Question

A pulse tracer is introduced in an ideal CSTR (with a mean residence time of \( t \)) at time = 0. The time taken for the exit concentration of the tracer to reach half of its initial value will be

• A. \( 2t \)
• B. \( 0.5t \)
• C. \( \frac{t}{0.693} \)
• D. \( 0.693t \)

Hint:

In a CSTR, the time for the tracer to reach half of its initial concentration follows an exponential decay with a time constant of \( 0.693 \times \text{residence time} \).

Answer 1

The Correct Option is D

Step 1: Understanding the concept.
In an ideal Continuous Stirred Tank Reactor (CSTR), the residence time distribution follows an exponential decay. The time for the exit concentration of a pulse tracer to reach half of its initial value is linked to the residence time, and it can be derived from the first-order decay equation.
Step 2: Applying the formula.
For an ideal CSTR, the time taken for the concentration to drop to half is given by: \[ t_{\frac{1}{2}} = 0.693 \times t \] Where \( t \) is the mean residence time. Step 3: Conclusion.
The time for the exit concentration of the tracer to reach half of its initial value is \( 0.693t \). Thus, the correct answer is \(\boxed{0.693t}\).