The Derivation of the Equilibrium Constant Expression

The Derivation of the Equilibrium Constant Expression, KVirtually all modern high school and first-year university chemistry textbooks teach that the general chemical equation aA + bB ⇌ cC + dD has the equilibrium constant [C ]c [ D ]d K [ A]a [ B ]b These textbooks also commonly include a simple derivation of this expression, which involves setting the rate equations for the forward reaction and reverse reaction equal to each other. To illustrate, one book (Brown, LeMay Jr., & Bursten, 1997) considers a forward reaction A → B, which they say has rate = k f [A] , and its reverse reaction B → A, which they say has rate = k r [B] . “Eventually the reaction reaches a point at which the forward and reverse rates are the same […] compounds A and B are in equilibrium” (p. 540). And so at equilibrium, Rate of forward reaction = Rate of reverse reaction k f [ A]  k r [ B] The equation is rearranged: kf kr  [ B] [ A] And students are then told that this ratio is called K, the equilibrium constant. Despite its popularity, this kinetic approach to deriving the equilibrium has come under criticism from other educators. The first and most common criticism is given by Quílez (2004) when he writes “this approach should be avoided […] It is an example of obtaining a correct result (equilibrium constant) using an improper procedure that writes down two rate equations based on the stoichiometry of the reaction” (p. 81). According to this criticism, the problem is that this method Contributions followed in the 1870s. as the kinetic derivation assumes a reaction has only a single. as well as Harcourt and Esson. To this end. would have to go a b something like this: assume that the forward rate equals k f [ A] [ B] and the reverse rate equals k r [C]c [ D]d . and the same is true for the reverse rate. from there go on to consider of the nature of scientific explanation and the value of demonstrations in education. the critics claim that using this method with a multi-step reaction. In the Etudes. In other words. fixed rate law. which only works for reactions that occur in one-step. say. However. less common criticism states that kinetic derivations are undermined by the fact that rate laws change as equilibrium is approached (Ashmore. and rearrange to get K  [C ]c [ D] d . 55). Our purpose here is to evaluate these two criticisms. they say. 1965). aA + bB ⇌ cC + dD. 1904. 1985). as well as from van’t Hoff (Laidler. made important early contributions in the 1860s. from Guldberg and Waage.seems to require the assumption that we can write down rate equations directly from the chemical equation. The earliest studies of how rate of reaction depended on concentration occurred in the late 19th century. [ A] a [ B]b a b this cannot be correct because the forward rate is not necessarily equal to k f [ A] [ B] for multi- step reactions. the “epoch-making” work on chemical kinetics and equilibrium was made in 1884 by van’t Hoff’s book Etudes de dynamique chimique (Mellor. van’t Hoff determines the equilibrium constant K by setting the rate of the forward reaction equal to the rate of the reverse reaction and rearranging: k1C1 1  k 2 C 2 n n2 . A second. we will begin by consulting the history of deriving the equilibrium expression. p. Berthelot and Péan de Saint-Gilles. then set them equal to each other. But. p. to the relationship” (van’t Hoff. Walker.n k1 C 2 2  K k 2 C1 n1 In the 1899 textbook Introduction to Physical Chemistry. he is careful to point out that this is only true for equilibrium of gaseous systems. the constant can only be derived from thermodynamic considerations (van’t Hoff. van’t Hoff gave the kinetic derivation approval even when he was aware of a thermodynamic proof: “experiment. will always multiply together to yield an overall equilibrium constant where the exponents equal the corresponding coefficients in the overall equation. if we instead write a general multi-step equation with all the terms moved to the left-hand side. in certain cases at least. And all three – van’t Hoff. 1886. 257). For equilibria of dilute aqueous systems. James Walker offers a similar derivation. However. independently of one another. They show that. This would be shown later by Frost (1941) and then more rigorously by Koenig (1965). In other words. so too the equilibrium constants of the elementary steps. thermodynamics. and the same goes for Mellor’s 1904 textbook Chemical Statics and Dynamics. 1886). Further. and kinetic theory lead. Although van’t Hoff and others said that a kinetic approach could be used. so that all products have negative coefficients. just as the coefficients of elementary chemical equations necessarily add up to the coefficients of the overall chemical equation. and Mellor – were aware of the fact that the rate equation could not always be directly derived from the chemical equation. they did not show exactly how this worked. where the exponents equal the coefficients. x1X1 + x2X2 + x3X3 + … + xnXn⇌ 0 . j . This means there is no problem using the kinetic approach to derive the equilibrium constant of an elementary step: n K j  [ X i ] x j . we can safely assume that the exponents of its forward and reverse rates match its coefficients. s  x j .n For any reaction in this notation. we know s x j 1 Therefore.i n K   [ X i ] j 1 i 1 But since the elementary step coefficients must add up to coefficients of the overall reaction.i is the coefficient of xi in the jth elementary step i 1 To find the overall equilibrium constant. then K   K j  s j 1 s  n x K     [ X i ] j . K. where xj.i j 1  i 1    This can be rewritten as. Since an elementary reaction occurs in one-step. we simply i 1 consider the equilibrium constant for an elementary step j. K   [ X i ] xi and to show this with kinetic theory.i  xi .i . we simply multiply together the equilibrium constants of each elementary step. call it K j . If there are s elementary steps overall. From this. It only assumes that the exponents of the rates of elementary reactions always equal the coefficients of the chemical equation. the first criticism’s allegation). This being said. This is because “an equilibrium state does not depend on the path by which it is reached” (Mysels. and even though this is not true. it is evident that the first criticism of the kinetic derivation of the equilibrium expression is misguided.e.. as above. assume that the exponents of the forward and reverse rates always equal the coefficients of the chemical equation. p. 1956. The above derivation does not. which is a perfectly safe assumption. as the criticism alleges. For such a method implicitly assumes that every reaction is an elementary reaction. The second criticism notes that the rate law for a reaction often changes as the reaction proceeds. and it is clear that the above derivation assumes that the rate law is constant throughout the reaction. we might wonder why the derivation still works despite this simplifying assumption. . such an assumption does not affect the end result because.n K   [ X i ] xi i 1 which is the desired result. however. The second criticism of the derivation. is correct. This is also why we will get the correct result from a method which wrongly assumes that rate law exponents always match the coefficients of the chemical equation (i. the state of equilibrium does not depend on the path used to get there. 179) and so the correct equilibrium constant expression will be derived even when we (wrongly) assume that the reaction occurs in such a way that the rate law remains constant throughout. So it appears that the kinetic derivation falls short of proving with certainty. or even any sort of probability. nor do they all produce their products with the exact same speed. e. which was about the value of the kinetic derivation in education. thermodynamic derivations are “exact” (Koenig. the kinetic derivation requires simplifying assumptions about how rate laws change during the course of a reaction. Presumably the problem here is that not all “collisions” occur exactly in the same way. And. as mentioned above. 1965. it fails to account for the condition of equilibrium through its causes. 1886) do not depend on modeling the mechanism of how reactions actually take place. using only the concentrations of the . 227). van’t Hoff. which is an approximate model of reactions. thermodynamic derivations (e. Because it requires very simplifying assumptions about how reactions actually occur.g.Koenig (1965) calls his kinetic derivation “approximate” (p. the equilibrium constant) through its causes.. this does not answer the more important practical question. Therefore. p. because they are an account of a fact (in this case.g. the derivation helps make clear to students the meaning of the equilibrium constant expression.. but simply rest on evident ideas about matter. However. he remarks that his “definition suffers from vagueness owing to the terms “collision” and immediate”” (p. this gives us certain knowledge of what the equilibrium constant is. as Aristotle notes in the Posterior Analytics (I. 227). the equilibrium constant. but it also relies on collision theory.2). 1919). that they contain energy and that perpetual motion is impossible (Einstein. After defining an elementary reaction to be a reaction that occurs through a collision of particles which immediately yields the given product. Despite its flaws. The whole point of the equilibrium expression is that it allows one to check if a system is in equilibrium. 228). In contrast. This is not only because. chemistry teachers may still wisely avoid it if they think it will confuse their students. teachers should be aware of what exactly are – and are not – the limitations of the kinetic approach to the equilibrium constant. and that a thermodynamic derivation is more rigorous. then the system one is considering is at equilibrium.’ one therefore knows that the forward rate equals the reverse rate. and so the system is at equilibrium. . or make them think that it is acceptable to write rate equations directly from the stoichiometry of the chemical equation. This could be presented to students with the warning that the derivation is approximate. If the equilibrium expression ‘checks out. But to make the best decision. Ultimately. the equilibrium expression can be thought of as a rearrangement of the equation rateforward = ratereverse. whether or not the kinetic derivation should be shown to students is a matter of prudence. If one substitutes in these concentrations and it gives a number equal to the equilibrium constant at the appropriate temperature. Why does this work? It works because. as the kinetic derivation shows. Even if the derivation can be useful.products and reactants. 42. . 2. Reaction kinetics and the law of mass action. 160-166. L. Upper Saddle River. (1965). Koenig. Education in Chemistry. 227-230. Laidler. B. F. J.pdf Frost. (1997). Jr. Journal of Chemical Education. Retrieved from http://germanhistorydocs. 272-274. 18. Chemical kinetics and the origins of physical chemistry. Green. T. Einstein. Mellor. On the deduction of the equilibrium law from kinetics.. and Co. J. Chemical statics and dynamics. P. Journal of Chemical Education. London: Longmans. What is the theory of relativity? The London Times. A. K. LeMay. (1965). (1919. Archive for History of Exact Sciences.ghi-dc.References Ashmore. A.G. November 28). 32. O. (1941). E. W.. Brown. (1904). H.org/pdf/eng/EDU_Einstein_ENGLISH. & Bursten. Effect of concentration on reaction rate and equilibrium. NJ: Prentice Hall.). 43-75. (1985). A. E.. Chemistry: the central science (7th ed. 239-302. (1884). (1899). 69-87. 178-179. J. The laws of reaction rates and equilibrium. . Archives Néerlandaises des Sciences Exactes et Naturelles. Amsterdam: Frederik Muller & Co. 20. J. (1956). J.Mysels. Etudes de dynamique chimique. H. 5. Van’t Hoff. Van’t Hoff. (2004). K. J. (1886). Quílez. 33. J. L’équilibre chimique dans les systèmes gazeux ou dissous à l’état dilué. Introduction to physical chemistry. London: Macmillan and Co. Walker. 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