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Heterogeneous Equilibria:

Heterogeneous Equilibria: A homogenous reaction is one in which all the substances are in the same state. A heterogeneous reaction is one in which all the substances are not in the same state. CaCO 3 (s) CO 2 (g) + CaO (s) Calcium carbonate carbon dioxide lime.

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Heterogeneous Equilibria:

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  1. Heterogeneous Equilibria: A homogenous reaction is one in which all the substances are in the same state. A heterogeneous reaction is one in which all the substances are not in the same state. CaCO3(s) CO2(g) + CaO(s) Calcium carbonate carbon dioxide lime When writing equilibrium constant expressions for Heterogeneous equilibria, you don’t include pure solids or pure liquids. Their concentrations don’t change Keq = [CO2 ][ CaO ]Keq = [CO2 ] [ CaCO3 ] Try writing an Keq expression for this reaction: 2H2O (l) 2H2(g) + O2(g)

  2. Keq expression for this reaction: 2H2O (l) 2H2(g) + O2(g) K = [H2]2[O2] Keq expression for this reaction: CuSO4 . 5H2O (s)CuSO4 (s) + 5H2O (g) K = [H2O]5

  3. La Chatelier's Principle This states that when a change is imposed on a system at equilibrium, the position of the equilibrium shifts in a direction that tends to reduce the effect of that change. CONCENTRATION: When a reactant or product is added to a system at equilibrium, the system shifts away from the added component. (moves in a direction that uses up the excess component) But if a reactant or product is removed, the system shifts toward the removed component.

  4. N2(g) + 3H2(g) 2NH3(g) Equilibrium concentrations [N2] = 0.399 M [H2] = 1.197 M [NH3] = 0.202 M What will happen if 1.000 mol L-1 N2 is added to equilibrium? New Equilibrium concentrations [N2] = 1.348 M [H2] = 1.044 M [NH3] = 0.304 M Position I: Position II:

  5. VOLUME: When the volume of a gaseous system at equilibrium is decreased, the system shifts in the direction that gives the smaller number of gas molecules decreasing the volume of the following reaction will: CaCO3(s) CaO (s) + CO2(g) shifts the equilibrium to the left

  6. N2(g) + 3H2(g) 2NH3(g) shifts the equilibrium to the right PCl3(g) + 3NH3(g) P(NH2)3 (g) + 3HCl (g) has no effect on equilibrium position Take the equilibrium between 2NO2 (g) N2O4 (g) brown gas clear gas Which direction is favored by increasing the volume? The right!

  7. TEMPERATURE: When heat is added to a system at equilibrium, the system shifts in a direction that uses up the excess heat, the exothermic reaction. A reaction that absorbs heat is endothermic, a reaction that produces heat is exothermic. N2(g) + 3H2(g) 2NH3(g) + 92 kJ CaCO3(s) + 556 kJ  CaO (s) + CO2(g)

  8. REVIEW EXERCISE: N2O4(g) + energy  2NO2(g) Change: Addition of N2O4(g) Addition of NO2(g) Removal of N2O4(g) Removal of NO2(g) Decrease in container volume Increase in container volume Increase in temperature Decrease in temperature • Shift: • Right • Left • Left • Right • Left • Right • Right • Left

  9. Using the Equilibrium Constant: What can the size of Keq tell you? Take the equilibrium A (g) B (g) If the value of Keq is one then the equilibrium concentrations of A and B are the same. A (g) B (g) If the value of Keq is less than one, then the reaction at equilibrium consists mainly of reactants - the equilibrium position is far to the left. A (g) B (g) If the value of Keq is much larger than one, the reaction system mainly consists of products - the equilibrium position is far to the right. A (g) B (g)

  10. You can also use the equilibrium constant to find the concentrations of reactants and products. For example if you know the value of K and the concentration of all the reactants and products except one, we can calculate the missing concentration. Gaseous phosphorus pentachloride decomposes to chlorine gas and gaseous phosphorus trichloride. In a certain experiment, at a temperature where K = 8.96 10-2 , the equilibrium concentrations of PCl5 is 6.70 10-3 and PCl3 is 0.300 M. Calculate the concentration of Cl2 present at equilibrium. [Cl2] = 2.00 10-3 mol L-1

  11. Solubility Equilibria Often there is an equilibrium between a dissolving solid and its aqueous solution. CaF2 (s) Ca 2+ (aq) + 2F -(aq) Ksp = [Ca 2+] [F -]2 Where Ksp is the solubility product constant, or simply the solubility product.

  12. Calculating the solubility product: CuBr has a measured solubility of 2.0 10-4 at 25 C. Calculate the solids Ksp value CuBr (s) Cu + (aq) + Br -(aq) Ksp = [Cu +] [Br -] We know that 2.0 10-4 mol of solid CuBr dissolves per 1.0L of solution to come to equilibrium. CuBr (s) Cu + (aq) + Br -(aq) Ksp = [Cu +] [Br -] = (2.0 10-4 )(2.0 10-4) = 4.0 10-8

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