Chapter 14: Chemical Equilibrium CHE 124: General Chemistry II Dr. Jerome Williams, Ph.D. Saint Leo University

Overview Homogeneous Equilibrium – Equilibrium Constants for Gaseous Reactions Heterogeneous Equilibrium Example Problems

Homogeneous Equilibrium Homogeneous Equilibrium: When all reacting species are in the same phase, all reactants and products are included in the expression. Amounts of components are given as molarity or partial pressure of a gas.

4 Equilibrium Constants for Reactions Involving Gases the concentration of a gas in a mixture is proportional to its partial pressure therefore, the equilibrium constant can be expressed as the ratio of the partial pressures of the gases for aA(g) + bB(g)  cC(g) + dD(g) the equilibrium constant expressions are or

K c and K p in calculating K p, the partial pressures are always in atm the values of K p and K c are not necessarily the same – because of the difference in units – K p = K c when  n = 0 the relationship between them is:  n is the difference between the number of moles of reactants and moles of products

Deriving the Relationship between K p and K c

7 Deriving the Relationship Between K p and K c for aA(g) + bB(g)  cC(g) + dD(g) substituting

Ex 14.3 – Find K c for the reaction 2 NO(g) + O 2 (g)  2 NO 2 (g), given K p = 2.2 x 10 25°C K is a unitless number since there are more moles of reactant than product, K c should be larger than K p, and it is K p = 2.2 x KcKcKcKc Check: Solution: Concept Plan: Relationships: Given:Find: KpKp KcKc 2 NO(g) + O 2 (g)  2 NO 2 (g)  n = 2  3 = -1

Practice – Calculate the value of K p or K c for each of the following at 27 °C 2 SO 2 (g) + O 2 (g)  2 SO 3 (g)K c = 8 x N 2 (g) + 2 O 2 (g)  2 NO 2 (g)K p = 3 x 10 −17 9 Tro: Chemistry: A Molecular Approach, 2/e

Heterogeneous Equilibria Pure solids and pure liquids are materials whose concentration doesn’t change during the course of a reaction – its amount can change, but the amount of it in solution doesn’t because it isn’t in solution Because their concentration doesn’t change, solids and liquids are not included in the equilibrium constant expression For the reaction aA(s) + bB(aq)  cC(l) + dD(aq) the equilibrium constant expression is 10 Tro: Chemistry: A Molecular Approach, 2/e

Heterogeneous Equilibria The amount of C is different, but the amounts of CO and CO 2 remain the same. Therefore the amount of C has no effect on the position of equilibrium. 11 Tro: Chemistry: A Molecular Approach, 2/e

HNO 2(aq) + H 2 O (l)  H 3 O + (aq) + NO 2 − (aq) K = 4.6 x 10 −4 Ca(NO 3 ) 2(aq) + H 2 SO 4(aq)  CaSO 4(s) + 2 HNO 3(aq) K = 1 x 10 4 Practice – Write the equilibrium constant expressions, K, and predict the position of equilibrium for the following 12 Tro: Chemistry: A Molecular Approach, 2/e

Example: Homogeneous Equilibrium The following pictures represent mixtures of A molecules (red) and B molecules (blue), which interconvert according to the equation A B. If Mixture (1) is at equilibrium, which of the other mixtures is also at equilibrium?

Example: Homogeneous Equilibrium Write the K p and K c expressions for: 2 N 2 O 5 (g) 4 NO 2 (g) + O 2 (g) The equilibrium concentrations for the reaction between CO and Cl 2 to form carbonyl chloride (phosgene gas) CO(g) + Cl 2 (g) COCl 2 (g) at 74°C are: [CO] = 1.2 x 10 –2 M, [Cl 2 ] = M, and [COCl 2 ] = 0.14 M. Calculate K c and K p.

Example: Homogeneous Equilibrium Methane (CH 4 ) reacts with hydrogen sulfide to yield H 2 and carbon disulfide, a solvent used in manufacturing. What is the value of K p at 1000 K if the partial pressures in an equilibrium mixture at 1000 K are 0.20 atm of CH 4, 0.25 atm of H 2 S, 0.52 atm of CS 2, and 0.10 atm of H 2 ?

Example: Heterogeneous Equilibrium Write the equilibrium equation for each of the following reactions: (a)CO 2 (g) + C(s) 2 CO(g) (b)Hg(l) + Hg 2+ (aq) Hg 2 2+ (aq) (c)2 Fe(s) + 3 H 2 O(g) Fe 2 O 3 (s) + 3 H 2 (g) (d)2 H 2 O(l) 2 H 2 (g) + O 2 (g)