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Gaseous Equilibrium – Chapter 12

Gaseous Equilibrium – Chapter 12. 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 – Chapter 12

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  1. Gaseous Equilibrium – Chapter 12 • 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) *Reference Table 12.1 – page 322

  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 12.1 – Page 326 • 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 12.2 – Page 328 • 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. Determining K • When given the partial pressures, fairly simple, apply the equation previously discussed. • Example 12.3: 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. 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. Example: 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. • When K < 1 the reaction favors the reactants, the reverse reaction. • When K = 1 neither the forward nor the reverse is favored.

  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 Example 12.5 – page 332

  15. Example 12.6 – Page 333 • 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. • 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 – an equilibrium table is suggested. • 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. Effects of Concentration • 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.

  17. Effects of Pressure • 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 away from the side with the most moles.

  18. Effects of Temperature • 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

  19. 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: (see page 338) • Example 12.8 – page 339

  20. 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

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