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Understanding Chemical Equilibrium: Laws, Problems, and Calculations

Discover the concept of chemical equilibrium where reactions balance in reversible processes. Learn about the Law of Chemical Equilibrium and how to calculate Keq for homogeneous and heterogeneous equations. Practice problems for calculating Keq provided.

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Understanding Chemical Equilibrium: Laws, Problems, and Calculations

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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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