1 / 51

At equilibrium: rate forward = rate reverse

Kinetics applies to the speed of a reaction, the concentration of product that appears (or of reactant that disappears) per unit time. Equilibrium applies to the extent of a reaction, the concentration of product that has appeared after an unlimited time, or once no further change occurs.

rmackie
Download Presentation

At equilibrium: rate forward = rate reverse

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Kinetics applies to the speed of a reaction, the concentration of product that appears (or of reactant that disappears) per unit time. Equilibrium applies to the extent of a reaction, the concentration of product that has appeared after an unlimited time, or once no further change occurs. At equilibrium: rateforward= ratereverse A system at equilibrium is dynamic on the molecular level; no further net change is observed because changes in one direction are balanced by changes in the other.

  2. CHEMICAL EQUILIBRIUM - RATES OF REACTION kF Reactants  products kB Chemical reactions are a dynamic process, that is, reactions involve both forward and reverse processes. Chemical Equilibrium is reached by a reaction mixture when the rates of forward and reverse reactions becomes equal. kF = kB NO net change appears obvious although the system is still in constant motion.

  3. Rate Forward Rate Reverse Reaction Dynamics Initially, only the forward reaction takes place. As the forward reaction proceeds it makes products and uses reactants. Because the reactant concentration decreases, the forward reaction slows. As the products accumulate, the reverse reaction speeds up. Once equilibrium is established, the forward and reverse reactions proceed at the same rate, so the concentrations of all materials stay constant. Eventually, the reaction proceeds in the reverse direction as fast as it proceeds in the forward direction. At this time equilibrium is established. Rate Time

  4. N2O4(g)2NO2(g) Reaching equilibrium on the macroscopic and molecular levels.

  5. H2(g) + I2(g)  2 HI(g) at time 0, there are only reactants in the mixture, so only the forward reaction can take place [H2] = 8, [I2] = 8, [HI] = 0 at time 16, there are both reactants and products in the mixture, so both the forward reaction and reverse reaction can take place [H2] = 6, [I2] = 6, [HI] = 4 5

  6. H2(g) + I2(g)  2 HI(g) at time 32, there are now more products than reactants in the mixture − the forward reaction has slowed down as the reactants run out, and the reverse reaction accelerated [H2] = 4, [I2] = 4, [HI] = 8 at time 48, the amounts of products and reactants in the mixture haven’t changed – the forward and reverse reactions are proceeding at the same rate – it has reached equilibrium

  7. LAW OF MASS ACTION K  equilibrium constant aA + bBcC + dD K = Products Reactants K = [C]c [D]d = Q reaction quotient [B]b [A]a Write the equilibrium equation for: a. H2O + H2O  H3O+ + OH- b. 4NH3(g) + 302(g) 2N2(g) + 6H2O(g) c. N2(g) + 3H2(g) 2NH3(g)

  8. Workshop Question E1 HOMOGENEOUS EQUILIBRIA H2(g) + I2(g) 2HI(g) Kc = then Kp = _________ 2O3(g) 3O2(g) Kp = Kc = ________ [PHI]2 [PI2][PH2] (PO2)3 (PO3)2 HETEROGENEOUS EQUILIBRIA 2H2O(l) H3O+(aq)_ + OH-(aq) C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(g) Kc = [H3O+][OH-] Kc = _____________

  9. HETEROGENEOUS EQUILIBRIA Substance in more than one phase 1. CaCO3(s) CaO(s) + CO2(g) Kc = [CaO][CO2] [CaCO3] How do the [ ] of solid express? Pure solids & liquid have constant [conc. ] D = g/cm3 = mol MW g/mol cm3 Kc = constant [CO2] constant Kc = Kccon = [CO2] con ?

  10. The reaction quotient for a heterogeneous system. solids do not change their concentrations

  11. Q. Each of the mixtures listed below was placed in a closed container and allowed to stand. Which of these mixtures is capable of attaining the equilibrium, expressed by 1 a) pure CaCO3 b) CaO & PCO2 > Kp c) solid CaCO3 & PCO2 > Kp d) CaCO3 & CaO

  12. LE CHATELIER’S PRINICIPLE “If a system at equilibrium is disturbed by a change in temperature, pressure, or concentration of one of its components, that system will shift it’s equilibrium position as to ‘counteract’ the effect of the disturbance.” Equilibrium can be disturbed by: - adding or removing components - a change in pressure - a change in volume - a change in temperature

  13. PREDICTING THE DIRECTION OF THE SHIFT I. CATALYST A catalyst increases the rate at which equilibrium is achieved but not the composition of the equilibrium mixture. II. THE REACTION QUOTIENT Q < K: the reaction shifts to the products Q > K: the reaction shifts to the reactants III. CHANGES IN VOLUME Reducing the volume of a gas at equilibrium causes the system to shift in the direction that reduces the number of moles of gas.

  14. + lower P (higher V) more moles of gas higher P (lower V) fewer moles of gas The effect of pressure (volume) on an equilibrium system.

  15. When the pressure is decreased by increasing the volume, the position of equilibrium shifts toward the side with the greater number of molecules – the reactant side. Since there are more gas molecules on the reactants side of the reaction, when the pressure is increased the position of equilibrium shifts toward the products. The Effect of Volume Changes on Equilibrium 15

  16. PROBLEM: How would you change the volume of each of the following reactions to increase the yield of the products. (a) CaCO3(s) CaO(s) + CO2(g) (b) S(s) + 3F2(g) SF6(g) (c) Cl2(g) + I2(g) 2ICl(g) PLAN: When gases are present a change in volume will affect the concentration of the gas. If the volume decreases (pressure increases), the reaction will shift to fewer moles of gas and vice versa. Sample Problem Predicting the Effect of a Change in Volume (Pressure) on the Equilibrium Position SOLUTION:

  17. IV. CHANGES IN TEMPERATURE When heat is added to a system, the equilibrium shifts in the direction that absorbs heat. Cooling has the opposite effect and shifts the equilibrium towards the side which produces heat. Endothermic Reactions: heat + Reactants  Products An increase in temperature leads to a shift towards the products, a decrease leads to a shift towards the reactants. (Keq increases) Exothermic Reactions: Reactants  Products + heat An increase in temperature leads to a shift towards reactants. (Keq decreases)

  18. The Effect of Temperature Changes on Equilibrium

  19. PROBLEM: How does an increase in temperature affect the concentration of the underlined substance and Kc for the following reactions? (a) CaO(s) + H2O(l) Ca(OH)2(aq) H0 = -82kJ (b) CaCO3(s) CaO(s) + CO2(g) H0 = 178kJ (c) SO2(g) S(s) + O2(g) H0 = 297kJ PLAN: Express the heat of reaction as a reactant or a product. Then consider the increase in temperature and its effect on Kc. Sample Problem Predicting the Effect of a Change in Temperature on the Equilibrium Position SOLUTION:

  20. Practice - Le Châtelier’s Principle • The reaction 2 SO2(g) + O2(g)  2 SO3(g) with H° = -198 kJ is at equilibrium. How will each of the following changes affect the equilibrium concentrations of each gas once equilibrium is re-established? • adding more O2 to the container • condensing and removing SO3 • compressing the gases • cooling the container • doubling the volume of the container • warming the mixture • adding the inert gas helium to the container • adding a catalyst to the mixture

  21. Workshop question E2a N2O4(g) 2 NO2(g)H = 58 kJ In which direction will the equilibrium shift when each of the following changes are made to a system at equilibrium? A) add N2O4 B) remove NO2 C) increase the total pressure by adding N2 D) increase the volume of the container E) decrease the temperature

  22. Workshop Question E2b PCl5(g) PCl3(g) + Cl2(g)H = 88 kJ In which direction will the equilibrium shift when each of the following changes are made to a system at equilibrium? A) add Cl2 B) temperature is increased C) the volume of the reaction system is decreased D) PCl5 is added E) a catalyst is added

  23. [C]c[D]d Q = [A]a[B]b Q - The Reaction Quotient At any time, t, the system can be sampled to determine the amounts of reactants and products present. A ratio of products to reactants, calculated in the same manner as K tells us whether the system has come to equilibrium (Q = K) or whether the reaction has to proceed further from reactants to products (Q < K) or in the reverse direction from products to reactants (Q > K). We use the molar concentrations of the substances in the reaction. This is symbolized by using square brackets - [ ]. For a general reaction aA + bB cC + dD where a, b, c, and d are the numerical coefficients in the balanced equation, Q (and K) can be calculated as

  24. Reaction Progress Reaction Progress Equilibrium: no net change reactants products reactants products Reaction direction and the relative sizes of Q and K.

  25. The change in Q during the N2O4-NO2 reaction.

  26. Q, K, and the Direction of Reaction 26

  27. Predicting the direction of reaction: Q > K forms more reactants  Q = K equilibrium Q < K forms more products  Note: 1. Kf = 1 Kr 2. K = Kn If the balanced equation is multiplied by a factor then the K (& Q) is multiplied by the exponent.

  28. small K large K intermediate K The range of equilibrium constants

  29. UNDERSTANDING THE DIRECTION OF REACTIONS AND Keg The following reaction is a means of “fixing” nitrogen: N2(g) + O2(g) 2 NO(g) A. If the value for Q at 25°C is 1 x 10-30, describe the feasibility of this reaction for Nitrogen fixation. B. Write the equilibrium expression, Kc C. Write the equilibrium expression for 2NO(g) N2(g) + O2(g) D. Determine the Kc for “C”

  30. Workshop Question E3 1) The equilibrium constant is given for one of the reactions below. Determine the value of the missing equilibrium constant. Which reaction is the most feasible at this temperature and explain why?   2 SO2(g) + O2(g) ⇌ 2 SO3(g) Kc = 1.7 × 106 SO3(g) ⇌ 1/2 O2(g) + SO2(g) Kc = ? 2) The equilibrium constant is given for two of the reactions below. Determine the value of the missing equilibrium constant.   A(g) + 2B(g) ⇌ AB2(g) Kc = 59 AB2(g) + B(g) ⇌ AB3(g) Kc = ? A(g) + 3B(g) ⇌ AB3(g) Kc = 478 3) Consider the following reaction and its equilibrium constant: I2(g) ⇌ 2I(g) Kp = 0.209  Write the equilibrium expression. If a reaction mixture contains 0.89 atm I2 and 1.77 atm I. Which of the following statements is TRUE concerning this system? A) The reaction will shift in the direction of reactants. B) The reaction quotient will increase. C) The reaction will shift in the direction of products. D) The equilibrium constant will decrease. E) The system is at equilibrium.

  31. K as either Kc or Kp Kc = the equilibrium constant using concentrations. Kp = the equilibrium constant using pressure Kp = Kc(RT) n gas n is the difference between the number of moles of products and moles of reactants R = 0.082 L atm / mol K

  32. Kc vs. Kp Kp = Kc(RT) n gas R = 0.082 L atm/ mol K For the following write Kp then Kc for: N2(g) + 3H2(g) 2NH3(g) Calculate Kp if Kc = 0.105

  33. Workshop E4 on Kc vs. Kp • Write Kp then Kc for: • 1. N2O4(g) 2NO2(g) • 2. Calculate Kc if Kp = 0.245 for Q#1 • 2SO3(g) 2SO2(g) + O2(g) if Kc= 4.07x10-3, what is Kp?

  34. Predicting the direction of reaction: Q > K forms more reactants  Q = K equilibrium Q < K forms more products  Note: 1. Kf = 1 Kr 2. K = Kn If the balanced equation is multiplied by a factor then the K (& Q) is multiplied by the exponent.

  35. small K large K intermediate K The range of equilibrium constants

  36. K as either Kc or Kp Kc = the equilibrium constant using concentrations. Kp = the equilibrium constant using pressure Kp = Kc(RT) n gas n is the difference between the number of moles of products and moles of reactants R = 0.082 L atm / mol K

  37. Calculating Variations on Q and K [C]c[D]d Qc = aA + bB cC + dD [A]a[B]b Q’ = 1/Qc cC + dD aA + bB Qc’ = (Qc)n n aA + bB cC + dD For a sequence of equilibria, Koverall = K1 x K2 x K3 x …

  38. CALCULATING THE Keq at Equilibrium 1. In one experiment, Haberintroduced a mixture of H2 & N2 into a reaction vessel and allowed the system to attain chemical equilibrium at 472°C. The equilibrium mixture of gases were analyzed and found to contain 0.1207M H 2, 0.0402M N2, and 0.00272M NH3. Calculate Keq. ?

  39. CALCULATING THE Keqat Equilibrium 2. For the Haber process: N2(g) + 3H2(g) 2NH3(g) Kp = 1.45 x 10-5 at 500°C a) If an equilibrium mixture of the three gas have partial pressures of 0.928 atm for H2 and 0.432 atm for N2, what is the partial pressure of NH3? b) If an equilibrium mixture of the three gas started with partial pressures of 0.928 atm for H2 and 0.432 atm for N2, what is the equilibrium partial pressure of NH3?

  40. Workshop Eb1 on CALCULATING THE Keq at Equilibrium 1. In another experiment, Haberintroduced ammonia into a reaction vessel and allowed the system to attain chemical equilibrium at 472°C. The equilibrium mixture of gases were analyzed and was found to contain 1.56 M H 2, 9.23 M N2, and 0.0776 M NH3. Calculate Keq. 2. Nitryl Chloride, NO2Cl, is in equilibrium in a closed container with NO2 and Cl2. 2 NO2Cl(g)  2 NO2(g) + Cl2(g) Calculate Keq if [NO2Cl]eq= 0.00106M, [NO2]eq= 0.0108M & [Cl2]eq = 0.00538M ?

  41. PREDICTING DIRECTION OF REACTION 1. Predicting the direction of reaction Q = reaction quotient at equil Q = K Q > K  species on Rt (prod) (no net Rx) react to form left K = [Equil] Q = [Non Equil] Q < K  forms more products Goal: Calculate Q to determine state of Rx, equil, more product or more reaction

  42. I.C.E. & PREDICTING DIRECTION OF REACTION 3a. Sulfur Trioxide decomposes at High temperature in a sealed container. 2 SO3(g) 2 SO2(g) + O2 (g) Initially the vessel is filled at 1000K with SO3(g) at a concentration of 6.09 x 10-3M. At equilibrium, the [SO3] is 2.44 x 10-3M. Calculate Kc. 3b. Calculate the value for Q and predict the direction in which the reaction will proceed towards equilibrium if the initial concentrations are: [SO3] = 2.0 x 10-3M [SO2] = 5.0 x 10-3M [O2] = 3.0 x 10-2M

  43. Workshop Eb2 on I.C.E. & Predicting Direction 1. A mixture of 5.0 x 10-3mol of H2 and 1.0 x 10-2mol of I2 is placed in a 5.0L container at 448°C and allowed to come to equilibrium. Analysis of this equilibrium mixture shows that the [HI] is 1.87 x 10-3 M. Calculate Kc: H2(g) + I2(g) ↔ 2HI(g) 2. At 448°C the equilibrium constant Kc for the reaction below is 50.5 H2(g) + I2(g) ↔ 2HI(g) Predict how the reaction will proceed to reach equilibrium if the initial amount of HI is 2.0 x 10-2mol, H2 is 1.0 x 10-2mol, and I2 is 3.0 x 10-2mol in a 2.00 L container.

  44. Finding Equilibrium Concentrations When Given the Equilibrium Constant and Initial Concentrations or Pressures • first decide which direction the reaction will proceed • compare Q to K • define the changes of all materials in terms of x • use the coefficient from the chemical equation for the coefficient of x • the x change is + for materials on the side the reaction is proceeding toward • the x change is  for materials on the side the reaction is proceeding away from • solve for x • for 2nd order equations, take square roots of both sides or use the quadratic formula • may be able to simplify and approximate answer for very large or small equilibrium constants

  45. Calculating Keq Quadratic formula 4. A 3.00 L flask is filled with 1.00 mol of H2 and 2.00 mol I2 at 448°C and Kc is 50.5, what are the equilibrium concentrations of H2, I2 & HI? Successive Approximation 2B. For the Haber process: N2(g) + 3H2(g) 2NH3(g) Kp = 1.45 x 10-5 at 500°C If an equilibrium mixture of the three gas started with partial pressures of 0.928 atm for H2 and 0.432 atm for N2, what is the partial pressure of NH3 (eq)?

  46. Calculating Keq 5. The equilibrium constant for the Haber process at 472°C is Kc = 0.105. A 2.00 L flask is filled with 0.500 mol of ammonia and is then allowed to reach equilibrium at 472°C. What are the equilibrium concentrations? N2(g) + 3 H2(g) 2 NH3(g) 6. For the reaction PCl5(g) PCl3(g) + Cl2(g) at a certain temperature Kc equals 450. What will happen when 0.10 mol of PCl5, 1.0 mol of PCl3, and l.5 mol of Cl2 are added to a 2.0-L container and the system is brought to the temperature at which Kc=450. What are the equilibrium concentrations?

  47. Calculating Keq 7. A mixture of 0.005 mol of hydrogen and 0.01 mol iodine is placed in a 5.0 L container at 448 oC and allowed to reach equilibrium. Analysis of the equilibrium mixture shows the [HI] to be 1.87x10-3M. Calculate K at this temperature. 8. In the decomposition of HI into its elements; what would be the equilibrium partial pressures of each constituent, if the initial pressure of each introduced into the reaction chamber was 1.00 M? Kp = 0.0108 at this temperature.

  48. Workshop Eb3 on Calculating Keq 1) The decomposition of ammonia is: 2 NH3(g) → N2 (g) + 3 H2 (g). If Kp is 1.5 × 103 at 400°C, what is the partial pressure of ammonia at equilibrium when N2is 0. 10 atm and H2is 0.15 atm? 2) At a certain temperature, Kc equals 1.4 × 102for the reaction: 2 CO(g) + O2 (g) ⇌ 2 CO2 (g). If a 3.00-L flask contains 0.400 mol of CO2and 0.100 mol of O2at equilibrium, how many moles of CO are also present in the flask? 3) At a certain temperature the equilibrium constant, Kc, equals 0.11 for the reaction: 2 ICl(g) ⇌ I2 (g) + Cl2 (g). What is the equilibrium concentration of ICl if 0. 45 mol of I2and 0. 45 mol of Cl2are initially mixed in a 2.0-L flask? 4) Kc is 1.67 × 1020 at 25°C for the formation of iron(III) oxalate complex ion: Fe3+ (aq) + 3 C2O42-(aq) ⇌ [Fe(C2O4)3]3-(aq). If 0. 0200 M Fe3+ is initially mixed with 1.00 M oxalate ion, what is the concentration of Fe3+ ion at equilibrium? Assume all of the initial Fe3+ converts to complex ion, then calculate the amount of Fe3+ that returns to its original form.

  49. EFFECT OF VARIOUS DISTURBANCES ON AN EQUILIBRIUM SYSTEM DISTURBANCE NET DIRECTION OF REACTION EFFECT ON VALUE OF K Concentration Increase (reactant) Toward formation of product None Decrease (reactant) Toward formation of reactant None Pressure (volume) Increase P Toward formation of lower amount (mol) of gas1 None Decrease P Toward formation of higher amount (mol) of gas None Temperature Increase T Toward absorption of heat Increases H°rxn>0 Decreases if H°rxn<0 Decrease T Toward release of heat Increases H°rxn<0 Decreases H°rxn>0 Catalyst added None; rates of forward and reverse reactions increase equally None

  50. K2 H0rxn 1 1 = - - ln T2 T1 R K1 Ea 1 1 k2 Hvap 1 1 P2 - - = - ln = - ln T2 T1 R k1 T2 T1 R P1 H0rxn 1 1 K2 - = - ln T2 T1 R K1 The van’t Hoff Equation The Effect of T on K R = universal gas constant = 8.314 J/mol*K K1 is the equilibrium constant at T1 Temperature Dependence

More Related