Here's the solution for Question 2, covering both parts.
2.1. Set up a stoichiometric table and express concentrations.
The reactions are:
- A→B (desired)
- A→C (undesired)
Given that the reactions are elementary and the rate constant kB=2dm3/mol⋅min (units for a second-order reaction) and kC=1min−1 (units for a first-order reaction), we assume the reaction orders are dictated by these units.
Therefore:
- Rate of formation of B: rB=kBCA2
- Rate of formation of C: rC=kCCA
The total rate of disappearance of A is:
−rA=−rA1+(−rA2)=kBCA2+kCCA
The reactor operates isothermally and with constant density (ν=ν0).
Let X be the overall conversion of A.
Let XB be the fractional conversion of A to B, and XC be the fractional conversion of A to C.
The total conversion of A is X=XB+XC.
Stoichiometric Table (in terms of concentrations):
| Component | Initial Concentration (Ci0) | Change in Concentration | Final Concentration (Ci) |
| :-------- | :------------------------------- | :---------------------- | :-------------------------- |
| A | CA0 | −CA0X | CA=CA0(1−X) |
| B | 0 | +CA0XB | CB=CA0XB |
| C | 0 | +CA0XC | CC=CA0XC |
The concentrations as a function of overall conversion X and initial concentration CA0 are:
CA=CA0(1−X)
CB=CA0XB
CC=CA0XC
2.2. Evaluate whether a CSTR or PFR will provide a higher yield of B.
To determine which reactor provides a higher yield of B, we need to calculate the yield for each reactor type at 90% conversion of A (X=0.9).
The yield of B (YB) is defined as the moles of B formed per mole of A fed, which is YB=CA0CB=XB.
Given values:
- CA0=4mol/dm3
- X=0.9
- kB=2dm3/mol⋅min
- kC=1min−1
The instantaneous selectivity of B with respect to A reacted is:
SBA=−rArB=kBCA2+kCCAkBCA2=kBCA+kCkBCA
Since SBA depends on CA, it is not constant. To maximize SBA, we need to maximize CA.
For a Plug Flow Reactor (PFR):
Step 1: Calculate the conversion of A to B (XB) for a PFR.
For a PFR, the differential conversion of B is related to the instantaneous selectivity: dXB=SBAdX.
XB=∫0XSBAdX=∫0XkBCA+kCkBCAdX
Substitute CA=CA0(1−X):
XB=∫0XkBCA0(1−X)+kCkBCA0(1−X)dX
Plug in the given values: kB=2, kC=1, CA0=4, and X=0.9.
XB=∫00.92×4(1−X)+12×4(1−X)dX=∫00.98(1−X)+18(1−X)dX
Let u=1−X, so du=−dX. The limits change from X=0⇒u=1 to X=0.9⇒u=0.1.
XB=∫10.18u+18u(−du)=∫0.11(1−8u+11)du
XB=[u−81ln(8u+1)]0.11
XB=(1−81ln(8(1)+1))−(0.1−81ln(8(0.1)+1))
XB=(1−81ln(9))−(0.1−81ln(1.8))
XB=0.9−81(ln9−ln1.8)=0.9−81ln(1.89)
XB=0.9−81ln(5)
XB=0.9−81.6094379=0.9−0.2011797=0.69882
The yield of B for a PFR is YBPFR=0.6988.
For a Continuous Stirred-Tank Reactor (CSTR):
Step 2: Calculate the conversion of A to B (XB) for a CSTR.
In a CSTR, the concentration of A is uniform throughout the reactor and equal to the outlet concentration.
CA=CA0(1−X)=4mol/dm3(1−0.9)=4×0.1=0.4mol/dm3
The instantaneous selectivity SBA is constant throughout the CSTR, evaluated at this outlet concentration:
SBACSTR=kBCA+kCkBCA=2×0.4+12×0.4=0.8+10.8=1.80.8=94
For a CSTR, the overall conversion of A to B is XB=X×SBACSTR.
XB=0.9×94=0.4
The yield of B for a CSTR is YBCSTR=0.4000.
Conclusion:
Step 3: Compare the yields.
Comparing the yields:
- YBPFR=0.6988
- YBCSTR=0.4000
Since 0.6988>0.4000, the Plug Flow Reactor (PFR) will provide a higher yield of B. This is because the selectivity to B increases with higher concentrations of A, and a PFR maintains a higher average concentration of A throughout the reactor compared to a CSTR for the same overall conversion.
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