Consider a stationary flame: all the time derivatives are equal to zero, it is a stationary system, but obviously the reactions and the rates of reaction are not equal to zero. Equa-
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tion 6 is limited to closed reactors. In order to extend it, let us write the material balance in an open volume perfectly mixed V with an input flow FSin of a substance S and an output flow FSout (amount of substance per time).
FSin
FSout
nS V
There can be a difference between FSin and FSout for two reasons: (i) The amount of S in the volume V changes. This gives a term dnSdt in the material balance equation. (ii) A reaction in this volume produces or consumes the substance S. With a single reaction, the rate of production or consumption of S is νSr per unit volume and the contribution to the material balance equation is νSrV. Thus, we have FSout = FSin −
d nS dt
+ νS r V
(8)
If the volume V were closed, we would have FSout = FSin = 0 and eq 8 would reduce to eq 4 for the substance S. The relationship r = (dnSdt )(νSV ) is a particular case of a material balance equation, not a definition. Let us now consider an open reactor and suppose it has reached a steady state. Then dnSdt = 0 and eq 8 reduces to
r =
FSout − FSin νS V
(9)
The physical meaning of r must be the same in eq 9 as in eq 4 and the concept of reaction rate is not strictly linked with any time derivative. In order to extend it, a transformation of eq 9 is useful. Let Q be the total flow through the reactor (volume per time). The ratio FSoutQ has the dimension amount-of-substance-per-volume and is equal to the concentration in the output flow. As this concentration is equal to the concentration in the perfectly mixed volume V, we get FSoutQ = nSV. The ratio VQ has the dimension of a time. It is the mean time the molecules spend in the reactor and is called the residence time τ. With this definition we obtain FSout = nSτ. If we had no reaction, the composition in the reactor would be identical to the composition of the input flow. Noting nS° is the amount of S in the reactor without reaction and assuming constant density we obtain in a similar way FSin = nS°τ. With these expressions, eq 9 becomes r =
n S − n S ° νS τ V
(10)
This equation shows some analogy with eq 6 but is clearly different. For example, the integration of eq 6 for a reaction of first order gives an exponential relationship between nS and t while eq 10 gives a simple algebraic relationship between nS and τ. The equations are different but the concept
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Research: Science and Education
of reaction rate is the same in a closed and in an open reactor and a good definition should apply to both. We propose the following one. The rate of a reaction is the amount of substance this reaction produces or consumes per unit time and unit volume, divided by the corresponding stoichiometric coefficient.
In a closed reactor we mean an elapsed time and in an open reactor we mean a residence time, but in both cases it is simply the time spent by the reactants in the reactor. With this comment, “per unit time” applies to both kinds of reactors in contrast with “differentiated with respect to time”. A definition is not necessarily an equation. Equations come after, when we apply the definition to a given system. A given reaction i produces or consumes a substance S at a rate νS,iri, where νS,i is the stoichiometric coefficient of the substance S for the reaction i. In a volume V the rate of conversion is equal to νS,iriV. If the substance S appears in different reactions it is produced or consumed at a global rate V ∑νS,iri. When we consider an open reactor in a steady state the chemical amounts in the reactor are constant. The substances are brought in or taken out by the flow at the same rate as they are produced or consumed by the reaction. This gives a generalization of eq 9, V ∑νS,iri = FSout − FSin. The rate term appears like a chemical input flow of products or
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output flow of reactants and the material balance equation expresses that these chemical flows are equal to the net physical flow. On the other hand, in a closed reactor the global rate of conversion is equal to the rate of change of the amounts of substances in the reactor. This gives a generalization of eq 4, V ∑νS,iri = dnSdt. If the intensive variables ri have not the same value in each point of the reactor, we just have to apply these equations to infinitesimal volumes. Our definition is always true and can be used to write the material balance equations under any particular circumstance. Note 1. This is the usual convention in the mathematical equations. The positive coefficients for the reactants usually used in the chemical equations are the absolute values |νi |.
Literature Cited 1. 2. 3. 4. 5. 6. 7.
Le Vent, S. J. Chem. Educ. 2003, 80, 89–91. Schmitz, G. Phys. Chem. Chem. Phys. 1999, 1, 1909–1914. Cvitas, T. J. Chem. Educ. 1999, 76, 1574–1577. Croce, A. E. J. Chem. Educ. 2002, 79, 506–509. Toby, S.; Tobias, I. J. Chem. Educ. 2003, 80, 520–523. Lee, J. Y. J. Chem. Educ. 2001, 78, 1283–1284. Rasiel, Y.; Freeman, W. A. J. Chem. Educ. 1970, 47, 159–160.
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