Question

c) Initially, a tank is filled with 130 kmol of an aqueous sulfuric acid solution (H2SO4) which contains 5.0 mol% of H2SO4. The solution in the tank is concentrated by adding pure sulfuric acid at a rate of 20.0 L/min. i) Derive an equation to relate the mole balance of sulfuric acid with time. ii) Evaluate the time taken to concentrate the solution to 5%, 35%, 65%, 90% and 95% of H2SO4 and sketch a graph incorporating the values obtained. iii) Predict the time taken for the solution to reach 100% concentration and discuss the answer based on the solution properties. Given: Density H2SO4 = 1.84 g/mL Molecular weight of H = 1 g/mol Molecular weight of S = 32 g/mol Molecular weight of O = 16 g/mol Inn

Answer 1

Moles of sulfuric acid at any time will be sum of initial mole of sulfuric acid in tank and moles of sulfuric acid added in tank. A mole of solution containing 5 mol% sulfuric acid filled in tank is 130 kmole. Moles of sulfuric acid initially filled = 5 % of 130 = 0.05 X 130 kmole = 6.5 kmole Calculation for molar flow of sulfuric acid to the solution Volumetric flow x density molar flow of sulfuric acid = mol/min molar mass Volumetric flow of sulfuric acid = 20 L/min density of sulfuric acid = 1.84 g ml 1.84 kg/L g molar mass of sulfuric acid = 2H + S + 40 = 2 + 32 +64 = 98- mol = 98kg/kmol molar flow of sulfuric acid 20 X 1.84 kmol/min 98 molar flow of sulfuric acid = 0.3755 kmol/min At any time t min sulfuric acid added will be 0.3756t By taking mole balance of sulfuric acid Sulfuric acid in tank at any time (N1) = initially filled + sulfuric acid added N1 = 6.5 +3.755t

Part b Moles of water in solution = 130 -6.5 = 123.5 kmole Total moles in tank at any time t = 6.5 +3.755t + 123.5 = 130 + 3.755t kmole Total moles in tank at any time t = 130 + 3.755t kmole Mole fraction of sulfuric acid in solution at any time t is X moles of sulfuric acid Mole percent of sulfuric acid = Total moles X 6.5 +3.755t 130 + 3.755t By solving fort 130X + 3.755Xt = 6.5 + 3.755t 130X – 6.5 = 3.755t - 3.755Xt 3.755t(1 – X) = 130X – 6.5 t 130X – 6.5 3.755(1-x) Time required for 5% sulfuric acid Mole fraction = 5/100 = 0.05 t = 130 x 0.05 -6.5 3.755(1 – 0.05) t=0 min

Time required for 35% sulfuric acid Mole fraction = 35/100 = 0.35 t = 130 x 0.35 – 6.5 3.755(1 – 0.35) t = 15.9787 min Time required for 65% sulfuric acid Mole fraction = 65/100 = 0.65 t = 130 x 0.65 – 6.5 3.755(1 – 0.65) t = 59.3494 min Time required for 90% sulfuric acid Mole fraction = 90/100 = 0.9 t 130 x 0.9 - 6.5 3.755(1 – 0.9) t = 294.2743min Time required for 95% sulfuric acid Mole fraction = 95/100 = 0.95 t 130 x 0.95 -6.5 3.755(1 – 0.95) t = 623. 1691 min

700 600 500 400 Time in min 300 200 100 0 0 20 40 60 80 100 Mol % sulfuric acid Part C Time taken to reach 100% sulfuric acid will be infinity. As the solution filled in tank contain the 95% water i.e. 135.5 kmole. We know that concentration of any component say component A in solution is 100% when the component is pure and does not contain any other component. So having water in it is impossible to form 100% sulfuric acid solution.