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  1. Topics in CHM 1046 Unit 11 Intermolecular Forces, Liquids & Solids (T1) Unit 12 Properties of Solutions (T1) • Intermolecular forces (IMF) • Themodynamics • Chemical Kinetics • Chemical Equilibrium Combination of above: Unit 13 Thermochemistry (T1) Unit 14 Chemical Thermodynamics (T2) Unit 15 Chemical Kinetics (T2) Unit 16 Chem. Equilibrium Gases & Heterogeneous (T3) Unit 17.Acid Base Equilibria (T3) Unit 18 Acid Base Equilibria Buffers & Hydrolysis (T3) Unit 19 Acid Base Equilibria Titrations (T4) Unit 20 Aqueous Equilibria Solubility Product (T4) Unit 21 Electrochemistry (T4)

  2. CHM 1046: General Chemistry and Qualitative Analysis Unit 16Chemical Equilibrium: Gases & Heterogeneous Dr. Jorge L. Alonso Miami-Dade College – Kendall Campus Miami, FL • Textbook Reference: • Chapter # 17 • Module # 5 (& Appendix 2)

  3. N2O4 (g) 2 NO2 (g) The Concept of Equilibrium {Non-Equilibrium Reactions}: proceed in one direction, A  B 2 NO (g) + O2(g) 2 NO2(g) (clear gases) (red-brown gas) {Equilibrium Reactions}: proceed in both directions, A ↔ B  (red-brown gas) (clear gas) 2X Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate. This does not mean chemicals are found in same concentration!

  4. [NO2]2 [N2O4] kf kr = N2O4 (g) 2 NO2 (g) The Equilibrium Constant (Keq) • Therefore, at equilibrium Ratef = Rater kf [N2O4] = kr [NO2]2 • Rewriting this, it becomes Keq = The Equilibrium Constant is the ratio of the rate constants at a particular temperature.

  5. aA(aq) + bB(aq) cC(aq) + dD(aq) [C]c[D]d [A]a[B]b Kc = The Equilibrium Constant (K) • To generalize this expression, consider the reaction • The equilibrium expression for this reaction would be where [X] = concentration of each chemical in moles/ liter (M).

  6. aA + bB aA + bB cC + dD cC + dD [C]c[D]d [A]a[B]b [C]c[D]d [A]a[B]b Kc = Kc = What Does the Value of K Mean? 10 x Kc =10 = 10 1 • IfK >> 1,the reaction isproduct-favored; product predominates at equilibrium. 10 x Kc = 1 = 0.1 10 • If K << 1, the reaction is reactant-favored; reactant predominates at equilibrium.

  7. aA(g) + bB(g) cC(g) + dD(g) (PC)c (PD)d (PA)a (PB)b Kp = [C]c[D]d [A]a[B]b Kc = The Equilibrium Constant for Gases Because pressure is proportional to concentration for gases in a closed system, the equilibrium expression can also be written Partial press: (75 torrs) (150 torrs) (300 torrs) (600 torrs) = total (1125 torrs) where Kp = equilibrium constant for gases, and Px = partial pressure of each gas.

  8. ( )c ( )d ()a ()b (PC)c (PD)d (PA)a (PB)b Kp = [C]c[D]d [A]a[B]b n V n V n V n V n V Kc = RT RT RT P = RT RT Relationship between Kcand Kp PV = nRT • ideal gas law: = Plugging this into the expression for Kp ☺ Kp= Kc(RT)n Where n = (moles of gaseous product) − (moles of gaseous reactant)

  9. [NO2]2 [N2O4] = Equilibrium Can Be Reached from Either Direction @ 100°C KC The ratio of [NO2]2 to [N2O4] remains constant at this temperature no matter what the initial concentrations of NO2 and N2O4 are.

  10. Equilibrium Can Be Reached from Either Direction N2O4 (g) 2 NO2 (g)

  11. KII = KI = = 0.212 at 100C N2O4(g) 2 NO2(g) 2 NO2(g) N2O4(g) [NO2]2 [N2O4] [N2O4] [NO2]2 Manipulating Equilibrium Constants RULE #1: Reciprocal Rule The equilibrium constantof areverse reaction is the reciprocal of the equilibrium constant of the forward reaction. 1 = 0.212 = 4.72 at 100C

  12. KI = = 0.212 at 100C KIII = = (0.212)2 at 100C KIV = N2O4(g) 2 NO2(g) ½N2O4(g) NO2(g) 2N2O4(g) 4 NO2 (g) [NO2]4 [N2O4]2 [NO2]2 [N2O4] Manipulating Equilibrium Constants RULE #2: Coefficient Rule If the coefficients of a chemical equation arechanged (increased or decreased) by a factor n, then the value of K is raised to that power.

  13. Manipulating Equilibrium Constants RULE #3: Multiple Equilibria Rule When two or more equations are added to obtain a final resultant equation, the equilibrium constant for the resultant equation is product of the constants of the added equations. 2 A(aq) + B(aq)↔ 4 C(aq) 4 C(aq) + E(aq)↔ 2 F(aq) 2 A(aq) + B(aq) + E ↔ 2 F(aq)

  14. N2O4(g) 2 NO2(g) Reactions involving only gasses or only solutions [NO2]2 [N2O4] Kc = = 0.212 at 100C Homogeneousvs.HeterogeneousEquilibria Reactions involving different phases (s, l, g, aq) of matter CO2 (g) + CaO(s) CaCO3 (s)

  15. PbCl2 (s) Pb2+ (aq) + 2 Cl−(aq) The Concentrations of Solids and Liquids Are Essentially Constant The concentrations of solids and liquids do not appear in equilibrium expressions because their concentration does not change. Write the equilibrium expression! Kc = [Pb2+] [Cl−]2 Concentration of both solids and liquids can be obtained by dividing the density of the substance by its molar mass—and both of these are constants at constant temperature.

  16. What are the Equilibrium Expressions for these Heterogeneous Equilibria?

  17. CaCO3 (s) CO2 (g) + CaO(s) As long as some CaCO3 or CaO remain in the system, the amount of CO2 above the solid will remain the same. How does the concentration of CO2 differ in the two bell jars?

  18. [A] + [B] [C] + [D] (1) Calculating Kcfrom experimental data: [ ]i & [ ]e Equilibrium Calculations (2) Calculating Equilibrium Concentrations [ ]ewhen Kc & [ ]i are known ICE Tables: Initial Change Equilibrium

  19. Kc= [HI]2 [H2] [I2] H2 (g) + I2 (g) 2 HI (g) (1) Calculating Kcfrom experimental data: [ ]i & [ ]e A closed system initially containing 1.000 x 10−3 M H2 and 2.000 x 10−3 M I2 at 448C is allowed to reach equilibrium. Analysis of the equilibrium mixture shows that the concentration of HI is 1.87 x 10−3 M. Calculate Kc at 448C for the reaction taking place, which is [HI] Increases by 1.87 x 10-3M Stoichiometry tells us [H2] and [I2] decrease by half as much

  20. Kc= [HI]2 [H2] [I2] (1.87 x 10-3)2 (6.5 x 10-5)(1.065 x 10-3) = = 51 …and, therefore, the equilibrium constant

  21. [NO2]2 [N2O4] = N2O4 (g) 2 NO2 (g) Calculating Kcfrom experimental data: [ ]i & [ ]e @ 100°C  x + x - 2x + x - 2x + x - 2x - x + 2x Kc The ratio of [NO2]2 to [N2O4] remains constant at this temperature no matter what the initial concentrations of NO2 and N2O4 are.

  22. 2006B Q5 Practice Problem { }

  23. Equilibrium Calculations (1) Calculating Kcfrom experimental data: [ ]i & [ ]e (2) Calculating Equilibrium Concentrations when K & [ ]i are known H2(g) + I2(g)↔ 2HI(g) 51 =

  24. aA + bB cC + dD (2) Calculating Equilibrium Concentrations [ ]ewhen K & [ ]i are known Solve for x by using the quadratic equation:

  25. Calculating Equilibrium Concentrations when K & [ ]i are known Problem: for the reaction H2(g) + I2(g) 2HI(g) , K= 54.3 @ 430ºC. Suppose that the initial concentrations of H2, I2, and HI are 0.00623 M, 0.00414 M, and 0.0224 M, respectively. Calculate the concentrations of these species at equilibrium. H2(g) + I2(g) 2HI(g) Deciding if x is negligible:

  26. Problem: for the reaction H2(g) + I2(g) 2HI(g) , K= 54.3 @ 430ºC. Suppose that the initial concentrations of H2, I2, and HI are 0.00623 M, 0.00414 M, and 0.0224 M, respectively. Calculate the concentrations of these species at equilibrium. Deciding if x is negligible: Take the largest [A]i and multiply it by 100 and also divide it by 100. Compare the answers of above to the value of K If comparison are not significant as compared to K, then x is negligible and you can delete the x that is subtracted from [A]I If these calc values are significant compared to K, then x is not negligible and you must use the quadratic to solve for x. YOU MUST USE QUADRATIC!

  27. Calculating Equilibrium Concentrations when K & [ ]i are known [H2]i = 0.00623 [I2]i = 0.00414 First answer is physically impossible, since conc. of H2 and I2 would be more than original conc. x calc without quadratic: [H2]e = 0.00623 – 0.00156) = 0.00467 M [I2]e = (0.00414 – 0.00156) = 0.00258 M [HI]e = (0.0242 + 2 x 0.00156) = 0.0255 M

  28. Equilibrium Calculations (1) Calculating Kcfrom experimental data: [ ]i & [ ]e (2) Calculating Equilibrium Concentrations when K & [ ]iare known H2(g) + I2(g) 2HI(g) 51 = Deciding if x is negligible:

  29. [N2O4] [NO2]2 = 2 NO2 (g) N2O4(g) Le Châtelier’s Principle {Equi.Intro1} {Equi.Intro2} “Ifa system at equilibrium is disturbedby a change in temperature, pressure, or the concentration of one of the components,the system will shift its equilibrium position so as to counteract the effect of the disturbance.” KC The effect of (1) Concentration Henri Louis Le Chatelier 1850-1936  (red-brown gas) (clear gas) The effect of (2) Pressure & Volume (3) Temperature {L.C.&PressVol}

  30. 2 NO2 (g) N2O4(g) Le Châtelier’s Principle The effect of (3) Temperature + HEAT  H = -58.0 kJ (red-brown gas) (clear gas) Why does heat favor the formation of NO2 over N2O4? {Le Châtelier’s&Temp} {Molecular Explanation} Keq at different Temp:

  31. Co(H2O)62+(aq) + 4 Cl(aq) CoCl4 (aq) + 6 H2O (l) The Effect of Temperature Changes Add heat (), which way will the equlibrium be shifted?  H= + (pink) + HEAT (blue)  {CoCl}

  32. Briggs-Rauscher reaction {ClockReaction} IO3− + 2H2O2 + CH2(COOH)2 + H+→ ICH(COOH)2 + 2O2 + 3H2O

  33. Le Châtelier’s Principle & the Haber Process It was not until the early 20th century that this method was developed to harness the atmospheric abundance of nitrogen to create ammonia, which can then be oxidized to make the nitrates and nitrites essential for the production of fertilizers and ammunitions. ∆H = -92.4 kJ/n @ 250 C NH3 + HNO3 NH4NO3(explosives & fertilizers)

  34. Le Châtelier’s Principle & the Haber Process Shift equilibrium to the right by increasing press & conc. of N2 + H2 and by removing NH3from system Fe N2 (g) + 3H2 (g) 2NH3 (g) This apparatus helps push the equilibrium to the right by removing the ammonia (NH3) from the system as a liquid Removed by liquefaction Fe

  35. The Effect of Catalysts on Equilibrium Increase the rate of both the forward and reverse reactions. How does the addition of a catalyst affect the equilibrium between subs. A and B? catalyst • Equilibrium is achieved faster, but the equilibrium composition remains unaltered.

  36. aA(aq) + bB(aq) cC(aq) + dD(aq) [C]c[D]d [A]a[B]b [C]ic[D]id [A]ia[B]ib K = Q = Equilibrium Constant (K) vs. Reaction Quotient (Q) • To calculate Q, one substitutes the initial concentrations on reactants and products into the equilibrium expression. • Q gives the same ratio the equilibrium expression gives, but for a system that is not at equilibrium. At Equilibrium: At Non-Equilibrium:

  37. aA(aq) + bB(aq) cC(aq) + dD(aq) [C]c[D]d [A]a[B]b [C]ic[D]id [A]ia[B]ib K = Q = If Q = K, the system is at equilibrium. If Q < K, there is too much reactant, and the equilibrium shifts to the right. If Q < K? If Q > K, there is too much product and the equilibrium shifts to the left. If Q > K?

  38. N2(g) + 3H2(g) 2NH3(g) [NH3]i2 [N2]i[H2]i [NH3]2eq [N2]eq[H2]eq Q = KC = Kc, Q and Spontaneity (ΔG) ΔG= - Reverse Process ΔG = + ΔG= - Reverse Process ΔG = + @ Equilibrium K = Q, ΔG = 0

  39. Free Energy (G) & Reaction Quotient (Q) under non-Standard Conditions ☺ G = G + RT ln Q = G + RT 2.303 log Q Non-standard conditions Non-standard conditions G = - (spontaneity), implies COMPLETE conversion of all reactants to → products @ standard conditions (25ºC, 1 atm, 1M). Gimplies DIFFERENT CONCENTRATIONSof both reactants ↔ products and under non-standard conditions of concentration, temperature and pressure. • If [Reactants]↑, then G = - • If [Products]↑, then G = + • If at Equilibrium , then G = 0

  40. Free Energy (G) & Equilibrium Constant (K) at Equilibrium At equilibrium, G= 0and Q = K so…… G = G + RT ln Q 0 = G + RT ln K K = eG/RT Rearranging …… ☺ G = - RT ln K =- 2.303 RT log K

  41. 2 NO2 (g) N2O4(g) Free Energy (ΔG), the Equilibrium Constant (ΔK),and Temperature How would T ↓or ↑ affect K? Low Temp. H = -58.0 kJ High Temp. (clear gas) (red-brown gas) @ T ↓, K ↑ G°= - G = - RT ln K @ T ↑, K ↓ G°= + Low Temperature: High Temperature: G°= - G°= + Q = K Reactants only Equilibrium mixture Products only {G, Q & T↑NO2 ← N2O4}HighT {G, Q & T↓ NO2→ N2O4}LowT

  42. 2003B Q1 2 HIH2 + I2

  43. Kp = Kc(RT)Δn 2003B Q1

  44. 2000A Q1

  45. 2000A Q1

  46. 2004B Q1