17.5 Equilibrium Calculations

Example 17.5b

Calculation of an Equilibrium Constant

Iodine molecules react reversibly with iodide ions to produce triiodide ions.

If a solution with the concentrations of I₂ and I⁻ both equal to 1.000 × 10⁻³M before reaction gives an equilibrium concentration of I₂ of 6.61 × 10⁻⁴M, what is the equilibrium constant for the reaction?

Solution

We will begin this problem by calculating the changes in concentration as the system goes to equilibrium. Then we determine the equilibrium concentrations and, finally, the equilibrium constant. First, we set up a table with the initial concentrations, the changes in concentrations, and the equilibrium concentrations using −x as the change in concentration of I₂.

Since the equilibrium concentration of I₂ is given, we can solve for x. At equilibrium the concentration of I₂ is 6.61 × 10⁻⁴M so that

Now we can fill in the table with the concentrations at equilibrium.

We now calculate the value of the equilibrium constant.

Example 17.5c

Calculation of a Missing Equilibrium Concentration

Nitrogen oxides are air pollutants produced by the reaction of nitrogen and oxygen at high temperatures. At 2000 °C, the value of the equilibrium constant for the reaction, [latex]\text{N}_2(g)\;+\;\text{O}_2(g)\;{\rightleftharpoons}\;2\text{NO}(g)[/latex], is 4.1 × 10⁻⁴. Find the concentration of NO(g) in an equilibrium mixture with air at 1 atm pressure at this temperature. In air, [N₂] = 0.036 mol/L and [O₂] 0.0089 mol/L.

Solution

We are given all of the equilibrium concentrations except that of NO. Thus, we can solve for the missing equilibrium concentration by rearranging the equation for the equilibrium constant.

Thus [NO] is 3.6 × 10⁻⁴ mol/L at equilibrium under these conditions.

We can check our answer by substituting all equilibrium concentrations into the expression for the reaction quotient to see whether it is equal to the equilibrium constant.

The answer checks; our calculated value gives the equilibrium constant within the error associated with the significant figures in the problem.

Example 17.5d

Calculation of Concentration Changes as a Reaction Goes to Equilibrium

Under certain conditions, the equilibrium constant for the decomposition of PCl₅(g) into PCl₃(g) and Cl₂(g) is 0.0211. What are the equilibrium concentrations of PCl₅, PCl₃, and Cl₂ if the initial concentration of PCl₅ was 1.00 M?

Solution

Use the stepwise process described earlier.

1. Determine the direction the reaction proceeds.

   The balanced equation for the decomposition of PCl₅ is

   [latex]\text{PCl}_5(g)\;{\rightleftharpoons}\;\text{PCl}_3(g)\;+\;\text{Cl}_2(g)[/latex]

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

2. Determine the relative changes needed to reach equilibrium, then write the equilibrium concentrations in terms of these changes.

   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{array}{lcccc} \text{PCl}_5(g) & {\rightleftharpoons} & \text{PCl}_3(g) & + & \text{Cl}_2(g) \\[0.5em] -x & & x & & x \end{array}[/latex]

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

3. Solve for the change and the equilibrium concentrations.

   Substituting the equilibrium concentrations into the equilibrium constant equation gives

   [latex]K_c = \frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]} = 0.0211[/latex]
   [latex]= \frac{(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 = \frac{(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 = \frac{-b\;{\pm}\;\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 = \frac{-0.0211\;{\pm}\;\sqrt{(0.0211)^2\;-\;4(1)(-0.0211)}}{2(1)}[/latex]
   [latex]= \frac{-0.0211\;{\pm}\;\sqrt{(4.45\;\times\;10^{-4})\;+\;(8.44\;\times\;10^{-2})}}{2}[/latex]
   [latex]= \frac{-0.0211\;{\pm}\;0.291}{2}[/latex]

   Hence

   [latex]x = \frac{-0.0211\;+\;0.291}{2} = 0.135[/latex]

   or

   [latex]x = \frac{-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][\text{PCl}_5] = 1.00\;-\;0.135 = 0.87\;M[/latex]
   [latex][\text{PCl}_3] = x = 0.135\;M[/latex]
   [latex][\text{Cl}_2] = x = 0.135\;M[/latex]
4. Check the arithmetic.

   Substitution into the expression for K_c (to check the calculation) gives

   [latex]K_c = \frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]} = \frac{(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.

Example 17.5e

Approximate Solution Starting Close to Equilibrium

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

Solution

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

The exact solution may be obtained by solving for x in

which, once multiplied through and rearranged, becomes

Thus [H⁺] = [CN^–] = x = 8.6 × 10^–6M 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\;{\ll}\;0.15[/latex], then 0.15 – x ≈ 0.15. If this assumption is true, then it simplifies obtaining x

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

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

Example 17.5f

Approximate Solution After Shifting Starting Concentration

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

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

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

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.