Question

For the irreversible unimolecular first-order type reaction:

• A. The fractional conversion at any time is same for constant volume systems and variable systems
• B. The fractional conversion at any time is more for constant volume systems than variable-volume systems
• C. The fractional conversion at any time is less for constant volume systems than variable-volume systems
• D. None of the options are correct

Hint:

The independence of conversion from system volume is a unique fingerprint of first-order kinetics. For any other reaction order (\(n \neq 1\)), the volume term will not cancel out, meaning conversion will explicitly depend on whether the system is constant or variable volume.

Answer 1

The Correct Option is A

Concept: In chemical reaction engineering, we evaluate reactions under two physical scenarios: constant-volume batch reactors (where the system density remains constant) and variable-volume batch reactors (where the volume changes linearly with conversion due to mole changes or temperature effects). The fractional conversion \(X_A\) represents the fraction of the reactant converted into products.

Step 1: Setting up the rate equation for a first-order system.
For a first-order irreversible reaction \(A \rightarrow \text{Products}\), the rate of disappearance of reactant \(A\) based on moles is: \[ -\frac{1}{V} \frac{dN_A}{dt} = k C_A \] We define concentration as \(C_A = \frac{N_A}{V}\). Substituting this into the equation yields: \[ -\frac{1}{V} \frac{dN_A}{dt} = k \left(\frac{N_A}{V}\right) \] Notice that the volume term \(V\) cancels out completely on both sides of the differential equation: \[ -\frac{dN_A}{dt} = k N_A \]

Step 2: Tracking moles using fractional conversion.
The number of moles of reactant remaining at any time \(t\) can be represented using fractional conversion \(X_A\) as: \[ N_A = N_{A0}(1 - X_A) \] Differentiating both sides with respect to time \(t\) gives: \[ \frac{dN_A}{dt} = -N_{A0} \frac{dX_A}{dt} \] Substituting these components back into our simplified differential mole balance: \[ -\left(-N_{A0} \frac{dX_A}{dt}\right) = k N_{A0}(1 - X_A) \] Dividing both sides by the initial moles \(N_{A0}\): \[ \frac{dX_A}{dt} = k(1 - X_A) \]

Step 3: Integrating to find \(X_A\).
Separating variables and integrating from the initial state (\(t=0, X_A=0\)) to an arbitrary time \(t\): \[ \int_{0}^{X_A} \frac{dX_A}{1 - X_A} = k \int_{0}^{t} dt \quad \Rightarrow \quad -\ln(1 - X_A) = kt \quad \Rightarrow \quad X_A = 1 - e^{-kt} \] Because the volume parameter \(V\) completely dropped out of our governing equations during Step 1, this exact mathematical relation holds true regardless of whether the system volume is fixed or varies over time. Thus, the fractional conversion remains identical for both configurations.