Topic Extension Equilibrium AcidBase Equilibrium Solubility Equilibrium ComplexIons

Topic Extension Equilibrium Acid-Base Equilibrium Solubility Equilibrium Complex-Ions Equilibrium Qualitative Analysis

Chemical Reactions Two Types A. COMPLETE Reactants Products time B. EQUILIBRIUM Reactants Products time

Chapter 17: Equilibrium The Extent of Chemical Reactions 17. 1 The Dynamic Nature of the Equilibrium State 17. 2 The Reaction Quotient and the Equilibrium Constant 17. 3 Expressing Equilibria with Pressure Units: Relation Between Kc and Kp 17. 4 Reaction Direction: Comparing Q and K 17. 5 How to Solve Equilibrium Problems 17. 6 Reaction Conditions and the Equilibrium State: Le Châtelier’s Principle

Reaching Equilibrium on the Macroscopic and Molecular Level 2 NO 2 (g) Fig. 17. 1 N 2 O 4 (g)

Law of Mass Action • At a given T, a chemical system reaches a state in which the ratio of reactants and products has a constant value. • Ratio called REACTION QUOTIENT, Q Q = [final] [initial] = [Products]p [ Reactants]R At equilibrium Q = K

Writing the Mass-Action Expression for Equilibrium For the general reaction: a. A + b. B Rate forward c [D]d [C] Kc = [A]a [B]b c. C + d. D = Rate reverse When not at equilibrium use Q [ ] = Molarity Example: The Haber process for ammonia production: N 2 (g) + 3 H 2 (g) 2 NH 3 (g) K =

Example: 3. 0 - M o l 2. 0 e s 1. 0 - 0 Write the balanced chemical reaction Write the equilibrium constant, Kc Moles H 2 Equilibrium amounts Moles CO Moles CH 4 = moles H 2 O time

Properties of Kc A. Can be based on: Molarity = [ ] identified as KC Partial pressure = atm identified as Kp details next slide B. Constant for given temperature C. Kc = 1 Kc details on next slide reverse D. Homogeneous or Heterogeneous Included in K expression: (aq) and (g) species only No solids, liquids in expression - M is constant.

Expressing K with Pressure Units For gases, PV=n. RT can be rearranged to give: P = n RT V n P n or: = Since = Molarity, and R is a constant if we V RT V keep the temperature constant then the molar concentration is directly proportional to the pressure. Therefore for an equilibrium between gaseous compounds we can express the reaction quotient in terms of partial pressures. For: Qp = 2 NO(g) + O 2 (g) P 2 NO 2 P 2 NO x PO 2 2 NO 2 (g) If there is no change in the number of moles of reactants and products then n = 0 then Kc = Kp , or if there is a change in the number of moles of reactants or products then: Kp = Kc(RT) ngas

The Form of K for a Forward and Reverse Reaction The production of sulfuric acid depends upon the conversion of sulfur dioxide to sulfuric trioxide before the sulfur trioxide is reacted with water to make the sulfuric acid. 2 SO 2 (g) + O 2 (g) 2 SO 3 (g) 2 [SO ] 3 Qc(fwd) = [SO 2]2[O 2] For the reverse reaction: Qc(rev) = and: 2 SO 3 (g) [SO 2]2[O 2] [SO 3]2 Kc(fwd) = 1 Kc(rev) = = 1 Qc(fwd) 1 261 2 SO 2 (g) + O 2 (g) at 1000 K Kc(fwd) = 261 = 3. 83 x 10 -3

Applications Of Equilibrium Qualitative Value of KC large number favors ______ small number favors ______ Examples: Kc = 5. 79 x 10 -7 Kc = 0. 00469 Kc = 2. 37 x 1012 Favors ______________

Quantitative Example Problem 1: Calculating Kc from concentrations data. Class Practice Problem #2 Example Problem 2: Find equilibrium concentration of one species given Kc and the other concentrations. Class Practice Problem #3

Example Problem 3: Predict extent of reaction Use reaction quotient - Qc Used when we do not know if the system is at equilibrium. Problem solving strategy Given: Kc and [ ] step 1: calculate Qc using concentrations step 2: compare Qc with Kc [see below] If Qc Kc reaction need to go toward products If Qc Kc reaction need to go toward reactants If Qc = Kc reaction is at equilibrium

Example Problem 4: I. C. E. • Problem: Given the that the reaction to form HF from molecular hydrogen and fluorine has a reaction quotient of 115 at a certain temperature. If 3. 000 mol of each component is added to a 1. 500 L flask, calculate the equilibrium concentrations of each species. • Write the balanced reaction and Kc expression Plan: 1. Calculate the concentrations of each component. 2. Figure out the changes and express in terms of x 3. Express the final or equil. Conc. In terms of x 4. Solve the equilibrium equation for x 5. Determine the actual concentration. Concentration (M) Initial Change Equilibrium H 2 + F 2 2 HF

Changes Qualitative Quantitative 1. Calculate actual concentrations after shifts. 2. Calculate new Kc after temperature change. Le Chatelier’s Principle ln K 2 K 1 - H = R 1 T 2 1 - T 1 R = 8. 31 J/mol-K

The Effect of a Change in Concentration Given an equilibrium equation such as : CH 4 (g) + NH 3 (g) Add NH 3 CH 4 (g) + NH 3 (g) HCN(g) + 3 H 2 (g) Forces equilibrium to produce more product. HCN(g) + 3 H 2 (g) Remove NH 3 Forces the reaction equilibrium to go back to the left and produce more of the reactants.

Summary: Effect of Various Disturbances on an Equilibrium System Disturbance Net Direction of Reaction Effect on Value of K Concentration Increase [reactant] Toward formation of product Decrease [reactant] Toward formation of reactant Pressure (volume) Increase P Decrease P Toward formation of lower amount (mol) of gas Toward formation of higher amount (mol) of gas None

Disturbance Temperature Increase T Decrease T Catalyst added Net Direction of Reaction Effect on Value of K Toward absorption of heat K Increases if H 0 rxn> 0 K Decreases if H 0 rxn< 0 Toward release of heat K Increases if H 0 rxn< 0 K Decreases if H 0 rxn> 0 None; rates of forward and reverse reactions increase equally None