M14Q2: Equilibrium Expressions and Equilibrium Constants

Learning Objectives

  • Write an equilibrium constant expression, such as Kc, in terms of reactants and products and the stoichiometry of the reaction.
  • Explain why an equilibrium constant is numeric only and has no units.
  • Determine whether a reaction is reactant-favored or product-favored at equilibrium from the value of the equilibrium constant.

| Key Concepts and Summary | Key Equations | Glossary | End of Section Exercises |

Now that we have a symbol (⇌) to designate reversible reactions, we will need a way to express mathematically how the amounts of reactants and products affect the equilibrium of the system. A general equation for a reversible reaction may be written as follows:

aA + bB  ⇌  cC + dD

We can derive the equilibrium constant (Keq) for this reaction by using the definition of equilibrium:

Rateforward = Ratereverse

and

Rateforward = k1[A]a[B]b

Ratereverse = k-1[C]c[D]d

Setting the rates equal to each other and combining like terms, we can see that:

k1[A]a[B]b = k-1[C]c[D]d

\dfrac{k_{1}}{k_{-1}} = \dfrac{[\text{C}]^{c}[\text{D}]^{d}}{[\text{A}]^{a}[\text{B}]^{b}} = Keq

Thus we can see that the equilibrium constant is equal to both a ratio of rate constants, as well as the ratio of concentrations within a given reaction. When evaluated using concentrations, it is called Kc, where the subscript “c” refers to concentration. We use brackets to indicate molar concentrations of reactants and products. The equilibrium constant, Kc, is equal to the molar concentrations of the products of the chemical equation (multiplied together) over the reactants (also multiplied together), with each concentration raised to the power of the coefficient of that substance in the balanced chemical equation. Solids and liquids are not included when writing Keq. For the reaction above, we can write Kc as:

Kc = \dfrac{[\text{C}]^{c}[\text{D}]^{d}}{[\text{A}]^{a}[\text{B}]^{b}}

Example 1

Writing Equilibrium Expressions
Write the Kc expression for the reaction quotient for each of the following reactions:

  1. 3 O2(g)  ⇌  2 O3(g)
  2. N2(g) + 3 H2(g)  ⇌  2 NH3(g)
  3. 4 NH3(g) + 7 O2(g)  ⇌  4 NO2(g) + 6 H2O(g)

Solution

  1. Kc = \dfrac{[\text{O}_{3}]^{2}}{[\text{O}_{2}]^{3}}
  2. Kc = \dfrac{[\text{NH}_{3}]^{2}}{[\text{N}_{2}][\text{H}_{2}]^{3}}
  3. Kc = \dfrac{[\text{NO}_{2}]^{4}[\text{H}_{2}\text{O}]^6}{[\text{NH}_{3}]^{4}[\text{O}_{2}]^{7}}

Check Your Learning
Write the equilibrium expression for each of the following reactions:

  1. 2 SO2(g) + O2(g)  ⇌  2 SO3(g)
  2. C4H8(g)  ⇌  2 C2H4(g)
  3. 2 C4H10(g) + 13 O2(g)  ⇌  8 CO2(g) + 10 H2O(g)

Answer:

  1. Kc = \dfrac{[\text{SO}_{3}]^{2}}{[\text{SO}_{2}]^{2}[\text{O}_{2}]}
  2. Kc = \dfrac{[\text{C}_{2}\text{H}_{4}]^{2}}{[\text{C}_{4}\text{H}_{8}]}
  3. Kc = \dfrac{[\text{CO}_{2}]^{8}[\text{H}_{2}\text{O}]^{10}}{[\text{C}_{4}\text{H}_{10}]^{2}[\text{O}_{2}]^{13}}

The magnitude of an equilibrium constant is a measure of the yield of a reaction when it reaches equilibrium. A large value for Kc (much greater than 1) indicates that equilibrium is attained only after the reactants have been largely converted into products. A small value of Kc (much less than 1) indicates that equilibrium is attained when only a small proportion of the reactants have been converted into products.

The correct units for each reactant/product in the equilibrium expression are Molar (for solutions) and partial pressure (for gases). It is important to remember that pure liquids and solids are always omitted from the equilibrium expression and the value of K is always unitless. Going more in depth into this is outside the scope of this course.

Homogeneous Equilibria

A homogeneous equilibrium is one in which all of the reactants and products are present in a single solution (by definition, a homogeneous mixture). In this chapter, we will concentrate on the most common type of homogeneous equilibria which occurs in liquid-phase solutions. Reactions between solutes in liquid solutions belong to one type of homogeneous equilibria. The chemical species involved can be molecules, ions, or a mixture of both. Several examples are provided here.

C2H2(aq) + 2 Br2(aq)  ⇌  C2H2Br4(aq)      Kc = \dfrac{[\text{C}_{2}\text{H}_{2}\text{Br}_{4}]}{[\text{C}_{2}\text{H}_{2}][\text{Br}_{2}]^{2}}

I2(aq) + I–(aq)  ⇌  I3–(aq)      Kc = \dfrac{[\text{I}_{3}^{-}]}{[\text{I}_{2}][\text{I}^{-}]}

HF(aq) + H2O(ℓ)  ⇌  H3O+(aq) + F–(aq)      Kc = \dfrac{[\text{H}_{3}\text{O}^{+}][\text{F}^{-}]}{[\text{HF}]}

NH3(aq) + H2O(ℓ)  ⇌  NH4+(aq) + OH–(aq)      Kc = \dfrac{[\text{NH}_{4}^{+}][\text{OH}^{-}]}{[\text{NH}_{3}]}

In each of these examples, the equilibrium system is an aqueous solution, as denoted by the aq annotations on the solute formulas. Since H2O(l) is the solvent for these solutions, its concentration does not appear as a term in the Kc expression, as discussed earlier, even though it may also appear as a reactant or product in the chemical equation.

Whenever gases are involved in a reaction, the partial pressure of each gas can be used instead of its concentration in the equation for the equilibrium expression because the partial pressure of a gas is directly proportional to its concentration at constant temperature. Using the partial pressures of the gases, we can write the equilibrium expression, denoted as Kp because it is the result of pressures, and calculate for the system. For example:

2 NH3(g)  ⇌  N2(g) + 3 H2(g)    Kp = \dfrac{P_{\text{N}_{2}}(P_{\text{H}_{2}})^{3}}{(P_{\text{NH}_{3}})^{2}}

Heterogeneous Equilibria

A heterogeneous equilibrium is a system in which the reactants and products are found in two or more phases. The phases may be any combination of solid, liquid, gas, and aqueous solutions. When dealing with these equilibria, remember that solids and pure liquids do not appear in equilibrium constant expressions.

Some heterogeneous equilibria involve chemical changes; for example:

PbCl2(s)  ⇌  Pb2+(aq) + 2 Cl–(aq)      Kc = [Pb2+][Cl–]2

CaO(s) + CO2(g)  ⇌  CaCO3(s)     Kc = \dfrac{1}{[\text{CO}_{2}]}

C(s) + 2 S(g)  ⇌  CS2(g)      Kc = \dfrac{[\text{CS}_{2}]}{[\text{S}]^{2}}

Other heterogeneous equilibria involve phase changes, for example, the evaporation of liquid bromine, as shown in the following equation:

Br2(ℓ)  ⇌  Br2(g)      Kc = [Br2]

Key Concepts and Summary

In this section, we learn about the mathematical way to evaluate a chemical system at equilibrium, known as the equilibrium constant, Keq, which is the ratio of the product and reactant amounts at equilibrium. If Keq >> 1, the system has more products than reactants at equilibrium. If Keq << 1, the system has more reactants than products at equilibrium. If Keq ~ 1, the system has about equal amounts of products and reactants at equilibrium.

Key Equations

  • aA + bB ⇌ cC + dD
  • Keq = \dfrac{k_{1}}{k_{-1}} = \dfrac{[\text{C}]^{c}[\text{D}]^{d}}{[\text{A}]^{a}[\text{B}]^{b}}

Glossary

equilibrium constant (Keq)
a measure of whether a chemical system is reactant- or product-favored at equilibrium. Equal to the ratio of products to reactants, as well as the rate constant of the forward direction divided by the reverse direction
heterogeneous equilibrium
a system in which the reactants and products are found in two or more phases
homogeneous equilibrium
a system where all of the reactants and products are present in a single phase

Chemistry End of Section Exercises

  1. Write the mathematical expression for the equilibrium constant, Kc, for each of the following reactions:
    1. CH4(g) + Cl2(g)  ⇌  CH3Cl(g) + HCl(g)
    2. N2(g) + O2(g)  ⇌  2 NO(g)
    3. 2 SO2(g) + O2(g)  ⇌  2 SO3(g)
    4. BaSO3(s)  ⇌  BaO(s) + SO2(g)
    5. P4(g) + 5 O2(g)  ⇌  P2O10(s)
    6. Br2(g)  ⇌  2 Br(g)
    7. CH4(g) + 2 O2(g)  ⇌  CO2(g) + 2 H2O(ℓ)
    8. Br2(ℓ) + 2 O3(g)  ⇌  O2(g) + 2 BrO2(s)
  2. Write the mathematical expression for the equilibrium constant, Kc, for each of the following unbalanced reactions:
    1. N2(g) + H2(g)  ⇌  NH3(g)
    2. NH3(g) + O2(g)  ⇌  NO(g) + H2O(g)
    3. N2O4(g)  ⇌  NO2(g)
    4. CO2(g) + H2(g)  ⇌  CO(g) + H2O(g)
    5. NH4Cl(s)  ⇌  NH3(g) + HCl(g)
    6. Pb(NO3)2(s)  ⇌  PbO(s) + NO2(g) + O2(g)
    7. H2(g) + O2(g)  ⇌  H2O(ℓ)
    8. S8(g)  ⇌  S(g)
  3. What is the value of the equilibrium constant at 500 °C for the formation of NH3 according to the following equation?
    N2(g) + 3 H2(g)  ⇌  2 NH3(g)

    An equilibrium mixture of NH3(g), H2(g), and N2(g) at 500°C was found to contain 1.35 M H2, 1.15 M N2, and 0.412 M NH3. Based on the value you calculated for Kc, is this a product-favored or reactant-favored reaction?

  4. Hydrogen is prepared commercially by the reaction of methane and water vapor at elevated temperatures.
    CH4(g) + H2O(g)  ⇌  3 H2(g) + CO(g)

    What is the equilibrium constant for the reaction if a mixture at equilibrium contains gases with the following concentrations: CH4, 0.126 M; H2O, 0.242 M; CO, 0.126 M; H2, 1.15 M, at a temperature of 760°C? Based on the value you calculated for Kc, is this a product-favored or reactant-favored reaction?

  5. Calculate Kc for the reaction:
    2 HI(g)  ⇌  H2(g) + I2(g)

    given that the concentrations of each species at equilibrium are as follows: [HI] = 0.85 M, [H2] = 0.27 M, and [I2] = 0.60 M.

  6. Write the mathematical expression for the equilibrium constant, Kp, for the following unbalanced reaction:
    BaSO3(s)  ⇌  BaO(s) + SO2(g)
  7. The equilibrium expression for a chemical reaction is: Kc = \dfrac{[\text{FeCl}_2]^2 [\text{SnCl}_4]}{[\text{FeCl}_3]^2 [\text{SnCl}_2]}.  What is the balanced chemical equation for the reaction?
    1. FeCl2(aq) + SnCl4(aq)  ⇌  FeCl3(aq) + SnCl2(aq)
    2. 2 FeCl2(aq) + SnCl4(aq)  ⇌  2 FeCl3(aq) + SnCl2(aq)
    3. FeCl3(aq) + SnCl4(aq)  ⇌  FeCl2(aq) + SnCl2(aq)
    4. 2 FeCl3(aq) + SnCl2(aq)  ⇌  2 FeCl2(aq) + SnCl4(aq)
    5. FeCl2(aq) + 2 SnCl4(aq)  ⇌  2 FeCl3(aq) + 2 SnCl2(aq)

Answers to Chemistry End of Section Exercises

  1. (a) Kc = \dfrac{[\text{CH}_{3}\text{Cl}][\text{HCl}]}{[\text{CH}_{4}][\text{Cl}_{2}]}
    (b) Kc = \dfrac{[\text{NO}]^{2}}{[\text{N}_{2}][\text{O}_{2}]}
    (c) Kc = \dfrac{[\text{SO}_{3}]^{2}}{[\text{SO}_{2}]^{2}[\text{O}_{2}]}
    (d) Kc = [SO2]
    (e) Kc = \dfrac{1}{[\text{P}_{4}][\text{O}_{2}]^5}
    (f) Kc = \dfrac{[\text{Br}]^2}{[\text{Br}_{2}]}
    (g) Kc = \dfrac{[\text{CO}_{2}]}{[\text{CH}_{4}][\text{O}_{2}]^2}
    (h) Kc = \dfrac{[\text{O}_2]}{[\text{O}_3]^2}
  2. (a) Kc = \dfrac{[\text{NH}_3]^{2}}{[\text{N}_2][\text{H}_2]^3}
    (b) Kc = \dfrac{[\text{NO}]^4[\text{H}_2\text{O}]^5}{[\text{NH}_3]^4[\text{O}_2]^5}
    (c) Kc = \dfrac{[\text{NO}_2]^2}{[\text{N}_2\text{O}_4]}
    (d) Kc = \dfrac{[\text{CO}][\text{H}_2\text{O}]}{[\text{CO}_2][\text{H}_2]}
    (e) Kc = [NH3][HCl]
    (f) Kc = [NO2]4[O2]
    (g) Kc = \dfrac{1}{[\text{H}_2]^2[\text{O}_2]}
    (h) Kc = \dfrac{[\text{S}]^8}{[\text{S}_8]}
  3. Kc = 6.00 × 10-2; reactant-favored
  4. Kc = 6.28; product-favored
  5. Kc = 0.22
  6. Kp = PSO2
  7. D
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