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Equilibrium Constants

Key Concepts

Equilibrium Constants (K):

  • are derived from experimental data

  • are temperature dependent

  • provide a measure of the equilibrium position
  1. if K is large (> 102) products of the reaction are favoured

  2. if K is small, reactants are favoured.

  • remain the same value for a constant temperature even if the equilibrium concentrations of the reactants or products are altered

  • do NOT change in the presence of a catalyst (a catalyst changes the time taken to reach equilibrium but does not alter the equilibrium position or equilibrium constant)

  • For the reaction:
    aA + bB cC + dD

    where A, B, C and D are the chemical species and
    a, b, c and d are the stoichiometric coefficients,

    K1 = [C]c[D]d

    [A]a[B]b

  • For the same reaction written in reverse:
    cC + dD aA + bB

    where A, B, C and D are the chemical species and
    a, b, c and d are the stoichiometric coefficients,

    K2 = [A]a[B]b

    [C]c[D]d
    K2 = 1/K1 and K1 = 1/K2

  • If the concentration of species in the reaction is used to calculate the equilibrium constant then it can be given the symbol Kc. If the equilibrium constant has no subscript, that is it is given as K, this refers to Kc and the concentration of each species in the reaction is used to calculate the equilibrium expression.

  • If a solvent takes part in a reaction, its concentration is said to remain constant and is incorporated into the value of K.

  • If a solid is present in a reaction, its concentration is said to remain constant and is incorporated into the value of K.

Writing Equilibrium Expressions

Reactions Involving Aqueous Species

  1. All species are present in aqueous solution eg:

    Ag+(aq) +2NH3(aq) Ag(NH3)2+(aq)

    K = [Ag(NH3)2+]

    [Ag+][NH3]2

  2. Solvent takes part in reaction, eg:

    NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)

    K = [NH4+][OH-]

    [NH3]

    The concentration of the water as solvent is said to be constant and is incorporated into the value of K.

  3. A solid is present in the reaction, eg

    Pb(NO3)2(aq) + 2KI(aq) PbI2(s)

    K = 1

    [Pb(NO3)2][KI]2

    The concentration of the solid is said to be constant and is incorporated into the value of K.

Reactions Involving Gases

  1. All species are present as gases, eg:

    CO(g) + 2H2(g) CH3OH(g)

    K = [CH3OH]

    [CO][H2]2

  2. A solid is present in the reaction, eg:

    CaCO3(s) CaO(s) + CO2(g)

    K = [CO2]

    The concentration of solids are said to be constant and are incorporated into the value of K.

Some Special Special Cases

Calculating Equilibrium Constant : Concentration of All Species Known

For the reaction COCl2(g) CO(g) + Cl2(g) at 900oC at equilibrium,
the concentration of COCl2(g) is 3.0 x 10-6 mol L-1 while that of CO(g) and Cl2(g) is 4.97 x 10-4 mol L-1.
Calculate the value of K.

  1. Write the equilibrium expression using the balanced chemical equation:
    K = [CO][Cl2]

    [COCl2]

  2. Extract all the relevant data from the question:
    [CO] = 4.97 x 10-4 mol L-1
    [Cl2] = 4.97 x 10-4 mol L-1
    [COCl2] = 3.0 x 10-6 mol L-1
    K = ?

  3. Substitute the values into the equilibrium expression and solve:
    K = [4.97 x 10-4][4.97 x 10-4]

    [3.0 x 10-6]

    K = 0.082

Calculating Equilibrium Constant : Initial Concentrations and Equilibrium Concentration of One Species Known

A mixture of 5.0 mol H2(g) and 10.0 mol I2(g) are placed in a 5L container at 450oC and allowed to come to equilibrium. At equilibrium the concentration of HI(g) is 1.87 mol L-1.
Calculate the value for K for this reaction.
  1. Write the balanced chemical equation for the reaction:
    H2(g) + I2(g) 2HI(g)

  2. Write the equilibrium expression for the reaction:
    K = [HI]2

    [H2][I2]

  3. Calculate the equilibrium concentrations for all reactants and products:
    [H2] [I2] [HI]
    initial concentration
    mol L-1 = moles/Volume
    5.0mol/5L
    = 1.0 mol L-1
    10.0/5L
    = 2.0 mol L-1
    0 mol L-1
    equilibrium concentration expression
    Let x=amount used in reaction
    1.0 - x 2.0 - x 1.87 mol L-1
    calculate x
    using mole ratio from equation
    x = ½ x 1.87
    x = 0.935
    x = ½ x 1.87
    x = 0.935
    1.87 mol L-1
    calculate equilibrium concentrations
    1.0 - 0.935
    = 0.065 mol L-1
    2.0 - 0.935
    = 1.065 mol L-1
    1.87 mol L-1

  4. Substitute the equilibrium concentrations into the equation and solve for K:
    K = [1.87]2

    [0.065][1.065]

    K = 50.5

Calculating Equilibrium Constant : Initial Reactant Concentration and %Dissociation Known

4.0 moles of NOCl(g) were placed in a 2L vessel at 460oC. The following dissociation reaction occurred:
2NOCl(g) 2NO(g) + Cl2(g)

At equilibrium the NOCl(g) was 33% dissociated. Calculate the equilibrium constant for this reaction.

  1. Write the equilibrium expression for the reaction:
    K = [NO]2[Cl2]

    [NOCl]2

  2. Calculate the equilibrium concentrations for all reactants and products:
    [NOCl] [NO] [Cl2]
    initial concentration
    mol L-1 = moles/Volume
    4.0mol/2L
    = 2.0 mol L-1
    0 mol L-1 0 mol L-1
    equilibrium concentration
    expression

    Let x=amount used in reaction
    2.0 - 2x 2x x
    calculate x
    using 33% dissociation of NOCl
    2x = 33/100 x 2.0
    2x = 0.66 mol L-1
    and using mole ratio from equation 2x = 0.66 mol L-1 x = ½ x 0.66
    x = 0.33 mol L-1
    calculate equilibrium
    concentrations

    2.0 - 0.66
    = 1.34 mol L-1
    0.66 mol L-1 0.33 mol L-1

  3. Substitute the equilibrium concentrations into the equation and solve for K:
    K = [0.66]2[0.33]

    [1.34]2

    K = 0.08


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