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Ethyl acetate is an extensively used solvent and can be formed by the vapor-phas

ID: 495170 • Letter: E

Question

Ethyl acetate is an extensively used solvent and can be formed by the vapor-phase esterification of acetic acid and ethanol. The reaction was studied using a microporous resin as a catalyst in a packed bed reactor [Ind. Eng. Chem. Res., 26 (2), 198 (1987)]. The reaction is first-order in ethanol and pseudo-first-order in acetic acid. For an equal molar feed rate of acetic acid and ethanol, the specific reaction rate constant is 1.2 dm^3/g cat/min. The total molar feed rate is 10 mol/min, the initial pressure is 10 atm and the temperature is 113 degree C. a) Calculate the weight of catalyst needed to achieve 80% conversion if the pressure drop throughout the reactor can be considered negligible? b) Calculate the maximum weight of catalyst needed to achieve 80% conversion if there is pressure drop throughout the reactor and the pressure drop parameter, alpha = 0.001 g^-1?

Explanation / Answer

b) Specific reation rate constant (kCA0) = 1.2 dm3/g cat/min

Total molar feed rate = 10 mol/min, No. of moles (n) = 10 mol

Pressure (P)=10 atm, Gas constant (R)=0.0821 L-atm/mol.K, T=113°C=273+113=386 K

Use Ideal gas law to find Volumetric feed rate (FA0)

PV=nRT

V=nRT/P = 10X0.0821X386 / 10 = 31.69 L = 31.69 dm3

Volumetric feed rate (FA0) = 31.69 dm3

Pressure drop () = 0.001 g-1

Mole balance, dX/dW = -rA/FA0 -----1

Rate Law, -rA = kCA ------2

Stoichiometry, CA = CA0(1-X)*P/P0 ------3

P/P0 = (1-W)1/2 ------4

Substitute equations 2, 3 & 4 in 1

dX/dW = [kCA0(1-X)/FA0](1-W)1/2

Integrate from X = 0 to X = 0.8 (80% conversion)

ln (1/1-0.8) = 1.2 X 2 [1-(1-0.0001W)3/2] / (31.69 X 3 X 0.001)

Wieght of catalyst, W=4 g

Integrate from X = 0 to X = 0.8 (80% conversion)

ln (1/1-0.8) = 1.2 X 2 [1-(1-0.0001W)3/2] / (31.69 X 3 X 0.001)

Wieght of catalyst, W=4 g

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