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Equilibrium Constant, K, and Extent of Reaction Tutorial

Key Concepts

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K = a large number

When K is large, the forward reaction occurs to a greater extent than the reverse reaction.

Consider the following reaction in which reactants A(g) and B(g) react to produce the product C(g):

A(g) + B(g) ⇋ C(g)

Imagine placing 1 mole of A(g) and 1 mole of B(g) in a 1 L sealed vessel.

If the reaction proceeds to a large extent, then most of the reactants will react to produce the product.

When the system reaches equilibrium, the concentration of each reactant will be low, say 0.01 mole in the 1 L container, while the concentration of the product will be much greater, say 0.99 moles of C(g) in the 1 L container.

We can summarise these observations in a table2:

  reactants   product
Reaction A(g) + B(g) C(g)
Initial Concentration
(mol L-1)
1   1   0
Equilibrium Concentration
(mol L-1)
0.01   0.01   0.99

We can use the concentration of each reactant and product at equilibrium to calculate the equilibrium constant, Kc, for this reaction under these conditions:

1. Write the equilibrium constant expression: Kc =     [C(g)]    
[A(g)][B(g)]
2. Substitute in the values: Kc =     0.99    
0.01 × 0.01
3. Solve: Kc = 9900

The value of the equilibrium constant, Kc, is very large because the equilibrium position for this reaction lies very much to the right, that is, the forward reaction occurs to a much greater extent than the reverse reaction.

If K is large, the forward reaction is favoured.

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K = a small number

When K is small, the forward reaction occurs to a less extent than the reverse reaction

Consider a different reaction in which reactants X(g) and Yg react to produce the product Z(g):

X(g) + Y(g) ⇋ Z(g)

Imagine placing 1 mole of X(g) and 1 mole of Y(g) in a 1 L sealed vessel.

If the reaction proceeds to only a small extent, then most of the reactants will NOT react to produce the product.

When the system reaches equilibrium, the concentration of each reactant will be high, say 0.99 mole in the 1 L container, while the concentration of the product will be much less, say 0.01 moles of Z(g) in the 1 L container.

We can summarise these observations in a table:

  reactants   product
Reaction X(g) + Y(g) Z(g)
Initial Concentration
(mol L-1)
1   1   0
Equilibrium Concentration
(mol L-1)
0.99   0.99   0.01

We can use the concentration of each reactant and product at equilibrium to calculate the equilibrium constant, Kc, for this reaction under these conditions:

1. Write the equilibrium constant expression: Kc =     [Z(g)]    
[X(g)][Y(g)]
2. Substitute in the values: Kc =     0.01    
0.99 × 0.99
3. Solve: Kc = 0.01

The value of the equilibrium constant, Kc, is small because the equilibrium position for this reaction lies very much to the left, that is, the forward reaction occurs to a much less extent than the reverse reaction.

If the value of K is small, the reverse reaction is favoured.

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Worked Example

(Problem solving based on the StoPGoPS approach to problem solving.)

Question: Consider the reaction below for the formation of the complex ion Zn(CN)42-(aq) in aqueous solution:

Zn2+(aq) + 4CN-(aq) ⇋ Zn(CN)42-(aq)

The value of the equilibrium constant for this reaction at 25°C is 1019.

Does the equilibrium position lie to the left or right for this reaction at 25°C?

Solution:

  1. What have you been asked to do?

    Determine whether the equilibrium position lies to the left or right for this reaction at 25oC.

  2. What information (data) have you been given?

    Reaction: Zn2+(aq) + 4CN-(aq) ⇋ Zn(CN)42-(aq)

    Kc = 1019 (at 25°C)

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

    The larger the value of the equilibrium constant, Kc, the greater the extent of the forward reaction.
    That is, the larger the value of Kc, the further to the right the equilibrium position lies.

  4. Determine the equilibrium position for this reaction at 25°C

    1019 is a very large number!

    K is therefore very large.

    Since K is large, the forward reaction is favoured and the equilibrium position must lie very much to the right.

  5. Is your answer plausible?

    Imagine the concentration of all the species at equilibrium are in an exact stoichiometric ratio (mole ratio).
    If the hypothetical concentration of the reactants were:
        [Zn2+] = 1 mol L-1
        [CN-] = 4 mol L-1
    calculate the [Zn(CN)42-(aq)]

    K =     [Zn(CN)42-(aq)]    
    [Zn2+][CN-]4
    1019 =     [Zn(CN)42-(aq)]    
    [1][4]4
    1019 × [4]4 = [Zn(CN)42-(aq)]
    2.56 × 1021 = [Zn(CN)42-(aq)]

    This shows that the concentration of the product, [Zn(CN)42-(aq)], is much, much, larger, (orders of magnitude larger), than the concentration of the reactants at equilibrium, therefore the equilibrium position must lie very much to the right.
    So, our answer, that the equilibrium position lies to the right, is plausible.
  6. State your solution to the problem:

    For this reaction at 25°C the equilibrium position lies to the right.

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1. Although this discussion will concern Kc, the same logic can be applied to KP, for this reason we have used K rather than Kc for generalisations regarding the extent of reaction and the magnitude of the equilibrium constant.

2. The table being used is a R.I.C.E. or I.C.E. table, the only difference is that we have neglected to show the Change in concentration row as it is not required since we have been told the equilibrium concentrations.