课程名称︰化学反应工程 课程性质︰必修 课程教师︰吴纪圣 开课学院：工学院 开课系所︰化工系 考试日期（年月日）︰103.05.28 考试时限（分钟）：110 是否需发放奖励金：是 （如未明确表示，则不予发放） 试题 : The 2nd Midterm of Reaction Engineering (Class 2) May 28, 2014 10:20~12:10 普301 1. The elemental gas-phase reaction, A → B + 2C, is carried out isothermally in a flow reactor with no pressure drop. The specific reaction rate constant at 50℃ is 10^-3 min^-1, and the activation energy is 85 kJ/mol. Pure A enters the reactor at 10 atm and 127℃ and a molar flow rate of 2.5 mol/min. Calculate the reactor volume and space time to achieve 90% conversion in (a) a PFR (10%), (b) a CSTR (10%). 2. The liquid-phase reaction of methanol and triphenyl takes place in a batch reactor at 25℃. CH3OH + (C6H5)3CCl → (C6H5)3COCH3 + HCl For an equal molar feed, the following concentration-time data was obtained for the methanol. Cmethanol (mol/dm^3) 1.0 0.95 0.816 0.707 0.5 0.37 Time (h) 0 0.278 1.389 2.78 8.33 16.66 The following concentration-time data were obtained for an initial methanol concentration 0.1 mol/dm^3 and an initial triphenyl of 1.0 mol/dm^3. Cmethanol (mol/dm^3) 0.1 0.0847 0.0735 0.0526 0.0357 Time (h) 0 1 2 5 10 (a) Determine the order of reaction and rate constant of rate law (15%). (b) If you were to take more data points, what would be the resonable setting (e.g. Cmethanol, Ctriphenyl)? Why? (5%) 3. The gas-phase reaction, A + B → C + D, takes place isothermally at 300K in a packed-bed reactor in which the feed is equal molar in A and B with CA0 = 0.1 mol/dm^3. The reaction is second order in A and zero order in B. Currently, 50% conversion is achieved in a reactor with 100 kg of catalyst for a columetric flow rate 100 dm^3/min. The pressure drop parameter, α = 0.0099 kg^-1. What is the rate constant? 4. The decomposition of ozone in an inert gas M is by the following elementary steps. Derive the rate law of ozone decomposition by PSSH. k1 M + O3 ←→ O2 + O + M k2 O3 + O —→ 2O2 rate limiting step 5. Derive the rate law for the following enzymatic reaction, i.e., rate of product rP =? Where E: enzyne, S: substrate, C: cofactor k1 k2 E + S ←→ E‧S —→ P + E k3 E + C ←→ E‧C k4 E‧C + S —→ P + E‧C ------------------------------------------------------------------------------ Gas constant: R = 8314 Pa-dm^3/(mol-K) or 82.06 X 10^-3 dm^3-bar/(mol-K), or 1.987 cal/mol-K Batch: NA0 dX/dt = -rA V CSTR: V = FA0 (Xout - Xin)/(-rA)out PFR: FA0 dX/dV = -rA x (1 + εx) 1 ∫----------- dx = (1 + ε) ln----- - εx 0 (1 - x) 1-x x (1 + εx) (1 + ε)x 1 ∫----------- dx = ----------- - εln------- 0 (1 - x)^2 1 - x 1 - x --

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