# 15.5: Calculating Equilibrium Constants

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- 25171

There are two fundamental kinds of equilibrium problems:

- those in which we are given the concentrations of the reactants and the products at equilibrium (or, more often, information that allows us to calculate these concentrations), and we are asked to calculate the equilibrium constant for the reaction; and
- those in which we are given the equilibrium constant and the initial concentrations of reactants, and we are asked to calculate the concentration of one or more substances at equilibrium. In this section, we describe methods for solving both kinds of problems.

## Calculating an Equilibrium Constant from Equilibrium Concentrations

We saw in the exercise in Example 6 in Section 15.2 that the equilibrium constant for the decomposition of \(CaCO_{3(s)}\) to \(CaO_{(s)}\) and \(CO_{2(g)}\) is \(K = [CO_2]\). At 800°C, the concentration of \(CO_2\) in equilibrium with solid \(CaCO_3\) and \(CaO\) is \(2.5 \times 10^{-3}\; M\). Thus K at 800°C is \(2.5 \times 10^{-3}\). (Remember that equilibrium constants are unitless.)

A more complex example of this type of problem is the conversion of n-butane, an additive used to increase the volatility of gasoline, into isobutane (2-methylpropane).

This reaction can be written as follows:

and the equilibrium constant \(K = [\text{isobutane}]/[\text{n-butane}]\). At equilibrium, a mixture of n-butane and isobutane at room temperature was found to contain 0.041 M isobutane and 0.016 M n-butane. Substituting these concentrations into the equilibrium constant expression,

Thus the equilibrium constant for the reaction as written is 2.6.

Chemists are not often given the concentrations of all the substances, and they are not likely to measure the equilibrium concentrations of all the relevant substances for a particular system. In such cases, we can obtain the equilibrium concentrations from the initial concentrations of the reactants and the balanced chemical equation for the reaction, as long as the equilibrium concentration of one of the substances is known. Example \(\PageIndex{2}\) shows one way to do this.

## Calculating Equilibrium Concentrations from the Equilibrium Constant

To describe how to calculate equilibrium concentrations from an equilibrium constant, we first consider a system that contains only a single product and a single reactant, the conversion of n-butane to isobutane (Equation \(\ref{Eq1}\)), for which K = 2.6 at 25°C. If we begin with a 1.00 M sample of n-butane, we can determine the concentration of n-butane and isobutane at equilibrium by constructing a table showing what is known and what needs to be calculated, just as we did in Example \(\PageIndex{2}\).

\[\text{n-butane}_{(g)} \rightleftharpoons \text{isobutane}_{(g)}\]

ICE |
\([\text{n-butane}_{(g)} ]\) |
\([\text{isobutane}_{(g)}]\) |
---|---|---|

Initial |
||

Change |
||

Final |

The initial concentrations of the reactant and product are both known: [n-butane]i = 1.00 M and [isobutane]i = 0 M. We need to calculate the equilibrium concentrations of both n-butane and isobutane. Because it is generally difficult to calculate final concentrations directly, we focus on the change in the concentrations of the substances between the initial and the final (equilibrium) conditions. If, for example, we define the change in the concentration of isobutane (Δ[isobutane]) as \(+x\), then the change in the concentration of n-butane is Δ[n-butane] = \(−x\). This is because the balanced chemical equation for the reaction tells us that 1 mol of n-butane is consumed for every 1 mol of isobutane produced. We can then express the final concentrations in terms of the initial concentrations and the changes they have undergone.

\[\text{n-butane}_{(g)} \rightleftharpoons \text{isobutane}_{(g)}\]

ICE |
\([\text{n-butane}_{(g)} ]\) |
\([\text{isobutane}_{(g)}]\) |
---|---|---|

Initial |
1.00 | 0 |

Change |
\(−x\) | \(+x\) |

Final |
\((1.00 − x)\) | \((0 + x) = x\) |

Substituting the expressions for the final concentrations of n-butane and isobutane from the table into the equilibrium equation,

Rearranging and solving for \(x\),

\[x=2.6(1.00−x)= 2.6−2.6x\]

\[x+2.6x= 2.6\]

\[x=0.72\]

We obtain the final concentrations by substituting this \(x\) value into the expressions for the final concentrations of n-butane and isobutane listed in the table:

\[[\text{n-butane}]_f = (1.00 − x) M = (1.00 − 0.72) M = 0.28\; M\]

\[[\text{isobutane}]_f = (0.00 + x) M = (0.00 + 0.72) M = 0.72\; M\]

We can check the results by substituting them back into the equilibrium constant expression to see whether they give the same \(K\) that we used in the calculation:

\[K=\dfrac{[\text{isobutane}]}{[\text{n-butane}]}=\left(\dfrac{0.72\; \cancel{M}}{0.28\;\cancel{M}}\right) =2.6\]

This is the same \(K\) we were given, so we can be confident of our results.

Example \(\PageIndex{3}\) illustrates a common type of equilibrium problem that you are likely to encounter.

In Example \(\PageIndex{3}\), the initial concentrations of the reactants were the same, which gave us an equation that was a perfect square and simplified our calculations. Often, however, the initial concentrations of the reactants are not the same, and/or one or more of the products may be present when the reaction starts. Under these conditions, there is usually no way to simplify the problem, and we must determine the equilibrium concentrations with other means. Such a case is described in Example \(\PageIndex{4}\).

In many situations it is not necessary to solve a quadratic (or higher-order) equation. Most of these cases involve reactions for which the equilibrium constant is either very small (\(K ≤ 10^{−3}\)) or very large (\(K ≥ 10^3\)), which means that the change in the concentration (defined as \(x\)) is essentially negligible compared with the initial concentration of a substance. Knowing this simplifies the calculations dramatically, as illustrated in Example \(\PageIndex{5}\).

Another type of problem that can be simplified by assuming that changes in concentration are negligible is one in which the equilibrium constant is very large (\(K \geq 10^3\)). A large equilibrium constant implies that the reactants are converted almost entirely to products, so we can assume that the reaction proceeds 100% to completion. When we solve this type of problem, we view the system as equilibrating from the products side of the reaction rather than the reactants side. This approach is illustrated in Example \(\PageIndex{6}\).

## Summary

Various methods can be used to solve the two fundamental types of equilibrium problems: (1) those in which we calculate the concentrations of reactants and products at equilibrium and (2) those in which we use the equilibrium constant and the initial concentrations of reactants to determine the composition of the equilibrium mixture. When an equilibrium constant is calculated from equilibrium concentrations, molar concentrations or partial pressures are substituted into the equilibrium constant expression for the reaction. Equilibrium constants can be used to calculate the equilibrium concentrations of reactants and products by using the quantities or concentrations of the reactants, the stoichiometry of the balanced chemical equation for the reaction, and a tabular format to obtain the final concentrations of all species at equilibrium.