5.6.4

Calculating Equilibrium Concentrations

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Calculating Equilibrium Concentrations

You can calculate the concentrations or partial pressures of substances in a reversible reaction at equilibrium if you have some information about the equilibrium.

What you need to know

What you need to know

  • To complete calculations with the equilibrium constant, you need to know three of the following:
    • The balanced equation for the reaction.
    • The initial concentrations of the substances.
    • The equilibrium concentrations of the reactants or products.
    • The equilibrium constant (K).
Example question

Example question

  • Iodine molecules react reversibly with iodide ions to produce triiodide ions.
    • I2(aq) + I(aq) \rightleftharpoons I3(aq)
  • A solution has concentrations of I2 and I both equal to 1.000 × 10−3 M before the reaction gives an equilibrium.
  • At equilibrium, the concentration of I2 is 6.61 × 10−4 M.
  • What is the equilibrium constant for the reaction?
Solutions - step 1

Solutions - step 1

  • The equilibrium constant for this reaction can be calculated using
    • Kc=[I3][I2]  [I]K_c = \large\frac{[I_3^-]}{[I_2]\; [I^-]}
  • Equilibrium concentrations are needed for all the reactants and products to calculate the equilibrium constant (K).
Solution - step 2

Solution - step 2

  • You are told the initial concentrations of the reactants and the equilibrium concentration of the product in the question.
  • Create a table to show the initial concentrations.
Solution - step 3

Solution - step 3

  • As the reaction reaches equilibrium:
    • The reactants each lose a concentration of xx.
    • The product gains a concentration of xx.
  • Fill in another row of the table to show the change in concentrations.
Solution - step 4

Solution - step 4

  • Add another row which shows the concentrations of the reactants and products at equilibrium in terms of xx.
Solution - step 5

Solution - step 5

  • You know from the question that the concentration of I2 at equilibrium is 6.61 × 10−4 M. So:
    • 1 × 10−3xx = 6.61 × 10−4
    • You can work out xx = 3.39 × 10−4 M
Solution - step 6

Solution - step 6

  • You can then update the equilibrium concentrations in your table by substituting in the values of xx.
Solution - step 7

Solution - step 7

  • Finally, substitute the equilibrium concentrations into the equilibrium constant (K) expression and solve to find K.
    • Kc=[I3][I2]  [I]K_c = \large\frac{[I_3^-]}{[I_2]\; [I^-]}
    • Kc=3.39×1046.61×104  ×  6.61×104K_c = \large\frac{3.39\times10^{-4}}{6.61\times10^{-4}\;\times\;6.61\times10^{-4}} = 776
Jump to other topics
1

Structure - Models of the Particulate of Matter

2

Structure - Models of Bonding & Structure

3

Structure - Classification of Matter

3.1

The Periodic Table: Classification of Elements

3.2

Periodic Trends

3.3

Group 1 Alkali Metals

3.4

Halogens

3.5

Noble gases, group 18

3.6

Functional Groups: Classification of Organic

3.7

Functional Group Chemistry

3.8

Alkanes

3.9

Alcohols

3.10

Halogenoalkanes

4

Reactivity - What Drives Chemical Reaction?

5

Reactivity - How Much, How Fast & How Far?

6

Reactivity - The Mechanisms of Chemical Change

7

Measurement, Data Processing & Analysis

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