Ch13.1 Chemical Equilibrium

A chemical reaction is usually written in a way that suggests it proceeds in one direction, the direction in which we read, but all chemical reactions are reversible, and both the forward and reverse reaction occur to one degree or another depending on conditions. In a chemical equilibrium, the forward and reverse reactions occur at equal rates, and the concentrations of products and reactants remain constant. If we run a reaction in a closed system so that the products cannot escape, we often find the reaction does not yield 100% products. Instead, some reactants remain after the concentrations stop changing. At this point, when there is no further change in concentrations of reactants and products, we say the reaction is at equilibrium.

For example, when we place a sample of dinitrogen tetroxide (N₂O₄, a colorless gas) in a glass tube, it forms nitrogen dioxide (NO₂, a brown gas) by the reaction

The color becomes darker as N₂O₄ is converted to NO₂. When the system reaches equilibrium, both N₂O₄ and NO₂ are present (Figure 1).

The formation of NO₂ from N₂O₄ is a reversible reaction, which is identified by the equilibrium arrow (⇌). In a reversible reaction, the reactants can combine to form products and the products can react to form the reactants. Thus, not only can N₂O₄ decompose to form NO₂, but the NO₂ produced can react to form N₂O₄. As soon as the forward reaction produces any NO₂, the reverse reaction begins and NO₂ starts to react to form N₂O₄. At equilibrium, the concentrations of N₂O₄ and NO₂ no longer change because the rate of NO₂ formation is exactly equal to the rate of NO₂ consumption, and the rate of formation of N₂O₄ is exactly equal to the rate of consumption of N₂O₄. Chemical equilibrium is a dynamic process: the numbers of reactant and product molecules remain constant, but there is a flux back and forth between them (Figure 2).

In a chemical equilibrium, the forward and reverse reactions do not stop, rather they continue to occur at the same rate, leading to constant concentrations of the reactants and the products. Plots showing how the reaction rates and concentrations change with respect to time are shown in Figure 1.

We can detect a state of equilibrium because the concentrations of reactants and products do not appear to change. However, it is important that we verify that the absence of change is due to equilibrium and not to a reaction rate that is so slow that changes in concentration are difficult to detect.

We use an equilibrium arrow when writing an equation for a reversible reaction. Such a reaction may or may not be at equilibrium. When we wish to speak about one particular component of a reversible reaction, we use a single arrow. For example, in the equilibrium shown in Figure 1, the rate of the forward reaction

is equal to the rate of the backward reaction

Equilibrium and Soft Drinks

The connection between chemistry and carbonated soft drinks goes back to 1767, when Joseph Priestley (1733–1804; mostly known today for his role in the discovery and identification of oxygen) discovered a method of infusing water with carbon dioxide to make carbonated water. In 1772, Priestly published a paper entitled “Impregnating Water with Fixed Air.” The paper describes dripping oil of vitriol (today we call this sulfuric acid, but what a great way to describe sulfuric acid: “oil of vitriol” literally means “liquid nastiness”) onto chalk (calcium carbonate). The resulting CO₂ falls into the container of water beneath the vessel in which the initial reaction takes place; agitation helps the gaseous CO₂ mix into the liquid water.

[latex]\text{H}_2\text{SO}_4(l)\;+\;\text{CaCO}_3(s)\;{\longrightarrow}\;\text{CO}_2(g)\;+\;\text{H}_2\text{O}(l)\;+\;\text{CaSO}_4(aq)[/latex]

Carbon dioxide is slightly soluble in water. There is an equilibrium reaction that occurs as the carbon dioxide reacts with the water to form carbonic acid (H₂CO₃). Since carbonic acid is a weak acid, it can dissociate into protons (H⁺) and hydrogen carbonate ions [latex](\text{HCO}_3^{\;\;-})[/latex].

[latex]\text{CO}_2(aq)\;+\;\text{H}_2\text{O}(l)\;{\rightleftharpoons}\;\text{H}_2\text{CO}_3(aq)\;{\rightleftharpoons}\;\text{HCO}_3^{\;\;-}(aq)\;+\;\text{H}^{+}(aq)[/latex]

Today, CO₂ can be pressurized into soft drinks, establishing the equilibrium shown above under a higher pressure condition. Once you open the beverage container, however, a cascade of equilibrium shifts occurs. First, the CO₂ gas in the air space on top of the bottle escapes, causing the equilibrium between gas-phase CO₂ and dissolved or aqueous CO₂ to shift, lowering the concentration of CO₂ in the soft drink. Less CO₂ dissolved in the liquid leads to carbonic acid decomposing to dissolved CO₂ and H₂O. The lowered carbonic acid concentration causes a shift of the final equilibrium. The net result is that the CO₂ bubbles up out of the beverage, releasing the gas into the air (Figure 3), results in a soft drink with a much lowered CO₂ concentration, often referred to as “flat.”

Let us consider the evaporation of bromine as a second example of a system at equilibrium.

An equilibrium can be established for a physical change—like this liquid to gas transition—as well as for a chemical reaction. Figure 4 shows a sample of liquid bromine at equilibrium with bromine vapor in a closed container. When we pour liquid bromine into an empty bottle in which there is no bromine vapor, some liquid evaporates: the amount of liquid decreases and the amount of vapor increases. If we cap the bottle so no vapor escapes, the amount of liquid and vapor will eventually stop changing and an equilibrium between the liquid and the vapor will be established. If the bottle were not capped, the bromine vapor would escape and no equilibrium would be reached.

Ch13.2 Equilibrium Constant

A general equation for a reversible reaction may be written as follows:

We can write the equilibrium constant (K) for this equation. When evaluated using concentrations, it is called K_c. We use brackets to indicate molar concentrations of reactants and products.

The equilibrium constant is equal to the molar concentrations of the products of the chemical equation (multiplied together) over the reactants (also multiplied together), with each concentration raised to the power of the coefficient of that substance in the balanced chemical equation.

The magnitude of an equilibrium constant is a measure of the yield of a reaction when it reaches equilibrium. A large value for K_c indicates that equilibrium is attained after the reactants have been largely converted into products. A small value of K_c—much less than 1—indicates that equilibrium is attained when only a small proportion of the reactants have been converted into products.

Ch13.3 Working with Equilibrium Constants

In Example 5, it was mentioned that the common practice is to omit units when evaluating equilibrium constants. It should be pointed out that using concentrations in these computations is a convenient but simplified approach that sometimes leads to results that seemingly conflict with the law of mass action. For example, equilibria involving aqueous ions often exhibit equilibrium constants that vary quite significantly (are not constant) at high solution concentrations. This may be avoided by computing K_c values using the activities of the reactants and products in the equilibrium system instead of their concentrations. The activity of a substance is a measure of its effective concentration under specified conditions. While a detailed discussion of this important quantity is beyond the scope of an introductory text, it is necessary to be aware of a few important aspects:

• Activities are dimensionless (unitless) quantities and are in essence “adjusted” concentrations.
• For relatively dilute solutions, a substance’s activity and its molar concentration are roughly equal.
• Activities for pure condensed phases (solids and liquids) are equal to 1.

As a consequence of this last consideration, K_c expressions do not contain terms for solids or liquids (being numerically equal to 1, these terms have no effect on the expression’s value).

A homogeneous equilibrium is one in which all of the reactants and products are present in a single solution (by definition, a homogeneous mixture). Reactions between solutes in liquid solutions belong to one type of homogeneous equilibria. Several examples are provided here.

In each of these examples, the equilibrium system is an aqueous solution, as denoted by the aq annotations on the solute formulas. H₂O(l) is the solvent for these solutions, its concentration does not appear as a term in the K_c expression, as discussed above, even though it may also appear as a reactant or product in the chemical equation.

A heterogeneous equilibrium is a system in which reactants and products are found in two or more phases. The phases may be any combination of solid, liquid, or gas phases, and solutions. Again, solids and pure liquids do not appear in equilibrium constant expressions (the activities of pure solids, pure liquids, and solvents are 1).

Some heterogeneous equilibria involve chemical changes; for example:

Other heterogeneous equilibria involve phase changes, for example, the evaporation of liquid bromine, as shown in the following equation:

The equilibrium constant expression must be manipulated if a reaction changes stoichiometry, is reversed or split into elementary steps.

When all the coefficients in a balanced chemical equation are multiplied by some factor n, then the new equilibrium constant is the original equilibrium constant raised to the n^th power. For example:

When a reaction’s direction is reversed, the equilibrium constant for the new reaction is simply the inverse of that for the original reaction. For example:

When two reactions are added together to obtain a new reaction, the equilibrium constant for the new reaction is the product of the equilibrium constants for the original reactions. For example:

Ch13.4 Equilibrium Constant and Partial Pressure

Reactions in which all reactants and products are gases represent another class of homogeneous equilibria. We use molar concentrations in the following examples, but partial pressures of the gases may be used as well.

Note that the concentration of H₂O(g) has been included in the last example because water is not the solvent in this gas-phase reaction and its concentration (and activity) changes.

Whenever gases are involved in a reaction, the partial pressure of each gas can be used instead of its concentration in the equation for the equilibrium constant because the partial pressure of a gas is directly proportional to its concentration at constant temperature. This relationship can be derived from the ideal gas equation, where M is the molar concentration of gas, [latex]\frac{n}{V}[/latex].

Thus, at constant temperature, the pressure of a gas is directly proportional to its concentration.

Using the partial pressures of the gases, we can write the equilibrium constant for the system [latex]\text{C}_2\text{H}_6(g)\;{\rightleftharpoons}\;\text{C}_2\text{H}_4(g)\;+\;\text{H}_2(g)[/latex] by following the same guidelines for deriving concentration-based expressions:

where K_P designates an equilibrium constant derived using partial pressures instead of concentrations, [latex]P_{\text{C}_2\text{H}_6}[/latex] is the partial pressure of C₂H₆; [latex]P_{\text{H}_2}[/latex], the partial pressure of H₂; and [latex]P_{\text{C}_2\text{H}_6}[/latex], the partial pressure of C₂H₄.

The equation relating K_c and K_P is derived as follows. For the gas-phase reaction [latex]m\text{A}\;+\;n\text{B}\;{\rightleftharpoons}\;x\text{C}\;+\;y\text{D}[/latex]:

In this equation:

Δn is the difference between the sum of the coefficients of the gaseous products and the sum of the coefficients of the gaseous reactants in the reaction (the change in moles of gas between the reactants and the products). For the gas-phase reaction [latex]m\text{A}\;+\;n\text{B}\;{\leftrightharpoons}\;x\text{C}\;+\;y\text{D}[/latex], we have

Note that the gas constant, R, can be expressed in a variety of units with varying values corresponding to the unit conversions. Use the R value and associated units that match the partial pressure units used in the K_p expression. Oftentimes, the R used is 0.082057 [latex]\frac{\text{L}{\cdot}\text{atm}}{\text{mol}{\cdot}\text{K}}[/latex].

For heterogeneous equilibria that involve gases, equilibrium constants can also be expressed using partial pressures instead of concentrations. Two examples are:

Ch13.5 Calculating Equilibrium Constants

Changes in concentrations or pressures of reactants and products occur as a reaction system approaches equilibrium. We can relate these changes to each other using the coefficients in the balanced chemical equation describing the system. We will use the decomposition of ammonia as an example. On heating, ammonia reversibly decomposes into nitrogen and hydrogen:

If a sample of ammonia decomposes in a closed system and the concentration of N₂ increases by 0.11 M, the change in the N₂ concentration, Δ[N₂], the final concentration minus the initial concentration, is 0.11 M. The change is positive because the concentration of N₂ increases.

The concentration of H₂ also increases as ammonia decomposes. The balanced chemical equation tells us that Δ[H₂] is three times Δ[N₂] because for each mole of N₂ produced, 3 moles of H₂ are produced. Hence, Δ[H₂] = 0.33 M.

The change in concentration of NH₃, Δ[NH₃], is twice that of Δ[N₂]—2 moles of NH₃ must decompose for each mole of N₂ formed. However, the change in the NH₃ concentration is negative because the concentration of ammonia decreases as it decomposes.

We can relate these relationships directly to the coefficients in the equation

Note that all the changes on one side of the arrows are of the same sign and that all the changes on the other side of the arrows are of the opposite sign.

If we did not know the magnitude of the change in the concentration of N₂, we could represent it by the symbol x, Δ[N₂] = x M. The changes in the other concentrations would then be represented as:

The simplest way for us to find the coefficients for the concentration changes in any reaction is to use the coefficients in the balanced chemical equation. The sign of the coefficient is positive when the concentration increases; it is negative when the concentration decreases.

Ch13.6 Calculation of an Equilibrium Constant

If concentrations of reactants and products at equilibrium are known, we can solve the equation for K_c, as it will be the only unknown.

The following example shows how to use the stoichiometry of the reaction and a combination of initial concentrations and equilibrium concentrations to determine an equilibrium constant. This technique, commonly called an ICE table—for Initial, Change, and Equilibrium–will be helpful in solving many equilibrium problems. A table is generated beginning with the equilibrium reaction in question. Underneath the reaction the initial concentrations of the reactants and products are listed—these conditions are usually provided in the problem and no shift toward equilibrium have happened yet. The next row of data is the change that occurs as the system shifts toward equilibrium—do not forget to consider the reaction stoichiometry as we’ve just described. The last row contains the concentrations once equilibrium has been reached.

Ch13.7 Calculation of a Missing Equilibrium Concentration

If we know the equilibrium constant for a reaction and know the concentrations at equilibrium of all reactants and products except one, we can calculate the missing concentration.

Ch13.8 Calculation of Changes in Concentration

If we know the equilibrium constant for a reaction and a set of concentrations of reactants and products that are not at equilibrium, we can calculate the changes in concentrations as the system comes to equilibrium, as well as the new concentrations at equilibrium. The typical procedure can be summarized in four steps.

1. Balance the chemical reaction equation and determine the direction the reaction proceeds to come to equilibrium.
2. Determine the relative changes needed to reach equilibrium—we typically represent the smallest change with the symbol x and express the other changes in terms of x, then write the equilibrium concentrations in terms of these changes.
3. Substitute the equilibrium concentrations into the expression for the equilibrium constant, solve for x, and check any assumptions used to find x. Then calculate the equilibrium concentrations.
4. Check the arithmetic by substituting the calculated equilibrium concentrations into the equilibrium expression and determining whether they give the equilibrium constant

Sometimes a particular step may be more complex in some problems and less complex in others. However, every calculation of equilibrium concentrations from a set of initial concentrations will involve these steps. An ICE table is useful in these calculations.

Sometimes it is possible to find solutions to equilibrium problems without actually solving a quadratic (or more complicated) equation. First, however, it is useful to verify that equilibrium can be obtained starting from two extremes: all (or mostly) reactants and all (or mostly) products.

Consider the ionization of 0.150 M HA, a weak acid.

The most obvious way to determine the equilibrium concentrations would be to start with only reactants. This can be called the “all reactant” starting point. Using x for the amount of acid ionized at equilibrium, the ICE table is

Setting up and solving the quadratic equation gives

Using the positive (physically reasonable) root, the equilibrium concentrations are

A less obvious way to solve the problem would be to assume all the HA ionizes first, and then the system comes to equilibrium. This can be called the “all product” starting point. Assuming all of the HA ionizes gives

Using these as initial concentrations and “y” to represent the concentration of HA at equilibrium, the ICE table is

Setting up and solving the quadratic equation gives

Retain a few extra significant figures to minimize rounding problems.

Rounding each solution to three significant figures gives

Using the physically significant root (0.140 M) gives the equilibrium concentrations as

Thus, the two approaches give the same results (to three decimal places), and show that both starting points lead to the same equilibrium conditions. The “all reactant” starting point resulted in a relatively small change (x) because the system was close to equilibrium, while the “all product” starting point had a relatively large change (y) that was nearly the size of the initial concentrations. It can be said that a system that starts “close” to equilibrium will require only a ”small” change in conditions (x) to reach equilibrium.

Recall that a small K_c means that very little of the reactants form products and a large K_c means that most of the reactants form products. If the system can be arranged so it starts “close” to equilibrium, then if the change (x) is small compared to any initial concentrations, it can be neglected. Small is usually defined as resulting in an error that does not change the answer given the significant figures involved. The following two examples demonstrate this.

The second example requires that the original information be processed a bit, but it still can be solved using a small x approximation.

Ch13.9 Reaction Quotient

We can write the reaction quotient (Q) for a generic reaction:

as (Q_c is used to indicate usage of concentrations):

The numeric value of Q_c for a given reaction depends on the concentrations of products and reactants present at the time when Q_c is determined. When pure reactants are mixed, Q_c = 0 because there are no products present at that point. As the reaction proceeds, the value of Q_c increases as the concentrations of the products increase and the concentrations of the reactants decrease (Figure 5). When the reaction reaches equilibrium, the value of the reaction quotient no longer changes because the concentrations no longer change, and Q_c = K_c.

That a reaction quotient always assumes the same value at equilibrium can be expressed as:

This equation is a mathematical statement of the law of mass action: When a reaction has attained equilibrium at a given temperature, the reaction quotient for the reaction always has the same value.

Once a value of K_c is known for a reaction, it can be used to predict directional shifts when compared to the value of Q_c. A system that is not at equilibrium will proceed in the direction that establishes equilibrium. The data in Figure 6 illustrate this. When heated to a consistent temperature, 800 °C, different starting mixtures of CO, H₂O, CO₂, and H₂ react to reach compositions adhering to the same equilibrium (the value of Q_c changes until it equals the value of K_c).

It is important to recognize that an equilibrium can be established starting either from reactants or from products, or from a mixture of both. For example, equilibrium was established from Mixture 2 in Figure 6 when the products of the reaction were heated in a closed container. In fact, one technique to determine whether a reaction is truly at equilibrium is to start with only reactants in one experiment and start with only products in another. If the same value of the reaction quotient is observed when the concentrations stop changing in both experiments, then we may be certain that the system has reached equilibrium.

If we have a mixture of reactants and products that have not yet reached equilibrium, the changes necessary to reach equilibrium may not be obvious. In such a case, we can compare the values of Q and K for the system to predict the changes: when Q_c < K_c, the reaction will proceed to the right (product side); when Q_c > K_c, the reaction will proceed to the left (reactant side).