Chem Equilibrium Review Key

Name: Hour: Date: CHEMICAL EQUILIBRIUM REVIEW 1. Given the fact that the concentrations of reactants and products are not changing, why is the word dynamic used to describe chemical equilibrium? 1   1. quizlet.com quizlet.com Reactants and products are continuously being formed (dynamic) but at a constant rate (equilibrium) 2. Why does a large Keq value mean that the products are favored in an equilibrium system? Keq = [products] / [reactants] large numerator (products) = large Keq 3. Does the following equation represent a homogeneous equilibrium or heterogeneous equilibrium? Explain your answer. H₂O(s) + H₂O(l) More than one state of matter 4. Write equilibrium constant expressions for these homogeneous equilibria. a. 2N₂H₄(g) + 2NO₂(g) ⇌ 3N₂(g) + 4H₂O(g) Keq = [N₂]³[H₂O]⁴ / [N₂H₄]²[NO₂]² b. 2NbCl₄(g) ⇌ NbCl₃(g) + NbCl₅(g) Keq = [NbCl₃][NbCl₅] / [NbCl₄]² c. Write the chemical equation for Keq = [CH₄][H₂O] / [CO][H₂]³ CO + 3H₂ ⇌ CH₄ + H₂O 5. Write equilibrium constant expressions for these heterogeneous equilibria. a. CaCO₃(s) ⇌ CaO(s) + CO₂(g) Keq = [CO₂] b. C₆H₆(l) ⇌ C₆H₆(g) Keq = [C₆H₆] c. 2Fe(s) + 3H₂O(g) ⇌ Fe₂O₃(s) + 3H₂(g) Keq = [H₂]³ / [H₂O]³ 6. Use this equation to answer #6-7. H₂(g) + I₂(g) ⇌ 2HI(g) A mixture of H₂ and I₂ is allowed to react at 448°C. When equilibrium is established, the concentrations of the substances are found to be [H₂] = 0.46 M, [I₂] = 0.39 M, and [HI] = 3.0 M. Calculate the value of Keq at 448°C: Keq = [HI]² / [H₂][I₂] = (3.0)² / (0.46)(0.39) = 50.2 7. The equilibrium constant (Keq) for the reaction below is 50. What is the concentration of hydrogen gas if [I₂] = 1 M, and [HI] = 2.5 M? Keq = [HI]² / [H₂][I₂] 50 = (2.5)² / x(1) 50x = 6.25 x = 0.125 [H₂] = 0.125 M 8. What is meant by a stress on a reaction at equilibrium? A change that shifts to one side of the reaction (concentration, volume/pressure, temperature) 9. Why does removing a reactant cause an equilibrium shift to the left? The system attempts to regain equilibrium by replacing the reactant 10. Why does adding a product cause equilibrium to shift to the left? The system attempts to regain equilibrium by consuming the added product 11. When equilibrium shifts to the right, what happens to each of the following? a. The concentration of the reactants: [reactants] decrease b. The concentration of the products: [products] increase 12. How would each of the following changes affect the equilibrium position of the system used to produce methanol from carbon monoxide and hydrogen? CO(g) + 2H₂(g) ⇌ CH₃OH(g) + heat a. Adding CO to the system: ← b. Adding a catalyst to the system: no change c. Removing CH₃OH from the system: → d. Increasing the volume of the system: ← 13. How would these equilibria be affected by decreasing the temperature? a. 2O₃(g) ⇌ 3O₂(g) + heat → to replace heat b. heat + H₂(g) + F₂(g) ⇌ 2HF ← 14. Phosphorus pentachloride decomposes to form phosphorus trichloride according to the following equation: PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) + heat How would you accomplish each of the following? a. Increase the amount of PCl₃: Add PCl₅, Remove Cl₂, remove heat, ↑ volume (↓ pressure) b. Increase the concentration of PCl₅: Add PCl₅, Add Cl₂, add heat, ↓ volume (↑ pressure) 15. You place 0.750 mol of nitrogen, and 2.250 mol of hydrogen into a reaction vessel at 450°C. What is the composition of the equilibrium mixture if you obtain 0.060 mol of ammonia (NH₃)? (Practice 1) N₂(g) + 3H₂(g) ⇌ 2NH₃(g) Start 0.750 2.250 0 Change -x -3x +2x Eq 0.750-x 2.250-3x 2x = 0.060 x = 0.030 N₂ = 0.750 - 0.030 = 0.72 mol H₂ = 2.250 - 3(0.030) = 2.16 mol NH₃ = 0.060 mol 16. Obtain the value of Keq at 500K for the reaction below if 0.1 M CO and 0.2 M H₂ are placed in a reaction vessel at 500K. When this mixture came to equilibrium, the vessel contained 0.08 M CO. (Practice 2) CO(g) + 2H₂(g) ⇌ CH₃OH(g) Start 0.1 0.2 0 Change -x -2x +x Eq 0.1-x 0.2-2x x Keq = [CH₃OH] / [CO][H₂]² = (0.02) / (0.08)(0.16)² 17. The following reaction has a Keq equal to 3.07 x 10^-4 at 24°C. For each of the compositions below, decide whether or not the reaction mixture is at equilibrium. If not, decide which direction the reaction should go. (Practice 1 3) 2NOBr(g) ⇌ 2NO(g) + Br₂(g) a. [NOBr] = 0.0720 M, [NO] = 0.0162 M, [Br₂] = 0.0123 M Q = [NO]²[Br₂] / [NOBr]² = 6.23 x 10^-4 K < Q, shift to the left (reactants) b. [NOBr] = 0.121 M, [NO] = 0.0159 M, [Br₂] = 0.0139 M Q = [NO]²[Br₂] / [NOBr]² = 2.4 x 10^-4 K > Q, shift to the right (products) 18. Write the solubility product expressions for the a. Ag₂S ⇌ 2Ag⁺ + S²⁻ Ksp = [Ag⁺]²[S²⁻] b. SrSO₄ ⇌ Sr²⁺ + SO₄²⁻ Ksp = [Sr²⁺][SO₄²⁻] 19. Review the value of K and the solubility of a substance (Table A-10 in notes) 20. Calculate the solubility of the following salts considering the solubility product constant. (Practice 5) a. PbI₂, Ksp = 1.39 x 10^-8 PbI₂ ⇌ Pb²⁺ + 2I⁻ x 2x Ksp = [Pb²⁺][I⁻]² = 1.39 x 10^-8 (x)(2x)² = 1.39 x 10^-8 4x³ = 1.39 x 10^-8 x³ = 3.48 x 10^-9 x = 0.0015 M b. BaCO₃, Ksp = 8.10 x 10^-9 BaCO₃ ⇌ Ba²⁺ + CO₃²⁻ x x Ksp = [Ba²⁺][CO₃²⁻] = 8.10 x 10^-8 (x)(x) = 8.10 x 10^-8 x² = 8.10 x 10^-8 x = 2.85 x 10^-4 M 21. Calculate Ksp for the following compounds from their solubility. (Practice 6) a. CuCO₃, solubility = 1.6 x 10^-5 M CuCO₃ ⇌ Cu²⁺ + CO₃²⁻ x x Ksp = [Cu²⁺][CO₃²⁻] = (x)(x) = (1.6 x 10^-5)² = 2.56 x 10^-10 b. Mg(OH)₂, solubility = 1.1 x 10^-4 M Mg(OH)₂ ⇌ Mg²⁺ + 2OH⁻ x 2x Ksp = [Mg²⁺][OH⁻]² = (x)(2x)² = 4x³ = 4(1.1 x 10^-4)³ = 5.32 x 10^-12 22. What is the molar solubility of strontium sulfate (SrSO₄) in 0.20 M sodium sulfate (Na₂SO₄)? The Ksp of SrSO₄ is 2.5 x 10^-7. (Practice 7) SrSO₄ ⇌ Sr²⁺ + SO₄²⁻ x 0.20 + x Ksp = [Sr²⁺][SO₄²⁻] = 2.5 x 10^-7 (x)(0.20 + x) = 2.5 x 10^-7 0.20x = 2.5 x 10^-7 x = 1.25 x 10^-6 M