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Chemical Equilibrium Chapter 18 Modern Chemistry

Chemical Equilibrium Chapter 18 Modern Chemistry. Sections 1 & 2 The Nature of Chemical Equilibrium Shifting Equilibrium. Section 18.1. The Nature of Chemical Equilibrium. Vocabulary. Reversible Reaction Chemical Equilibrium Equilibrium Expression Equilibrium Constant

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Chemical Equilibrium Chapter 18 Modern Chemistry

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  1. ChemicalEquilibriumChapter 18 Modern Chemistry Sections 1 & 2 The Nature of Chemical Equilibrium Shifting Equilibrium Chapter 18 Section 1 ChemicaL Equilibrium p. 589-597

  2. Section 18.1 The Nature of Chemical Equilibrium Chapter 18 Section 1 ChemicaL Equilibrium p. 589-597

  3. Vocabulary • Reversible Reaction • Chemical Equilibrium • Equilibrium Expression • Equilibrium Constant • LeChatelier’s Principle Insert Holt Disc 2 Chapter 18 Section 1 ChemicaL Equilibrium p. 589-597

  4. Reversible Reactions Insert Holt Disc 2 Insert Glencoe Disc 1 Chapter 18 Section 1 ChemicaL Equilibrium p. 589-597

  5. Reversible Reactions • Products can react to re-form the reactants. • Must occur in a “closed” system • 2HgO(s) 2Hg(l) + O2(g) • 2Hg(l) + O2(g) 2HgO(s) • Both of these reactions occur simultaneously • 2HgO(s) 2Hg(l) + O2(g) Chapter 18 Section 1 ChemicaL Equilibrium p. 589-597

  6. Chemical Equilibrium • Rate of its forward reaction equals the rate of its reverse reaction …. • and the concentrations of its products and reactants remain unchanged • Eventually all reversible reactions will reach eq. if the system is closed and conditions don’t change. • Eq. is dynamic – always in motion. Chapter 18 Section 1 ChemicaL Equilibrium p. 589-597

  7. Reaction Rate vs time p. 591 Chapter 18 Section 1 ChemicaL Equilibrium p. 589-597

  8. Equilibrium Demonstration Chapter 18 Section 1 ChemicaL Equilibrium p. 589-597

  9. Reaction Rate vs. Time p. 591 Rateforward = Ratereverse Chapter 18 Section 1 ChemicaL Equilibrium p. 589-597

  10. What is “favored” at Eq? • At equilibrium equal rates! reactants are favored neither is favored products are favored R P Chapter 18 Section 1 ChemicaL Equilibrium p. 589-597

  11. Equilibrium Expression [ ] = concentration in mol/L • n A + m B x C + y D • Dependant on temperature • Independent of initial concentrations [C]x [D]y products Keq = = [A]n[B]m reactants x, y, n, m = coefficients Chapter 18 Section 1 ChemicaL Equilibrium p. 589-597

  12. Equilibrium Constant PRODUCTS • If Keq is large (>1) then products are favored at eq. • If Keq is small (<1) then reactants are favored at eq. Keq = REACTANTS PRODUCTS Keq = REACTANTS Pure liquids and solid are omitted. Chapter 18 Section 1 ChemicaL Equilibrium p. 589-597

  13. Equilibrium Constants Table Chapter 18 Section 1 ChemicaL Equilibrium p. 589-597

  14. Sample Problem p. 594 • An equilibrium mixture of N2, O2 , and NO gases at 1500 K is determined to consist of 6.4x10–3 mol/L of N2, 1.7x10–3 mol/L of O2, and 1.1x10–5 mol/L of NO. What is the equilibrium constant for the system at this temperature? Keq = 1.1 x 10−5 Chapter 18 Section 1 ChemicaL Equilibrium p. 589-597

  15. Practice Problems p. 595 #1 • At equilibrium a mixture of N2, H2, and NH3 gas at 500°C is determined to consist of 0.602 mol/L of N2, 0.420 mol/L of H2, and 0.113 mol/L of NH3.What is the equilibrium constant for the reaction N2(g) + 3H2(g) 2NH3(g) at this temperature? 0.286 Chapter 18 Section 1 ChemicaL Equilibrium p. 589-597

  16. Practice Problems p. 595 #2 • The reaction AB2C(g) B2(g) + AC(g) reached equilibrium at 900 K in a 5.00 L vessel. At equilibrium 0.084 mol of AB2C, 0.035 mol of B2, and 0.059 mol of AC were detected. What is the equilibrium constant at this temperature for this system? (Don’t forget to convert amounts to concentrations.) 4.9 x 10−3 Chapter 18 Section 1 ChemicaL Equilibrium p. 589-597

  17. Practice Problems p. 595 #3 • A reaction between gaseous sulfur dioxide and oxygen gas to produce gaseous sulfur trioxide takes place at 600°C.At that temperature, the concentration of SO2 is found to be 1.50 mol/L, the concentration of O2 is 1.25 mol/L, and the concentration of SO3 is 3.50 mol/L. Using the balanced chemical equation, calculate the equilibrium constant for this system. 4.36 Chapter 18 Section 1 ChemicaL Equilibrium p. 589-597

  18. Equilibrium Concentrationsand Keq values p. 593 Chapter 18 Section 1 ChemicaL Equilibrium p. 589-597

  19. Eq. Concentration Problem • For the decomposition reaction of ammonia, 0.75 M of ammonia are added to an empty 1 L flask. When the reversible reaction has achieved equilibrium the concentration of nitrogen in the flask is 0.15 M. Find the equilibrium concentrations of hydrogen and ammonia. Also find the Keq. Chapter 18 Section 1 ChemicaL Equilibrium p. 589-597

  20. Determining Eq. Concentrations N2 (g)+3 H2 (g) 2NH3(g) 1/3 = 0.15/x x = 0.45M 0 M 0 M 0.75 M +0.15 M +0.45 M 0.15 M 0.45 M Chapter 18 Section 1 ChemicaL Equilibrium p. 589-597

  21. Determining Eq. Concentrations N2 (g)+3 H2 (g) 2NH3(g) 1/2 = 0.15/x x = 0.30M 0 M 0 M 0.75 M +0.15 M +0.45 M -0.30 M 0.15 M 0.45 M 0.45 M If this side is + then the other side is -. Chapter 18 Section 1 ChemicaL Equilibrium p. 589-597

  22. Eq. Concentration Problem • For the synthesis reaction of hydrogen and iodine, 0.20M of hydrogen and 0.30M of iodine are added to an empty 1 L flask. When the reversible reaction has achieved equilibrium the concentration of hydrogen in the flask is 0.10 M. Find the equilibrium concentrations of iodine and hydrogen iodide. Also find the Keq. Chapter 18 Section 1 ChemicaL Equilibrium p. 589-597

  23. Determining Eq. Concentrations H2 (g) + I2 (g)  2HI(g) 1/1 = 0.10/x x = 0.10M 0.20 M 0.30 M 0 M -0.10 M -0.10 M 0.10 M 0.20 M Chapter 18 Section 1 ChemicaL Equilibrium p. 589-597

  24. Determining Eq. Concentrations H2 (g) + I2 (g)  2HI(g) 1/2 = 0.10/x x = 0.20M 0.20 M 0.30 M 0 M -0.10 M -0.10 M +0.20 M 0.10 M 0.20 M 0.20 M If this side is - then the other side is +. Chapter 18 Section 1 ChemicaL Equilibrium p. 589-597

  25. Eq Concentration Problem • A 0.20 M solution of HC2H3O2 is 5.0% ionized. Find the equilibrium concentrations of H+, C2H3O21- and HC2H3O2, also find the Keq. answer Chapter 18 Section 1 ChemicaL Equilibrium p. 589-597

  26. Eq Conc. & Keq Problems • At a specific temperature and pressure, 1.2 moles of hydrogen, 0.40 moles of nitrogen and 1.3 moles of ammonia are put into a closed one liter flask. When allowed to reach equilibrium the amount of ammonia is 1.6 moles. Find the Keq for this system. • N2(g) + 3H2(g) 2NH3(g) answer Chapter 18 Section 1 ChemicaL Equilibrium p. 589-597

  27. Eq Conc. & Keq Problems • When .56 moles of SO3 is placed in a liter container, some of it decomposes. The equilibrium concentrations of SO2 is 0.42 moles / liter. Calculate the equilibrium concentration of O2 and SO3 and the Keq. O2 = 0.21 M SO3 = 0.14 M Keq = .53 Chapter 18 Section 1 ChemicaL Equilibrium p. 589-597

  28. Ch 18 Sec 1 Homework Page 595 # 1-9 Chapter 18 Section 1 ChemicaL Equilibrium p. 589-597

  29. Ch 18 Sec 1 Homework Equilibrium Concentrations and Keq Worksheet Chapter 18 Section 1 ChemicaL Equilibrium p. 589-597

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