Question

If the area under the curve of \(1/-r_A\) versus \(X_A\) for PFR used to carry out a reaction is \(5\text{ m}^3\text{s/mol}\), then the ratio of change in volume to molar feed rate is _______

• A. 5
• B. 10
• C. 15
• D. 20

Hint:

For a Plug Flow Reactor (PFR), the value of \(V / F_{A0}\) is always equal to the area under the \(1/-r_A\) vs \(X_A\) curve.
For a Continuous Stirred Tank Reactor (CSTR), the value of \(V / F_{A0}\) is equal to the area of a rectangle with width \(X_A\) and height \(1/-r_A\).
Understanding this graphical distinction is highly useful for comparing reactor sizes in competitive exams.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question asks for the ratio of reactor volume (change in volume) to the molar feed rate of the reactant for a Plug Flow Reactor (PFR).
We are given the area under the Levenspiel plot, which is a graphical representation of the design equation of a chemical reactor.
Specifically, the Levenspiel plot features \(1/-r_A\) on the vertical axis plotted against fractional conversion \(X_A\) on the horizontal axis.

Step 2: Key Formula or Approach:
The performance design equation for a Plug Flow Reactor (PFR) operating at steady state is given by:
\[ \frac{V}{F_{A0}} = \int_{0}^{X_A} \frac{dX_A}{-r_A} \] where:
\(V\) is the volume of the plug flow reactor (\(\text{m}^3\)).
\(F_{A0}\) is the molar feed rate of reactant A entering the reactor (mol/s).
\(X_A\) is the fractional conversion of reactant A.
\(-r_A\) is the rate of reaction of A (\(\text{mol}/(\text{m}^3 \cdot \text{s})\)).

Step 3: Detailed Explanation:
The ratio of the change in volume to the molar feed rate is mathematically expressed as \(V / F_{A0}\).
From the PFR design equation, this ratio is equivalent to the definite integral of \(1/-r_A\) with respect to \(X_A\) from \(0\) to \(X_A\):
\[ \frac{V}{F_{A0}} = \int_{0}^{X_A} \left(\frac{1}{-r_A}\right) dX_A \] In calculus, the definite integral of a function plotted on a graph represents the area under the curve of that function within the specified limits.
Therefore, the value of the integral is exactly equal to the area under the curve of \(1/-r_A\) versus \(X_A\).
The problem states that this area is equal to \(5\text{ m}^3\text{s/mol}\).
Consequently, the ratio of the reactor volume to the molar feed rate is:
\[ \frac{V}{F_{A0}} = 5\text{ m}^3\text{s/mol} \] This matches the value given in Option (A).

Step 4: Final Answer
Thus, the ratio of change in volume to molar feed rate is 5, corresponding to option (A).