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314 Chapter 15 Chemical Equilibrium M = + dRT + = 0.750 atm = 34.71 g/mol = + PNOMNO + (0.750)(34.7) = (0.750 3x)(46.0) + 2x(30.0) + x(32.0) x = 0.184 Kₚ = = (0.750 3x) = (0.750 3(0.184))² = 0.6356 = 0.6356 Check: The units (none) are correct. The magnitude (0.01) is reasonable since there are higher concentrations of the products than the reactants. 15.99 Given: = 3.0 atm, mole fraction O₂ = 0.12, T = 600 K Find: Kₚ Conceptual Plan: mole fraction Kₚ mole fraction = Solution: = (0.12)(3.0) = 0.36 atm = = 2(0.36 = 0.72 atm = 3.0 0.36 0.72 = = = 0.0506 = 5.1 X 10⁻² Check: The units (none) are correct. The magnitude (0.05) is reasonable since there are higher concentrations of the products than the reactants. Conceptual Problems 15.101 Yes, the direction will depend on the volume. If the initial moles of A and B are equal, the initial concentrations of A and B are equal regardless of the volume. Because Kc = [B]² [A] = 1, if the volume is such that the [A] = [B] 1.0, then Q > K and the reaction goes to the left to reach equilibrium. 15.103 An examination of the data shows that when = 1.0, = 1.0; therefore, Kₚ = = (1.0)b = 1.0. Therefore, the value of the numerator and denominator must be equal. We see from the data that = so = PA. Because the stoichiometric coefficients become exponents in the equilibrium expression, a = 1 and b = 2. 15.105 Looking at the plot, at equilibrium [A] = 0.59 M and [B] = 0.78 M. Since A(g) 11 2 B(g), K = [A] = 0.59 = 1.0. This value is reasonable, since the concentrations of the reactant and product are similar. Questions for Group Work 15.107 (a) (b) Adding H₂ will increase the denominator and not change the numerator. Copyright © 2017 Pearson Education, Inc.