Chemistry Lab/Equilibrium

This page is related to the 2013 focus on Equilibrium for Chem Lab.

Basic Information
The term "equilibrium" in chemistry refers to the ability of a reaction to proceed in either direction. When a balance between reactants and products is met, the equation is said to be in equilibrium. Although many reactions are simplified by assuming they run to completion, in reality, this is not the case. In equilibrium, the forward (reactants --> products) and reverse (products --> reactants) are occurring at equal but opposite rates (hence, equilibrium). The ratio of reactants to products at equilibrium can be defined as a constant, generally k.

The standard example of chemical equilibrium given in most textbooks is the decomposition of N2O4 (g) to 2NO2 (g) (probably because the color changes, making it easy to see changes in equilibrium). Because of its simplicity, this example will be used to introduce calculation of the equilibrium constant.

The Equilibrium Constant
K, the equilibrium constant, is derived from chemical kinetics. It can be defined as the concentration of the products to the power of their coefficients divided by the concentration of the reactants to the power of their coefficients. To use the example given earlier:

K for the decomposition of N2O4 (g) to yield 2NO2 (g) is given as follows: $$[NO2]^2/[N2O4]$$. Because the coefficient of NO2 is 2, the concentration of this gas is squared. As an example, suppose the concentration of NO2 at equilibrium (not initially) is 0.0172 M and the concentration of N2O4 at equilibrium is 0.00140 M. To calculate the equilibrium constant, square the concentration of NO2 and divide it by the concentration of N2O4:

$$(0.0172)^2/(0.001400) = 0.211$$

Notice the lack of units because the M in the numerator and the M in the denominator cancel.

For a slightly more complex example, consider the Haber process of synthesizing ammonia at high temperatures and pressures from nitrogen and hydrogen gas. The equation for this reaction is N2 (g) + 3H2 (g) --> 2NH3 (g). To calculate this equilibrium constant, take the concentration of N2 times the concentration of hydrogen cubed divided by the concentration of ammonia squared, like so:

$$[N2][H2]^3/[NH3]^2$$.

Heterogeneous Equilibia
All of the reactants and products in the above examples were gases. This is because when calculating an equilibrium constant, solids and liquids are not taken into account (aqueous reactants or products, however, are). The concentrations of solids and liquids remain constant, unlike the concentrations of gases (if the mass of a solid is doubled, its volume is also doubled). For example, in the equilibrium constant for the dissociation of PbCl2 in water (PbCl2 (s) --> Pb2+ (aq) + 2Cl- (aq)), the solid lead chloride will not be factored into the calculation. The expression then becomes $$[Pb2+][Cl-]^2$$.

LeChatelier's Principle
LeChatelier's principle says that changes in the system that an equilibrium reaction occurs in will cause a change in the overall equilibrium. That is, if a stress is applied to a system, the system will react in such a way as to minimize or counter the stress. Examples of these changes or stresses can include volume, temperature, pressure, or concentration.

Volume
If the volume of a system is reduced, the system will react by shifting in the direction that reduces the number of moles of gas. To return to the example of N2O4 and NO2, if the volume of this system is reduced, the reaction will be driven toward N2O4. Recall the equation: $$N2O4 (g) --> 2NO2 (g)$$. Because there are fewer moles of N2O4 than there are moles of NO2, if the volume decreases, the reaction will shift toward N2O4, converting more products into reactants than reactants into products. The opposite is also true: if the volume is increased, the reaction will shift toward NO2 because there will be more moles of gas to occupy the larger volume.

Pressure
The effect of a change in pressure is the same as the effect of a change in volume. According to the ideal gas law, if the pressure of a system is increased, its volume will decrease and vice versa.

Concentration
Simply put, if the concentration of the reactants is increased, the system will favor products to try to attain equilibrium again. In contrast, if the concentration of the products is increased, the system will shift toward the reactants.

Temperature
This is more complicated than the above examples. To predict the effect of a temperature change, one must first know if the reaction in question is endothermic or exothermic. The standard model of an endothermic reaction is reactants + heat --> products. For an exothermic reaction, reactants --> products + heat. To tell if a reaction is endo- or exothermic, look at the sign of the enthalpy (the change in heat of a reaction - how much heat is given off or used): if it is positive the reaction is endothermic. If it is negative, the reaction is exothermic.

If a reaction is endothermic, an increase in temperature will cause the system to shift toward the products because heat is being added - the right side of the equation has a net increase. If a reaction is exothermic, however, the reaction will shift the other way, to the left, because there is a net increase in the terms of the right side of the equation.

Catalysts
Although catalysts can change the rate at which equilibrium is attained, it will not affect K, because K is calculated from concentration, not rate.

Acid-Base Equilibria
There are two main definitions of acids and bases: the Arrhenius definition and the Brønsted-Lowry definition. According to Arrhenius, an acid is any substance that increases the concentration of the hydronium (H3O+) in water and a base any substance that increases the concentration of the hydroxide (OH-) ion. According to the Brønsted-Lowry definition, acids are proton donors and bases proton acceptors; that is, when an acid dissociates in water, it gives off an H+ ion, which is simply a proton. Bases give off OH- ions, which combine with the free protons to form water.

In the process of titration, the concentrations of acids and bases can be found. This is a typical lab in this event. As an example, consider the titration of sodium hydroxide with sulfuric acid. Given the concentration of either the acid or the base, the concentration of the other can be determined. Assuming the concentration of the sulfuric acid is 0.2690 M and 25.0 mL of acid and 13.83 mL of base are used in the titration, concentration of the sodium hydroxide can be determined.

To solve such a titration problem, first write a balanced equation:

$$H2SO4 + 2NaOH --> Na2SO4 + H2O$$.

Next, calculate how many moles of sulfuric acid were used:

$$0.2690 M * 0.0250 L = 0.00673 moles of H2SO4$$.

Third, find the mole ratios of the acid and the base. In this example, 1 mole of sulfuric acid is stoichiometrically equal to 2 moles of sodium hydroxide (they react in a 1:2 ratio). This can be found from the balanced equation.

The number of moles of NaOH used in this titration is equal to $$0.0673 moles * 2 = 0.0135 moles$$.

Finally, to find the concentration of NaOH, divide the moles titrated by the volume titrated. In this problem, the volume is given; however, in the event, you will need to find the volume used in the titration by subtracting the final burette reading from the initial burette reading.

$$0.0135 moles/0.01383 L = 0.976 M$$.

From here, the equilibrium constant can be calculated using the technique demonstrated earlier.

Titration Tips
Given that titration labs may be used in competition, it is helpful to have a few tips for correctly titrating.

1. Do not overshoot the endpoint of the reaction. If you are using phenolphthalein, the typical indicator for a titration, you must make sure the color is not too pink at the end of your titration. If it is, you know you added too much acid or base. You want a light pink color that does not fade after a few seconds of swirling. One drop can make all the difference!

2. Make sure to get rid of all air bubbles. This can greatly affect your results.

3. If both partners are working on the lab at once, have one swirl the flask into which you are titrating while the other works with the burette. Swirling ensures equal mixing.

Autoionization of Water
Water can act as a Brønsted-Lowry acid or base; some water molecules will donate protons to other water molecules, resulting in hydronium (H3O+) and hydroxide ions. Because this is an equilibrium process, we can find the equilibrium constant for this reaction, which turns out to be 1.0 * 10^-14.

pH
From the equilibrium constant for water, the pH of a solution can be calculated. pH (potential hydrogen) is equal to $$-log[H3O+]$$, the negative logarithm of the concentration of H3O+ (or the concentration of the H+ ion - this can be written either way, with different textbooks using either H3O+ or H+. This is because free protons usually do not remain free in solution: they tend to be attracted to water molecules, forming H3O+). For example, if the concentration of H3O+ is 1.0 * 10^-4, calculate the pH as follows:

$$-log(1.0 * 10^-4) = 4$$.

This shows the solution is acidic: pH values less than 7 indicate acidity, whereas values above 7 indicate alkalinity (a value of 7 means a solution is neutral: the concentrations of H3O+ and OH- are equal).

If the pOH (potential hydroxide) of a solution is desired, subtract the pH from 14 or take the negative log of the concentration of OH-. In the above example, the pOH of the solution is 10 because 14 - 4 = 10.

Strong Acids
Strong acids dissociate completely in water. The seven most common strong acids are hydrochloric acid (HCl), hydrobromic acid (HBr), hydriodic acid (HI), nitric acid (HNO3), chloric acid (HClO3), perchloric acid (HClO4), and sulfuric acid (H2SO4).

Strong Bases
Like strong acids, strong bases dissociate completely. The most common strong bases are the hydroxides of the alkali metals (for example, sodium hydroxide, NaOH) and the hydroxides of the alkaline earth metals (such as calcium hydroxide, CaOH2).

Common Ion Effect
The basic idea of the common ion effect is that if a solute is added that contains an ion already in solution, it will affect the equilibrium of the solution. For example, if hydrochloric acid (HCl) is added to a solution of NaCl, this will drive the reaction (dissociation of solid NaCl) toward the reactant, causing NaCl to precipitate out of solution. Because a stress has been applied to the system (the concentration of the Cl- ion has been increased), Le Chatelier's principle dictates that the reaction will shift in such a way as to counter the stress.

Solubility Product
The solubility product is an equilibrium constant that expresses how much a compound dissolves in solution. Because these are examples of heterogeneous equilibria - a solid is involved in the reaction - the solubility product is equal to the concentration of the first ion to the power of its coefficient times the concentration of the second ion to the power of its coefficient. For example, for the dissociation of barium sulfate (BaSO4), the solubility product is equal to $$[Ba2+][SO42-]$$.

Solubility and the solubility product describe similar phenomena; however, they are not the same. Solubility refers to the quantity of a compound that dissolves to create a solution; solubility product, the equilibrium constant for the equilibrium between an ionic compound that dissolves in solution and the saturated solution it forms. The magnitude of the solubility product is a measure of the extent to which a compound dissolves to form a solution.

Some factors that can affect solubility include pH and the common ion effect. Basic ions will dissolve more readily in acidic solutions than acidic ions will. When acidic ions dissolve, they increase the concentration of the H3O+ ion, which will affect solubility according to the common ion effect.