Chapters 8, 9 and 18

Submitted by:Chua, PaulineClimaco, Mary AnnZablan, Bianca

Problem 8.3Problem 8.21Problem 18.5Problem 18.23

Problem 8.3

Pure A (CAo=100) is fed to a mixed flow reactor, R and S are formed, and the following outlet concentrations are rerecorded. Find a kinetic scheme to fit this data.Run CA CR CS

1 75 15 102 25 45 30

Problem 8.3

Assuming A→R→S. In Figure 8.14

Results don’t fit the kinetics

Problem 8.3

Assuming A→R; A→S -rR=k1CA -rS=k2CA

For mixed flow: τ1=(CR-CRo)/k1CA

τ2=(CR-CRo)/k2CA

Dividing, CR/Cs = k1/k2

Therefore the assumed scheme is correct with k1=1.5k2

Run CA CR CS CR/CS

1 75 15 10 1.5

2 25 45 30 1.5

Problem 8.21

Chemical A reacts to form R (k1=6 hr-

1) and R reacts away to form S (k2=3 hr-1). In addition, R slowly decomposes to form T (k3= 1 hr-1). If a solution containing 1.0 mol/L of A is introduced into a batch reactor, how long would it take to reach CR

max, and what would be CR max?

Problem 8.21

Given:

k1=6 hr-1

k2=3 hr-1

k3=1 hr-1

From the units of k, it can be concluded that all reactions are 1st order

CAo=1.0 mol/L

Problem 8.21

1. rA=-k1CA

2. rR=k1CA-k2CR-k3CR

3. rS=k2CR

4. rT=k3CR

By equation 2

rR=k1CA-(k2+k3)CR = k1CA-k23CR =dCR/dt and CA=CAoe-k1t

Problem 8.21

dCR/dt = k1CAoe-k1t-k23CR

dCR/dt + k23CR = k1CAoe-k1t

dCR/dt + P(t)y = Q(t)

Problem 8.21

To get the maximized CR

Problem 8.21

t=12.16395 min (time to reach CRmax )

Problem 8.21

To get CRmax

CRmax=0.4444 mol/L

Problem 18.5

A gas containing A (2 mol/m3) is fed (1 m3/hr) to a plug flow reactor with a recycle loop (0.02 m3 loop volume, 3 kg of catalyst), and the output composition from the reactor system is measured (0.5 mol A/m3). Find the rate equation for the decomposition of A for the following cases. Be sure to give the units of –rA’, CA, and k’ in your final expressions:

Very large recycle A→3R, n=1, 50% A-50% inerts in feed

Problem 18.5

A→3R, n=1 (1st order) ε=[1+3-(1+1)]/(1+1)=1

Mixed Flow Reactor

Thus,

Problem 18.23

The first-order decomposition of A is run in an experimental mixed flow reactor. Find the role played by pore diffusion in these runs; in effect, determine whether the runs were made under diffusion free, strong resistance, or intermediate conditions.

dP W CAo ν XA

3 1 100 9 0.412 4 300 8 0.6

Problem 18.23

Run

dp

W CAo

ν XA

CA k’

1 3 1 100 9 0.4

60 6

2 12 4 300 8 0.6

120

3

For no diffusion resistance: k2’/k1’ = 1

For strong resistance k2’/k1’ = dp1/dp2 = 3/12 =

1/4 From the data

k2’/k1’ =3/6 = ½ Thus, the runs were

made in intermediate conditions between strong resistance and no diffusion resistance.

For mixed flow