2.5: Equilibrium Calculations
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We know that at equilibrium, the value of the reaction quotient of any reaction is equal to its equilibrium constant. Thus, we can use the mathematical expression for Q to determine a number of quantities associated with a reaction at equilibrium or approaching equilibrium. While we have learned to identify in which direction a reaction will shift to reach equilibrium, we want to extend that understanding to quantitative calculations. We do so by evaluating the ways that the concentrations of products and reactants change as a reaction approaches equilibrium, keeping in mind the stoichiometric ratios of the reaction. This algebraic approach to equilibrium calculations will be explored in this section.
Changes in concentrations or pressures of reactants and products occur as a reaction system approaches equilibrium. In this section we will see that we can relate these changes to each other using the coefficients in the balanced chemical equation describing the system. We use the decomposition of ammonia as an example.
On heating, ammonia reversibly decomposes into nitrogen and hydrogen according to this equation:
\[\ce{2NH3}(g)⇌\ce{N2}(g)+\ce{3H2}(g)\]
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 change in the H2 concentration, Δ[H2], is also positive—the concentration of H2 increases as ammonia decomposes. The chemical equation tells us that the change in the concentration of H2 is three times the change in the concentration of N2 because for each mole of N2 produced, 3 moles of H2 are produced.
\[ \begin{align*} \ce{Δ[H2]} &=3×\ce{Δ[N2]} \\[4pt] &=3×(0.11\:M) \\[4pt] &=0.33\:M \end{align*}\]
The change in concentration of NH3, Δ[NH3], is twice that of Δ[N2]; the equation indicates that 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.
\[ \begin{align*} Δ[\ce{NH3}] &=−2×Δ[\ce{N2}] \\[4pt] &=−2×(0.11\:M) \\[4pt] &=−0.22\:M \end{align*}\]
We can relate these relationships directly to the coefficients in the equation
&\phantom{Δ[NH3}\ce{2NH3}(g)
&&⇌
&&\phantom{Δ[N2}\ce{N2}(g)
&&+
&&\phantom{Δ[H2]}\ce{3H2}(g)\\
&Δ[\ce{NH3}]=−2×Δ[\ce{N2}]&& && Δ[\ce{N2}]=0.11\:M && && Δ[\ce{H2}]=3×Δ[\ce{N2}]
\end{align}\]
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.
The changes in the other concentrations would then be represented as:
\(Δ[\ce{H2}]=3×Δ[\ce{N2}]=3x\)
The coefficients in the Δ terms are identical to those in the balanced equation for the reaction.
\(\begin{alignat}{3}
&\ce{2NH3}(g)⇌\:&&\ce{N2}(g)+\:&&\ce{3H2}(g)\\
&−2x &&x &&3x
\end{alignat}\)
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.
Calculations Involving Equilibrium Concentrations
Because the value of the reaction quotient of any reaction at equilibrium is equal to its equilibrium constant, we can use the mathematical expression for Qc (i.e., the law of mass action) to determine a number of quantities associated with a reaction at equilibrium. It may help if we keep in mind that Qc = Kc (at equilibrium) in all of these situations and that there are only three basic types of calculations:
- Calculation of an equilibrium constant. If concentrations of reactants and products at equilibrium are known, the value of the equilibrium constant for the reaction can be calculated.
- Calculation of missing equilibrium concentrations. If the value of the equilibrium constant and all of the equilibrium concentrations, except one, are known, the remaining concentration can be calculated.
- Calculation of equilibrium concentrations from initial concentrations. If the value of the equilibrium constant and a set of concentrations of reactants and products that are not at equilibrium are known, the concentrations at equilibrium can be calculated.
A similar list could be generated using QP, KP, and partial pressure. We will look at solving each of these cases in sequence.
Calculation of an Equilibrium Constant
Since the law of mass action is the only equation we have to describe the relationship between Kc and the concentrations of reactants and products, any problem that requires us to solve for Kc must provide enough information to determine the reactant and product concentrations at equilibrium. Armed with the concentrations, we can solve the equation for Kc, as it will be the only unknown.
Example \(\PageIndex{1}\) showed us how to determine the equilibrium constant of a reaction if we know the concentrations of reactants and products at equilibrium. 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 chart 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 we consider no shift toward equilibrium to have happened. 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 described in a previous section of this chapter. The last row contains the concentrations once equilibrium has been reached.
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.
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.
- Determine the direction the reaction proceeds to come to equilibrium.
- Write a balanced chemical equation for the reaction.
- If the direction in which the reaction must proceed to reach equilibrium is not obvious, calculate Qc from the initial concentrations and compare to Kc to determine the direction of change.
- Determine the relative changes needed to reach equilibrium, then write the equilibrium concentrations in terms of these changes.
- Define the changes in the initial concentrations that are needed for the reaction to reach equilibrium. Generally, we represent the smallest change with the symbol x and express the other changes in terms of the smallest change.
- Define missing equilibrium concentrations in terms of the initial concentrations and the changes in concentration determined in (a).
- Solve for the change and the equilibrium concentrations.
- Substitute the equilibrium concentrations into the expression for the equilibrium constant, solve for x, and check any assumptions used to find x.
- Calculate the equilibrium concentrations.
- Check the arithmetic.
- Check the calculated equilibrium concentrations by substituting them into the equilibrium expression and determining whether they give the equilibrium constant.
Sometimes a particular step may differ from problem to problem—it 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.
In solving equilibrium problems that involve changes in concentration, sometimes it is convenient to set up an ICE table, as described in the previous section.
Sometimes it is possible to use chemical insight 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 could be called the “all reactant” starting point. Using x for the amount of acid ionized at equilibrium, this is the ICE table and solution.
Setting up and solving the quadratic equation gives
Using the positive (physical) root, the equilibrium concentrations are
A less obvious way to solve the problem would be to assume all the HA ionizes first, then the system comes to equilibrium. This could 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, this is the ICE table for this starting point.
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 Kc means that very little of the reactants form products and a large Kc 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 of less than 5%. 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.
Summary
The ratios of the rate of change in concentrations of a reaction are equal to the ratios of the coefficients in the balanced chemical equation. The sign of the coefficient of X is positive when the concentration increases and negative when it decreases. We learned to approach three basic types of equilibrium problems. When given the concentrations of the reactants and products at equilibrium, we can solve for the equilibrium constant; when given the equilibrium constant and some of the concentrations involved, we can solve for the missing concentrations; and when given the equilibrium constant and the initial concentrations, we can solve for the concentrations at equilibrium.
Contributors
Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/85abf193-2bd...a7ac8df6@9.110).