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CHAPTER 14 Chemical Equilibrium
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14.1: Equilibrium Constant, K eq Objective: (1) To write the equilibrium constant expression for a chemical reaction.
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Reversible Reactions and Equilibrium Reversible Reaction: A chemical reactions in which products re-form the original reactants. Arrows that point in opposite directions are used to indicate a reaction is reversible. Example: H 2 (g) + I 2 (g) 2HI(g) Chemical Equilibrium: A state of balance in which the rate of a forward reaction equals the rate of the reverse reactions and the concentrations of products and reactants remain unchanged.
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Equilibrium Constant, K eq Equilibrium Constant, K eq : a number that relates that concentrations of starting materials and products of a reversible chemical reaction to one another at a given temperature. aA + bB cC + dD concentration coefficient
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Writing an Equilibrium Constant Expression Step 1: Balance the chemical equation. Step 2: Set up your K eq expression with the products on the top of a fraction and the reactants on the bottom of a fraction. Step 3: Raise each substance's concentration to the power equal to the substance’s coefficient in the balanced equation. Note: Solids (s) and pure liquids (l) are not used in the expression because their concentrations do not change.
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Example Write the equilibrium constant expression for the following reaction: CaCO 3 (s) + CO 2 (aq) + H 2 O(l)Ca 2+ (aq) + 2HCO 3 - (aq)
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Practice Write the equilibrium constant expression for the following chemical reactions at equilibrium (don’t forget to balance the equation): 1.) H 2 CO 3 (aq) + H 2 O(l) HCO 3 - (aq) + H 3 O + (aq) 2.) COCl 2 (g) CO(g) + Cl 2 (g) 3.) CO(g) C(s) + CO 2 (g)
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Answers 1.) H 2 CO 3 (aq) + H 2 O(l) HCO 3 - (aq) + H 3 O + (aq) 2.) COCl 2 (g) CO(g) + Cl 2 (g) 3.) 2CO(g) C(s) + CO 2 (g)
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14.1: Equilibrium Constant, K eq Objective: (1) To calculate the equilibrium constant.
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What does the Keq tell us? Keq < 1Favors Reactants Keq = 1Same amount of Reactants and Products Keq > 1Favors Products Practice: Determine if the following K eq values favor the reactants, products, or neither. 1.) K eq = 0.02 2.) K eq = 13.) K eq = 50
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Calculating K eq Step 1: Write the balanced chemical equation. Step 2: Set up your K eq expression. Step 3: Substitute concentrations. Step 4: Calculate!
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Example An aqueous solution of carbonic acid reacts to reach equilibrium as described below: H 2 CO 3 (aq) + H 2 O(l) HCO 3 - (aq) + H 3 O + (aq) The solution contains the following solute concentrations: H 2 CO 3 = 3.3 x 10 -2 M; HCO 3 - = 1.19 x 10 -4 M; H 3 O + = 1.19 x 10 -4 M. Determine the K eq. Note: K eq does not have units!
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Practice 1.a. Calculate the equilibrium constant for the following reaction: COCl 2 (g) CO(g) + Cl 2 (g) [CO] = 0.0178 M [Cl 2 ] = 0.0178 M [COCl 2 ] = 0.00740 M b. Are the reactants for products favored?
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Practice 2.a. For the system involving dinitrogen tetraoxide and nitrogen dioxide at equilibrium at a temperature of 100 ⁰ C, the product concentration of N 2 O 4 is 4.0 x 10 -2 M and the reactant concentration of NO 2 is 1.4x 10 -1 M. What is the K eq value for this reaction? NO 2 (g) N 2 O 4 (g) b. Are the reactants or products favored?
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Practice 3.a. An equilibrium mixture at 852 K is found to contain 3.61 x 10 -3 M of SO 2, 6.11 x 10 -4 M of O 2, and 1.01 x 10 -2 M of SO 3. Calculate the equilibrium constant for the reaction. SO 2 (g) + O 2 (g) SO 3 (g) b. Are the reactants or products favored?
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Calculating Concentrations from K eq 4. K eq for the equilibrium below is 1.8 x 10 -5 at a temperature of 25 ⁰ C. Calculate [NH 3 ] when [NH 4 + ] and [OH - ] are 3.5 x 10 -4 M. NH 3 (aq) + H 2 O(l) NH 4 + (aq) + OH - (aq) [NH 3 ] = 6.8 x10 -3 M
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Practice 5. a. If the equilibrium constant is 1.65 x 10 -3 at 2027 ⁰ C for the reaction below, what is the equilibrium concentration of NO when [N 2 ] = 1.8 x 10 -3 M and [O 2 ] = 4.2 x 10 -3 M. N 2 (g) + O 2 (g) NO(g) b. Are the reactants for products favored?
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Practice 6.a. At 600 ⁰ C, the K eq for the reaction below is 4.32 when [SO 3 ] = 0.260 M and [O 2 ] = 0.045 M. Calculate the equilibrium concentration for sulfur dioxide. SO 2 (g) + O 2 (g) SO 3 (g) b. Are the reactants or products favored?
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14.2 Solubility Product Constant, K sp Objective: (1) To calculate the solubility product constant, K sp.
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Solubility The maximum concentration of a salt in an aqueous solution is called the solubility of the salt in water.
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Solubility Product Constant, K sp Solubility Product Constant, K sp : the equilibrium constant for a solid that is in equilibrium with the solid’s dissolved ions. How much of a partially soluble salt will dissolve? A a B b (s) aA (aq) + bB (aq)
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Calculating K sp The lower the value of K sp, the less soluble the substance. Practice: Rank the following substances from least soluble to most soluble: SaltK sp Ag 2 CO 3 8.4 x 10 -12 BaSO 4 1.1 x 10 -10 Ca 3 (PO 4 ) 2 2.1 x 10 -33 CuS1.3 x 10 -36
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Calculating K sp The lower the value of K sp, the less soluble the substance. Practice: Rank the following substances from least soluble to most soluble: SaltK sp Ag 2 CO 3 8.4 x 10 -12 BaSO 4 1.1 x 10 -10 Ca 3 (PO 4 ) 2 2.1 x 10 -33 CuS1.3 x 10 -36 CuSLeast soluble Ca 3 (PO 4 ) 2 Ag 2 CO 3 BaSO 4 Most soluble
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Calculating K sp Step 1: Write and Balance the equation. Step 2: Determine the concentration of the ions. Step 3: Write the solubility product expression. Step 4: Substitute values and calculate.
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Example Most parts of oceans are nearly saturated with calcium fluoride. A saturated solution of CaF 2 at 25 ⁰ C has a solubility of 3.4 x 10 -4 M. Calculate the solubility product constant for CaF 2. CaF 2 (s) Ca 2+ (aq) + F - (aq)
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Solution 1. Balance equation: CaF 2 (s) Ca 2+ (aq) + 2F - (aq) 2. Determine Concentrations: CaF 2 (s) Ca 2+ (aq) + 2F - (aq) 3.4 x 10 -4 3.4 x 10 -4 6.8 x 10 -4 3. Write solubility product expression: 4. Substitute values and calculate: Note: K sp does not have units!
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Practice 1. Copper(I) bromide is dissolved in water to saturation at 25 ⁰ C. The concentration of Cu + and Br - ions in solution is 7.9 x 10 -5 M. Calculate the K sp for copper(I) bromide at this temperature.
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Practice 2. What is the K sp value for calcium phosphate at 298 K if the concentrations in a solution at equilibrium with excess solid are 3.42 x 10 -7 M for Ca 2+ and 2.28 x 10 -7 M for PO 4 3- ions?
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Practice 3. If a saturated solution of silver chloride contains an AgCl concentration of 1.34 x 10 -5 M, what is the solubility product constant?
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Practice 4. A saturated solution of magnesium fluoride contains a MgCl 2 concentration of 1.19x10 -3 M. What is the K sp for magnesium fluoride?
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Calculating Concentration from K sp 5. What is the concentration of Ca 2+ in a saturated solution of CaF 2 if the concentration of F - is 2.20 x 10 -3 M and K sp = 5.30 x 10 -9.
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Practice 6. What is the concentration of Al 3+ in a saturated solution of Al(OH) 3 if the OH - concentration is 7.90 x 10 -9 M. K sp = 1.30 x 10 -33.
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Practice: Chem 331 7. The K sp for lead(II) iodide is 7.08 x 10 -9 at 25 ⁰ C. What is the molar concentration of PbI 2 in a saturated solution?
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The K sp for lead(II) iodide is 7.08 x 10 -9 at 25 ⁰ C. What is the molar concentration of PbI 2 in a saturated solution? Step 1: Write and Balance Equation PbI 2 (s) Pb 2+ (aq) + 2I - (aq)
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The K sp for lead(II) iodide is 7.08 x 10 -9 at 25 ⁰ C. What is the molar concentration of PbI 2 in a saturated solution? Step 1: Write and Balance Equation PbI 2 (s) Pb 2+ (aq) + 2I - (aq) Step 2: Write the K sp expression
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The K sp for lead(II) iodide is 7.08 x 10 -9 at 25 ⁰ C. What is the molar concentration of PbI 2 in a saturated solution? Step 1: Write and Balance Equation PbI 2 (s) Pb 2+ (aq) + 2I - (aq) Step 2: Write the K sp expression Step 3: Assign x values to concentrations PbI 2 (s) Pb 2+ (aq) + 2I - (aq) x x 2x
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The K sp for lead(II) iodide is 7.08 x 10 -9 at 25 ⁰ C. What is the molar concentration of PbI 2 in a saturated solution? Step 1: Write and Balance Equation PbI 2 (s) Pb 2+ (aq) + 2I - (aq) Step 2: Write the K sp expression Step 3: Assign x values to concentrations PbI 2 (s) Pb 2+ (aq) + 2I - (aq) x x 2x Step 4: Solve x = [PbI 2 ] = 1.21 x 10 -3 M
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Practice 8. The K sp of calcium sulfate is 9.1 x 10 -6. What is the molar concentration of calcium sulfate in a saturated solution?
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Practice 9. The K sp of CdF 2 is 6.4 x 10 -3. What is the molar concentration of cadmium fluoride in a saturated solution?
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14.3 LeChatelier’s Principle Objective: (1) To use LeChatelier’s Principle to determine how a system at equilibrium will respond to an external stress.
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LeChatelier’s Principle LeChatelier’s Principle: When a system at equilibrium is disturbed, the system adjusts in a way to reduce the change. There are 3 possible disturbances: Change in (1) concentration, (2) temperature, or (3) pressure
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1. Change in Concentration Increase concentration of reactant Equilibrium shifts toward products Decrease concentration of reactant Equilibrium shifts toward reactants Increase concentration of product Equilibrium shifts toward reactants Decrease concentration of product Equilibrium shifts toward products
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Use the following reaction to answer the questions below: H 2 (g) + I 2 (g) 2HI (g) In which direction (left or right) does the equilibrium shift in each of the following situations: 1.) Increase H 2 2.) Decrease I 2 3.) Increase HI 4.) Decrease HI
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Use the following reaction to answer the questions below: H 2 (g) + I 2 (g) 2HI (g) In which direction (left or right) does the equilibrium shift in each of the following situations: 1.) Increase H 2 RIGHT 2.) Decrease I 2 LEFT 3.) Increase HILEFT 4.) Decrease HIRIGHT
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2. Change in Temperature Think of heat as a reactant or product Exothermic: heat is a product Endothermic: heat is a reactant For an exothermic reaction: Increasing temperature equilibrium favors reactants Decreasing temperature equilibrium favors products For an endothermic reaction Increasing temperature equilibrium favors products Decreasing temperature equilibrium favors reactants
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Use the following reaction to answer the questions below: 2SO 3 (g) + CO 2 (g) + heat CS 2 (g) + 4O 2 (g) In which direction (left or right) does the equilibrium shift in each of the following situations: 1.) Increase the temperature 2.) Decrease the temperature
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Use the following reaction to answer the questions below: 2SO 3 (g) + CO 2 (g) + heat CS 2 (g) + 4O 2 (g) In which direction (left or right) does the equilibrium shift in each of the following situations: 1.) Increase the temperatureRIGHT 2.) Decrease the temperatureLEFT
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3. Change in Pressure Only affects gases! Increasing pressure Equilibrium shifts toward the side with fewer moles of gas Decreasing pressure Equilibrium shifts toward the side with more moles of gas
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Use the following reaction to answer the questions below: 2SO 3 (g) + CO 2 (g) + heat CS 2 (g) + 4O 2 (g) In which direction (left or right) does the equilibrium shift in each of the following situations: 1.) Increase the pressure 2.) Decrease the pressure
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Use the following reaction to answer the questions below: 2SO 3 (g) + CO 2 (g) + heat CS 2 (g) + 4O 2 (g) In which direction (left or right) does the equilibrium shift in each of the following situations: 1.) Increase the pressureLEFT (3 moles gas) 2.) Decrease the pressureRIGHT (5 moles gas)
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Use the following reaction to answer the questions below: H 2 (g) + I 2 (g) 2HI (g) In which direction (left or right) does the equilibrium shift in each of the following situations: 1.) Increase Pressure 2.) Decrease Pressure
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Use the following reaction to answer the questions below: H 2 (g) + I 2 (g) 2HI (g) In which direction (left or right) does the equilibrium shift in each of the following situations: 1.) Increase Pressure NO CHANGE 2.) Decrease Pressure NO CHANGE
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Practice What direction will the equilibrium shift (left or right) in the reaction: ___POCl 3(g) ___PCl 3(g) + ___O 2 (g) + heat 1.) Add PCl 3 2.) Increase Pressure 3.) Increase Temperature
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Practice What direction will the equilibrium shift (left or right) in the reaction: _2_POCl 3(g) _2_PCl 3(g) + _1_O 2 (g) + heat 1.) Add PCl 3 LEFT 2.) Increase PressureLEFT 3.) Increase Temperature LEFT
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