Question 1 (2 marks) The following consecutive first order reaction in series were conducted in a batch reactor: Via Mathematica computation show the graphically relationship between the concentration...

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Question 1 (2 marks) The following consecutive first order reaction in series were conducted in a batch reactor: Via Mathematica computation show the graphically relationship between the concentration of each species with time (show the concentration variation of all 4 species in one graph) Also show graphically the effect of i) temperature and ii) activation energy (Copy and paste your mathematica plots on your submission) Question 2 (2 marks) The following 1st order reaction is carried in a number of adiabatic CSTR reactors with volume of 1, 2, 5 and 10 litres: [mol/lit/s] The concentration reactant A in solvent S is 5 mol.l-1. The solution is flowing into the reactor at a rate of 10 ml per second. The molar flow rate of solvent S is 10 mol per min and the molar flow rate of A is 3 mol. per min. The product R is susceptible to degradation at temperature greater than 120 oC. The reactant enters the reactor at 40oC. Determine the optimum reactor to use in terms of maximum throughput of pristine product R. Physical Data: Molar heat capacity of A = 20.0 J.mol-1.K-1 Molar heat capacity of R = 25.0 J.mol-1.K-1 Molar heat capacity of solvent S: 35.0 J.mol-1.K-1 Heat of formation of A at TR (25oC): -80 kJ.mol-1. Heat of formation of R at TR : -145 kJ.mol-1 Use Mathematica to solve this problem. Question 3 (2 marks) A CSTR of volume V first filled with water is to conduct a first order reaction involving reactant A: . Solution of reactant A of concentration CAO, is fed into the reactor a at molar feed rate FAO , set up the governing equation via material balance, for this transient phase assuming a constant temperature for the reaction in the reactor. Use Mathematica to solve this differential equation and obtain an analytical and graphical relationship between CA and time, t, for this transient process. A RT A C e r / 100000 12 10 - = - R A k ¾ ® ¾ T S R A k k k ¾ ® ¾ ¾ ® ¾ ¾ ® ¾ 3 2 1 R A A k ¾ ® ¾