1. Equilibrium Chapter 13

2. What Is It? The state where concentrations of all reactants and products remain constant with time. At the molecular level, the reaction continues. Macroscopically, the reaction appears static.

3. Why Is It? Reaction rate depends on concentration. Collision Theory says more collisions = faster reactions. As the reaction progresses, the concentration of products begins to increase.

4. Nothing appears to change because the rates of reaction are so naturally slow. A catalyst is used to begin the process for commercial production of ammonia. Haber applied Le Chatelier’s Principle to maximize the forward reaction. Equilibrium Position is Determined By: Initial Concentrations Energies of reactants and products Organization (disorder) of reactants and products N 2(g) + 3H 2(g) 2NH 3(g)

5. Le Chatelier’s Principle When a stress is applied to a system at equilibrium, the reaction shifts to relieve the stress.

6. What stresses out a Reaction? Heat (heat is measured in kJ or kcal) Pressure (related to # of moles on each side of the reaction) Concentration (how much of each component is added) …When any of these increase or decrease from original conditions, that is stress.

7. The Equilibrium Constant The equilibrium constant is represented by K Where: wA + xB yC + zD

8. Try Me! Write the equilibrium expression for the following reaction: 4 NH 3 + 7O 2 4NO 2 + 6H 2 O Calculate the value of the equilibrium constant if the concentrations of the reactants and products are as follows: [NH 3 ] = 3.1 x 10 -2 mol/L [O 2 ] = 5.4 x 10 -2 mol/L [NO 2 ] =3.1 x 10 -2 mol/L [H 2 O] = 4.7 x 10 -2 mol/L

9. Number of equilibrium positions availableNumber of values for K Equilibrium Position A set of concentrations that indicate whether products or reactants dominate while the rxn is at equilibrium.

10. Equilibrium and Pressures Pressure and concentration are interchangeable K Kp vs In General: Kp = K(RT)  n Figure out the relationship between K and Kp for the Haber Process. Do Now!  n =  products -  reactants PV = nRT P = (n/V)RT n/V = concentration

11. Heterogeneous Equilibria Pure liquids and pure solids are not included in the equilibrium expression for a reaction. Concentrations of PURE liquids and solids cannot change.

12. Applications of the Equilibrium Constant The Equilibrium constant gives information that will allow us to… –Decide how likely it is that the reaction will occur. –Determine if a reaction is at equilibrium given a set of concentrations. –Determine which direction the reaction must shift in order to reach an equilibrium position. If K is greater than 1 The reaction is much More likely to occur Spontaneously (the equilibrium lies farthest to the right) If K is greater than 1 The reaction is much More likely to occur Spontaneously (the equilibrium lies farthest to the right) If K is very small The reaction is not likely to be spontaneous (the equilibrium is near the reactants) If K is very small The reaction is not likely to be spontaneous (the equilibrium is near the reactants) Spontaneous Does Not Mean Fast! Spontaneous Does Not Mean Fast!

13. The Reaction Quotient Obtained by replacing initial concentrations into the concentrations of the equilibrium expression. Three possible cases: Q > K Shift Left Q = K No Shift Q < K Shift Right

14. Try Me!! For the synthesis of ammonia at 500 o C, the equilibrium constant is 6.0 x 10 -2. Predict the direction in which the system will shift to reach equilibrium: [NH 3 ] o = 1.0 x 10 -3 M [N 2 ] o = 1.0 x 10 -5 M [H 2 ] o = 2.0 x 10 -3 M Q = 1. 3 x 1 0 7 S h i f t t o t h e L e f t

15. Try Me Again and Again! Same problem, different conditions: [NH 3 ] o = 2.0 x 10 -4 M [N 2 ] o = 1.5 x 10 -5 M [H 2 ] o = 3.54 x 10 -1 M [NH 3 ] o = 1.0 x 10 -4 M [N 2 ] o = 5.0 M [H 2 ] o = 1.0 x 10 -2 M Q = 2. 0 x 1 0 - 3 S h i f t t o t h e R i g h t Q = 6. 0 1 x 1 0 - 2 N o S h i f t

16. Solving for Concentrations and Pressures Several Types of problems and Solving methods. Plug and Chug ICE ICE with Stoichiometry ICE with the quadratic equation

17. Type 1: Plug and Chug Consider an experiment in which gaseous N 2 O 4 was placed in a flask and allowed to reach equilibrium at a temperature where K p = 0.133. At equilibrium, the pressure of N 2 O 4 was found to be 2.71 atm. Calculate the equilibrium pressure of NO 2. N 2 O 4(g) 2NO 2(g)

18. Type 2: Initial, Change,Equilibrium At a certain temperature a 1.00L flask initially contained 0.298 mol PCl 3(g) and 8.70 x10 -3 mol PCl 5(g). After the system had reached equilibrium, 2.00 x10 -3 mol Cl 2(g) was found in the flask. Gaseous PCl 5 decomposes according to the reaction PCl 5(g) PCl 3(g) + Cl 2(g) Calculate the concentrations of all species and the value of k.

19. Type 3: ICE with Stoichiometry Hydrofluoric acid vapor decomposes into its elemental components hydrogen and fluorine. At 700K the equilibrium constant is 5.10. Calculate the equilibrium concentrations of all species if 1.00 mol of each component is mixed in a 1.00L flask.

20. Type 4: ICE with Quadratic A 1.00L flask is filled with 1.0 mol H 2 gas and 2.0 mol I 2 gas at 448 o C. The value of the equilibrium constant K c for the reaction H 2 + I 2 2HI at 448 o C is 50.5. What are the equilibrium concentrations of H 2, I 2 and HI in mol/L?