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18 Chemistry of the Environment Solutions to Exercises 18.89 (a) 17 8.5 pairs Owing to its lower electronegativity, N is more likely to be electron deficient and to accommodate the odd electron. (b) The fact that NO₂ is an electron deficient molecule indicates that it will be highly reactive. Dimerization results in formation of a N-N single bond which completes the octet of both N atoms. NO₂ and exist in equilibrium in a closed system. The reaction is exothermic, Equation [22.48]. In an urban environment, NO₂ is produced from hot automobile combustion. At these temperatures, equilibrium favors the monomer because the reaction is exothermic. (c) + 4CO(g) NO₂ + CO(g) NO₂ is an oxidizing agent and CO is a reducing agent, so we expect products to contain N in a more reduced form, NO or and C in a more oxidized form, (d) No. Because it is an odd-electron molecule, NO₂ is very reactive. We expect it to undergo chemical reactions or photodissociate before it can migrate to the stratosphere. The expected half-life of an NO₂ molecule is short. 18.90 (a) Holding one reactant concentration constant and changing the other, evaluate the effect this has on the initial rate. Use these observations to write the rate law. Compare Experiments 1 and 3. [O₃] is constant, [H] doubles, initial rate doubles. The reaction is first order in [H]. Compare Experiments 2 and 1. [H] is constant, [O₃] doubles, initial rate doubles. The reaction is first order in [O₃]. rate = k[O₃][H] (b) Calculate a value for the rate constant for each experiment, then average them to obtain a single representative value. rate = k[O₃][H]; k = k₁ = M/s = 1.1293 = k₂ = 9.44 = 1.1215 = k₃ = = 1.1245 = = (1.1293 + 1.1215 + 1.1245 1044)/3 = 1.1251 = 1.13 M⁻¹ 571