Example 8-1 Maximizing the Selectivity for the Famous Trambouze Reactions Reactant A decomposes by three simultaneous reactions to form three products, one that is desired, B, and two that are undesired, X and Y. These gas-phase reactions, along with the appropriate rate laws, are called the Trambouze reactions (AIChE J., 5,384). k 1) A W→X -1A==k₁ = 0.0001- mol dm³-s (zero order) k₂ 2) WB -12A=7B= k₂CA= (0.0015s CA (first order) 3) A WY dm³ 2 -13A=7y=k³CA= -0.008- mol-s CA (second order) The specific reaction rates are given at 300 K and the activation energies for reactions (1), (2), and (3) are E₁ = 10,000 kcal/mole, Ez = 15,000 kcal/mole, and E3 = 20,000 kcal/mole. (a) How, and under what conditions (e.g., reactor type(s), temperature, concentrations), should the reaction be carried out to maximize the selectivity of species B for an entering concentration of species A of 0.4 M and a volumetric flow rate of 2.0 dm³/s? (b) How could the conversion of B be increased and still keep selectivity relatively high? (c) Extract the example drawings using Polymath Professional??