# Equilibrium

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```					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] < [reactants]  K larger than 1 indicates that at equilibrium the reaction mixture is almost all product. Sources: Rhonda Alexander, Kotz & Purcell, 2 Ed. and Zumdahl, 5 ed. AP Chemistry Special Focus Page 1 of 6
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K:     does not indicate anything about the rate varies only with temperature, constant at a given temperature independent of initial concentrations value is given without units

Changing the coefficients will change K: Multiplying coefficients - new K is equal to the old K raised to the power that the coefficients were  .# multiplied by; Knew = Kold Reversing the equation - new K is the reciprocal of old K; Knew = 1/Kold  Adding several equations to get a net equation – new K is the product of the old Ks of the  reactions added, the new K is the K for the net reaction; Knew = (Kold)(Kold)… Example 2: At a given temperature K = 278 for the reaction: 2SO 2+ O2 ↔ 2SO3 (all gases) Calculate the K for: a. SO2 + ½ O2 ↔SO3 b. 2 SO3 ↔ 2 SO2 + O2 Kc and Kp The equilibrium constant given in terms of concentrations may be written as K or K c. Molarity of gases and pressure are proportional but not equal. Partial pressures of gases can be used instead of [ ] in the equilibrium expression and Kc is replaced by Kp, which is the equilibrium constant in terms of pressure. Kp = Kc (RT) ∆n = moles of gas products – moles of gas reactants R = gas law constant with pressure units T = Kelvin temperature Example 3: Consider the reaction: 2 NOCl ↔ 2 NO + Cl2 (all gases). At 25ºC, a particular experiment had the following equilibrium pressures: PNOCl= 1.20 atm -2 PNO = 1.25 x 10 atm -1 PCl2 = 3.00 x 10 atm Calculate Kp and Kc.
∆n

Reaction Quotient = Q The reaction quotient is used to determine the direction the reaction will proceed to establish equilibrium. It is used when the initial concentration of the products are not assumed to be zero. Q is equal to the value obtained when concentrations not known to be the equilibrium concentrations are plugged into the equilibrium expression for the reaction. aA + bB ↔ cC + dD

[C ]c [ D]d Q [ A]a [ B]b
Q = K : reaction is at equilibrium Q > K : reaction proceeds towards reactants to reach equilibrium Q < K : reaction proceeds towards product to reach equilibrium Le Chatelier’s Principle: If a change is imposed on a system at equilibrium the position of the equilibrium will shift in a direction that tends to reduce the change. The reaction proceeding towards the products = shift right. The reaction proceeding faster towards the reactants = shift left. The reaction shifts the direction that counteracts the imposed change. Sources: Rhonda Alexander, Kotz & Purcell, 2 Ed. and Zumdahl, 5 ed. AP Chemistry Special Focus Page 2 of 6
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The addition of a reactant or product will shift the reaction away from that reactant or product to reduce the affect of the addition. The removal of a reactant or product will shift the reaction towards the removed reactant or product to produce more of the removed species. How do we remove a substance from the reaction?  Neutralization  Precipitation  Complex ion formation Changing the pressure will affect the equilibrium only if there are gases involved in the reaction. An increase in pressure results in the reaction shift to the side with fewer moles of gas. The effect of a temperature change is dependent upon the enthalpy change of the reaction. If the reaction is endothermic (+∆H) adding heat will favor the products. If the reaction is exothermic (-∆H) adding heat will favor the reactants. Adding a catalyst has NO EFFECT on the equilibrium position of a reaction! It lowers the activation energy for both the forward and reverse reaction equally.

Example 4: What will be the change in equilibrium position on each of the following reactions for each of the changes listed? 1. 2 HI ↔ H2 + I2 (all gases) exothermic 2. CO + 2 H2 ↔ CH3OH (all gases) endothermic a. hydrogen added b. I2 is removed c. Ar gas added d. the volume of the container doubled e. temperature decreased RICE Method Calculations to determine the equilibrium position or to determine K when the equilibrium concentrations or pressures are not given. Method: R. = write Reaction (do stoichiometry if required) and Relationship (eq expression) I. = Initial concentrations C. = Change in concentration to reach equilibrium (in terms of x if solving for equilibrium position) E. = Equilibrium concentration Example 5: Assume that for the reaction for the formation of gaseous hydrogen fluoride from hydrogen and fluorine gas has an equilibrium constant of 115. Calculate the equilibrium concentration of all the species when three moles of each reactant is added to a 1.5 liter container.

Example 6: Assume that gaseous hydrogen iodide is synthesized from hydrogen and iodide gas at a temperature where Kp = 100. Suppose .5 atm of HI, .01 atm of H2 and .01 atm of I2 were mixed in a 5 liter flask. Calculate the equilibrium pressures of all species.

Example 7: Gaseous NOCl decomposes to form the gases NO and Cl2. At 35°C the equilibrium constant -5 is 1.6 x 10 . In an experiment to in which 1.0 mol NOCl is placed a in 2.0-L flask, what are the equilibrium concentrations? Rules for Ignoring x 1. Show the –x and then put a line through it indication that a simplifying assumption has been made. 2. Calculate the percent ionization to prove the assumption may be made. Percent ionization should be < 5% for the –x value to be dropped. Percent ionization = (value of x / initial concentration of solution) * 100 Sources: Rhonda Alexander, Kotz & Purcell, 2 Ed. and Zumdahl, 5 ed. AP Chemistry Special Focus Page 3 of 6
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Typical Equilibrium Problem Formats 1. Determine the value of an equilibrium constant given equilibrium concentrations or partial pressures. Example 8: Given the following reaction: H2 (g) + CO2 (g)  H2O (g) + CO (g) When H2 (g) is mixed with CO2 (g) at 2,000K, equilibrium is achieved according to the equation above. In one experiment, the following equilibrium concentrations were measured: [H2] = 0.20 mol/L [CO2] = 0.30 mol/L [H2O] = [CO] = 0.55 mol/L (b) Using the equilibrium concentrations given above, calculate the value of K c, the equilibrium constant for the reaction. 2. Determine the equilibrium constant, K, given the initial concentrations of all reactants and products and the equilibrium concentration of one of the species. Example 8, cont. (d) When the system is cooled from 2,000K to a lower temperature, 30.0% of the CO (g) is converted back to CO2 (g). Calculate the value of Kc at this lower temperature. 3. Determine equilibrium concentrations given the K and an initial set of concentrations. Example 8, cont. (e) In a different experiment, 0.50 mol of H2 (g) is mixed with 0.50 mol if CO2 (g) in a 3.0-liter reaction vessel at 2,000 K. Calculate the equilibrium concentration, in moles per liter, of CO (g) at this temperature. Example 9: A sample of solid ammonium chloride was placed in an evacuated container and heated so it decomposed to ammonia gas and hydrogen chloride gas. After heating, the equilibrium partial pressure of ammonia was determined to be 2.2 atm. Calculate Kp for the reaction. Example 10: At 25°C, K = 0.090 for the reaction H2O(g) + Cl2O(g)  2HOCl(g) Calculate the concentration of all species at equilibrium for each of the following cases: a. 1.0g H2O and 2.0g of Cl2O are mixed in a 1.0L flask. b. 1.0 mol pure HOCl is placed in a 2.0 L flask.

Applying Stress to a System at Equilibrium Determining the new equilibrium concentrations after a change in the concentration of one of the substances in the system at equilibrium occurred. Example 11: At equilibrium the equilibrium constant at a given temperature for the reaction N 2O4(g) ↔ -3 NO2(g) is 5 X 10 . a. If 1 mole of N2O4 is injected into a 1 liter container and allowed to reach equilibrium, how many moles of each gas would be present at equilibrium? b. If the volume were halved at a constant temperature, how many moles of each gas would be present after equilibrium is established? Example 12: Consider the reaction H2(g) + I2(g) ↔ 2 HI(g) at 229ºC in a 1 liter container. When equilibrium is established, the following concentrations are present: [HI] = 0.49 M, [H2] = 0.08 M, [I2] = 0.08 M. If an additional 0.300 moles of HI is added at equilibrium, what concentrations will be present when the new equilibrium is established? Sources: Rhonda Alexander, Kotz & Purcell, 2 Ed. and Zumdahl, 5 ed. AP Chemistry Special Focus Page 4 of 6
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Solubility Equilibrium The Solubility Product Expression Solubility – the amount of solute that will dissolve in a given amount of solvent. All ionic compounds (even insoluble ones) ionize to some extent. For the insoluble salt silver chloride -5 only 0.002 g of the solid will dissolve in a liter of water. The solution is saturated but has only 1.3 x 10 moles of AgCl per liter. AgCl (s)  Ag (aq) + Cl (aq) The equilibrium constant expressions for this reaction: + K = [Ag ][Cl ] The product of the concentrations of the Ag and Cl ions at equilibrium is equal to a constant (K). Since this constant is proportional to the solubility of the salt, it is called the solubility product equilibrium constant for the reaction, or Ksp. + Ksp = [Ag ][Cl ] The Ksp expression for a salt is the product of the concentrations of the ions, with each concentration raised to a power equal to the coefficient of that ion in the balanced equation for the solubility equilibrium. Example 13: Write the Ksp expression for a saturated solution of CaF2 in water. Uses of the Solubility Product Constant 1. Determining the solubility of a salt in a saturated solution. Ksp is called the solubility product because it is literally the product of the solubilities of the ions in moles per liter. The solubility product of a salt can therefore be calculated from its solubility, or vice versa. Unfortunately, there is no simple way to predict the relative solubilities of salts from their Ksp’s if the salts produce different numbers of positive and negative ions when they dissolve in water. Example 14: Determine which salt – CaCO3 or Ag2CO3 – is more soluble in water in units of -9 -12 moles per liter. (CaCO3: Ksp = 2.8 x 10 and Ag2CO3: Ksp = 8.1 x 10 ) 2. Predicting whether a precipitate will form when two solutions combine. This calculation typically involves ions from two different sources. The concentration of each ion in any solution is directly related to the salt that is the source of the ion. Therefore to begin the problem you must calculate the initial concentrations of each ion in the mixture of the two solutions. Once the concentrations of the ions are known the ion product, Q sp, is calculated and the value of Q is compared with that of the Ksp. Qsp is the product of the concentrations of the ions at any moment in time (not necessarily at equilibrium). You can consider it a “trial K.” Considering the reverse reaction will allow us to determine if a precipitate will form when given the amounts of each ion in solution. If Qsp > 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 there are not enough ions in the solution. In order to return to equilibrium, more of the solid salt must dissociate into its ions. o Qsp = Ksp : This means that the system is at equilibrium. 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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