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15 Chemical Equilibrium Solutions to Exercises 15.52 [Br₂] = 0.25 mol/3.0 L = 0.08333 = 0.083 M; [Cl₂] = 0.55 mol/3.0 L = 0.1833 = 0.18 M Br₂(g) + Cl₂(g) 11 2BrCl(g) initial 0.083 M 0.18 M 0 change -x -x +2x equil. (0.083 x) (0.18 +2x ; = 7.0(0.0153 0.2666x + x²); 0 = 0.1069 1.8662x + = = 0.06387 = 0.064 M 2(3) (The 0.56 M quadratic solution is not chemically meaningful.) [BrCl] = 2x = 0.1277 = 0.13 M; [Br₂] = 0.08333 0.06387 = 0.01946 = 0.019 M [Cl₂] = 0.1833 0.06387 = 0.1195 = 0.12 M Check. Kc = = 7.0 15.53 Analyze/Plan. Write the expression, substitute the stated pressure relationship, and solve for Solve. When = these terms cancel and = = 0.416 atm. This is true for all cases where = 15.54 Kc = [NH₃][H₂S] = 1.2 10⁻⁴ Because of the stoichiometry, equilibrium concentrations of H₂S and will be equal; call this quantity y. Then, = 1.2 15.55 (a) 11 At equilibrium, [Ca²⁺] = = (b) A saturated solution of CaSO₄ (aq) is 4.9 10⁻³ M. 1.4 L of this solution contain: A bit more than 1.0 g CaSO₄ is needed in order to have some undissolved CaSO₄ (s) in equilibrium with 1.4 L of saturated solution. 15.56 (a) Analyze/Plan. If only PH₃BCl₃(s) is present initially, the equation requires that the equilibrium concentrations of PH₃(g) and BCl₃(g) are equal. Write the Kc expres- sion and solve for = [PH₃] = Solve. Kc = 1.87 = = 0.043243 = 0.0432 M PH₃ and BCl₃ 447