Discussion
A first order reaction with rate constant 0.5 per minute is run in two equal CSTRs in series, each of space time 2 minutes. What overall conversion results?
Answer: C
Correct answer: 75 per cent
For a first order reaction each equal tank multiplies the concentration by \( 1/(1+k\tau) \), and here \( k\tau = 0.5\times2 = 1 \): \[ \frac{C_2}{C_0} = \left(\frac{1}{1+1}\right)^{2} = \frac{1}{4}, \qquad X = 1-\tfrac14 = 75\%. \] A single tank of the same total space time, \( \tau=4 \), would give \( 1-1/3 = 67 \) per cent - which is the distractor offered, and the measure of what the split is worth.
No comment is present. Be the first to comment.
Loading…
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development