CH1810-Lecture #8 Chemical Equilibrium: LeChatlier s Principle and Calculations with K eq

Transcription

1 CH1810-Lecture #8 Chemical Equilibrium: LeChatlier s Principle and Calculations with K eq

2 LeChatlier s Principle A system at equilibrium responds to a stress in such a way that it relieves that stress. At equilibrium, the concentration of all reactants and products remain the same. If conditions change, concentrations of all components will change until equilibrium is re-established Factors that will change the relative rates of forward/reverse reactions, or change the value of Q compared to K, will cause a shift in the position of equilibrium.

3 Concentration Stress Adding a reactant will decrease the amount of other reactants and increase the amount of products until a new equilibrium is reached. Removing a product will decrease the amount of other products and reactants until a new equilibrium is reached.

4 A + B C + D Add A A + B C + D Equilibrium Adjusts A + B C + D

5 Concentration Stress A + B C + D Remove C A + B C + D Equilibrium Adjusts A + B C + D Equilibrium shifts toward the side with removed chemicals.

6 Effects of Pressure and Volume For a gas phase reaction, increasing the pressure (or decreasing the volume) shifts the equilibrium in the direction toward the production of fewer gas particles.

7 Effects of Pressure and Volume Consider the equilibrium: 2 NO2(g) N2O4(g) Changing volumes will change partial pressures, change Q relative to K. For example, a decrease volume by factor of 2 will increase partial pressure by 2. Q < K, reaction shifts right toward products

8 Effects of Pressure and Volume When volume increases, the equilibrium shifts to the side with more gas molecules. When volume decreases, the equilibrium shifts to the side with fewer gas molecules.

9 Effects of Pressure and Volume Decreasing the size of a container increases the concentration of all the gases in the container. The total pressure in the container will increase. According to LeChatlier s Principle, the equilibrium should shift to relieve the stress. The equilibrium shifts to decrease the pressure by reducing the number of gas molecules in the container.

10 Effects of Pressure and Volume An example of a gaseous reaction not influenced by volume of the container: H2 + Cl2 2 HCl An example of a gaseous reaction influenced by volume of the container: 2 H2 + O2 2 H2O

11 Effect of Temperature Exothermic reactions release heat energy and endothermic reactions absorb heat energy. We can write HEAT as product or reactant in a chemical reaction: A + B C + D + HEAT An exothermic reaction HEAT + A + B C + D An endothermic reaction Adding HEAT to an equilibrium, therefore, will shift an equilibrium in the same manner as changing the amount of a product or reactant. Keep in mind, CHANGING TEMPERATURE CHANGES THE EQUILIBRIUM CONSTANT!!

12 Effect of Temperature An endothermic reaction HEAT ENERGY+

13 Effect of Temperature N2 (g) + 3 H2 (g) 2 NH3 (g) + HEAT H = -92 kj/mol (at 450ºC)

14 Effect of Temperature and Pressure on Yield of Ammonia N2 (g) + 3 H2 (g) 2 NH3 (g) H = -92 kj/mol (at 450ºC)

15 COMMERCIAL PRODUCTION OF AMMONIA N2 (g) + 3 H2 (g) 2 NH3 (g) H = -92 kj/mol (at 450ºC)

16 6) Consider the following reaction at equilibrium: H2 (g) + I2 (g) 2 HI (g) What happens when the volume is increased? a) Reaction shifts right b) Reaction shifts left c) There is no effect d) Can t tell

17 7) Consider the following reaction at equilibrium: + energy 2 KClO3 (s) 2 KCl (s) + 3 O2 (g) ΔH < 0 What happens when the reaction mixture is heated? a) Reaction shifts right b) Reaction shifts left c) There is no effect d) Can t tell

18 8) Given the following reaction, how could you increase the concentration of products? N2 (g) + 3 H2 (g) 2 NH3 (g) ΔH + energy = kj a) Increase volume, add N2, or increase temperature b) Increase volume, add N2, or decrease temperature c) Decrease volume, add N2, or decrease temperature d) Decrease volume, remove N2, or decrease temperature e) Decrease volume, remove N2, or increase temperature

19 Effect of Catalysts Catalysts provide an alternative, more efficient mechanism for a reaction. Catalysts work for both forward and reverse reactions. Catalysts affect the rate of both forward and reverse reactions by the same factor. Catalysts do not affect the position of equilibrium.

20 Effect of Catalysts Systems reach equilibrium faster in a catalyzed reaction, but there is no change in K or position of equilibrium.

21 Calculations Based on Keq

22 1) Consider the following reaction: Cl2 (g) + PCl3 (g) PCl5 (g) Kp = 24.2 If a sealed container contains initially only PCl5 with a pressure of 1.35 atm, determine the equilibrium partial pressures of all three components. Write the reverse reaction, calculate Kp for the reverse reaction, make a RICE table, and determine the pressures. PCl5 (g) PCl3 (g) + Cl2 (g) Kp = Kp -1 = R PCl5 (g) PCl3 (g) + Cl2 (g) I C - x + x + x E x x x

23 [PCl 3 ] [Cl 2 ] Kp = = [PCl 5 ] [x] [x] [1.35-x] = [x 2 ] [1.35-x] = [x 2 ] = x [x 2 ] x = 0 x = ± (4)(1)(0.0558) x = x = = atm = atm

24 PCl5 (g) PCl3 (g) + Cl2 (g) Kp = Kp -1 = R PCl5 (g) PCl3 (g) + Cl2 (g) I C E x + x + x atm -x x atm x atm Kp = [0.215] [0.215] [1.13] =

25 PCl5 (g) PCl3 (g) + Cl2 (g) Kp = Kp -1 = [x 2 ] [1.35-x] = [PCl5] = 1.35 atm assuming x <<< 1.35 x = [x 2 ] [1.35] = x = error = 20 %

26 2) Consider the following reaction: Cl2 (g) + Br2 (g) 2 BrCl (g) Kp = 4.7 x 10-2 If a sealed container contains initially only Cl2 and Br2,each at a prressure of atm, determine the equilibrium partial pressures of all three components. Cl2 (g) + Br2 (g) 2 BrCl (g) Kp = = 4.7 x 10-2 [BrCl] 2 [Cl 2 ][Br 2 ] R I C Cl2 (g) + Br2 (g) 2 BrCl (g) x - x + 2x E x x 2x

27 [BrCl] 2 Kp = = [Cl 2 ][Br 2 ] [2x] 2 [0.100-x] 2 = 4.7 x 10-2 [4x 2 ] = 4.7 x [0.100-x] [4x 2 ] = 4.7 x [0.100-x] [2x] [0.100-x] = or x = x = or 2x = atm

28 2) Consider the following reaction: Cl2 (g) + Br2 (g) 2 BrCl (g) Kp = 4.7 x 10-2 If a sealed container contains initially only Cl2 and Br2, each at a prressure of atm, determine the equilibrium partial pressures of all three components. Cl2 (g) + Br2 (g) 2 BrCl (g) Kp = = 4.7 x 10-2 [BrCl] 2 [Cl 2 ][Br 2 ] R I C Cl2 (g) + Br2 (g) 2 BrCl (g) E

29 Cl2 (g) + Br2 (g) 2 BrCl (g) Kp = 4.7 x 10-2 [2x] [0.100-x] = [Cl2], [Br2] = atm assuming x <<< x = [2x] [0.100] = x = error = 56%

30 3) Consider the following reaction: 2 ICl(g) Cl2 (g) + I2 (g) Kp = 1.4 x 10-5 If a sealed container contains initially only ICl, at a pressure of 1.37 atm, determine the equilibrium partial pressures of all three components. 2 ICl(g) Cl2 (g) + I2 (g) Kp = = 1.4 x 10-5 [Cl 2 ][Br 2 ] [ICl] 2 R I C 2 ICl (g) Cl2 (g) + 2 I2 (g) x + ½ x + ½ x E x ½ x ½ x

31 [I 2 ][Cl 2 ] Kp = = [ICl] 2 [x/2] 2 [1.37-x] 2 = 1.4 x 10-5 x 2 (4)[1.37-x] 2 = 1.4 x 10-5 assuming x <<< 1.37 x 2 (4)(1.37) 2 = 1.4 x 10-5 x 2 = 1.05 x 10-4 x = 1.02 x 10-2 ½ x = 5.1 x x = 1.36

32 3) Consider the following reaction: 2 ICl(g) Cl2 (g) + I2 (g) Kp = 1.4 x 10-5 If a sealed container contains initially only ICl, at a prressure of 1.37 atm, determine the equilibrium partial pressures of all three components. 2 ICl(g) Cl2 (g) + I2 (g) Kp = = 1.4 x 10-5 [Cl 2 ][Br 2 ] [ICl] 2 R I C 2 ICl (g) Cl2 (g) + 2 I2 (g) E

33 2 ICl(g) Cl2 (g) + I2 (g) Kp = 1.4 x 10-5 [ICl] = 1.37 atm x = [x/2] 2 [1.37-x] 2 = 1.4 x 10-5 do not assume x <<< 1.37 use quadratic formula x = error = 6 %