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Chapter 14

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  1. Chapter 14 Chemical Equilibrium Chemistry, 7e Thinkwell vs. Chang CH 14: 14-30(even), 38-46(even), 52-60(even), 66, 67, 72, 74, 75, 80, 84, 90, 102

  2. Thinkwell vs. Chang

  3. 14.1 The Concept of Equilibrium and the Equilibrium Constant • Most reactions are reversible • this is denoted by… reactant qe products • With reversible reactions, a state of chemical equilibrium is achieved • Chemical equilibrium - the rates of the forward and reverse reactions are equal and the concentrations of the reactants and products remain constant.

  4. Introduction to CPS Remote UseQuestion 1 Question 1 • I am present and my remote is working properly. • (a) True • (b) False (if so tell me now)

  5. Introduction to CPS Remote UseQuestion 1 Answer Question 1 • I am present and my remote is working properly. • (a) True Correct Answer • (b) False (if so tell me now)

  6. N2O4 colorless gas NO2 dark brown

  7. No matter how much of the substances you start with, eventually equilibrium will be reached. • Although there doesn’t appear to be a connection between the concentrations of the three above...

  8. ...THERE IS! For the reaction: aA + bB qe cC + dD Equilibrium constant (a dimensionless quantity) [ ] represents the equilibrium concentration in mol/L

  9. N2O4qe 2NO2 0.670M 0.2000M 0.448M 0.446M 0.643M 0.0500M 0.0547M 0.0898M 0.0457M 0.0204M 0.000M 0.000M

  10. N2O4qe 2NO2 The value obtained for K = 4.63 x 10-3 • What does the size of K tell us about the equilibrium? • If K«1, the denominator is much larger than the numerator. Therefore the reactants are favored. • If K»1, the numerator is much larger than the denominator. Therefore the products are favored. • Which case do we have for the above reaction? • The equilibrium favors the reactants. We say that “it lies to the left.” Look at the graphs once again.

  11. Question 2 Question 2 (Feel free to discuss with a neighbor) • When K = 4.63 x 10-3 for the reaction above, which is true? • Products are favored • Reactants are favored • (c) Neither is favored

  12. Question 2 Answers Question 2 (Feel free to discuss with a neighbor) • When K = 4.63 x 10-3 for the reaction above, which is true? • Products are favored • Reactants are favored Correct Answer • (c) Neither is favored

  13. Is the reactant of N2O4 favored in each case? Yes

  14. 14.2 Writing Equilibrium Constant Expressions • We must distinguish between two types of equilibria: • Homogeneous equilibria - all reacting species are in the same phase. Examples: • all are gasses • all are aqueous • Heterogeneous equilibria - reacting species exist in different phases. Examples: • solids and gases • solids and liquids

  15. Include all species in the reaction. • For the reaction: • aA qe bB • Can be expressed in terms of concentrations as in our previous example or • Can be expressed in terms of partial pressure if gases. HOMOGENEOUS EQUILIBRIA partial pressure is in atm

  16. What is the relationship between Kc and Kp? See derivation pg 564. R=0.08206 Latm/mol K Temperature in? Kelvin HOMOGENEOUS EQUILIBRIA Kp=Kc(RT)n n = moles of gaseous products - moles of gaseous reactants When n = 0 Kp = Kc

  17. Question 3 An equilibrium mixture of the reaction 2H2S(g) qe 2H2(g) + S2(g) on analysis,was found to contain 1.0 mol H2S, 4.0 mol H2 and 0.80 mol S2 in a 4.0 L vessel. Calculate the equilibrium constants Kc and Kp for this reaction. (Temp = 25oC) Discuss with neighbor • Kc = 78.3 and Kp = 3.2 • Kc = 3.2 and Kp = 78.3 • (c) Kc = 12.8 and Kp = 19.6

  18. An equilibrium mixture of the reaction 2H2S(g) qe 2H2(g) + S2(g) on analysis,was found to contain 1.0 mol H2S, 4.0 mol H2 and 0.80 mol S2 in a 4.0 L vessel. Calculate the equilibrium constants Kc and Kp for this reaction. (Temp = 25oC) HOMOGENEOUS EQUILIBRIA T = 25ºC V = 4.0 L • Kc = 78.3 and Kp = 3.2 • Kc = 12.8 and Kp = 19.6 • (c) Kc = 3.2 and Kp = 78.3 Correct Answer

  19. 2H2S(g) qe 2H2(g) + S2(g) Kc = [H2]2[S2]1 = (4/4)2(0.8/4)1 = 3.2 the units of M are [H2S]2 (1/4)2 omitted by convention Since PV = nRT P = nRT/V = partial pressure of a gas at constant V and T = [n ]2[n ]1 [n ]2 (RT)3 V3 Kp = (P )2(P )1 (P )2 V2 (RT)2 S2 H2 S2 H2 • • H2S H2S = (16)(0.8)(0.08206 L·atm·mol-1·K)(298 K) (1)(4.0 L) = 78.3 (no units)

  20. Consider the reaction 2NO(g) + O2(g) qe 2NO2(g) At 430oC, an equilibrium mixture consists of 0.020 mol of O2, 0.040 mol of NO, and 0.96 mol of NO2. Calculate Kp for the reaction, given that the total pressure is 0.20 atm. HOMOGENEOUS EQUILIBRIA T = 430ºC PT = 0.20 atm Ans: Kp=1.5 x 105

  21. Solving for Kp Given: 2NO(g) + O2(g) qe 2NO2(g) Via Dalton’s Law of partial pressures (CH 5): n n + n + n 0.040 1.02 NO  = = = 3.92 x 10-2 NO NO O2 NO2 0.020 1.02 n n + n + n O2  = = = 1.96 x 10-2 O2 NO O2 NO2 n n + n + n 0.96 1.02 NO2  = = 0.941 = NO2 NO O2 NO2

  22. Solving for Kp P =  PT = 7.84 x 10-3 atm NO NO P =  PT = 3.92 x 10-3 atm O2 O2 P =  PT = 0.188 atm NO2 NO2 Kp = (P )2 (P )1(P )2 = (0.188)2 (3.92 x 10-3)(7.84 x 10-3)2 NO2 = 1.5 x 105 O2 NO

  23. Why Assumptions Can be Bad From last time………recall that……. Kp = (P )2 (P )1(P )2 = (0.188)2 (3.92 x 10-3)(7.84 x 10-3)2 NO2 = 1.5 x 105 O2 NO PTV = nTRT nTRT PT (1.02 mol)(0.08206 L • atm • mol-1 • K-1)(273 + 430)K (0.20 atm) = V = V = 294.2 L (very different from 1 L)

  24. Another way to solve last time’s example Given: 2NO(g) + O2(g) qe 2NO2(g) P = nRT V Insert V = 294.2 L, n’s, and T provided n2 n2• n (RT)2 V3 V2 (RT)3 NO2 Kp = (P )2 (P )2 (P ) NO2 = • • NO O2 NO O2 Kp = 1.5 x 105

  25. What about Kc ? Given: 2NO(g) + O2(g) qe 2NO2(g) KP = Kc(RT)Dn Kc = KP (RT)Dn = (1.5 x 105) [(0.08206)(430 + 273)](2 – 3) = = 8.7 x 106

  26. Another way to solve for Kc n2 n2• n [NO2]2 [NO]2[O2] 1 V2 V V2 1 1 NO2 Kc = = • • • NO O2 Kc = 8.7 x 106

  27. Ammonia is placed in a reactor, and the temperature is increased to 745oC, where some will decompose to nitrogen and hydrogen. The initial concentration of ammonia was 0.0240 M. After equilibrium is attained, the concentration of ammonia is 0.0040 M. Calculate Kc at 745oC for 2NH3(g) qeN2(g) + 3H2(g) HOMOGENEOUS EQUILIBRIA Via ICE table (a.k.a. Initial…Change…..Equilibrium) Conc. (M) NH3 N2 H2 Initial 0.0240 0 0 Change -0.020 +? +? Equilib. 0.0040 ? ? ans: 1.7 x 10-2

  28. NH3 (g) qe N2 (g) + 3H2 (g 0.02 mol NH3 1 mol N2 1 2 mol NH3 = 0.01 mol (0.01 M if 1 L) • 0.02 mol NH3 3 mol H2 1 2 mol NH3 = 0.03 mol (0.03 M if 1 L) • Kc = (0.01) (0.03)3 (0.004)2 = 1.7 x 10-2

  29. Let’s look at the following example: • CaCO3(s) qe CaO(s) + CO2(g) • What would the equilibrium constant be?  HETEROGENEOUS EQUILIBRIA I’ll call this constant... • Concentration of a solid is “intensive” - it is a constant for that substance. • Because [CaO] and [CaCO3] are constants, I can join them with the other constant, giving...

  30. cancel... HETEROGENEOUS EQUILIBRIA giving... What is a constant, divided and/or multiplied by a constant?

  31. Also, for the above example... HETEROGENEOUS EQUILIBRIA

  32. Heterogeneous Equilibria Expressions (1) 2SO2 (g) + O2 (g) qe 2SO3 (g) Kc = [ SO3 ]2 [O2][SO2]2 Kp = (P )2 (P )1(P )2 SO3 O2 SO2 (2) CaCO3 (s) qe CaO + CO2 (g) Kp = P Kc = [CO2] CO2 (3) H2O (l) CO2 (g) qe H2CO3 (aq) Kp = (P )-1 Kc = [ H2CO3 ] [CO2] CO2 (4) S (s) = H2SO3 (aq) qe H2S2O3 (aq) Kp = undefined (no gases) Kc = [ H2S2O3] [H2SO3]

  33. HETEROGENEOUS EQUILIBRIA What is the bottom line? Do not include solids and liquids in the equilibrium constant expression.

  34. At equilibrium, the pressure of the reacting mixture CaCO3(s) qe CaO(s) + CO2(g) is 0.105 atm at 350oC. Calculate Kp and Kc for this reaction. HETEROGENEOUS EQUILIBRIA Kp = P Kc = [CO2] CO2 N = mol = P = Kc V L RT Kp =0.105 and Kc = 2.05 x 10-3

  35. Let us consider what to do when the product molecules of one equilibrium system are involved in a second equilibrium. A + B qe C + D C + D qe E + F (1) (2) (3) MULTIPLE EQUILIBRIA overall reaction: A + B qe E + F

  36. Therefore: If a reaction can be expressed as the sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions. MULTIPLE EQUILIBRIA

  37. HOW MANIPULATIONS TO THE EQUATION CHANGES K • Reversing the direction of the equilibrium: • Take the reciprocal of K • How the equation is balanced will determine the value for K. The K must be accompanied by a balanced equation! • For example, if you double the coefficients, what will happen to the K expression • Let’s look at examples of the above on the overhead.

  38. 14.3 The Relationship between Chemical Kinetics and Chemical Equilibrium The following reaction takes place as a single elementary step: kf kr A + B qwe C ratef = kf[A][B] rater = kr[C] At equilibrium, ratef = rater. Therefore kf[A][B] = kr[C], giving…

  39. Since we know that the rate constant if affected by temperature, we can see that the value of the equilibrium constant will be affected by temperature.

  40. 14.4 What does the Equilibrium Constant Tell Us? We can use the equilibrium constant to: 1. Predict the direction of a reaction. 2. Calculate equilibrium concentrations.

  41. We know that in order to calculate the equilibrium constant, we simply must plug in equilibrium concentrations into the expression. • We can obtain the reaction quotient (Qc) by substituting the initial concentrations into the equilibrium constant expression. • Once you have a value of Qc, a judgement of the direction can be made according to the following: • Qc>Kc system proceeds from rt to left. • Qc=Kc system is at equilibrium • Qc<Kc system proceeds from left to right PREDICTING THE DIRECTION OF A RXN

  42. PREDICTING THE DIRECTION OF A RXN

  43. At 700 K, the reaction 2SO2(g) + O2(g) qe 2SO3(g) has Kc = 4.3 x 106, and the following concentrations are present: [SO2] = 0.10 M; [SO3] = 10 M; [O2] = 0.10 M Is the mixture at equilibrium, if not, in which direction will the reaction proceed? PREDICTING THE DIRECTION OF A RXN Q < Kc And……?

  44. You will be given the equilibrium constant value and the initial concentrations. The following is an outline of the procedure to follow. 1. Use an “ICE table” to expression the equilibrium concentrations in terms of the initial concentration and a variable x. 2. Write the equilibrium constant expression in terms of the equilibrium concentrations. 3. Solve for x. 4. Relate the value for x to the concentrations of all species. CALCULATING EQUILIBRIM CONCENTRATIONS

  45. At 250o C, the equilibrium constant, Kp for the reaction below is 1.80. PCl5(g) qe PCl3(g) + Cl2(g) Sufficient PCl5 is put into a reaction vessel to give an initial pressure of 2.74 atm at 250oC. Calculate the pressure of PCl5 after the system has reached equilibrium. Ans. 1.24 atm CALCULATING EQUILIBRIM CONCENTRATIONS Kp = P P P Cl2 PCl3 PCl5 Via ICE table (a.k.a. Initial…Change…..Equilibrium) P (atm) PCl5 PCl3 Cl2 Initial 2.74 0 0 Change - x + x + x Equilib. 2.74 - x x x

  46. = x2 (2.74 – x) Kp = P P P Cl2 PCl3 PCl5 0 = (1)x2 + Kp(x) - (2.74)Kp and x = -b ± [ b2 – 4ac]0.5 2(a) And x = 1.50 atm Pf (PCl5) = 2.74 – 1.50 = 1.24 atm

  47. Hydrogen iodide decomposes according to the equation: 2HI(g) qe H2(g) + I2(g) Kc= 0.0156 at 400oC 0.550 mol HI was injected into a 2.0 L reaction vessel at 400oC. Calculate the concentration of HI at equilibrium. CALCULATING EQUILIBRIM CONCENTRATIONS Solve for next time Ans. 0.220M

  48. Solution to Last Time’s Problem 2HI(g) qe H2(g) + I2(g) Kc= 0.0156 at 400oC Kc = [H2][I2] = (x)(x) [HI]2 ( [HI]o – 2x ) , where [HI]o = 0.550 mol 2.0 L Kc = x2 = x2 (0.275 – 2x)2 (0.275 – 2x)(0.275 – 2x) = x2 and with rearrangement……. (0.0756 – 1.1x + 4x2) 0 = -x2 + Kc[0.0756 – 1.1x + 4x2] = -(0.938)x2 – (0.0172)x + 0.00118

  49. Still Working…. Via quadratic equation and solving for x…Where x = -b ± [b2 – 4ac]0.5 2a x = (0.0172) ± [(0.0172)2 – 4(-0.938)(0.00118)]0.5 2(-0.938) = 0.0172 ± [(2.94 x 10-4) + (4.43 x 10-3)]0.5 = 0.0172 – 6.87 x 10-2 -1.88 -1.88 = 2.74 x 10-2 M = [H2] = [I2] (at equilibrium) Since [HI] = [HI]o – 2x = [0.274 – 2(2.74 x 10-2)]M [HI] = 0.22 M (Finally)

  50. 14.5 Factors that Affect Chemical Equilibrium • The balance between the forward and reverse reaction, once in equilibrium, is quite delicate. • When conditions change, the equilibrium will shift to the left or right to reestablish equilibrium. • Le Chatelier’s Principle - If an external stress is applied to a system at equilibrium, the system adjusts in such a way that the stress is partially offset.