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Chapter 18 Equilibrium

Chapter 18 Equilibrium. A + B  AB We may think that all reactions change all reactants to products, or the reaction has gone to completion But in reality, products may start to change back into reactants; the reaction is reversible . A + B ↔ AB. CO 2 + H 2 O  C 6 H 12 O 6 + O 2

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Chapter 18 Equilibrium

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  1. Chapter 18 Equilibrium A + B  AB • We may think that all reactions change all reactants to products, or the reaction has gone to completion • But in reality, products may start to change back into reactants; the reaction is reversible. A + B ↔ AB

  2. CO2 + H2O  C6H12O6 + O2 O2 + C6H12O6  CO2 + H2O

  3. Equilibrium • Equilibrium is the state where the forward and reverse reactions balance because they are occurring at equal rates. • NO OVERALL CHANGE!

  4. The Law of Chemical Equilibrium • the point in the reaction where a ratio of reactant and product concentration has a constant value, Keq

  5. Homogeneous equilibrium • States of all compounds are the ____________ H2(g) + I2(g)↔ 2HI (g) • Heterogeneous equilibrium • States of compounds are ________________ CaCO3(g)↔ CaO (s) + CO2(g) Remember we only pay attention to the GASES when we calculate Keq!

  6. What is the correct the Keq? N2(g) + 3H2(g)↔ 2NH3(g)

  7. What is the correct the Keq? 2 NbCl4(g)↔ NbCl3(g) + NbCl5(g)

  8. What is the correct Keq? H2O (l)↔ H2O(g)

  9. If Keq = 2.4, which is more favored? A. products B. reactants C. neither is favored D. not enough information

  10. Chemical Equilibrium Problems I • Write the equilibrium constant expression (Keq) for these homogeneous equations. 1. N2O4 (g) ↔ 2 NO2 (g)

  11. 2. CO (g) + 3H2(g)↔ CH4(g)+ H2O (g) 3. 2 H2S (g)↔ 2 H2(g) + S2(g)

  12. Write the equilibrium constant expression (Keq) for these heterogeneous equations. • C10H8(s)↔ C10H8(g) • CaCO3(s) ↔ CaO (s) + CO2(g)

  13. 3. C (s) + H2O(g) ↔ CO(g) + H2(g) 4. FeO(s) + CO(g) ↔ Fe(s) + CO2(g)

  14. 0.98 • 15.59 • 0.064 • 1.02 • 2.05 1. Calculate the Keq for the following equation using the data: [N2O4] = 0.0613 mol/L [NO2] = 0.0627 mol/L N2O4(g)↔ 2NO2(g)

  15. A. 0.25 B. 0.72 C. 0.04 D. 3.93 E. 0.02 2. [CO] = 0.0613 mol/L [H2] = 0.1839 mol/L [CH4]=0.0387 mol/L [H2O]=0.0387 mol/L CO (g) + H2 (g)↔ CH4(g)+ H2O (g)

  16. 3. [H2]=1.5 mol/L [N2]=2.0 mol/L [NH3]=1.8 mol/L 3H2(g) + N2(g)↔ 2NH3(g) • 0.48 • 0.24 • 2.5 • 0.40

  17. 3. If the Keq = 0.48, we know that the equilibrium favors… a. reactants b. products c. neither

  18. 4. [Mg]=2 mol/L [HCl]=3 mol/L [MgCl]=6 mol/L [H2]=3 mol/L 2 Mg (s) + 2 HCl (g)↔ 2 MgCl (g) + H2(g) • 3 • 12

  19. 4. Since Keq = 3, we know that the equilibrium favors a. reactants b. products c. neither

  20. 5. [H2]=0.52 mol/L [I2]=0.23 mol/L [HI]=1.7 mol/L H2(g) + I2(g)↔ 2HI (g) • 202 • 0.04 • 24.16

  21. 5. Since Keq= 24.16, we know that the equilibrium favors a. reactants b. products c. neither

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