Example 2

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.

Check Your Learning
Ethanol and acetic acid react and form water and ethyl acetate, the solvent responsible for the odor of some nail polish removers.

When 1 mol each of C₂H₅OH and CH₃CO₂H are allowed to react in 1 L of the solvent dioxane, equilibrium is established when 0.33 mol of each of the reactants remains. Calculate the equilibrium constant for the reaction. (Note: Water is not a solvent in this reaction. This time, it is a reactant dissolved in an organic solvent and therefore should be included in the equilibrium expression.)

Example 3

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, N₂(g) + O₂(g) ⇌ 2 NO(g), 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.

Check Your Learning
The equilibrium constant for the reaction of nitrogen and hydrogen to produce ammonia at a certain temperature is 6.00 × 10⁻². Calculate the equilibrium concentration of ammonia if the equilibrium concentrations of nitrogen and hydrogen are 4.26 M and 2.09 M, respectively.

Example 4

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

   PCl₅(g)   ⇌   PCl₃(g)  +  Cl₂(g)

   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.
   PCl₅(g)  ⇌   PCl₃(g)  +  Cl₂(g)
    -x                  +x             +x

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

   	PCl5(g)   ⇌   PCl3(g)   +   Cl2(g)
   I: Initial concentration (M)	1.00	0	0
   C: Change (M)	– x	+ x	+ x
   E: Equilibrium concentration (M)	1.00 – x	x	x

    

3. Solve for the change and the equilibrium concentrations.

   Substituting the equilibrium concentrations into the equilibrium constant equation gives

   K_c = [latex]\frac{[PCl_{3}][Cl_{2}]}{[PCl_{5}]}[/latex] = 0.0211
   = [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.

   0.0211 = [latex]\frac{(x)(x)}{(1.00 - x)}[/latex]
   0.0211(1.00 – x) = x²
   x² + 0.0211x – 0.0211 = 0

   Appendix B shows us an equation of the form ax² + bx + c = 0 can be rearranged to solve for x:

   x = [latex]\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:

   x = [latex]\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]
   x = [latex]\frac{-0.0211 + 0.201}{2}[/latex] = 0.135 or
   x = [latex]\frac{-0.0211 - 0.201}{2}[/latex] = -0.156

   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

   [PCl₅] = 1.00 – 0.135 = 0.87 M
   [PCl₃] = x = 0.135 M
   [Cl₂] = x = 0.135 M
4. Check the arithmetic.

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

   K_c = [latex]\frac{[PCl_{3}][Cl_{2}]}{[PCl_{5}]} = \frac{(0.135)(0.135)}{0.87}[/latex] = 0.021

   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.

Check Your Learning
Acetic acid, CH₃CO₂H, reacts with ethanol, C₂H₅OH, to form water and ethyl acetate, CH₃CO₂C₂H₅.

The equilibrium constant for this reaction with dioxane as a solvent is 4.0 (Note: Water is not a solvent in this reaction. This time it is a reactant dissolved in an organic solvent and therefore should be included in the equilibrium expression). What are the equilibrium concentrations when a mixture that is 0.15 M in CH₃CO₂H, 0.15 M in C₂H₅OH, 0.40 M in CH₃CO₂C₂H₅, and 0.40 M in H₂O are mixed in enough dioxane to make 1.0 L of solution?

Check Your Learning
A 1.00-L flask is filled with 1.00 moles of H₂ and 2.00 moles of I₂. The value of the equilibrium constant for the reaction of hydrogen and iodine reacting to form hydrogen iodide is 50.5 under the given conditions. What are the equilibrium concentrations of H₂, I₂, and HI in moles/L?

Example 5

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 using the quadratic formula with

solving

Thus [H₃O⁺] = [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  x << 0.15, 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.

Check Your Learning
What are the equilibrium concentrations in a 0.25 M NH₃ solution?

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

Example 6

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.

Check Your Learning
What are the equilibrium concentrations when 0.25 mol Ni²⁺ is added to 1.00 L of 2.00 M NH₃ solution?

With such a large equilibrium constant, first form as much product as possible, then assume that only a small amount (x) of the product shifts left. Calculate the error in your assumption.