Module 7

Lesson 2.2  Predicting Equilibrium Concentrations



Key Concepts


To calculate the equilibrium constant (Kc), you need a balanced chemical equation for the reaction system and the equilibrium concentrations of each species that occurs in the equilibrium expression (or enough information to calculate the concentrations).



  1. Calculating Kc when equilibrium concentrations are provided


    In some instances, you are provided with the equilibrium concentrations of both the reactants and the products. In these cases, calculating percent yield or Kc is straightforward. Simply substitute the equilibrium concentrations into the equilibrium law expression and calculate Kc.


    For example, consider the data provided for the Hydrogen-Iodine System 1 at equilibrium.




  1.  
  2. Calculating Kc when equilibrium concentrations are not provided


    What if you are not provided with the equilibrium concentrations? In this case, you always will be given sufficient information to calculate equilibrium concentrations on your own. This process can be simplified using a tool known as an ICE table.


    The letters "ICE" stand for Initial, Change, and Equilibrium.

    • Initial concentrations represent quantities present before the reaction commences. Often, zero product will be present at the beginning of the reaction.

    • Changes in concentration of substances will be either negative (indicating a decrease in concentration) or positive (indicating an increase in concentration). To determine the change for one substance, both initial and final concentrations must be known. By using the mole ratio from the balanced equation, the change in the concentration of one substance can be used to determine the change in concentration for another substance.

    • Equilibrium concentrations can be calculated after initial concentrations and changes are known. The determination of equilibrium concentrations is essential because only equilibrium values can be substituted into equilibrium law expressions to calculate equilibrium constants (Kc).



Watch


 

Read pages 686 and 687 of the textbook. Study Sample Problem 15.2.

Check Your Understanding


Complete Practice Question 6 on page 682 of your textbook. Also, complete the Section 15.1 Questions 7, 9 and 10 on page 689. Check your answers by clicking the link below.

Page 682 Practice Question 6

  1. The graph shows the equilibrium concentration of ethene is 2.5 mol/L. The change in ethene concentration is 4.0 mol - 2.5 mol = 1.5 mol. Because the coefficients in the balanced chemical equation are all 1, all species undergo a change in concentration of 1.5 mol/L. The equilibrium concentrations for the other compounds in the reaction are shown below:

    \( \mathrm { Br_2(g)~ } \) \( \mathrm { 2.50~\frac{mol}{L} - 1.5~ mol/L = 1.0 mol/L } \)
    \( \mathrm { C_2H_3Br(g)~ } \) \( \mathrm { 0~\frac{mol}{L}+ 1.50~\frac{mol}{L} = 1.50~\frac{mol}{L} } \)



  2. Concentration C2H4(g) (mol/L) Br2(g) (mol/L) C2H3Br(g) (mol/L)
    Initial 4.0 2.50 0
    Change -1.5 -1.5 +1.5
    Equilibrium 2.5 1.0 1.5

  3. The volume of the container is one litre. This is determined by the information provided on the graph that shows an initial concentration of ethane of 4.0 mol/L. Because 4.0 mol of ethane were used, the volume of the container must be 1.0 L.


  4. Because all coefficients are 1, the moles of bromine are lower than that of ethane. Therefore, bromine is the limiting reagent. The maximum yield of bromoethene would be 2.50 mol. Therefore, % yield = (1.5 mol / 2.5 mol) x 100 = 60%.


Page 689 Section 15.1 Question 7 a-g

  1. \( \mathrm { 2 HBr(g) \leftrightharpoons H_2(g) + Br_2(g) } \)





  3. \( \mathrm { \text{Moles}_{HBr~at~equilibrium} = 0.100~\frac{mol}{L} \times 2.00~ L = 0.200~ mol } \)


  4. \( \mathrm { Change~in~amount~of~HBr = 1.00~ mol - 0.200~ mol = 0.800~ mol } \)


  5. \( \mathrm { Change~in~H_2(g) = 0.800~mol \times (\dfrac{1 ~H_2}{ 2~ HBr}) = 0.400~ mol } \)


    Because H2 and Br2 are produced in 1:1 proportions, the chemical amount of Br2 formed is 0.400 mol.


  6. Concentration at equilibrium:

    \( \mathrm { HBr(g) = 0.100~ \frac{mol}{L} } \)

    \( \mathrm { H_2(g) = \frac{0.400~mol}{ 2.00 ~L} = 0.200~ \frac{mol}{L} } \)

    \( \mathrm { Br_2(g) = \frac{0.400~mol}{ 2.00 ~L} = 0.200~ \frac{mol}{L} } \)


    g.  

     


Page 689 Section 15.1 Question 9

Assume flask contains no other substances at start of experiment except CO(g) and H2O(g) and that the reaction chamber has a volume of 1.00 L.

Concentration CO2(g) (mol/L) H2(g) (mol/L) CO(g) (mol/L) H2O(g) (mol/L)
Initial 0 0 0.20 0.25
Change +0.10 +0.10 -0.10 -0.10
Equilibrium 0.10 0.10 0.10 0.15



Page 689 Section 15.1 Question 10



  1. Concentration 2HBr(g) (mol/L) H2(g) (mol/L) Br2(g) (mol/L)
    Initial 0 0.50 0.50
    Change +2x -x -x
    Equilibrium 2x 0.50 - x 0.50 - x