Ch 18: Chemical Equilibrium

Section 18.2 Shifting Equilibrium

Standards and Objectives Students know how to use LeChatelier’s principle to predict the effect of changes in concentration, temperature, and pressure. Students know equilibrium is established when forward and reverse reaction rates are equal. Objective We will discuss the factors that disturb equilibrium, discuss reactions that go to equilibrium, & describe the common ion effect.

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 SO2(g) + O2(g) ⇄ 2 SO3(g)

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

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.

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 you add heat (a product) the reaction shifts to the reactants If you take away heat (a product), the reaction shifts toward the products. 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.

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 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) + AgNO3(aq)  NaNO3(aq) + AgCl(s) Na+(aq) + Cl-(aq) + Ag+(aq) + NO3-(aq)  Na+(aq) + NO3-(aq) + AgCl(s)

Formation of a Slightly Ionized Product This occurs with the neutralization reactions of acids and bases. HCl(aq) + NaOH(aq) H3O+(aq) + Cl-(aq) + Na+(aq) + OH-(aq)  Na+(aq) + Cl-(aq) + 2H2O(l) H3O+(aq) + OH-(aq)  2H2O(l) Water only slightly ionizes, so it exists as mainly H2O molecules.

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.

Homework 18.2 pg 604 #1-5

Section 18.1 The Nature of Chemical Equilibrium

Standards and Objectives Standard 9.c. Students know how to write and calculate an equilibrium constant expression for a reaction. Objective We will write chemical equilibrium expressions and carry out calculations involving them.

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

Write Equilibrium Expressions: H2 + I2 ↔ 2HI 2HgO ↔ 2Hg + O2 2SO2 + O2 ↔ 2SO3 N2 + 3H2 ↔ 2NH3

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

Calculating Keq A liter of a gas mixture at equilibrium at 10°C contains 0.0045 mol of N2O4 and 0.030 mol of NO2. Write the expression for the equilibrium constant and calculate Keq. N2O4(g) ⇄ 2NO2(g) Keq = [NO2]2 = (0.030 mol/L)2 [N2O4] = 0.0045 mol/L Keq = 0.20

Homework 18.1 pg 595 #1, 3, 6-9