GATE 2023 CH – Question 60

Chemical Reaction Engineering · Non-ideal reactors and residence time distribution · 2 marks · Numerical answer

An irreversible liquid-phase second-order reaction $A \xrightarrow{k} B$ with rate constant $k = 0.2$ liter mol$^{-1}$min$^{-1}$, is carried out in an isothermal non-ideal reactor. A tracer experiment conducted on this reactor resulted in a residence time distribution ($E$-curve) as shown in the figure below. The areas of the rectangles (i), (ii), and (iii) are equal. Pure $A$ at a concentration of 1.5 mol liter$^{-1}$ is fed to the reactor. The segregated model mimics the nonideality of this reactor. The percentage conversion of $A$ at the exit of the reactor is ______ (rounded off to the nearest integer).

$E(t)$ in min⁻¹ against $t$ in min. $E$ is zero until 5 min. Between 5 and 10 min the curve is at a height equal to rectangles (i) and (ii) stacked, and between 10 and 15 min at the height of rectangle (iii) alone. The three rectangles (i) (the top half of the left block), (ii) (the bottom half of the left block) and (iii) (the right block) have equal areas.

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Correct answer: 72

Explanation

Let the height of rectangle (iii) be $h$. Its area is $5h$, and rectangle (ii), also 5 wide with the same height, has the same area. Rectangle (i) sits on top of (ii) with the same width and the same area, so its height is also $h$. So $E = 2h$ for $5 < t < 10$ and $E = h$ for $10 < t < 15$. The total area is $5(2h) + 5h = 15h = 1$, so $h = \frac{1}{15}$. For a segregated model, $X = \int E(t)X_{batch}(t)dt$, with the batch conversion of a second-order reaction $X_{batch} = \frac{kC_0t}{1 + kC_0t}$ and $kC_0 = 0.3$ min⁻¹. Using $\int\frac{0.3t}{1 + 0.3t}dt = t - \frac{\ln(1 + 0.3t)}{0.3}$: from 5 to 10, $5 - 3.333\ln\frac{4}{2.5} = 3.433$; from 10 to 15, $5 - 3.333\ln\frac{5.5}{4} = 3.938$. Then $X = \frac{2}{15} \times 3.433 + \frac{1}{15} \times 3.938 = 0.458 + 0.263 = 0.720$, which is 72%.