II.1 EQUILIBRIUM REVERSIBLE REACTIO2S

Transcription

1 II.1 EQUILIBRIUM REVERSIBLE REACTIO2S Many reactions can go in reverse and have separate activation energies. Reactants Products or Reactants Products Reactants form Products Products form Reactants Example: i) N 2 O 4(g) is heated in a CLOSED flask. ii) 2NO 2(g) molecules will build up and collide. iii) This reaction may go forward and reverse (depends on the conditions). We can write both the forward and reverse reactions on the same line using a double arrow. iv) As long as there is N 2 O 4 present, the forward reaction will keep on happening and as long as there is NO 2 present, the reverse reaction will keep on happening. 1

2 DY2AMIC EQUILIBRIUM Read p and Try #2 II.2 CHARACTERISTICS OF EQUILIBRIUM How do we recognize a reaction is in equilibrium? What does equilibrium look like on a graph? 2 2 O 4 2 2O 2 Example 1:! Equilibrium DOES NOT mean concentration of reactants and products are equal.! Equilibrium DOES mean rates of forward and reverse reactions are the same. Given the reaction: 2O 2(g) + CO (g)! 2O (g) + CO 2(g) a. If one mole of NO 2 and one mole of CO are mixed in a 1.0 litre container, the rate of the forward reaction will initially be (fast/slow) b. While the forward rate > the reverse rate, the [NO] and the [CO 2 ] will be (increasing/decreasing) c. After the initial mixing, the rate of the forward reaction will be (increasing/decreasing) d. The rate of the reverse reaction will be (increasing/decreasing) e. Once equilibrium is established, what can be said about the rates of the forward and reverse reaction? f. NO 2 is a dark brown colour. All of the other gases are colourless. Describe what will happen to the colour of the gas mixture from when the NO 2 and the CO are mixed until equilibrium is established. 2

3 Hebden: Do Set 7 #3 6, 11, 12 (p ) II. 3 PREDICTI2G THE SPO2TA2EITY OF A REACTIO2 A SPONTANEOUS reaction is a reaction that occurs by itself, without outside assistance. What causes a reaction to: a) proceed forward b) not proceed (proceed in reverse) c) form an equilibrium E2THALPY Endothermic C + heat " D Exothermic A " B + heat 3 In exothermic reactions, the have minimum enthalpy, it will favour the, and tend to 3 we should expect exothermic reactions to 3 In endothermic reactions, the reactants have minimum enthalpy, therefore, it will favour the, and tend to 3 we should expect endothermic reactions to (without outside assistance like increasing temperature) BUT some endothermic reactions do occur spontaneously. For example, cold packs mixing the chemicals absorbs heat (making it very cold) 3

4 ENTROPY is the amount of randomness in a system. E2TROPY Probability is high that events occurring in life will lead to How do we decide which side of a reaction has the most entropy? PHASE Draw the KMT model of the phases. Which phase has the most entropy? Least entropy? Why? gas solution liquid solid < > So you can look at the subscripts for the phases to tell which side (reactant/product) has maximum entropy. Example 2: Label the side that has maximum entropy. Is the entropy increasing or decreasing? C 2 H 2 Cl 4(l)! C 2 H 2(g) + 2 Cl 2(g) But what if both reactant and product are the same phase? The side having the 2UMBER OF MOLECULES Example 3: Label the side that has maximum entropy. 3 H 2(g) + N 2(g)! 2 NH 3(g) Example 4: a. Given the equation: Cl 2(g)! Cl 2(aq) H = 25 kj Enthalpy favours the (reactant/product) and entropy favours the. 4

5 b. If the reaction CO (g) + 3 H 2(g)! CH 4(g) + H 2 O (g) kj was proceeding to the right, the enthalpy would be (increasing/decreasing) and the entropy is If we consider both enthalpy and entropy, how do we decide which side is favoured? In nature, there is a tendency toward enthalpy and entropy. If enthalpy favours reactants and entropy favours reactants then If enthalpy favours products and entropy favours products then If enthalpy favours reactants and entropy favours products then If enthalpy favours products and entropy favours reactants then Example 5: Determine the spontaneity of the reaction: CaCO 3 (s) + energy! CaO (s) + CO 2 (g) Hebden: Do Set 8 #14, 15 (a, b, d, f, h), 16 (a, b, c, e) (p.48 49) LE CHATELIER S PRINCIPLE: II.4 LE CHÂTELIER S PRI2CIPLE APPLICATIO2 OF LE CHÂTELIER S PRI2CIPLE TEMPERATURE CHA2GES N 2 O 4(g) + heat! 2NO 2(g) Temperature changes affect all equilibrium (s, l, aq, g) LCP: If we ADD HEATto the equilibrium, the reaction will shift in way that tends to N 2 O 4(g) + heat! 2NO 2(g) LCP: If we REMOVE HEAT from the equilibrium, the reaction will shift in way that tends to N 2 O 4(g) + heat! 2NO 2(g) CO2CE2TRATIO2 A2D PARTIAL PRESSURE CHA2GES In a mixture of gases, partial pressure is the pressure exerted by ONE gas. e.g. In a mixture containing NO and CO 2 gases: total pressure = partial pressure of NO + partial pressure of CO 2. If you increase P NO, it s like increasing the concentration of NO. 5

6 CO 2(g) + NO (g)! CO (g) + NO 2 (g). LCP: If we I2CREASE THE CO2CE2TRATIO2 of a certain substance in an equilibrium mixture, the reaction will shift in way that tends to If you increase P NO, it s like increasing the concentration of NO. CO 2(g) + NO (g)! CO (g) + NO 2 (g). LCP: If we DECREASE THE CO2CE2TRATIO2 of a certain substance in an equilibrium mixture, the reaction will shift in way that tends to If you decrease P NO CO 2(g) + NO (g)! CO (g) + NO 2 (g). Example 6: What is the effect of adding FeCl 3(aq) to this equilibrium: Fe(NO 3 ) 3(aq) + KSCN (aq)! FeSCN 2+ (aq) + KNO 3(aq) Be careful with spectator ions. Complete ionic equation: Net ionic equation: Also, changing solids or liquids have no effect on the equilibrium. Example 7: Changing which species will have an effect on the equilibrium: CaCO 3(s)! CaO (s) + CO 2(g) VOLUME OF THE CO2TAI2ER A2D TOTAL PRESSURE CHA2GES Consider: N 2 O 4(g) + heat! 2 NO 2(g) Compare the reactant side and product side, which side has more pressure? vs. 6

7 If you decreased the volume of the balloon, which side will it affect more? (Which side will feel more concentrated?) LCP: If we I2CREASE THE TOTAL PRESSURE (decrease volume) on an equilibrium system involving gases, the reaction will shift in way that tends to LCP: If we DECREASE THE TOTAL PRESSURE (increase volume) on an equilibrium system involving gases, the reaction will shift in way that tends to Example 8: 2C 2 H 6(g) + 7O 2(g)! 4CO 2(g) + 6H 2 O (g) a) Increasing the total pressure on this system, will cause a shift to the side with moles of gas, which in this case is the side. b) Decreasing the total pressure on this system, will cause a shift to the side with moles of gas, which in this case is the side. c) Increasing the total volume on this system (the same as the total pressure) will cause a shift to the side with moles of gas, which in this case is the side. Consider: CATALYSTS LCP: ADDI2G A CATALYST speed up forward and reverse reactions; therefore, the GRAPH OF EQUILIBRIUM CHA2GES 7

8 TEMPERATURE N 2 O 4(g) + heat! 2NO 2(g) Suppose we increase the temperature at 2 min., according to LCP the equilibrium will shift to the. The [N 2 O 4 ] will and the [NO 2 ] will originally at equilibrium conc. are constant CONCENTRATION N 2 O 4(g) + heat! 2NO 2(g) Suppose we inject more NO 2 into the container at 2 min., so the [NO 2 ] will suddenly increase. According to LCP the equilibrium will shift to the. originally at equilibrium conc. are constant 8

9 PRESSURE N 2 O 4(g) + heat! 2NO 2(g) Suppose we decrease the volume of the container suddenly at 2 min., so the [N 2 O 4 ] and [NO 2 ] will suddenly. According to LCP in order to decrease the pressure, the equilibrium will shift to the side with moles of gas, that is to the. originally at equilibrium conc. are constant CATALYST N 2 O 4(g) + heat! 2NO 2(g) Adding a catalyst speeds up BOTH the forward and reverse reactions, so the concentrations of the species originally at equilibrium conc. are constant add catalyst 9

10 Hebden: Do Set 9 #17, 19, 21, 23, 24, 26, 27 (p.54 55) II.5 I2DUSTRIAL APPLICATIO2S OF EQUILIBRIUM PRI2CIPLES Hebden: Read The Haber Process on p. 56 and Do #29 N 2(g) + 3H 2(g)! 2NH 3(g) + 92 kj 29. a) For highest yield of NH 3(g), use pressure. b) For highest yield of NH 3(g), use temperature. c) For fastest reaction rate, use temperature. d) Problem: Solution: e) Speed up the reaction by 2OTE: Shift in equilibrium means when a new equilibrium is reached, the concentration of reactants or products is more or less. It does not say anything about the rate of the reaction. f) Increasing the temperature will the rate of reaction. The equilibrium will shift (more/less than) Hebden: Do #30 (p.56). This means that when the new equilibrium is reached, the amount of reactants will be the amount of products. II.6 EQUILIBRIUM EXPRESSIO2 We can quantify (attach a numerical value to) the equilibrium for most reactions by creating a RATIO of PRODUCTS TO REACTANTS For a reaction: aa + bb! cc + dd The EQUILIBRIUM EXPRESSION, K eq = Example 9: Write the equilibrium expression for the reaction for : 2HI (g)! H 2(g) + I 2(g) H 2(g) + I 2(g)! 2HI (g) 10

11 EQUILIBRIUM CO2STA2T The RATIO obtained from the equilibrium expression is a unique to a particular reaction. This ratio is called the EQUILIBRIUM CONSTANT or K eq The concentration of solids and liquids are fairly constant. The do not expand or contract much. Since they are constants, their values are already incorporated with K eq. OMIT solids and pure liquids from the K eq expression. Example 10: The equilibrium constant expression for the reactions: BaSO 4(s)! Ba 2+ (aq) + SO 4 2- (aq) CH 3 COCH 3(l) + Cl 2(g)! CH 3 COCH 2 Cl (l) + HCl (g) Do Hebden: Set 10 # 31 (a, c, e, f, h, i), 32, 34, 35 (a, c, e, h) (p.60) II.7 THE MEA2I2G OF K EQ K eq = [Products] [Reactants] K eq = [Products] [Reactants] Large K eq means at equilibrium, there is lots of and very little Small K eq means at equilibrium, there is lots of and very little Temperature FACTORS AFFECTI2G K EQ Consider the original reaction: If we increase the temperature, reactants! products + heat reactants! products + heat The equilibrium will shift reactants! products + heat 11

12 Write the K eq expression K eq = So K eq (remains the same/becomes larger/becomes smaller) Concentration Consider the original reaction: If we increase reactants, reactants! products reactants! products The equilibrium will shift Write the K eq expression K eq = So K eq (remains the same/becomes larger/becomers/smaller) Pressure Consider the original reaction: If we decrease the volume (or increase the pressure), N 2(g) + 3H 2(g)! 2NH 3(g) N 2(g) + 3H 2(g)! 2NH 3(g) The equilibrium will shift Write the K eq expression K eq = So K eq (remains the same/becomes larger/becomes smaller) Changes in concentration, pressure, surface area, and adding a catalyst ONLY CHANGING affects Keq. Example 11: 2 NO (g) + Cl 2(g)! 2 NOCl (g) + 76 kj What happens to the K eq in the above reaction if the temperature is decreased? Why? Hebden: Do Set 11 #36, 38, 39, 44, 45 (p.62) Chemical systems can either be II. 8 EQUILIBRIUM CALCULATIO2S 12

13 AT EQUILIBRIUM or NOT AT EQUILIBRIUM type 1 type 2 type 3 type 4 Method: 1) find eqm concentrations 2) write K eq expression 3) plug in eqm concentrations into K eq Method: 1) set up an ICE box 2) fill in what you are given. I = initial conc., C = change according to mole ratio, E = equilibrium conc. 3) write K eq expression 4) plug in eqm conc. into K eq Method: 1) set up an ICE box 2) fill in what you are given. I, C, E. Let x = unknown. 3) write K eq expression 4) plug in eqm conc. into K eq Method: 1) plug in intial conc. into Trial K eq expression 2) compare trial K eq with K eq : Trial K eq < K eq = shift right Trial K eq > K eq = shift left Trial K eq = K eq = no shift Type 1: System is AT EQUILBRIUM Example 12: What is the K eq for the reaction, 2 HI (g)! H 2(g) + I 2(g) if AT EQUILIBRIUM, there is 2.0 mol of HI, 3.0 mol of I 2, and 3.5 mol of H 2 in a 5.0 L closed flask? (Answer K eq = 2.6) 13

14 Type 2: System is CHANGING TO EQUILIBRIUM Calculate K eq Example 13: 2 NO (g) + O 2(g)! 2 NO 2(g) 4.0 mol of NO 2 is INTIALLY ADDED to a 2.0 L flask. After a while, equilibrium is reached. At equilibrium, 0.50 mol of NO is found. What is K eq? 2 NO (g) + O 2(g)! 2 NO 2(g) I C E Example 14: PCl 5(g)! PCl 3(g) + Cl 2(g) 8.0 moles of PCl 3 and 6.0 moles of Cl 2 are placed in a 2.0 L flask. After equilibrium, 0.50 moles of Cl 2 are found. What is K eq? (answer: K eq = 0.11) PCl 5(g)! PCl 3(g) + Cl 2(g) I C E Hebden: Do Set 12 #47 49 (p.70) 14

15 Type 3: System is CHANGING TO EQUILIBRIUM Calculate [species] Example 15: 2 A (g) + B (g)! 2 C (g) An unknown amount of C was INITIALLY PLACED into a 3.0 L flask. When equilibrium was reached, the concentration of A was 0.60 M. If K eq has a value of 34.0, how many MOLES of C were placed into the flask originally? 2 A (g) + B (g)! 2 C (g) I C E Example 16: A 2(g) + B 2(g)! 2AB (g) If 6.0 mol of A 2 and 6.0 mol of B 2 are placed in a 1.0 L bulb and allowed to come to equilibrium, what will be the concentration of all the species at equilibrium? K eq = 2.5. (answer: [A 2] = [B 2] = 3.4 M, [AB] = 5.3 M) 15

16 Type 4: PREDICTING REACITON SHIFT We can also use the K eq to predict which direction a reaction will go (left or right) for any initial concentrations of reactants and products. We use REACTION QUOTIENT (Q) or TRIAL K eq. It has the same form as K eq EXCEPT: K eq = [product] Trial K eq = [product] [reactant] [reactant] Trial K eq > K eq reaction will shift Trial K eq > K eq reaction will shift Trial K eq < K eq reaction will shift Trial K eq < K eq reaction will shift Trial K eq = K eq reaction will Trial K eq = K eq reaction will Example 19: 2SO 2(g) + O 2(g)! 2SO 3(g) The K eq is 0.14 for the above reaction. If a 1.0 L flask is filled with 0.10 mol of SO 3 and with 0.20 mol of SO 2 and O 2, is the reaction at equilibrium? If not, in which direction does it proceed? Hebden: Do Set 13 #52, 53, 55 59, (60 63) [Type 3] and 50, 54 [Type 4] (p.71 72) 16