CHAPTER 14 Chemical Equilibrium

14.1: Equilibrium Constant, K eq  Objective: (1) To write the equilibrium constant expression for a chemical reaction.

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.

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

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.

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)

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)

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)

14.1: Equilibrium Constant, K eq  Objective: (1) To calculate the equilibrium constant.

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 = ) K eq = 13.) K eq = 50

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!

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 M; HCO 3 - = 1.19 x M; H 3 O + = 1.19 x M. Determine the K eq. Note: K eq does not have units!

Practice 1.a. Calculate the equilibrium constant for the following reaction: COCl 2 (g) CO(g) + Cl 2 (g) [CO] = M [Cl 2 ] = M [COCl 2 ] = M b. Are the reactants for products favored?

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 M and the reactant concentration of NO 2 is 1.4x 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?

Practice 3.a. An equilibrium mixture at 852 K is found to contain 3.61 x M of SO 2, 6.11 x M of O 2, and 1.01 x 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?

Calculating Concentrations from K eq 4. K eq for the equilibrium below is 1.8 x at a temperature of 25 ⁰ C. Calculate [NH 3 ] when [NH 4 + ] and [OH - ] are 3.5 x M. NH 3 (aq) + H 2 O(l) NH 4 + (aq) + OH - (aq) [NH 3 ] = 6.8 x10 -3 M

Practice 5. a. If the equilibrium constant is 1.65 x at 2027 ⁰ C for the reaction below, what is the equilibrium concentration of NO when [N 2 ] = 1.8 x M and [O 2 ] = 4.2 x M. N 2 (g) + O 2 (g) NO(g) b. Are the reactants for products favored?

Practice 6.a. At 600 ⁰ C, the K eq for the reaction below is 4.32 when [SO 3 ] = M and [O 2 ] = M. Calculate the equilibrium concentration for sulfur dioxide. SO 2 (g) + O 2 (g) SO 3 (g) b. Are the reactants or products favored?

14.2 Solubility Product Constant, K sp  Objective: (1) To calculate the solubility product constant, K sp.

Solubility  The maximum concentration of a salt in an aqueous solution is called the solubility of the salt in water.

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)

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 x BaSO x Ca 3 (PO 4 ) x CuS1.3 x

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 x BaSO x Ca 3 (PO 4 ) x CuS1.3 x CuSLeast soluble Ca 3 (PO 4 ) 2 Ag 2 CO 3 BaSO 4 Most soluble

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.

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 M. Calculate the solubility product constant for CaF 2. CaF 2 (s) Ca 2+ (aq) + F - (aq)

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 x x Write solubility product expression: 4. Substitute values and calculate: Note: K sp does not have units!

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 M. Calculate the K sp for copper(I) bromide at this temperature.

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 M for Ca 2+ and 2.28 x M for PO 4 3- ions?

Practice 3. If a saturated solution of silver chloride contains an AgCl concentration of 1.34 x M, what is the solubility product constant?

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?

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 M and K sp = 5.30 x

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 M. K sp = 1.30 x

Practice: Chem The K sp for lead(II) iodide is 7.08 x at 25 ⁰ C. What is the molar concentration of PbI 2 in a saturated solution?

The K sp for lead(II) iodide is 7.08 x 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)

The K sp for lead(II) iodide is 7.08 x 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

The K sp for lead(II) iodide is 7.08 x 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

The K sp for lead(II) iodide is 7.08 x 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 M

Practice 8. The K sp of calcium sulfate is 9.1 x What is the molar concentration of calcium sulfate in a saturated solution?

Practice 9. The K sp of CdF 2 is 6.4 x What is the molar concentration of cadmium fluoride in a saturated solution?

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.

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

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

 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

 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

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

 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

 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

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

 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

 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)

 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

 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

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

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