13.4 Equilibrium Calculations

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 $\times$ 10⁻³ M before reaction gives an equilibrium concentration of I₂ of 6.61 $\times$ 10⁻⁴ M, what is the equilibrium constant for the reaction?

Solution

To calculate the equilibrium constants, equilibrium concentrations are needed for all the reactants and products:

Provided are the initial concentrations of the reactants and the equilibrium concentration of the product. Use this information to derive terms for the equilibrium concentrations of the reactants, presenting all the information in an ICE table.

At equilibrium the concentration of I₂ is 6.61 $\times$ 10⁻⁴ M so that

The ICE table may now be updated with numerical values for all its concentrations:

Finally, substitute the equilibrium concentrations into the K expression and solve:

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 $\frac{1}{3}$ mol of each of the reactants remains. Calculate the equilibrium constant for the reaction. (Note: Water is a solute in this reaction.)

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 K_c for the reaction, ${\text{N}}_2(g)+{\text{O}}_2(g)\rightleftharpoons 2\text{NO}(g),$ is 4.1 $\times$ 10⁻⁴. Calculate the equilibrium concentration of NO(g) in air at 1 atm pressure and 2000 °C. The equilibrium concentrations of N₂ and O₂ at this pressure and temperature are 0.036 M and 0.0089 M, respectively.

Solution

Substitute the provided quantities into the equilibrium constant expression and solve for [NO]:

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

To confirm this result, it may be used along with the provided equilibrium concentrations to calculate a value for K:

This result is consistent with the provided value for K within nominal uncertainty, differing by just 1 in the least significant digit’s place.

Check Your Learning

The equilibrium constant K_c for the reaction of nitrogen and hydrogen to produce ammonia at a certain temperature is 6.00 $\times$ 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.

Calculation of Equilibrium Concentrations

Under certain conditions, the equilibrium constant K_c 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 in a mixture that initially contained only PCl₅ at a concentration of 1.00 M?

Solution

Use the stepwise process described earlier.

1. Step 1.

   Determine the direction the reaction proceeds.

   The balanced equation for the decomposition of PCl₅ is

   $${\text{PCl}}_5(g)\rightleftharpoons {\text{PCl}}_3(g)+{\text{Cl}}_2(g)$$

   Because only the reactant is present initially Q_c = 0 and the reaction will proceed to the right.

2. Step 2.

   Develop an ICE table.

3. Step 3.

   Solve for the change and the equilibrium concentrations.

   Substituting the equilibrium concentrations into the equilibrium constant equation gives

   $$K_c=\frac{[{\text{PCl}}_3][{\text{Cl}}_2]}{\left[{\text{PCl}}_5\right]}=0.0211$$

   $$=\frac{(x)(x)}{(1.00-x)}$$

   $$0.0211=\frac{(x)(x)}{(1.00-x)}$$

   $$0.0211(1.00-x)=x^2$$

   $$x^2+0.0211x-0.0211=0$$

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

   $$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$

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

   $$x=\frac{-0.0211\pm \sqrt{{(0.0211)}^2-4(1)(\mathrm{-0.0211})}}{2(1)}$$

   $$=\frac{-0.0211\pm \sqrt{(4.45\times {10}^{\mathrm{-4}})+(8.44\times {10}^{\mathrm{-2}})}}{2}$$

   $$=\frac{-0.0211\pm 0.291}{2}$$

   The two roots of the quadratic are, therefore,

   $$x=\frac{-0.0211+0.291}{2}=0.135$$

   and

   $$x=\frac{-0.0211-0.291}{2}=\mathrm{-0.156}$$

   For this scenario, only the positive root is physically meaningful (concentrations are either zero or positive), and so x = 0.135 M.

   The equilibrium concentrations are

   $$\left[{\text{PCl}}_5\right]=1.00-0.135=0.87M$$

   $$\left[{\text{PCl}}_3\right]=x=0.135M$$

   $$\left[{\text{Cl}}_2\right]=x=0.135M$$
4. Step 4.

   Confirm the calculated equilibrium concentrations.

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

   $$K_c=\frac{[{\text{PCl}}_3][{\text{Cl}}_2]}{[{\text{PCl}}_5]}=\frac{(0.135)(0.135)}{0.87}=0.021$$

   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).

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. What are the equilibrium concentrations for a mixture that is initially 0.15 M in CH₃CO₂H, 0.15 M in C₂H₅OH, 0.40 M in CH₃CO₂C₂H₅, and 0.40 M in H₂O?

Check Your Learning

A 1.00-L flask is filled with 1.00 mole 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?

Calculation of Equilibrium Concentrations Using an Algebra-Simplifying Assumption

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.

Substitute the equilibrium concentration terms into the K_c expression

rearrange to the quadratic form and solve for x

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

Note in this case that the change in concentration is significantly less than the initial concentration (a consequence of the small K), and so the initial concentration experiences a negligible change:

This approximation allows for a more expedient mathematical approach to the calculation that avoids the need to solve for the roots of a quadratic equation:

The value of x calculated is, indeed, much less than the initial concentration

and so the approximation was justified. If this simplified approach were to yield a value for x that did not justify the approximation, the calculation would need to be repeated without making the approximation.

Check Your Learning

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