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13.16 For the cracking reaction,

C3H8(g)→ C2H4(g) + CH4(g) The equilibrium conversion is negligible at 300 K, but becomes appreciable at temperaturesabove 500 K. For a pressure of 1 bar, determine : a. The fractional conversion of propane at 625 K. b. The temperature at which the fractional conversion is 85%. Solution : C3H8(g)→ C2H4(g) + CH4(g) v = 1 nC3 H 8  1  

Basis: 1 mole C3H8feed. By Eq. (13.4) Fractional conversion of C3H8=

By Eq. (13.5) yC3H8 

1  1 

n0  nC3 H8



n0

yC2 H 4 

 1 

1  (1   )  1 yCH 4 

 1 

From data in Table C.4.

H 298  82670 −1 V := ( 1 ) 1

J mol

G298  42290

J mol

1.213 −8.824 28.785 𝐴 ∶= (1.424) B := (14.394) .10-3 C := (−4.392) .10-6 1.702 −2.164 9.081

end := rows(A)

i := 1.. end

A   vi. Ai

B   vi.Bi

C   vi.Ci

i

i

i

A  1.913

B  5.31x10 3

a. T := 625 kelvin T



T0

C po R

dT  AT0   1 

C  2.268 x10 6

T0 := 298.15 kelvin B 2 2 C 3 3 D    1  T0   1  T0   1    2 3 T0   









(Van Ness, Eq. 4.19)

 C po dT  D    1  2     A ln    BT   CT     1  0 0 2 2  T R T   T  0  2     0 T

(Van Ness, Eq. 5.15)

Where,   625



298,15

T T0

3   625  2  2.268  106   625 3   625   5.31  10 dT  1,913  298.15    1  298.152    1  298.153    1    298.15     298.15   R 2 3  298.15      o 625 C p dT   11.2997  298,15 R

C po

   625      1   C p0 dT 625  5.31  10 3  298.15  2.268  10 6  298.152   298.15   625  1  1 . 913  ln     298.15  T 298.15  2 298,15 R       625 C p0 dT  R T  0.0022506 298,15 625

T T C p dT G 0 G00  H 00 H 00 1 C p     dT   RT RT0 RT T T0 R R T T0 0

0

(Van Ness, Eq. 13.18) G 0 42290  82670 82670 1    (11.2997)  0.0022506 RT (8.314)(625) (8.314)(625) 625

G 0  2187.9 RT (Van Ness, Eq.13.11a)   G 0    1.52356 K  exp   RT  (Van Ness, Eq.13.28)  := 0.5 (guess)

Given

 = 0.777

2 (1   ).(1   )

b.  = 0.777

 G 0  ln K RT

K

K

 := Find (  )

2 (1   ).(1   )

G 0  4972.3

K  2.604

J mol

The problem is now to find the T which generates this value. Its not dificult to find T by trial. This lead to the value T = 646.8 K

13.21 For the methanol synthesis reaction,

CO(g) + 2H2(g) → CH3OH(g) The equilibrium conversion to methanol is large at 300 K, but decreases rapidly with increasing T . However,reaction rates become appreciable only at higher temperatures. For a feed mixture of carbon monoxide and hydrogen in the stoichiometric proportions. a. What is the equilibrium mole fraction of methanol at 1 bar and 300 K b. At what temperature does the equilibrium mole fraction of methanol equal 0.50 for a pressure of 1 bar? c. At what temperature does the equilibrium mole fraction of methanol equal 0.50 for a pressure of 100 bar. Assuming the equilibrium mixture is an ideal gas ? d. At what temperature does the equilibrium mole fraction of methanol equal 0.50 for a pressure of 100 bar, assuming the equilibrium mixture is an ideal solution of gases?

Solution : CO(g) + 2H2(g) = CH3OH(g) The sthochiometric numbers are

 CH 3OH  1  2 H 2  2  CO  1  gas  1  (2)  1  2 Basis : 1 mol CO , 2 mol H2 feed ( From the data of Table C. 4 ,) 𝐉

∆H298 = - 90135 𝑴𝒐𝒍

n0 = 3 𝐉

∆G298 = - 24791 𝑴𝒐𝒍

This is the reaction of Ex. 4.6, Pg. 139, from which : ∆A = -7.663 ∆B = 10.815x10-3 ∆C = -3.45x10-6 (a) T = 300 Kelvin T



T0

C po R

dT  AT0   1 

(Van Ness, Eq. 4.19)

∆D = -0.135x10-5

T0 = 298.15 B 2 2 C 3 3 D    1  T0   1  T0   1    2 3 T0   









 C po dT  D    1  2 T R T  A ln   BT0   CT0   2T02  2   1   0 T

(Van Ness, Eq. 5.15) T Where,   T0 923,15



298,15

923,15



298,15

 300   1 2 3 6 5      300  3 . 45  10 300  0 . 135  10  300  10.815  10 3     298 , 15   dT  -7.663  298,15    1  298,152   298,153     1    1    298,15     298,15   R 2 3 298,15  300   298,15         298,15 

C po

C po R

dT  - 9.043

    300         1  5 C dT    298,15   300 300  0,135  10   3 6 2  7.663  ln  10.815  10  298,15    3.45  10  298,15   2   298,15  1 R T 298 , 15 2   300  298,15      298,152     298 , 15       

923,15

0 p

C p0 dT  R T  -0.03024 298,15

923,15

T T C p dT G 0 G00  H 00 H 00 1 C p     dT   RT RT0 RT T T0 R R T T0 0

0

(Van Ness, Eq. 13.18) G 0 - 24791  (- 90135 ) - 90135 1    (9.403)  (0.03024) RT (8,413)( 298,15) (8,413)(300) 300

G 0  -2.439 x 104 RT G 0 ln K    2.439 x 104 RT (Van Ness, Eq.13.11a)   G 0 K  exp   RT

   1.762 x 104 

By Equation (13.5), with the species numbered in the order in which they appear in the reaction, 1− ℰ 2− 2 .ℰ ℰ y1 = y2 = y3 = 3−2 .ℰ 3−2 .ℰ 3−2 .ℰ By Equation (13.28), ℰ = 0.8 (guess) → P = 1 ; P0 = 1 ℰ.( 3−2 .ℰ )2 P 2 Given = ( ) .K ℰ = Find (ℰ) ; ℰ = 0.9752 4.( 1− ℰ)3 P0 ℰ y3 = → y3 = 0.9291 3−2 .ℰ

(b) y3 = 0.5

By the preceding equation

3.𝑦3

→ ℰ = 0.75 2.𝑦3+ 1 Solution of the equilibrium equation for K gives ℰ.( 3−2 .ℰ )2 K= → K = 27 4.( 1− ℰ)3 ℰ=

Find by trial the value of T for which this is correct. It turns out to be : T = 364.47 Kelvin (c) For P = 100 bar , the preceding equation Becomes ℰ.( 3−2 .ℰ )2

. 100-2 → K = 2.7 x 10-3 4.( 1− ℰ)3 Another solution by trial for T yields, T = 516.48 kelvin K=

(d) Equation (13.27) applies, and requires fugacity coefficients. Since iteration will be necessary, assume a starting T of 528 K, for which :

© For CO(1) :

Tr = Pr =

528 132.9 100 34.99

;

Tr = 3.973

;

Pr = 2.858

By using the formula: Tr = T / Tc Pr = P / Pc B0 = 0.083-(0.422/(Tr1.6)) B1 = 0.139-(0.172/(Tr4.2)) ln = (Pr/Tr) x (B0 + (ω x B1))

The results of calculations are tabulated in the following table Tc(K) Pc(atm) CO

Tri

Pri

B0

B1

ɸi

132,9

34,99

33,19

13,13

0,048 3,965388 7,550729 0,036435 0,138472 1,085494 0,216 15,87828 20,12186 0,077941 0,138998 1,062606

512,6

80,97

0,564 1,028092 3,262937 -0,3207

2H2 CH3OH

ωi*

* Price ωi obtained from Appendix B (Smith & Van Ness ed-4)

-0,01411

0,352364

© For CH3OH(3) :

Tr = Pr =

528 512.6 100 34.99

;

Tr = 1.03

;

Pr = 1.235

By equation (11.64) and data from Tables E.15 & E.16. ɸ3 = 0.6206 x (0.9763)0.564 ɸ3 = 0.612 For H2(2), The reduced temperature is so large that it may be assumed ideal ; ɸ = 1. Therefore ; i = 1....3 𝟏. 𝟎𝟑𝟐 −𝟏 ∏𝒊 ﴾ɸ ﴿vi = 0.593 ɸ = (𝟏. 𝟎𝟎𝟎) ; v = (−𝟐) ; 𝟎. 𝟔𝟏𝟐 𝟏 The expression used for K in part (c) now becomes : ℰ.( 3−2 .ℰ )2 K= . 100-2 . 0.593 4.( 1− ℰ)3 K= 1.6011 x 10-3 Another solution by trial for T yields : T = 528.7 Kelvin