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Centeno, Robenjoy C.

ChE 416

Problem Set

1. A steam of pure gaseous reactant A (CAo=660 mol/liter) enters a plug flow reactor at a flow rate of FAo=540 mmol/min and polymerizes there as follows 3A↔R, -rA=54mmol/liter/min How large a reactor is needed to lower the concentration of A in the exit stream to CA=330 mmol/liter? Sol’n: 𝑛𝑥−1 − 𝑛𝑥−0 1 − 3 2 = =− 𝑛𝑥−0 3 3 𝐶𝐴𝑜 (1 − 𝑋𝐴 ) 𝐶𝐴 = (1 + 𝜀𝐴 𝑋𝐴 ) 660𝑚𝑚𝑜𝑙 )(1 − 𝑋𝐴 ) 330𝑚𝑚𝑜𝑙 ( 𝑙 = 2 𝑙 [1 + (− 3)(𝑋𝐴 )] 𝑋𝐴 = 0.75 𝜀𝐴 =

0.75

𝑑𝑥𝐴 −𝑟𝐴 0 660𝑚𝑚𝑜𝑙 0.75 𝑑𝑥𝐴 𝜏=( )∫ 𝑚𝑚𝑜𝑙 𝑙 0 54 /𝑚𝑖𝑛 𝑙 𝜏 = 𝐶𝐴𝑜 ∫

660𝑚𝑚𝑜𝑙/𝑙 [𝑋𝐴 ]|0.75 0 54𝑚𝑚𝑜𝑙 /𝑚𝑖𝑛 𝑙 𝜏 = 9.17 𝑚𝑖𝑛 𝜏=

𝑉 = 𝜏𝑉1 𝑉 = 9.17 min (

540 ) 𝑙/𝑚𝑖𝑛 660

V=7.503L 2. Acetone is produced from 2-propanol in the presence of dioxygen and the photocatalyst Ti02 when the reactor is irradiated with ultraviolet light. For a reaction carried out at room temperature in 1.0 mol of liquid 2-propano1 containing 0.125 g of catalyst, the following product concentrations were measured as a function of irradiation time (J. D. Lee, M.S. Thesis, Univ. of Virginia, 1993.) Reaction time(min) 20 40 60 80 100 120 140 160 180 Acetone produced 1.9 3.9 5.0 6.2 8.2 10 11.5 13.2 14.0 (gacetone/g2-propanol)× 104

Calculate the first-order rate constant. 16 14

y = 0.0776x + 0.4528

Conc.

12 10 8 6 4 2 0 0

50

100

150

200

time

k = 0.0776 min-1

3. A first-order homogeneous reaction of A going to 3B is carried out in a constant pressure batch reactor. It is found that starting with pure A the volume after 12 min is increased by 70 percent at a pressure of 1.8 atm. If the same reaction is to be carried out in a constant volume reactor and the initial pressure is 1.8 atm, calculate the time required to bring the pressure to 2.5 atm. Sol’n: 𝐴 → 3𝐵 Volume increased by 70%: Final Volume (V) = 0.7 Initial Volume (V0) Variable volume system,

𝑉 = 𝑉0 (1 + 𝜉𝐴 𝜒𝐴 ) 2𝜒𝐴 = 0.7 𝜒𝐴 = 0.35

t =? 𝜒𝐴

𝑡 = 𝑁𝐴0 ∫ 0

𝑑𝜒𝐴 (−𝑟𝐴 )𝑉 1−𝜒𝐴 ) 𝑉0 (1 + 𝜉𝐴 𝜒𝐴 ) 1+𝜉𝐴 𝜒𝐴

(−𝑟𝐴 )𝑉 = 𝑘𝐶𝐴0 (

𝑉𝜒 −1 − 𝑉𝜒𝐴 −0 3 − 1 𝜉𝐴 = 𝐴 = =2 𝑉𝜒𝐴 −0 1 Calculate 𝜒𝐴 1.7𝑉0 = 𝑉0 (1 + 2𝜒𝐴 ) 1 + 2𝜒𝐴 = 1.7

Time taken increase in volume, T=12min, and 𝜒𝐴 = 0.35 𝜒

𝑡 = 𝑁𝐴0 ∫0 𝐴

𝑑𝜒𝐴 1−𝜒𝐴 𝑘𝐶𝐴0 ( )𝑉 (1+𝜉𝐴 𝜒𝐴 ) 1+𝜉𝐴 𝜒𝐴 0 𝜒𝐴

𝑁𝐴 1 𝑑𝜒𝐴 = 0× ∫ (−𝛾 𝑉0 𝑘𝐶𝐴0 𝐴 )𝑉 0

𝜒𝐴

@Constant Temperature: k =0.036 min-1

0

Therefore, 1 1 𝑡 = [𝑙𝑛 ] 𝑘 (1 − 𝜒𝐴 )

1 𝑑𝜒𝐴 = ∫ 𝑘 (−𝛾𝐴 )𝑉 1 𝜒 = [ln (1 − 𝜒𝐴 )]0 𝐴 𝑘 1 1 = [𝑙𝑛 ] 𝑘 (1 − 𝜒𝐴 ) 1 1 𝑘 = [𝑙𝑛 ] 𝑡 (1 − 𝜒𝐴 ) 1 1 = [𝑙𝑛 ] 12 (1 − 0.35) = 0.036 𝑚𝑖𝑛−1 @Constant Volume: P0 = 1.8 atm

@P= 2.5 atm, 𝜒𝐴 = 0.1935 1

1

𝑡 = 0.036 [𝑙𝑛 (1−0.1935)] = 5.97 𝑚𝑖𝑛

4. Consider the reversible, elementary, gas phase reaction shown below that occurs at 300 K in a constant volume reactor of 1.0 L.

a. For an initial charge to the reactor of 1.0 mol of A, 2.0 mol of B, and no C, find the equilibrium conversion of A and the final pressure of the system. Plot the composition in the reactor as a function of time. b. Consider the reversible, elementary, gas phase reaction of A and B to form C occurring at 300 K in a variable volume (constant pressure) reactor with an initial volume of 1.0 L. For a reactant charge to the reactor of 1.0 mol of A, 2.0 mol of B, and no C, find the equilibrium conversion of A. Plot the composition in the reactor and the reactor volume as a function of time. c. Consider the effect of adding an inert gas, I, to the reacting mixture. For a reactant charge to the reactor of 1.0 mol of A, 2.0 mol of B, no C, and 3.0 mol of I, find the equilibrium conversion of A. Plot the composition in the reactor and the reactor volume as a function of time. Sol’n: a. Reversible elementary gas phase reaction Temperature (T) = 300K Volume of Reactor (V) = 1L k1=6.0 L/mol h k2=3.0 L/mol h 𝑑𝐶𝐴 𝑑𝑥𝐴 = 𝐶𝐴𝑜 = 𝑘1 𝐶𝐴 𝐶𝐵 − 𝑘𝐶𝐶 𝑑𝑡 𝑑𝑡 𝑑𝑥𝐴 𝐶𝐴𝑜 = 𝑘1 𝐶𝐴𝑜 (1 − 𝑥)(𝐶𝐵𝑜 − 𝐶𝐵𝑜 𝑥𝐵 ) − 𝑘2 𝐶𝐶 𝑑𝑡 For equilibrium conversion, −

𝑑𝑥𝐴 =0 𝑑𝑡

𝑘1 𝐶𝐴𝑜 (1 − 𝑥𝑒 )(𝐶𝐵𝑜 − 𝐶𝐵𝑜 𝑥𝐵 ) − 𝑘2 𝐶𝐶 = 0 𝑘1 𝐶𝐴𝑜 (1 − 𝑥𝑒 )(𝐶𝐵𝑜 − 𝐶𝐴𝑜 𝑥𝑒 ) − 𝑘2 (𝐶𝐶𝑜 + 𝐶𝐴𝑜 + 𝑥𝑒 ) = 0 Here Cc=0 𝑘1 𝐶𝐴𝑜 (1 − 𝑥𝑒 )(𝐶𝐵𝑜 − 𝐶𝐴𝑜 𝑥𝐵 ) = 𝑘2 𝐶𝐴𝑜 𝑥𝑒 𝑘1 (1 − 𝑥𝑒 )(𝐶𝐵𝑜 − 𝐶𝐴𝑜 𝑥𝑒 ) = 𝑘2 𝑥𝑒 (6)(1 − 𝑥𝑒 )(2 − 𝑥𝑒 ) = 𝑘2 𝑥𝑒 (6)(2 − 𝑥𝑒 − 2𝑥𝑒 + 𝑥𝑒 2 ) = 3𝑥𝑒 2𝑥𝑒 2 − 7𝑥𝑒 + 4 = 0 On solving the equation, 𝑥𝑒 = 0.719

So product contains, 𝑛𝐴𝑜 (1 − 𝑥) = 1 − 0.719 = 0.281 𝑚𝑜𝑙/ℎ 𝑉 𝐶𝐵 = 𝐶𝐵𝑜 (1 − 𝑥) = (2)(1 − 0.719) = 0.562 𝑚𝑜𝑙/ℎ 𝐶𝐶 = 0.719 𝑚𝑜𝑙/ℎ Total= 0.281+0.562+0.719= 1.562 mol/h 𝐶𝐴 = 𝐶𝐴𝑜 (1 − 𝑥) =

Final pressure is 𝑛𝑅𝑇 𝑃= = 𝐶𝑅𝑇 = (1.562)(8.314)(300) × 103 = 3895.94 𝑘𝑃𝑎 𝑉

b. Variable Volume reactor Feed to the reactor: NAo= 1mol NBo= 2mol NCo= 0mol Xeq can be calculated by:

Equilibrium conversion (Xeq)=?

[𝐶𝐶𝑒 ] @ 𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 [𝐶𝐴𝑒 ][𝐶𝐵𝑒 ] 𝑁𝐶 𝑁𝐴𝑜 𝑋 𝐶𝐶𝑒 = = 𝑉 𝑉 𝑁𝐴 𝑁𝐴𝑜 (1 − 𝑋) 𝐶𝐴𝑒 = = 𝑉 𝑉 𝑁𝐵 𝑁𝐵𝑜 (1 − 𝑋) 𝐶𝐵𝑒 = = 𝑉 𝑉 𝐾𝐶 =

𝑁 𝑋 [ 𝐴𝑜 𝑉 ] 𝐾𝐶 = 𝑁 (1 − 𝑋) 𝑁𝐵𝑜 (1 − 𝑋) [ 𝐴𝑜 𝑉 ][ ] 𝑉 𝑋 𝑉 )( ) → 𝑒𝑞. 1 1 − 𝑋 𝑁𝐵𝑜 (1 − 𝑋) Substituting the values in eq.1 we get 𝑋 1 2=( )( ) 1 − 𝑋 2(1 − 𝑋) 𝐾𝐶 = (

𝑋 (1 − 𝑋)2 4 + 4𝑋 2 − 8𝑋 = 𝑋 4𝑋 2 − 9𝑋 + 4 = 0 X= 1.64 X= 0.6096 Since conversion cannot be greater than 1, therefore we neglect the root, X= 1.64 4=

∴ The conversion attained X= 0.6096 ≈ 60.96% c. Adding inert Feed (in moles) A= 1mol, B= 2mol, C= 0, Inert= 3mol 𝐴+𝐵 ⇌𝐶 @equilibrium [𝐶𝐶𝑒 ] 𝐾𝐶 = [𝐶𝐴𝑒 ][𝐶𝐵𝑒 ] 𝑁𝐶 𝑁𝐴𝑜 𝑋 𝐶𝐶𝑒 = = 𝑉 𝑉 𝑁𝐴 𝑁𝐴𝑜 (1 − 𝑋) 𝐶𝐴𝑒 = = 𝑉 𝑉 𝑁𝐵 𝑁𝐵𝑜 (1 − 𝑋) 𝐶𝐵𝑒 = = 𝑉 𝑉 But inert is added, 𝑉 = 𝑉𝑜 (1 + 𝜖𝑋) 𝜖 = fractional change in conversion 4−3 𝜖 = 3 = 1⁄3 [𝐶𝐶𝑒 ] [𝐶𝐴𝑒 ][𝐶𝐵𝑒 ] 𝑁𝐴𝑜 𝑋 ( ) 𝑉𝑜 (1 + 𝜖𝑋) = 𝑁 (1 − 𝑋) 𝑁𝐵𝑜 (1 − 𝑋) ( 𝐴𝑜 )( ) 𝑉𝑜 (1 + 𝜖𝑋) 𝑉𝑜 (1 + 𝜖𝑋) 𝑘1 𝑋 𝑉𝑜 (1 + 𝜖𝑋) 𝐾𝐶 = = [( ) ] 𝑘2 1 − 𝑋 𝑁𝐵𝑜 (1 − 𝑋) 1(1 + 1⁄3 𝑋) 6 𝑋 = = [( ) ] 3 1−𝑋 2(1 − 𝑋) 𝐾𝐶 =

2=

𝑋(1 + 1⁄3 𝑋) 2(1 − 𝑋)2

4(1 − 𝑋)2 = 𝑋(1 + 1⁄3 𝑋) 4(1 + 𝑋 2 − 2𝑋) = 𝑋 + 1⁄3 𝑋 2 4 + 4𝑋 2 − 8𝑋 = 𝑋 + 1⁄3 𝑋 2

1 (4 − ) 𝑋 2 − 9𝑋 + 4 = 0 3 11 2 𝑋 − 9𝑋 + 4 = 0 3 11𝑋 2 − 27𝑋 + 12 = 0 𝑋 = 1.87 𝑋 = 0.5824 Since the conversion cannot be greater than 1, we reject the value of X=1.87 Hence, conversion at equilibrium: X=0.5824 ≈ 58.24% 5. The irreversible reaction 2A → B takes place in the gas phase in a constant temperature plug flow reactor. Reactant A and diluent gas are fed in equimolar ratio, and the conversion of A is 85 percent. If the molar feed rate of A is doubled, what is the conversion of A assuming the feed rate of diluent is unchanged? Solution: 2A → B 𝑋𝐴 = 0.85 = 85% For first order reaction For PFR: 𝑉 𝜏 𝑋𝐴 = = 𝐹𝐴𝑜 𝐶𝐴𝑜 −𝑟𝐴 𝑉 𝜏 𝑋𝐴 = = 𝐹𝐴𝑜 𝐶𝐴𝑜 𝑘𝐶𝐴𝑜 (1 − 𝑋𝐴 ) 𝑉 𝑋𝐴 = 𝜏𝑘 = 𝐹𝐴𝑜 1 − 𝑋𝐴 For 1st order reaction: −𝑙𝑛(1 − 𝑋𝐴 ) = 𝑘𝜏 =

𝑉 𝐹𝐴𝑜

−𝑙𝑛(1 − 𝑋𝐴1 ) 𝐹𝐴2 = −𝑙𝑛(1 − 𝑋𝐴2 ) 𝐹𝐴1 𝐹𝐴2 = 2𝐹𝐴1 −𝑙𝑛(1−0.85) −𝑙𝑛(1−𝑋𝐴2 )

=2

−𝑙𝑛(1 − 𝑋𝐴2 ) = 0.9485 1 − 𝑋𝐴2 = 0.3873 𝑋𝐴2 = 0.6127

6. Titanium dioxide particles are used to brighten paints. They are produced by gas-phase oxidation of TiCl4 vapor in a hydrocarbon flame. The dominant reaction is hydrolysis, 𝑇𝑖𝐶𝑙4 + 2𝐻2 𝑂 ⇒ 𝑇𝑖𝑂2 (𝑠) + 4𝐻𝐶𝑙 The reaction rate is first-order in TiC14 and zero-order in H20. The rate constant for the reaction is: 88000𝐽/𝑚𝑜𝑙 −1 𝑘 = 8.0 × 104 𝑒𝑥𝑝 [− ]𝑠 𝑅𝑔 𝑇 The reaction takes place at 1200 K in a constant pressure flow reactor at 1 atm pressure (1.01 X 105 Pa). The gas composition at the entrance to the reactor is: CO2 8% H2O 8% O2 5% TiCl4 3% N2 Remainder a. What space time is required to achieve 99 percent conversion of the TiC14 to TiO2? b. The reactor is 0.2 m diameter and 1.5 m long. Assuming that the reactor operates 80 percent of the time, how many kilograms of TiO2 can be produced per year? (The molecular weight of TiO2 is 80 g/mol.) Rg = 8.3144 J/mol/K Given: TiCl4 gas phase oxidation of TiCl4 vapor in hydrocarbon flow 𝑇𝑖𝐶𝑙4 + 2 𝐻2 𝑂 ⇒ 𝑇𝑖𝑂2 (𝑠) + 4 𝐻𝐶𝑙 The reaction rate is first order in TiCl4 and zero order in H2O. 88000𝐽

The reaction constant 𝑘 = 8 × 104 exp (𝑚𝑜𝑙(8.314)𝑇) Reaction Temperature (T) =1200K

@ constant pressure flow reactor at P = 1atm (8 × 104 𝑃𝑎)

The gas composition at entrance to the reactor: CO2 -8%

H2O -8%

O2 -5%

TiCl4 -3%

N2 -76%

Other data given: Reactor diameter (d) = 0.2 m, length (L)= 1.5 m, gas constant (R)= 8.3144 J/ mole –K Reactor operates 80% time Mol.wt of TiO2= 80 g/mol

a) Solution: 88000

Rate constant (k) =8 × 104 exp (8.314(1200)) = 8 × 104 exp(8.82) log 𝑘 = log(8 × 104 ) + log(8.82) = log(8) + log(104 ) + log(8.82) ∴ 𝑘 = 706412𝑠 −1

log 𝑘 = 0.9032 + 4 + 0.9455 ; Atomic wt. of Ti = 48 g/atom

Mol.wt of TiCl4=48 +35.5 (4)= 190 g/mole The average molecular wt. of the gas mixture =

8(44) 8(18) 5(32) 3(190) 76(28) + + + + 100 100 100 100 100

= 33.54𝑔/𝑚𝑜𝑙

Only 3% of TiCl4 is available in the reaction, 99

3

And 99% of 3% of TiCl4= 100 (100) = 0.0297𝑔 𝑜𝑓 𝑇𝑖𝐶𝑙4 𝑎𝑣𝑎𝑖𝑙𝑎𝑏𝑙𝑒 𝑖𝑛 𝑡ℎ𝑒 𝑟𝑥𝑛 𝑡𝑜 𝑡𝑎𝑘𝑒 𝑝𝑎𝑟𝑡 Volume of the reactor (V)=

𝜋𝑑 2 𝐿 2

=

22 0.2 2 ( 2 ) (1.5) 7

= 0.0471𝑚3 = 47.1 𝑙𝑖𝑡𝑒𝑟𝑠

For the first order reaction: Reaction rate (rA) =

𝑑𝐶 𝑑𝑡

= 𝑘𝐶𝑎 𝑓

For this reaction, separating and integrating ,we get −(𝐶𝑎 − 𝐶𝑎𝑜 ( − ln

𝑑𝐶𝑎 ) = 𝑘 ∫ 𝑡 − 0𝑑𝑡 𝐶𝑎

𝐶𝑎 = 𝑘𝑡 → 𝑒𝑞. 1 𝐶𝑎𝑜 𝑃𝑉 = 𝑛𝑅𝑇

𝑛 𝑃 105 = 𝐶𝑎 = = = 10.02 𝑉 𝑅𝑇 (8.314)(1200) Or,

𝑑𝑋𝐴 𝑑𝑡

= 𝑘(1 − 𝑋𝐴 ) ln(1 − 𝑋𝐴 ) = 𝑘𝑡

ln(1 − 0.0297) = (706412)𝑡 0.030 = (706412)𝑡 For Zero order reaction:

;

∴ 𝑡 = 21192.36 𝑠𝑒𝑐

𝑡=

𝐶𝑎𝑜 0.031 = = 21192.36 𝑠𝑒𝑐 𝑘 706412

Space time (𝜏) =

𝑉𝐶𝑎𝑜 𝐹𝑎𝑜

= 𝐶𝑎𝑖

𝑋𝐴 𝑓−𝑋𝐴𝑖 −𝑟𝐴

0.0297 ) 0.03

= 10.023 (

= 9.93 𝑠𝑒𝑐

b) Solution: Production of TiO2= 9.93 = 47.1(1 − 𝑋𝐴 ) 9.93 47.1

= (1 − 𝑋𝐴 )

0.21 = (1 − 𝑋𝐴 ) 𝑋𝐴 = 0.79(80) = 63.2𝑔/𝑠 𝑋𝐴 =

63.2𝑔 𝑠(

3600𝑠 24ℎ 365 𝑑𝑎𝑦𝑠 )( )( ) 1ℎ 1𝑑𝑎𝑦 1𝑦𝑟

=

1993075200𝑔 𝑦𝑟

= 1993𝑡𝑜𝑛/𝑦𝑟