The Drive to Chemical Reaction EquilibriumHow chemical reaction equilibrium is attained and why energy (or enthalpy) and entropy both play a role. One of the simplest amino acids is alanine, C3H7NO2. As do all amino acids, alanine exists in two forms that are mirror images of each other. The form found in nature is called L-alanine, and its mirror image form is called D-alanine. (The L and D notation comes from a similar effect found in sugars. When you take organic chemistry, you will learn that modern organic notation calls L-alanine (S)-alanine and D-alanine is called (R)-alanine, but the L- and D- notation is still very common, especially in biochemical literature.) Pictures of these two molecules are shown below (C is grey, H is white, N is blue, and O is red):
These two molecules are physically different, and one can separate a mixture of them into pure L- or pure D- forms. Because they differ only by being mirror images, they must have the same energy, enthalpy, and entropy. If we start with pure L-alanine, we can ask if it will spontaneously turn into D-alanine. There is certainly no energetic reason to do so—both forms have the same energy, and in fact, the First Law of Thermodynamics would be happy if none, all, or any fraction of L- turned into D-. In fact, pure L- or D- left to itself just sits there and does not convert into the other form. But the reason why nothing happens is actually a tricky one: any amino acid is too complicated a molecule to invert itself into its mirror image all by itself. As is often the case, nature comes to the rescue. There is a naturally occurring enzyme called alanine racemase which, when mixed with either form of alanine, causes either form to turn into the other form. The enzyme itself is not changed (enzymes are catalysts, and this is one attribute of all catalysts). If one waits until no further change is observed (i.e., until chemical equilibrium is attained), one finds that the reaction mixture is 50/50 in each form, no matter which form one starts with. In other words, we can state that the equilibrium constant for this reaction is 1:
What drives this reaction to this equilibrium state? It is not enthalpy (∆H for this reaction must be zero), so it must be entropy even though ∆S is also zero for this reaction! Our goal here is to see how entropy plays a perhaps unexpected role in attaining chemical reaction equilibrium, and along the way, we will understand better the role enthalpy plays. We will leave alanine behind for now and instead focus on a reaction that is very similar, but one that has an enthalpy change. The reaction we will choose is the conversion of hydrogen cyanide, HCN, into hydrogen isocyanide, HNC. Both of these species are gases at ordinary temperatures, and they differ only by bonding. HCN is a linear molecule with atoms bonded as written: H—C—N. HNC, however, is an isomer of HCN (an isomer of one molecule is a new molecule with rearranged bonding). It is also linear, but the three atoms are bonded in the order H—N—C. Bonding changes indicate enthalpy changes, and HNC has a molar enthalpy some 61 kJ mol–1 higher than HCN. In other words,
On the other hand, these two molecules obviously have the same mass, are both gases, and both have quite similar structures: linear with H on one end. This tells us that they must have almost identical molar entropies, so we can write ∆S = 0 as a very good approximation. We now consider free energy. We know ∆G = ∆H – T∆S in general, so if ∆S = 0, as it does here, then ∆G = ∆H = 61 kJ mol–1. Recall that ∆G for a reaction is defined to be the difference between the free energy of the products (here, only HNC) and that of the reactants (only HCN) when the products are pure, the reactants are pure, and the two are not mixed. In other words, a reaction ∆G (often written ∆Gr to emphasize this fact) generally does not tell us everything we might want to know about the chemical reaction equilibrium state of the reactants and products. (To return to our alanine example, ∆Gr = 0, which tells us K = 1 because K = exp(–∆Gr/RT) = exp(0) = 1, but it does not really tell us why the equilibrium mixture is 50/50.) To get to the heart of the matter, we must consider the total free energy of the reaction mixture and see how it changes as the amounts of reactants and products change as we approach equilibrium. At equilibrium, the total free energy must be at a minimum. Recall that spontaneous processes are those that lower free energy; so, if we are not at equilibrium, something spontaneous will happen—chemical reaction in one direction or another—until we reach the lowest possible total free energy. It is easy to write an expression for Gtot for our reaction. It is just the number of moles of HCN times the molar free energy of HCN, GHCN, plus a similar product for HNC. Moreover, we know how to write the molar free energy of a gas in terms of its standard state free energy, Go, and its partial pressure P: G = Go + RT ln P. Thus, we can write
We can simplify this expression if we do three things. First, because this is an endothermic reaction with only a single reactant and a single product, we can choose our zero of free energy to be that for HCN in its standard state (making Go for HCN zero), and this choice makes Go for HNC equal to ∆Gr, which is 61 kJ mol–1. The second thing we can do is to specify simple initial conditions for this reaction. Let's choose 1 mol of pure HCN and no HNC at the start, and let's pick a starting pressure of 1 atm. Finally, let's introduce the degree of advancement variable (the "extent of reaction" variable we introduced early in the course) into the story. For this reaction, the extent of reaction variable x is defined in general as
For our specific conditions, this becomes
Moreover, our choice for a total pressure of 1 atm (which will not change even if all the HCN turns into HNC, because the reaction produces one mole of gaseous product for every mole of gaseous reactant consumed) along with our choice for a total of 1 mol of gas, again no matter how the reaction advances, allows us to write a very simple expression for the total free energy, Gtot. We look first at the free energies of each gas, recalling that the partial pressure of a gas is its mole fraction times the total pressure (and, to save space and clutter, let's do a bad thing: suppress the atm and mol unit symbols). We find
We substitute these expressions into our expression for the grand total free energy, and find
This is our final working formula. It gives us Gtot as a function of x, and we can graph this function as x goes from zero (pure HCN) to 1 mol (pure HNC). Before we do, however, note that the first term is linear in x and the second term contains T. Note as well that the second term must be negative at any non-zero T because x and 1 – x are both always between 0 and 1 mol, and the natural log of a fraction is a negative number. At low enough T, the linear term dominates. At high enough T, the second, negative term dominates. The graph below shows Gtot as a function of x at a low and at a high T:
At 300 K, equilibrium is totally dominated by the first term, linear in x, which shows up on the graph as the (nearly) straight blue line. The equilibrium constant is very small, and the equilibrium mixture is essentially pure HCN; the total free energy is straight up from x = 0 to x = 1 mol, and production of HCN can't lower the free energy at any x > 0. (Actually, K is never exactly zero, but at 300 K, K = exp(–∆Gr/RT) = exp[–61,000 J/(8.314 J mol–1 K–1) (300 K)] = 2.4 x 10–11, which is very, very close to zero!) Now consider the red line for 4000 K. (This line was drawn assuming ∆Hr was constant from 300 K to 4000 K, which it isn't, but the correction is probably small.) At 4000 K, the equilibrium constant has grown (remember: K increases with T for an endothermic reaction) to K = 0.16 (a factor of almost 10 billion for a bit more than a factor of 10 increase in T). This line has a definite minimum, located by the dark red arrow. This tells us that at 4000 K, the reaction equilibrium mixture contains a significant amount of product HNC. The minimum occurs at x = nHNC = 0.14 mol, and this must be the equilibrium value for x, or xeq. We can find a general expression for xeq if we find a general expression for the minimum in the total free energy as a function of x. The minimum is given by dGtot/dx = 0. Some calculus and algebra (give it a try!) shows
We have recovered a familiar expression. If we had started with the usual equilibrium constant expression for this reaction, written under our initial conditions, we would have written
These results show the importance of the temperature-dependent term in our expression for Gtot. This term is called the entropy of mixing term. It is not a part of the reaction entropy change, ∆Sr, (which we argued was close to zero for the HCN/HNC reaction and was exactly zero for the alanine case). Rather it is a real entropy effect that lowers the total entropy whenever energetic considerations allow for products to be produced that can mix with the reactants. For the alanine case, that's the only "driving force" in the reaction, and it accounts for the 50/50 mixture, the K = 1 value, for that reaction: the only term in Gtot is the entropy of mixing term, and it has a minimum at x = 0.5 mol for all temperatures. Finally, we can note that the entropy of mixing term can always dominate if the temperature is high enough. As T approaches infinity, energetic (or enthalpic) differences no longer matter. The free energy is controlled by mixing entropy!
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