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Chapter 13

Chapter 13. Equilibrium. Unit Essential Question. How do equilibrium reactions compare to other reactions?. Lesson Essential Question (sections 1-3). What is an equilibrium expression and how can an equilibrium expression be determined?. Section 13.1. The equilibrium condition.

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Chapter 13

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  1. Chapter 13 Equilibrium

  2. Unit Essential Question How do equilibrium reactions compare to other reactions?

  3. Lesson Essential Question(sections 1-3) What is an equilibrium expression and how can an equilibrium expression be determined?

  4. Section 13.1 The equilibrium condition

  5. Reactions are Reversible • Most reactions do NOT proceed to completion and then stop. • Most reactions proceed in both the forward and reverse directions: A + B C + D ( forward) C + D A + B (reverse)

  6. Reactions are Reversible • A + B C + D ( forward) • C + D A + B (reverse) • Initially only A and B are present, so only the forward reaction is possible. • As C and D are produced the reverse reaction speeds up, while the forward reaction slows down. • Eventually the rates become equal.

  7. Chemical equilibrium: concentrations of all species remain constant with time.

  8. What is equal at Equilibrium? • Rates are equal. • Concentrations are not. • Rates are determined by concentrations and activation energy. • The concentrations do not change at equilibrium. • Recall equilibrium is dynamic! • Reactions keep going at equilibrium! • A + B C + D

  9. Section 1 Homework Pg. 614 # 9, 12

  10. Section 13.2 The equilibrium constant

  11. Reaction Quotient, Q • A ratio between the amount of reactants and products can be made at any time during the reaction. This is called the reaction quotient (Q). • For aA + bBcC + dD the reaction quotient is: Q = [C]c [D]d [A]a [B]b • Notice the effect of coefficients! • Q may be written using concentrations (Qc) or partial pressures (Qp).

  12. Equilibrium Constant, K • The ratio between the amount of reactants and products at equilibrium is called the equilibrium constant (K). • K = [C]c [D]d [A]a [B]b • Again, may be Kc or Kp. • K has no units- it represents a ratio! • Side note: P = CRT. Derived from ideal gas law since C = n/V (C = concentration). equilibrium expression

  13. Equilibrium Constant, K • Pure solids and liquids are NOT included in the equilibrium expression; their concentrations are assumed to be one. Why? • Concentrations of pure substances don’t change! • Also, K is constant at any temperature. * If T changes then K will change!

  14. Example Write the equilibrium expression (Kp) for the following reaction: SiH4(g) + 2Cl2(g) SiCl4(g) + 2H2(g) Remember to use partial pressures! Kp = (PSiCl4)(PH2)2 (PSiH4)(PCl2)2

  15. Altering K • If we write the reaction in reverse: cC + dD aA + bB • Then the new equilibrium constant is: K’ = [A]a [B]b = 1/K [C]c [D]d

  16. Example At a given temperature, K = 1.3 x 10-2 for the following reaction: N2(g) + 3H2(g) 2NH3(g) What is the value of K for the following: 2NH3(g) N2(g) + 3H2(g) K = 1/(1.3 x 10-2) = 77

  17. Altering K • If we multiply the reaction by some number n: (aA + bB cC + dD)n naA + nbB ncC + ndD • Then the new equilibrium constant is: K’ = [C]nc [D]nd = ([C]c [D]d)n = Kn [A]na [B]nb ([A]a [B]b)n

  18. Example At a given temperature, K = 1.3 x 10-2 for the following reaction: N2(g) + 3H2(g) 2NH3(g) What is the value of K for the following: 1/2N2(g) + 3/2H2(g) NH3(g) K = (1.3 x 10-2)1/2 = 0.11

  19. Example- Calculating K • N2(g) + 3H2(g) 2NH3(g) InitialEquilibrium • [N2]0 =1.000 M [N2] = 0.921M • [H2]0 =1.000 M [H2] = 0.763M • [NH3]0 =0 M [NH3] = 0.157M • K = _ [NH3]2_ = 0.0603 [N2] [H2]3

  20. Example- Calculating K • N2(g) + 3H2(g) 2NH3(g) InitialEquilibrium • [N2]0 = 0 M [N2] = 0.399 M • [H2]0 = 0 M [H2] = 1.197 M • [NH3]0 = 1.000 M [NH3] = 0.203M • K = 0.0602 • K is the same no matter what the amount of starting materials.

  21. What does K tell us? By looking at the size of K (and also Q), you can determine relative amounts of reactants and products present. Large values of K indicate greater amounts of products compared to reactants (forward reaction is favored/reaction lies to the right). Small values of K indicate more reactants are present (reverse reaction is favored/reaction lies to the left).

  22. Kc and Kp • Values of K calculated from concentrations or partial pressures can be related: Kp = Kc(RT)Δn • R = ideal gas constant: 0.0821Latm/molK • Make sure to choose the correct R value- look at units of P! Also T must be in K! • Δn = change in # moles of gas in the reaction: mol product – mol reactant

  23. Example For the reaction C(s) + CO2(g) 2CO(g) Kp=1.90 at 25°C. What is Kc? 1.90 = Kc[(0.0821Latm/molK)(298K)]1 Kc = 0.0776

  24. Sections 2&3 Homework Pg. 614 # 17(a&b), 18 (a&b),19(c&d), 21, 23, 25, 29

  25. Lesson Essential Question(sections 5-7) How can equilibrium pressures and concentrations be calculated?

  26. Section 13.5 Applications of the equilibrium constant

  27. Using Q • Recall that it’s calculated the same way as the equilibrium constant, but for a system not at equilibrium (or if you’re not sure if a reaction is at equilibrium or not). • By calculating Q you can compare the value to K to get an idea of where a reaction is in terms of reaching equilibrium.

  28. What Q tells us: • If Q<K • Not enough products (has not yet reached equilibrium). • Shift to the right to reach equilibrium. • If Q>K • Too many products (has gone past equilibrium). • Shift to the left to reach equilibrium. • If Q=K system is at equilibrium

  29. Example For the reaction 2NO(g) N2(g) + O2(g) K = 2.4 x 103 at a certain temperature. If there are 0.024mol NO, 2.0mol N2, and 2.6mol O2 in a 1.0L flask, is the reaction at equilibrium? If not, in which direction will the reaction shift to reach equilibrium? Q = (2.0)(2.6) = 9.0 x 103 Q > K (0.024)2 rxn. shifts left

  30. Section 5 Homework Pg. 615 # 33(b&c), 34

  31. Section 13.6 Solving equilibrium problems

  32. Procedure Write balanced equation. Write equilibrium expression. Calculate Q, determine shift direction. Start ICE table– determine I (initial concentrations). Define change, C – add to I to find E (equilibrium concentrations). Substitute E into expression and solve Check values in expression, see if they equal K.

  33. Example If K = 5.10 and there is 1.000mol of each substance present in a 1.000L flask, what are the equilibrium concentrations? CO(g) + H2O(g) CO2(g) + H2(g) Initial concentrations: all 1.000M Determine Q: Q = (1.000M)(1.000M) = 1.000 Q<K, so (1.000M)(1.000M) rxn. shifts right

  34. Example Cont. • Now make your ICE table! Write ICE in a column below and to the left of the eqn. • Use balanced equation coefficients and the direction the reaction shifts to find change. • Add I and C to get E. • ICE table helps you keep track of everything: CO(g) + H2O(g) CO2(g) + H2(g) I 1.000 1.000 1.000 1.000 C -x -x +x +x E 1.000-x 1.000-x 1.000+x 1.000+x

  35. Example Cont. Now plug the E values into the equilibrium expression and solve for x: 5.10 = [CO2][H2]_ = (1.000+x)(1.000+x) [CO] [H2O] (1.000-x)(1.000-x) 5.10 = (1.000 + x)2 solve for x: x = 0.387 mol/L (1.000 – x)2

  36. Example Cont. [CO] & [H2O] = 1.000 – 0.387 = 0.613M [CO2] & [H2] = 1.000 + 0.387 = 1.387M Verify: K = (1.387)(1.387) = 5.12 (0.613)(0.613)

  37. Practice K = 115 for a reaction where 3.000mol of all species were added to a 1.500L flask. What are the equilibrium concentrations? H2(g) + F2(g) 2HF(g) Q = 1.000, so rxn. will shift right. 115 = (2.000 + 2x)2 => x = 1.528 (2.000 – x)2 [H2] = [F2] = 0.472M ; [HF] = 5.056M

  38. Using the Quadratic Formula • You may have wondered when you would use this! • Need it to solve some equilibrium problems (form: ax2 + bx + c). • Previous problems worked out nicely- just had to take the square root of both sides. • Can use graphing calculator program (if you have it) to solve. Otherwise know it:

  39. Example #1 – Quadratic Formula K = 115 for a reaction where 3.000mol of H2 and 6.000mol F2 were added to a 3.000L flask. What are the equilibrium concentrations? H2(g) + F2(g) 2HF(g) Q = _[HF]2_ [H2][F2] Reaction will shift right.

  40. Example #1 Cont. a = 111 b = -345 c = 230. H2(g) + F2(g) 2HF(g) I 1.000 2.000 0 C -x -x +2x E 1.000-x 2.000-x 2x K = 115 = _____(2x)2_____ (1.000-x)(2.000-x) (1.000-x)(2.000-x)(115) = 4x2 (2.000 -3.000x+x2)(115) = 4x2 230. – 345x + 115x2 = 4x2 111x2 – 345x + 230. = 0

  41. Example #1 Cont. a = 111 b = -345 c = 230. Can’t have a negative concentration! Use quadratic formula: x = 345 ±√(-345)2 – 4(111)(230.) 2(111) x = 2.14M and x = 0.968M Only one value of x is correct: [H2] = 1.000M – x [H2] = 1.000M – 2.14M = -1.14M [H2] = 1.000M – 0.968M = 0.032M [F2] = 2.000M – 0.968M = 1.032M [HF] = 2(0.968) = 1.936M

  42. What if K is very small? • If K is < 10-3 we can consider it to be negligible with respect to concentrations. • Can make calculations simpler. • Can ignore x only where it would make little change in the concentration. • MUST check to see if the calculated equilibrium concentration is less than 5% different from the initial – if it’s more than 5%, then you can’t disregard x.

  43. Example #2- Small K Values Exactly 1.0mol NOCl is placed in a 2.0L flask. The value of K = 1.6x10-5. What are the equilibrium concentrations? 2NOCl(g) 2NO(g) + Cl2(g) *Rxn. will shift right b/c there are no products initially. *The equilibrium expression will be: K = [NO]2[Cl] = 1.6x10-5 [NOCl]2

  44. Example #2- Small K Values 2NOCl(g) 2NO(g) + Cl2(g) I 0.50 0 0 C -2x +2x +x E 0.50-2x 2x x *This would result in a complicated math equation to solve. *Simplification: K is small (<10-3), so x is negligible only with respect to NOCl. -Small K means reaction barely proceeds right, so 0.50M NOCl won’t change noticeably.

  45. Example #2- Small K Values 2NOCl(g) 2NO(g) + Cl2(g) I 0.50 0 0 C -2x +2x +x E 0.50-2x 2x x *Simplification: 0.50 – 2x ≈ 0.50M 1.6x10-5 = (2x)2(x) = _4x3_ (0.50)2 (0.50)2 x3 = (1.6x10-5)(0.50)2 => x = 0.010 4

  46. Example #2- Small K Values Verify assumption that x is negligible with respect to NOCl: [NOCl] = 0.50 – 2(0.010) = 0.48 *This must be less than 5% different from the initial concentration: 0.50-0.48 = 0.02 0.02/0.50 x 100 = 4% Assumption was ok!

  47. Example #2- Small K Values Calculate the rest of the equilibrium concentrations: [NOCl] = 0.50M (this was the simplification!) [NO] = 2x = 2(0.010) = 0.020M [Cl] = x = 0.010M

  48. Section 6 Homework Pg. 616 # 45, 46, 48, 51

  49. Lesson Essential Question How does Le Châtelier’s Principle effect the equilibrium of a reaction?

  50. Section 13.7 Le CHÂTELIER’S PRINCIPLE

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