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Equilibrium

Equilibrium

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Equilibrium

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  1. Equilibrium

    Unit 10
  2. Reaction Dynamics If the products of a reaction are removed from the system as they are made, then a chemical reaction will proceed until the limiting reactants are used up. However, if the products are allowed to accumulate; they will start reacting together to form the original reactants - called the reverse reaction. We show this reverse reaction by using a double arrow (H2(g) + I2(g)2HI(g)) The forward reaction slows down as the amounts of reactants decreases because the reactant concentrations are decreasing At the same time the reverse reaction speeds up as the concentration of the products increases.
  3. Reaction Dynamics Eventually the forward reaction is using reactants and making products as fast as the reverse reaction is using products and making reactants. This is called chemical equilibrium. rateforward = ratereverse Note: This equilibrium is dynamic and will only occur in a closed system. When a system reaches equilibrium, the amounts of reactants and products in the system stays constant. The forward and reverse reactions still continue, but because they go at the same rate the amounts of materials don't change.
  4. [C]c[D]d [A]a[B]b = Keq Equilibrium Expression There is a mathematical relationship between the amounts of reactants and products at equilibrium regardless of the amounts of reactant and product that you start with aA + bBÛcC + dD Capital letters (A,B,C,D) – reactants or products Lowercase letter (a,b,c,d) – coefficients from the equation NOTE – products on top, reactants on bottom In this expression, Keq is a number called the equilibrium constant. ratio of product concentration to reactant concentration at equilibrium
  5. Equilibrium Expression Do not include solids or liquids, only solutions and gases. Things whose concentration cannot change have no effect on the Keq The value of Keq depends on temperature of the reaction – if temperature changes then the value of Keqchanges. A value of K which depends on concentration in M is known as Kc A value of K which depends on gas partial pressure is known as Kp A value of K which depends on solubility and ion concentration, in M, in aqueous solution is known as Ksp
  6. Example – Determine the value of the Equilibrium Constant for the Reaction 2 SO2(g) + O2(g)2 SO3(g) [SO2] = 1.50M; [O2] = 1.25 M; [SO3] = 3.50 M Determine the Equilibrium Expression Plug the equilibrium concentrations into to Equilibrium Expression Solve the Equation
  7. Position of Equilibrium The size of the equilibrium constant shows whether products or reactants are favored at equilibrium. Keq > 1, products are favored at equilibrium Keq < 1, reactants are favored at equilibrium Keq >> 1, reaction goes virtually to completion Keq << 1, reactants react almost not at all
  8. Value of K (graphically) K > 1 K < 1
  9. Example – Determining concentration from the value of the Equilibrium Constant Phosgene, COCl2, used in the manufacture of plastics, is prepared from CO and Cl2: CO(g) + Cl2(g)  COCl2(g) An equilibrium mixture at 395 0C contains 0.012 mol CO and 0.025 mol Cl2 per liter as well as some COCl2. If the Kc at 395 0C is 1.23 x 103, what is the concentration of COCl2?
  10. Equilibrium Constant Expression for Gases Kc - Kp In addition to concentration, equilibrium can be represented for homogeneous gaseous systems in terms of partial pressures. This produces a special equilibrium constant, Kp. Consider the reaction for the Haber process: N2(g) + 3H2 2NH3 Kp = (PNH3 )2 since P = CRT (Ideal Gas Law) then the previous (PH2)3(PN2) equation can be rewritten as Kp = {[NH3](RT )}2 = [NH3]2 x 1 = Kc ; {[H2](RT)}3 {[N2](RT) [H2]3 [N2] (RT)2 (RT)2 This could also be written as Kp = Kc(RT)-2 ; So in general, Kp = Kc(RT)Δn where Δn = total moles of gaseous products – total moles of gaseous reactants. In this case 2 – 4 = -2
  11. Example - Kp from Kc The Kc for the formation of PCl5(g) at 25 0C is 11.2 PCl3(g) + Cl2(g)  PCl5(g) Calculate Kp for the reaction at this temperature.
  12. Balanced Equations and the K Calculate the value for Kc for the formation of ammonia having the following equilibrium concentrations: [H2] = .15 M; [N2] = 0.25 M; and [NH3] = 0.00015 M 2.7 x 10-5 Now calculate the value for Kc for the reverse (decomposition of ammonia) reaction. 3.8 x 104 What do you notice? The equilibrium constants for a reaction and its reverse are the reciprocals of one another.
  13. Balanced Equations and the K Now calculate the constants for the same reaction written using different sets of stoichiometric coefficients, with the same concentrations. 2 N2(g) + 6H2(g)  4 NH3(g) Kc= 7.1 x 10-10 ½ N2(g) + 3/2 H2(g)  NH3(g) Kc= 5.2 x 10-3 Notice that when we doubled the coefficients in the first example, the value for K is squared. When we multiply by ½ in the second, the value is the square root of our original K (K1/2). When the stoichiometric coefficients of a balanced equation are multiplied by some factor, the equilibrium constant for the new equation (Knew) is the old K raised to the power of the multiplication factor.
  14. Balanced Equations and the K Remember Hess’s Law? K for an overall reaction can be determined using the K for each equation that is added to give the overall. Consider these 2 hypothetical reaction equations: A  2 B K1 = [B]2 [A] 2 B  3 C K2 = [C]3 [B]2 A  2 B 2 B  3 C A  3 C Koverall = [C]3 [A] Notice that K1 x K2 = [B]2 x [C]3 = [C]3 = Koverall [A] [B]2 [A] If you add two or more individual chemical equations to obtain an overall equation, multiply the corresponding equilibrium constants by each other to obtain the overall K.
  15. The Reaction Quotient - Q Used to determine the direction a reaction must proceed to reach equilibrium. Calculated the same way as K 3 possibilities Q < K must proceed right Q > K must proceed left Q = K at equilibrium
  16. The Reaction Quotient - Q Kp= 81.9 I2(g) + Cl2(g)  2 ICl(g) If the reaction mixture given above contains PI2 = 0.114 atm; PCl2 = 0.102 atm; PICl = 0.355 atm; which way will the reaction proceed to reach equilibrium? Q = 10.8; since Q < K there is not enough product to be at equilibrium, so the reaction must proceed to the right.
  17. EQUILIBRIUM CONCENTRATION “ICE” PROBLEMS Kc = 1.00 x 102 H2(g) + I2(g)  2HI(g) If 1.00 mol of HI is introduced into a 2.00 L container what are the equilibrium concentrations?
  18. At 3900 0C Kc for the reaction N2(g) + O2(g)  2NO(g) is 0.0123 If 0.750 mol NO is placed in a 3.00L vessel, what are the equilibrium concentrations?
  19. Gaseous hydrogen iodide is synthesized from hydrogen gas and iodine vapor at a temperature where Kp= 100. If the partial pressures are initially 0.500 atm for HI, 0.100 atm for H2 and 0.00500 atm for I2, in a closed flask. Calculate the equilibrium pressures for each gas.
  20. At 35 0C Kc for the reaction 2 NOCl(g)  2NO(g) + Cl2(g) is 1.6 x 10-5 If 1.00 molNOClis placed in a 2.00L vessel, what are the equilibrium concentrations?
  21. Under the proper conditions phosphorus trichloride gas will react with chlorine gas to produce gaseous phosphorus pentachloride. At 30 0C the Kc for the decomposition of PCl5 is 1.000 x 10-3 M. If 2.00 mol of PCl5 is placed in a 2.00 L vessel and allowed to come to equilibrium, what are the equilibrium concentrations?
  22. Solving Equilibrium Problems Write a balanced equation for the reaction. Write the equilibrium expression. List the initial concentrations. Calculate Q. Determine the direction of the shift. Define the change needed to reach equilibrium. Apply change (x) to initial concentrations (ICE). Substitute concentrations into equilibrium expression and solve for unknown. Check to see that calculated equilibrium concentrations give the correct value of K.
  23. A special example, [Equilibrium] with Ksp The equilibrium constant for the dissociation of solid silver chloride (Ksp) is 1.8 x 10-10 M2. If the concentration of Ag+ is 1.0 x 10-5M, what is the concentration of Cl- that can be in equilibrium with Ag+ at that concentration?
  24. Le Châtelier’s Principle Le Châtelier's Principle guides us in predicting the effect various changes have on the position of equilibrium it says that if stress is applied to a system in dynamic equilibrium, the system will change to relieve the stress. The position of equilibrium moves to counteract the change. Three common stressors: Concentration Temperature Pressure
  25. Concentration Changes and Le Châtelier’s Principle A + B ↔ C + D Adding a reactant – equilibrium shifts right Removing a reactant – equilibrium shifts left Adding a product – equilibrium shifts left Removing a product – equilibrium shifts right
  26. Changing Pressure and Le Châtelier’s Principle Only affects a reaction involving gases with an unequal number of mole of reactants & products. Increasing the pressure on the system causes the position of equilibrium to shift toward the side of the reaction with the fewer gas molecules Decreasing pressure causes a shift toward the side with more gas molecules Example: 3H2(g) + N2(g) 2NH3(g) ↑ Pressure – shifts to the right ↓ Pressure – shift to the left
  27. Changing Temperature and Le Châtelier’s Principle Increasing the temperature causes the reaction to shift away from the heat. For exothermic reactions - Think of heat as a product of the reaction Therefore shift the position of equilibrium toward the reactant side For endothermic reactions - Think of heat as a reactant The position of equilibrium will shift toward the products Cooling an exothermic or endothermic reaction will have the opposite effects.
  28. Examples – Le Chatelier’s Principle What effect to the following changes have on the equilibrium position for the following reaction? 1. PCl5(g) + heat ↔ PCl3(g) + Cl2(g) a. addition of Cl2 b. increase in pressure c. removal of heat d. removal of PCl3 as formed 2. C(s) + H2O(g) + heat ↔ CO(g) + H2(g) a. Lowering the temperature b. Increasing the pressure c. Removal of H2 as formed
  29. Examples – Le Chatelier’s Principle 3. At 425 K – Fe3O4(s) + 4H2(g)↔ 3Fe(s) + 4H2O(g) How would the equilibrium concentration of H2O be affected by the following: a. Adding more H2 b. Adding more Fe(s) c. Decreasing the pressure d. Adding a catalyst