Tutorial Chemical Reaction Engineering · Tutorial CRE: Chemical Thermodynamics II Summary and...

1Institute of Process Engineering, G25-217, [email protected]

Tutorial Chemical Reaction Engineering:

Dipl.-Ing. Andreas Jörke1

24-May-16

4. Chemical Thermodynamics II

Tutorial CRE: Chemical Thermodynamics II

Dehydrogenation of methane

Ideal gases, initial composition: pure methane

Questions:

Equilibrium constant Kp ?

Extend of reaction ξ eq and conversion X eq at equilibrium?

Impact of a temperature rise of 100 K on the equilibrium state?

Impact of a pressure increase of 1 bar on the equilibrium state?

24-May-16 CRE: Chemical Thermodynamics II 2

4 2 2 22CH C H 3H 1.0 bar 1300 Kp T

o

F iH o / kJ/mol iS p,

4

2 2

2

/ J/ mol K / J/ mol K

CH 74.52 186.38 64.62

C H 228.20 200.81 62.75

H 0 130.60 30.17

ic


Enthalpy of reaction / heat of reaction? Hess´s law

At elevated temperatures? Kirchhoff´s law!


2 2 2 2 4

o o

R F i

o o o

F C H F H F C H3 2

i

i

H T H T

H T H T H T

o o

F Fi iH T H

p,

298 K

o o

F F

ˆd

T

i

i i

c T

H T H

0 o

p, Fi ic T T H p, 1300 K 298 Kic

o

R 1300 K 228.20 62.87 3 0 30.23 2 74.52 64.75 kJ/mol

401.3 kJ/mol

H


Entropy change of reaction? Hess´s law



o o

i iS T S

p,

298 K

o o

ˆdˆ

Ti

i i

cT

T

S T S

o

p, 0lni i

Tc S

T

p,

1300 Kln

298 Kic

o

R 1300 K 200.81 92.43 3 130.60 44.44 2 186.38 95.19 J/ mol K

255.22 J/ mol K

S

2 2 2 2 4

o o

R i

o o o

C H H C H3 2

i

i

S T S T

S T S T S T


Gibbs free energy change of reaction? Hess´s law or legendre transformation of the Gibbs-Helmholtz-Equation

We use the Gibbs-Helmholtz-Equation with the already calculated thermodynamic quantities!


o o o

R F i i

o o o

R R R

i i

i i

G T G T T

G T H T T S T

o o o

R R R

-3

1300 K 1300 K 1300 K 1300 K

401.3 kJ/mol 1300 K 255.22 10 kJ/ mol K

69.51 kJ/mol

G H S

o o

R R 3

p

1300 Kexp exp 1.61 10

R R 1300 K

G T GK

T




2 4 2

4

2 3

C H H 3

p o o o o 2

CH

1.61 10

i i i

iii i

i

i i i

x xp x p p pK x

p p p p x

ii

j

j

nx

n

with 0

i i in n and

2 2

2

4

C H

H

CH

2 mol 2

3

2 mol 2

2 mol 2

2 mol 2

x

x

x

follows

2 mol 2 3 2 mol 2j

j

n

2 4 2

4

2 2 23 4C H H

p 4 2o 2 o

CH

2

o

27 1 mol

2 mol 2 1 mol

x xp pK

p x p

p

p

1 2427 1 mol

4

16 1 mol 2

3

21.61 10

1 mol




43

p 2 2

271.61 10

16 1 mol 1 molK

2

peq

p

1 mol0.1731 mol

27

16

K

K

4 4

4 4

0 eq eqCH CHeq

0 0

CH CH

2

0.1731 17.31%

n nX

n n


Enthalpy of reaction / heat of reaction? Hess´s law



2 2 2 2 4

o o

R F i

o o o

F C H F H F C H3 2

i

i

H T H T

H T H T H T

o o

F Fi iH T H

p,

298 K

o o

F F

ˆd

T

i

i i

c T

H T H

0 o

p, Fi ic T T H p, 1400 K 298 Kic

o

R 1400 K 228.20 69.15 3 0 33.25 2 74.52 71.21 kJ/mol

403.7 kJ/mol

H


Entropy change of reaction? Hess´s law



o o

i iS T S

p,

298 K

o o

ˆdˆ

Ti

i i

cT

T

S T S

o

p, 0lni i

Tc S

T

p,

1400 Kln

298 Kic

o

R 1400 K 200.81 97.08 3 130.60 46.68 2 186.38 99.98 J/ mol K

257.01 J/ mol K

S

2 2 2 2 4

o o

R i

o o o

C H H C H3 2

i

i

S T S T

S T S T S T


Gibbs free energy change of reaction? Hess´s law or legendre transformation of the Gibbs-Helmholtz-Equation

We use the Gibbs-Helmholtz-Equation with the already calculated thermodynamic quantities!


o o o

R F i i

o o o

R R R

i i

i i

G T G T T

G T H T T S T

o o o

R R R

-3

1400 K 1400 K 1400 K 1400 K

403.7 kJ/mol 1400 K 257.01 10 kJ/ mol K

43.89 kJ/mol

G H S

o o

R R 2

p

1400 Kexp exp 2.305 10

R R 1400 K

G T GK

T


Extend of reaction ξ eq and conversion X eq at equilibrium (1400K)?


42

p 2 2

272.305 10

16 1 mol 1 molK

2

peq

p

1 mol0.3235 mol

27

16

K

K

4 4

4 4

0 eq eqCH CHeq

0 0

CH CH

2

0.3235 32.35%

n nX

n n

Increasing conversion! Endothermic reaction!


Extend of reaction ξ eq and conversion X eq at equilibrium (2 bar)?


43

p 2 2

2 bar 271.61 10

1 bar 16 1 mol 1 molK

2

peq

p

1 mol0.1233 mol

27

4

K

K

4 4

4 4

0 eq eqCH CHeq

0 0

CH CH

2

0.1233 12.33%

n nX

n n

Decreasing conversion! Expanding reaction!


Summary and Recipe:

1. Find thermodynamic properties of formation for pure substances in thermodynamic textbooks / tables

2. Calculate heat of reaction, entropy and Gibbs free energy change of reaction

3. Calculate equilibrium constant and formulate mass action law

4. Introduce equilibrium conversion / extend of reaction to reduce the number of unknowns solve equation and find equilibrium conversion


o

FG o

F, H o, S

o o o

R R RG T H T T S T

o

R

p o o oexp

R

i i i

iii i

i

i i i

G Tp x p pK x

p p p T

0

0

i ii

i

n nX

n

ji

i j

nn

p influence T influence

ii

j

j

nx

n


Sulfur dioxide oxidation sulfuric acid production

Ideal gases

Questions:

Conversion of SO2 and molar fraction of SO3 at equilibrium?

Test two different initial conditions:

a) SO2:O2 = 2:1

b) SO2:Air = 1:5


2 2 32SO O 2SO 1.0 bar 600,700,800,900,1000 Kp T

2 3 2

o o o

F SO F SO F O / K / kJ/mol / kJ/mol / kJ/mol

600 300.29 342.66 0

700 299.45 332.39 0

800 298.38 321.95 0

900 295.97 310.20 0

1000 288.64 293.57 0

T G G G

[1] Barin, Ihsan et al., Thermochemical data of pure substances, VCH-Wiley, 2nd Ed., 1993


Gibbs free energy change of reaction? Hess´s law


o o o

R F i ii i

i i

G T G T T

o

R

p expR

G TK

T

o

R p

7

4

3

1

/ K / kJ/mol

600 84.73 2.38 10

700 65.88 8.25 10

800 47.12 1.19 10

900 28.45 4.48 10

1000 9.86 3.27

T G K


SO2:O2 = 2:1


3

2 2

1 2

SO

p o o o o 2

SO O

i i i

iii i

i

i i i

xp x p p pK x

p p p p x x

ii

j

j

nx

n

with 0 1i i in n X and follows

2 2 2 2

2 2 2 2 2 2

2 2

3 3 3 2 2 2

3 2

2 2 2 2

0

SO SO SO SO

0 0

O O O SO SO SO

O SO

0 0

SO SO SO SO SO SO

SO SO

SO SO SO SO

1 2 1

1 1 2 2

1

2 2 2 2

2

2 1 1 2 3j

j

n n X X

n n n n n X

n X

n n n n n X

n X

n X X X X

2

2

2

2

2

2

2

3

2

SO

SO

SO

SO

O

SO

SO

SO

SO

12

3

1

3

2

3

Xx

X

Xx

X

Xx

X

ji

i j

nn

and


SO2:O2 = 2:1

Solve numerically/graphically for all temperatures T


1

p o

pK

p

2 23

2 22

122SO ,eq SO ,eqSO

32

SO OSO ,eq

3

1

X Xx

x x X

2 3SO ,eq SO ,eq/ K / / %-mol

600 99.6 99.3

700 97.4 96.2

800 88.9 84.2

900 70.8 61.8

1000 45.5 35.8

T X x


SO2:Air = 1:5 20% O2, 80% N2


3

2 2

1 2

SO

p o o o o 2

SO O

i i i

iii i

i

i i i

xp x p p pK x

p p p p x x

ii

j

j

nx

n

with 0 1i i in n X and follows

2 2 2 2

2 2 2 2 2 2

2 2

3 3 3 2 2 2

3 2

2 2 2 2

0

SO SO SO SO

0 0

O O O SO SO SO

O SO

0 0

SO SO SO SO SO SO

SO SO

SO SO SO SO

1 2 1

1 1 2 2

2

2 2 2 2

2

2 1 2 2 8 12j

j

n n X X

n n n n n X

n X

n n n n n X

n X

n X X X X

2

2

2

2

2

2

2

3

2

SO

SO

SO

SO

O

SO

SO

SO

SO

12

12

2

12

2

12

Xx

X

Xx

X

Xx

X

ji

i j

nn

and

2 2 2 3 2 22SO 2O 8N 2SO O 8N


SO2:Air = 1:5

Solve numerically/graphically for all temperatures T


1

p o

pK

p

2 23

2 22 2

122SO ,eq SO ,eqSO

22

SO OSO ,eq SO ,eq

12

1 2

X Xx

x x X X

2 3SO ,eq SO ,eq/ K / / %-mol

600 100 18.2

700 99.0 18.0

800 91.6 16.5

900 70.0 12.4

1000 40.8 7.0

T X x


Summary:

Using pure oxygen vs. air makes no difference from the thermodynamic point of view

Oxygen is more expensive than air

Using air results in a diluted product Costly separation tasks! Bigger apparatuses, compressors, electrical power etc. Trade-off!



Explanation for no significant difference in using oxygen/air:

1) The O2 excess shifts the equilibrium to the product side X eq increases

2) Adding inert diluent (N2) has the same effect like decreasing the total/partial pressures reaction with decreasing amount of substance shifts equilibrium to reactand side X eq decreases

Both effects cancel out each other no net effect on X eq !


X eq

T

equilibrium kinetics

feasible region

Tutorial Chemical Reaction Engineering · Tutorial CRE: Chemical Thermodynamics II Summary and...

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