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

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

3. 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 K c and K p 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

4. Reaching Equilibrium on the Macroscopic and Molecular Level 2 NO 2 (g) N 2 O 4 (g) Fig. 17.1

5. 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 [initial] [ Reactants] R [final] [Products] p = Q = At equilibrium Q = K

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

7. 3.0 - 2.0 - 1.0 - 0 MolesMoles time Moles H 2 Moles CO Moles CH 4 = moles H 2 O Equilibrium amounts Example: Write the balanced chemical reaction Write the equilibrium constant, K c

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

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

10. 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) Q c(fwd) = [SO 3 ] 2 [SO 2 ] 2 [O 2 ] 2 SO 3 (g) 2 SO 2 (g) + O 2 (g) For the reverse reaction: Q c(rev) = = [SO 2 ] 2 [O 2 ] [SO 3 ] 2 1 Q c(fwd) and: K c(fwd) = = = 3.83 x 10 -3 at 1000K K c(fwd) = 261 1 K c(rev) 1 261

11. Applications Of Equilibrium Qualitative Value of K C large number favors ____________ small number favors ____________ Examples: Favors K c = 5.79 x 10 -7 ___________________ K c = 0.00469 _____________ K c = 2.37 x 10 12 ______________

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

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

14. 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 K c 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) H 2 + F 2  2 HF Initial Change Equilibrium

15. Changes Qualitative Le Chatelier’s Principle Quantitative 1. Calculate actual concentrations after shifts. 2. Calculate new K c after temperature change. ln = - K 1 RT 2 T 1 K 2 -  H 1 1 R = 8.31 J/mol-K

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

17. 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 None Decrease [reactant] Toward formation of reactant None Pressure (volume) Increase P Toward formation of lower amount (mol) of gas None Decrease P Toward formation of higher amount (mol) of gas None

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