Module 7 

Lesson 2.1  Measuring Chemical Equilibrium



Key Concepts


The extent of a reversible reaction is described by the position of its equilibrium. Two values can be used to describe a state of equilibrium -  percent yield and the equilibrium constant.

  1. Percent Yield


    The percent yield is the yield of product measured at equilibrium compared with the hypothetical maximum percent yield of product. Percent yield is useful in establishing position of equilibrium.


    To calculate the percent yield of an equilibrium system, use the stoichiometric method to calculate the hypothetical maximum percent yield of product, assuming no reverse reaction.


    For example, consider System 1 in Table 1 on page 679 of the textbook. In this example, 5 moles of hydrogen and 5 moles of iodine are reacting together to produce hydrogen iodide. Based on the balanced equation, you would expect 10 moles of HI to be formed. However, once equilibrium is reached, only 7.8 moles of product are in the reaction vessel.


    \( \mathrm { \%~yield = \frac{actual~yield}{theoretical~yield} \times 100 } \)

    \( \mathrm { \%~yield = \frac{7.8~mol}{10~mol}\times 100 } \)

    \( \mathrm { \%~yield = 78\% } \)


    In other words, the percent yield of product for this particular system is 78%, favouring the products.




    Percent yield can be used to group equilibrium systems into four categories as shown in the table below.





Check Your Understanding


Complete Questions 4 and 7 on page 682. Check your work by clicking the link below.

Page 682 Practice Question 4

  1. \( \mathrm { CH_4(g) + Cl_2(g) \leftrightharpoons CH_3Cl(g) + HCl(g) } \)
  2. The maximum possible yield of CH3Cl(g) is 2.00 mol because 2.00 mol of CH4(g) are used and CH4(g) is the limiting reagent. As well, all reactants combine in 1:1 molar proportions.
  3. % reaction = (1.40 mol / 2.00 mol) x100 = 70.0%. The position of the equilibrium favours the products.

Page 682 Practice Question 7

  1. \( \mathrm { Cu(s) + 2 Ag^+(aq) \rightarrow Cu^{2+}(aq) + 2 Ag(s) } \)
  2. \( \mathrm { Ca^{2+}(aq) + SO_4^{2-}(aq) \overset{>50\%}{\leftrightharpoons} CaSO_4(s) } \)
  3. \( \mathrm { CH_3COOH(aq) + H_2O(l) \overset{<50\%}{\leftrightharpoons} CH_3COO^-(aq) + H_3O^+(aq) } \)


  1.  
  2. The Equilibrium Constant, Kc


    The equilibrium constant Kc is another method of describing a state of equilibrium. The equilibrium constant does not change as the concentrations of the various species change. However, the equilibrium constant will change if the temperature of the system changes.


    The equilibrium constant can be calculated using the equilibrium law expression provided that the equilibrium concentrations of reactants and products are known. The expression can be used to find the equilibrium concentration of one of the reactants or one of the products, provided the Kc and all other concentrations are known.


    The equilibrium law expression relates the concentration of the products relative to the reactants in a chemical system at equilibrium. A, B, C, and D represent equilibrium concentrations and a, b, c, and d represent the coefficients of the balanced equation.





If Kc = 1, then [products] = [reactants].

If Kc > 1, then [products] > [reactants].

If Kc < 1, then [products] < [reactants].


When we write equilibrium law expressions, states are important. Depending on the states of the reactants and products, equilibrium systems are classified as either homogeneous or heterogeneous.



A. Homogeneous Equilibrium Systems


Products and reactants are in the same state of matter (either liquid, gaseous, or aqueous). See the examples listed below.


  1. All substances are liquids.


     

  2. All substances are aqueous.




  3. All substances are gases.






B. Heterogeneous Equilibrium Systems


Heterogeneous systems contain more than one state of matter. In heterogeneous systems, pure substances in solid or liquid state are not included in the equilibrium expression. Only substances in dissolved or gaseous state are included. Pure substances in solid or liquid state are omitted from the equilibrium law expression because their concentrations are essentially fixed. See the examples below:




 Read pages 684 and 685 of the textbook, giving close attention to Communication examples 1 to 4. Notice that, in Communication examples 3 and 4, solid substances are not included in the equilibrium law expression.

Check Your Understanding


Complete Section 15.1 Questions 1 (a, b, e, f), 2 and 4 (a, b) on pages 688 of the textbook. Check your work by clicking the link below.

Page 688 Section 15.1 Question 1 (a, b, e, and f)

1a. Equation \( \mathrm { H_2(g) + Cl_2(g) \leftrightharpoons 2 HCl(g) } \)
Equilibrium Law Expression

1b. Equation \( \mathrm { N_2(g) + 3 H_2(g) \leftrightharpoons 2 HN_3(g) } \)
Equilibrium Law Expression

1e. Equation \( \mathrm { CaCO_3(s) \leftrightharpoons CaO(s) + CO_2(g) } \)
Equilibrium Law Expression

1f. Equation \( \mathrm { Na_2SO_4(aq) + CaCl_2(aq) \leftrightharpoons 2 NaCl(aq) + CaSO_4(s) } \)
Net ionic equations: \( \mathrm { SO_4^{2-}(aq) + Ca^{2+}(aq) \leftrightharpoons CaSO_4(s) } \)
Equilibrium Law Expression

Page 688 Section 15.1 Question 2

Alkenes tend to undergo addition reactions with halogens readily. We can expect that the reaction is stoichiometric, or greater than 99% complete.


Pages 688  Section 15.1 Questions 4a and 4b

4a.

4b.