8.5.2: Equilibrium and Non-Equilibrium States

Let us look at a chemical system macroscopically. If we consider a reaction system that begins to change when the reactants are mixed up (that is, it occurs spontaneously), we will eventually see that the change slows down and then stops. It would not be unreasonable to think that the system is static and assume that the molecules in the system are stable and no longer reacting. However, as we discussed earlier, at the molecular level we see that the system is still changing and the molecules of reactants and products are still reacting in both the forwards and reverse reactions. In the case of our acetic acid example, there are still molecules of acetic acid, (AcOH), acetate (AcO^-), and hydronium ion (H₃O⁺) colliding with solvent water molecules and each other. Some of these reactions will have enough energy to be productive; molecules of acetate will transfer protons to water molecules and the reverse reaction will also occur. What has changed is that the rate of acetate (AcO^-) and hydronium ion (H₃O⁺) formation is equal and opposite to the rate of acetic acid deprotonation (transfer of the proton to water). Although there is no net change at the macroscopic level, things are happening at the molecular level. Bonds are breaking and forming. This is the dynamic equilibrium we discussed earlier.

Now what happens when we disturb the system. At equilibrium, the acetic acid–water system contains acetic acid (AcOH), protons (H₃O⁺), and acetate ion (AcO^–). We know that a 0.10-M solution of acetic acid has concentrations of [H₃O⁺] = [AcO^–] = 1.3 x 10^–3 M. Now we add enough acetate¹⁶⁷ to make the acetate concentration 0.10 M? One way to think about this new situation is to consider the probabilities of the forward and backward reactions. If we add more product (acetate), the rate of the backward reaction must increase (because there are more acetate ions around to collide with). Note that to do this, the acetate must react with the hydronium ion, so we predict that the [H₃O⁺] will decrease and the acetate will increase. But as we saw previously, as soon as more acetic acid is formed, the probability of the forward reaction increases and a new equilibrium position is established, where the rate of the forward reactions equal the rate of the backward reactions. Using this argument we might expect that at the new equilibrium position there will be more acetic acid, more acetate, and less hydronium ion than there was originally. We predict that the position of equilibrium will shift backwards towards acetic acid.

This probability argument gives us an idea of what will happen when an reaction at equilibrium is disturbed, but it doesn’t tell us exactly where it will restabilize. For that we have to look at Q and K_eq. If we take the new initial reaction conditions (0.10 M AcOH, 0.10 M AcO^–, and 1.3 x 10^–3 M H₃O⁺) and analyze them to determine the concentrations of all participating species, we can calculate Q and compare it to K_eq:

Q = [H₃O⁺][AcO^–]/[AcOH] = (1.3 x 10^–3)(0.1)/(0.1)

This generates a value for Q as 1.3 x 10^–3. Now, if we compare Q and K_eq, we see that Q is larger than K_eq (1.3 x 10^–3 > 1.8 x 10^–5). To re-establish equilibrium, the system will have to shift so that Q becomes smaller or equal to K_eq (at which point ΔG = 0). To do this, the numerator (products) must decrease, while the denominator (reactants) must increase.¹⁶⁸ In other words, the reaction must go backwards in order to reestablish an equilibrium state. This approach leads us to the same conclusion as our earlier probability argument.

If we recalculate the [H₃O⁺] under the new equilibrium conditions (that is 0.10 M AcOH and 0.10 M acetate), we find that it has decreased considerably from its initial value of 1.3 x 10^–3, down to the new value of 1.8 x 10^–5 M.¹⁶⁹ Using this to calculate the pH, we discover that addition of sodium acetate causes the pH to rise from 2.9 to 4.5. This may not seem like much, but remember that each pH unit is a factor of 10, so this rise in pH actually indicates a drop in hydronium ion concentration of a bit less than a hundredfold. In order to regain the most stable situation, the system shifts to the left, thereby reducing the amount of product:

AcOH + H₂O ⇄ H₃O⁺ + AcO^–

There are a number of exercises that will allow you to better understand the calculations involved in defining the effects of perturbations (changes in conditions, concentrations, and temperature) on the equilibrium state of a system. (Many chemistry books are full of such buffer and pH problems.) What is really important to note is that a system will return to equilibrium upon perturbation. This is where the system is most stable. And once the system is at equilibrium, further perturbations will lead to a new equilibrium state.

References

¹⁶⁷ Of course, there is no such thing as acetate (CH3COO–) alone. There must also be a counter-ion present. Typically, we use ions such as Na+ or K+, stable monovalent cations that will not participate in any further reaction. So when we say we add acetate to the solution, we really mean we add sodium acetate—the sodium salt of acetic acid (just like sodium chloride is the sodium salt of hydrochloric acid).

¹⁶⁸ If you think about it for a moment you will see that if the concentration of any species changes in a closed system, then the concentrations of all the other species must also change.

¹⁶⁹ You might be wondering if there is some trick here. There is—we are ignoring several side reactions that in fact tend to cancel each other out. If you are interested, there are a number of helpful sites that can assist you with the more complex calculations required.