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Set up a stoichiometric table in terms of symbols and express the concentration as a function of conversion and concentration. Evaluate whether a Continuous Stirred-Tank Reactor (CSTR) or a Plug Flow Reactor (PFR) will provide a higher yield of B.

Chemistry

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2 parts
2.1

CA0(1−X)C_{A0}(1-X)

CA0XBC_{A0}X_B

CA0XCC_{A0}X_C

2.2
0.6988
0.4000

Step-by-step solution

3 steps

Here's the solution for Question 2, covering both parts.

Part 2.1

Set up a stoichiometric table and express concentrations.

The reactions are:

  1. A→BA \rightarrow B (desired)
  2. A→CA \rightarrow C (undesired)

Given that the reactions are elementary and the rate constant kB=2 dm3/mol⋅mink_B = 2 \text{ dm}^3/\text{mol} \cdot \text{min} (units for a second-order reaction) and kC=1 min−1k_C = 1 \text{ min}^{-1} (units for a first-order reaction), we assume the reaction orders are dictated by these units.
Therefore:

  • Rate of formation of B: rB=kBCA2r_B = k_B C_A^2
  • Rate of formation of C: rC=kCCAr_C = k_C C_A

The total rate of disappearance of A is:

−rA=−rA1+(−rA2)=kBCA2+kCCA-r_A = -r_{A1} + (-r_{A2}) = k_B C_A^2 + k_C C_A

The reactor operates isothermally and with constant density (ν=ν0\nu = \nu_0).
Let XX be the overall conversion of A.
Let XBX_B be the fractional conversion of A to B, and XCX_C be the fractional conversion of A to C.
The total conversion of A is X=XB+XCX = X_B + X_C.

Stoichiometric Table (in terms of concentrations):

ComponentInitial Concentration (Ci0C_{i0})Change in ConcentrationFinal Concentration (CiC_i)
ACA0C_{A0}−CA0X-C_{A0}XCA=CA0(1−X)C_A = C_{A0}(1-X)
B0+CA0XB+C_{A0}X_BCB=CA0XBC_B = C_{A0}X_B
C0+CA0XC+C_{A0}X_CCC=CA0XCC_C = C_{A0}X_C

The concentrations as a function of overall conversion XX and initial concentration CA0C_{A0} are:

CA=CA0(1−X)C_A = \boxed{C_{A0}(1-X)} CB=CA0XBC_B = \boxed{C_{A0}X_B} CC=CA0XCC_C = \boxed{C_{A0}X_C}

Part 2.20.6988

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=0.9X=0.9).
The yield of B (YBY_B) is defined as the moles of B formed per mole of A fed, which is YB=CBCA0=XBY_B = \frac{C_B}{C_{A0}} = X_B.

Given values:

  • CA0=4 mol/dm3C_{A0} = 4 \text{ mol/dm}^3
  • X=0.9X = 0.9
  • kB=2 dm3/mol⋅mink_B = 2 \text{ dm}^3/\text{mol} \cdot \text{min}
  • kC=1 min−1k_C = 1 \text{ min}^{-1}

The instantaneous selectivity of B with respect to A reacted is:

SBA=rB−rA=kBCA2kBCA2+kCCA=kBCAkBCA+kCS_{BA} = \frac{r_B}{-r_A} = \frac{k_B C_A^2}{k_B C_A^2 + k_C C_A} = \frac{k_B C_A}{k_B C_A + k_C}

Since SBAS_{BA} depends on CAC_A, it is not constant. To maximize SBAS_{BA}, we need to maximize CAC_A.

For a Plug Flow Reactor (PFR):

  1. Calculate the conversion of A to B (XBX_B) for a PFR.

    For a PFR, the differential conversion of B is related to the instantaneous selectivity: dXB=SBAdXdX_B = S_{BA} dX.

    XB=∫0XSBAdX=∫0XkBCAkBCA+kCdXX_B = \int_0^X S_{BA} dX = \int_0^X \frac{k_B C_A}{k_B C_A + k_C} dX

    Substitute CA=CA0(1−X)C_A = C_{A0}(1-X):

    XB=∫0XkBCA0(1−X)kBCA0(1−X)+kCdXX_B = \int_0^X \frac{k_B C_{A0}(1-X)}{k_B C_{A0}(1-X) + k_C} dX

    Plug in the given values: kB=2k_B = 2, kC=1k_C = 1, CA0=4C_{A0} = 4, and X=0.9X = 0.9.

    XB=∫00.92×4(1−X)2×4(1−X)+1dX=∫00.98(1−X)8(1−X)+1dXX_B = \int_0^{0.9} \frac{2 \times 4 (1-X)}{2 \times 4 (1-X) + 1} dX = \int_0^{0.9} \frac{8(1-X)}{8(1-X) + 1} dX

    Let u=1−Xu = 1-X, so du=−dXdu = -dX. The limits change from X=0⇒u=1X=0 \Rightarrow u=1 to X=0.9⇒u=0.1X=0.9 \Rightarrow u=0.1.

    XB=∫10.18u8u+1(−du)=∫0.11(1−18u+1)duX_B = \int_1^{0.1} \frac{8u}{8u+1} (-du) = \int_{0.1}^1 \left( 1 - \frac{1}{8u+1} \right) du XB=[u−18ln⁡(8u+1)]0.11X_B = \left[ u - \frac{1}{8} \ln(8u+1) \right]_{0.1}^1 XB=(1−18ln⁡(8(1)+1))−(0.1−18ln⁡(8(0.1)+1))X_B = \left( 1 - \frac{1}{8} \ln(8(1)+1) \right) - \left( 0.1 - \frac{1}{8} \ln(8(0.1)+1) \right) XB=(1−18ln⁡(9))−(0.1−18ln⁡(1.8))X_B = \left( 1 - \frac{1}{8} \ln(9) \right) - \left( 0.1 - \frac{1}{8} \ln(1.8) \right) XB=0.9−18(ln⁡9−ln⁡1.8)=0.9−18ln⁡(91.8)X_B = 0.9 - \frac{1}{8} (\ln 9 - \ln 1.8) = 0.9 - \frac{1}{8} \ln \left( \frac{9}{1.8} \right) XB=0.9−18ln⁡(5)X_B = 0.9 - \frac{1}{8} \ln(5) XB=0.9−1.60943798=0.9−0.2011797=0.69882X_B = 0.9 - \frac{1.6094379}{8} = 0.9 - 0.2011797 = 0.69882

    The yield of B for a PFR is YBPFR=0.6988Y_B^{PFR} = \boxed{0.6988}.

    For a Continuous Stirred-Tank Reactor (CSTR):

  2. Calculate the conversion of A to B (XBX_B) 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)=4 mol/dm3(1−0.9)=4×0.1=0.4 mol/dm3C_A = C_{A0}(1-X) = 4 \text{ mol/dm}^3 (1-0.9) = 4 \times 0.1 = 0.4 \text{ mol/dm}^3

    The instantaneous selectivity SBAS_{BA} is constant throughout the CSTR, evaluated at this outlet concentration:

    SBACSTR=kBCAkBCA+kC=2×0.42×0.4+1=0.80.8+1=0.81.8=49S_{BA}^{CSTR} = \frac{k_B C_A}{k_B C_A + k_C} = \frac{2 \times 0.4}{2 \times 0.4 + 1} = \frac{0.8}{0.8 + 1} = \frac{0.8}{1.8} = \frac{4}{9}

    For a CSTR, the overall conversion of A to B is XB=X×SBACSTRX_B = X \times S_{BA}^{CSTR}.

    XB=0.9×49=0.4X_B = 0.9 \times \frac{4}{9} = 0.4

    The yield of B for a CSTR is YBCSTR=0.4000Y_B^{CSTR} = \boxed{0.4000}.

    Conclusion:

  3. Compare the yields.

    Comparing the yields:

    • YBPFR=0.6988Y_B^{PFR} = 0.6988
    • YBCSTR=0.4000Y_B^{CSTR} = 0.4000

    Since 0.6988>0.40000.6988 > 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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