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22 Chemistry of the Nonmetals Solutions to Exercises 22.92 (a) First calculate the molar solubility of Cl₂ in water. 1 atm (0.310 L) n = 0.08206 = 0.01384 = 0.0138 mol Cl₂ atm K M = 0.01384 0.100 mol L = [CI⁻] = [HOCI] = [H⁺] Let this quantity = X. Then, (0.1384-x) = 4.7 X 10⁻⁴ Assuming that is small compared with 0.1384: = (0.1384)(4.7 = 6.504 X = 0.0402 = 0.040 M We can correct the denominator using this value, to get a better estimate of = 4.7 10⁻⁴; = 0.0359 = 0.036 M 0.1384-0.0402 One more round of approximation gives = 0.0364 = 0.036 M. This is the equilibrium concentration of HCIO. (b) From the equilibrium reaction in part (a), [H⁺] = 0.036 M. pH = = 1.4 The HOCI produced by this equilibrium will ionize slightly to produce additional H⁺(aq). However, the value for HOCI is small, 3.0 10⁻⁸, and the acid ionization will be suppressed by the presence of H⁺(aq) from the solubility equilibrium. [H⁺] from ionization of HOCI will be small compared to 0.036 M and will not significantly impact the pH. 22.93 (a) 2NH₄ClO₄(s) (b) = Σ prod - Σ react = + 3/2 + 1/2 N₂ + 5/4 - = = (c) The aluminum reacts exothermically with O₂(g) and HCl(g) produced in the decomposition, providing additional heat and thrust. (d) There are (1/2 + + 3/2 + 5/4) = 4.25 mol gas per mol decomposed 1lb 453.6 1 lb 117.49 1 mol g 1 mol 4.25 NH₄CIO₄ mol gas = 16.408 = 16.4 mol gas V = nRT P = 16.408 mol gasx mol K 273 1 atm = 367.57 = 368 L 22.94 (a) (b) = = + 2(-285.83) 95.40 = kJ 687