Additional Homework Problems

CDP6-B_B Solution

Part A

To solve this problem, we must follow the usual algorithm for multiple reactions with catalysis.

Mole Balance

Since there are four components, there are four mole balances for this problem. They are:

IN - OUT + ACCUM = 0

v_oC_Ao - v_oC_A + r_AW = 0
-v_oC_B + r_BW = 0
-v_oC_C + r_CW = 0
-v_oC_D + r_DW = 0

Remember, from the problem statement, that the volume change with reaction can be neglected since the reactions are carried out in excess air. Therefore, v = v_o.

Rate Laws

Next, the rate laws must be formed, using the reaction sequence diagram and noting that Reaction 1 is second-order while Reactions 2 and 3 are first-order.

The rate laws become:

r_A = -k₁C_A² - k₃C_A
r_B = k₁C_A² - k₂C_B
r_C = k₂C_B
r_D = k₃C_A

Polymath Solution

This set of equations, along with the reaction parameters in the problem statement, may be entered into Polymath to find the exiting concentrations. The equations entered in polymath are as follows:

Polymath then solves the equations and gives us an output file:

The final concentration values are listed in this results table. Therefore, the exit concentrations from a CSTR in this problem are:

C_A = 2.96 x 10^-3 mol/dm³
C_B = 1.94 x 10^-3 mol/dm³
C_C = 1.55 x 10^-3 mol/dm³
C_D = 3.55 x 10^-3 mol/dm³

Part B

Selectivity is a ratio of reaction rates. Therefore, the selectivity of B to C is the ratio of r_B and r_C, and the selectivity of B to D is the ratio of r_B and r_D.

Part C

To solve this problem, we must set the up equations needed for a PBR with multiple reactions. The mole balances are:

These, combined with the same rate laws as in Part A, give the following Polymath equations:

These equations result in the following plot of concentrations as a function of catalyst weight in the PBR.

From this plot, we can see that the concentration of B is a maximum at a catalyst weight of approximately 2000 kg. To get the exact answer, we display C_B as a function of W in a table of values:

This table shows that the concentration of B is a maximum at a catalyst weight of 1937.5 kg.

Part D

To model these changes in C_Ao for the PFR, we just replace the original value of C_A in the Polymath equations above. The new product distribution, with C_Ao as a larger value (0.5 mol/dm³) is:

Also, if we change the entering concentration of C_A to a very low value, such as 0.001 mol/dm³, we get this product distribution:

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