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 \(\boxed{0.693t}\).

Answer 2

In an ideal CSTR, a pulse of tracer entering at time zero mixes instantly with the entire tank contents, so the tracer concentration inside (and therefore at the exit) decays exponentially with time according to the mass balance \( C(t) = C_0 e^{-t/\tau} \), where \( \tau \) is the mean residence time. We want the time at which \( C(t) = C_0/2 \), and we can check each option by substituting it back into this decay law.

1. \( 2t \): Substituting gives \( C/C_0 = e^{-2} \approx 0.135 \), meaning the concentration would have fallen to about 13.5% of its initial value, far below half, so this time is too long.
2. \( 0.5t \): Substituting gives \( C/C_0 = e^{-0.5} \approx 0.607 \), so the concentration is still well above half at this time, too short.
3. \( \frac{t}{0.693} \): This is roughly \( 1.44t \), and substituting gives \( C/C_0 = e^{-1.44} \approx 0.237 \), which is well below half, so this overshoots the half-life point.
4. \( 0.693t \): Substituting gives \( C/C_0 = e^{-0.693} \approx 0.500 \), exactly the half-concentration point being asked about.

Only \( t = 0.693\tau \) satisfies \( C(t)/C_0 = 0.5 \) for the exponential decay of an ideal CSTR.

Therefore, the correct answer is \( 0.693t \).