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Chapter 15

Chapter 15. Chemical Equilibrium. Review Section of Chapter 15 Test. Calculating an Empirical Formula Stoichiometry (mass – mass) Empirical vs. Molecular Formulas Bonds Comparison of single/double/triple bonds Resonance VSEPR theory Molecular Shape Bond Angles.

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Chapter 15

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  1. Chapter 15 Chemical Equilibrium

  2. Review Section of Chapter 15 Test • Calculating an Empirical Formula • Stoichiometry (mass – mass) • Empirical vs. Molecular Formulas • Bonds • Comparison of single/double/triple bonds • Resonance • VSEPR theory • Molecular Shape • Bond Angles

  3. Reversible Reactions- most chemical reactions are reversible under the correct conditions

  4. Reversible Reactions In a reversible reaction, there is both a forward and a reverse reaction. • Suppose SO2 and O2 are present initially. As they collide, the forward reaction begins. 2SO2(g) + O2(g) → 2SO3(g) • As SO3 molecules form, they also collide in the reverse reaction that forms reactants. A reversible reaction is written with a double arrow. forward 2SO2(g) + O2(g) 2SO3(g) reverse

  5. Chemical Equilibrium At equilibrium • The rate of the forward reaction becomes equal to the rate of the reverse reaction. • The forward and reverse reactions continue at equal rates in both directions. Figure 15.5 Page 454

  6. Chemical Equilibrium When equilibrium is Reached: • There is no further change in the amounts (concentrations) of reactant and product.

  7. N2(g) + O2(g) ↔ 2NO(g) At equilibrium the concentrations remain constant however the forward and reverse reactions are still occuring.

  8. Writing Equilibrium Constant-Expressions

  9. The Equilibrium Constant, Keq • Two methods for describing the equilibrium constant of a reaction. • Kc- describes the equilibrium of a reaction where the concentrations of the materials is known. • Kp- describes the equilibrium of a gaseous reaction using partial pressures instead of concentrations.

  10. Kcc = concentration aA + bBcC + dD Coefficients in the chemical equation become exponents in the equilibrium constant expression. Include only substances in the gas or aqueous phase. Solid’s and liquid’s concentrations do not change during a chemical reaction.

  11. Write the equilibrium expression (Kc) for the reaction below.Ni (s) + 4CO (g)  Ni(CO)4(g)

  12. Write the equilibrium expression (Kc) for the reaction below.Ni (s) + 4CO (g)  Ni(CO)4(g) [Ni(CO)4] Kc = [CO]4

  13. Determine the equilibrium constant.A ↔ B

  14. Kpp= pressure • Equilibrium is described in terms of the partial pressures of the reactants and products. 2CO2 (g) 2CO (g) + O2 (g)

  15. Write the equilibrium expression (Kp) for the following mixture of gases at equilibrium. 2NO (g) + O2 (g)  2NO2 (g) (PNO2)2 Kp = (PNO)2 (PO2)

  16. Properties of the Equilibrium Constant • Keq is constant for a particular reaction at a specific temperature. • We generally omit the units of the equilibrium constant. • Note that the equilibrium constant expression has products over reactants. • K>>1 implies products are favored, and Keq lies to the right (towards the products). • K<<1 implies reactants are favored, and Keq lies to the left. (towards the reactants).

  17. The Magnitude of Equilibrium Constants

  18. The equilibrium constant of a reaction in the reverse direction is the inverse of the equilibrium constant of the reaction in the forward direction… Keq(forward) = 1/Keq(reverse) For example: At 100 ºC, Keq(forward) = 6.49 At 100 ºC, Keq(reverse) = 1/6.49 = 0.154 More Facts About Keq

  19. The equilibrium constant for a net reaction made up of two or more steps is the product of the equilibrium constants for the individual steps. For example: A + B ↔ X + C Keq(1) = 2.0 X + B ↔ D Keq(2) = 5.0 A + 2B ↔ C + D Keq = K1 x K2 = 10.0 When this is written in terms of concentrations at equilibrium… Keq=10.0 = [C][D]/[A][B]2 More Facts About Keq

  20. The equilibrium expression only contains the concentrations of gases or aqueous substances and NEVER solids or pure liquids. Why? - Consider the decomposition of calcium carbonate: CaCO3(s)↔ CaO(s) + CO2(g) - The concentrations of solids and pure liquids are constant. Therefore. Keq = Kp = PCO2 Or Keq = Kc = [CO2] • This is an example of a heterogeneous equilibrium. More Facts About Keq

  21. Note: Although the concentrations of these species are not included in the equilibrium expression, they do participate in the reaction and must be present for an equilibrium to be established!

  22. What is the equilibrium-constant expression for this reaction? Kc = [Pb2+] [Cl−]2

  23. The equilibrium constant (K) is related to the rate constant (k)

  24. We can show for a reversible reaction having equal rates in both directions that….

  25. The Reaction Quotient- Q • There are times when a chemist may be given information about a chemical reaction that may or may not be at equilibrium. • A “Q” calculation can be used to determine the direction the reaction is heading to reach equilibrium.

  26. Predicting Direction of Reaction • We define Q, the reaction quotient, for a general reaction

  27. The Reaction Quotient- Q • If Q > K, then the reaction has too many products and is heading towards the reactant side. • If Q < K, then the reaction has too many reactants and is heading towards the product side. • If Q = K, then the reaction is at equilibrium.

  28. At 448 oC the equilibrium constant, Keq, for the reaction: H2(g) + I2(g) 2HI(g) is 50.5. Predict how the reaction will proceed to reach equilibrium at 448 oC if we start with 2.0 x 10-2 mol of HI, 1.0 x 10-2 mol of H2, and 3.0 x 10-2 mol of I2 in a 2.0-L container. Reaction Shifts

  29. At 2000°C, the equilibrium constant for the reaction N2(g) + O2(g) ↔ 2NO(g) is 4.1 x 10-4. Find the concentration of NO(g) in a mixture of NO(g), N2(g), and O2(g) in which, at equilibrium, [N2] = 0.036 mol L-1 and [O2] = 0.0089 mol L-1. Example 15.6 Page 459

  30. The equilibrium constant for the reaction of nitrogen and hydrogen to produce ammonia at a certain temperature is 6.00 x 10-2. Calculate the equilibrium concentration of ammonia if the equilibrium concentrations of nitrogen and hydrogen are 4.26M and 2.09M, respectively. Example 15.7 Page 459

  31. Iodine molecules react reversibly with iodide ions to produce triiodide ions. I2(aq) + I-(aq) ↔ I3-(aq) If a solution is prepared with the concentrations of both I2 and I- equal to 1.000 x 10-3M before reaction and if the concentration of I2 changes to 6.61 x 10-4M at equilibrium, what is the equilibrium constant for the reaction? The “ICE” Table Example 15.5 Page 457

  32. Iodine molecules react reversibly with iodide ions to produce triiodide ions. I2(aq) + I-(aq) ↔ I3-(aq) If a solution is prepared with the concentrations of both I2 and I- equal to 1.000 x 10-3M before reaction and if the concentration of I2 changes to 6.61 x 10-4M at equilibrium, what is the equilibrium constant for the reaction? Example 15.5 Page 457

  33. Find the concentration of NO(g) at equilibrium when a mixture O2(g) with a concentration of 0.50M and N2(g) with a concentration of 0.75Mis heated at 700°C. THE 5% RULE N2(g) + O2(g) ↔ 2NO(g) K = 4.1 x 10-9. Example 15.10 Page 466

  34. Example 15.8 Page 462 THE 5% RULE Under certain conditions the equilibrium constant for the decomposition of PCl5(g) into PCl3(g) and Cl2(g) is 0.0211. What are the equilibrium concentrations of all of three of these gases when the initial concentration of PCl5 was 1.00M?

  35. Example 15.8. The Quadratic Equation(See Appendix A.4)

  36. See the graphing calculator solutions method in notebook.

  37. Acetic acid, CH3CO2H, reacts with ethanol to form water and ethyl acetate. CH3CO2H + C2H5OH ↔ CH3CO2C2H5 + H2O The equilibrium constant for this reaction (when done with dioxane as a solvent) is 4.0. What are the equilibrium concentrations of all chemical species when 0.15 mol of CH3CO2H, 0.15 mol of C2H5OH, 0.40 mol of CH3CO2C2H5, and 0.40 mol of H2O are mixed in enough dioxane to make 1.0L of solution. Example 15.9 Page 464

  38. Effect of a Catalyst on Equilibrium • A catalyst speeds up the forward and reverse reactions equally and therefore equilibrium is reached faster. • However a catalyst has no effect on the value of the equilibrium constant or on the equilibrium concentrations.

  39. Le Chatelier’s Principle • Le Chatelier's principle states that if a stress is applied to a system at equilibrium, the system will shift to minimize the stress. • Stresses include changes in: • Concentration • Temperature • Pressure • Reactions can shift: • Right (toward products) • Left (towards reactants)

  40. Stress: Change in concentration

  41. Stress: Change in concentration

  42. Stress: Change in concentration SCN- is thiocyanate

  43. Stress: Change in concentration Would Adding Water Create a Stress?

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