5.4: Equilibrium Calculations

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.

1. 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 Q_c from the initial concentrations and compare to K_c to determine the direction of change.
2. 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).
3. 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.
4. 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.

Because we have no products initially, Q_c = 0 and the reaction will proceed to the right.

• Let us represent the increase in concentration of PCl₃ by the symbol x. The other changes may be written in terms of x by considering the coefficients in the chemical equation.
  [latex]\begin{alignat}{3}
  &\ce{PCl5}(g)⇌\:&&\ce{PCl3}(g)+\:&&\ce{Cl2}(g)\\
  &-x &&x &&x
  \end{alignat}[/latex]

• The changes in concentration and the expressions for the equilibrium concentrations are:

  

• Substituting the equilibrium concentrations into the equilibrium constant equation gives
  [latex]K_c=\ce{\dfrac{[PCl3][Cl2]}{[PCl5]}}=0.0211[/latex]
  [latex]=\dfrac{(x)(x)}{(1.00−x)}[/latex]

  This equation contains only one variable, x, the change in concentration. We can write the equation as a quadratic equation and solve for x using the quadratic formula.

  [latex]0.0211=\dfrac{(x)(x)}{(1.00−x)}[/latex]
  [latex]0.0211(1.00−x)=x^2[/latex]
  [latex]x^2+0.0211x−0.0211=0[/latex]

  An equation of the form ax² + bx + c = 0 can be rearranged to solve for x:

  [latex]x=\dfrac{−b±\sqrt{b^2−4ac}}{2a}[/latex]

  In this case, a = 1, b = 0.0211, and c = −0.0211. Substituting the appropriate values for a, b, and c yields:

  [latex]x=\dfrac{−0.0211±\sqrt{(0.0211)^2−4(1)(−0.0211)}}{2(1)}[/latex]
  [latex]=\dfrac{−0.0211±\sqrt{(4.45×10^{−4})+(8.44×10^{−2})}}{2}[/latex]
  [latex]=\dfrac{−0.0211±0.291}{2}[/latex]

  Hence

  [latex]x=\dfrac{−0.0211+0.291}{2}=0.135[/latex]

  or

  [latex]x=\dfrac{−0.0211−0.291}{2}=−0.156[/latex]

  Quadratic equations often have two different solutions, one that is physically possible and one that is physically impossible (an extraneous root). In this case, the second solution (−0.156) is physically impossible because we know the change must be a positive number (otherwise we would end up with negative values for concentrations of the products). Thus, x = 0.135 M.

  The equilibrium concentrations are

  [latex]\ce{[PCl5]}=1.00−0.135=0.87\:M[/latex]
  [latex]\ce{[PCl3]}=x=0.135\:M[/latex]
  [latex]\ce{[Cl2]}=x=0.135\:M[/latex]

• Substitution into the expression for K_c (to check the calculation) gives
  [latex]K_c=\ce{\dfrac{[PCl3][Cl2]}{[PCl5]}}=\dfrac{(0.135)(0.135)}{0.87}=0.021[/latex]

  The equilibrium constant calculated from the equilibrium concentrations is equal to the value of K_c given in the problem (when rounded to the proper number of significant figures). Thus, the calculated equilibrium concentrations check.

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 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 of less than 5%. The following two examples demonstrate this.

Example 5.4.5: Approximate Solution Starting Close to Equilibrium

What are the concentrations at equilibrium of a 0.15 M solution of HCN?

[latex]\ce{HCN}(aq)⇌\ce{H+}(aq)+\ce{CN-}(aq) \hspace{20px} K_c=4.9×10^{−10} \nonumber[/latex]

Solution

Using “x” to represent the concentration of each product at equilibrium gives this ICE table.

The exact solution may be obtained using the quadratic formula with

[latex]K_c=\dfrac{(x)(x)}{0.15−x} \nonumber[/latex]

solving

[latex]x^2+4.9×10^{−10}−7.35×10^{−11}=0 \nonumber[/latex]

[latex]x=8.56×10^{−6}\:M\textrm{ (3 sig. figs.)}=8.6×10^{−6}\:M\textrm{ (2 sig. figs.)} \nonumber[/latex]

Thus [H⁺] = [CN^–] = x = 8.6 × 10^–6 M and [HCN] = 0.15 – x = 0.15 M.

In this case, chemical intuition can provide a simpler solution. From the equilibrium constant and the initial conditions, x must be small compared to 0.15 M. More formally, if [latex]x≪0.15[/latex], then 0.15 – x ≈ 0.15. If this assumption is true, then it simplifies obtaining x

[latex]K_c=\dfrac{(x)(x)}{0.15−x}≈\dfrac{x^2}{0.15} \nonumber[/latex]

[latex]4.9×10^{−10}=\dfrac{x^2}{0.15} \nonumber[/latex]

[latex]x^2=(0.15)(4.9×10^{−10})=7.4×10^{−11} \nonumber[/latex]

[latex]x=\sqrt{7.4×10^{−11}}=8.6×10^{−6}\:M \nonumber[/latex]

In this example, solving the exact (quadratic) equation and using approximations gave the same result to two significant figures. While most of the time the approximation is a bit different from the exact solution, as long as the error is less than 5%, the approximate solution is considered valid. In this problem, the 5% applies to IF (0.15 – x) ≈ 0.15 M, so if

[latex]\dfrac{x}{0.15}×100\%=\dfrac{8.6×10^{−6}}{0.15}×100\%=0.006\% \nonumber[/latex]

is less than 5%, as it is in this case, the assumption is valid. The approximate solution is thus a valid solution.

Assume that x is much less than 0.25 M and calculate the error in your assumption.

Answer

[latex]\ce{[OH- ]}=\ce{[NH4+]}=0.0021\:M[/latex]; [NH₃] = 0.25 M, error = 0.84%

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

Example 5.4.6: Approximate Solution After Shifting Starting Concentration

Copper(II) ions form a complex ion in the presence of ammonia

[latex]\ce{Cu^2+}(aq)+\ce{4NH3}(aq)⇌\ce{Cu(NH3)4^2+}(aq) \hspace{20px} K_c=5.0×10^{13}=\ce{\dfrac{[Cu(NH3)4^2+]}{[Cu^2+(aq)][NH3]^4}} \nonumber[/latex]

If 0.010 mol Cu²⁺ is added to 1.00 L of a solution that is 1.00 M NH₃ what are the concentrations when the system comes to equilibrium?

Solution The initial concentration of copper(II) is 0.010 M. The equilibrium constant is very large so it would be better to start with as much product as possible because “all products” is much closer to equilibrium than “all reactants.” Note that Cu²⁺ is the limiting reactant; if all 0.010 M of it reacts to form product the concentrations would be

[latex]\ce{[Cu^2+]}=0.010−0.010=0\:M \nonumber[/latex]

[latex]\ce{[Cu(NH3)4^2+]}=0.010\:M \nonumber[/latex]

[latex]\ce{[NH3]}=1.00−4×0.010=0.96\:M \nonumber[/latex]

Using these “shifted” values as initial concentrations with x as the free copper(II) ion concentration at equilibrium gives this ICE table.

Since we are starting close to equilibrium, x should be small so that

[latex]0.96+4x≈0.96\:M[/latex]

Select the smallest concentration for the 5% rule.

This is much less than 5%, so the assumptions are valid. The concentrations at equilibrium are

By starting with the maximum amount of product, this system was near equilibrium and the change (x) was very small. With only a small change required to get to equilibrium, the equation for x was greatly simplified and gave a valid result well within the 5% error maximum.