go to the AUS-e-TUTE homepage

Equilibrium Constant Calculations Chemistry Tutorial

Key Concepts

Equilibrium Constants (K):

Please do not block ads on this website.
No ads = no money for us = no free stuff for you!

Introduction

Consider the chemical reaction in which a pale yellow aqueous solution of iron(III) ions (Fe3+(aq)) reacts with a colourless aqueous solution of thiocyantate ions (SCN-(aq)) to produce bright red Fe(SCN)2+(aq):

Fe3+(aq) + SCN-(aq) ⇋ Fe(SCN)2+(aq)

The concentration of the species can be determined by measuring the absorption of light in the sample.

Three separate experiments are performed at the same temperature.
In each experiment the initial concentrations of Fe3+(aq) and SCN-(aq) are different.
The reaction is allowed to proceed to equilibrium.

The concentration of each species at equilibrium is measured and recorded as shown below:

Experiment Initial Concentrations
mol L-1		 Equilibrium Concentrations
mol L-1
[Fe3+(aq)] [SCN-(aq)] [Fe(SCN)2+(aq)] [Fe3+(aq)] [SCN-(aq)] [Fe(SCN)2+(aq)]
1. 0.110 0.110 0.000 0.0100 0.0100 0.100
2. 0.600 0.505 0.000 0.100 0.00500 0.500
3. 0.275 0.260 0.000 0.0250 0.0100 0.2500

Can you see a pattern in the data?

We can see that the:

But can you see another pattern?
Remember that we can write a mass-action expression for this reaction:

Q =     [Fe(SCN)2+(aq)]    
[Fe3+(aq)][SCN-(aq)]

Let's use the values for the concentration of each species at equilibrium to calculate the value of the mass-action expression, Q

Experiment Equilibrium Concentrations
mol L-1
Q =   [Fe(SCN)2+(aq)]  
[Fe3+(aq)][SCN-(aq)]
[Fe3+(aq)] [SCN-(aq)] [Fe(SCN)2+(aq)]
1. 0.0100 0.0100 0.100
    [0.100]    
[0.0100][0.0100] 		= 1000
2. 0.100 0.00500 0.500
    [0.500]    
[0.100][0.00500]		 = 1000
3. 0.0250 0.0100 0.2500
    [0.2500]    
[0.0250][0.0100]		 = 1000

At equilibrium, the value of the mass-action expression, Q, is a constant.
This constant is known as the equilibrium constant and is given the symbol K.
At equilibrium, Q = K and is known as the equilibrium condition.
The value of the equilibrium constant for a given reaction at a specified temperature does not change if the concentration of a reactant or product changes.

Consider the equilibrium reaction in which ammonia gas is synthesised from gaseous nitrogen and hydrogen:

N2(g) + 3H2(g) ⇋ 2NH3(g)

Three experiments are carried out at the same temperature.
In each experiment the initial concentrations of each species was changed, the system allowed to come to equilibrium, the concentration of each species was measured and recorded, then the value of the mass-action expression calculated for the equilibrium concentrations:

Experiment Initial
Concentrations
mol L-1		 Equilibrium
Concentrations
mol L-1
Q =   [NH3(g)]2  
[N2(g)][H2(g)]3
[N2(g)] [H2(g)] [NH3(g)] [N2(g)] [H2(g)] [NH3(g)]
1. 1.00 3.00 0 0.325 0.975 1.350
    [1.350]2    
[0.325][0.975]3		 = 6
2. 1.000 1.000 0 0.781 0.343 0.438
    [0.438]2    
[0.781][0.343]3		 = 6
3. 1.000 1.000 1.000 0.885 0.655 1.230
    [1.230]2    
[0.885][0.655]3		 = 6

At equilibrium, Q is a constant, Q = K, and the value of the equilibrium constant for this reaction at this temperature is 6.
Notice that even if we start off with an initial mixture that includes both reactant and product molecules, when the reaction reaches equilibrium, the value of Q is still the same as when the initial reaction mixture contained only reactant molecules and no product molecules.

The value of the equilibrium constant for a particular reaction at a specified temperature does not change when the equilibrium concentrations of reactant and/or product molecules is changed.

So, if we know the concentrations of all the species in a given chemical reaction at equilibrium at a certain temperature, we can calculate the value of the equilibrium constant, K, for that temperature.

We can then use this value of the equilibrium constant to calculate the concentrations of reactants and/or products for this reaction in other experiments performed at the same temperature.

Do you know this?

Join AUS-e-TUTE!

Play the game now!

Some Special Special Cases

The equilibrium constant for some types of reactions is given a special name, and a special symbol, because of the importance of these reactions.
Some examples are given in the table below:

Reaction Name for Equilibrium Constant Symbol for
Equilibrium Constant
Self-dissociation of water
(self-ionisation of water)
(auto-dissociation of water)		 Ion Product for Water Kw
Dissociation of acids
(ionisation of acids)
(ionization of acids)		 Acid Dissociation Constants Ka
Dissociation of bases
(ionisation of bases)
(ionization of bases)		 Base Dissociation Constants Kb
Dissolving an ionic compound Solubility Products Ksp
gaseous reactions (calculated using partial pressures not concentrations) KP

Do you understand this?

Join AUS-e-TUTE!

Take the test now!

Worked Example of Calculating Equilibrium Constant:
Concentration of All Species Known

(Based on the StoPGoPS method for problem solving)

Question: For the reaction shown below:

COCl2(g) ⇋ CO(g) + Cl2(g)

at 900oC at equilibrium the concentration of COCl2(g) is 3.0 × 10-6 mol L-1 while that of CO(g) and Cl2(g) is 4.97 × 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 information (data) have you been given?

    (i) COCl2(g) ⇋ CO(g) + Cl2(g)

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

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

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

  3. 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]

  4. Substitute the values into the equilibrium expression and solve:
    K = [4.97 × 10-4][4.97 × 10-4]
    [3.0 × 10-6]

    K = 0.082

  5. 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 × 5 = 25
    and 25 ÷ 3 ≈ 8
    in which case our "rough" calculation would become 8 × 0.01 = 0.08
    which is in very good agreement with our carefully calculated value of 0.082

  6. State the solution:

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

Can you apply this?

Join AUS-e-TUTE!

Do the drill now!

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

(Based on the StoPGoPS method for problem solving)

Question: 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 × 10-3 mol L-1, while the equilibrium concentration of CH3OH(g) was found to be 4.25 × 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 (information) 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 × 10-3 mol L-1
    (iv) at equilibrium, [CH3OH(g)] = 4.25 × 10-8 mol L-1

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

    Write the expression for the equilibrium constant:

    K =   [CH3OH]  
    [CO][H2]2

    Rearrange the equation so you can find [H2]:

    [H2]2 =   [CH3OH]  
    K[CO]

    Take the square root of both sides of the equation, so:

    [H2] = √( [CH3OH]
    K[CO]    		 )

  4. Substitute in the values and solve:

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

  5. 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 × 10-8]        
    [3.75 × 10-3][4.35 × 10-3]2
      =         [4.25 × 10-8]        
    [3.75 × 10-3]1.89 × 10-5
      = 4.25 × 10-8
    7.09 × 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.

  6. State the solution to the problem:

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

Can you apply this?

Join AUS-e-TUTE!

Take the exam now!

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