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Chapter 15 Chemical Equilibrium 309 (0.621 (0.101 + x)(0.621 + = X 10⁻⁴ (0.101 (0.621 - + 2x) x) = 2.182 X 10⁻² x = 0.04325 = 0.0433 PH₂ = = 0.621 + x = 0.621 + 0.0433 = 0.664 atm PHI = 0.101 - 2x = 0.101 2(0.0433) = 0.0144 atm Check: Plug the values into the equilibrium expression: Kₚ = = 4.703 X 10⁻⁴ = 4.70 X 10⁻⁴; this value is close to the original equilibrium constant. 15.83 Given: 200.0 L container; 1.27 kg N₂; 0.310 kg H₂; 725 K; Kₚ = 5.3 X 10⁻⁵ Find: mass in g of NH₃ and % yield Conceptual Plan: Kₚ Kc and then kg g mol M and then prepare an ICE table. Represent = g mol kg molar mass vol the change with x, sum the table, determine the equilibrium values, put the equilibrium values in the equilibrium expression, and solve for x. Determine Then M mol g and then determine M vol mol X molar mass theoretical yield % yield. actual yield determine limiting reactant % yield = X 100% theoretical yield Solution: = = 0.1876 = 1.27 1000 X 1 mol = 45.325 mol N₂ [N₂] = 45.325 mol = 0.22663 M nH₂ = X 1 mol H₂ = 153.77 mol H₂ [H₂] = 153.77 = 0.76885 M N₂(g) + Reaction 1: [N₂] [H₂] [NH₃] Initial 0.2266 0.7689 0.0 Change -x -3x +2x Equil 0.2266 x 0.7689 - 3x 2x Reaction shifts to the right. = [NH₃]² = (0.2266 x)(0.7689 = 0.1876 Assume that x is small compared to 0.2268 and 3x is small compared to 0.7689. = 0.1877 x = 0.06951 Check assumptions: 0.06951 0.2268 X 100% = 30.6, which is not valid, and 3(0.06951) X 100% = 27.1%. 0.7689 Use method of successive substitution to solve for x. This yields x = 0.0460. [NH₃] = 2x = 2(0.0460) = 0.0920 M Check: Plug the values into the equilibrium expression: = (0.2266 3(0.0460))³ = 0.1866; this value is close to the original equilibrium constant. Copyright © 2017 Pearson Education, Inc.