Equilibrium When the rates of 2 opposing processes are equal, the condition is called equilibrium. 5 Conditions of Equilibrium 1. The reaction must be reversible. If one driving force favors the forward process & the other driving force favors the reverse process, the process is reversible. 2. The forward and reverse reactions must occur simultaneously. The process is dynamic. 3. The forward and reverse reactions must occur at the same rate. The nature of equilibrium is the same no matter which direction equilibrium is approached. 4. There is no visible change. The concentrations of reactants & products are constant (not equal) at equilibrium. 5. The reaction occurs in a closed system. The change in concentrations with time for the reaction: H2O (g) + CO (g) ↔ H2 (g) + CO2 (g) when equimolar quantities of water and carbon monoxide are mixed.
Characteristics of Equilibrium Constants the equilibrium constant (K) is temperature dependent the equilibrium position (equilibrium concentrations of reactants and products) can be changed at constant temperature but K will remain constant Equilibrium Expression aA + bB ↔ cC + dD
[C ]c [ D]d Kc [ A]a [ B]b
Include gases & aqueous solutions; do not include solids and liquids because their concentrations are constant. Example 1: Write the equilibrium expression for each of the following. a. 4 NH3(g) + 7 O2(g) ↔ 4 NO2(g) + 6 H2O(g) b. P4(s) + 5 O2(g) ↔ P4O10(s) c. NH4NO3(s) ↔ N2O(g) + 2 H2O(g) Equilibrium Constant K The value of K conveys the extent of the reaction: large value for K: the reaction produces a substantial amount of product before reaching equilibrium [products] > [reactants] small value for K: the reaction produces a small amount of product before reaching equilibrium [products] K : reaction proceeds towards reactants to reach equilibrium Q Ksp : we will have precipitate. When Qsp = Ksp the solution has just become saturated. Any addition of either ion will result in precipitation. Example 15: If 100 mL of 0.00075 M sodium sulfate, Na2SO4, and 50 mL of 0.015 M barium chloride, BaCl2, -10 solutions are mixed will a precipitate form? (Ksp BaSO4 = 1.1 x 10 ) Example 16: + 22+ Calculate the final concentrations of K , C2O4 , Ba , and Br in a solution prepared by adding 0.100 L of 0.20 M K2C2O4 to 0.150 L of 0.25 M BaBr2. Sources: Rhonda Alexander, Kotz & Purcell, 2 Ed. and Zumdahl, 5 ed. AP Chemistry Special Focus Page 5 of 6
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3. Predicting the effect of the addition of a common ion to a solution. The solubility of an insoluble salt can be decreased even further by the addition of a common ion - an ion present in the insoluble salt from another (soluble) salt. The common ion effect is based on Le Chatelier’s Principle. Imagine we have a saturated solution of AgCl. The equilibrium reaction for the dissociation of this salt is: + AgCl (s) Ag (aq) + Cl (aq) [Ag ] is equal to [Cl ] at equilibrium because the mole ratio of Ag to Cl is 1:1. What would happen to the solution if a tiny bit of AgNO3 (a soluble salt) were added? Since AgNO3 is soluble, it dissociates + + completely to give Ag and NO3 ions. There would now be 2 sources of the Ag ion, from the AgCl and from the AgNO3: + AgCl (s) Ag (aq) + Cl (aq) + AgNO3 (s) Ag (aq) + NO3 (aq) + Adding AgNO3 increases the Ag concentration and the solution is no longer at equilibrium. The ion product (Qsp) at that moment is bigger than the solubility product (K sp). The reaction will eventually return + + to equilibrium but when it does, the [Ag ] is no longer equal to the [Cl ]. Instead, the [Ag ] will be larger than the [Cl ]. Let’s go back to the saturated AgCl solution. What would happen this time if a tiny bit of NaCl (a soluble + salt) were added? Since NaCl is soluble, it dissociates completely to give Na and Cl ions. There would now be two sources of the Cl ion, from AgCl and from NaCl: + AgCl (s) Ag (aq) + Cl (aq) + NaCl (s) Na (aq) + Cl (aq) Adding NaCl increases the Cl concentration and the solution is no longer at equilibrium. The ion product (Qsp) at that moment is bigger than the solubility product (Ksp). The reaction will eventually return to + equilibrium but when it does, the [Ag ] is no longer equal to the [Cl ]. Instead, the [Cl ] will be larger than + the [Ag ]. The ion product (Qsp) can be used to determine in which direction a system must shift in order to reach equilibrium. There are 3 possible situations: o Qsp Ksp : This means that there are too many ions in the solution. In order to return to equilibrium, the excess ions will precipitate to form more solid. Example 17: The molar solubility of MgF2 is 0.0012 M at 25ºC. Calculate the molar solubility of MgF2 in 0.10 M sodium -9 fluoride. (Ksp of MgF2 = 6.4 x 10 ) Selective Precipitation A technique in which one ion is selectively removed from a mixture of ions by precipitation. Example 18: Solid silver nitrate is slowly added to a solution that is 0.0010 M each in NaCl, NaBr, and NaI. Calculate the grams of silver nitrate required to add to a liter of this solution to initiate precipitation of -16 -10 -13 each silver halide. (Ksp AgI: 1.5 x 10 , Ksp AgCl = 1.8 x 10 , Ksp AgBr = 3.3 x 10 ) Example 19: A solution contains 1.0 x 10 M Cu and 2.0 x 10 M Pb . If a source of I is added -8 -12 gradually to this solution, will PbI2 (Ksp = 1.4 x 10 ) or CuI (Ksp = 5.3 x 10 ) precipitate first? Specify the concentration of I necessary to begin precipitation of each salt.
-4 + -3 2+ + + -
Sources: Rhonda Alexander, Kotz & Purcell, 2 Ed. and Zumdahl, 5 ed. AP Chemistry Special Focus Page 6 of 6
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