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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 227Β°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 Β°C (20)(200 πΏ)
ππ
PV=nRT, ππ΄0 = π¦π΄0 = (0.75) = 72.6 πππ π
π 0.0826 (500)
i.
ii.Bimolecular, β
ππΆπ΄ ππ‘
=-rA= kCA2
CA0= 72.6/200=0.363 mol/L π₯
π‘ = πΆπ΄0 β« 0
ππ₯ βππ΄
βππ΄ = ππΆπ΄2 , πΆπ΄ πΆπ΄ = πΆπ΄0
= πΆπ΄0
1βπ₯ 1+ππ₯
, Ι= yA0Ξ΄ , Ξ΄= Β½ - 1= -0.5, yA0= 1
1βπ₯ 1βπ₯ = 0.363 1 + ππ₯ 1 β 0.5π₯ π₯
π‘ = πΆπ΄0 β« 0
ππ₯ 0.7 (0.363
π₯ 1 π‘= β« (0.7)(0.363) 0
1βπ₯ 2 ) 1 β 0.5π₯ ππ₯
1βπ₯ 2 ( ) 1 β 0.5π₯
0.8
= 3.93 β«
πππ. π’ππππππππ’πππ, βππ΄ = ππΆπ΄ , πΆπ΄
0
= πΆπ΄0
1 β 0.5π₯ 2 ( ) . ππ₯ = 7.86 π ππ 1βπ₯
1βπ₯ 1+ππ₯
, Ι= yA0Ξ΄, Ξ΄= 1 - 1= 0, yA0= 1
π₯ 0.99 1 ππ₯ ππ₯ π‘= β« = 27.54 β« = 126.6 π ππ (0.1)(0.363) 0 1 β π₯ 1βπ₯ 0
βππ
πΆπ΄ = ππ‘, πΆπ΄0
βππ
0.363(1 β 0.99) = 0.1 Γ π‘ 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. βππ»2 = π=
1 πππ»2 1 πππ2 = π ππ‘ π ππ‘
π 2 0.6 (0.75) = 0.211 π3 4
1 π»2 + π2 β π»2 π 2 2
16
18
H2O produced =108/18= 6 kmol/s H2 used= 6 kmol/s O2 used= 3 kmol/s βππ»2 =
1 6 ππππ ππππ = 28.43 3 . π ππ 0.211π3 π ππ π
βππ2 =
1 3 ππππ ππππ = 14.15 3 . π ππ 3 0.211π π ππ π
Q2: a. Letβs consider the production of ethyl benzene; 2 Ethylene + Tolueneβ 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 227Β°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) βNO(g) + CO2(g). Suggest a mechanism consistent with the rate law. NO2+ NO2βN+NO3 Slow NO3+COβNO2+CO2
Fast
Q3: a. Suppose the following data were obtained for the homogeneous gas-phase reaction 2A + 2B βC + 2D carried out in a rigid 2-L vessel at 800Β°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 π = πΆπ
π
-rA= β
ππΆπ΄
=β
ππ‘
1 πππ΄ π
π ππ‘
π
π½
= ((π
π)π΄πΌ+π½) ππ΄πΌ ππ΅
πππ΄ ππ΄ π½ =( ) ππ΄πΌ ππ΅ β1+πΌ+π½ (π
π) ππ‘ πππ΄ π½ βππ΄π = β = ππ΄π ππ΄πΌ ππ΅ ππ‘ β
π
π΄ Where; kAP= ((π
π)β1+πΌ+π½ )
Values of
πππ΄
are calculated from (dP/dt)0 data as at any instant;
ππ‘
πππ π
π πππ‘ π
π πππ΄ ( ) = ( ) = ( ) ππ‘ 0 π ππ‘ 0 2π ππ‘ 0
=
1 πππ΄ ( ) 2 ππ‘ 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: πππ΄ ( ) ππ‘ 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
πΌ π½
ππ΄π2 π (ππ΄ ππ΅ )2 = ππ΄π1 π (ππΌ ππ½ ) π΄ π΅ πΌ
14.4 36 =( ) 1.6 12 9 = 3Ξ±, Ξ± =2
1
Use initial rate method, experiments 1 and 3; πΌ π½
ππ΄π3 π (ππ΄ ππ΅ )3 = ππ΄π1 π (ππΌ ππ½ ) π΄ π΅ π½
1
3.2 68 =( ) 1.6 34
2 = 2Ξ², Ξ² =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 Γ 10-4 kPa-2. min-1 From equation kA = (RT)2kAP = (8.314 L.kPa/K.mol)2(3.27 Γ 10-4) = 2.26Γ10-2 L.K.min-1/mol2
βππ΄ = 2.26 Γ 10β2 ππ΄2 ππ΅ Q3:b: The reaction of triphenyl methyl chloride (trityl) (A) and methanol (B); (C6H5)3CCl + CH3OH β (C6H5)3COCH3 + HCl was carried out in a solution of benzene and pyridine at 25 Β°C. Pyridine reacts with HCl that then precipitates as pyridine hydrochloride thereby making the reaction irreversible. The concentration β 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.