go to the AUS-e-TUTE homepage
home test Join AUS-e-TUTE game contact
 

 

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

Example: 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 the equilibrium constant, K, for this reaction under these conditions.

  1. What have you been asked to do?

    Calculate the value of the equilibrium constant, K

  2. What data have you been given?

    (i) COCl2(g) CO(g) + Cl2(g)

    (ii) at equilibruim, [COCl2(g)] = 3.0 x 10-6 mol L-1

    (iii) at equilibrium, [CO(g)] = 4.97 x 10-4 mol L-1

    (iv) at equilibrium, [Cl2(g)] = 4.97 x 10-4 mol L-1

    What is the relationship between the data you have been given and what you need to find?
    Write the equilibrium expression using the balanced chemical equation:

    K = [CO][Cl2]

    [COCl2]

  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

  4. Is your answer plausible?

    Perform a "rough" calculation using round numbers to make sure that your answer is "about" right.

    K = [10-4][10-4] = 10-8 = 10-2 = 0.01


    [10-6] 10-6

    Our answer of 0.082 is of the same order of magnitude as the "rough" calculation, so we have some confidence in our answer.
    We could also note that 4.97 is "about" 5, so 5 x 5 = 25 and 25 ÷ 3 ≈ 8 in which case our "rough" calculation would become 8 x 0.01 = 0.08 which is in very good agreement with our carefully calculated value of 0.082

  5. State the solution:

    The value of the equilibrium constant, K, for this reaction under these conditions is 0.082

Calculating the Equilibrium Concentration of a Species Given the Value of K

Example: Methanol can be made by reacting carbon monoxide gas with hydrogen gas:

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

The value of the equilibrium constant, K, for this reaction at 100oC was found to be 0.600.
The concentration of CO(g) at equilibrium was determined to be 3.75 x 10-3 mol L-1, while the equilibrium concentration of CH3OH(g) was found to be 4.25 x 10-8 mol L-1.
Calculate the concentration of H2(g) in mol L-1 at equilibrium.

  1. What have you been asked to do?

    Calculate concentration of H2(g) in mol L-1 at equilibrium:

    [H2(g)] = ? mol L-1

  2. What data have you been given in the question?

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

    (ii) K = 0.600

    (iii) at equilibrium, [CO(g)] = 3.75 x 10-3 mol L-1

    (iv) at equilibrium, [CH3OH(g)] = 4.25 x 10-8 mol L-1

    What is the relationship between what you have been given and what you need to find?

    K =   [CH3OH]  
    [CO][H2]2

    Rearrange the equation:

    [H2]2 =   [CH3OH]  
    K[CO]

    So:

    [H2] = √( [CH3OH]
    K[CO]
    )

  3. Substitute in the values and solve:

    [H2] = √( [CH3OH]
    K[CO]
    )
          = √(     [4.25 x 10-8]    
    0.600[3.75 x 10-3]
    )
          = √( 4.25 x 10-8
    2.25 x 10-3
    )
      = √(1.889 x 10-5)  
      = 4.35 x 10-3 mol L-1  

  4. Is your answer plausible?

    Use the equilibrium concentrations given, and your calculated value of [H2(g)] to see if you get the same value for the equilibrium constant, K, as that given in the question:

    K =   [CH3OH]  
    [CO][H2]2
      =         [4.25 x 10-8]        
    [3.75 x 10-3][4.35 x 10-3]2
      =         [4.25 x 10-8]        
    [3.75x 10-3]1.89 x 10-5
      = 4.25 x 10-8
    7.09 x 10-8
      = 0.600

    By using our calculated value for [H2] we calculate the same value for K as was given in the question, so we are confident our answer is correct.

  5. State the solution to the problem:

    The concentration of hydrogen gas at equilibrium is 4.35 x 10-3 mol L-1

Calculating the Value of the Equilibrium Constant, K, Given Initial Concentration Data

This requires you to calculate the equilibrium concentration of each species before you can calculate the value of the equilibrium constant.
You will find a full discussion of this process in the R.I.C.E. Tables and Calculating K Tutorial


What would you like to do now?
advertise on the AUS-e-TUTE website and newsletters
 
 

Search this Site

You can search this site using a key term or a concept to find tutorials, tests, exams and learning activities (games).
 

Become an AUS-e-TUTE Member

 

AUS-e-TUTE's Blog

 

Subscribe to our Free Newsletter

Email email us to
subscribe to AUS-e-TUTE's free quarterly newsletter, AUS-e-NEWS.

AUS-e-NEWS quarterly newsletter

AUS-e-NEWS is emailed out in
December, March, June, and September.

 

Ask Chris, the Chemist, a Question

The quickest way to find the definition of a term is to ask Chris, the AUS-e-TUTE Chemist.

Chris can also send you to the relevant
AUS-e-TUTE tutorial topic page.

 
 
 

Share this Page

Bookmark and Share
 
 

© AUS-e-TUTE