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Chemical Equilibrium: The Extent of Chemical Reactions

Chemical Equilibrium: The Extent of Chemical Reactions. N 2 O 4 ( g ). 2 NO 2 ( g ). The State of Chemical Equilibrium. Chemical Equilibrium : The state reached when the concentrations of reactants and products remain constant over time. Colorless. Brown. N 2 O 4 ( g ). 2 NO 2 ( g ).

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Chemical Equilibrium: The Extent of Chemical Reactions

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  1. Chemical Equilibrium: The Extent of Chemical Reactions

  2. N2O4(g) 2 NO2(g) The State of Chemical Equilibrium Chemical Equilibrium: The state reached when the concentrations of reactants and products remain constant over time. Colorless Brown

  3. N2O4(g) 2 NO2(g) The State of Chemical Equilibrium

  4. The State of Chemical Equilibrium

  5. a A + b B N2O4(g) c C + d D 2 NO2(g) [C]c[D]d [NO2]2 [A]a[B]b [N2O4] Concentration is Molarity [ ]: units of mol/L The Equilibrium Constant Kc For a general reversible reaction: Products Equilibrium equation: Kc = Reactants Equilibrium constant Equilibrium constant expression For the following reaction: Kc = = 4.64 x 10–3 (at 25 °C)

  6. (0.0125)2 (0.0141)2 [NO2]2 0.0337 0.0429 [N2O4] The Equilibrium Constant Kc Experiment 1 Experiment 5 Kc = = 4.64 x 10–3 = 4.63 x 10–3

  7. L atm 0.082 06 K mol a A + b B c C + d D (PC)c(PD)d (PA)a(PB)b Pressure is like concentration of a gas: units are atm The Equilibrium Constant Kp What is the relationship of mol/L to P? P = nRT/V well isn’t n/V = M? Products Equilibrium equation: KP= Reactants Equilibrium constant Equilibrium constant expression Kp = Kc(RT)∆n R is the gas constant, T is the absolute temperature (Kelvin). ∆n is the number of moles of gaseous products minus the number of moles of gaseous reactants. *If the substance is a pure solid or a liquid, put a 1 in its place.

  8. [CaO][CO2] (1)[CO2] [CaCO3] CaCO3(s) CaO(s) + CO2(g) (1) Kp = P CO2 Heterogeneous Equilibria Limestone Lime Kc = = = [CO2] Pure solids and pure liquids are not included. Kc = [CO2]

  9. Heterogeneous Equilibria

  10. (at 500 K) Kc = 4.2 x 10–48 2 H2(g) + O2(g) H2(g) + I2(g) 2 H2O(g) 2 H2(g) + O2(g) 2 H2O(g) 2 HI(g) (at 700 K) Kc = 57.0 (at 500 K) Kc = 2.4 x 1047 Using the Equilibrium Constant

  11. Write the equilibrium expressions for Kc for the following: 2H2O(l) 2H2(g) + O2(g) 2HCl(aq)  H2(g) + Cl2(g) Example #1

  12. The following equilibrium concentrations were observed for the Haber process at 127oC N2(g) + 3H2(g) 2NH3 (g) [NH3] = 0.031M [N2] = 0.85M [H2] = 0.0031M Calculate Kc and Kp. Example Problem #2

  13. The reaction below occurs at 25oC. 2NO(g) + Cl2(g) 2NOCl(g) The equilibrium partial pressures are: PNOCl = 1.2atm PNO = 0.05atm PCl2 = 0.30atm Calculate Kp and Kc. Example Problem #3

  14. Reaction Quotient Predicting shifts in equilibrium mathematically Qc = [C]c[D]dQp = (PC)c(PD)d [A]a[B]b (PA)a(PB)b Q<K: shifts right Q>K: shifts left Q=K: at equilibrium

  15. Using the Equilibrium Constant net reaction goes from left to right (reactants to products). • If Qc < Kc net reaction goes from right to left (products to reactants). • If Qc > Kc no net reaction occurs. • If Qc = Kc

  16. Using the Equilibrium Constant

  17. The equilibrium constant, Kp, is 2.33 for the reaction C(s) + CO2(g) 2CO(g) at 50oC. If 0.75g of C, 3.5atm of CO2 and 4.5 atm CO are present in a container, is the reaction at equilibrium? If not, which way will it shift? Example #1

  18. The equilibrium constant, Kp, for the reaction 2H2O(l) 2H2(g) + O2(g) is 5 x 10-6. If 50g of hydrogen, 25g of oxygen and 100g of water are placed in a 5L flask at 100oC, will the reaction be at equilibrium? If not, which way will it shift? Example #2

  19. Consider the reaction: N2(g) + 3H2(g) 2NH3 (g) The equilibrium constant, Kc, at 25oC is 0.082. If 0.4mol of nitrogen, 0.098mol of hydrogen and 2.32mol of ammonia are present in a 4L container, is the system at equilibrium? If not, which way will it shift? Example #3

  20. The ICE method aA+ bB cC + dD I # # 0 0 C -ax -bx +cx +dx E #-ax #-bx cx dx I= initial concentrations, C= change during reaction process E = equilibrium concentrations. Determining Equilibrium concentrations

  21. H2(g) + I2(g) 2 HI(g) Using the Equilibrium Constant At 700 K, 0.500 mol of HI is added to a 2.00 L container and allowed to come to equilibrium. Calculate the equilibrium concentrations of H2, I2, and HI. Kc is 57.0 at 700 K.

  22. (0.250 – 2x)2 x2 H2(g) + I2(g) 2 HI(g) [HI]2 Kc = [H2][I2] Using the Equilibrium Constant Set up a table: Substitute values into the equilibrium expression: 57.0 =

  23. (0.250 – 2x)2 x2 Using the Equilibrium Constant Solve for “x”: 57.0 = x = 0.0262 Determine the equilibrium concentrations: H2: 0.0262 M I2: 0.0262 M HI: 0.250 – 2(0.0262) = 0.198 M

  24. For the reaction N2O4(g) 2NO2(g) the equilibrium constant, Kp, is 0.133. If 2.71 atm of N2O4 are introduced into a container, what will the equilibrium partial pressure be for all the species in the reaction? Example Problem #1

  25. For the reaction: CO(g) +H2O(g) CO2(g) + H2 (g) the equilibrium constant, Kc, is 5.10 at 700K. If 3 moles of CO and 3 moles of H2O are introduced into a 5L flask, what will the equilibrium concentrations be for all the species present in the reaction? Example Problem #2

  26. For the reaction 2NOCl(g) 2NO(g) + Cl2(g) the equilibrium constant, Kc, is 1.6 x 10-5 at 35oC. If 3 moles of NOCl are placed in a 6L container, what will the concentrations of all the species reach at equilibrium? Example #3

  27. In a 2L flask 2 moles of HF are introduced. The reaction is allowed to come to equilibrium. At equilibrium the concentration of the hydrogen gas reaches 0.064M. Determine the equilibrium constant. 2HF (g) H2(g) + F2 (g) Example #4

  28. Initially the system contains 0.4M A and 1.2M B. After the system reaches equilibrium, 26% of the A has reacted. Determine the equilibrium constant for this reaction. 4A + 2B  3C Example #5

  29. Reverse ICE

  30. The equilibrium constant for the reaction below at 448oC is 50.5. H2 (g) + I2(g) 2HI (g) Determine the equilibrium concentrations if 3 moles of HI are introduced into a 4 L reaction vessel. Example #1

  31. For the reaction below at 500oC the equilibrium constant, Kp, is 68956.52atm2. 2NH3(g) N2(g) + 3H2 (g) If 2.5 atm of N2 and 5.5 atm of H2 are placed in a 10 L reaction vessel, what are the equilibrium partial pressures of all the gases? Example #2

  32. Fifteen grams of solid NH4HS(s) is placed in an evacuated flask at 24oC. At equilibrium the total pressure in the flask is 0.614atm. What is the equilibrium constant, Kp, for the reaction below? NH4HS(s) NH3(g) + H2S(g) Example #3

  33. For the reaction 2SO2(g)+O2(g)  2SO3 (g) The equilibrium constant, Kc, is 3 x 104. If 240 grams of SO3 are introduced into a 5L reaction vessel, what will the equilibrium concentrations be for all the species present? Example #4

  34. For the reaction: 2H2O(l) 2H2 (g) + O2(g) If 5000g of H2O are introduced into a 5L flask, the total pressure at equilibrium is 4.5atm. Calculate the Kp value for this reaction. Example #5

  35. If a stress is placed on a system at equilibrium, that system will shift to relieve the stress. LeChatelier’s Principle

  36. Add a substance, the arrow points away from what you added. Remove a substance, the arrow points toward what you removed. You cannot change the concentration of solids or liquids (adding solid or liquid has no effect on equilibrium) Change concentration

  37. N2(g) + 3 H2(g) 2 NH3(g) The Effect of Concentration Changes on an Equilibrium Mixture

  38. 2H2(g) + O2(g) 2H2O (l) Add H2 Remove O2 Add H2O Remove H2 Example of Concentration changes

  39. DH: positive, energy is a reactant (endothermic) DH: negative, energy is a product (exothermic) Heat adds energy Cooling removes energy Changing Temperature

  40. C(s) + O2(g) CO2(g)DH = -394kJ/mol Heat Cool Example of temperature change

  41. Changing pressure by changing volume: -increase pressure, shifts to the side with less gas particles. -decrease pressure, shifts to the side with more gas particles. *Change pressure by adding a gas not present in the reaction, no effect. Changing pressure

  42. N2(g) + 3 H2(g) 2 NH3(g) The Effect of Pressure and Volume Changes on an Equilibrium Mixture

  43. CaCO3 (s) CaO(s) + CO2(g) Increase volume Decrease volume Increase pressure by adding neon. Example for pressure changes

  44. A catalyst will increase the rate of both the forward and reverse reactions at the same time, thus, no effect on equilibrium. Add a Catalyst

  45. The Effect of a Catalyst on Equilibrium

  46. Na2CO3(s) Na2O(s) + CO2(g) DH >0 Add Na2O Heat Add CO2 Increase volume Cool Remove Na2CO3 Example Problem

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