Reactors HW1 Solutions

8/7/2019 Reactors HW1 Solutions

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CHE 4151 ChemE Reactor Design HW # 1: Introduction and Review

(due Thursday, 1/27)

1.1) [Atkins & Jones, Problem 18-29, 18-49 modified]

From experiment (i.e. batch) you determine that dinitrogen pentoxide, N2O5,decomposes by first order kinetics with a rate constant of 3.7 x 10 -5 (units) at298K.

a) Give proper units on the rate constant.

b) How many minutes will it take for the N2O5 concentration to decreasefrom 0.0273 M to 0.0175 M?

c) You also found that the rate constant at 350K is 5.3 x 10-5 (units). Findthe activation energy of the N2O5 decomposition reaction.

ANSWER:

a) For first order reaction the rate equation is r = k CA, thus the unitsfor k are 1/s:

r (mol/vol*time) = k (1/s) CA

(mol/vol)

b) CA indicates the concentration of N2O5,

(rA ) =dCA

dt= kCA Batch

dCA

CACAi

CAf

= kdt0

t

lnCAfCAi

= kt

t=1

klnCAf

CAi

=

1

3.7x105s

1ln

0.0175M

0.0273M

=12,000s = 200min


2/9

c) Lets refer to k at T1 = 298K as k1 and k at T2 = 350K as k2,

lnk1

k2

=

Ea

R

1

T1

1

T2

Ea=

R lnk1

k2

1

T1

1

T2

= 8.314

J

mol*K

ln

3.7x105s1

5.3x105s1

1

298K

1

350K

= 6000J/mol= 6.0KJ/mol


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1.2) [Jennings 2009, modified]

A reaction, A B, is to be carried out in a constant volume batch reactorwith an initial concentration of reactant A, CA0. The reaction follows second

order kinetics. If the conversion of A is 20% at a 25

o

C and 80% conversionat 50 oC after the same amount of time and with the same initialconcentration, calculate the activation energy of the reaction?ANSWER:First solve for an expression for k using the reaction rate equation

(rA ) = kCA2

Batch

(rA ) =dCA

dt= kCA

2

dCA

CA2

CA 0

CAf

= kdt0

t

1

CA 0

1

CAf

= kt

For constant Volume, we know that

CAf=CA0(1 xA )

1

CA 01

1

(1 xA )

=

1

CA 0

xA

(1 xA )

= kt

k=xA

tCA 0(1 xA )

Lets refer to:k1 for conditions 1conversion of A is 20% after t minutes at a temperature of 25 oC

xA1 = 0.20, T1 = 298 K,

k1 =

xA1

tCA0(1 x

A1)

k2 for conditions 2 conversion of A is 80% after t minutes at atemperature of 50 oC

xA2 = 0.20, T1 = 323 K,

k2=

xA2

tCA0(1 x

A 2)


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lnk1

k2

=

Ea

R

1

T1

1

T2

lnk1

k2

= ln

xA1

tCA 0(1 x

A1)

xA2tC

A0(1 x

A2)

= ln

xA1

(1 xA1)

xA 2

(1 xA 2)

= lnxA1(1 xA2)

xA2(1 xA1)

Ea=

R lnxA1(1 x

A2)

xA2(1 x

A1)

1

T1

1

T2

=

8.314J

mol*K

ln

0.20(1 0.80)

0.80(1 0.20)

1

298K

1

323K

= 88751J/mol= 89KJ/mol


5/9

1.3) [pg. 102 Himmelblau, 6th edition]A common method used in manufacturing sodium hypochlorite bleach is bythe reaction:

Cl2+ 2NaOH NaCl+ NaOCl+ H

2O

Chlorine gas is bubbled through an aqueous solution of sodium hydroxide,after which the desired product is separated from the sodium chloride (a by-product of the reaction). A water-NaOH solution that contains 1145 lb of pureNaOH is reacted with 851 lb of gaseous chlorine. The NaOCl formed weighs618 lb.

a) What is the limiting reactant?b) What is the degree of completion of the reaction?c) What was the percent excess of the reactant used?

ANSWER:

MWNaOH

= 40lb / lbmol

MWCl2

= 70.9lb / lbmol

MWNaOCl

= 74.5lb / lbmol

a) Calculate how much NaOCl can be formed if all of the reactant is

consumed, the reactant yielding the smaller amount of product isthe limiting reactant,

NNaOH

vNaOH

=

1145lblbmol

40lb

2=14.3lbmol

NCl2

vCl2

=

851lblbmol

70.9lb

1=12.0lbmol

NNaOH

vNaOH

>

NCl2

vCl2

, therefore Cl2 is the limiting reactant


6/9

b) Or the conversion of the limiting reactant

NNaOCl=618lb

74.5lb / lbmol= 8.30lbmol= NCl2 (used)

=N

i0

Ni

Nk0

vkv i

=

NCl2

0 NCl2

f

NCl2 0

1

1

=

8.30lbmol

12.0lbmol= 0.69

69% degree of completion

c) How much of NaOH will remain in excess if 100% of Cl2 gas is used,

NNaOH consumed =

vNaOH

vCl2

NCl2 0 =

2

1

(12.0lbmol) = 24lbmol

NNaOH remaining =

NNaOH0

vNaOHvCl2

NCl2 0 = 28.6lbmol

2

1

(12.0lbmol) = 4.6lbmol

% Excess =

Ni0

vi

vk

Nk0

vi

vk

Nk0

(100) =4.6lbmol

24lbmol

(100) =19% excess of NaOH


7/9

1.4) [Problem 3-49, pg. 239 Himmelblau, 6th edition]A gas with the following molar flow rates (mol/s) is burned in 50% excess air(50% more oxygen) in a furnace. What is the composition of the flue gas (mol/s)?Assume complete combustion and air composition of 21% O2and 79% N2byvolume.

CH4 60

C2H6 20CO 5

O2 5

N2 10

ANSWER:Draw a diagram

Write the equation for complete combustion for each enteringcomponent (except N2 and O2):

CH4+ 2O

2CO

2+ 2H

2O

C2H

6+

7

2O

2 2CO

2+ 3H

2O

CO+1

2O

2CO

2

Find the total flow rate of water produced in the reactions (sinceexcess oxygen is provided we can assume that all of the carboncontaining compounds will completely combust, i.e. CH4, C6H6 and COare the limiting reactants in their respective equations):


8/9

60mol

CH4

s

2molH2O

1molCH4

=120mol

O2

s

20mol

C2H6

s

3molH2O

1molC2H6

= 60mol

O2

s

FH2O

= (120+ 60)mol

s=180

molH2O

s F

H2O_OUT

Find the total flow rate of CO2 produced:

60mol

CH4

s

1molCO2

1molCH4

= 60mol

CO2

s

20mol

C2H6

s

2molCO2

1molC2H6

= 40mol

CO2

s

5mol

CO

s

1molCO2

1molCO

= 5mol

CO2

s

FCO2

= (60+ 40 + 5)mol

s=105

molCO2

s F

CO2_OUT


9/9

Find the total flow rate of oxygen needed for complete combustion ofthe entering gas:

60molCH4

s

2molO21molCH4

=120molO2s

20molC2H6

s

72molO2

1molC2H6

= 70molO2s

5molCO

s

1

2molO2

1molCO= 2.5

molO2

s

FO2 = (120 + 70 + 2.5)mol

s=192.5

molO2s

for_50%_excess :

O2 _ IN:192.5molO2s

(1.5) = 288.75molO2 _ IN

s

O2 _OUT: 288.75molO2 _ IN

s 192.5( )

molO2s

+ 5molO2s

=101.25molO2 _OUT

s FO2 _OUT

N2 _ IN&OUT: 288.75molO2

s

0.79molN20.21molO2

+10molN2

s=1096.25

molN2

s FN2 _OUT

Reactors HW1 Solutions

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