Question

Carbon monoxide (CO) and hydrogen (H2) are fed to a continuous catalytic reactor operating at steady state. There are no other components in the feed. The outlet stream contains unconverted CO and H2, along with the products methanol (CH30H), ethanol (C2HsOH), isopropanol (C3H70H), and carbon dioxide (C02). These are the only species in the product stream.

The reactions occuring are:

CO+ 2H2 → CH3OH
3CO+ 3H2 → C2H5OH+ CO2
5CO+ 4H2 → C3H7OH + 2CO2

The feed rates of CO and H2 to the reactor are 100 mol/h(each). the rates in the stream that leaves the reactor (in mols/h) are H2-30; CO-30; C2H5OH-5. What’s the mole fraction of each species in the product steam?

Answer 1

Step-by-step explanation:

Let the products in the outlet streams be P(i.e. CH₃OH), Q(i.e. C₂H₅OH) and R(i.e. C₃H₇OH) respectively

Then the mass balance for CO, H and C₂H₅OH can be computed as follows:

For CO, The mass balance is;

100 - P - 3Q - 5R = 30 --- (1)

For H₂, The mass balance is;

100 - 2P - 3Q - 4R = 30 --- (2)

For C₂H₅OH, The mass balance is;

Q = 5 --- (3)

Replacing the value of equation(3) into equation (1) and (2); we have:

From equation (1):

100 - P - 3(5) - 5R = 30 --- (1)

100 - 15 - P - 5R = 30

85 - P - 5R = 30

85 - 30 = P + 5R

P + 5R = 55 ----- (4)

From equation (2):

100 - 2P - 3Q - 4R = 30 --- (2)

100 - 2P - 3(5) - 4R = 30

100 - 15 - 30 = 2P +4R

55 = 2P + 4R

2P + 4R = 55 ----- (5)

From equation (4) and (5); we have:

P + 5R = 55 ----- (4)

2P + 4R = 55 ----- (5)

From equation (4);

Let P = 55 - 5R

Then, replace P = 55 - 5R in equation (5)

2(55 - 5R) + 4R = 55

110 - 10R + 4R = 55

110 - 6R = 55

6R = 110 - 55

6R = 55

R = 55/6

R = 9.17

Substitute, the value of R in equation (5); we have:

2P + 4R = 55

2P + 4(9.17) = 55

2P = 55 - 4(9.17)

2P = 55 - 36.68

2P = 18.32

P = 18.32/2

P = 9.16

Therefore, the outlet stream rates are as follows:

Mass Feed Rate ( mol/hr) Molar ratio

CO 30 0.281

H₂ 30 0.281

CH₃OH 9.16 0.085

C₂H₅OH 5 0.047

C₃H₇OH 9.17 0.086

CO₂ 23.34 0.219

Total: 106.67