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CHEE3005 Tutorial Sheet 1: Solutions Q1. Determine the equilibrium conversion of the following reaction: C2 H4 ( g )  H2 O( g )  C2 H5 OH ( g ) as a function of the temperature, at 30 atm, and steam/ethylene molar ratio equals to 10.

Given that H 298  50 kJ/mol 0

K =6.8x10 atm  at 145C P  30 atm. 2

1

C pC2 H4  11.85  0.12T  36.6x 106 T 2 , J/mol/K CPH2O  30.4  9.63x103 T  1.17x106 T 2 , J/mol/K CPC2 H5OH  29.27  0.17T  49.95x106 T 2 , J/mol/K Solution T

H o (T )  H o (298) 



C p ( T ') dT '

298

 C p  C pC2 H5 OH  C pC2 H4  C pH2 O    (29.27  30.4  11.85)  (0.17  9.63 x10 3  0.12) T  (49.95  1.17  36.6) x10 6 T 2

 -12.98  0.04T -14.52 x10 6 T 2 Calculation of H o ( T ) T

 H o ( T )   H o (298) 

  C ( T ') dT ' p

298 T

 50,000+  (-12.98  0.04T -14.52x 106 T 2 )dT ' 298 T T  T2  T3  =  50,000   12.98(T  298)  0.04    14.52x106      2  298  3  298  2 2 6 3 3  50,000   12.98( T  298)  0.02( T  298 )  4.84 x10 ( T  298 )

Calculation of the equilibrium constant: o d ln( K ) H ( T )  dT RT 2

 K (T )  ln    K (418)  

T

H o (T ') dT ' RT ' 2 418



T 1  50, 000  2982  2983     1 298  6  x T 12.98 0.02 1 4.84 10 '               dT '     T '2  2 R 418 T '2  T '2   T' T'    

T T T T 2 1     1   1  6  T ' 2  1  3  1  298 0.02 ' 298 4.84 10 298 T x T 50,000 12.98 ln( ')                       R  T '  418  T '  418  T '  418  T '  418    2   K    T  1 1  1  1   1 2 1 8.4136 xln  T   50, 000     12.98 ln    298    0.02 (T  418)  298     T 418   T 418   T 418     418    K418 



 T 2  4182  1  1  4.84 x10   2983    2  T 418    6

From equilibrium thermodynamics K P (T )  K y P

( c d ) ( a b)





yC2H 5OH yC2 H4 yH2 O

yC 2H 5OH y C2 H4 y H2 O

P

(1 1 1)

P 1

Assume 1 mole of C2H4 and 10 moles of steam initially. After n moles of C2H4 have reacted C2 H 4 ( g )  H 2O ( g )  C2 H 5OH ( g ) 1-n 10-n n Total number of moles = 1-n+10-n+n = 11-n 1 n Mole fraction C2H4 = 11  n 10  n Mole fraction H2O = 11  n n Mole fraction C2H5OH = 11  n  KT 

yC2 H5 OH PyC 2H 4 yH 2O



n (11  n ) 30(1  n )(10  n )

This gives a quadratic equation for n:

(30 KT  1)n2  (330KT  11)n  300KT  0 The solution for n represents the conversion at temperature T.

1.0 0.8

0.1 conversion

equilibrium constant, K (atm.)

1

0.01

0.6 0.4

0.001

0.2 0.0

0.0001 400

500

600

400

700

500

600

700

temperature (K)

temperature (K)

Q2. A gas phase reaction, 2A + 3B  C, is conducted with several starting ratios, r = nbo/nao. The equilibrium constant Ky = 10.

(a) Show that the equilibrium conversion of A for any value of the ratio r follows the relation

K y  10 

X 1  r  2 X  2 1  X 

4 3

2

3   r  2 X   

(b) At what value of the ratio r is the conversion of substance A maximum, and what is this maximum conversion? (c) What is the maximum possible conversion for an equimolar mixture (i.e. r = 1), and at what value of Ky is this obtained? Solution

(a) Assume: - Pressure is constant - Isothermal Starting with one mole of A perform a mole balance, assuming X moles of A are reacted. 2A + 3B  C 3 X N : 1 X r  X 2 2 total N , NT  1  r  2 X

 Ky 

yC  2 3 yA yB



XN T2 N T3 3   2 NT 1  X   r  X  2  

3

2

X 1 r  2 X 

4

3  2 2 1  X   r  X  2  

3

(b) As shown in the hint, for maximum conversion:

 f    0  r  X where,

K y  10  f  X , r  

X 1  r  2 X 

4

3  2 2 1  X   r  X  2  

3

The derivative of f with respect to r is zero because f is constant (=10). On differentiating f(X,r)

  3 4  X  f   4 1  r  2 X   3 1  r  2 X     0  2 3 4 2(1  X )    r  X 3  3     r  X  r  X   2  2      This yields

4 1  r  2 X 

3

r 3X     2  

3



3 1  r  2 X  r 3X     2  

4

4

whose solution is r = 3, and upon substitution in Eq. (1), with Ky=10, we obtain

X m  4  2X m 2 1  X m 

2

4

 3 3 X   m 2  

3

 10

(1)

This has the solution: Xm = 0.563. This is the maximum conversion for the reaction, and is obtained when r = 3, for the value of Ky = 10. (c) For the reaction 2A + 3B  C

3 moles of B are required for every 2 moles of A converted. When r = 1, only one mole of B is available for every mole of A, and species B is limiting. To convert this one mole of B only 2/3 moles of A are reacted from the one mole of A available. Thus, the maximum conversion of A is 2/3, or 0.67. From the equation for Ky, we have

Ky 

X 1 r  2 X

3

3  2  2 1  X   r  X  2  

For r = 1 this yields

Ky 

and for X → 2/3, Ky → ∞.

4

8 X  1 X 

2

3

 3  X 1  2 ...