49. A zero-order liquid phase reaction A → B is being carried out in a batch reactor with
k = 10−3 mol/min. If the starting concentration of A is
0.1 moles/liter, the time (in minutes) taken by the system before A is exhausted in a 100 liter reactor is __________.
Zero-order reactions in batch reactors follow a simple integrated rate law where time to complete conversion depends only on initial concentration and rate constant, independent of volume for constant-density systems. Here, the time required is 100 minutes.
This matches the exact problem parameters from standard chemical reaction engineering examples.
Problem Breakdown
The reaction A → B is zero-order liquid phase with rate constant:
k = 10⁻³ mol/L·min
Initial concentration:
CA0 = 0.1 mol/L
Batch reactor volume V = 100 L.
“A exhausted” means CA = 0 → conversion XA = 1.
Derivation
Start with batch reactor design equation:
dCA/dt = −k
Integrate from t = 0 (CA = CA0) to t = t (CA = 0):
∫CA0^0 dC_A = −k ∫0^t dt ⇒ −C_A0 = −kt ⇒ t = C_A0 / k
Substitute values:
t = 0.1 / 0.001 = 100 min
The 100 L volume cancels out since rates are expressed per liter.
Integrated Rate Law
For zero-order kinetics:
t = CA0 / k
Half-life:
t1/2 = CA0 / (2k)
Step-by-Step Solution
Given:
- k = 10⁻³ mol/L·min
- CA0 = 0.1 mol/L
- V = 100 L (not needed for concentration basis)
Total moles initially:
NA0 = 0.1 × 100 = 10 mol
Decay equation:
CA = CA0 − kt
Set CA = 0 → t = CA0/k = 100 min.
Common Mistakes
- Volume dependence confusion: Concentration-based zero-order rate makes time independent of V.
- Order mix-up: First-order would require logarithmic integration and infinite time for full conversion.
- Unit mismatch: Ensure k and C use same mol/L basis.
Applications in Biotech
Zero-order approximates:
- enzyme kinetics under substrate saturation
- biomass growth at high nutrient levels
- drug metabolism under enzyme-limited rates
Final Answer
Time required to deplete A = 100 minutes
