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Gaseous Equilibrium

Gaseous Equilibrium. Equilibrium systems are reaction systems that are reversible. Reactants are not completely consumed causing the reaction system to try to reach a state of “equilibrium”.

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Gaseous Equilibrium

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  1. Gaseous Equilibrium • Equilibrium systems are reaction systems that are reversible. Reactants are not completely consumed causing the reaction system to try to reach a state of “equilibrium”. • Reversible means that the reaction may occur in the forward direction (thus favoring the products) or in the reverse reaction (thus favoring the reactants). • At equilibrium, the rate of the forward reaction is equal to rate of the reverse reaction. This means that the amounts of all the species at equilibrium remain constant.

  2. Equilibrium Systems 2N2 (g) + 3H2 (g) ⇌ 2NH3 (g) N2O4(g) ⇌ 2NO2 (g)

  3. The Equilibrium Constant • When partial pressures (in atm) remain constant (independent of the original composition, the volume of the container, or the total pressure) a constant for this system can be calculated and is called the equilibrium constant (Kp). Or the equilibrium constant can be calculated using the concentration of the products and reactants at equilibrium (Kc). • When solving for Kc or Kp you must first decipher the equilibrium constant expression from the chemical equation: aA + bB ⇌ cC + dD

  4. Calculating Keq Kc = [C]c[D]d [A]a[B]b or Kp = (PC)c(PD)d (PA)a(PB)b Kp = Kc(RT)Δng • R = 0.0821 L atm/mol K • Δng = the change in the # of moles of gas in the equation (products – reactants).

  5. The Coefficient Rule • If the coefficients in a balanced equation are multiplied by a factor than the equilibrium constant for that equation is raised to that power: K’ = Kn For Example:N2O4(g) ⇌ 2NO2 (g) K = 11 then for…. • ½ N2O4(g) ⇌ NO2 (g)K’ = ?

  6. The Reciprocal Rule • The equilibrium constants for the forward and reverse reactions are reciprocals of each other: K” = 1/K For Example:N2O4(g) ⇌ 2NO2 (g) K = 11 then for…. • 2NO2 (g) ⇌ N2O4 (g)K’’ = ?

  7. The Rule of Multiple Equilibria • If a reaction can be expressed as the sum of 2 or more reactions, then K for the overall reaction is the product of the equilibrium constants for those reactions added. K1 x K2… = K3 SO2 (g) + NO2 (g)⇌NO (g) + SO3 (g) K3 = ? SO2 (g) + ½ O2 (g)⇌SO3 (g) K1 = 2.2 NO2 (g)⇌ NO (g) + ½ O2 (g) K2 = 4.0

  8. Example 1 • Consider the reaction by which the air pollutant nitrogen monoxide is made from the elements nitrogen and oxygen in an automobile engine. • (a) Write the equilibrium constant expression for the reaction. • (b) At 25 oC, K for this reaction is 4.2 x 10-31. Calculate K for the formation of 1mole of this product from its foundational elements. • (c) Find K for the reaction below at 25 oC N2 (g) + 2O2 (g)⇌ 2NO2 (g) K = 1.0 x 10-8 2NO (g) + O2 (g) ⇌ 2NO2 (g) K = ?

  9. Heterogeneous Equilibrium • This occurs when there is more than one phase present in a chemical equation. • The position of the equilibrium is independent of the amount of solid or pure liquid – so they are not included in the equilibrium constant expression. • For example, write the equilibrium constant expression for the following: CO2 (g) + H2 (g)⇌CO (g) + H2O (l) K = I2 (s)⇌ I2 (g) K =

  10. Example 2 • Write the expression for K for • (a) the reduction of black solid copper (II) oxide with hydrogen to form copper metal and steam. • (b) the reaction of steam with red hot coke to form a mixture of hydrogen and carbon monoxide, called water gas.

  11. Example 3 - Determining K • When given the partial pressures, fairly simple, apply the equation previously discussed. • Solid ammonium chloride is sometimes used as a flux in soldering because it decomposes on heating into ammonia gas and hydrogen chloride gas. The HCl formed removes oxide films from metals to be soldered. In a certain equilibrium system at 400 oC, 22.6 g of ammonium chloride is present; the partial pressures of ammonia and hydrogen chloride are 2.5 atm and 4.8 atm, respectively. Calculate the K.

  12. Example 4 - An Equilibrium Table • However, it is a bit more difficult when your given only one partial pressure @ equilibrium. This is when an equilibrium table will be useful: Consider the equilibrium system 2HI (g) ⇌ H2 (g) + I2 (g) Originally, a system contains only HI at a pressure of 1.00 atm at 520 oC. The equilibrium partial pressure of H2 is found to be 0.10 atm. Calculate: (a) PI2 at equilibrium (b) PHI at equilibrium (c) K

  13. Understanding K • When K > 1 the reaction favors the products, the forward reaction (ΔGo < 0). • When K < 1 the reaction favors the reactants, the reverse reaction (ΔGo > 0). • When K = 1 neither the forward nor the reverse is favored (ΔGo = 0).

  14. The reaction quotient, Q • This is basically calculated the same way as K. Except the values used for Q are not values at equilibrium. This value will allow you to predict which direction the reaction will move towards: • If Q < K then the reaction will move in the forward direction • If Q > K then the reaction will move in the reverse reaction • If Q = K then the reaction is already at equilibrium

  15. Solving for equilibrium values from initial values – ICE Tables • More commonly K is used to determine the equilibrium partial pressures (concentrations) of reactants and products from the original partial pressures (or concentrations). To do this: • Using the balanced chemical equation for the reaction, write the expression for K. • Prepare a table that will allow you to write the Initial amounts, the Change that occurs, and the Equilibrium amounts. This will help organize your information – ICE tables. • Express the equilibrium partial pressures of all species in terms of a single unknown, x. Remember that the changes in partial pressures are related through the coefficients of the balanced equation. • Substitute the equilibrium terms into the expression for K. This will give an algebraic equation that must be solved for x. • Once x is found, refer back to the table and calculate the equilibrium partial pressures.

  16. Example 5 • Assume that the reaction for the formation of gaseous hydrogen fluoride from hydrogen and fluorine gases has an equilibrium constant of 1.15 x 102 at a certain temperature. In a particular experiment, 3.000 mol of each component was added to a 1.500 L flask. • a. Write the equilibrium constant expression for the reaction. • b. What direction will the reaction shift to reach equilibrium? Explain. • c. Calculate the equilibrium concentrations of all species.

  17. Example 6 • Gaseous NOCl decomposes to form nitrogen monoxide and chlorine gases. At 35 oC the equilibrium constant is 1.6 x 10-5. In an experiment, 1.0 mol of NOCl is placed in a 2.0 L flask. • a. Write the equilibrium constant expression for this reaction. • b. What are the equilibrium concentrations of all species?

  18. Le Chatelier’s Principle:Effects of Concentration, Pressure, & Temperature • Concentration – adding or removing reactants or products: • If a species is added the reaction will shift away from that species, in order to restore equilibrium. • If a species is removed the reactions will shift towards that species in order to restore equilibrium.

  19. Pressure – compressing or expanding the system: • When a system is compressed (increase in pressure) the reaction will shift towards the side with the least amount of moles. • When the system is expanded (decrease in pressure) the reaction will shift towards the side with more moles.

  20. Temperature – increasing or decreasing the temperature: • When temperature is increased the reaction will shift away from the heat. • So for the following endothermic reaction which way will the reaction shift? Will the K value decreases or increase? Why? N2O4 (g) ⇌2NO2 (g) H = 57.2 kJ • When the temperature is decreased the reaction will shift towards the heat. • So for the following endothermic reaction which way will the reaction shift? Will the K value decrease or increase? Why? N2O4 (g) ⇌2NO2 (g) H = 57.2 kJ

  21. van’t Hoff Equation • The equilibrium constant changes with temperature (this is the only one of the three “stresses” that changes the value of K), to determine this change the van’t Hoff equation maybe used:

  22. Equilibrium & Gibbs Free Energy • Relationship between standard Gibbs free energy and the equilibrium constant (or reaction quotient): • ΔGo = -RT lnK • Gibbs free energy and standard Gibbs free energy: • ΔG = ΔGo + RT lnQ

  23. Example 7 • Consider the equilibrium reaction for the formation of gaseous ammonia from gaseous nitrogen and hydrogen, where ΔGo = -33.3 kJ/mol. For each of the following mixtures of reactants and products below at 25 oC, predict the direction in which the system will shift to reach equilibrium. If the reaction is at equilibrium then solve for the equilibrium constant. • a.Pammonia = 1.00 atm, Pnitrogen = 1.47 atm, Phydrogen = 1.00 x 10-2 atm • b.Pammonia = Pnitrogen = Phydrogen = 1.00 atm

  24. MC #1 CuO(s) + H2(g) <===> Cu(s) + H2O(g) ΔH = - 2.0 kJ/mol When the substances in the equation above are at equilibrium at pressure P and temperature T, the equilibrium can be shifted to favor the products by (A) increasing the pressure by means of a moving piston at constant T(B) increasing the pressure by adding an inert gas such as nitrogen(C) decreasing the temperature(D) allowing some gases to escape at constant P and T(E) adding a catalyst

  25. MC #2 HgO(s) + 4 I¯ + H2O  HgI42¯ + 2 OH¯ ΔH < 0 Consider the equilibrium above. Which of the following changes will increase the concentration of HgI42¯ (A) Increasing the concentration of OH¯(B) Adding 6 M HNO3(C) Increasing the mass of HgO present(D) Increasing the temperature(E) Adding a catalyst

  26. MC #3 • In which of the following systems would the number of moles of the substances present at equilibrium NOT be shifted by a change in the volume of the system at constant temperature? (A) CO(g) + NO(g) <===> CO2(g) + 1/2 N2(g)(B) N2 (g) + 3 H2 (g) <===> 2 NH3(g)(C) N2 (g) + 2 O2 (g) <===> 2 NO2(g)(D) N2O4 (g) <===> 2 NO2(g)(E) NO(g) + O3 (g) <===> NO2(g) + O2(g)

  27. MC #4 PCl3(g) + Cl2(g)  PCl5(g) + energy Some PCl3 and Cl2 are mixed in a container at 200 °C and the system reaches equilibrium according to the equation above. Which of the following causes an increase in the number of moles of PCl5 present at equilibrium? I. Decreasing the volume of the container II. Raising the temperature III. Adding a mole of He gas at constant volume (A) I only(B) II only(C) I and III only(D) II and III only(E) I, II, and III

  28. MC #5 4 HCl(g) + O2(g)  2 Cl2(g) + 2 H2O(g) Equal numbers of moles of HCl and O2 in a closed system are allowed to reach equilibrium as represented by the equation above. Which of the following must be true at equilibrium? I. [HCl] must be less than [Cl2]. II. [O2] must be greater than [HCl]. III. [Cl2] must equal [H2O]. (A) I only(B) II only(C) I and III only(D) II and III only(E) I, II, and III

  29. FRQ #1 Answer the following questions regarding the decomposition of arsenic pentafluoride, AsF5(g). (a) A 55.8 g sample of AsF5(g) is introduced into an evacuated 10.5 L container at 105°C. (i) What is the initial molar concentration of AsF5(g) in the container? (ii) What is the initial pressure, in atmospheres, of the AsF5(g) in the container? At 105°C, AsF5(g) decomposes into AsF3(g) and F2(g) according to the following chemical equation. AsF5(g) AsF3(g) + F2(g) (b) In terms of molar concentrations, write the equilibrium-constant expression for the decomposition of AsF5(g). (c) When equilibrium is established, 27.7 percent of the original number of moles of AsF5(g) has decomposed. (i) Calculate the molar concentration of AsF5(g) at equilibrium. (ii) Using molar concentrations, calculate the value of the equilibrium constant, Keq, at 105°C. (d) Calculate the mole fraction of F2(g) in the container at equilibrium.

  30. FRQ #2 C(s) + CO2 (g) 2 CO(g) Solid carbon and carbon dioxide gas at 1,160 K were placed in a rigid 2.00 L container, and the reaction represented above occurred. As the reaction proceeded, the total pressure in the container was monitored. When equilibrium was reached, there was still some C(s) remaining in the container. Results are given below. Time (hours) Total Pressure of Gases in Container at 1,160 K (atm) 0.0 5.00 2.0 6.26 4.0 7.09 6.0 7.75 8.0 8.37 10.0 8.37 (a) Write the expression for the equilibrium constant, Kp for the reaction. (b) Calculate the number of moles of CO2(g) initially placed in the container. (Assume that the volume of the solid carbon is negligible.) (c) For the reaction mixture at equilibrium at 1,160 K, the partial pressure of the CO2(g) is 1.63 atm. Calculate (i) the partial pressure of CO(g) (ii) the value of the equilibrium constant, Kp (d) If a suitable solid catalyst were placed in the reaction vessel, would the final total pressure of the gases at equilibrium be greater than, less than, or equal to the final total pressure of the gases at equilibrium without the catalyst? Justify your answer. (Assume that the volume of the solid catalyst is negligible.) In another experiment involving the same reaction, a rigid 2.00 L container initially contains 10.0 g of C(s), plus CO(g) and CO2(g), each at a partial pressure of 2.00 atm at 1,160 K. (e) Predict whether the partial pressure of CO2(g) will increase, decrease, or remain the same as this system approaches equilibrium. Justify your prediction with a calculation.

  31. FRQ #3 2 H2S(g) 2 H2(g) + S2(g) When heated, hydrogen sulfide gas decomposes ac­cording to the equation above. A 3.40 g sample of H2S(g) is introduced into an evacuated rigid 1.25 L container. The sealed container is heated to 483 K, and 3.72×10–2 mol of S2(g) is present at equilibrium. (a) Write the expression for the equilibrium constant, Kc, for the decomposition reaction represented above. (b) Calculate the equilibrium concentration, in molL-1, of the following gases in the container at 483 K. (i) H2(g) (ii) H2S(g) (c) Calculate the value of the equilibrium constant, Kc, for the decomposition reaction at 483 K. (d) Calculate the partial pressure of S2(g) in the container at equilibrium at 483 K. (e) For the reaction H2(g) + S2(g) H2S(g) at 483 K, calculate the value of the equilibrium constant, Kc.

  32. FRQ #4 C(s) + H2O(g) CO(g) + H2(g) Hº = +131kJ A rigid container holds a mixture of graphite pellets (C(s)), H2O(g), CO(g), and H2(g) at equilibrium. State whether the number of moles of CO(g) in the container will increase,decrease, or remain the same after each of the following disturbances is applied to the original mixture. For each case, assume that all other variables remain constant except for the given disturbance. Explain each answer with a short statement. (a) Additional H2(g) is added to the equilibrium mixture at constant volume. (b) The temperature of the equilibrium mixture is increased at constant volume. (c) The volume of the container is decreased at constant temperature. (d) The graphite pellets are pulverized.

  33. FRQ #5 CO2(g) + H2(g) H2O(g) + CO(g) When H2(g) is mixed with CO2(g) at 2,000 K, equilibrium is achieved according to the equation above. In one experiment, the following equilibrium concentrations were measured. [H2] = 0.20 mol/L [CO2] = 0.30 mol/L [H2O] = [CO] = 0.55 mol/L (a) What is the mole fraction of CO(g) in the equilibrium mixture? (b) Using the equilibrium concentrations given above, calculate the value of Kc, the equilibrium constant for the reaction. (c) Determine Kp in terms of Kc for this system. (d) When the system is cooled from 2,000 K to a lower temperature, 30.0 percent of the CO(g) is converted back to CO2(g). Calculate the value of Kc at this lower temperature. (e) In a different experiment, 0.50 mole of H2(g) is mixed with 0.50 mole of CO2(g) in a 3.0-liter reaction vessel at 2,000 K. Calculate the equilibrium concentration, in moles per liter, of CO(g) at this temperature.

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