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Equilibrium

Equilibrium. UNIT 12. Overview. Concept of Equilibrium Equilibrium constant Equilibrium expression Heterogeneous vs homogeneous equilibrium Solving for equilibrium constant Solving for K Manipulating equilibrium constant Using gases. Equilibrium Calculations ICE method

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Equilibrium

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  1. Equilibrium UNIT 12

  2. Overview • Concept of Equilibrium • Equilibrium constant • Equilibrium expression • Heterogeneous vs homogeneous equilibrium • Solving for equilibrium constant • Solving for K • Manipulating equilibrium constant • Using gases • Equilibrium Calculations • ICE method • ICE Stoichiometry • Reaction Quotient (Q) • Relationship K and Q • Le Chatlier’s Principle • Pressure • Volume • Concentration • Temperature • Catalysts

  3. The Concept of Equilibrium • We’ve already used the phrase “equilibrium” when talking about reactions. • In principle, every chemical reaction is reversible ... capable of moving in the forward or backward direction. 2 H2 + O2 2 H2O Some reactions are easily reversible ... Some not so easy ...

  4. The Concept of Equilibrium • Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate.

  5. The Concept of Equilibrium • As a system approaches equilibrium, both the forward and reverse reactions are occurring. • At equilibrium, the forward and reverse reactions are proceeding at the same rate.

  6. A System at Equilibrium • Once equilibrium is achieved, the amount of each reactant and product remains constant.

  7. A System at Equilibrium Rates become equal Concentrations become constant

  8. Depicting Equilibrium • In a system at equilibrium, both the forward and reverse reactions are running simultaneously. We write the chemical equation with a double arrow:

  9. Equilibrium Constant • Forward reaction: Reverse reaction: Rate Law Rate law

  10. Equilibrium Constant • At equilibrium • Rearranging gives: r

  11. Equilibrium Constant • The ratio of the rate constants is a constant (as long as T is constant). The expression becomes

  12. Equilibrium Expression (Law of Mass Action) To generalize, the reaction: Has the equilibrium expression: This expression is true even if you don’t know the elementary reaction mechanism.

  13. Equilibrium Expression • [A], [B], [C], [D] = molar concentrations or partial pressures at equilibrium • Products = numerator and reactants = denominator • Coefficients in balanced equation = exponents

  14. Equilibrium Expression • Equilibrium constant can have different subscripts • Kc = concentration • Moles, liters, or molarity • Kp = partial pressure • Atmospheres • Ksp = solubility product • Reactant is a solid so only products are in expression

  15. Equilibrium Expression • Does not depend on initial concentration of reactants and products • Solids and pure liquids are not included in equilibrium expression (only gases and solutions) • Keq does not have units (describes activity of reaction) • Depends only on the particular reaction and the temperature

  16. PbCl2 (s) Pb2+ (aq) + 2 Cl−(aq) Equilibrium Expression • The Concentrations of Solids and Liquids Are Essentially Constant • Therefore, the concentrations of solids and liquids do not appear in the equilibrium expression Kc = [Pb2+] [Cl−]2

  17. Sample Exercise 1 • Write the equilibrium expression for Kc for the following reactions:

  18. Sample Exercise 1 – Answers

  19. Homogeneous vs Heterogeneous • Homogeneous equilibrium • Reactants and products are all in the same phase • Example: 2NOBr(g) ↔ 2NO(g) + Br2(g) • Heterogeneous equilibrium • Reactants and products are occur in different phases • Example: PbCl2 (s) ↔ Pb2+(aq) + 2Cl-(aq)

  20. Equilibrium - Summary • At equilibrium… • Concentrations of reactants and products no longer change with time • Neither reactants nor products can escape from the system • The equilibrium constant comes from a ratio of concentration

  21. Solving for Keq Kc, the final ratio of [NO2]2 to [N2O4], reaches a constant no matter what the initial concentrations of NO2 and N2O4 are.

  22. Solving for Keq [0.0172]2 [0.00140] = = 0.211

  23. Solving for Keq This graph shows data from the last two trials from the table.

  24. What does the value of K mean? • If K >> 1, the reaction is product-favored; contains mostly products at equilibrium • If K << 1, the reaction is reactant-favored; contains mostly reactants at equilibrium.

  25. Manipulating Equilibrium Constants • The equilibrium constant of a reaction in the reverse reaction is the reciprocal of the equilibrium constant of the forward reaction.

  26. Manipulating Equilibrium Constants • The equilibrium constant of a reaction that has been multiplied by a number is the equilibrium constant raised to a power that is equal to that number.

  27. Manipulating Equilibrium Constants • If two reactions can be added together to create a third reaction, then the Keq for the two reactions can be multiplied together to get the Keq for the third reaction. If A + B ↔ C Keq = K1 and C ↔ D + E Keq = K2 then A + B ↔ D + E Keq = K1K2

  28. Manipulating Equilibrium Constants (Summary) • The equilibrium constant in the reverse direction is the inverse of the equilibrium constant in the forward direction. • The equilibrium constant of a reaction that has been multiplied by a number is the equilibrium constant raised to a power equal to that number. • The equilibrium constant for a net reaction made up of multiple steps is the product of the equilibrium constants for individual steps.

  29. Equilibrium Constant and Gases • Because pressure is proportional to concentration for gases, the equilibrium expression can also be written in terms of partial pressures (instead of concentration): Mixed versions are also used sometimes:

  30. Relationship between Kc and Kp From the ideal gas law we know that = Pressure in terms of concentration

  31. Relationship between Kc and Kp • Substituting P=[A]RT into the expression for Kp for each substance, the relationship between Kc and Kp becomes Kp= Kc(RT)n Where: n = (moles of gaseous product) – (moles of gaseous reactant) R = 0.0821 L∙atm/mol∙K GIVEN TO YOU ON AP CHEAT SHEET!!

  32. Relationship between Kc and Kp • In the synthesis of ammonia from nitrogen and hydrogen, N2(g) + 3H2(g) ↔ 2NH3(g). Given Kc of 9.60 at 300°C, calculate Kp. Kp= Kc(RT)n n = (moles gaseous products – moles gaseous reactants) = 2-4 = -2 Kp= 9.60(0.0821×573K)-2= 4.34×10-3

  33. The ICE Method

  34. Br2 (g) 2Br (g) ICE • 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. Comparing initial concentrations to equilibrium concentrations

  35. [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 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. 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 14.4

  36. -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] = 0.062 – x = 0.0648 M 14.4

  37. Example Problem: Calculate Concentration Note the moles into a 10.32 L vessel stuff ... calculate molarity. Starting concentration of HI: 2.5 mol/10.32 L = 0.242 M 2 HI H2 + I2 0.242 M 0 0 Initial: Change: Equil: -2x +x +x 0.242-2x x x

  38. Example Problem: Calculate Concentration And yes, it’s a quadratic equation. Doing a bit of rearranging: x = 0.00802 or –0.00925 Since we are using this to model a real, physical system, we reject the negative Root. The [H2] at equil. is 0.00802 M.

  39. Initial Concentration of I2: 0.50 mol/2.5L = 0.20 M I2 2 I 0.20 0 -x +2x 0.20-x 2x Initial change equil:

  40. Approximating If Keq is really small the reaction will not proceed to the right very far, meaning the equilibrium concentrations will be nearly the same as the initial concentrations of your reactants. 0.20 – x is just about 0.20 is x is really dinky. (Example: A million dollars minus a nickel is still a million dollars) If the difference between Keq and initial concentrations is around 3 orders of magnitude or more, go for it. Otherwise, you have to use the quadratic.

  41. Initial Concentration of I2: 0.50 mol/2.5L = 0.20 M I2 2 I 0.20 0 -x +2x 0.20-x 2x Initial change equil: More than 3 orders of mag. between these numbers. The simplification will work here.

  42. Initial Concentration of I2: 0.50 mol/2.5L = 0.20 M I2 2 I 0.20 0 -x +2x 0.20-x 2x Initial change equil: More than 3 orders of mag. between these numbers. The simplification will work here. x = 3.83 x 10-6 M

  43. Equilibrium Stoichiometry Given initial concentrations of reactants AND Equilibrium concentration of products (or vice versa)

  44. H2 (g) + I2 (g) 2 HI (g) Equilibrium Stoichiometry A closed system initially containing1.000 x 10-3 M H2 and 2.000 x 10-3 M I2 at 448C is allowed to reach equilibrium. Analysis of the equilibrium mixture shows that the concentration of HI is 1.87 x 10-3 M. Calculate Kc at 448C for the reaction:

  45. Equilibrium Stoichiometry ICEmethod: Write what we know.

  46. Equilibrium Stoichiometry Calculate the change in [HI]. [HI] Increases by 1.87 x 10-3M

  47. Equilibrium Stoichiometry Use stoichiometry to solve for the change in concentration of [H2] and [I2] from the given change in [HI].

  48. Equilibrium Stoichiometry Now calculate concentrations at equilibrium

  49. Kc= [HI]2 [H2] [I2] (1.87 x 10-3)2 (6.5 x 10-5)(1.065 x 10-3) = = 51 …and, therefore, the equilibrium constant

  50. The Reaction Quotient (Q) • To calculate Q, one substitutes the initial concentrations of reactants and products into the equilibrium expression. • Q gives the same ratio the equilibrium expression gives, but for a system that is not at equilibrium. [A], [B], [C], [D] = initial molar concentrations or partial pressures Q

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