Give step-by-step solution with explanation and final answer:
(give me answer without using bullet-points)Chemical Engineering — Non-Isothermal Batch Reactor Energy Balance
A batch reactor is used t

Question

Give step-by-step solution with explanation and final answer:
(give me answer without using bullet-points)Chemical Engineering — Non-Isothermal Batch Reactor Energy Balance
A batch reactor is used to carry out an irreversible liquid-phase reaction
AB
The reactor initially contains 5 m* of reactant A at a concentration of 2.2 kmol/m? and a temperature of 300
K. The reaction is exothermic with a heat of reaction of -85,000 kJ/kmol of A reacted. The average specific
heat capacity of the reacting mixture is 4.0 kJ/kg-K, and the density is 950 kg/m”. The reactor is perfectly
mixed and insulated.
(a) Calculate the initial number of kilomoles of A in the reactor.
(b) Determine the total heat released if 60% conversion of A is achieved.
(c) Calculate the total mass of reacting mixture in the reactor.
(d) Determine the temperature rise of the reactor contents due to the reaction.
(e) Calculate the final reactor temperature after 60% conversion.
(f) If the reactor initially operates at 320 K instead of 300 K, determine the new final temperature.

VIPSolutions · Accepted Answer

Let's solve each part step by step.

---

### (a) Initial Number of Kilomoles of A

The initial number of kilomoles of A, $ N_{A,} $, is given by the product of the initial concentration and volume:

$$
N_{A,} = C_{A,} \times V_
$$

Given $ C_{A,} = 2.2 $ kmol/m³ and $ V_ = 5 $ m³,

$$
N_{A,} = 2.2 \text{ kmol/m}^3 \times 5 \text{ m}^3 = 11 \text{ kmol}
$$

---

### (b) Total Heat Released at 60% Conversion

60% conversion means the amount of A reacted is $ .6 \times N_{A,} $:

$$
\text{Moles reacted} = .6 \times 11 = 6.6 \text{ kmol}
$$

Given the reaction enthalpy, $ \Delta H = -85,000 $ kJ/kmol,

$$
\text{Total heat released} = (\text{Moles reacted}) \times (\Delta H)
$$
$$
= 6.6 \text{ kmol} \times (-85,000 \text{ kJ/kmol}) = -561,000 \text{ kJ}
$$

(The negative sign denotes exothermic reaction; heat is released.)

---

### (c) Total Mass of Reacting Mixture

Total mass ($ m $) is given by density ($ \rho $) times volume ($ V_ $):

$$
m = \rho \times V_ = 950 \text{ kg/m}^3 \times 5 \text{ m}^3 = 475 \text{ kg}
$$

---

### (d) Temperature Rise Due to Reaction

Heat released raises the temperature of the mixture, calculated by:

$$
q = m C_p \Delta T \implies \Delta T = \frac{q}{m C_p}
$$

From (b), $ q = 561,000 $ kJ (absolute value used for temperature calculation), $ m = 475 $ kg, $ C_p = 4. $ kJ/kg·K

$$
\Delta T = \frac{561,000 \text{ kJ}}{475 \text{ kg} \times 4. \text{ kJ/kg·K}}
$$
$$
= \frac{561,000}{19,000} = 29.53 \text{ K}
$$

---

### (e) Final Reactor Temperature after 60% Conversion

Initial temperature is 300 K. The temperature rise is 29.53 K.

$$
T_{final} = 300 \text{ K} + 29.53 \text{ K} = 329.53 \text{ K}
$$

---

### (f) New Final Temperature if Initial Temperature is 320 K

Use the same temperature rise $ \Delta T = 29.53 $ K.

$$
T_{final, new} = 320 \text{ K} + 29.53 \text{ K} = 349.53 \text{ K}
$$

---

## Final Answers

(a) **The initial number of kilomoles of A is 11 kmol.**

(b) **The total heat released at 60% conversion is 561,000 kJ.**

(c) **The total mass of the reacting mixture is 475 kg.**

(d) **The temperature rise of the reactor contents is 29.53 K.**

(e) **The final reactor temperature after 60% conversion is 329.53 K.**

(f) **If the reactor initially operates at 320 K, the new final temperature is 349.53 K.**

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