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310 Chapter 15 Chemical Equilibrium Determine grams 0.0922 L X 200.0 X X 17.03 = 313.8 g = 3.1 X 10² g Determine the theoretical yield and the limiting reactant: 1.27 X 1000 kg X 28.02 1 X 2 1 X 17.03 NH₃ = 1544 g NH₃ 0.310 X 1000 kg X 2.016 1 X 2 3 X 17.03 g = 1745 g N₂ produces the least amount of therefore, it is the limiting reactant, and the theoretical yield is 1.54 X 10³ g % yield = 1.54 3.1 X X 10² 10³ X 100% = 20.% 15.85 Given: at equilibrium: Pco = 0.30 atm; = 0.10 atm; = 0.60 atm, add 0.40 atm Cl₂ Find: when system returns to equilibrium Conceptual Plan: Use equilibrium partial pressures to determine Kₚ. For the new conditions, prepare an ICE table, represent the change with x, sum the table, determine the equilibrium values, put the equilibrium values in the equilibrium expression, and solve for x. Determine Solution: CO(g) + 1L Condition 1: Pco 0.30 0.10 0.60 Kₚ = = (0.30)(0.10) (0.60) = 20 CO(g) + 1L Condition 2: Pco Initial 0.30 0.10 + 0.40 0.60 Change -x -x +x Equil 0.30 x 0.50 - x Reaction shifts to the right because the concentration of Cl₂ was increased. = = (0.30 x) = 20 (0.60 + x) 20x² 17x + 2.4 = 0 4ac -(-17) ± 4(20)(2.4) = 2a 2(20) x = 0.67 or 0.18 so x = 0.18 = 0.30 - 0.18 = 0.12 atm; = 0.50 0.18 = 0.32; = 0.60 + 0.18 = 0.78 atm Check: Plug the values into the equilibrium expression: = (0.12)(0.32) (0.78) = 20.3 = 20.; this is the same as the original equilibrium constant. Copyright © 2017 Pearson Education, Inc.