example ch 8

8/12/2019 Example Ch 8

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Consider the reaction: C

Example 8.1

2H4(g) + H2O(g)→ C2H5OH(g). If an equimolar mixture of

ethylene and water vapor is fed to a reactor which is maintained at 500 K and 40 bar

determine the degree of conversion, assuming that the reaction mixture behaves like an idealgas. Assume the following ideal gas specific heat data: C p

ig = a + bT + cT2 + dT3 + eT –2

Species

(J/mol); T(K).

a bx10 cx10 dx10 ex10-

C2H 20.6914 205.346 – 99.793 18.825 -

H2 4.196O 154.565 – 81.076 16.813 -

C2H5 28.850OH 12.055 - - 1.006

From standard tables ΔH0R,298 ≈ -52.7 KJ; ΔG0

R,298

Now

= 14.5 KJ

0 0 0

298

298

T

T P H H C dT ∆ = ∆ + ∆∫ ---------------------- (A)

0 0 2 3 2

, /

P i P iC C a bT CT dT e T α ∆ = ∑ = ∆ + ∆ + ∆ + ∆ + ∆ ------------- (B)

Where:

, ,i i i ia a b b etcα α ∆ = ∑ ∆ = ∑

For example Δa = 20.691 – 4.196 -28.850 = -12.355Δb = (205.346 – 154.565 – 12.055) x 10-3 = 3.8726 x 10-2

Similarly Δc = -1.8717 x 10

-5 ; Δd = 2.012 x 10-9 ; Δe = -1.006 x 10

Putting B in A and integrating A we get:

5

2 50 2 3

94 5

3.8726 10 1.8717 1050.944( ) 12.355

2 3

2.012 101.006 10

4

T H KJ T T T

T T

− −

−

× ×∆ = − − + − +

×+ ×

…………..(C)

By Vant Hoff equation:0 0

2

( )T T d G RT H

dT RT ∆ ∆= −

00 0

298

2

298(298)

T

T T GG H

dT RT R RT

∆∆ ∆∴ − = − ∫ ------------------------------------------------- D

We already know 0

T H ∆ from (C); putting C in D and integrating we obtain:

2 5 90 2 3 4

5

3.8726 10 1.8718 10 2.012 1050.944 12.355 ln

2 6 12

1.006 10

56.6812

T G T T T T T

T T

− − −× × ×∆ = − + − + −

×

+ +



Putting T=500k, 0

T G∆ = 11.43kJ

Consider the following reaction: A(g) + B(g) = C(g) + 3D(g) CH

Example 8.2

4 + H2O→ CO + 3H

Intially the following number of moles are introduced in the reactor. Obtian the mole fraction

expressions in terms of reaction coordinate.

2

0, An = 2 mol,0, Bn = 1 mol,

0,C n = 1 mol0, Dn = 4 mol

i = -i = -1-1+1+3 = 2α α ∑

nooi

i

n∑= = 2+ 1 +1 + 4 = 8

oi iii

o

nn yn n

α ξ αξ

+= =+

∴∴ 2

8 2

A y

ξ

ξ

−=

+;

1

8 2 B

y ξ

ξ

−=

+;

1

8 2C

y ξ

ξ

+=

+;

2

4+ 8 2

H y

ξ

ξ =

+

Consider the following simultaneous reactions. Express the reaction mixture composition as

function of the reaction co-ordinates. All reactants and products are gaseous.

Example 8.3

A + B = C + 3D ..(1)

A + 2B = E + 4D ..(2)

Initial number of moles:

0, An = 2 mol;0, Bn = 3 mol

Let the reaction co-ordinates for each reaction be1 2

and ξ ξ respectively.

j A B C D E α j=Σαi,j

1 - 1 -1 1 3 0 2

2 -2 -2 0 4 1 2

,

io j i j j

i

o j j j

n y

n

α ξ

α ξ

+=

+

∑∑

; no

∴

= 2+3 = 5

1 2 1 2 1

1 2 1 2 1 2

2 3 2; ;

5 2 2 5 2 2 5 2 2 A B C

y y yξ ξ ξ ξ ξ

ξ ξ ξ ξ ξ ξ

− − − −= = =

+ + + + + +



1 2 2

1 2 1 2

3 4 ;5 2 2 5 2 2

D E y y

ξ ξ ξ

ξ ξ ξ ξ

+= =

+ + + +

Consider the reaction : C

Example 8.4

2H4(g) + H2O(g) → C2H5OH(g). If an equimolar mixture of


determine the equilibrium constant, assuming that the reaction mixture behaves like an ideal

gas. Assume the following ideal gas specific heat data: C pig = a + bT + cT2 + dT3 + eT –2

Species

(J/mol); T(K).

a bx10 cx103

dx106 ex10

9

-5

C2H 20.691

4205.346 – 99.793 18.825 -

H2 4.196O 154.565 – 81.076 16.813 -

C2H5 28.850OH 12.055 - - 1.006

From standard tables ΔH0R,298 ≈ -52.7 KJ; ΔG0

R,298

Now

= 14.5 KJ

0 0 0

298

298

T

T P H H C dT ∆ = ∆ + ∆∫ ---------------------- (A)

0 0 2 3 2

, /P i P iC C a bT CT dT e T α ∆ = ∑ = ∆ + ∆ + ∆ + ∆ + ∆

------------- (B)

Where:

, ,i i i ia a b b etcα α ∆ = ∑ ∆ = ∑

For example Δa = 20.691 – 4.196 -28.850 = -12.355

Δb = (205.346 – 154.565 – 12.055) x 10-3 = 3.8726 x 10-2


-5 ; Δd = 2.012 x 10-9 ; Δe = -1.006 x 10


5

2 50 2 3

94 5

3.8726 10 1.8717 1050.944( ) 12.3552 3

2.012 101.006 10

4

T H KJ T T T

T T

− −

−

× ×∆ = − − + − +

×+ ×

…………..(C)

By Vant Hoff equation:0 0

2

( )T T

d G RT H

dT RT

∆ ∆= −

00 0

298

2

298(298)

T

T T GG H

dT RT R RT

∆∆ ∆∴ − = − ∫ ------------------------------------------------- D

We already know 0T H ∆ from (C); putting C in D and integrating we obtain:



2 5 90 2 3 4

5

3.8726 10 1.8718 10 2.012 1050.944 12.355 ln

2 6 12

1.006 1056.681

2

T G T T T T T

T T

− − −× × ×∆ = − + − + −

×+ +

Putting T=500k, 0T G∆ = 11.43kJ

Hence K 500

= exp [ -1143/8.314 x 1000] = 0.064

Consider the reaction: C

Example 8.5

2H4(g) + H2O(g)→ C2H5OH(g). If an equimolar mixture of


determine the degree of conversion, assuming that the reaction mixture behaves like an ideal

gas. Assume the following ideal gas specific heat data: C pig = a + bT + cT2 + dT3 + eT –2

Species

(J/mol); T(K).

a bx10 cx10 dx10 ex10-

C2H 20.6914 205.346 – 99.793 18.825 -

H2 4.196O 154.565 – 81.076 16.813 -

C2H5 28.850OH 12.055 - - 1.006

From standard tables ΔH0R,298 ≈ -52.7 KJ; ΔG0R,298

Now

= 14.5 KJ

0 0 0

298

298

T

T P H H C dT ∆ = ∆ + ∆∫ ---------------------- (A)

0 0 2 3 2

, /

P i P iC C a bT CT dT e T α ∆ = ∑ = ∆ + ∆ + ∆ + ∆ + ∆ ------------- (B)

Where:

, ,i i i ia a b b etcα α ∆ = ∑ ∆ = ∑

For example Δa = 20.691 – 4.196 -28.850 = -12.355

Δb = (205.346 – 154.565 – 12.055) x 10-3 = 3.8726 x 10-2


-5 ; Δd = 2.012 x 10-9 ; Δe = -1.006 x 10


5

2 50 2 3

94 5

3.8726 10 1.8717 1050.944( ) 12.355

2 3

2.012 101.006 10

4

T H KJ T T T

T T

− −

−

× ×∆ = − − + − +

×+ ×

…………..(C)

By Vant Hoff equation:

0 0

2

( )T T d G RT H

dT RT

∆ ∆

= −



00 0

298

2

298(298)

T

T T GG H

dT RT R RT

∆∆ ∆∴ − = − ∫ ------------------------------------------------- D

We already know 0

T H ∆ from (C); putting C in D and integrating we obtain:

2 5 9

0 2 3 4

5

3.8726 10 1.8718 10 2.012 1050.944 12.355 ln2 6 12

1.006 1056.681

2

T G T T T T T

T T

− − −

× × ×∆ = − + − + −

×+ +

Putting T=500k, 0

T G∆ = 11.43kJ

Hence K 500

α = Σα

= exp [ -1143/8.314 x 1000] = 0.064

i

∴

= 1-1-1 = -1

K = K ϕK yPα ; K ϕ

P = 40bar

=1(since ideal gas assumption is made)

Component ni0 n(exit) yi

C2H4 1 1-ε (1-ε )/(2-ε)

(exit)

H2O 1 1-ε (1-ε )/(2-ε)

C2H5OH 0 ε (ε )/(2-ε)

nt

( ) ( ){ }

( )

( )2 5 2 4 2 2 2

2(2 )

11 2 y C H OH C H H O

K K y y yξ ξ ξ ξ

ξ ξ ξ

−−∴ = = = =

−− −

(at exit) = 2-ε

Now K = K ϕK yPα = (1) K y P-1

( ) ( )2

40 ; 2 1 y yK K K ξ ξ ξ ∴ = = − −

( )

( ) 5002

240 40 0.064 2.56

1 yK K

ξ ξ

ξ

−∴ = = × = × =

−

On solving ξ = 0.47

Thus ( ) ( )2 4 2

1 2 0.3464C H H O y yξ ξ = − − = =

( )2 5

2 0.3072C H OH

y ξ ξ = − =

The following two independent reactions occur in the steam cracking of methane at 1000 K

and 1 bar: CH

Example 8. 6

4(g) + H2O(g) → CO(g) + 3H2(g); and CO(g) + H2O(g) → CO2(g) + H2(g).

Assuming ideal gas behaviour determine the equilibrium composition of the gas leaving the



reactor if an equimolar mixture of CH4 and H2

Let ξ

O is fed to the reactor, and that at 1000K, the

equilibrium constants for the two reactions are 30 and 1.5 respectively.

1 and ξ2

Comp n

be the reaction co-ordinate for the two reactions, we have

i0 nexit y

CH

exit

4 1 1 - ξ1 (1 - ξ1) / 2(1 + ξ1

H

)

2O 1 1 – ξ 1 – ξ2 (1 – ξ1 – ξ2) / 2(1 + ξ1

CO 0 ξ

)

1 – ξ2 (ξ1 – ξ2) / 2(1 + ξ1

CO

)

2 0 ξ2 ξ2 / 2(1 + ξ1

H

)

2 0 3 ξ1 + ξ2 (3 ξ1 + ξ2) / 2(1 + ξ1

)

Total moles at equilibrum: 2(1 + ξ1

)

K = K ϕK yPα (for each reaction); K ϕ

Thus

= 1.0 (ideal gas assumption); P = 1 bar

( )( )

( )( )

( )( )

( )( )

3

1 2 1 2 1 2 1 2

1

1 1 1 1

3 3

2 1 2 1 2 1 2 1K

ξ ξ ξ ξ ε ξ ξ ξ

ξ ξ ξ ξ

− + − +=

+ + + +

( ) ( )

( ) ( ) ( )

3

1 2 1 2

2

1 1 1 2

330

4 1 1 1

ξ ξ ξ ξ

ξ ξ ξ ξ

− += =

+ − − − ……………………………. A

Similarly( )

( )( )1 2 2

2

1 2 1 2

31.5

1K

ξ ξ ξ

ξ ξ ξ ξ

+= =

− − − ………………………………. B

A and B needs to be solved simultaneously; a simple way to do this is to

(i) Assume2ε , calculate

1ε using B

(ii) Use2

ξ and1ξ in A to check if K 1

(iii) If

= 30

1 30,K ≠ assume new2

ε and go to step 1

Using the above algorithm, one finally obtains:1ξ = 0.7980,

2ξ = 0.0626.

Thus:2 4 2 2

0.0174, 0.0562, 0.0388, 0.2045, 0.6831CO CH H O CO H

y y y y y= = = = =

The gas n-pentane (1) is known to isomerise into neo-pentane (2) and iso-pentane (3)

according to the following reaction scheme:

Example 8.7

1 2 2 3 3 1; ;P P P P P P

. 3 moles of



pure n-pentane is fed into a reactor at 400o

K and 0.5 atm. Compute the number of moles of

each species present at equilibrium.

We use here the method of undetermined Lagrangian Multipliers.

The set of equation to be solved are:

15, 36C H

A A= =

For P11

5 129600ln 0C H

i

n

RT n RT RT

λ λ + + + =

∑ :

For P22

5 128900ln 0C H

i

n

RT n RT RT

λ λ + + + =

∑ :

For P3 3 5 128200

ln 0C H

i

n

RT n RT RT

λ λ + + + =

∑

:

Atomic mass balance for C: ( )1 2 35 15n n n+ + =

For H: ( )1 2 312 36n n n+ + =

n1 + n2 + n3

Alternately:

= 10

1

5 129600ln 0C H y

RT RT RT

λ λ + + + = …………………. (A)

2

5 128900ln 0C H y

RT RT RT

λ λ + + + = …………………. (B)

T = 400K

3

5 128200ln 0

RT

C H y RT RT

λ λ + + + = …………………. (C)

and y1 + y2 + y3

It follows from (A) – (C), y

= 1 …………………. (D)

2 / y1 = 2.41; y3 / y2

Using eqn. (D) y

= 2.41

1 = 0.108, y2 = 0.29, y3

= 0.63

Species

Example 8.8

0

f G∆ at 400o

K (Cal/mol)

P 96001

P 89002

P 82003



Consider the liquid phase reaction: A(l) + B(l) → C(l) + D(l). At 50oC, the equilibrium

constant is 0.09. Initial number of moles, nA,0 = 1 mole; nB,0

∴K =∴

= 1 mol Find the equilibrium

conversion. Assume ideal solution behaviour.

( ) ( )

1.0

-1expi i i

i iP V x

RT

γ γ γ

∑

Also, γi

Hence K = π

= 1 (ideal solution)

( ) i

i x

γ

xA = xB = (1-ξ)/2 ; xC = xD

∴ K = x

= ξ / 2

C xD / xA xB = [ξ/(1- ξ)]

⇒ 0.09 = [ξ/(1- ξ)]

2

Thus, ξ

2

e

= 0.23

Consider the following reaction: A(s) + B(g) → C(s) + D(g). Determine the equilibrium

fraction of B which reacts at 500

Example 8.9

oC if equal number of moles of A and B are introduced into

the reactor initially. The equilibrium constant for the reaction at 500o

The reaction is:

C is 2.0.

( ) ( ) ( ) ( ) A s B g C s D g+ → + ; basis 1 mole of A & B each initially

C D A BK a a a a=

For solids: 1a =

Thus:

; 0, 1 D B yK a a K K P and K α

φ φ α = = = =

yK K ∴ =

If one assumes equimolar feed of reactants:

(1 ); B D

y yξ ξ = − =

2.0 0.671

yK K

ξ ξ

ξ ∴ = = = ⇒ =

−

Thus 67% of B reacts.

example ch 8

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