On taking 60 g `CH_(3)`COOH and 46g `CH_(3)CH_(2)`OH in a 5 lit. flask in tle presence of `H_(3)O^(+)` (catalyst), at 298K 44 g of `CH_(3)COOC_(2)H_(5)` is formed at equilibrium. If amount of `CH_(3)`COOH is doubled without affecting amount of `CH_(3)CH_(2)OH` then, `CH_(3)"COOC"_(2)H_(5)` formed is

Options

A. 20.33 g
B. 22g
C. 44 g
D. 58.66 g

Answer

To solve the problem step by step, we will follow the process of calculating the equilibrium concentrations and then determine the amount of `CH3COOC2H5` formed when the amount of `CH3COOH` is doubled.

### Step 1: Calculate the initial moles of `CH3COOH` and `CH3CH2OH`

1. **Molecular weight of `CH3COOH` (acetic acid)** = 60 g/mol
- Given mass = 60 g
- Moles of `CH3COOH` = mass / molecular weight = 60 g / 60 g/mol = 1 mol

2. **Molecular weight of `CH3CH2OH` (ethanol)** = 46 g/mol
- Given mass = 46 g
- Moles of `CH3CH2OH` = mass / molecular weight = 46 g / 46 g/mol = 1 mol

### Step 2: Calculate the equilibrium moles of `CH3COOC2H5`

3. **Given that 44 g of `CH3COOC2H5` is formed at equilibrium**:
- Molecular weight of `CH3COOC2H5` = 88 g/mol
- Moles of `CH3COOC2H5` = mass / molecular weight = 44 g / 88 g/mol = 0.5 mol

### Step 3: Set up the equilibrium expression

4. The reaction is:
\[ CH3COOH + CH3CH2OH \rightleftharpoons CH3COOC2H5 + H2O \]

5. Initial concentrations (before reaction):
- `CH3COOH`: 1 mol in 5 L = 0.2 M
- `CH3CH2OH`: 1 mol in 5 L = 0.2 M
- `CH3COOC2H5`: 0 M

6. Change in concentrations at equilibrium:
- Let `x` be the amount of `CH3COOH` and `CH3CH2OH` that reacts.
- At equilibrium:
- `CH3COOH`: 0.2 - x
- `CH3CH2OH`: 0.2 - x
- `CH3COOC2H5`: x

7. From the information given, we know that at equilibrium, `x` = 0.1 (since 0.5 mol of `CH3COOC2H5` is formed).

### Step 4: Calculate the equilibrium constant (K)

8. The equilibrium constant expression is:
\[ K = \frac{[CH3COOC2H5]}{[CH3COOH][CH3CH2OH]} \]

9. Substituting the equilibrium concentrations:
\[ K = \frac{0.1}{(0.2 - 0.1)(0.2 - 0.1)} = \frac{0.1}{(0.1)(0.1)} = 10 \]

### Step 5: Doubling `CH3COOH` and finding new equilibrium

10. When the amount of `CH3COOH` is doubled:
- New concentration of `CH3COOH`: 0.4 M
- Concentration of `CH3CH2OH` remains 0.2 M

11. Let `y` be the amount of `CH3COOH` and `CH3CH2OH` that reacts at the new equilibrium:
- At equilibrium:
- `CH3COOH`: 0.4 - y
- `CH3CH2OH`: 0.2 - y
- `CH3COOC2H5`: y

12. The equilibrium constant remains the same:
\[ K = \frac{y}{(0.4 - y)(0.2 - y)} = 10 \]

### Step 6: Solve for `y`

13. Rearranging gives:
\[ y = 10(0.4 - y)(0.2 - y) \]

14. Expanding and simplifying:
\[ y = 10(0.08 - 0.4y - 0.2y + y^2) \]
\[ y = 0.8 - 6y + 10y^2 \]
\[ 10y^2 - 7y - 0.8 = 0 \]

15. Using the quadratic formula:
\[ y = \frac{-(-7) \pm \sqrt{(-7)^2 - 4 \cdot 10 \cdot (-0.8)}}{2 \cdot 10} \]
\[ y = \frac{7 \pm \sqrt{49 + 32}}{20} \]
\[ y = \frac{7 \pm \sqrt{81}}{20} \]
\[ y = \frac{7 \pm 9}{20} \]
\[ y = \frac{16}{20} \text{ or } \frac{-2}{20} \]
\[ y = 0.8 \text{ (only positive value is valid)} \]

### Step 7: Calculate the mass of `CH3COOC2H5` formed

16. Moles of `CH3COOC2H5` formed:
\[ \text{Mass} = y \times \text{Molecular weight} = 0.8 \times 88 = 70.4 \text{ g} \]

### Final Answer

The amount of `CH3COOC2H5` formed when the amount of `CH3COOH` is doubled is **70.4 g**.

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