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Gaseous Chemical Equilibrium. The Dynamic Nature of Equilibrium A. What is equilibrium? a state of balance; no net change in a dynamic process.

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Presentation on theme: "Gaseous Chemical Equilibrium. The Dynamic Nature of Equilibrium A. What is equilibrium? a state of balance; no net change in a dynamic process."— Presentation transcript:

1 Gaseous Chemical Equilibrium

2 The Dynamic Nature of Equilibrium A. What is equilibrium? a state of balance; no net change in a dynamic process

3 1. Chemical equilibrium No net change with the total amount of reactants and products remaining constant, while the reaction continues Eek = equilibrium

4 2. General Characteristics Double Headed Arrow Beginning of Rxn  Form lots of products (before eek is established) Moments Later  Forming both products and reactants (eek) ** Not Necessarily equal proportions of both sides of the reaction!

5 B. Dynamic Equilibrium - Characteristics two opposing processes occur at exactly the same rate No net change Dynamic Eek will not occur as soon as the reversible reaction begins, it takes time.

6 C. Graph The N 2 O 4  2NO 2 Equilibrium System

7 D. The 2NO 2  N 2 O 4 Equilibrium System 1. Description Dynamic, reversible, no net change 2. Equilibrium Conditions Specific for a Reaction (will be unique) MUST have the Balanced Written Equation Coefficients will matter! Temperature Dependent CONSTANT regardless of concentration

8 II.The Equilibrium Expression, K eq (Equilibrium Law or Law of Mass-Action) A. Writing Expressions for K eq aA + bB cC + dD K eq = [C] c [D] d _ for aqueous solutions use [A] a [B] b Molarities

9 The Equilibrium Expression Example: Write the K eq for: 4NH 3 (g) + 5O 2 (g)  4NO(g) + 6H 2 O(g) N 2 (g) + 3H 2 (g)  2NH 3 (g)

10 B. Characteristics of K eq 1. Independent of : Pure solids and pure liquids as long as some of the substance is present 2. Dependent on: Gaseous substances and solutions (aqueous)

11 C. The meaning of K eq 1. Keq = 1; reactants and products are present in equal amounts at eek 2. Keq > 1; products are present in greater proportion at eek 3. Keq < 1; reactants are present in greater proportion at eek

12 4. Examples: N 2 (g) + O 2 (g)  2 NO(g) K eq =55 N 2 (g) + 3H 2 (g)  2NH 3 (g) K eq =5x10 -6

13 III. Modifying Equilibrium Constant Expressions A. Reversible reactions 1. Rule: The Keq value for the reverse of a reaction will be the reciprocal

14 III. Modifying Equilibrium Constant Expressions Example: write the Keq for the following equation and its reverse A(g) + B(g)  C(g)

15 III. Modifying Equilibrium Constant Expressions B. Summation of Reactions and modifying coefficients Rule: the Keq of summed reactions will be the product of the Keq’s; the factor you adjust coefficients by becomes the power

16 Example: A(g) + B(g)  C(g)Keq=3 A(g) + B(g)  D(g)Keq=6 What is the Keq for C(g)  D(g)

17 Example: ½ N 2 (g) +1/2O 2 (g)  NO(g) K eq =6.9x10 -16 NO 2 (g)  NO(g) + ½ O 2 (g) K eq =6.7x10 -7 N 2 O 4 (g)  2NO 2 (g) K eq =0.15 What is the K eq for N 2(g) +2O 2(g)  N 2 O 4(g)

18 C. Heterogeneous Systems Rule: solid and liquid substances = 1 Example: Write the K eq for: Zn (s) + 2H + (aq)  Zn 2+ (aq) + H 2(g) CaCO 3(s)  CaO (s) + CO 2(g)

19 IV. Determination of K A. Calculating K eq from experimental values. Write the expression for the reversible reaction between solid ammonium chloride and gaseous products, hydrogen chloride and ammonia (NH 3 ). At equilibrium in a 1 liter container, the following amounts are present: 12.0 mol ammonium chloride, 3.0 mol of ammonia and 5.0 mol of hydrogen chloride Determine the K eq.

20 B. Example When 4.29 moles of PCl 3 (g) and 4.29 moles of Cl 2 (g) are placed in a 1.00 Liter container at 250 o C, the following equilibrium is established: PCl 3 (g) + Cl 2 (g)  PCl 5 (g) The equilibrium concentration of phosphorus pentachloride is 2.59 mol/L. What are the equilibrium concentrations of the other two gases? Calculate K eq for the above reaction system.

21 C. Example: Consider the equilibrium system: 2NO(g) + Br 2 (g)  2NOBr(g) At a given temperature, 1.6 mol of NO and 1.6 mol of Br 2 are added to a 1.00 Liter flask and the equilibrium concentration of NOBr is found to be 0.53 M. Calculate the equilibrium concentrations of the other 2 gases and the value of K eq.

22 B. IRE Problems with unknown equilibrium concentrations Description of problems Define both the R and E lines in terms of “x”

23 2. Example: Carbon dioxide and hydrogen gas at concentrations of 1.00 M each are introduced into a container and the following system is established: CO 2 (g) + H 2 (g)  CO(g) + H 2 O(g) K c = 0.64 What are the equilibrium concentrations of all the species?

24 3. Example: For the system: H 2 (g) + I 2 (g)  2HI(g) at 425 o C, K = 55.5. If 2.5 atm of hydrogen and 2.5 atm of iodine are placed in a vessel and heated to 425 o C and the system reaches equilibrium, what are the pressures of all species?

25 14.4 Qualitative Treatment of Equilibrium: Le Chatelier A. Statement of Le Chatelier’s Principle “When a system is stressed (changes in concentration, temp, gas pressure, or volume of container), the system will respond by attaining new equilibrium conditions that counteract the change”

26 14.4 Qualitative Treatment of Equilibrium: Le Chatelier B. For the system, N 2 O 4 (g)  2NO 2 (g)  H = 57.2 kJ 1. Changes in amounts of species a. Adding or removing reactant Add Reactant  Eek shifts toward products (use it up) Remove Reactant  Eek shifts toward reactants (make more)

27 14.4 Qualitative Treatment of Equilibrium: Le Chatelier b. Adding or removing product Add Product  Eek shifts toward reactants (use it up) Remove Product  Eek shifts toward products (make more) c. Adding or removing pure solid or solvent Has NO effect on equilibrium (no shift) d. Adding inert substance (inert = non-reactive) Has NO effect on equilibrium (no shift)

28 14.4 Qualitative Treatment of Equilibrium: Le Chatelier 2. Changes in pressure Pressure increased  Eek shifts in the direction producing the smaller number of moles of gas Pressure decreased  Eek shifts in the direction producing the larger number of moles of gas Moles of gas equal on both sides = pressure does not effect Eek.

29 14.4 Qualitative Treatment of Equilibrium: Le Chatelier 3. Changes in volume Decreases in volume = increase in pressure Increases in volume = decrease in pressure

30 14.4 Qualitative Treatment of Equilibrium: Le Chatelier 4. Changes in temperature Increase Temp  Eek shifts in the direction of the endothermic rxn; away from heat Decrease Temp  Eek shifts in the direction of the exothermic rxn; toward heat

31 14.4 Qualitative Treatment of Equilibrium: Le Chatelier Example 14.12 List all of the ways to shift the equilibrium of the following systems to the right: a. N 2 (g) + 3H 2 (g)  2NH 3 (g)

32 14.4 Qualitative Treatment of Equilibrium: Le Chatelier b. CO(g) + H 2 O(g)  CO 2 (g) + H 2 (g)

33 End of Chapter 14!


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