EXAMPLE 15.1 Writing Equilibrium constant Expressions for Chemical Reactions Write an equilibrium expression for the chemical equation. CO(g) + 2 H2(g) CH3OH(g) SOLUTION The equilibrium expression is the concentration of the products raised to their stoichiometric coefficients divided by the concentration of the reactants raised to their stoichiometric coefficients. Notice that the expression is a ratio of products over reactants. Notice also that the coefficients in the chemical equation are the exponents in the equilibrium expression. SKILLBUILDER 15.1 | Writing Equilibrium Expressions for Chemical Reactions Write an equilibrium expression for the chemical equation. H2(g) + F2(g) 2 HF(g) Answer: FOR MORE PRACTICE ​Example 15.10; Problems 43, 44.

4 HCl(g) + O2(g) 2 H2O(l) + 2 Cl2(g) EXAMPLE 15.2 Writing Equilibrium Expressions for Reactions Involving a Solid or a Liquid Write an equilibrium expression for the chemical equation. CaCO3(s) CaO(s) + CO2(g) SOLUTION Since CaCO3(s) and CaO(s) are both solids, they are omitted from the equilibrium expression. Keq = [CO2] SKILLBUILDER 15.2 | Writing Equilibrium Expressions for Reactions Involving a Solid or a Liquid Write an equilibrium expression for the chemical equation. 4 HCl(g) + O2(g) 2 H2O(l) + 2 Cl2(g) Answer: FOR MORE PRACTICE ​Problems 45, 46.

EXAMPLE 15.3 Calculating Equilibrium Constants Consider the reaction: 2 CH4(g) C2H2(g) + 3 H2(g) A mixture of CH4, C2H2, and H2 is allowed to come to equilibrium at 1700 °C. The measured equilibrium concentrations are [CH4] = 0.0203 M, [C2H2] = 0.0451 M, and [H2] = 0.112 M. What is the value of the equilibrium constant at this temperature? You are given the concentrations of the reactants and products of a reaction at equilibrium. You are asked to find the equilibrium constant. Write the expression for Keq from the balanced equation. To calculate the value of Keq, substitute the correct equilibrium concentrations into the expression for Keq. GIVEN: [CH4] = 0.0203 M [C2H2] = 0.0451 M [H2] = 0.112 M FIND: Keq SOLUTION

EXAMPLE 15.3 Calculating Equilibrium Constants Continued SKILLBUILDER 15.3 | Calculating Equilibrium Constants Consider the reaction: CO(g) + 2 H2(g) CH3OH(g) A mixture of CO, H2, and CH3OH is allowed to come to equilibrium at 225 °C. The measured equilibrium concentrations are [CO] = 0.489 M, [H2] = 0.146 M, and [CH3OH] = 0.151 M. What is the value of the equilibrium constant at this temperature? Answer: Keq = 14.5 SKILLBUILDER PLUS Suppose that the preceding reaction is carried out at a different temperature and that the initial concentrations of the reactants are [CO] = 0.500 M and [H2] = 1.00 M. Assuming there is no product at the beginning of the reaction and at equilibrium [CO] = 0.15 M, find the equilibrium constant at this new temperature. Hint: Use the stoichiometric relationships from the balanced equation to find the equilibrium concentrations of H2 and CH3OH. Answer: Keq = 26 FOR MORE PRACTICE ​Example 15.11; Problems 51, 52, 53, 54, 55, 56.

H2(g) + I2(g) 2 HI(g) Keq = 69 at 340 °C EXAMPLE 15.4 Using Equilibrium Constants in Calculations Consider the reaction: H2(g) + I2(g) 2 HI(g) Keq = 69 at 340 °C In an equilibrium mixture, the concentrations of H2 and I2 are both 0.020 M. What is the equilibrium concentration of HI? SORT You are given the equilibrium concentrations of the reactants in a chemical reaction and also the value of the equilibrium constant. You are asked to find the concentration of the product. STRATEGIZE Draw a solution map showing how the equilibrium constant expression gives the relationship between the given concentrations and the concentration you are asked to find. GIVEN: [H2] = [I2] = 0.020 M Keq = 69 FIND: [HI] SOLUTION MAP

EXAMPLE 15.4 Using Equilibrium Constants in Calculations Continued SOLVE Solve the equilibrium expression for [HI] and substitute in the appropriate values to calculate it. Since the value of [HI] is squared, you must take the square root of both sides of the equation to solve for [HI] because CHECK You can check your answer by substituting it back into the expression for Keq. SOLUTION The calculated value of Keq is about equal to the given value of Keq (which was 69), indicating that your answer is correct. The slight difference is due to rounding error, which is common in problems like these.

EXAMPLE 15.4 Using Equilibrium Constants in Calculations Continued SKILLBUILDER 15.4 | Using Equilibrium Constants in Calculations Diatomic iodine (I2) decomposes at high temperature to form I atoms according to the reaction: I2(g) 2 I(g) Keq = 0.011 at 1200 °C In an equilibrium mixture, the concentration of I2 is 0.10 M. What is the equilibrium concentration of I? Answer: 0.033 M FOR MORE PRACTICE ​Example 15.12; Problems 57, 58, 59, 60.

EXAMPLE 15.5 The Effect of a Concentration Change on Equilibrium Consider the following reaction at equilibrium: CaCO3(s) CaO(s) + CO2(g) What is the effect of adding more CO2 to the reaction mixture? What is the effect of adding more CaCO3? SOLUTION Adding more CO2 increases the concentration of CO2 and causes the reaction to shift to the left. Adding more CaCO3 does not increase the concentration of CaCO3 because CaCO3 is a solid and thus has a constant concentration. It is therefore not included in the equilibrium expression and has no effect on the position of the equilibrium. Skillbuilder 15.5 | The Effect of a Concentration Change on Equilibrium Consider the following reaction in chemical equilibrium: 2 BrNO(g) 2 NO(g) + Br2(g) What is the effect of adding more Br2 to the reaction mixture? What is the effect of adding more BrNO? Answer: Adding Br2 causes a shift to the left; adding BrNO causes a shift to the right.

EXAMPLE 15.5 The Effect of a Concentration Change on Equilibrium Continued SKILLBUILDER PLUS What is the effect of removing some Br2 from the preceding reaction mixture? Answer: Removing Br2 causes a shift to the right. FOR MORE PRACTICE ​Example 15.13a, b; Problems 63, 64, 65, 66.

EXAMPLE 15.6 The Effect of a Volume Change on Equilibrium Consider the reaction at chemical equilibrium: 2 KClO3(s) 2 KCl(s) + 3 O2(g) What is the effect of decreasing the volume of the reaction mixture? Increasing the volume of the reaction mixture? SOLUTION The chemical equation has 3 mol of gas on the right and 0 mol of gas on the left. Decreasing the volume of the reaction mixture increases the pressure and causes the reaction to shift to the left (toward the side with fewer moles of gas particles). Increasing the volume of the reaction mixture decreases the pressure and causes the reaction to shift to the right (toward the side with more moles of gas particles). Skillbuilder 15.6 | The Effect of a Volume Change on Equilibrium Consider the reaction at chemical equilibrium: 2 SO2(g) + O2(g) 2 SO3(g) What is the effect of decreasing the volume of the reaction mixture? Increasing the volume of the reaction mixture? Answer: Decreasing volume causes a shift to the right; increasing volume causes a shift to the left.

EXAMPLE 15.6 The Effect of a Volume Change on Equilibrium Continued FOR MORE PRACTICE ​Example 15.13a, b; Problems 63, 64, 65, 66.

Heat + CaCO3(s) CaO(s) + CO2(g) EXAMPLE 15.7 The Effect of a Temperature Change on Equilibrium The following reaction is endothermic: CaCO3(s) CaO(s) + CO2(g) What is the effect of increasing the temperature of the reaction mixture? Decreasing the temperature? SOLUTION Because the reaction is endothermic, we can think of heat as a reactant. Heat + CaCO3(s) CaO(s) + CO2(g) Raising the temperature is adding heat, causing the reaction to shift to the right. Lowering the temperature is removing heat, causing the reaction to shift to the left. Skillbuilder 15.7 | The Effect of a Temperature Change on Equilibrium The following reaction is exothermic. 2 SO2(g) + O2(g) 2 SO3(g) What is the effect of increasing the temperature of the reaction mixture? Decreasing the temperature?

EXAMPLE 15.7 The Effect of a Temperature Change on Equilibrium Continued Answer: Increasing the temperature shifts the reaction to the left; decreasing the temperature shifts the reaction to the right. FOR MORE PRACTICE ​Example 15.13d; Problems 71, 72, 73, 74.

EXAMPLE 15.8 Writing Expressions for Ksp Write expressions for Ksp for each ionic compound. (a) BaSO4 (b) Mn(OH)2 (c) Ag2CrO4 SOLUTION To write the expression for Ksp, first write the chemical reaction showing the solid compound in equilibrium with its dissolved aqueous ions. Then write the equilibrium expression based on this equation. (a) BaSO4(s) Ba2+(aq) + SO42– (aq) Ksp = [Ba2+][SO42–] (b) Mn(OH)2(s) Mn2+(aq) + 2 OH– (aq) Ksp = [Mn2+][OH–]2 (c) Ag2CrO4(s) 2 Ag+(aq) + CrO42– (aq) Ksp = [Ag+]2[CrO42–]

EXAMPLE 15.8 Writing Expressions for Ksp Continued SKILLBUILDER 15.8 | Writing Expressions for Ksp Write expressions for Ksp for each ionic compound. (a) AgI (b) Ca(OH)2 Answers: (a) Ksp = [Ag+][I–] (b) Ksp = [Ca2+][OH–]2 FOR MORE PRACTICE ​Example 15.14; Problems 77, 78.

EXAMPLE 15.9 Writing Expressions for Ksp Calculate the molar solubility of BaSO4. SOLUTION Begin by writing the reaction by which solid BaSO4 dissolves into its constituent aqueous ions. Next, write the expression for Ksp. Define the molar solubility (S) as [Ba2+] or [SO42–] at equilibrium. Substitute S into the equilibrium expression and solve for it. BaSO4(s) Ba2+(aq) + SO42–(aq) Ksp = [Ba2+][SO42–] S = [Ba2+] = [SO42–] = S × S = S2 Therefore

EXAMPLE 15.9 Writing Expressions for Ksp Continued Finally, look up the value of Ksp in Table 15.2 and calculate S. The molar solubility of BaSO4 is 1.03 × 10–5 mol/L.

EXAMPLE 15.9 Writing Expressions for Ksp Continued SKILLBUILDER 15.9 | Calculating Molar Solubility from Ksp Calculate the molar solubility of CaSO4. Answer: 8.43 × 10–3 M FOR MORE PRACTICE ​Example 15.15; Problems 85, 86, 87, 88.