Question

A first-order liquid-phase reaction (k = 0.1 min$^{-1}$) is carried out in a CSTR at 80% conversion. The space time required is:

• A. 40 min
• B. 16 min
• C. 10 min
• D. 80 min

Hint:

For a first-order reaction in a CSTR, the relationship can also be written in terms of the dimensionless Damköhler number (\( Da = k\tau \)):
\[ X_A = \frac{Da}{1 + Da} \quad \implies \quad Da = \frac{X_A}{1 - X_A} \]
Here, \( Da = \frac{0.8}{0.2} = 4 \). Since \( Da = k\tau \), we find \( \tau = 4 / 0.1 = 40 \text{ min} \).

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question asks us to calculate the space-time (\( \tau \)) required to achieve an \( 80\% \) conversion of a reactant undergoing a first-order, constant-density liquid-phase reaction in a Continuous Stirred Tank Reactor (CSTR).

Step 2: Key Formula or Approach:
The performance equation for a CSTR is:
\[ \tau = \frac{C_{A0} \cdot X_A}{-r_A} \]
For a first-order liquid-phase reaction:
\[ -r_A = k \cdot C_A = k \cdot C_{A0} \cdot (1 - X_A) \]
Substituting the rate law into the performance equation gives:
\[ \tau = \frac{C_{A0} \cdot X_A}{k \cdot C_{A0} \cdot (1 - X_A)} = \frac{X_A}{k \cdot (1 - X_A)} \]

Step 3: Detailed Explanation:

• Identify the given values:
Rate constant, \( k = 0.1 \text{ min}^{-1} \)
Fractional conversion, \( X_A = 80\% = 0.80 \)

• Substitute these values into the derived CSTR equation:
\[ \tau = \frac{0.80}{0.1 \cdot (1 - 0.80)} \]

• Perform the arithmetic:
\[ \tau = \frac{0.80}{0.1 \cdot 0.20} \]
\[ \tau = \frac{0.80}{0.02} \]
\[ \tau = 40 \text{ minutes} \]


Step 4: Final Answer:
The required space-time for the CSTR is 40 minutes.