tag:blogger.com,1999:blog-52829252656529066362017-06-21T19:36:46.702-07:00Investigations into Basic Problems"Every day my method led me to discover truths that seemed to me to be quite important and not widely known; my pleasure in this so filled my mind that nothing else mattered to me." - René Descartes, Discourse on the Method, §3.Georgios Scholarios / Georgehttp://www.blogger.com/profile/04093146576088231265noreply@blogger.comBlogger1125tag:blogger.com,1999:blog-5282925265652906636.post-5443527880816059082015-07-17T23:41:00.000-07:002016-09-04T11:21:08.360-07:00Where does the equilibrium expression come from?<div dir="ltr" style="text-align: left;" trbidi="on"><div style="text-align: justify;"><span style="font-family: Helvetica Neue, Arial, Helvetica, sans-serif;"><b>Preface (Added August 2016)</b><p>Since I've published this article back in July 2015, I have done more research into the question of where the equilibrium expression comes from. From this I have learned that the derivation I give below is only approximate and does not work for the majority of reactions whose rate laws are more complicated than what is shown in high school or first year university. Nonetheless, it is still very useful in showing the meaning of the equilibrium constant, something often lost on students. The whole point of the equilibrium expression is that it allows one to check if a system is in equilibrium, using only the concentrations of the products and reactants. If you measure the concentrations for a chemical system and plug them into the equilibrium expression, calculate and get a number equal to the equilibrium constant at the appropriate temperature, then the system you are considering is at equilibrium. Why does this happen? As the kinetic derivation shows, the equilibrium expression can be thought of as simply a rearrangement of the equation \(rate_{forward} = rate_{reverse}\), so if the equilibrium expression calculation comes out right, this is just a different way of saying the forward rate equals the reverse rate, i.e., the system is at equilibrium.<p><b>Introduction</b><p>Most high school chemistry students know the term "equilibrium expression" and will be able to recognize that the general chemical equation \(aA + bB \rightleftharpoons cC + dD\) has the equilibrium expression $$K = \frac{[C]^c[D]^d}{[A]^a[B]^b}$$ <p>But what they don't know is where this comes from. Considering that even most introductory mathematics courses skip derivations of formulas (such as the quadratic formula), it is not terribly surprising that introductory chemistry courses do the same. While there are valid objections to doing so, I feel that it can increase students' interest in the subject in addition to making the course feel less like memorization.</p> <p>So, in this post I intend to attempt to show where the general equilibrium expression comes from. In other words, I will try to show why the equilibrium expression is specifically \(\frac{[C]^c[D]^d}{[A]^a[B]^b}\) and <i>not</i>, for example, \(\frac{c[C] \times d[D]}{a[A] \times b[B]}\). I will start by (1) trying to derive the equilibrium expression for reactions that occur in one step. Then I will move onto (2) reactions that occur in multiple steps.</p> <b>1. One-Step Reactions</b><p>Assume the reaction \(aA + bB \rightleftharpoons cC + dD\) is a one-step reaction.</p> <p> For <i>any</i> reaction at equilibrium, the rate of the forward reaction equals the rate of the reverse reaction. $$rate_{forward} = rate_{reverse}$$ And for any <i>one-step</i> reaction, the rate equals that reaction's <i>rate constant</i> multiplied by the concentration of each product to the power of its coefficient. So, for our one-step reaction above, $$rate_{forward} = k_1[A]^a[B]^b$$ $$rate_{reverse} = k_2[C]^c[D]^d$$ From the above three mathematical equations, it easily follows that $$k_1[A]^a[B]^b = k_2[C]^c[D]^d$$ Now I will rearrange the equation with the aim of coming up with the equilibrium expression. Dividing both sides by \(k_2\) yields, $$\frac{k_1}{k_2}[A]^a[B]^b = [C]^c[D]^d$$ Then I will divide both sides by \([A]^a[B]^b\) to get, $$\frac{k_1}{k_2}=\frac{[C]^c[D]^d}{[A]^a[B]^b}$$ Since \(k_1\) and \(k_2\) are constants, \(\frac{k_1}{k_2}\) is also a constant, which we can alternatively call \(K\). So, $$K=\frac{[C]^c[D]^d}{[A]^a[B]^b}$$ Voilà, I have (hopefully successfully) shown where the general equilibrium expression for one-step reactions comes from.</p> <b>2. Multi-Step Reactions</b><p>Assume the reaction \(aA + bB \rightleftharpoons cC + dD\) is a multi-step reaction. This makes the problem considerably more difficult, since now the rate equations of the forward and reverse reactions are not simply the concentrations with the coefficients as exponents (that can <i>only</i> be done for one-step reactions).</p> <p>The reaction occurs in two steps, step I and step II. $$\textbf{I)} \enspace aA + bB \rightleftharpoons xX$$ $$\textbf{II)} \enspace xX \rightleftharpoons cC + dD$$ To proceed, I will first note that, if the overall reaction is at equilibrium, then each step of the reaction is also at equilibrium.<sup>1</sup> I admit that, for me, this observation was not immediately obvious. But I was convinced of it when I found it impossible to imagine an overall reaction at equilibrium with one of its step not at equilibrium. (If you are unsure, try doing this for yourself). Besides, every reversible reaction will reach equilibrium eventually<sup>2</sup>, so <i>even if</i> it is possible for a reaction to be at equilibrium when one of its step isn't, all reversible reactions will eventually reach a state where both it and all of its steps are at equilibrium.</p> <p>Noting that each step is at equilibrium allows us to write the equilibrium expression for each step. Since each step is always a one-step reaction, we can easily do this using the method in the above section (§1). The equilibrium expression for steps I and II will be called \(K_I\) and \(K_{II}\), respectively. Therefore,</p> $$K_I = \frac{[X]^x}{[A]^a[B]^b}$$ $$K_{II} = \frac{[C]^c[D]^d}{[X]^x}$$ If you multiply \(K_I\) and \(K_{II}\) together, you end up with $$ \begin{align} K_I \times K_{II} &= \frac{[X]^x}{[A]^a[B]^b} \times \frac{[C]^c[D]^d}{[X]^x} \\ & = \frac{[C]^c[D]^d}{[A]^a[B]^b} \\ \end{align} $$ Notice that what you end up with is the equilibrium expression of the overall reaction! However, we cannot stop here because all I have shown you is <i>that</i> multiplying the equilibrium expressions of the steps gives you the equilibrium expression of the overall reaction. But I haven't shown you <i>why</i> this happens, which I will attempt to do next.</p> <p>It is a basic fact about multi-step reactions that whenever you add together the equations of the steps, you end up with the equation of the overall reaction. This works because adding the equations allows like products (or reactants) on the same side of the equation arrow to be combined, and it allows like products (or reactants) on the opposite sides of the equation arrow to be cancelled out. This allows (for instance) intermediates to be cancelled out, which gives you the equation of the overall reaction.</p> <p>Essentially the same process is in play when you multiply the equilibrium expressions together. In the equilibrium expressions, the products are always multiplied together on the bottom of the fraction (i.e., in the denominator), and the reactants are always multiplied together on the top (i.e., in the numerator). So multiplying together the equilibrium expressions allows like products (or reactants) on the same side of the fraction bar to be combined, and it allows like products (or reactants) on the opposite sides of the fraction bar to be cancelled out. This allows (for instance) intermediates to be cancelled out (which is what happened to \(X\) above), giving you the overall reaction equilibrium expression.</p> <p>Since this process works exactly the same in both adding and multiplying, and since adding together the steps always gives you the equation of the overall reaction, it follows that multiplying together the equilibrium expressions of the steps will always give you the equilibrium expression of the overall reaction.</p> <b>Conclusion</b><p>Having now shown where the equilibrium expression comes from for both one- and multi-step reactions, I can say that I have successfully shown where the equilibrium expression comes from for every reaction.</p></span></div><hr><sup>1</sup>http://www2.ucdsb.on.ca/tiss/stretton/CHEM4/APEquilibrium.html. <br><sup>2</sup>M.S. Silberberg, <i>Chemistry: the Molecular Nature of Matter and Change</i> (5th ed.), New York, 2009, p. 173. <br></div>Georgios Scholarios / Georgehttp://www.blogger.com/profile/04093146576088231265noreply@blogger.com3