Worksheet#2

Chemical Equilibrium, the Equilibrium Constant (Keq), and the Reaction Quotient (Q)

As a reaction proceeds in a closed system, the concentrations of reactants and products change until, at equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction. At that point there is no net change in the concentrations of reactants or products. For example, for the simple reaction: A = B:

Regardless of the initial concentrations, the concentrations of reactants and products present at equilbrium are related through the equilibrium constant (Keq) as follows: For the reaction: aA + bB ⇌ cC + dD

If all reactants and products are in solution then: Keq = Kc =

(The subscript “eq” indicates “equilibrium” concentration. The subscript “c” stands for “concentration”. The exponents are the coefficients from the balanced equation.)

Important points about equilibrium constants:

1.  Every reaction has a characteristic equilibrium constant that depends only on temperature.

  1. Equilibrium constants have no units.

The concentrations of pure solids or pure liquids do not appear explicitly in the equilibrium constant expression. The concentrations of pure solids and liquids are constant (density divided by molar mass). These constant concentrations are simply incorporated into the equilibrium constant.

  1. The numerical value of the equilibrium constant expresses the tendency for reactants to be converted to products. If Keq > 1 we say that “products are favored”. If Keq < 1 we say that “reactants are favored”. A very large equilibrium constant ( >1000) means that the reaction will nearly go to completion. A very small equilibrium constant (<0.001) means very little product will be present at equilibrium (or that the reverse reaction goes to completion).

4.  A catalyst does NOT change the equilibrium constant. A catalyst increases the rates of the forward and reverse reactions by the same factor.

5.  Changing the temperature DOES change the equilibrium constant. Increasing the temperature increases the equilibrium constant for an endothermic reaction and decreases the equilibrium constant for an exothermic reaction. Increasing the temperature shifts equilibrium in the endothermic direction. (See optional document on my website for details.)

Example 1: Calculate the equilibrium constant Kc at 25 oC for the reaction

2 NOCl(g) « 2 NO(g) + Cl2(g)

using the following information. In one experiment 2.00 mol of NOCl is placed in a 1.00 -L flask, and the concentration of NO after equilibrium is achieved is 0.66 mol/L.

Method:

(a) Calculate the initial concentration of NOCl from the information given. C = n/v

(b) Form an equilibrium table and fill in all known quantities. Represent unknown quantities with variables. (c) Form the Kc equation and fill in all known information. Calculate Kc.

Solution:

Step(a): The initial concentration [NOCl]0 =

Step(b): The equilibrium table is:

Balanced Equation / 2 NOCl(g) « / 2 NO(g) + / Cl2(g)
Initial Concentrations (M) / 2.00 / 0 / 0
Change (M) / - 2x / 2x / x
Equilibrium Concentrations (M) / (2.00 - 2x) / 2x = 0.66 / x

From the table we have: [NO] = 2x = 0.66 M thus x = 0.33 M

[NOCl] = (2.00 - 2x) M = (2.00 - 2(0.33)) M = 1.34 M

[Cl2] = x = 0.33 M

Step(c): We now have the information we need to calculate Kc:

Writing Equilibrium Constant Expressions: Do the following:

1. H2 (g) + CO2 (g) ↔ CO (g) + H2O (g)

2. N2 (g) + 3 H2 (g) ↔ 2 NH3 (g)

3. HC2H3O2 (aq) ↔ H+ (aq) + C2H3O2¯ (aq)

4. HNO2 (aq) ↔ H+ (aq) + NO2¯ (aq)

5. HC2H3O2 (aq) + H2O (l) ↔ > H3O + (aq) + C2H3O2¯ (aq)

6. HNO2 (aq) + H2O (l) ↔ H3O + (aq) + NO2¯ (aq)

7. AgBr (s) ↔ Ag+ (aq) + Br¯ (aq)

8. FeS (s) ↔ Fe2+ (aq) + S2¯ (aq)

9. PCl5 (g) ↔ Cl2 (g) + PCl3 (g)

10. 4 NH3 (g) + 5 O2 (g) ↔ 6 H2O (g) + 4 NO (g)

11. Choose the correct answer in the brackets.

a. For reaction #4, K = 160. This reaction favors ( reactants / products )

b. For reaction #9, K = 1.1x10-11. This equilibrium lies to the ( left / right )

c. For reaction #3, K = 2.3 x104. ( The reaction is nearly complete / the reaction barely occurs )

d. For reaction #6, K = 7.1x10-4. Which would there be more of at equilibrium? ( HNO2 / H+ and NO2¯ )

e. For reaction #11, K = 0.035. Which reaction is favored? ( forward / reverse )

12. Find the value of the equilibrium constant:

H2(g) + CO2(g) ↔ CO(g) + H2O(g)

The equilibrium concentrations are found to be:

[H2] = 0.060 M;[CO2] = 0.075 M;[CO] = 0.100 M;[H2O] = 0.025M ANS: 0.56

13. Find the value of the equilibrium constant for H2(g) + I2(g) ↔ 2 HI (g) given equilibrium concentrations:

[H2] = 1.2x10-3 M; [I2] = 3.4x10-3 M; [HI] = 4.0x10-8 M.

ANS: 3.9x10 -10

14. Find the value of the equilibrium constant for N2(g)+ 3H2(g)↔2 NH3(g) given equilibrium concentrations:

[H2] = 0.1 M [N2] =0.06 M; [NH3] = 6.0 M.

ANS: 6.0 x10 5

15. Find the equilibrium value of [CO] if Kc=14.5 : CO (g) + 2H2 (g) ↔ CH3OH (g)

Equilibrium concentrations: [H2] = 0.322 M and [CH3OH] = 1.56 M.

ANS: 1.04

16. For the reaction A (g) ↔ B (g) + 2 C (g), complete the following table and find (Kc).

ANSWER = 0.023

A / B / 2 C
initial / 0.75 / 0 / 0
change due to reaction
equilibrium / 0.30

17. For the reaction 2 A (g) + 3 B (g) ↔ C (g) + 2 D (g), complete the following table and find (K)

ANSWER = 2.03

2 A / 3 B / C / 2 D
initial / 0.40 / 0.50 / 0 / 0
change due to reaction
equilibrium / 0.26

18. For the reaction C (g) + 3 D (g) ↔ A (g) + 2 B (g), complete the following table and find (Kc)

ANSWER = 146.3

C / 3 D / A / 2 B
initial / 0.50 / 0.60 / 1.00 / 1.00
change due to reaction
equilibrium conc (M) / 0.40

19. Find the value of the equilibrium constant for N2O4 (g) ↔ 2 NO2 (g) given initial concentrations: [N2O4]o = 1.00 atm; [NO2]o = 0; the equilibrium conc. of NO2 is 0.40 M. ANSWER k= 0.20

20. Find the value of the equilibrium constant: CO (g) + 2H2 (g) ↔ CH3OH (g) Initial values: [CO]o = 1.75 M; [H2]o = 0.80 M; [CH3OH]o = 0.65 M. The equilibrium value of CO is 1.60 M ANSWER k = 2.0