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Chapter 16

Chapter 16. Chemical Equilibrium. Overview. Describing Chemical Equilibrium Chemical Equilibrium – A Dynamic Equilibrium (the link to Chemical Kinetics) The Equilibrium Constant. Heterogeneous Equilibria; solvents in homogeneous equilibria. Using the Equilibrium Constant

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Chapter 16

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  1. Chapter 16 Chemical Equilibrium

  2. Overview • Describing Chemical Equilibrium • Chemical Equilibrium – A Dynamic Equilibrium (the link to Chemical Kinetics) • The Equilibrium Constant. • Heterogeneous Equilibria; solvents in homogeneous equilibria.

  3. Using the Equilibrium Constant • Qualitatively Interpreting the Equilibrium Constant • Predicting the Direction of a Reaction • Calculating Equilibrium Concentrations

  4. Changing Reaction Conditions; Le Châtelier’s Principle. • Removing Products or Adding Reactants • Changing the Pressure or Temperature • Effect of a Catalyst.

  5. The Nature of the Equilibrium State Reaction Reversibility • Few chemical reactions proceed in only one direction; most are, at least to some extent reversible. • Because of reversibility, reactions in general do not go completely to products. • Reactions in general result in a mixture of reactants and products.

  6. catalytic methanation/steam reforming • CO(g) + 3H2(g) ----------> CH4(g) + H2O(g) • CH4(g) + H2O(g) ----------> CO(g) + 3H2(g)

  7. Formation of Stalactites and Stalagmites • CaCO3(s) + CO2(aq) + H2O(l) ----------> Ca2+(aq) + 2HCO3-(aq) • Ca2+(aq) + 2HCO3-(aq) ---------------> CaCO3(s) + CO2(aq) + H2O(l)

  8. Equilibrium • When the rates of the forward and reverse processes are equal, a state of chemical equilibrium exists and the concentrations of reactant and products no longer change with time. • An equilibrium such as this is called a dynamic equilibrium because the forward and reverse reactions (in a closed system) occur at the same rate.

  9. Chemical Equilibrium – A Dynamic Equilibrium • Upon addition of reactants and/or products, they react until a constant amount of reactants and products are present = equilibrium. • Equilibrium is dynamic since product is constantly made (forward reaction), but at the same rate it is consumed (reverse reaction).

  10. The Equilibrium State • Not all reactants are completely converted to product. • Reaction equilibria deal with the extent of reaction. • Arrows between reactants and products separate them and qualitatively indicate the extent of reaction. • Single arrow points to dominant side: H2(g) + O2(g)  H2O(g) • Double arrow indicates both reactants and products present after equilibrium obtained: N2O4 (g)  2NO2(g).

  11. Equilibrium exists when rates of forward and reverse reaction are the same. Equilibrium can be obtained from any mixture of reactants and products.

  12. The Equilibrium Constant • For the general reaction: aA(g) + bB(g) <----------> cC(g) + dD(g) • Equilibrium Constant Expression: expression obtained by multiplying the concentrations of products together, dividing by the concentrations of reactants and raising each concentration term to a power equal to the coefficient in the chemical equation.

  13. Equilibrium Constant Kc: • Value obtained for the equilibrium constant expression the value of which is constant for a particular reaction at a given temperature (Law of Mass Action)

  14. Manipulating Equilibrium Expressions • For a reaction that is the reverse of another C(s) + 1/2O2(g) CO(g)

  15. For a reaction that is twice that of another C(s) + 1/2O2(g) CO(g)

  16. For a reaction that is the sum of two others H2(g) + Br2(l)  2HBr(g) H2(g)  2H(g) Br2(l)  2Br(l) H(g) + Br(g)  HBr(g)

  17. Calculating an Equilibrium Constant Determine the equilibrium constant for the formation of HI(g) if the equilibrium concentration of H2, I2 and HI are 0.0060 M, 0.106 M, and 0.189 M, respectively. H2(g) + I2(g)  2HI(g) Kc = ?

  18. Calculating an Equilibrium Constant Determine the equilibrium constant for the reaction below: ½ H2(g) + ½ I2(g)  HI(g) Kc = ?

  19. Calculating an Equilibrium Constant Determine the equilibrium constant for the reaction: ½N2(g) + 3/2 H2(g)  NH3(g) given N2(g) + 3H2(g)  2NH3(g) Kc = 1.7x102

  20. Calculating an Equilibrium Constant CO(g) + 3H2(g) ----------> CH4(g) + H2O(g) • Initial 1.00 mol CO + 3.00 mol H2 in 10.0 L at 1200 K • Condense H2O = 0.387 mol

  21. The ICE Table • Initial • Change • Equilibrium

  22. CO(g) + 3H2(g) ----------> CH4(g) + H2O(g) initial 1.00 3.00 0 0 change -x -3x +x +x equilibrium 1.00 - x 3.00 - 3x +x +x X = 0.0387 M

  23. Equilibrium Constant Kc, Kp • For gases Kc and Kp are used. • KP same format as Kc except pressures used instead of concentrations. • Liquids and solids do not appear • Independent of starting amounts, container volume or Ptot • Varies with temperature

  24. Write out the equilibrium expression for KP using the reaction below: N2(g) + 3H2(g)  2NH3(g) KP = ?

  25. What is equilibrium expression (KP) for the reaction below? ½ N2(g) + 3/2 H2(g)  NH3(g) KP = ?

  26. In general Kp = Kc(0.0821 T) Dn(g) • (where Dng is the change in numbers of moles of gas)

  27. Determine the equilibrium constant (KP) for the formation of one mole of ammonia if at 500K, PNH3 = 0.15 atm, PN2 = 1.2 atm. and PH2 = 0.81 atm.

  28. Heterogeneous Equilbria • Do not include a solvent or solid in the equilibrium expression. Their composition is constant and included in the equilibrium constant. • Water concentration is 55.5 M; very high compared with reactants and products. • The concentration of a solid such as CaCO3 stays the same as long as some solid is present.

  29. Determine the equilibrium expression for the reaction: CaCO3(s) + C(gr)  CaO(s) + 2CO(g).

  30. Determine the equilibrium expression for the reaction of acetic acid with water. CH3COOH(aq) + H2O(l)  CH3COO(aq) + H3O+(aq)

  31. Applications of the Equilibrium Constant • Extent of reaction: The magnitude of the equilibrium constant allows us to predict the extent of the reaction. • Very large K (e.g. 1010)  mostly products. • Very small K (e.g. 1010)  mostly reactants. • When K is around 1, a significant amount of reactant and product present in the equilibrium mixture.

  32. Using the Equilibrium Constant • Direction of Reaction: the reaction quotient can be used to determine the direction of a reaction with certain initial concentrations.

  33. Reaction Quotient: For the general reaction, aA + bB  cC + dD,. where the t refers to the time that concentrations of the mixture are measured; not necessarily at equilibrium.

  34. Comparison of Qc with Kc reveals direction of reaction. When only reactants Qc = 0; leads to If Qc < Kc, products will form. When only products present, Qc. If Qc > Kc, reactants will form. When Qc = Kc, no net reaction.

  35. Determine direction of reaction: H2(g) + I2(g)  2HI(g). Assume that [H2]o = [I2]o = [HI]o = 0.0020M at 490oC for which Kc = 46.

  36. Calculating Equilibrium Concentrations • Using initial concentrations, stoichiometry and Kc, equilibrium concentrations of all components can be determined.

  37. Determine equilibrium amounts of hydrogen and iodine if you started with exactly 1.00 mol H2(g) and 2.00 mol I2 in a 1.00 L flask; after equilibrium had been attained, [HI] = 1.86 M. H2(g) + I2(g)  2HI(g)

  38. [H2]o = [I2]o = 2.4 M, were mixed and heated to 490oC in a container. Calculate equil. composition. Given that Kc = 46. H2(g) +I2(g)2HI(g)

  39. For the reaction: PCl5(g)  PCl3(g) + Cl2(g), Kc = 0.800 M at 340oC. Find the equilibrium amounts of these compounds if they all start out at 0.120 M. Solution: First determine Qc so that you know which direction the reaction will proceed.

  40. Factors that Alter the Composition of an Equilibrium Mixture • A change to the system, which is initially at equilibrium, can cause a change in the equilibrium composition.

  41. Le Châtelier’s Principle: “If a stress is applied to a reaction mixture at equilibrium, reaction occurs in the direction that relieves the stress.”

  42. Types of stress on equilibrium: • Concentration of reactants or products. You can add or remove one or more components in a reaction mixture. • With gases changing the pressure or volume is a way of changing the concentrations of all components in the mixture. • Change temperature.

  43. Changes in Concentrations • Addition or removal of either reactant or product shifts the equilibrium to reduce the excess compound. • Removal of product or addition of reactant has the same effect; they shift the equilibrium to the right.

  44. Stress and Changes in total Volume (or related pressure) • N2(g) + 3 H2(g) <--------> 2 NH3(g) + 92.00 kJ • A stress imposed by a decrease in volume is actually a stress caused by the increase in the concentration. The stress is relieved when the system reduces the number of molecules. • ie. 4 molecules ------> 2 molecules

  45. A decrease in total pressure (which is caused by an increase in volume) will shift the reaction to the reverse. A decrease in to total volume of a gaseous system (or the accompanying increase in pressure) shifts an equilibrium in the direction of the fewer molecules as shown by the equation for the reaction.

  46. Stress and Changes in Temperature • In any reaction a temperature increase favors the reaction that absorbs heat. i.e. the endothermic reaction. • eg. N2(g) + 3 H2(g) <-------> 2 NH3(g) + 92.00 kJ • This reaction is exothermic as written. That means the reverse reaction or backward reaction is endothermic. Adding heat (demonstrated as an increase in temperature) will shift the reaction left and more reactants will be formed as the products get used up. A decease in temperature will favor the reaction that produces heat. i.e. the exothermic reaction. The reaction above would switch to the forward direction to produce more heat and product.

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