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 of these

Answer

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

### Step-by-Step Solution:

1. **Write the Reaction**:
The reaction of zinc with hydrochloric acid can be represented as:
\[ \text{Zn (s)} + 2 \text{HCl (aq)} \rightarrow \text{ZnCl}_2 (aq) + \text{H}_2 (g) \]
Here, 1 mole of zinc reacts with 2 moles of hydrochloric acid to produce 1 mole of zinc chloride and 1 mole of hydrogen gas.

2. **Determine the Change in Moles of Gas (\(\Delta N_G\))**:
\(\Delta N_G\) is calculated as the difference between the moles of gaseous products and the moles of gaseous reactants.
- Products: 1 mole of \(\text{H}_2\) (g)
- Reactants: 0 moles of gas (since HCl is in aqueous form)

Therefore,
\[ \Delta N_G = \text{moles of products} - \text{moles of reactants} = 1 - 0 = 1 \]

3. **Calculate the Work Done**:
The work done in an open vessel can be calculated using the formula:
\[ W = -\Delta N_G \cdot R \cdot T \]
where:
- \(R\) is the universal gas constant, \(8.314 \, \text{J/mol·K}\)
- \(T\) is the temperature in Kelvin. Convert \(25^\circ C\) to Kelvin:
\[ T = 25 + 273 = 298 \, \text{K} \]

Plugging in the values:
\[ W = -1 \cdot 8.314 \, \text{J/mol·K} \cdot 298 \, \text{K} \]

4. **Calculate for 2 Moles**:
Since we have 2 moles of zinc, we multiply the work done for 1 mole by 2:
\[ W = -1 \cdot 8.314 \cdot 298 \cdot 2 \]

5. **Convert to kJ**:
Since the answer is required in kJ, we convert Joules to kilojoules by dividing by 1000:
\[ W = \frac{-1 \cdot 8.314 \cdot 298 \cdot 2}{1000} \]

6. **Final Calculation**:
Performing the calculation:
\[ W = \frac{-1 \cdot 8.314 \cdot 298 \cdot 2}{1000} = -4.955 \, \text{kJ} \]

### Final Answer:
The work done in an open vessel when 2 moles of zinc are dissolved in HCl at 25°C is approximately:
\[ \boxed{-4.955 \, \text{kJ}} \]