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Chemical Equilibrium

Chapter 13. Chemical Equilibrium. Chemical Equilibrium. When a reaction is not totally converted from reactants to products, a condition can be set up known as chemical equilibrium.

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Chemical Equilibrium

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  1. Chapter 13 Chemical Equilibrium

  2. Chemical Equilibrium When a reaction is not totally converted from reactants to products, a condition can be set up known as chemical equilibrium. In this state, the concentrations of all reactants and products remain constant with time, as both forward and reverse reactions are occurring at the same rate. On the molecular level, there is frantic activity. Equilibrium is not static, but is a highly dynamicsituation.

  3. Chemical Equilibrium: occurs when opposing reactions are proceeding at equal rates and the concentrations of products and reactants no longer change with time. Ratef = Rater

  4. Chemical Equilibrium For this to occur, the system must be closed, no products or reactants may escape the system.

  5. For Reaction: AB Forward Reaction: AB Rate = kf[A] kf= forward rate constant Reverse reaction: BA Rate = kr[B] kr= reverse rate constant

  6. As forward reaction occurs, • [A] decreases and forward rate slows. • [B] increases and the rate of reverse reaction increases. • Eventually the forward and reverse reactions reach the same rate. (Chemical Equilibrium)

  7. kf[A]= kr[B] • Once equilibrium is established, the concentrations of A and B do not change.

  8. Writing equilibrium expressions Keq = [products] [reactants] NOTE: [at equilibrium] Note: eq is the general subscript for an equilibrium constant. Your textbook uses Keq throughout. Please look at the equilibrium constants given on the AP exam. These are the same as Keq, but are more specific to a type of reaction (examples: gaseous reaction or acid base reaction).

  9. Two common ways of describing equilibrium: Kc = Concentration of substances in the reaction are known. Kp = Units of partial pressure are used instead of concentration. *Kp is used when the reactants and products are in the gaseous state. *Solids and liquids are left out of equilibrium expressions because their concentrations do not change during chemical reactions.

  10. Equilibrium Expressions Summary • The K value for a reaction that is reversed is the reciprocal of that K for the forward reaction • There are no units for K because of corrections to the nonideal behavior of substances in the reaction.

  11. Law of Mass Action: Expresses the relationship between concentrations of reactants and products at equilibrium. Example: aA + bB  cC + dD Kc = [C]c[D]d [A]a[B]b

  12. Kc = [C]c[D]d [A]a[B]b • The expression above is known as the equilibrium-constant expression. • Kcis the equilibrium constant. • Subscript c indicates concentration in molarity.

  13. Write the equilibrium constant expressions for the following reaction in the forward direction. 2SO2(g) + O2(g) 2SO3(g)

  14. Homework #22 • K = = = 3.2 × 1011

  15. Example Calculations can be made if the equilibrium concentrations of reactants and products are known. If [NH3] = 3.1 x 10-2 mol/L, [N2]=8.5 x 10-1 mol/L, and [H2] = 3.1 x 10-3 mol/L, N2 (g) + 3H2 (g) 2NH3 (g) a. What is the value of K? b. What is the value of K’ (for the reverse reaction)? c. What would be the K for this reaction? ½ N2 (g) + 3/2 H2 (g) NH3 (g) (3.1 x 10-2 )2 = (8.5 x 10-1 )(3.1 x 10-3 )3 = 3.8 x 104 K’ = 1/K = 1/ 3.8 x 104 =2.6 x 10-5 For the reaction in (c), Kc = K ½ = (3.8 x 104 L2/mol2) ½ = 1.9 x 102

  16. The Haber Process N2(g) + 3H2(g) 2NH3(g) • This process for making ammonia is done at high temperature and pressure. • When complete all 3 components are present in the closed tank.

  17. Regardless of the starting concentrations, at equilibrium the relative concentrations of the 3 gases are the same.

  18. Equilibrium constant for Haber Process • The equilibrium-constant depends only on the stoichiometry (3,2,1) of the reaction, not on its mechanism.

  19. The following equilibrium process has been studied at 230ºC: 2NO(g)+ O2(g)2NO2(g) In one experiment theconcentration of the reacting species at equilibrium are found to be [NO]=0.0542M [O2]=0.127M [NO2]=15.5M. Calculate Kc of the reaction at this temperature.

  20. 6.44 x 105Kc has no units

  21. Kc in terms of pressure Example: aA + bB  cC + dD PA is partial pressure of A in atm, PB is the partial pressure of B, etc.

  22. Example The reaction forming nitrosyl chloride is 2 NO (g) + Cl2 (g) 2 NOCl (g) If the pressures at equilibrium are PNOCl = 1.2 atm, PNO=5.0 x 10-2 atm, and PCl2 = 3.0 x 10-1 atm, what is the value of Kp? Kp = PNOCl2 = (1.2)2= 1.9 x 103 atm-1 (PNO )2(PCl2) (5.0 x 10-2)2(3.0 x 10-1)

  23. Values for Kc and Kp are usually different. • It is possible to calculate one from the other. • n is the change in the number of moles of gas. Δn = nproducts - nreactants

  24. N2O4(g) 2NO2(g) • n = 2-1 = 1 • For the above reaction, Kp=Kc(RT)

  25. For the equilibrium 2SO3(g) 2SO2(g) + O2(g) at temperature 1000K, Kc has a value 4.08 x 10-3. Calculate the value for Kp. Kp = 4.08 x 10-3 (.0821 x 1000)1 0.335

  26. A 3:1 starting mixture of hydrogen, H2, and Nitrogen, N2 comes to equilibrium at 500 °C. The mixture at equilibrium is 3.506% NH3, 96.143% N2, and 0.3506% H2 by volume. The total pressure in the reaction vessel was 50.0 atm. What is the value of Kp and Kc for the reaction? N2 + 3H2 2NH3 Hint: Use Dalton’s Law of partial pressures to determine the partial pressure of each component in the reaction. PA = XA PT Since we know we have 100%, Partial Pressure = (% of substance /100 X total pressure)

  27. P NH3 = 0.03506 x 50.0 atm = 1.75 atm P N2 = 0.96143 x 50.0 atm = 48.1 atm P H2 = 0.003506 x 50.0 atm = 0.175 atm (P NH3)2 (1.75)2 Kp = = =11.9 (P N2) (P H2)3 (48.1) (0.175)3 Δn ( 1 RT ) Kc = Kp -2 ( 1 0.0821 x 773 ) = 4.79 x 104 Kc = Kp

  28. Homework #28 • KP = K(RT)Δn, • K = Δn = 2 - 3 = -1; • K = 0.25 × (0.08206 × 1100) = 23

  29. Magnitude of Equilibrium constants • When Kc is very large the numerator is much larger than the denominator. • The equilibrium lies to the right, or to the product side of the equation (mostly products formed). K =[products] [reactants]

  30. When the equilibrium constant is very small the equilibrium lies to the left, or the reactant side (mostly reactants). • K>>1 Products favored. • K<<1 Reactants favored.

  31. Heterogenous Equilibria • Substances that are in equilibria and are in different phases. CaCO3(s) CaO(s)+ CO2(gas) Equilibrium constant

  32. The concentration of a pure solid or liquid is a constant. • If a pure solid or pure liquid are involved in a heterogeneous equilibrium, its concentration is not included in the equilibrium-constant expression for the Rx.

  33. Write the equilibrium expression Kc and Kp for the following reaction: 3Fe(s)+ 4H2O(g) Fe3O4(s) + 4H2(g)

  34. Applications of Equilibrium Constants Magnitude of K determines whether reaction will proceed. Using K we can predict: • Direction of reaction • Calculate concentrations of reactants and products at equilibrium.

  35. Reaction Quotient (Q) • This is used when you are given information about reactants and products when they have not yet reached equilibrium. • Allows us to determine direction reaction WILL go. --may be given concentrations before reaction begins. --may be given concentrations determined during a reaction. • Calculate Q as you do K. N2(g) + 3H2(g) 2NH3(g) • Compare Q and K as follows.

  36. Reaction Quotient (Q) and reaction direction • Q=K when system is at equilibrium • Q>K reaction goes to left(reactants) • Q<K reaction goes to right(products)

  37. Direction of a Reaction Step 1: Calculate the reaction quotient, Qc Step 2: Compare Qc to Kc

  38. case 1: Qc > Kc Too much product! Large numerator. At equilibrium case 2: Qc = Kc Too much reactants! case 3: Qc < Kc Large denominator.

  39. N2(g)+ 3H2(g) 2 NH3(g) • At the start of a reaction, there are 0.249 mol N2, 3.21 x 10-2 mol H2, and 6.42 x 10-4 mol NH3 in a 3.50 L reaction vessel at 200º C. If the equilibrium constant (Kc) for the reaction is 0.65 at this temperature, decide whether the system is at equilibrium. If it is not, predict which way the net reaction will proceed.

  40. Initial Concentrations

  41. N2(g)+ 3H2(g) 2 NH2(g) Kc= 0.65 Qc is smaller than Kc. Not at equilibrium; too much reactants. Net Rx goes to right.

  42. Problem N2(g) + 3H2(g) 2NH3(g) In the reaction above, the value of Kc at 500°C is 6.0 x 10-2. At some point during the reaction, the concentrations of each material were measured. At this point, the concentrations of each substance were [N2] = 1.0 x 10-5 M, [H2] = 1.5 x 10-3 M, and [NH3] = 1.5 x 10-3 M. Determine the direction that the reaction was most likely to proceed when the measurements were taken. (.0015)3 = (1.0 x 10-5) (1.5 x 10-3 )3 Qc=1.0 x105 Most likely will go to the left (reactant side)

  43. Extent of reaction • Kc >> 1 (any # > 10) Forward rxn goes to near completion Kc << 1 (any # < 0.1) Reverse rxn goes to near completion

  44. Homework #34 • Determine Q for each reaction, and compare this value to Kp (2.4 × 103) to determine which direction the reaction shifts to reach equilibrium. Note that, for this reaction, K = Kp since Δn = 0. • Q = = 2.2 × 103 Q < Kp so the reaction shifts right to reach equilibrium.

  45. Homework #34 Cont. b. Q = = 4.0 × 103 > Kp  Reaction shifts left to reach equilibrium. c. Q = = 2.4 × 103 = Kp; at equilbrium

  46. Calculating Equilibrium Constant “When all Equilibrium Concentrations are not known.”

  47. Calculating Equilibrium Constants • Equilibrium concentrations are often unknown. • If one equilibrium concentration is known, we can use stoichiometry to find the equilibrium concentration of the other species.

  48. Equilibrium Concentrations Initial concentration Change in concentration Equilibrium concentration Called an ICE diagram

  49. When 2.00 mole each of hydrogen (H2) and (I2) are mixed in an evacuated 1.00 L vessel, 3.50 moles of HI are produced. What is the value of the equilibrium constant, Kc?Procedure for calculating equilibrium concentrations. H2(g) + I2(g) 2HI(g) 1) First, we design our table so there is a column for each component in the equation, and rows for start, Δ, and finish.

  50. H2(g) + I2(g) 2HI(g) • Next, we fill in what we know. • *Since HI is a product, it is 0 at the start of the reaction. • *We also know that at equilibrium, 3.50 mol of HI exist.

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