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Chapter 18: Chemical Equilibrium

Chapter 18: Chemical Equilibrium. A State of Dynamic Balance. What is Equilibrium?. A reversible reaction is a reaction that can take place in both the forward and reverse directions. When a reaction is reversible a double-headed arrow will be used. 

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Chapter 18: Chemical Equilibrium

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  1. Chapter 18: Chemical Equilibrium A State of Dynamic Balance

  2. What is Equilibrium? • A reversible reaction is a reaction that can take place in both the forward and reverse directions. • When a reaction is reversible a double-headed arrow will be used.  • Chemical equilibrium occurs when the forward and reverse reactions occur at the same speed. At equilibrium, the concentrations of reactants and products remain the same.

  3. Law of Chemical Equilibrium • This law states that at a given temperature, a chemical system may achieve a state in which a ratio of reactant and product concentrations has a constant value. • The general equation for an equilibrium reaction is aA + bB cC + dD

  4. Equilibrium Constant • The equilibrium constant expression for an equilibrium system is given as Keq = [C]c[D]d [A]a[B]b • The brackets refer to molar concentrations.

  5. Homogeneous vs Heterogeneous Equilibria • In homogeneous equilibria, all of the compounds are in the same state. • When more than one state is present, pure solids and liquids are not included in the equilibrium expression.

  6. Homogeneous Example • Write the equilibrium constant expression for the reaction of hydrogen sulfide and water vapor to form sulfur dioxide and hydrogen. H2S (g) + 2H2O (g)  SO2 (g) + 3H2 (g)

  7. Heterogeneous Example • Write the equilibrium constant expression for the high-temperature reaction of carbon dioxide and solid carbon to form carbon monoxide. CO2 (g) + C (s)  2 CO (g)

  8. Practice Problems • Write equilibrium constant expressions for the following equilibria. a. C2H4O (g)  CH4 (g) + CO (g) b. 3O2 (g)  2O3 (g) c. 2N2O (g) + O2 (g)  4NO (g) d. 4NH3 (g) + 3O2  2N2 (g) + 6H2O (g)

  9. Practice Problem • Write equilibrium constant expressions for the following equilibria. a. C4H10 (l)  C4H10 (g) b. NH4HS (s)  NH3 (g) + H2S (g) c. CO (g) + Fe3O4 (s)  CO2 (g) + 3 FeO (s) d. (NH4)2CO3 (s)  2NH3 (g) + CO2 (g) + H2O (g)

  10. Calculating Keq • The value of Keqis a constant for a given reaction at a given temperature. • If Keq is greater than 1, then products are favored at equilibrium. • If Keq is less than 1, reactants are favored. • If you know the concentrations you can find the Keq.

  11. Example • Nitrogen monoxide and bromine react to form nitrosyl bromide according to the following equation. 2NO (g) + Br2 (g)  2NOBr (g) • Calculate Keq for this equilibrium using the data [NOBr] = 0.0474 M, [NO] = 0.312 M and [Br2] = 0.259 M.

  12. Practice Problems • The following is the chemical equation for the decomposition of formamide. HCONH2 (g)  NH3 (g) + CO (g) Calculate Keq using the equilibrium data [HCONH2] = 0.0637 M, [NH3] = 0.518 M and [CO] = 0.518 M.

  13. Practice Problems • Hydrogen and carbon disulfide react to form methane and hydrogen sulfide according to this equation. 4H2 (g) + CS2 (g)  CH4 (g) + 2H2S (g) Calculate Keq if the equilibrium concentrations are [H2] = 0.205 M, [CS2] = 0.0664 M, [CH4] = 0.0196 M and [H2S] = 0.0392 M.

  14. Le Châtelier’s Principle • This principle states that if a stress is applied to a system at equilibrium, the system shifts in the direction that relieves the stress. • Situations: • Adding or removing components • Changes in volume • Changes in temperature

  15. Example • Use Le Chatelier’s principle to predict how each of the following changes would affect this equilibrium. 2NO (g) + Br2 (g)  2NOBr (g) • Adding more NOBr • Adding more NO • Removing Br2 • Decreasing the volume

  16. Changes in Temperature • A change in temperature changes both the equilibrium position and the value of Keq. • For example, consider H2 (g) + Cl2 (g)  2HCl (g) ΔH°= -185 kJ • The forward reaction releases heat, so heat is a product as seen as H2 (g) + Cl2 (g)  2HCl (g) + heat

  17. Temperature Changes (cont’d) • Raising the temperature causes the value of Keq to decrease, since heat is removed. • If the temperature is lowered, the equilibrium relieves the stress by shifting to the right, increasing the concentration of product and the Keq.

  18. Practice Problem • Phosphorus pentachloride decomposes exothermically to form phosphorus trichloride and chlorine. PCl5 (g)  PCl3 (g) + Cl2 (g) + heat How would you regulate the temperature of this equilibrium in order to do the following? • Increase the concentration of PCl5 • Decrease the concentration of PCl3 • Increase the amount of Cl2 in the system • Decrease Keq

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