thermo-chemistry enthalpy changes in chemical process part 2

Faculty-Wide Courses 2009-2010 Thermo-Chemistry Enthalpy Changes in Chemical Process

Eng. ISLAM IBRAHIM FEKRY Mob. 0109790568 Mail. [email protected]

SHEET No.2

PART 2 13. A sample of methane gas having a volume of 2.8 L at C and 1.65 atm was mixed

with a sample of oxygen gas having a volume of 35.0 L at C and 1.25 atm. The

mixture was then ignited to form carbon dioxide and water. Calculate

a- The volume of CO2 formed at a pressure of 2.5 atm and a temperature of C.

b- The % excess of O2.

sample of methane gas having a volume of 2.8 L at C and 1.65 atm

P V = N R T

( atm ) ( Lit ) = ( g mole ) (

) ( oK )

( 1.65 atm ) ( 2.8 L ) = ( ?? g mole ) ( 0.082

)( 298 oK )

( ?? g mole ) = (4.62)/(298*0.082)

( ?? g mole ) = (4.62)/(24.436)

( ?? g mole ) = 0.189 mole methane

sample of oxygen gas having a volume of 35.0 L at C and 1.25 atm

P V = N R T

( atm ) ( Lit ) = ( g mole ) (

) ( oK )

( 1.25 atm ) ( 35 L ) = ( ?? g mole ) ( 0.082

)( 304 oK )

( ?? g mole ) = (43.75)/(304*0.082)

( ?? g mole ) = (43.75)/(24.928)

( ?? g mole ) = 1.755 mole oxygen



The mixture was then ignited to form carbon dioxide and water

CH4 + O2 CO2 + 2 H2O

INPUT 0.189 1.755 0 0

REACTED 1X 2X X 2X

OUTPUT (0.189-1X) (1.755-2X) (X) (2X)

LIMITING COMPUND ( α= N/STICHOMETRIC ) THE SMALLEST α IS THE LIMITING

COMPUND 0.189 0.895

OUTPUT (0.189-1X= ZERO) (1.755-2X) (X) (2X)

0.189-1X=ZERO

X=0.189

OUTPUT (0.189-0.189) (1.755-0.378) (0.189) (0.378)

OUTPUT (ZERO) (1.377) (0.189) (0.378)

A- The volume of CO2 formed at a pressure of 2.5 atm and a temperature of C.

P V = N R T

( atm ) ( Lit ) = ( g mole ) (

) ( oK )

(2.5 atm ) ( ?? L) = (0.189 g mole) (0.082

) (398 oK)

(?? L) = (0.189 * 298 * 0.082)/ (2.5)

(?? L) = (6.168)/ (2.5)

(?? L) = 2.467 Liter

B- The % excess of O2.

The % excess of O2=

=

The % excess of O2= 1.377 / 1.755 = 78.46%



14. A compound containing only C, H and N yields the following data.

i. Complete combustion of 35.0 mg of the compound produced 33.5 mg of CO2 and

41.1 mg of H2O.

ii. A 62.5 mg sample of the compound was analyzed for nitrogen by the Dumas

method, giving 35.6 ml of N2 at 740 torr and C.

iii. The effusion rate of compound as a gas was measured and found to be 24.6

mL/min. The effusion rate of argon gas under identical conditions is 26.4 ml/min.

What is the molecular formula of compound?


41.1 mg of H2O.

% C in CO2 =

=

% C in Compund =

% H in H2O =

=

% H in Compund =

= 0.1304 = 13.04 %



No. mole N2 =

Mass N2 = No. mole N2 * M.wt N2 =0.001418*28 =0.03971 gm

% N in Compund

Carbon Hydrogen Nitrogen

% 26.10% 13.04% 63.53%

Bases of 100 gm 26.10 13.04 63.53

No. mole 26.10/12 13.04/1 63.53/14

2.175 13.04 4.5378

Empirical formula 2.175/2.175 13.04/2.175 4.5378/2.175

1 5.99 2.08

≈ 1 6 2



Which will give us an empirical formula of C1H6N2

Molecular Wight of the empirical formula is = ( (1*12)+(6*1)+(2*14) )=46

iii. The effusion rate of compound as a gas was measured and found to be (R2)24.6

ml/min. The effusion rate of argon gas under identical conditions is (R1) 26.4

ml/min.

Graham’s Law of Effusion

Mult. Factor =


The molecular formula of compound is C1H6N2



ANOTHER SOLUTION FOR Q. No. 14

I will assume that the formula of combusted fuel is CxHyNz

Make balance for complete combustion for CxHyNz

1- Balance the Carbon CxHyNz + O2 x CO2 + H2o + NO2

2- Balance the Hydrogen CxHyNz + O2 x CO2 +

H2o + NO2

3- Balance the Nytrogen CxHyNz + O2 x CO2 +

H2o + NO2

4- Balance the Oxygen CxHyNz +

O2 x CO2 +

H2o + NO2

CxHyNz +

O2 x CO2 +

H2o + NO2

1

x


41.1 mg of H2O.

A- Combustion of 35.0 mg of CxHyNz

No. mole CxHyNz = Mass CxHyNz / M.wt CxHyNz

=

CxHyNz +

O2 x CO2 +

H2o + NO2

reacted mole M

B- Produced 33.5 mg of CO2

No. mole CO2= Mass CO2/ M.wt CO2

=

=

=

=

=

Equan No. 1 =



C- Produced 41.1 mg of H2O

No. mole H2O = Mass H2O / M.wt H2O

=

=

=

=

=

=

Equan No. 2 =



No. mole CxHyNz = Mass CxHyNz / M.wt CxHyNz

=

CxHyNz

N2

reacted mole

No. mole N2= ??

P V = N R T

( atm ) ( Lit ) = ( g mole ) (

) ( oK )

(

atm ) ( L) = (??? g mole) (0.082

) (298 oK)

(??? g mole) = (

* )/ (298 * 0.082)

(??? g mole) = (0.03466)/ (24.436)

(??? g mole) = 1.4184 gmole =

1.4184 gmole =

1.4184 =

=

=

Equan No. 3 =



Note

Equan No. 1 =

Equan No. 2 =

Equan No. 3 = =

iii. The effusion rate of compound (CxHyNz) as a gas was measured and found to be

(R2)24.6 ml/min. The effusion rate of argon gas under identical conditions is (R1)

26.4 ml/min.

M. wt CxHyNz=M2=

Graham’s Law of Effusion

=

So That

Equan No. 1 =

Equan No. 2 =

Equan No. 3 = =


The molecular formula of compound is C1H6N2



15. An organic compound contains C, H, N and O. Combustion of 0.1023 g of the

compound in excess oxygen yielded 0.2766 g of CO2 and 0.0991 g of H2O. A sample

of 0.4831 g of the compound was analyzed for nitrogen by the Dumas method. At

STP 27.6 ml of dry N2 was obtained. In a third experiment, the density of the

compound as a gas was found to be 4.02 g/L at C and 256 torr. What are the

empirical and molecular formulas of the compound?

Solution: In order to determine the empirical formula, we need to first determine the mass percent of

each element in the compound from the information given above. To determine the % C and H, we need

to use the % of each in CO2 and H2O respectively, so that we can determine the actual mass of each from

the compound:

% C in CO2 =

=

% H in H2O =

=

% C in Compund =

% H in Compund =

= 0.1076 = 10.76 %

Use the ideal gas low to determine the moles of nitrogen from the sample, convert to grams, and

determine the % of N in the sample. STP is 1 atm, and 273 K

No. mole N2 =

Another solution to get No. mole N2

At STP, 27.6 mL of dry N2 was obtained,

so using the molar volume of gases at STP

0.0276 litres @ 1 mole / 22.4 litres = 0.001232 moles of N2

Mass N2 = No. mole N2 * M.wt N2 =0.001232*28 =0.03449 gm

% N in Compund

% O in Compund = 100 – (73.73+10.76+7.14) = 8.37%



finally, the % mass of oxygen in the compound is the remainder, which ends up being 8.37 %. Now, we

assume we have 100 gm of the compound, and determine the moles of each we have to determine what

the empirical formula is, after dividing everything through by the smallest number of moles:

Carbon Hydrogen Oxygen Nitrogen

% 73.73% 10.76% 8.37% 7.14%

Bases of 100 gm 73.73 10.76 8.37 7.14

No. mole 73.73/12 10.76/1 8.37/16 7.14/14

6.144 10.76 0.523 0.51

Empirical formula 6.144/0.51 10.76/0.51 0.523/0.51 0.51/0.51

12.04 21.09 1.025 1

≈ 12 21 1 1

Which will give us an empirical formula of C12H21NO

Now, we’ll use the density measurement to determine the molecular formula. We can rearrange the idal

gas law to get the M.wt of compound from its density

PV=NRT

PV=

R T

P =

R T

P = R T

=

=

Mult. Factor =

So that the molecular formula is C24H42N2O2



ANOTHER SOLUTION FOR Q. No. 15

I will assume that the formula of combusted fuel is CxHyOzNf

Make balance for complete combustion for CxHyOzNf

1- Balance the Carbon CxHyOzNf + O2 x CO2 + H2o + NO2

2- Balance the Hydrogen CxHyOzNf + O2 x CO2 +

H2o + NO2

3- Balance the Nytrogen CxHyOzNf + O2 x CO2 +

H2o + NO2

4- Balance the Oxygen CxHyOzNf +

O2 x CO2 +

H2o + NO2

CxHyOzNf +

O2 x CO2 +

H2o +

NO2

1

x

Combustion of 0.1023 g of the compound in excess oxygen yielded 0.2766 g of CO2

and 0.0991 g of H2O

A- Combustion of 0.1023 g of CxHyOzNf

No. mole CxHyOzNf = Mass CxHyOzNf / M.wt CxHyOzNf

=

CxHyNz +

O2 x CO2 +

H2o + NO2

reacted mole M

B- Produced 0.2766 g of CO2

No. mole CO2= Mass CO2/ M.wt CO2

=

=

=

=

Equan No. 1 =



C- Produced 0.0991 g of H2O

No. mole H2O = Mass H2O / M.wt H2O

=

=

=

=

Equan No. 2 =

A sample of 0.4831 g of the compound was analyzed for nitrogen by the Dumas

method. At STP 27.6 ml of dry N2 was obtained

No. mole N2 =

No. mole CxHyOzNf = Mass CxHyOzNf / M.wt CxHyOzNf

=

CxHyNz

N2

reacted mole

0.001232 gmole =

0.001232 =

=

=

Equan No. 3 =



Now, we’ll use the density measurement to determine the molecular formula. We can rearrange the idal

gas law to get the M.wt of compound from its density

PV=NRT

PV=

R T

P =

R T

P = R T

=

=

=

So That

Equan No. 1 =

Equan No. 2 =

Equan No. 3 = =

The amount of Oxygen from any equation from the three previous substite the

values of x, y, and f to get the value of z

Equan No. 1 =

=

=

=

=

=

=


The molecular formula of compound is C24H42N2O2



16. Consider a sample of a hydrocarbon ( a compound consisting of only carbon and

hydrogen ) “1st stat” at 0.959 atm and 298 K. upon combustion the entire sample in

oxygen, you collect a mixture of a gaseous carbon dioxide and water vapor at “2nd

stat” 1.51 atm and 375 K. This mixture has a density of 1.391 g/L and occupies a

volume four times as large as that of the pure hydrocarbon. Determine the

molecular formula of hydrocarbon.

I will assume that the formula of combusted fuel is CxHyNz

Make balance for complete combustion for CxHyNz

1- Balance the Carbon CxHy + O2 x CO2 + H2o

2- Balance the Hydrogen CxHy + O2 x CO2 +

H2o

3- Balance the Oxygen CxHy +

O2 x CO2 +

H2o

Bases 1 Mole of fuel

PV=NRT

PV=

R T

P =

R T

“1st stat” “2st stat”

P1=0.959 atm P2=1.51 atm

T1=298 K T2=375 K

N1=1 (assumed) N2= ??

V1=

V1=25.48lit V2=4*25.48=101.92 lit

V2=

=

Equation No. 1

ρ1=?? ρ 2= 1.391 g/l

= Mass/volume

= (x*44+

)/101.92=1.391

= (x*44+

) =141.770

= (x*44+ ) =141.770 Equation No.2



From the two previous Equation I can get the value of X and Y

Equation No. 1 Equation No. 2 =141.770

So That X≈2 and Y≈6

The molecular formula of compound is C2H6



17. A plane uses C8H10 as a fuel. Assume complete combustion, calculate the % excess

air needed such that water vapor in the combustion products is about to condense

at C and total pressure 736 mmHg. Calculate the density of the combustion

products and the partial pressure of the CO2. (Saturated vapor pressure of water at

C is 92.4 mmHg).



18. According to a law that limits the % CO2 in combustion product on dry basis to 10%,

Calculate the % excess of air needed for the complete combustion of C5 H12.

complete combustion of C5 H12

Theoretical amount of Oxygen : C5 H12 + 8 O2 5 CO2 + 6 H2O

Theoretical amount of Air : C5 H12 + 8 O2 + 32 N2 5 CO2 + 6 H2O +32N2

Excess amount of Air :

C5 H12 + (8+x) O2 + (32+4x) N2 5 CO2 + 6 H2O + (32+4x)N2 + x O2

% CO2 in combustion product on dry basis to 10%

Excess amount of Air :

C5 H12 + (8+2.6)O2 + (32+4*2.6)N2 5 CO2 + 6 H2O + (32+4*2.6)N2 + 2.6 O2

C5 H12 + (10.6)O2 + (42.4)N2 5 CO2 + 6 H2O + (42.4)N2 + 2.6 O2

Calculate the % excess of air needed for the complete combustion of C5 H12

The % excess of Air=

=

The % excess of O2= =

=



19. Calculate the quantity of air needed for the combustion of 220 gm of propane (C3H8)

in a furnace that uses 20% excess air. Calculate the volume of air at a pressure of

750 mmHg and a temperature o , Calculate the volume and density of the

combustion products and the partial pressure of CO2

1st step you have to write a balanced combustion reaction (assume that we use Oxygen only in

combustion)

C3H8 + 5 O2 3CO2 + 4H2O

from previous reaction we notes that every 1 mole of propane need 5 mole of Oxygen to make a

complete combustion.

2nd step you have to write a balanced combustion reaction using theoretical amount of air

(ratio between Oxygen and Nitrogen in air is 21:79 ≈ 1:4, this mean that every 1 mole of

Oxygen have 4 mole of nitrogen)

C3H8 + 5 O2 + 20 N2 3CO2 + 4H2O +20 N2

3rd step add the 20% excess to the theoretical amount of air ( ideal amount + 0.2 * ideal

amount )

C3H8 + 5 O2 + 20 N2 + 0.2 (5 O2 + 20 N2 ) 3CO2 + 4H2O +20 N2+0.2 (5 O2 + 20 N2)

C3H8 + 5 O2 + 20 N2 + 1 O2 + 4 N2 3CO2 + 4H2O +20 N2+1 O2 + 4 N2

C3H8 + 6 O2 + 24 N2 3CO2 + 4H2O +24 N2+1 O2

4th step Get the No. of mole of 220 gm of propane (C3H8)

No. of mole = Mass / M.Wt

= 220/(3*12+8)

= 220/(44)

= 5

5th step Calculate the amount of air I need it to burn 5 mole of propane

C3H8 + 6 O2 + 24 N2 3CO2 + 4H2O +24 N2+1 O2

1 6 24 3 4 24 1

5 30 120 15 20 120 5

No. mole of air required to make complete combustion = 25+100= 125 gmole

No. mole of air entered to furnace = 30+120= 150 gmole



Calculate the volume of air at a pressure of 750 mmHg and a temperature o

P V = N R T

( atm ) ( Lit ) = ( g mole )

) ( oK )

atm ) ( ?? Lit) = ( 150 gmole ) (0.08

) ( 305oK )

( ?? Lit) = (150 gmole ) (0.08

) ( 305oK ) / (0.986 atm)

( ?? Lit) = (3,751.5 atm. lit) / (0.986 atm)

(?? Lit) = (3,751.5 atm. lit) / (0.986 atm)

(?? Lit) = 3,804.766 lit

(Vair entered Lit) = 3,804.766 lit

Calculate the volume and density of the combustion products and the partial

pressure of CO2 at a pressure of 730 mmHg and a temperature of

A- Volume of the combustion products

P V = N R T

( atm ) ( Lit ) = ( g mole )

) ( oK )

atm ) ( ?? Lit) = (15+20+120+5 gmole ) (0.08

) ( 393oK )

( ?? Lit) = (160 gmole ) (0.08

) ( 393oK ) / (0.960 atm)

( ?? Lit) = (5,156.16 atm. lit) / (0.960 atm)

(?? Lit) = (5,156.16 atm. lit) / (0.960 atm)

(?? Lit) = 5,371 lit

(VCombustion Product Lit) = 5,371 lit

B- Density of the combustion products

Density =

C-Partial pressure of CO2

P CO2 = Mole Fraction * Pressure Total

P CO2 =

P CO2 =



20. Methane (CH4) gas flow into a combustion chamber at a rate of 200 lit/min at 1.50

atm and ambient temperature. Air is added to the chamber at 1.00 atm and the same

temperature, and the gases are ignited.

a- Amount of oxygen to ensure complete combustion of CH4 to CO2 (g) and H2O (g),

three times as much oxygen as is necessary is reacted. Assuming air is 21 mole

percent O2 and 79 mole percent N2 Calculate the flow rate of air necessary to deliver

the require.

b- Under the conditions in part a, combustion of methane was not complete as a

mixture of CO2 (g) and CO (g) was produced. It was determined that 95.0 % of the

carbon in the exhaust gas was present in CO2. The remainder was present as carbon

in CO. Calculate the composition of exhaust gas in terms of mole fraction of CO, CO2,

O2, N2 and H2O. Assume CH4 is completely reacted and N2 is unreached.

Calculate the Mole flow rate to get Mole flow rate of air from the balanced Combustion

reaction

Methane flow conditions ; Pressure =1.50 atm

Temperature= C=25+273= 298

Rate of Methane=

P V = N R T

( atm ) ( Lit/min ) = ( g mole/min ) (

) ( oK )

( atm ) ( 200 Lit/min) = (??? gmole/min ) (0.082

) ( 298oK )

(??? gmole/min ) = ( atm) * (200 Lit/min)/ (0.082

) ( 298oK )

(??? gmole/min) = (300 atm. Lit/min) / (24.436

)

(??? gmole/min) = (12.277 g mole/min)

CH4 + 2 O2 CO2 + 2 H2O

1 2 1 2

Theoretical O2 12.277 2*12.277 12.277 2*12.277

CH4 + 6 O2 CO2 + 2 H2O + 4 O2

Real O2 12.277 3*24.54 12.277 2*12.277 49.108

Real O2 12.277 73.662 12.277 24.554 49.108



Calculate the flow rate of air necessary to deliver the require

CH4 + 6 O2 + 22.57 N2 CO2 + 2 H2O + 22.57 N2 + 4 O2

Real Air 12.277 73.662 12.277 24.554 277.109 49.108

Moler flow rate of air needed is 73.662+277.109=350.771 g mole/min

Air flow conditions ; Pressure =1.00 atm

Temperature= C=25+273= 298

P V = N R T

( atm ) ( Lit/min ) = ( g mole/min ) (

) ( oK )

( atm ) ( ??? Lit/min) = (350.771 gmole/min ) (0.082

) ( 298oK )

(???Lit/min) = (350.771 gmole/min ) (0.082

) ( 298oK )/ (1 atm)

(???Lit/min) = (8,571.4 atm. Lit/min) / (1 )

(???Lit/min) = (8,571.4 Lit/min)

A- Amount of oxygen

Theoretical O2 is 24.554 gmole/min

Real O2 is 73.662 gmole/min

B- Calculate the composition of exhaust gas in terms of mole fraction of CO, CO2, O2,

N2 and H2O. Assume CH4 is completely reacted and N2 is unreached

CH4 + O2 + N2 CO+ CO2 + 2 H2O + 8 N2

thermo-chemistry enthalpy changes in chemical process part 2

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