Chem 1B Dr. White 1 Chapter 13: Chemical Equilibrium Outline Chemical Equilibrium. A. Definition:

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1 Chem 1B Dr. White 1 Chapter 13: Chemical Equilibrium Outline Chemical Equilibrium A. Definition: B. Consider: N 2 O 4 (g, colorless) 2NO 2 (g, brown) C. 3 Main Characteristics of Equilibrium The Equilibrium Constant (K) A. K 1. Law of Mass Action 2. General Description: ja (aq) + kb (aq) <-> lc (aq) + md (aq)

2 Chem 1B Dr. White 2 a. K b. K p i. CO (g) + 3H 2 (g) CH 4 (g) + H 2 O(g) ii. K p = iii. K = c. Heterogeneous Equilibrium i. S 8 (s) + 24F 2 (g) 8SF 6 (g) 3. Lecture Example: Write the equilibrium constant expression (K) for the following reactions. a. N 2 (g) + 3H 2 (g) 2NH 3 (g) b. 3 H 2 O (l) + H 3 PO 4 (aq) 3H 3 O + (aq) + PO 4 3- (aq) c. AgCl (s) Ag + (aq) + Cl - (aq)

3 Chem 1B Dr. White 3 4. Lecture Example: Calculate K for the reaction in reaction a above if the equilibrium concentrations were determined at 127 C to be: [NH 3 ] = 3.50 M, [N 2 ] = 1.25 M, and [H 2 ] = 1.50 M B. Equilibrium Positions Consider: N 2 O 4 (g, colorless) 2NO 2 (g, brown) C. More about K

4 Chem 1B Dr. White 4 a. K<1 b. K>1 c. Intermediate K D. Converting between K and K p General equation: Example: The equilibrium constant K for the following reaction at 2127 o C is 2.5 x What is K p at this temperature? N 2 (g) + O 2 (g) 2 NO(g) Example: The K p value for the following reaction at 25 C is 2.2 x What is K at this temperature? 2NO (g) + O 2 (g) 2NO 2 (g)

5 Chem 1B Dr. White Applications of the Equilibrium Constant A. Predicting the Reaction Direction 1. The Reaction Quotient (Q) Lecture Example: Write the Q expression for the following reaction: 2 SO 2 (g) + O 2 (g) 2 SO 3 (g)

6 Chem 1B Dr. White 6 Lecture Example: For the reaction, COCl2 CO + Cl2, K = 8.3 X 10 4 (at 360 C). If at a given time in the reaction, [COCl2] = 0.18 M, [CO] = 0.35 M, [Cl2] = 0.12 M which direction will the reaction proceed? Lecture Example: A 50.0 L reaction vessel at 400 C contains 1.00 mol N 2, 3.00 mol H 2, and mol NH 3. Will more NH 3 be formed or will it dissociate when the mixture goes to equilibrium at 400 C. N 2 (g) + 3 H 2 (g) 2 NH 3 (g) K = at 400 C B. Calculating Equilibrium Pressures and Concentrations from Initial Conditions 1. Use the ICE Box Method a. I: b. C: c. E:

7 Chem 1B Dr. White 7 2. ICE Box Lecture Examples: a. CALCULATING K FROM INITIAL & EQUILIBRIUM CONCENTRATIONS In a 2.00 L vessel, 3.8 mol CO are combined with 12.0 mol H2. At equilibrium, there are 0.60 mol CO. What are the concentrations of the other substances at equilibrium and what is the equilibrium constant? CO(g) + 3 H2(g) CH4(g) + H2O(g)

8 Chem 1B Dr. White 8 b. DETERMINING EQUILIBRIUM CONCENTRATIONS FROM INITIAL CONCENTRATIONS. ( Perfect Square ) The equilibrium constant, K, for the reaction of H 2 gas with I 2 gas to form HI gas is 57.0 at 700 K. If 1.00 mol of H 2 is allowed to react with 1.00 mol I 2 in a 10.0 L vessel at 700 K, what are the concentrations of H 2, I 2, and HI at equilibrium? c. DETERMINING EQUILIBRIUM PRESSURES FROM INITIAL PRESSURES. ( Perfect Square ) CO gas and water vapor react to form CO 2 gas and H 2 gas. At 1000 C the K p for this reaction is If atm of CO and water vapor are placed in a vessel what are the pressures of each substance at equilibrium?

9 Chem 1B Dr. White 9 d. DETERMINING EQUILIBRIUM CONCENTRATIONS FROM INITIAL CONCENTRATIONS. (Starting with reactants and products) What are the equilibrium concentrations for the reaction below when you start with mol each of CO2 and H2 and mol each of CO and H2O in 1.00 L vessel? CO2(g) + H2(g) CO(g) + H2O(g) K = e. DETERMINING EQUILIBRIUM CONCENTRATIONS FROM INITIAL CONCENTRATIONS. ( Quadratic ) The equilibrium constant K for the reaction of H 2 gas with I 2 gas to form HI gas is 49.7 at 458 C. If 1.00 mol of H 2 is allowed to react with 2.00 mol I 2 in a 1.00 L vessel at 700 K, what are the concentrations of H 2, I 2, and HI at equilibrium?

10 Chem 1B Dr. White 10 f. DETERMINING EQUILIBRIUM CONCENTRATIONS FROM INITIAL CONCENTRATIONS. ( Simplification ) Calculate the equilibrium concentration of silver ion in a 1.00 L of solution with mol AgCl (s) and mol Cl - in a solution with the equilibrium reaction of: AgCl (s) Ag + (aq) + Cl - (aq) K = 1.8 x 10-10

11 Chem 1B Dr. White Le Chatelier s Principle A. Concentration Changes Example: Consider the following reaction: CH 3 OH(g) + O 2 (g) HCOOH(g) + H 2 O(g) a. What direction does the equilibrium shift if more oxygen is added? b. What direction does the equilibrium shift is water is removed? Example: Consider the following reaction: AgCl(s) Ag + (aq) + Cl (aq) a. What happens to the concentration of the ions if AgCl is added? b. What happens to the concentration of the ions if Ag + is added? B. Pressure Changes

12 Chem 1B Dr. White 12 Lecture Example: How does a decrease in volume affect the concentration of the first reactant in the following reactions? a. C 2 H 4 (g) + H 2 (g) C 2 H 6 (g) b. XeF 6 (g) Xe(g) + 3 F 2 (g) c. C(s) + 2 F 2 (g) CF 4 (g) d. H 2 S(g) + Hg(l) HgS(s) + H 2 (g) D. Temperature Changes 1. Endothermic:

13 Chem 1B Dr. White Exothermic: Lecture Example: For the following reactions, how does the concentration of the first reactant change if the temperature decreases? a. 2 NOCl (g) 2 NO (g) + Cl 2 (g) (endothermic) b. C 2 H 4 (g) + H 2 (g) C 2 H 6 (g) (ΔH<0) 13.2 (Revisited) The Equilibrium Constant A. Manipulating K 1. Direction of reactions a. A B K a = b. B A K b = c. Therefore,

14 Chem 1B Dr. White Reactions as a sum of steps a. A B K a = b. B C K b = c. A C K c = d. K overall = 3. Coefficients multiplied by n a. aa + bb cc + dd K a = b. n (aa + bb cc + dd) K b = c. Therefore, d. SO 2 (g) + ½ O 2 (g) SO 3 (g) K = 56 at 900K 2SO 2 (g) + O 2 (g) 2SO 3 (g) K =

15 Chem 1B Dr. White Lecture Examples: Given: CO (g) + 3 H 2 (g) CH 4 (g) + H 2 O (g) K = 3.92 at 1200 K CH 4 (g) + 2 H 2 S (g) CS 2 (g) + 4 H 2 (g) K = 3.3 x 10 4 at 1200 K a. Determine K at 1200 K for: CO (g) + 2H 2 S (g) CS 2 (g) + H 2 O (g) + H 2 (g) b. Determine K c at 1200 K for: 1/3CO (g) + H 2 (g) 1/3 CH 4 (g) + 1/3 H 2 O (g) c. Determine K c at 1200 K for: 2 CS 2 (g) + 8 H 2 (g) 2 CH 4 (g) + 4 H 2 S (g)