First Mid Term Exam, Solution [PDF]

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Solution



Tikrit University-College of Engineering Chemical Engineering Department



Fall 2018 First midterm exam



Reactor Design



Q1: a: A 200 dm3 constant volume batch reactor is pressurized to 20 atm with a mixture of 75% A and 25% inert. The gas phase reaction is carried out isothermally at 227°C.



i.



How many moles of A are in the reactor initially? What is the initial concentration of A?



ii.



If the reaction is bimolecular, single reactant and elementary, k = 0.7 (units are missing) Calculate the time to consume 80% of A.



iii.



If the reaction is unimolecular, single reactant and elementary, k = 0.1 (units are missing) Calculate the time necessary to consume 99% of A.



yA0 = 0.75, V= 200 dm3, P= 20 atm, T= 227 °C (20)(200 𝐿)



𝑃𝑉



PV=nRT, 𝑛𝐴0 = 𝑦𝐴0 = (0.75) = 72.6 𝑚𝑜𝑙 𝑅𝑇 0.0826 (500)



i.



ii.Bimolecular, −



𝑑𝐶𝐴 𝑑𝑡



=-rA= kCA2



CA0= 72.6/200=0.363 mol/L 𝑥



𝑡 = 𝐶𝐴0 ∫ 0



𝑑𝑥 −𝑟𝐴



−𝑟𝐴 = 𝑘𝐶𝐴2 , 𝐶𝐴 𝐶𝐴 = 𝐶𝐴0



= 𝐶𝐴0



1−𝑥 1+𝜀𝑥



, ɛ= yA0δ , δ= ½ - 1= -0.5, yA0= 1



1−𝑥 1−𝑥 = 0.363 1 + 𝜀𝑥 1 − 0.5𝑥 𝑥



𝑡 = 𝐶𝐴0 ∫ 0



𝑑𝑥 0.7 (0.363



𝑥 1 𝑡= ∫ (0.7)(0.363) 0



1−𝑥 2 ) 1 − 0.5𝑥 𝑑𝑥



1−𝑥 2 ( ) 1 − 0.5𝑥



0.8



= 3.93 ∫



𝑖𝑖𝑖. 𝑢𝑛𝑖𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟, −𝑟𝐴 = 𝑘𝐶𝐴 , 𝐶𝐴



0



= 𝐶𝐴0



1 − 0.5𝑥 2 ( ) . 𝑑𝑥 = 7.86 𝑠𝑒𝑐 1−𝑥



1−𝑥 1+𝜀𝑥



, ɛ= yA0δ, δ= 1 - 1= 0, yA0= 1



𝑥 0.99 1 𝑑𝑥 𝑑𝑥 𝑡= ∫ = 27.54 ∫ = 126.6 𝑠𝑒𝑐 (0.1)(0.363) 0 1 − 𝑥 1−𝑥 0



−𝑙𝑛



𝐶𝐴 = 𝑘𝑡, 𝐶𝐴0



−𝑙𝑛



0.363(1 − 0.99) = 0.1 × 𝑡 0.363



4.6= 0.1t, t= 46 unit time Q1.b. A rocket engine burns a stochiometric mixture of fuel (liquid hydrogen) in oxidant (liquid oxygen). The combustion chamber is cylindrical, 75 cm long, and 60 cm in diameter, and the combustion process produces 108 kg/s of exhaust gases. If combustion is complete, find the rate of reaction with respect to hydrogen and oxygen. −𝑟𝐻2 = 𝑉=



1 𝑑𝑁𝐻2 1 𝑑𝑁𝑂2 = 𝑉 𝑑𝑡 𝑉 𝑑𝑡



𝜋 2 0.6 (0.75) = 0.211 𝑚3 4



1 𝐻2 + 𝑂2 → 𝐻2 𝑂 2 2



16



18



H2O produced =108/18= 6 kmol/s H2 used= 6 kmol/s O2 used= 3 kmol/s −𝑟𝐻2 =



1 6 𝑘𝑚𝑜𝑙 𝑘𝑚𝑜𝑙 = 28.43 3 . 𝑠𝑒𝑐 0.211𝑚3 𝑠𝑒𝑐 𝑚



−𝑟𝑂2 =



1 3 𝑘𝑚𝑜𝑙 𝑘𝑚𝑜𝑙 = 14.15 3 . 𝑠𝑒𝑐 3 0.211𝑚 𝑠𝑒𝑐 𝑚



Q2: a. Let’s consider the production of ethyl benzene; 2 Ethylene + Toluene→ Ethyl benzene + propylene The gas feed consists of 25% toluene and 75% ethylene. Write the rate of reaction solely as a function of conversion. Assume the reaction is elementary with kT=250(dm6/mol2s). The entering pressure is 8.2 atm and the entering temperature is 227°C and the reaction takes place isothermally with no pressure drop.



Q1.b. (10 points): 1) Patients diagnosed with depression have a decreased concentration of serotonin in their brain. To help increase this concentration, doctors administer Selective Serotonin Reuptake Inhibitors (SSRIs) to the patients. Normal levels of serotonin in a healthy individual range from 101-283 ng/ml. Serotonin is metabolized by the body at a specific rate of 9.63 x 10-6 s-1. Assume that when the SSRI is administered, the serotonin concentration increases to 175.0 ng/ml. How long will it take for the serotonin concentration to fall below 100.0 ng/ml?



Q2. b. 2) Rate law: r=k[NO2]2 Overall reaction: NO2(g) + CO(g) →NO(g) + CO2(g). Suggest a mechanism consistent with the rate law. NO2+ NO2→N+NO3 Slow NO3+CO→NO2+CO2



Fast



Q3: a. Suppose the following data were obtained for the homogeneous gas-phase reaction 2A + 2B →C + 2D carried out in a rigid 2-L vessel at 800°C and the stochiometric quantities.



P0, kPa



xA0



46 70 80



0.261 0.514 0.150



(dP/dt)0, kPa.min-1 -0.8 -7.2 -1.6



Assuming that at time zero no C or D is present, obtain the rate law for this reaction, stating the value and units of the rate constant in terms of L, mol, s.



In terms of A and initial rates and conditions, and an assumed form of the rate law, we write 𝑃 = 𝐶𝑅𝑇



-rA= −



𝑑𝐶𝐴



=−



𝑑𝑡



1 𝑑𝑃𝐴 𝑅𝑇 𝑑𝑡



𝑘



𝛽



= ((𝑅𝑇)𝐴𝛼+𝛽) 𝑃𝐴𝛼 𝑃𝐵



𝑑𝑃𝐴 𝑘𝐴 𝛽 =( ) 𝑃𝐴𝛼 𝑃𝐵 −1+𝛼+𝛽 (𝑅𝑇) 𝑑𝑡 𝑑𝑃𝐴 𝛽 −𝑟𝐴𝑝 = − = 𝑘𝐴𝑃 𝑃𝐴𝛼 𝑃𝐵 𝑑𝑡 −



𝑘



𝐴 Where; kAP= ((𝑅𝑇)−1+𝛼+𝛽 )



Values of



𝑑𝑃𝐴



are calculated from (dP/dt)0 data as at any instant;



𝑑𝑡



𝑑𝑃𝑇 𝑅𝑇 𝑑𝑛𝑡 𝑅𝑇 𝑑𝑛𝐴 ( ) = ( ) = ( ) 𝑑𝑡 0 𝑉 𝑑𝑡 0 2𝑉 𝑑𝑡 0



=



1 𝑑𝑃𝐴 ( ) 2 𝑑𝑡 0



Values of PA0 and PB0 can be calculated from the given values of PT and yA. The results for the three experiments are as follows: 𝑑𝑃𝐴 ( ) 𝑑𝑡 0 0.261 12 34 -1.6 0.514 36 34 -14.4 0.150 12 68 -3.2 Use initial rate method, experiments 1 and 2; yA0



PA0



PB0



𝛼 𝛽



𝑟𝐴𝑃2 𝑘 (𝑃𝐴 𝑃𝐵 )2 = 𝑟𝐴𝑃1 𝑘 (𝑃𝛼 𝑃𝛽 ) 𝐴 𝐵 𝛼



14.4 36 =( ) 1.6 12 9 = 3α, α =2



1



Use initial rate method, experiments 1 and 3; 𝛼 𝛽



𝑟𝐴𝑃3 𝑘 (𝑃𝐴 𝑃𝐵 )3 = 𝑟𝐴𝑃1 𝑘 (𝑃𝛼 𝑃𝛽 ) 𝐴 𝐵 𝛽



1



3.2 68 =( ) 1.6 34



2 = 2β, β =1 The overall order, is therefore 3. Substitution of these results into rate equation for any one of the three experiments gives; kAP = 3.27 × 10-4 kPa-2. min-1 From equation kA = (RT)2kAP = (8.314 L.kPa/K.mol)2(3.27 × 10-4) = 2.26×10-2 L.K.min-1/mol2



−𝑟𝐴 = 2.26 × 10−2 𝑃𝐴2 𝑃𝐵 Q3:b: The reaction of triphenyl methyl chloride (trityl) (A) and methanol (B); (C6H5)3CCl + CH3OH → (C6H5)3COCH3 + HCl was carried out in a solution of benzene and pyridine at 25 °C. Pyridine reacts with HCl that then precipitates as pyridine hydrochloride thereby making the reaction irreversible. The concentration – time data was obtained in a batch reactor:



The initial concentration of methanol was 0.5 mol dm3. Determine the reaction order with respect to triphenyl methyl chloride.



Thus, the reaction is second order w.r.t to triphenyl methyl chloride.