Question

A zero-order gas phase reaction \(A \to B\) with rate \((-r_A) = k = 100\) mol/(L min) is carried out in a mixed flow reactor of volume 1 L. Pure \(A\) is fed to the reactor at a rate of 1 mol/min. At time \(t = 0\), the outlet flow is stopped while the inlet flow rate and reactor temperature remain unchanged. Assume that the reactor was operating under steady state before the flow was stopped (\(t<0\)). The rate of consumption of \(A\), \(-\frac{dC_A}{dt}\), in mol/(L min), at \(t = 1\) min is:

• A. 63.2
• B. 36.8
• C. 90.6
• D. 99.0

Hint:

In zero-order reactions, the rate of reaction is independent of the concentration of the reactant.

Answer 1

The Correct Option is D

Step 1: Identify the Rate Law.
For a zero-order reaction, the rate of consumption of \(A\) is constant: \[ -r_A = k = 100 \, {mol/(L·min)}. \] Step 2: Steady-State Conditions.
Before the outlet flow is stopped, at steady state, the rate of consumption of \(A\) equals the flow rate, so the steady-state concentration of \(A\) is: \[ C_A = \frac{F}{k} = \frac{1 \, {mol/min}}{100 \, {mol/(L·min)}} = 0.01 \, {mol/L}. \] Step 3: Consumption Rate at \(t = 1\).
Since the reaction is zero-order, the rate of consumption at any time, including \(t = 1\), remains constant at \(100 \, {mol/(L·min)}\).