CH 18: CHEMICAL EQUILIBRIUM

SECTION 18.2 SHIFTING EQUILIBRIUM

REVERSIBLE REACTIONS Reversible Reactions: one in which the conversion of reactants to products and the conversion of products to reactants occur simultaneously.  The double arrow tells you that the reaction is reversible. 2 SO 2 (g) + O 2 (g) ⇄ 2 SO 3 (g)

Chemical Equilibrium: when the rates of the forward and reverse reactions are equal, the reaction has reached a state of balance. At chemical equilibrium, no net change occurs in the actual amounts of the components of the system. The relative concentrations of the reactants and products at equilibrium constitute the equilibrium position of a reaction.

Think of a 2-story mall with an escalator. You have 100 people on the top floor and 50 people on the bottom floor. If 10 people want to move to the top floor, how can this happen while still keeping 100 people on the top and 50 people on the bottom? You will have to have 10 people go down the escalator at the same time as 10 people go up the escalator. People are moving at the same RATE!

FACTORS AFFECTING EQUILIBRIUM: LE CHÂTELIER’S PRINCIPLE Le Châtelier’s Principle: If a stress is applied to a system in dynamic equilibrium, the system changes in a way that relieves the stress. There are three factors that we will talk about that can affect equilibrium.

CONCENTRATION If you add more of a reactant, the reaction goes toward the products. If you take away some of a reactant, the reaction goes toward the reactants. If you add more of a product, the reaction goes toward the reactants. If you take away some of a product, the reaction goes toward the products.

TEMPERATURE If heat is a product and you add heat, the reaction shifts toward the reactants. If heat is a reactant and you add heat, the reaction shifts toward the products. If heat is a product and you take away heat, the reaction shifts toward the products. If heat is a reactant and you take away heat, the reaction shifts toward the reactants.

PRESSURE – ONLY GASES!! If you increase pressure, the reaction shifts towards the side with less moles. If you decrease pressure, the reaction shifts towards the side with more moles.

VOLUME – ONLY GASES!! If you ­­­­­­­­­­­­increase volume, it is the same as decreasing pressure, so the reaction shifts towards the side with more moles. If you ­­­­­­­­­­­­decrease volume, it is the same as increasing pressure, so the reaction shifts towards the side with less moles. Because it is the same as changing pressure, we do not consider this to be a 4 th factor.

CATALYSTS Adding a catalyst has no effect on the equilibrium, it will only help to reach equilibrium faster.

REACTIONS TO COMPLETION A reaction is considered to “go to completion”, when almost all of the ions are removed from the solution. This depends on the solubility of the product formed, and if it is soluble, then on its degree of ionization.

FORMATION OF A GAS Gases are not very soluble, so when a gas is formed and the reaction container is open to the air, the gas will escape and the reaction will go almost to completion.

FORMATION OF A SLIGHTLY IONIZED PRODUCT This occurs with the neutralization reactions of acids and bases. HCl (aq) + NaOH (aq) H 3 O + (aq) + Cl - (aq) + Na + (aq) + OH - (aq)  Na + (aq) + Cl - (aq) + 2H 2 O (l) H 3 O + (aq) + OH - (aq)  2H 2 O (l) Water only slightly ionizes, so it exists as mainly H 2 O molecules.

FORMATION OF A PRECIPITATE If a product is insoluble (a precipitate), then when the product forms, it cannot dissolve to allow the reaction to go in the reverse direction. NaCl (aq) + AgNO 3(aq)  NaNO 3(aq) + AgCl (s)

COMMON ION EFFECT This is the phenomenon in which the addition of an ion common to two solute brings about precipitation or reduced ionization. If you have a NaCl solution and you add HCl acid, the extra Cl - ions from the acid will mix with the Na + ions to form NaCl precipitate.

SECTION 18.1 THE NATURE OF CHEMICAL EQUILIBRIUM

EQUILIBRIUM EXPRESSIONS Equilibrium Constant: K eq is the ratio of product concentrations to reactant concentrations at equilibrium. nA + mB ⇄ xC + yD K eq = [C] x [D] y [A] n [B] m

WRITE EQUILIBRIUM EXPRESSIONS: 1.H 2 + I 2 ↔ 2HI 2.2HgO ↔ 2Hg + O 2 3.2SO 2 + O 2 ↔ 2SO 3 4.N 2 + 3H 2 ↔ 2NH 3

EQUILIBRIUM CONSTANTS To find an Equilibrium Constant, plug in the concentrations of the reactants and products into the equilibrium expression and solve! K eq > 1, products favored at equilibrium K eq < 1, reactants favored at equilibrium K eq does not have any units.

CALCULATING K eq A liter of a gas mixture at equilibrium at 10°C contains mol of N 2 O 4 and mol of NO 2. Write the expression for the equilibrium constant and calculate K eq. N 2 O 4 (g) ⇄ 2NO 2 (g) K eq = [NO 2 ] 2 = (0.030 mol/L) 2 [N 2 O 4 ] = mol/L K eq = 0.20

CALCULATING K eq An equilibrium mixture of N 2, O 2, and NO gases is determined to consist of 6.4 mol/L of N 2, 1.7 mol/L of O 2, and 1.1 mol/L of NO. What is the K eq for this system? N 2 + O 2 ⇄ 2NO K eq = [NO] 2 [N 2 ] x [O 2 ] K eq = (1.1) 2 (6.4) x (1.7) K eq = 0.11