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Chemical Equilibrium

Chemical Equilibrium

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Chemical Equilibrium

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  1. Chemical Equilibrium

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

  3. H2O (l) H2O (g) N2O4(g) 2NO2(g) • Chemical equilibrium is achieved when: • the rates of the forward and reverse reactions are equal and • the concentrations of the reactants and products remain constant Physical equilibrium Chemical equilibrium brown colorless

  4. equilibrium equilibrium equilibrium N2O4(g) 2NO2(g) Demo Start with NO2 Start with N2O4 Start with NO2 & N2O4

  5. constant N2O4(g) 2NO2(g)

  6. Using Equilibrium Constants 01 • We can make the following generalizations concerning the composition of equilibrium mixtures: • If Kc > 103, products predominate over reactants. If Kc is very large, the reaction is said to proceed to completion. • If Kc is in the range 10–3 to 103, appreciable concentrations of both reactants and products are present. • If Kc < 10–3, reactants predominate over products. If Kc is very small, the reaction proceeds hardly at all.

  7. 2H2(g) + O2(g) H2(g) + I2(g) 2H2O(g) 2H2(g) + O2(g) 2H2O(g) 2HI(g) Using the Equilibrium Constant (at 500 K) Kc = 4.2 x 10-48 (at 500 K) Kc = 57.0 (at 500 K) Kc = 2.4 x 1047

  8. N2O4(g) 2NO2(g) K = aA + bB cC + dD [NO2]2 [C]c[D]d [N2O4] K = [A]a[B]b = 4.63 x 10-3 Law of Mass Action

  9. N2O4(g) 2NO2(g) 2NO2(g) N2O4(g) ‘ K = K = = 4.63 x 10-3 When the equation for a reversible reaction is written in the opposite direction, the equilibrium constant becomes the reciprocal of the original equilibrium constant. [NO2]2 [N2O4] [N2O4] [NO2]2 1 K = = 216

  10. [N2][H2]3 [NH3]2 [NH3]4 [N2]2[H2]6 [N2][H2]3 [NH3]2 2N2(g) + 6H2(g) N2(g) + 3H2(g) 2NH3(g) N2(g) + 3H2(g) 4NH3(g) 2NH3(g) 1 Kc Write the Equilibrium Constant Kc If you multiply both side of an equilibrium reaction by n the equilibrium constant should be raised to the power of n. Kc = ´ Kc = = ´´ Kc = = Kc 2

  11. N2O4(g) 2NO2(g) Kc = Kp = In most cases Kc Kp P2 [NO2]2 NO2 [N2O4] aA (g) + bB (g)cC (g) + dD (g) P N2O4 Homogenous equilibrium applies to reactions in which all reacting species are in the same phase. Kp = Kc(RT)Dn Dn = moles of gaseous products – moles of gaseous reactants = (c + d) – (a + b)

  12. CH3COOH (aq) + H2O (l) CH3COO-(aq) + H3O+ (aq) ‘ ‘ Kc = Kc [H2O] General practice not to include units for the equilibrium constant. [CH3COO-][H3O+] [CH3COO-][H3O+] = [CH3COOH][H2O] [CH3COOH] Homogeneous Equilibrium [H2O] = constant Kc = The concentration of pure liquids are not included in the expression for the equilibrium constant.

  13. The equilibrium concentrations for the reaction between carbon monoxide and molecular chlorine to form COCl2 (g) at 740C are [CO] = 0.012 M, [Cl2] = 0.054 M, and [COCl2] = 0.14 M. Calculate the equilibrium constants Kc and Kp. CO (g) + Cl2(g) COCl2(g) = 0.14 0.012 x 0.054 [COCl2] [CO][Cl2] = 220 Kc= Kp = Kc(RT)Dn Dn = 1 – 2 = -1 R = 0.0821 T = 273 + 74 = 347 K Kp= 220 x (0.0821 x 347)-1 = 7.7

  14. The equilibrium constant Kp for the reaction is 158 at 1000K. What is the equilibrium pressure of O2 if the PNO = 0.400 atm and PNO = 0.270 atm? 2 PNO PO = Kp 2 PNO 2 Kp = 2NO2 (g) 2NO (g) + O2 (g) PO PO = 158 x (0.400)2/(0.270)2 2 2 2 2 PNO PNO 2 2 2 = 347 atm

  15. CaCO3(s) CaO (s) + CO2(g) ‘ ‘ Kc = Kc x [CaO][CO2] [CaCO3] [CaCO3] [CaO] Kp = PCO 2 Heterogenous equilibrium applies to reactions in which reactants and products are in different phases. [CaCO3] = constant [CaO] = constant Kc= [CO2] = The concentration of puresolids and pure liquids are not included in the expression for the equilibrium constant.

  16. CaCO3(s) CaO (s) + CO2(g) PCO PCO 2 2 does not depend on the amount of CaCO3 or CaO = Kp

  17. Consider the following equilibrium at 295 K: The partial pressure of each gas is 0.265 atm. Calculate Kp and Kcfor the reaction? Kp = P NH3 NH4HS (s) NH3(g) + H2S (g) P H2S = 0.265 x 0.265 = 0.0702 Kp = Kc(RT)Dn Kc= Kp(RT)-Dn Dn = 2 – 0 = 2 T = 295 K Kc = 0.0702 x (0.0821 x 295)-2 = 1.20 x 10-4

  18. Writing Equilibrium Constant Expressions • The concentrations of the reacting species in the liquid phase are expressed in M. In the gaseous phase, the concentrations can be expressed in M or in atm. • The concentrations of pure solids and pure liquids, do not appear in the equilibrium constant expressions. • The equilibrium constant is a dimensionless quantity. • In quoting a value for the equilibrium constant, you must specify the balanced equation and the temperature. • If a reaction can be expressed as a sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions.

  19. equilibrium equilibrium equilibrium N2O4(g) 2NO2(g) Demo Review Start with NO2 Start with N2O4 Start with NO2 & N2O4

  20. Kc = Qc = [NO2]2 [NO2]02 [N2O4] [N2O4]0 The reaction quotient (Qc)is calculated by substituting the initial concentrations of the reactants and products into the equilibrium constant (Kc) expression. • IF • Qc>Kcsystem proceeds from right to left to reach equilibrium • Qc=Kc the system is at equilibrium • Qc< Kcsystem proceeds from left to right to reach equilibrium

  21. [NO ]= 2X mol/1 = 2X 2 [N O ] = X mol/1 = y-X 2 4 2 [NO ] 2 K = [N O ] 2 4 Use reaction quotient to predict the direction of shift when the volume is halved in the following equilibrium: N2O4(g) æ 2 NO2(g), • Consider the reaction: N2O4(g) æ 2 NO2(g), taking place in a cylinder with a volume = 1 unit.

  22. Le Châtelier’s Principle 10 • The Volume is then halved, which is equivalent to doubling the pressure. • Since Q > K, the [product] is too high and the reaction progresses in the reverse direction.

  23. At 12800C the equilibrium constant (Kc) for the reaction Is 1.1 x 10-3. If the initial concentrations are [Br2] = 0.063 M and [Br] = 0.012 M, calculate the concentrations of these species at equilibrium. [Br]2 [Br2] Kc = Br2 (g) 2Br (g) Br2 (g) 2Br (g) = 1.1 x 10-3 Kc = (0.012 + 2x)2 0.063 - x Let (x) be the change in concentration of Br2 Initial (M) 0.063 0.012 ICE Change (M) -x +2x Equilibrium (M) 0.063 - x 0.012 + 2x Solve for x

  24. -b ± b2 – 4ac x = 2a Br2 (g) 2Br (g) Initial (M) 0.063 0.012 Change (M) -x +2x Equilibrium (M) 0.063 - x 0.012 + 2x = 1.1 x 10-3 Kc = (0.012 + 2x)2 0.063 - x 4x2 + 0.048x + 0.000144 = 0.0000693 – 0.0011x 4x2 + 0.0491x + 0.0000747 = 0 ax2 + bx + c =0 x = -0.0105 x = -0.00178 At equilibrium, [Br] = 0.012 + 2x = -0.009 M or 0.00844 M At equilibrium, [Br2] = 0.063 – x = 0.0648 M

  25. Calculating Equilibrium Concentrations • Express the equilibrium concentrations of all species in terms of the initial concentrations and a single unknown x, which represents the change in concentration. • Write the equilibrium constant expression in terms of the equilibrium concentrations. Knowing the value of the equilibrium constant, solve for x. • Having solved for x, calculate the equilibrium concentrations of all species.

  26. Predicting the direction of a Reaction • The reaction quotient (Qc) is obtained by substituting initial concentrations into the equilibrium constant. Predicts reaction direction.Qc > Kc System proceeds to form reactants.Qc = Kc System is at equilibrium.Qc < Kc System proceeds to form products.

  27. Le Châtelier’s Principle 01 • Le Châtelier’s principle: If an external stress is applied to a system at equilibrium, the system adjusts in such a way that the stress is partially offset.

  28. Le Châtelier’s Principle 02 • Concentration Changes: • The concentration stress of an addedreactant or product is relieved by reaction in the direction thatconsumes the added substance. • The concentration stress of a removedreactant or product is relieved by reaction in the direction thatreplenishes the removed substance.

  29. Le Châtelier’s Principle: Haber process N2(g) + 3 H2(g) æ 2 NH3(g) Exothermic Cat:  iron or  ruthenium

  30. Le Châtelier’s Principle 06 • The reaction of iron(III) oxide with carbon monoxide occurs in a blast furnace when iron ore is reduced to iron metal: • Fe2O3(s) + 3 CO(g) æ 2 Fe(l) + 3 CO2(g) • Use Le Châtelier’s principle to predict the direction of reaction when an equilibrium mixture is disturbed by: • (a) Adding Fe2O3 (b) Removing CO2 (c) Removing CO

  31. Le Châtelier’s Principle 07 • Volume and Pressure Changes: Only reactions containing gases are affected by changes in volume and pressure. • Increasing pressure = Decreasing volume • PV = nRT tells us that increasing pressure or decreasing volume increases concentration.

  32. Le Châtelier’s Principle 08 • N2(g) + 3 H2(g) æ 2 NH3(g) Kc = 0.291 at 700 K

  33. Le Châtelier’s Principle 11 • Does the number of moles of reaction products increase, decrease, or remain the same when each of the following equilibria is subjected to a decrease in pressure by increasing the volume. • PCl5(g) æ PCl3(g) + Cl2(g) • CaO(s) + CO2(g) æ CaCO3(s) • 3 Fe(s) + 4 H2O(g) æ Fe3O4(s) + 4 H2(g)

  34. Le Châtelier’s Principle 13 • Temperature Changes: Changes in temperature can change the equilibrium constant. • Endothermic processes are favored when temperature increases. • Exothermic processes are favored when temperature decreases.

  35. Le Châtelier’s Principle 14 • Example: • The reaction N2(g) + 3 H2(g) æ 2 NH3(g) which is exothermic by 92.2 kJ.

  36. Le Châtelier’s Principle 15 • In the first step of the Ostwald process for synthesis of nitric acid, ammonia is oxidized to nitric oxide by the reaction: • 4 NH3(g) + 5 O2(g) æ 4 NO(g) + 6 H2O(g) ∆H° = –905.6 kJ • How does the equilibrium amount vary with an increase in temperature?

  37. Le Châtelier’s Principle 16 • The following pictures represent the composition of the equilibrium mixture at 400 K and 500 K for the reaction A(g) + B(g) æ AB(g). • Is the reaction endothermic or exothermic?

  38. Effect of Catalysis, Reduction of Activation Energy : No effect