課程名稱︰化學反應工程
課程性質︰必修
課程教師︰吳紀聖
開課學院：工學院
開課系所︰化工系
考試日期（年月日）︰103.05.28
考試時限（分鐘）：110
是否需發放獎勵金：是
（如未明確表示，則不予發放）

試題 :



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



--
※ 發信站: 批踢踢實業坊(ptt.cc), 來自: 140.112.250.1
※ 文章網址: http://www.ptt.cc/bbs/NTU-Exam/M.1403077192.A.644.html