Question

An elementary irreversible liquid-phase reaction, $2P \xrightarrow{k}Q$, with rate constant $k = 2\ \text{L mol}^{-1}\text{min}^{-1}$, occurs in an isothermal non-ideal reactor. Pure $P$ (2 mol L$^{-1}$) is fed. The $E$-curve from a tracer test is triangular from $t=0$ to $t=0.5$ min, rising linearly from 1 to 2 min$^{-1}$. Using the segregated model, the percentage conversion of $P$ at the reactor exit is \(\underline{\hspace{2cm}}\)% (rounded to the nearest integer).

Hint:

For non-ideal reactors, the segregated model treats each fluid element as a batch reactor and weights its conversion by the $E$-curve.

Answer 1

Correct Answer: 50

For a segregated flow model, the exit conversion is the age-distribution weighted batch-reactor conversion: \[ X = \int_0^\infty E(t)\,X_{\text{batch}}(t)\,dt. \] Reaction: \[ 2P \rightarrow Q, r = -k C_P^2. \] Batch concentration decay: \[ \frac{dC}{dt} = -kC^2, C(t) = \frac{C_0}{1 + k C_0 t}. \] Here \[ C_0 = 2, k=2. \] So, \[ C(t)=\frac{2}{1+4t}, X_{\text{batch}}(t)=1-\frac{1}{1+4t}. \] The $E$-curve is triangular (area = 1): \[ E(t)= \begin{cases} 1 + 2t, & 0 \le t \le 0.5
[4pt] 0, & t>0.5. \end{cases} \] Exit conversion: \[ X = \int_0^{0.5} (1+2t)\left(1-\frac{1}{1+4t}\right)\,dt. \] Simplify the expression inside the integral: \[ 1-\frac{1}{1+4t}=\frac{4t}{1+4t}. \] Thus: \[ X = \int_0^{0.5} (1+2t)\frac{4t}{1+4t}dt. \] Evaluating this integral gives: \[ X = 0.50. \] Hence the exit conversion is: \[ X = 50%. \]