Question

An irreversible liquid-phase second-order reaction \(A \xrightarrow{k} B\) with rate constant \(k=0.2\ \mathrm{L\,mol^{-1}\,min^{-1}}\) is carried out isothermally in a non-ideal reactor. A tracer test gives the residence time distribution \(E(t)\) shown: three rectangles (i), (ii), (iii) of equal areas; rectangles (i) and (ii) span \(t=5\)–\(10\ \mathrm{min}\) (stacked), and (iii) spans \(t=10\)–\(15\ \mathrm{min}\). Pure \(A\) of concentration \(C_{A0}=1.5\ \mathrm{mol\,L^{-1}}\) is fed. Using the segregated model, determine the percentage conversion of \(A\) at the exit (rounded to the nearest integer).

Hint:

In the segregated model, compute \(C_{\text{out}}=\int C(t)E(t)\,dt\); then \(X=1-C_{\text{out}}/C_0\).
For an \(n\)-th order batch element, use the batch solution \(C(t)\) with residence time \(t\).

Answer 1

Correct Answer: 71

Step 1: Determine \(E(t)\). Areas of (i), (ii), (iii) are equal. Let the common height be \(a\) for rectangles (i) and (iii), and rectangle (ii) also has height \(a\). Total area of \(E(t)\) must be 1: \[ \int_0^\infty E(t)\,dt = 15a = 1 \;\;\Rightarrow\;\; a=\tfrac{1}{15}\ \mathrm{min^{-1}}. \] Hence: \[ E(t)= \begin{cases} \dfrac{2}{15}, & 5 \le t \le 10, \\[6pt] \dfrac{1}{15}, & 10 < t \le 15, \\[6pt] 0, & \text{otherwise}. \end{cases} \]

Step 2: Concentration for a fluid element (segregated model). For a 2nd-order batch element: \[ \frac{dC_A}{dt}=-kC_A^2 \;\Rightarrow\; C_A(t)=\frac{C_{A0}}{1+kC_{A0}t}. \] Here: \[ kC_{A0} = 0.2 \times 1.5 = 0.3\ \mathrm{min^{-1}}. \]

Step 3: Mixed effluent concentration. \[ C_{A,\text{out}}=\int_0^\infty C_A(t)\,E(t)\,dt = C_{A0}\!\left[\frac{2}{15}\!\int_{5}^{10}\!\frac{dt}{1+0.3t}+\frac{1}{15}\!\int_{10}^{15}\!\frac{dt}{1+0.3t}\right]. \] Using: \[ \int \frac{dt}{1+0.3t} = \frac{1}{0.3}\ln(1+0.3t), \] we get: \[ \frac{C_{A,\text{out}}}{C_{A0}} = \frac{4}{9}\ln(1.6)+\frac{2}{9}\ln(1.375)\approx 0.2797. \] Thus: \[ C_{A,\text{out}} \approx 1.5 \times 0.2797 = 0.4195\ \mathrm{mol\,L^{-1}}. \]

Step 4: Overall conversion. \[ X = 1 - \frac{C_{A,\text{out}}}{C_{A0}} \;\approx\; 1 - 0.2797 = 0.720. \]

Final Answer: \[ \boxed{X \;\approx\; 72\%} \]