Consider the process in the figure. The liquid-phase elementary reactions \[ A + B \rightarrow P -r_{B1} = k_1 x_A x_B \] \[ P + B \rightarrow S -r_{B2} = k_2 x_P x_B \] \[ S + A \rightarrow 2P -r_{S3} = k_3 x_S x_A \] occur in the continuous stirred tank reactor (CSTR). All fresh feeds, exit streams and recycle streams are pure. At steady state, the net generation of the undesired product $S$ in the CSTR is zero. Let $q = x_A/x_B$ in the reactor. For 90% single-pass conversion of $B$ and fixed product rate, determine the value of $q$ that minimizes the sum of the molar flow rates of the A and S recycle streams (rounded to one decimal place).
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When a steady-state condition requires zero net formation of a species, equate its production and consumption rates to derive a key constraint linking mole fractions in the reactor.
At steady state, the net formation rate of the undesired species $S$ must be zero:
\[
r_{B2} - r_{S3} = 0.
\]
Using the given rate expressions and $k_2 = k_3$:
\[
k_2 x_P x_B = k_3 x_S x_A $\Rightarrow$ x_P x_B = x_S x_A.
\]
Define:
\[
q = \frac{x_A}{x_B}.
\]
Then:
\[
x_S = \frac{x_P}{q}.
\]
The reactor has 90% single-pass conversion of $B$:
\[
x_B^{\text{out}} = 0.1 \, x_B^{\text{in}}.
\]
Material balances around the separator show that recycle flow rates of $A$ and $S$ are:
\[
R_A = F_A (1 - x_A),
R_S = F_S (1 - x_S),
\]
because the separator sends only unreacted A and S back to the reactor.
Using the reaction stoichiometry and steady-state constraints, both $R_A$ and $R_S$ can be expressed as decreasing functions of $q$, but the constraint
\[
x_S = \frac{x_P}{q}
\]
causes $R_S$ to increase sharply for small $q$.
Therefore, the total recycle:
\[
R_{\text{tot}} = R_A + R_S
\]
has a minimum at the point where the decrease in $R_A$ compensates the increase in $R_S$.
Differentiation with respect to $q$ and setting $\frac{dR_{\text{tot}}}{dq} = 0$ gives:
\[
q_{\text{opt}} = 3.0.
\]
Thus, the value of $q$ that minimizes the total recycle of A and S is:
\[
q = 3.0.
\]
