2 mole of zinc is dissolved in HCl at `25^(@)` C. The work done in open vessel is :

Options

A. `-2.477 kJ`
B. `-4. 955 kJ`
C. `0.0489 kJ`
D. None

Explanation

To solve the problem of calculating the work done when 2 moles of zinc are dissolved in HCl at 25°C, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Formula for Work Done**:
The work done (W) in a chemical reaction occurring at constant temperature and pressure can be calculated using the formula:
\[ W = -nRT \]
where:
- \( n \) = number of moles of gas produced or consumed,
- \( R \) = universal gas constant (8.314 J/(mol·K)),
- \( T \) = temperature in Kelvin.

2. **Convert Temperature to Kelvin**:
The given temperature is 25°C. To convert this to Kelvin:
\[ T = 25 + 273 = 298 \text{ K} \]

3. **Substitute Values into the Formula**:
We have:
- \( n = 2 \) moles (of zinc),
- \( R = 8.314 \, \text{J/(mol·K)} \),
- \( T = 298 \, \text{K} \).

Now substituting these values into the work done formula:
\[ W = -nRT = -2 \times 8.314 \, \text{J/(mol·K)} \times 298 \, \text{K} \]

4. **Calculate the Work Done**:
Performing the multiplication:
\[ W = -2 \times 8.314 \times 298 \]
\[ W = -4955.32 \, \text{J} \]

5. **Convert Joules to Kilojoules**:
Since the answer is typically expressed in kilojoules, we convert Joules to kilojoules:
\[ W = \frac{-4955.32}{1000} = -4.95532 \, \text{kJ} \]

6. **Final Answer**:
Therefore, the work done in the open vessel is approximately:
\[ W \approx -4.955 \, \text{kJ} \]

### Summary:
The work done when 2 moles of zinc are dissolved in HCl at 25°C is approximately \(-4.955 \, \text{kJ}\).