ITK-234 Termodinamika Teknik Kimia 2

Chemical Equilibria

Dicky Dermawan www.dickydermawan.net78.net

[email protected]

Reaction Coordinate

e  1 A 1   2 A 2  ......

  3 A 3   3 A 3  ......

CH 4  CH 4   H 2 O   1 CO   3 H 2 H 2 O   1  CO   1  H 2   3 dn 1  1  dn  2 2  dn  3 3  dn 4  4  .....

 dn i  i  d e dn CH 4  1  dn H 2 O  1  dn CO 1  dn H 2 3  .....

 dn i  i  d e

Reaction Coordinate e n i  n i 0   i  e y i  n i n  n i 0 n 0    i    e e n n   n 0   n   e i n 0   n i 0     i If, initially there are 2 mol CH 4 , 1 mol H 2 O, 1 mol CO, and 4 mol H 2 : n CH n 0 4 , 0  n  2 CH 4 , 0  n H 2 O , 0 n H 2 O , 0  n CO , 0  1  n H 2 , 0  2 n CO , 0  1  1   1 4  8 n H 2 , 0  4 n CH  n   4   CH 4 n CH 4 2   e  H 2 O  n H 2 O  n n H 2 O   CO CO    n  H 2 H 2 1  e    8 1   2   1 e  n 1  CO 3   2 1  e n H 2  4  3  e y CH 4  8 2   2  e e y H 2 O  8 1   2 e  e y CO  8 1   2 e  e y H 2  4  3  e 8  2  e

Reaction Coordinate

Reaction Coordinate: Multiple Reactions  1 1 A 1 , 1   2 1 A 2 , 1  ......

  3 1 A 3 , 1   3 1 A 3 , 1  ......

 1 2 A 1 , 2   2 ..........

..........

2 A 2 , 2 ..........

 ......

.....

  3 2 A 3 , 2  ..........

  3 ..........

2 A 3 , 2 ..........

 ......

......

 1 j A 1 , j   2 j A 2 , j  ......

  3 j A 3 , j   3 j A 3 , j  ......

CH 4 CH 4   H 2 2 H O 2  O CO   CO 2 3 H 2  4 H 2

j

1 2

v CH 4

-1 -1

v H 2 O

-1 -2

v CO v CO 2

1 0 0 1

v H 2

3 4

ν j

2 2

Reaction Coordinate: Multiple Reactions dn 1 , 1  1 , 1  dn 2 , 1  2 , 1  dn 3 , 1  3 , 1  dn 4 , 1  4 , 1  .....

 dn i , 1  i , 1  d e 1 dn 1 , 2  1 , 2  dn 2 , 2  2 , 2  dn 3 , 2  3 , 2  dn 4 , 2  4 , 2  .....

 dn i , 2  i , 2  d e 2 dn 1 , j  1 , j  dn 2 , j  2 , j  dn 3 , j  3 , j  dn 4 , j  4 , j  .....

 dn i , j  i , j  d e j dn CH 4  1 , 1  dn H 2 O , 1  1  dn CO , 1 1  dn H 2 , 1 3  .....

 dn i , 1  i , 1  d e 1 dn CH 4  1 , 2  dn H 2 O , 2  2  dn CO 2 , 2 1  dn H 2 , 2 4  .....

 dn i , 2  i , 2  d e 2

n i Reaction Coordinate e  n i 0   j  i , j  e j y i  n i n  n i 0 n 0     j j   1 , i , j j  e  e j j n n   n 0   j  1 , j  n i  e j n 0   n i 0  j   i  i , j If, initially there are 2 mol CH 4 , 1 mol H 2 O, 1 mol CO, and 4 mol H 2 : y y CH 4 CO 2   8 8   2 2 2    e 1 e 1   e e 1 2  e 2 2  e 2 2  e 2 y y H 2 O H 2   4 8 8 1   2 e  1 e 1  2    3 2   e e 1 1   4 2    2 e  2 e e e 2 2 2 y CO  8  2  1  e 1 e  1 2  e 2

Reaction Coordinate

Reaction Coordinate & Equilibrium Condition

T/K

1000 1100 1200 1300 1400 1500

H 2 O

-192420 -187000 -181380 -175720 -170020 -164310 D

G f o /J .

mol-1 CO

-200240 -209110 -217830 -226530 -235130 -243740

CO 2

-395790 -395960 -296020 -396080 -396130 -396160   T , P  0

Reaction Coordinate & Equilibrium Condition (Cont’)

Method of Equilibrium Constant K   ( aˆ i )  i  exp    i R  T  G i 0 For gases: K   ( fˆ i )  i CH 4  H 2 O  CO  3 H 2 K  fˆ CH 4  1  H 2 O  1  fˆ CO  fˆ H 2 3 K  K y  K   P  For ideal gases: K  K y  P 

Method of Equilibrium Constant

Problems

Problem Hint: Different Way Leads to The Same Results

Problems

Calculation of Equilibrium Constant D G o 298 K     i  D G f o , 298 K  D G o 298 K   R  298  ln K 298 D H o 298 K D C o p       i



 i   D H C o p  f o , 298 K



D H T  D H o 298  T



298 D C o p  dT d ln K dT  D H T R  T 2 ln ln K K  d T 298 K ln K  D H T R  T 2  dT

D G f o ( 298 K )

D G f o ( 298 K )

D G f o ( 298 K )

D G f o ( 298 K )

D H f o ( 298 K )

D H f o ( 298 K )

D H f o ( 298 K )

D H f o ( 298 K )

Heat Capacity Data - Gases

Heat Capacity Data - Gases

Heat Capacity Data Solid

Heat Capacity Data - Liquid

Problems

Problems

Problems

Problems

Problems

Problems

Problems

Problems

Problems

Multiple Reactions

A bed of coal (assume pure carbon) in a coal gasifier is fed with steam and air and produce a gas stream containing H 2 , CO, O 2 , H 2 O, CO 2 , and N 2 . If the feed to the gasifier consists of 1 mol of steam and 2.38 mol of air, calculate the equilibrium composition of the gas stream at P = 20 bar.

T/K H 2 O

D

G f o /J .

mol-1 CO CO 2

1000 1100 1200 1300 1400 1500 -192420 -187000 -181380 -175720 -170020 -164310 -200240 -209110 -217830 -226530 -235130 -243740 -395790 -395960 -296020 -396080 -396130 -396160

Multiple Reactions

Problems

Problems

Problems

Lagrange Multiplier for Complex Reaction First Step: Atomic Balances  i n i  a ik  A k No. of eqn = no. of involved atom #2 Step: D G o fi  R  T  ln   n i n  i  P     k  k  a ik  0 No. of eqn = no. of species involved Use Computer to Solve All The Equation Simultaneously.

Problem

Calculate the equilibrium composition at 1000 K and 1 bar of a gas-phase system containing the species CH 4 , H 2 O, CO, CO 2 , and H 2 . In the initial unreacted state ther are present 2 mol of CH 4 and 3 mol of H 2 O. Assume ideal gases.

At 1000 K: D G f o CH 4  19720 J  mol -1 D G f o H 2 O   192420 J  mol 1 D G f o CO D G f o CO 2   200240 J  mol 1   395790 J  mol 1