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Equilibrium PowerPoint PPT Presentation

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Equilibrium. Quantitative vs. Reversible. So far we have studied quantitative reactions , reactions which proceed until the limiting reactant is consumed However, many reactions are not quantitative, they are reversible meaning they can proceed in both forward and reverse directions

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Quantitative vs reversible

Quantitative vs. Reversible

  • So far we have studied quantitative reactions, reactions which proceed until the limiting reactant is consumed

  • However, many reactions are not quantitative, they are reversible meaning they can proceed in both forward and reverse directions

    • A + B C + D

Dynamic equilibrium

Dynamic Equilibrium

  • When the rate of the forward reaction and the rate of the reverse reaction are equal, the system is in dynamic equilibrium

  • This does NOT mean that nothing is happening!

Properties of equilibrium systems

Properties of Equilibrium Systems

  • 1. Macroscopicproperties (e.g. colour, pressure, concentration, pH) are constant

    • I.e. It appears as if nothing is changing

  • 2. Can only be reached in a closed system

  • 3. Forward rate = Reverse rate

  • 4. Can be established from either direction

    At equilibrium, the concentrations of all reactants and products will remain constant

Types of equilibrium

Types of Equilibrium

  • Solubility Equilibrium: A dynamic

    equilibrium between a solute and solvent

    in a saturated solution in a closed system

  • Phase Equilibrium: a dynamic equilibrium between different physical states of a pure substance in a closed system

    • Ex: water/liquid in a sealed container

  • Chemical Reaction equilibrium: a dynamic equilibrium between reactants and products of a chemical reaction in a closed system



Sketch a graph of Concentration of Reactants and products vs. time for Experiment 1 and 2

More practice

More Practice

1. Consider the reaction:

  • Hydrogen gas and hydrogen iodide gas are both clear colourless, while iodine gas is purple.

  • Explain why when hydrogen gas and iodine gas are added to a closed flask and allowed to react, the colour never fades to colourless.

    2. Sketch a graph to illustrate the changes in concentration of each component over time as equilibrium is reached for Experiment 1


  • 3. If experiment 1 did not reach equilibrium, what would the theoretical yield of HI be?

  • 4. What is the actual yield of HI in experiment 1?

Solving equilibrium problems

Solving Equilibrium Problems

  • Use ICE Tables

  • I: Initial Concentration (mol/L)

  • C: Change in concentration

    • +x if increasing in concentration

    • -x if decreasing in concentration

    • Multiply x by the # of moles

  • E: Equilibrium Concentration (mol/L)

Practice 1

Practice #1

  • In a 4.00L container, 2.50 mol of carbon dioxide gas is decomposed. At equilibrium, [CO2(g)]eq = 0.125 mol/L. Use an ICE table to find [O2(g)]eq and [CO(g)]eq

Practice 2

Practice #2

  • 0.500 mol of NOCl(g) is decomposed in a closed 2.00 L container. [NO(g)]eq = 0.040 mol/L. Find [Cl2(g)]eq and [NOCl(g)]eq.

Practice 3

Practice #3

  • When 2.00 mol of ethene gas and 1.50 mol of bromine vapour come to equilibrium in a closed 1.00 L container, the equilibrium concentration of bromine vapour is measured at 0.150 mol/L. Find [C2H4(g)]eq and [C2H4Br2(g)]eq.

Practice 3 continued

Practice #3 Continued

  • Graph the equilibrium reaction for the addition of bromine to ethene.


At equilibrium, the rate of the forward and reverse reactions are equal


The equilibrium constant k eq

The Equilibrium Constant, Keq

For the reaction:

aA + bB  cC + dD

At equilibrium:

rfwd = rrev

kfwd[A]a[B]b = krev[C]c[D]d

kfwd = [C]c[D]d

krev [A]a[B]b

Keq = [C]c[D]d


Sub in rate law equation


Sub in Keq for


Equilibrium constant

Keq conditions

Keq Conditions

  • Note: the equilibrium constant expression only works with reactions that occur in a single step

  • Keq will remain the same as long as the temperature is kept constant(changing the temperature changes the forward and reverse reactions by different amounts and therefore change the Keq)



Write the equilibrium constant expression for the following two reactions

Heterogeneous equilibria

Heterogeneous Equilibria

  • Equilibrium systems can involve all states of matter

  • However, the concentration of a pure solid or liquid cannot change

  • Therefore, equilibrium constant expressions will not include solids and liquids

Magnitude of k eq

Keq = [C]c[D]d


Magnitude of Keq

Keq>> 1 At equilibrium there is more products than reactants. The reaction is product favoured

Keq = 1At equilibrium there is an equal amount of products and reactants

Keq << 1At equilibrium there is more reactants than products. The reaction is reactant favoured



Calculate the value of Keq for the following system

At Equilibrium:

[CO2] = 0.0954 mol/L

[H2] = 0.0454 mol/L

[CO] = [H2O] = 0.00460 mol/L



For each reaction and their respective equilibrium constant, predict whether reactants or products are favoured.

  • N2(g) + O2(g)2NO(g)Keq = 4.7 x 10-31

  • NO(g) + CO(g)2N2(g) + CO2(g)Keq = 2.2 x 1059

Practice 2 calculating equilibrium concentrations from initial concentrations

Practice 2 Calculating Equilibrium Concentrations from Initial Concentrations

  • Carbon monoxide reacts with water vapour to produce carbon dioxide and hydrogen. At 900oC, Keq is 4.200. Calculate the concentrations of all entities at equilibrium if 4.000 mol of each entity are initially place in a 1.00L closed container.

Practice 31

Practice #3

Simplifying Assumption: 100 rule (for small K values)

If: [reactant] > 100, you can simplify the Keq expression


Ex: 2CO2(g) 2CO(g) + O2(g)

If K = 6.40 x 10-7, determine the concentrations of all

substances at equilibrium if it starts with [CO2] = 0.250 mol/L

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