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Chemistry 102(001) Fall 2012

Chemistry 102(001) Fall 2012. CTH 328 10:00-11:15 am Instructor : Dr. Upali Siriwardane e-mail : upali@latech.edu Office : CTH 311 Phone 257-4941 Office Hours : M,W 8:00-9:00 & 11:00-12:00 am; Tu , Th , F 8:00 - 10:00am.. Exams: 10 :00-11:15 am, CTH 328.

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Chemistry 102(001) Fall 2012

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  1. Chemistry 102(001) Fall 2012 CTH 328 10:00-11:15 am Instructor: Dr. UpaliSiriwardane e-mail: upali@latech.edu Office: CTH 311 Phone 257-4941 Office Hours: M,W 8:00-9:00 & 11:00-12:00 am; Tu, Th, F 8:00 - 10:00am.. Exams: 10:00-11:15 am, CTH 328. September 27, 2012 (Test 1): Chapter 13 October 18, 2012 (Test 2): Chapter 14 &15 November 13, 2012 (Test 3):Chapter 16 &18 Optional Comprehensive Final Exam: November 15, 2012 : Chapters 13, 14, 15, 16, 17, and 18

  2. Chapter 14. Chemical Equilibrium 14.1 Characteristics of Chemical Equilibrium 14.2 The Equilibrium Constant 14.3 Determining Equilibrium Constants 14.5 The Meaning of Equilibrium Constant 14.6 Using Equilibrium Constants 14.7 Shifting a Chemical Equilibrium: Le Chatelier's Principle 14.8 Equilibrium at the Nanoscale 14.9 Controlling Chemical Reactions: The Haber-Bosch Process

  3. Law of mass Action Defines an equilibrium constant (K) for the process j A + k B l C + m D [C]l[D]m K = ----------------- ; [A], [B] etc are [A]j[B]k Equilibrium concentrations Pure liquid or solid concentrations are not written in the expression.

  4. 7) What is the difference between initial [A]iand equilibrium [A]eqconcentrations? 2NO2(g) Dark brown N2O4(g) colorless

  5. Equilibrium calculations [E]e [F]f [A]a [B]b Q = We can predict the direction of a reaction by calculating the reaction quotient. Reaction quotient, Q For the reaction: aA + bBeE + fF Q has the same form as Kc with one important difference. Q can be for any set of concentrations, not just at equilibrium.

  6. Equilibrium constant calculations • 1) Consider the reaction: • COCl2(g) CO(g) + Cl2(g) • At equilibrium, [CO] = 4.14 × 10-6 M; • [Cl2] = 4.14 × 10-6 M; and [COCl2] = 0.0627 M. • Calculate the value of the equilibrium constant.

  7. Terminology Initial concentration: concentration (M) of reactants and products before the equilibrium is reached. Equilibrium Concentration Concentration (M) of reactants and products After the equilibrium is reached.

  8. Reaction quotient • Any set of concentrations can be given and a Q calculated. By comparing Q to the Kc value, we can predict the direction for the reaction. • Q < Kc Net forward reaction will occur. • Q = Kc No change, at equilibrium. • Q > Kc Net reverse reaction will occur.

  9. Reaction quotient (Q) calculations • 2) Consider the reaction system: • C2H5OH (aq) + CH3COOH(aq) CH3COOC2H5 +H2O (l) • = ? • The concentrations of both ethanol and acetic acid are 0.45 M and the concentration of ethyl acetate is 1.1 M. Use the reaction quotient to determine whether the system is at equilibrium.

  10. Reaction quotient (Q) calculations • Consider the reaction system: • C2H5OH (aq) + CH3COOH(aq) CH3COOC2H5 +H2O (l) • The concentrations of both ethanol and acetic acid are 0.45 M and the concentration of ethyl acetate is 1.1 M. Use the reaction quotient to determine whether the system is at equilibrium. • = ? and Keq=0.95

  11. Determining Equilibrium ConstantsICE Method 1. Derive the equilibrium constant expression for the balanced chemical equation 2. Construct a Reaction Table with information (ICE) about reactants and products 3. Include the amounts reacted, x, in the Reaction Table 4. Calculate the equilibrium constant in terms of x

  12. Example: An equilibrium is established by placing 2.00 moles of N2O4(g) in a 5.00 L and heating the flask to 407 K. It was determined that at equilibrium the concentration of the NO2(g) is 0.525 mol/L. What is the value of the equilibrium constant? N2O4(g) 2 NO2(g) [NO2]2 Kc = [N2O4] N2O4(g) 2 NO2(g) [Initial] (mol/L) 0.40 0 [Change] -x 2x [Equilibrium] 0.40- 0.263= 0.138 0.525 -1/2 x x = 0 + x 0.40 - 1/2x

  13. What is the value of the equilibrium constant? 0.525 = 0 + x NO2(g ) N2O4(g) [NO2]2 Kc = [N2O4] 0.40 - 1/2x [NO2] = 0.40 - 1/2x x = 0.525 = 0.40 - 1/2(0525) [NO2]2 (0.525)2 Kc = = [N2O4] 0.138 = 2.00 = 0.138

  14. Equilibrium Calculations Hydrogen iodide, HI, decomposes according to the equation 2 HI(g)  H2(g) + I2(g) When 4.00 mol of HI placed in a 5.00-L vessel at 458ºC, the equilibrium mixture was found to contain 0.442 mol I2. What is the value of Kc for the reaction?

  15. 2 HI(g)  H2(g) + I2(g) Initial 4.00/5=.80 0 0 Change -2x x x Equilibrium 0.80-2x x x=0.442/5 x = 0.0884 Equilibrium concentrations [HI] = 0.80 - 2x = 0.8 - 2 x 0.0884 = 0.62 [H2] = x = 0.0884 [I2] = x = 0.0884 [H2] [I2] 0.0884 x 0.0884 Kc = ---------------- = ------------------------- = 0.0201 [HI]2 (0.62) 2

  16. Equilibrium calculations using ICE • 3) Consider the reaction A 2B, • where the value of Keqis 1.4 × 10-12. • At equilibrium, the concentration of B is 0.45 M. • What is the concentration of A?

  17. Equilibrium calculations using ICE • 4) Calculation of unknown concentration of reactants or products in an equilibrium mixture At 100o C the equilibrium constant (K) for the reaction: • H2(g) + I2(g) 2 HI(g) ; K = 1.15 x 102. • If 0.400 moles of H2 and 0.400 moles. If I2 are placed into a 12.0-liter container and allowed to react at this temperature, what is the HI concentration (moles/liter) at equilibrium?

  18. Equilibrium calculations using ICE • 4) H2(g) + I2(g) 2 HI(g) ; K = 1.15 x 102. • If 0.400 moles of H2 and 0.400 moles. If I2 are placed into a 12.0-liter container and allowed to react at this temperature, what is the HI concentration (moles/liter) at equilibrium?

  19. Types of Equilibrium • Heterogeneous Equilibrium • Heterogeneous Equilibrium • Acid Dissociation Constant- Ka • Base Dissociation Constant- Kb • Autoionization Constant- Kw • Solubility Product Constant-Ksp

  20. Heterogeneous Equilibrium CaCO3(s) CaO(s) + CO2(g) [CaO(s)][CO2(g)] Kc = [CaCO3(s)] concentrations of pure solids and liquids are constant are dropped from expression Kc = [CO2(g)]

  21. Acid Dissociation Constant HC2H3O2 (aq) + H2O(l) H3O+ (aq) + C2H3O2- (aq) [H3O+][C2H3O2-] K = [H2O][HC2H3O2] [H3O+][C2H3O2-] Ka = K  [H2O] = [HC2H3O2]

  22. Base Dissociation Constant NH3 + H2O(l) NH4+ + OH- [NH4+][OH-] K = [H2O][NH3] [NH4+][OH-] Kb = K  [H2O] = [NH3]

  23. Autoionization of Water H2O (l) + H2O (l) H3O+ + OH- [H3O+][OH-] K = [H2O]2 Kw = K [H2O]2 = [H3O+][OH-] = 1.0  10-14

  24. Solubility Product of Salts in Water AgCl(s) + H2O (l) Ag+(aq) + Cl- Ksp = [Ag+] [Cl-] Ksp (AgCl) = 1.77 × 10-10 Ksp (BaSO4) = 1.1 x 10-10

  25. What is K (Kc) and Kp Kc (K) - equilibrium constant calculated based on [A]-Concentrations. Kp- equilibrium constant calculated based on partial pressure Kp =

  26. Pressure Equilibrium Constants Kc & Kp N2 + 3H2 2NH3 [NH3]2 Kc = [N2][H2]3 (PNH3/RT)2 = (PN2/RT) (PH2/RT)3 (PNH3)2(1/RT)2 Kc = (PN2)(1/RT))(PH2)3(1/RT)3) (1/RT) 2 = Kp (1/RT)(1/RT)3 PNH32(1/RT)2 = PN2 PH23 (1/RT)(1/RT)3

  27. Kc vs. Kp N2 (g) + 3H2(g) 2NH3(g) (1/RT)2 Kc = Kp= Kp (1/RT)-2 (1/RT)(1/RT)3 • In General • Kc = Kp (1/RT)Dn • where Dn = #moles gaseous products • - # moles gaseous reactants

  28. What is K (Kc) and Kp Kc (K) - equilibrium constant calculated based on [A]-Concentrations. Kp- equilibrium constant calculated based on partial pressure (p) Kp = K(RT)Dn R = universal gas constant T = Kelvin Temperature, Dn = (sum of stoichiometric coefficients of gaseous products) - (sum of the stoichiometric coefficients of gaseous reactants)

  29. Partial pressure & Equilibrium Constants atm L mol K For the following equilibrium, Kc = 1.10 x 107 at 700. oC. What is the Kp? 2H2 (g) + S2 (g) 2H2S (g) Kp = Kc (RT)Dng T = 700 + 273 = 973 K R = 0.08206 Dng = ( 2 ) - ( 2 + 1) = -1

  30. Partial pressure & Equilibrium Constants atm L mol K • Kp = Kc (RT)Dng • = 1.10 x 107 (0.08206 ) (973 K) • = 1.378 x105 [ ] -1

  31. What is the reaction quotient, Q (Q) is constant in the equilibrium expression when initial concentration of reactants and products are used. SO2(g)+ NO2(g)  NO(g) +SO3(g) [NO][SO3] Q = ---------------- [SO2][NO2] comparing to K and Q provide the net direction to achieve equilibrium.

  32. Predicting the Direction of a Reaction

  33. Q Calculation • Consider the following reaction: • SO2(g) + NO2(g)  NO(g) + SO3(g) • (Kc = 85.0 at 460oC) • Given: 0.040 mole of SO2(g), 0.500 mole of NO2(g), 0.30 mole of NO(g),and 0.020 • mole of SO3(g) are mixed in a 5.00 L flask, Determine: • a) The net the reaction quotient, Q. • b) Direction to achieve equilibrium at 460oC.

  34. Q Calculation SO2(g) + NO2(g)  NO(g) + SO3(g) (Kc = 85.0 at 460oC)    [NO][SO3] Q = -------------         [SO2][NO2] 0.040 mole               0.500 mole                0.30 mole                 0.020 mole [SO2] = -------------; [NO2] = ----------- ; [NO] = ------------; [SO3] = -----------                   5.00 L                         5.00L                    5.00L                      5.00 L [SO2] = 8 x 10-3mole/L ; [NO2] =0.1mole/L; [NO] = 0.06 mole/L; [SO3] = 4 x 10-3mole/L 0.06 (4 x 10-3 ) Q = ---------------------- = 0.3           8.0 x 10-3 x 0.1 Therefore the equilibrium shift to right

  35. Equilibrium Calculation Example A sample of COCl2 is allowed to decompose. The value of Kc for the equilibrium COCl2 (g) CO (g) + Cl2 (g) is 2.2 x 10-10 at 100 oC. If the initial concentration of COCl2 is 0.095M, what will be the equilibrium concentrations for each of the species involved?

  36. X2 (0.095 - X) [ CO ] [ Cl2 ] [ COCl2 ] Equilibrium Calculation Example • COCl2 (g) CO (g) Cl2 (g) • Initial conc., M 0.095 0.000 0.000 • Change - X + X + X • in conc. due to reaction • Equilibrium M(0.095 -X) X X • Concentration, • Kc = =

  37. X2 (0.095 - X) Keq= 2.2 x 10-10 = Equilibrium calculation example Rearrangement gives X2 + 2.2 x 10-10 X - 2.09 x 10-11 = 0 This is a quadratic equation. Fortunately, there is a straightforward equation for their solution

  38. -b + b2 - 4ac 2a Quadratic Equations An equation of the form a X2 + b X + c = 0 Can be solved by using the following x = Only the positive root is meaningful in equilibrium problems.

  39. -b +b2 - 4ac 2a - 2.2 x 10-10 + [(2.2 x 10-10)2 - (4)(1)(- 2.09 x 10-11)]1/2 2 Equilibrium Calculation Example X2 + 2.2 x 10-10 X - 2.09 x 10-11 = 0 a b c X= X = X = 4.6 x 10-6 M X = -4.6 x 10-6 M

  40. Equilibrium Calculation Example Now that we know X, we can solve for the concentration of all of the species. COCl2 = 0.095 - X = 0.095 M CO = X = 4.6 x 10-6 M Cl2 = X = 4.6 x 10-6 M In this case, the change in the concentration of is COCl2 negligible.

  41. Le Chatelier’s principle Any stress placed on an equilibrium system will cause the system to shift to minimize the effect of the stress. You can put stress on a system by adding or removing something from one side of a reaction. N2(g) + 3H2 (g) 2NH3 (g) What effect will there be if you added more ammonia? How about more nitrogen?

  42. Predicting Shifts in Equilibria • Equilibrium concentrations are based on: • The specific equilibrium • The starting concentrations • Other factors such as: • Temperature • Pressure • Reaction specific conditions • Altering conditions will stress a system, resulting in an equilibrium shift.

  43. Increase in Concentrationor Partial Pressure for N2(g) + 3 H2(g) 2 NH3(g) an increase in N2 and/or H2 concentration or pressure, will cause the equilibrium to shift towards the production of NH3

  44. Shifts with Temperature N2O4(g) 2 NO2(g) ; D H=? (+or -) 2NO2(g) Dark brown N2O4(g) colorless

  45. Predicting Equilibrium Shifts • For the following equilibrium reactions: • H2(g) + CO2(g)  H2O(g) + CO(g) DH = 40 kJ • Predict the equilibrium shift if: • a) The temperature is increased • b) The pressure is decreased

  46. Shifting of Equilibrium N2O4(g)  2 NO2(g)

  47. Changes in pressure In general, increasing the pressure by decreasing volume shifts equilibrium towards the side that has the smaller number of moles of gas. H2 (g) + I2 (g) 2HI (g) N2O4 (g) 2NO2 (g) Unaffected by pressure Increased pressure, shift to left

  48. 5) How you would increase the products of following industrially important reactions:a) CO(g) +H2O (g) H2(g) +CO2 (g);DH= −41.2 kJ/mol

  49. Equilibrium Systems • product-favored if K > 1 • exothermic reactions favor products • increasing entropy in system favors products • at low temperature, product-favored reactions are usually exothermic • at high temperatures, product-favored reactions usually have increase in entropy

  50. Thermodynamics of Equilibrium • a) Enthalpy (DH) • Entropy (DS) • Free Energy (DG) • (DG is a combined term involving DH, DS and T)

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