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Chemistry 102(01) spring 2009

Chemistry 102(01) spring 2009. Instructor: Dr. Upali Siriwardane e-mail : upali@chem.latech.edu Office : CTH 311 Phone 257-4941 Office Hours : M,W, 8:00-9:00 & 11:00-12:00 a.m.; Tu,Th,F 9:00 - 10:00 a.m.  

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Chemistry 102(01) spring 2009

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  1. Chemistry 102(01) spring 2009 Instructor: Dr. Upali Siriwardane e-mail: upali@chem.latech.edu Office: CTH 311 Phone 257-4941 Office Hours: M,W, 8:00-9:00 & 11:00-12:00 a.m.; Tu,Th,F 9:00 - 10:00 a.m.   Test Dates: March 25, April 26, and May 18; Comprehensive Final Exam: May 20,2009 9:30-10:45 am, CTH 328. March 30, 2009 (Test 1): Chapter 13 April 22, 2009 (Test 2): Chapters 14 & 15 May 18, 2009 (Test 3):Chapters 16, 17 & 18 Comprehensive Final Exam: May 20,2009: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. Chemical equilibrium Different types of arrows are used in chemical equations associated with equilibria. Single arrow Assumes that the reaction proceeds to completion as written. Two single-headed arrows Used to indicate a system in equilibrium. Two single-headed arrows of different sizes. May be used to indicate when one side of an equilibrium system is favored.

  4. Chemical Equilibrium • Branch of chemistry dealing with reactions where reactants and products coexist in a dynamic equilibrium • the rates of forward and backward reactions have comparable rates reaction

  5. Kinetic Region Equilibrium Region HI Partial Pressure I2 H2 Time Chemical Equilibrium

  6. Complete Reaction Kinetic No change Region Concentration Time

  7. Dynamic Equilibrium Equilibrium • A state where the forward and reverse conditions occur at the same rate. I’m in static equilibrium.

  8. Chemical Equilibrium Equilibrium region. A point is finally reached where the forward and reverse reactions occur at the same rate. H2 + I2 2HI There is no net change in the concentration of any of the species.

  9. Forward and Backward Reactions This type of plot shows the energy changes during a reaction. Potential Energy activation energy H Reaction coordinate

  10. Value of K rate of forward Reaction k+ K = ------------------------------ = --- rate of backward Reactionk- K = a (infinity) -> Irreversible reactions K = 0 -> No reaction K = between 0 and 1 -> Equilibrium reactions

  11. 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.

  12. Equilibrium Expression An equilibrium expression could be written for any reaction [HI]2 K = ----------- = 16 L/mol [H2][I2] Keq >> 1 reaction will go mainly to products Keq ~ 1 reaction will produce roughly equal amounts of product and reactant Keq << 1 reaction will go mainly to reactants

  13. k is constant at a temperature 2NO2(g) Dark brown N2O4(g) colorless

  14. Calculating Stepwise Equilibrium Addtwo equations with K1 and K2 to get Keq • Keq = K1 x K2 Subtract one equations with K2 from another with K2 to get Keq • Keq = K1 / K2 DoublingK1 to get Keq • Keq = (K1)2 ;tripling Keq = (K1)3 etc. Reversing a reaction with K1 get Keq • Keq = (K1)½

  15. Stepwise Equilibrium [NO]2 Kc1 = [N2][O2] (1.) N2(g) + O2(g) 2NO(g) [NO2]2 Kc2 = [NO]2[O2] (2.) 2NO(g) + O2(g) 2NO2(g) Add to Combine (1.) & (2.) N2(g) + 2O2(g) 2NO2(g) [NO]2 [NO2]2 Kc =  = Kc1 Kc2 [N2][O2] [NO]2[O2]

  16. Stepwise Equilibrium • Consider the reactions • 2NO + O2 <===> 2 NO2 K = a • 2 NO2 <===> N2O4 K = b • The value of the equilibrium constant for the reaction • 2NO + O2 <===> N2O4 is • a + b • ab • c.(a/b)2 • d.(ab)2 • e.ab/2

  17. Stepwise Equilibrium • Consider the reactions • 2NO + O2 <===> 2 NO2 K = a • 2 NO2 <===> N2O4 K = b • The value of the equilibrium constant for the reaction • 4NO + 2O2 <===> 2 N2O4 is • a + b • ab • c.(a/b)2 • d.(ab)2 • e.ab/2

  18. Types of Equilibria • Homogenous equilibrium: Chemical equilibrium where reactants and products are in same phase. • Heterogeneous equilibrium: Chemical Equilibrium where at least one phase of a reactant or product is different from the rest.

  19. 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)]

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

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

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

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

  24. 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

  25. 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)

  26. 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

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

  28. 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

  29. 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.

  30. 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- 0.243= 0.138 0.525 x -1/2x = 0 + x 0.40 - 1/2x

  31. NO2(g ) N2O4(g) 0.525 = 0 + x [NO2]2 Kc = [N2O4] 0.40 - 1/2x x = 0.525 [NO2] = 0.40 - 1/2x = 0.40 - 1/2(0525) [NO2]2 (0.525)2 Kc = = [N2O4] 0.138 = 0.138 = 2.00

  32. 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?

  33. 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

  34. Selected Equilibrium Constants

  35. 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.

  36. 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 + bB eE + fF Q has the same form as Kc with one important difference. Q can be for any set of concentrations, not just at equilibrium.

  37. 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.

  38. Predicting the Direction of a Reaction

  39. 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.

  40. 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 equilirium shift to right

  41. 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?

  42. [ CO ] [ Cl2 ] [ COCl2 ] X2 (0.095 - X) 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 = =

  43. X2 (0.095 - X) Kc = 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

  44. -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.

  45. -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

  46. 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.

  47. 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?

  48. 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.

  49. 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

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

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