Give step-by-step solution with explanation and final answer:
A continuous stirred tank reactor (CSTR) is used to carry out an irreversible, liquid - phase, first-order reaction A→ B. A feed stream co

Question

Give step-by-step solution with explanation and final answer:
A continuous stirred tank reactor (CSTR) is used to carry out an irreversible, liquid - phase, first-order reaction A→ B. A feed stream containing reactant A enters the reactor at a volumetric flow rate of 2.0 m^3/h with an inlet concentration of 2.5 kmol/m^3. The reactor operates at steady state and constant temperature. The rate constant for the reaction at the operating temperature is 0.45 h^-1. The density of the reacting mixture remains constant throughout the process. The reactor volume is 5 m^3, and the system behaves ideally.

(a) Calculate the space time and space velocity of the reactor.

(b) Determine the fractional conversion of reactant A achieved in the reactor.

(c) Calculate the outlet concentration of reactant A and product B.

(d) Determine the rate of reaction of A within the reactor (in kmol/m^3-h).

(e) Calculate the molar flow rate of product B leaving the reactor.

(f) If the desired conversion is increased to 80%, determine the new reactor volume required while keeping all other operating conditions constant,

VIPSolutions · Accepted Answer

# CSTR First-Order Reaction Problem Solution

Given:

- Reaction: $ \text{A} \rightarrow \text{B} $ (irreversible, first order)
- Volumetric flow rate, $ q = 2.\, \text{m}^3/\text{h} $
- Inlet concentration of A, $ C_{A} = 2.5\, \text{kmol/m}^3 $
- Reactor volume, $ V = 5\, \text{m}^3 $
- Rate constant, $ k = .45\, \text{h}^{-1} $
- Steady state, isothermal, ideal CSTR

---

## (a) Space Time and Space Velocity

### Space Time ($ \tau $)
$$
\tau = \frac{V}{q}
$$
$$
\tau = \frac{5\, \text{m}^3}{2\, \text{m}^3/\text{h}} = 2.5\, \text{h}
$$

### Space Velocity ($ S $)
$$
S = \frac{q}{V}
$$
$$
S = \frac{2\, \text{m}^3/\text{h}}{5\, \text{m}^3} = .4\, \text{h}^{-1}
$$

---

## (b) Fractional Conversion of A

Fractional conversion in CSTR for first-order:

$$
X = \frac{k\tau}{1 + k\tau}
$$
$$
k\tau = .45 \times 2.5 = 1.125
$$
$$
X = \frac{1.125}{1 + 1.125} = \frac{1.125}{2.125} \approx .529
$$

**Fractional conversion:** $ X \approx .529 $ or **52.9%**

---

## (c) Outlet Concentration of A and B

### For A:
$$
C_A = C_{A}(1 - X)
$$
$$
C_A = 2.5 \times (1 - .529) = 2.5 \times .471 = 1.1775\, \text{kmol/m}^3
$$

### For B:
Since $ \text{A} \rightarrow \text{B} $ stoichiometry is 1:1,
$$
C_B = C_{A} X = 2.5 \times .529 = 1.3225\, \text{kmol/m}^3
$$

---

## (d) Rate of Reaction of A in the Reactor

For first-order:
$$
-r_A = k C_A
$$
$$
-r_A = .45 \times 1.1775 = .5309\, \text{kmol}/\text{m}^3\text{-h}
$$

---

## (e) Molar Flow Rate of Product B Leaving the Reactor

$$
F_B = q \times C_B
$$
$$
F_B = 2.\, \text{m}^3/\text{h} \times 1.3225\, \text{kmol/m}^3 = 2.645\, \text{kmol/h}
$$

---

## (f) Reactor Volume for 80% Conversion

Set $ X = .80 $:

CSTR first-order design equation:
$$
X = \frac{k\tau}{1 + k\tau} \implies \tau = \frac{X}{k(1 - X)}
$$
$$
\tau = \frac{.80}{.45 \times (1-.80)} = \frac{.80}{.45 \times .20} = \frac{.80}{.09} \approx 8.89\, \text{h}
$$
$$
V = \tau \times q = 8.89\, \text{h} \times 2\, \text{m}^3/\text{h} = 17.78\, \text{m}^3
$$

---

# **Summary Table**

| Part | Quantity                                | Value               |
|------|-----------------------------------------|---------------------|
| (a)  | Space time ($ \tau $)                 | 2.5 h               |
|      | Space velocity ($ S $)                | .4 h$^{-1}$      |
| (b)  | Fractional conversion ($ X $)         | .529 (52.9%)       |
| (c)  | Outlet conc. A ($ C_A $)              | 1.18 kmol/m$^3$   |
|      | Outlet conc. B ($ C_B $)              | 1.32 kmol/m$^3$   |
| (d)  | Rate of reaction ($ -r_A $)           | .53 kmol/m$^3$-h |
| (e)  | Outlet flow rate B ($ F_B $)          | 2.65 kmol/h         |
| (f)  | Required volume for 80% $ X $ ($ V $) | 17.8 m$^3$        |

---

**All calculations rounded to 2 decimal places where appropriate.**

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