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306 Chapter 15 Chemical Equilibrium 15.71 Given: C(s) + 2H₂(g) is exothermic. Find: Determine which will favor (a) Adding more C to the reaction mixture does not favor Adding C does not cause any change in the reac- tion. C is a solid, and the concentration remains constant; so the addition of more solid does not change the equilibrium concentration. (b) Adding more H₂ to the reaction mixture favors Adding H₂ increases the concentration of causing the reaction to shift to the right. (c) Raising the temperature of the reaction mixture does not favor CH₄. Because the reaction is exothermic, we can think of heat as a product. Raising the temperature is equivalent to adding a product, causing the reaction to shift to the left. (d) Lowering the volume of the reaction mixture favors CH₄. The chemical equation has 1 mole of gas on the right and 2 moles of gas on the left. Decreasing the volume of the reaction mixture increases the pressure and causes the reaction to shift to the right (toward the side with fewer moles of gas particles). (e) Adding a catalyst to the reaction mixture does not favor A catalyst added to the reaction mixture only speeds up the reaction; it does not change the equilibrium concentration. (f) Adding neon gas to the reaction mixture does not favor CH₄. Adding an inert gas to a reaction mixture at a fixed volume has no effect on the equilibrium. Cumulative Problems 15.73 (a) To find the value of K for the new equation, combine the two given equations to yield the new equation. Reverse equation 1 and use then add to equation 2. To find K for equation 3, use (1/K₁)(K₂). HbO₂(aq) + O₂(aq) K₁ = 1/1.8 + HbCO(aq) K₂ = 306 + CO(aq) HbCO(aq) + O₂(aq) K₃ = K₁K₂ = (1/1.8)(306) = 170 (b) Given: O₂ = 20%, CO = 0.10% Find: the ratio [HbCO] [HbO₂] [HbCO] Conceptual Plan: Determine the equilibrium expression and then determine Solution: then [HbCO] [HbO₂] = = 0.85 1.0 Because the ratio is almost 1:1, 0.10% CO will replace about 50% of the O₂ in the blood. The CO blocks the uptake of O₂ by the blood and is therefore highly toxic. 15.75 (a) Given: 4.45 g 10.0 1200 K, 2.00 g C, Kₚ = 5.78 Find: total pressure Conceptual Plan: g CO₂ mol and g C mol C and then determine limiting reactant 1 mol CO₂ 1 mol C 44.01 g 12.01 g C and then mol P Prepare an ICE table, represent the change with x, sum the table, PV = nRT determine the equilibrium values, put the equilibrium values in the equilibrium expression, and solve for x. Solution: 2CO(g) 4.45 X 1 mol = 0.1011 mol 2.00 1 mol = 0.1665 mol C Because the stoichiometry is 1:1, the CO₂ is the limiting reactant. = = 0.9956 atm + C(s) 2CO(g) Copyright © 2017 Pearson Education, Inc.