c
)
It is desired to produce an overhead product containing
8
0
mole percent benzene from a feed mixture of
6
8
mole percent benzene. The relative volatility of benzene to toluene is

Question

c
)
 ﻿It is desired to produce an overhead product containing 
8
0
 ﻿mole percent benzene from a feed mixture of 
6
8
 ﻿mole percent benzene. The relative volatility of benzene to toluene is 
2
.
5
 
.
 ﻿Continuous distillation in a still fitted with a partial condenser is performed, in which 
5
5
 ﻿mole percent of the entering vapour are condensed and returned to the still. The partial condenser is constructed so that vapour and liquid leaving it are in equilibrium.
i
.
 ﻿Calculate the moles of product per 
1
0
0
 ﻿kmol of feed.
[
4
]
ii
.
 ﻿Calculate the number of moles vaporized per 
1
0
0
 ﻿kmol feed.
[
2
]

Give step-by-step solution with explanation and final answer: 
Answer very exact and accurate answer. Check your calculations and formula twice before giving the output. Answer like a chemistry teacher explaining to student in a step by step manner ad not like a ai answero) Itis desired to produce an overhead product containing 80 mole percent benzene from a feed
mixture of 68 mole percent benzene. Ihe relative volatility of benzene to toluene is 2.5.
Continuous distillation in a still fitted with a partial condenser is performed, i

VIPSolutions · Accepted Answer

Absolutely, let's solve this step by step, exactly as a chemistry teacher would!

---

### **Given Data**

- **Feed composition** ($ x_F $): 68 mole% benzene $ = 0.68 $
- **Overhead product composition** ($ y_D $): 80 mole% benzene $ = 0.80 $
- **Relative volatility** ($ \alpha $): 2.5 (benzene to toluene)
- **Fraction of vapor condensed and returned** ($ L/V $): 55% $ = 0.55 $
- **Feed amount**: 100 kmol

---

## **Step 1: Understanding the Setup**

- Feed goes to a still, producing vapor.
- 55% of the vapor is condensed and **returned to the still** (this is the reflux, $ L $).
- The remaining 45% of the vapor is taken off as **overhead product** (distillate, $ D $).
- The partial condenser ensures **liquid and vapor are in equilibrium**.

---

## **Step 2: Relationship of Compositions in Equilibrium**

The equilibrium relation for binary distillation is:

$$
y = \frac{\alpha x}{1 + (\alpha - 1)x}
$$

Where:
- $ y $: mole fraction of benzene in vapor
- $ x $: mole fraction of benzene in liquid
- $ \alpha $: relative volatility

### **Here, the vapor leaving the still (before condenser) is in equilibrium with the liquid in the still.**

Let:
- $ x $ = mole fraction of benzene in liquid in the still (same as feed = 0.68, since it's a batch/continuous still and not a stage column)
- $ y $ = mole fraction of benzene in vapor leaving (to partial condenser)

So,

$$
y = \frac{2.5 \times 0.68}{1 + (2.5-1) \times 0.68}
$$
$$
= \frac{1.7}{1 + 1.02}
$$
$$
= \frac{1.7}{2.02}
$$
$$
= 0.8416
$$

---

## **Step 3: Split in the Partial Condenser**

- **Vapor entering condenser:** composition $ y = 0.8416 $
- **55% condensed and returned (reflux, liquid, composition $ x_L $)**
- **45% leaves as product (vapor, composition $ y_D $)**

But the **liquid and vapor leaving the condenser are in equilibrium**.

- **Vapor product composition**: $ y_D = 0.80 $
- **Liquid returned composition**: Let's call this $ x_L $

From the equilibrium equation:

$$
y_D = \frac{\alpha x_L}{1 + (\alpha - 1)x_L}
$$
$$
0.80 = \frac{2.5 x_L}{1 + 1.5 x_L}
$$

Solve for $ x_L $:

$$
0.80 (1 + 1.5x_L) = 2.5x_L
$$
$$
0.80 + 1.2x_L = 2.5x_L
$$
$$
0.80 = 2.5x_L - 1.2x_L
$$
$$
0.80 = 1.3x_L
$$
$$
x_L = \frac{0.80}{1.3} = 0.6154
$$

---

## **Step 4: Material Balances**

Let’s define:
- $ F $: Feed (100 kmol)
- $ D $: Distillate/overhead product (to find)
- $ V $: Vapor leaving the still (also to find)
- $ L $: Reflux returned = 0.55 $ V $

### **Benzene Balance Around the Still:**

Benzene in feed = Benzene in vapor out (to condenser)

$$
F \cdot x_F = V \cdot y
$$
$$
100 \cdot 0.68 = V \cdot 0.8416
$$
$$
V = \frac{68}{0.8416} = 80.80 \text{ kmol}
$$

---

### **Benzene Balance Around the Condenser:**

- **Vapor in:** $ V $ kmol, composition $ y = 0.8416 $
- **Vapor out (product):** $ D $ kmol, composition $ y_D = 0.80 $
- **Liquid out (reflux):** $ L = 0.55V $, composition $ x_L = 0.6154 $

**Total balance:**
$$
V = D + L
$$
$$
L = 0.55V \implies D = V - 0.55V = 0.45V
$$
$$
D = 0.45 \times 80.80 = 36.36 \text{ kmol}
$$

---

### **Boxed Final Answers**

#### **i. Moles of overhead product per 100 kmol feed:**

$$
\boxed{36.4 \text{ kmol}}
$$

#### **ii. Moles vaporized per 100 kmol feed:**

$$
\boxed{80.8 \text{ kmol}}
$$

---

## **Summary Table**

| Quantity              | Value        | Units       |
|-----------------------|-------------|-------------|
| Overhead Product ($D$) | 36.4        | kmol/100 kmol feed |
| Vaporized ($V$)        | 80.8        | kmol/100 kmol feed |

---

**If you have any more questions or need deeper explanation on any step, just ask!**

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