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Solution ManualChemical

1

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Introductory Chemical

Sticho Solutions 6th Edition David

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Solutions Manual to accompany

Introduction to Chemical Processes

Principles, Analysis, Synthesis Prepared by

Regina M. Murphy

Master your semester with ScribdAND CONFIDENTIAL PROPRIETARY Read Free Foron 30this Days Sign up to vote title & The New York Times  Useful  Not useful

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Cancel anytime. property of The McGraw-Hill Companies, (“McGraw-Hill”) and protected by copyright and other state state and federal federal laws

This Manual is the proprietary Special offer for students: Only $4.99/month.

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P1.1

SiO 2 + 3C  2CO + SiC P1.2

C 3H 5(NO 3) 3   2CO 2 +  3H 2O +  4 N 2 +  5O 2 From element balances on N, C, H, and O, we write 4 equations: 3 = 2 4 3 =  2 5 = 2 3 9 = 2 2 +  3 + 2 5 Solving, we find 5 3 1 C 3H 5(NO 3) 3  3CO2 + H 2O + N 2 + O 2 2 2 4 P1.3

(NH 4 ) 2 PtCl6   2Pt +  3NH 4Cl +  4N 2 +  5HCl Pt:  2 = 1 N:  3 + 2 4 = 2 H: 4 3 +  5 = 8 Cl:  3 +  5 = 6 Combine H and Cl balances and solve, than solve N balance: 2 1 2  3 = ,  5 = 5 ,  4 = 3 3 3

Master your semester with Scribd 2 2 1 (NH 4 ) 2 PtCl6  Pt + NH 4 Cl + N 2 + 5 HCl 3 3 & The New York Times 3 Special offer for students: Only $4.99/month. P1.4

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CaCO3:

2 gmol HCl gmol CaCO 3  gmol CaCO 3 100 g CaCO 3

MgCO3:

2 gmol HCl gmol MgCO 3

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36.5 g HCl 0.73g HCl = gmol HCl g CaCO 3

gmol MgCO 3 36.5 g HCl 0.869g HCl  = 84 g MgCO 3 gmol HCl g MgCO 3

MgCO3 has the best neutralizing ability, gram for gram. P1.5

Molar mass of urea (NH2)2CO = 2  14 + 4  1 + 12 + 16 = 60 g gmol. g 1 lb  = 1.3 lb 10 gmol  60 gmol 454 g 10 lbmol  60

lb 454 g  = 272,000 g lbmol lb

P1.6

Water is required to decompose the urea: (NH 2 ) 2 CO + H 2O  2NH 3 + CO 2 Fractional atom economy =

2 gmol NH 3  (17g gmol) You're Reading a Preview = 0.44 1 gmol urea  (60g gmol) + 1 gmol H 2O  (18g gmol) Unlock full access with a free trial.

(with only urea counted in the denominator, fractional atom economy is 0.57.) Download With Free Trial P1.7

Hexane: C 6H14 + 9.5O 2  6CO 2 + 7H 2O

Master your semester Scribd 6 gmol CO 2 44 gwith CO 2 /gmol CO 2 C6H14 86 g C 6H14 /gmol C 6H14 & The Newgmol York Times 

7 gmol H$4.99/month. 18 g H 2O /gmol H2O Special offer for students: Only 2O  gmol C H

86 g C H /gmol C H

=

=

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1.5g H 2O g C6H14

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P1.8

 lbmol NH 3  lbmol N 2  28 lb N 2     = 820 million lbs N 2  17 lb NH 3  2 lbmol NH 3  lbmol N 2 

(

)

(

 lbmol NH 3  3 lbmol H 2  2 lb H 2     = 180 million lbs H 2  17 lb NH 3  2 lbmol NH 3  lbmol H 2 

10 9 lb NH 3 

)

10 9 lb NH 3 

P1.9

Cl2:

$0.016 1 gmol 454 g 2000 lb $205    = gmol 71 g lb ton ton

NH3:

$0.0045 1 gmol 454 g 2000 lb $240    = gmol 17 g lb ton ton

P1.10

The conventional process has an atom economy of 0.45, which means that 0.55 lb reactants are shunted to waste per 0.45 lb of product made. At 300 million lb/yr 4-ADPA production, this amounts 367 million lb/yr waste. You're Reading a Preview The new process, with an atom economy of 0.84, produces 0.16 lb waste per 0.84 lb Unlock full access with a free trial. product. At 300 million lb/yr 4-ADPA production, this amounts 57 million lb/yr waste, o only 15% of the waste production of the conventional process. Download With Free Trial P1.11

Molar mass = 2 + 32 + 4(16) = 98 tons/tonmol

Master your with Scribd 45  semester 10 6 tons 1 tonmol = 4.6  10 5 tonmol yr  yr Times 98 tons & The New York 6 45  10 2000 lb Special offer for students: Only tons $4.99/month. yr



ton



454 g = 4.09  1013 g yr lb


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the cost would have to drop to $16,000. At 80,850 kg/day consumption of glucose, this converts to a glucose price of $0.198/kg.

The glucose-to-catechol process makes $49,200/day, but the benzene-to-catechol proces nets $89,300. The difference is $40,100. The glucose price would have to drop to $0.104/kg to be competitive with benzene. P1.13

Some possible explanations: greater number of reactions in pathway, more stringent product purity requirements, less pressure to trim costs by reducing wastes. P1.14

 $2.89  gal   = $0.36/lb : milk is a commodity chemical    gal  8 lb   $1.75 16 oz     = $2.33/lb : at this price, water is a specialty chemical!  12 oz  lb  P1.15

HNO 3 +  2CH 3OH   3C3H 7NO 2 +  4 CO 2 +  5H 2O You're Reading a Preview The element balance equations for N, C, H and O are

1 =  3

Unlock full access with a free trial.

= 3 3 +  4 Download With Free Trial 1 + 4 2 = 7 3 + 2 5 3 +  2 = 2 3 + 2 4 +  5  2

This is a set of 4 equations in 4 unknowns that we solve by substitution and elimination to find the balanced reaction:  Read Free Foron 30this Days Sign up to vote title

Master your semester with Scribd & The New York 1Times 1 HNOOnly CH 3OH  C 3H 7NO 2 3 + 3$4.99/month. Special offer for students: 3

2 3



+ CO 2 + 3 H 2O 3


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Grams of sodium oxalate required per gram of Freon-12 destroyed: 2 gmol Na 2C2O 4 gmol CF2Cl2 134 g Na 2C 2O 4   = 2.21 g Na 2C 2O 4 /g CF2Cl2 gmol CF2Cl2 121 g CF2Cl2 gmol Na 2C2O 4 Grams of solid products produced (includes NaF, NaCl and C):

 2 gmol NaCl 58.5 g NaCl   2 gmol NaF 42 g NaF   1 gmol C 12 g C +       +  gmol CF2Cl2 gmol NaCl   gmol CF2Cl2 gmol NaF   gmol CF2Cl2 gmol C 

gmol CF2Cl2 = 1.76 g solid products/g CF2Cl2 121 g CF2Cl2

P1.17

6 gmol H gmol C 2H 5OH 1 g H    100% = 13wt% H gmol C 2H 5OH 46 g C 2H 5OH gmol H 2 gmol H gmol H 2O 1 g H    100% = 11wt% H Water: gmol H 2O 18 g H 2O gmol H 12 gmol H gmol C 6H12O 6 1gH    100% = 6.7wt% H Glucose: gmol C 6H12O 6 180 g C 6H12O 6 gmol H 4 gmol H gmol CH 4 1gH You're Reading a Preview    100% = 25wt% H Methane: gmol CH 4 16 g CH 4 gmol H access with free trial. It does seem hard to believe thatUnlock they full achieved 50awt% H. Ethanol:

Download With Free Trial The reactions are balanced by writing element balance equations and solving them simultaneously. The balanced equations are given, along with a calculation of atom economy. P1.18

Master your semester with Scribd Hydrogenation: (a) conventional & The New York Times 1 C H COCH +

5 3 Special offer for students: Only 6$4.99/month.

4


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1Not useful Cancel anytime. NaBH 4 + H 2O  C 6H 5CH(OH)CH 3 + NaB(OH) 4 Useful

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Atom economy = 1.0! Oxidation: (a) conventional 2 1 C 6H 5CH(OH)CH 3 + CrO 3 + H 2SO 4  C 6H 5COCH 3 + Cr2 (SO 4 ) 3 + 2H 2O 3 3

Mi -1 122 -0.667 100 -1 98 +1 120 i

C6H5CH(OH)CH3 CrO3 H2SO4 C6H5 COCH3

Mi -122 -66.7 -98 120

i

Atom economy = 120/(122+66.7+98) = 0.42 (b) catalytic C 6H 5CH(OH)CH 3 + H 2O 2  C 6H 5COCH 3 + 2H 2O Atom economy = 120/(122+34) You're = 0.77Reading a Preview Unlock full access with a free trial.

C-C bond formation (a) conventional

Download With Free Trial C 6H 5CH(OH)CH 3 + Mg + CO 2 + 2HCl  C 6H 5CHCH 3COOH + MgCl2 + H 2O

Mi C6H5CH(OH)CH3 -1 122 Mg -1 24 CO2 -1 44 HCl Only $4.99/month. -2 36.5 Special offer for students: C6H5 CHCH3COOH +1 150 i

Mi -122 -24 -44 -73 150

i

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solve for stoichiometric coefficients – and check to see whether the coefficients for wate and/or carbon dioxide are nonzero. To balance the first reaction, we write C 6H 5OH +  2CH 3COCH 3   3C15H16O 2 +  4 CO 2 +  5H 2O The element balance equations for C, O and H are: 6 + 3 2 = 15 3 +  4 6 + 6 2 = 16 3 + 2 5 1 +  2 = 2 3 + 2 4 +  5 There are 3 equations and 4 stoichiometric coefficients. Thus, one of them is zero (in other words, that compound is NOT a byproduct.) We find a solution if we set 4 = 0: = 1/2, 3 = 1/2, 5 = 1/2. (There is not a reasonable solution if we assume no water is made.) We balance the remaining reactions in a similar fashion and find 4 balanced equations 1 1 1 C 6H 5OH + CH 3COCH 3 C15H16O 2 + H 2O 2 2 2 CH 4 + H 2O CO + 3H2 You're Reading a Preview Cl2 + CO COCl2 





Unlock full access with a free trial. C15H16O 2 + COCl2 + 2NaOH 

1 50

[

]n

+ 2NaCl + 2H 2O OC6H 4 C(CH OCOFree Trial Download 3) 2 C 6H 4With





To put together the generation-consumption analysis per mole of polycarbonate, we (a) multiply the 4th reaction by 50, (b) match phosgene consumption to phosgene generation by multiplying reaction 3 by 50, (c) match CO consumption to CO generation by  A multiplying reaction 2 by 50 and (d) match bisphenolRead A consumption toDays bisphenol Free Foron 30this Sign up to vote title  generation by multiplying reaction 1 by 100. The result is summarized in table form.


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Generation-consumption analysis for production of polycarbonate

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NaCl



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100



+100

P1.20

The balanced reactions are found from element balances on C, H, O and N. To determin if water is required as a reactant or product, we postulate that water is a product and then try to balance the equations. If the stoichiometric coefficient for water is zero, it is not a reactant or a product. If it is negative, water is a reactant, if positive, it is a product. The balanced chemical reactions are HCO 3 + NH 4 + + C 5H12O 2N 2 

C 6H13O 3N 3 + C 4 H 7O 4 N C10H18O 6N 4







C 6H13O 3N 3 + 2H 2O

C10H18O 6N 4 + H 2O

C 4 H 4 O 4 + C 6H14 O 2N 4

C 6H14 O 2N 4 + H 2O



CH 4 ON 2 + C 5H12O 2N 2

The generation-consumption table for this set of reactions is: R1 R2 R3 R4 net bicarbonate -1 -1 ammonium -1 -1 You're Reading a Preview ornithine -1 +1 0 Unlock full access citrulline +1 -1 0 with a free trial. water +2 +1 -1 +2 Download-1With Free Trial aspartic acid -1 arginosuccinate +1 -1 0 fumarate +1 +1 arginine +1 -1 0 urea +1 +1

Master your semester with Scribd overall reaction is: & The NewTheYork Times

Special offer for students: Only $4.99/month. HCO 3 + NH 4 + + C 4 H 7O 4 N  C4 H 4 O 4 


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+ CH 4 ON 2 + 2H 2O

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P1.22

The LeBlanc chemistry is given in Example 1.3. At a sodium carbonate production rate of 1000 ton/day, we complete the following process economy calculations. Compound NaCl H2SO4 HCl C CO2 CaCO3 Na2CO3 CaS sum

tons/day (SF = 1000/106) -2 58.5 -117 -1104 -1 98 -98 -927 +2 36.5 +73 +689 -2 12 -24 -226 +2 44 +88 +830 -1 100 -100 -943 +1 106 +106 +1000 +1 72 +72 +679 -2 (close enough to zero) i

Mi

i

Mi

$/ton $/day 95 80

-104,860 -74,160

87 105

-82,040 +105,000 -156,000

The LeBlanc process looks atrociously bad, at current prices. P1.23

The reactions are You're Reading a Preview 1 C 2H 5OH + O 2  CH 3Unlock CHOfull (R1) +H 2O with a free access trial. 2 1 CH 3CHO + O 2  CH 3COOH (R2) Download With Free Trial 2

Water is the only byproduct. The generation-consumption analysis is shown in the table.

Master your semester with Scribd Compound M M & The New York Times 1

2

Special offer for students: Only $4.99/month. C2H5OH -1

O

1/2

net

-1 1/2 1

i

46 32

i

-46 32

i

Read Free Foron 30this Days Sign up to vote title kg  Useful  Not useful (SF = 1/60) Cancel anytime. -0.77 0 53

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SiO2 C Si CO Cl2 SiCl4 H2 HCl sum

1

2

3

net

-1 -1 -2 -2 +1 -1 +1 +1 +2 +2 -2 -2 +1 -1 -2 -2 +4 +4


 Search document

SiO 2 + 2C  Si + 2CO Si + 2Cl2  SiCl 4 SiCl4 + 2H 2  Si + 4HCl Compound




(R1) (R2) (R3)

Mi

i

60 12 28 28 71 170 2 36.5

Mi

-60 -24 +28 +56 -142

Grams (SF = 3.57) -214 -86 +100 +200 -507

-4 -14 +146 +521 0

Reactant and byproduct quantities per 100 g Si produced are shown in the last column. The atom economy is 28/(60+24+142+4) = 0.12. You're Reading a Preview Water is a required byproduct in (R2) and (R3). The balanced reactions are: trial. (R1) C 4 H8 + CH 2O  C 5H10Unlock O full access with a free 1 C 5H10O + O 2  C 5H8Download O + H 2O With Free(R2) Trial 2 P1.25

C 5H10O + C 5H8O  C10H16O + H 2O

(R3)

We need to multiply (R1) by 2 to avoid making unwanted intermediates. The generation consumption analysis is:

Master your semester with Scribd M & The NewCompound York Times 1

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C4H8

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56

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Mi

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Read Free Foron 30this Days Sign up to vote title Grams Useful  Not useful Cancel anytime. (SF = 1000/152) -737 5

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P1.26

The reaction, written with unknown stoichiometric coefficients, is

(CaF)Ca 4 (PO 4 ) 3 +  2H 2SO 4 +  3H 2O   4CaH 4 (PO 4 ) 2 H 2O +  5CaSO 4 +  6HF We write the element balance equations to find the stoichiomeric coefficients: Ca: 5 =  4 +  5 F: 1 =  6 P: 3 = 2 4 O: 12 + 4 2 +  3 = 9 4 + 4 5 H: 2 2 + 2 3 = 6 4 +  6

Balances on F and P are readily solved, followed by the balance on Ca. Finally, H and O balances are solved. 1 2

3 2

3 2

1 2

(CaF)Ca 4 (PO 4 ) 3 + 3 H 2SO 4 + H 2O  CaH 4 (PO 4 ) 2 H 2O + 3 CaSO 4 + HF The process economy calculations are summarized in the table, per ton of monocalcium phosphate. You're Reading a Preview Compound Mi i Mi Tons $/ton $ i Unlock full access with a free trial. (SF = 1/378) Phosphate rock -1 504 -504 -1.33 128 -170 Download With Free Trial Sulfuric acid -3.5 98 -343 -0.91 80 -73 water -1.5 18 -27 -0.0714 Monocalcium phosphate +1.5 252 +378 +1 320 +320 Calcium sulfate +3.5 136 +476 +1.26 320 +403 Hydrogen fluoride +1 20 +20 +0.053  sum Read Free Foron 30+480 Days Sign up to vote this title

Master your semester with Scribd  & The NewRequired Yorkraw Times  Useful  Not useful materials and byproducts are listed in the “ tons” column. The fertilizer is a Special offer for students: Only $4.99/month.

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mix of monocalcium phosphate and calcium sulfate, per ton of mcp, we make 2.26 tons

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An overall reaction for reaction of glucose, oxygen, and ammonia to yeast, CO2 and water is: 1C 6H12O 6



+  2O 2 +  3NH 3   4 CH1.66O 0.5N 0.166 +  5CO 2 +  6H 2O

We know that 3.9 g CO2 are produced per gram of yeast. The molar mass of CO 2 is 44, and that of “yeast” is (12+1.66+0.5(16)+0.166(14)) = 23.98 g/gmol. Therefore, 3.9(23.98/44) or 2.1255 gmol CO2 are produced per gmol yeast. We will set 4 = 1 as ou basis, and  5 = 2.1255 from these data. Now we can complete the remaining element balances. C: 6 1 = 1 + 2.1255 H: 12 1 + 3 3 = 1.66 + 2 6 O: 6 1 + 2 2 = 0.5 + 2(2.1255) +  6 N:  3 = 0.166 The balanced reaction is: 0.521C 6H12O 6 + 2.085O 2 + 0.166NH 3  CH1.66O 0.5N 0.166 + 2.1255CO 2 + 2.545H 2 Of the 3.126 gmol C in glucose,You're 1 gmolReading is used atoPreview make yeast (or about 32%) and about 68% is used to make CO2. (This is probably the best measure of relative utilization of Unlock full access withmass a free trial. glucose for yeast vs. for CO2.) About 20% of the of carbon containing compounds is yeast, with the remainder as CO2. Download With Free Trial P1.28

A close examination of the first 3 reactions shows that only 2 are independent – if we reaction 1 and reaction 3 together, we get reaction 2. Therefore, we need to consider onl 2 of these 3 reactions. A generation-consumption table for reactions 1, 3, and 4 is shown (trial 1):  Read Free Foron 30this Days Sign up to vote title

Master your semester with Scribd & The New York Times i1

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Special offer for students: Cu2S Only $4.99/month. -1

Fe (SO )

1

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


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 CuS k  k

,

 



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

=0

k

and



 = 0 4 ,k k

 CuSO

k

From these restrictions, we find:

=  3  1 +  3  1



 4

=0

We can arbitrarily choose one multiplying factor, so we’ll set  1 = 1 =  3, which leaves us with  4 = 2. The revised generation-consumption table, along with calculations of mass requirements, is shown.

Cu2S Fe2(SO4)3 CuS CuSO4 FeSO4 S Fe Cu

i1

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159 400

-159 -800

Tons (SF=1/127) -1.25 -6.3

You're Reading a Preview

152 +912 +7.18 32 +32 +0.25 56 -112 -0.88 Download With Free Trial 63.5 +127 +1

Unlock full access with a free trial.

Per ton of metallic Cu, we need 1.25 tons chalcocite, but also 0.88 tons metallic Fe and an enormous 6.3 tons Fe2(SO4)3. 7.43 tons of byproducts are generated.

Master your semester with Scribd  P1.29 Read Free Foron 30this Days Sign up to vote title  first process, we use lactose to produce glucosewith the byproduct galactose. The & The NewIn the York Times Useful  Not useful economic evaluation is summarized in tabular form.


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P1.30

Sulfuric acid process Three reactions S + O 2  SO 2 1 SO 2 + O 2  SO 3 2 SO 3 + H 2O  H 2SO 4

(R1) (R2) (R3)

These reactions combine easily to an overall reaction of S + 1.5O 2 + H 2O  H 2SO 4 Nitric acid process: Three balanced reactions are 5 2NH 3 + O 2  2NO + 3H 2O 2 1 NO + O 2  NO 2 2 3NO 2 + H 2O  2HNO 3 + NO

(R1) (R2) (R3)

You're Reading a Preview The generation-consumption table gives: Unlock full access with a free trial.

Compound NH3 O2 NO H2O NO2 HNO3

R 1 R2 -2 -5/2 -1/2 +2 -1 +3 +1

R3 n et Download With Free Trial -2 -3 +1 +2 -1 +2 -3 -2 +2 +2 Read Free Foron 30this Days Sign up to vote title





Useful

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Special offer for students: Only $4.99/month. This doesn’t satisfy the restrictions on the solution, e.g., we have NO generated and NO

consumed, which are not allowed. To have no net generation or consumption of these tw

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NO H2O NO2 HNO3

+2 +3

 

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-3

+1 -1 -3 +2

+3



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+2 +2

For an overall reaction of NH 3 + 2O 2  HNO 3 + H 2O

The difference in value of nitric vs sulfuric acid is likely due to the difference in cost of ammonia vs sulfur. Sulfur is a byproduct of oil refining (desulfurization) and is available in very large quantities. Ammonia, on the other hand, is synthesized from nitrogen and methane in a high pressure, high temperature process. P1.31

Analysis of the process economy is summarized in the table. A multiplying factor of 3 was use in reaction R2 to eliminate generation/consumption of intermediates. compound

Lb $/lb $ (SF = 1/918) Glycerol stearate -1 -1 890 -890 -0.97 1.00 -0.97 You're Reading a Preview H2O -3 +3 18 Unlock Stearic acid +3 -3 284full access with a free trial. glycerol +1 +1 92 +92 0.100 1.10 +0.11 NaOH -3 -3 Download 40 -120 -0.13Trial 0.50 -0.065 With Free Sodium stearate +3 +3 306 +918 +1 x x To just break even, we need x - 0.97 + 0.11 - 0.065 = 0, or x = $0.925/lb soap. I found soap available in 18 lb quantities for about $2/pound on an internet site. You’ll spend about $2 for a 4 oz bar of soap at the drugstore. 1

2

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Master your semester with Scribd  Read Free Foron 30this Days Sign up to vote title P1.32  & The NewThisYork Times useful  Useful  Not problem is designed to encourage students to learn how to find and to use KirkSpecial offer for students: Only $4.99/month. Othmer and other reference books.

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HCN C6H8N2 H2 C6H16N2 sum

of 25

-2 -2 +1 -1 -4 -4 +1 +1

 

27

-54

2 -8 116 +116



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-54000

0.93 -50,220

-8000 +116,000 0

0.09 -720



-62,280

Raw material costs are $62,280/day at the desired production rate. Reaction pathway 2: The balanced chemical reactions are 2C 3H 3N + H 2  C6H8N 2 C 6H8N 2 + 4H 2  C 6H16N 2

(R1) (R2)

The process economy evaluation, at 116,000 lb/day, is summarized in a table. compound 1 C3H3N C6H8N2 H2 C6H16N2 sum

net

Mi

net Mi

-2 -2 +1 -1 -1 -4 -5 +1 +1

53

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Lb $/lb $ (SF = 1000) -106000 0.65 -68,900

2 -10 -10000 0.09 -900 116 You're +116Reading +116,000 a Preview 0 -69,800 Unlock full access with a free trial.

Raw material costs are $69,800/day at the desired production rate, or roughly 10% highe Download WithisFree Trial increasing the safety of the than in reaction pathway 1. However, no HCN required, process. The cost differential is insufficient to justify the increased risks associated with process 1. P1.34

Master your This semester problem requireswith studentsScribd to consider their own consumption patterns and to  Foron 30this Days Sign vote title estimate market size based on their own consumption,Read andupFree totolook up information on  & The Newmarket York sizeTimes in several common reference materials.  Useful  Not useful Special offer for students: Only $4.99/month. P1.35

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C3H3N +1 53 sum

+53

of 25

 

+100



 Search document



1.12 +112 -6.8

Pathway 1 loses money, and requires a highly toxic raw material (HCN). However, the atom economy is great (100%), and there are no byproducts to deal with. Pathway 2 Mi i Mi lb/day (SF = 100/53) C3H6 -1 42 -42 -79 NH3 -1 17 -17 -32 O2 -1.5 32 -48 -90.6 H2O +3 18 +54 +102 C3H3N +1 53 +53 +100 sum i

$/lb

$/day

0.19 -15 0.075 -2.4 0.015 -1.4 1.12

+112 +93.2

Pathway 2 has very favorable economics! Furthermore, the only byproduct is water, and ammonia and oxygen are relatively safe raw materials. However, the atom economy is poor (~50%). Pathway 3

You're Reading a Preview Unlock full access with a free trial.

Mi i Mi lb/day $/lb $/day (SF = Download 100/53) With Free Trial C2H4O -1 44 -44 -83 0.56 -46.5 HCN -1 27 -27 -51 0.60 -30.6 H2O +1 18 +18 +34 C3H3N +1 53 +53 +100 1.12 +112 sum +34.9 Read Free Foron 30this Days Sign up to vote title i

Master your semester with Scribd   & The NewTheYork Times Useful Not useful   process economics are pretty attractive, although not quite as much as Pathway 2. Cancel anytime.

Special offer for students: Only $4.99/month. The atom economy is better (75%) than Pathway 2 but not as good as Pathway 1.

However, pathway 3 does not avoid the use of the toxic reactant HCN.

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H 2NCHCH 3OH + HCN  H 2NCHCH 3CN + H 2O H 2NCHCH 3CN + 2H 2O  H 2NCHCH 3COOH + NH 3

(R2) (R3)

The process economy calculations are summarized in the table. I used values of $0.60 for HCN, and $0.455/lb for acetaldehyde. compound 1 CH3CHO NH3 C2H7NO HCN C3H6N2 H2O C3H7O2N

-1 -1 +1

2

3

net

-1

Mi net Mi Metric tons/yr $/ton $ (SF = 200/89) 44 -44 -98.9 1001 -99,000

+1 -1 -1 -1 +1 -1 +1 -2 -1 +1 +1

27

-27

-60.7

1320 -80,100

18 89

-18 +89

-40.5 +200

4750 +950,000

sum

0

+770,000

The atom economy is 100%, and the process economy is quite attractive. One disadvantage is the requirement for a highly toxic reactant, hydrogen cyanide. You're Reading a Preview Bucherer synthesis Unlock access withof a free trial. are: The 2 balanced chemical reactions forfull synthesis alanine CH 3CHO + NaCN + (NH 4 ) 2 CO 3  C 4 H 6O 2N 2 + NaOH + NH 3 + H 2O Download With Free Trial (R1) C 4 H 6O 2N 2 + 2H 2O  H 2NCHCH 3COOH + CO 2 + NH 3

(R2)

The process economy calculations are summarized in the table. I used values of $0.70 for NaCN, and $0.29/lb for ammonium carbonate.

Master your semester with Scribd Read Free Foron 30this Days Sign up to vote title   compound  M  M Metric tons/yr $/ton $ & The New York Times  Useful  Not useful (SF = 200/89) 1

2


CH3CHO

-1

net

i

net

i

Cancel anytime.

-1

44

-44

-98.9

1001

-99,000

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The atom economy is poor (43%). The process economy is quite attractive, although no as good as the Strecker synthesis. Handling all the byproducts would greatly increase the costs of running this process. Safety concerns are similar to the Strecker p rocess. P1.38

The aerobic decomposition of glucose to CO2 is a well-known reaction: C 6H12O 6 + 6O 2  6CO 2 + 6H 2O The unbalanced reaction from glucose to lactic acid is (guessing that water is a byproduct): C 6H12O 6 +  2O 2   3C 3H 6O 3 +  4 H 2O We can easily balance this: glucose-to-lactic acid conversion requires no oxygen (it is anaerobic – the reaction happens in the muscles during strenuous exercise). C 6H12O 6  2C3H 6O 3

Synthesis of bacteria from glucose will require ammonia and probably also oxygen. The unbalanced reaction is (guessing that water is a byproduct): C 6H12O 6 +  2O 2 +  3NH 3   4 CH1.666O 0.27N 0.20 +  5H 2O You're Reading a Preview We write element balances on C, H, O, N.with There Unlock fulland access a freeare trial.4 equations in 4 unknowns: 6 =  4

Download With Free Trial 12 + 3 3 = 1.666 4 + 2 5

6 + 2 2 = 0.27 4 +  5  3

= 0.20 4

Master your semester with Scribd We solve to find the stoichiometric coefficients & The New York Times C H O + 1.2NH  6CH O N 6 12 6 Special offer for students: Only $4.99/month.

3

1.666 0.27

Read Free Foron 30this Days Sign up to vote title Not useful  UsefulO +anytime.  0.79O 2 0.20 + 2.8H Cancel 2

product to balance this reaction!)

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8.8% of the glucose was consumed to grow bacteria, 20% was consumed to make lactic acid, and over 71% was oxidized to CO2. P1.39

Direct oxidation is the preferred method for converting ethylene-to-ethylene oxide. It would also be the preferred method for converting propylene-to-propylene oxide, but the reaction won’t “go” under typical processing conditions. Oxidation with hydrogen peroxide is too expensive. This problem considers whether a new process, in which hydrogen peroxide is generated from oxygen “in situ”, is economically attractive. Ethylene oxide by conventional process: 1 C 2H 4 + O 2  C 2H 4 O 2 Mi i Mi kg (SF = 1000/44) C2H4 -1 28 -28 -636.4 O2 -1/2 32 -16 -363.6 C2H4O +1 44 +44 +1000 sum i

$/kg

$

0.57 -363 0.033 -12 1.32 +1320 +945

You're Reading a Preview Propylene oxide by new process: Unlock full access with a free trial.

(R1) RAQ + O 2  RAHQ + H 2O 2 CH 4 + H 2O  CO + 3HDownload With Free(R2) Trial 2 (R3) RAHQ + H 2  RAQ (R4) C 3H 6 + H 2O 2  C 3H 6O + H 2O (Multiply stoichiometric coefficients for reaction R2 by 1/3 to avoid net generation/consumption of hydrogen.) Read Free Foron 30this Days Sign up to vote title

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The new process to make propylene oxide has very attractive economics, because propylene is cheaper than ethylene. A downside is the production of CO, which is a health hazard; however, CO can easily be oxidized to CO2. P1.40

Process with hydrogen peroxide: Balanced chemical equation is: C 6H10 + 4H 2O 2  C 6H10O 4 + 4H 2O The process economy is summarized as follows: kg/day (SF = 28100/146) -1 82 -82 -15,785 -4 34 -136 -26,175 +1 146 +146 +28,100 +4 18 +72 +13,860 i

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This is an attractive process based on environmental impact and on safety. Hydrogen peroxide is safer than nitric acid and produces no NO. Water is the only byproduct. You're Reading a Preview However, the economics are unfavorable, because hydrogen peroxide is a very expensiv oxidizing agent. Unlock full access with a free trial. British process Download With Free Trial Balanced chemical equation is: C 6H14 + 3O 2  C 6H10O 4 + 2H 2O The process economy is summarized as follows:

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desired product! We don’t know whether these byproducts are safe or toxic, whether the are useful or must be disposed of, whether they can easily or only with difficulty be separated from the desired product. These issues make this process much less appealing, unless the catalyst can be greatly improved so that more of the hexane reacts via the desired path. P1.41

Thinking about the raw material costs per mole is a useful way to begin this problem. $/lb lb/lbmol $/lbmol C2H4 0.27 28 7.56 C2H4Cl2 0.17 99 16.83 C2H2 1.22 26 31.72 Cl2 0.1 71 7.10 HCl 0.72 36.5 26.28 NaOH 1.13 40 45.20 C2H3Cl 0.22 62.5 13.75 Observations: Cl2 is cheaper per mole than HCl and would serve as a cheaper source of Cl. C2H2 and C2H4Cl2 are pricey; the best source of C for vinyl chloride would be ethylene. The price of HCl and C2H2 makes reaction 1 unattractive. NaOH costs more than the desired product; besides, reaction 5 hasavery poor atom economy. You're Reading Preview Unlock access with our a freecheaper trial. We could combine reactions 2 and 3, full which uses sources of C and Cl:

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Special offer for students: Only $4.99/month. The fractional atom economy is 62.5/(28+71) = 0.63. There is a loss of $0.91/lbmol of

vinyl chloride – and that’s before we consider operating costs or capital investment costs

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HCl O2 H2O

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+1 -2 -1 -0.5 -0.5 +1

This generation-consumption table would be greatly improved if we could find multiplying factors such that there was no net consumption of HCl, and no net generatio of dichloroethane. By inspection (or by working through the math) we find that multiplying reaction R3 by 2 will do the trick. 2

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The atom economy is 125/(56+71+16) = 0.87, which is quite good and an improvement over the combination of reactions 2 and 3. The costs above are for making 2 lbmol of vinyl chloride; dividing by 2 shows thatReading the profit is +$2.64/lbmol. Much more attractive You're a Preview than the earlier proposal! Unlock full access with a free trial.

Thus, the best pathway combines 3 reactions to take advantage of the cheaper raw materials, and to reduce byproduct generation. steps improve both atom and Download WithThese Free Trial process economy. P1.42

The two reactions for production of DSIDA from ammonia, formaldehyde, hydrogen cyanide and sodium hydroxide are  Read Free Foron 30this Days Sign up to vote title

Master your semester with Scribd & The New York NHTimes 3 + 2CH 2O + 2HCN  C 4 H 5N 3 + 2H 2O  Useful

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Special offer for students: Only C 4$4.99/month. H 5N 3 + 2NaOH + 2H 2O  C4 H 5O 4 NNa 2 + 2NH 3

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C4H5O4NNa2

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The cost for the raw materials is $828/day at a DSIDA production rate of 1770 lb/day. Synthesis of DSIDA from DEA and NaOH produces hydrogen as a byproduct: C 4 H11O 2N + 2NaOH  C 4 H 5O 4 NNa 2 + 4H 2

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The cost for the raw materials is $865/day at a DSIDA production rate of 1770 lb/day. The cost is very similar to the conventional process; the environmental and safety advantages of the new process are significant. Overall the new process is quite attractive Synthesis of DEA from ethylene, oxygen and ammonia proceeds via two reactions: You're Reading a Preview 1 C 2H 4 + O 2  C 2H 4 O (R4) Unlock full access with a free trial. 2 (R5) 2C 2H 4 O + NH 3  C 4 H11O 2N Download With Free Trial Multiplication of reaction R4 by a multiplying factor of 2 results in no net generation or consumption of ethylene oxide.  Master your semester Scribd   with    & The New York Times 4

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Chapter01

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