The work done in open vessel at 300 K, when 56 g iron reacts with dilute HCI is

Options

A. 600 cal
B. 300 cal
C. 150 cal
D. 100 cal

Answer

To solve the problem of calculating the work done when 56 g of iron reacts with dilute HCl at 300 K in an open vessel, we can follow these steps:

### Step-by-Step Solution:

1. **Determine the number of moles of iron (Fe)**:
- The molar mass of iron (Fe) is approximately 56 g/mol.
- Given mass of iron = 56 g.
- Number of moles (n) = Given mass / Molar mass = 56 g / 56 g/mol = 1 mole.

2. **Identify the reaction**:
- The reaction of iron with dilute hydrochloric acid (HCl) can be represented as:
\[ \text{Fe (s)} + 2 \text{HCl (aq)} \rightarrow \text{FeCl}_2 (aq) + \text{H}_2 (g) \]
- From the reaction, we see that 1 mole of iron produces 1 mole of hydrogen gas (H₂).

3. **Calculate the change in the number of moles of gas (Δn)**:
- In this reaction, 1 mole of iron produces 1 mole of hydrogen gas.
- Thus, Δn (change in moles of gas) = moles of products - moles of reactants = 1 - 0 = 1.

4. **Use the formula for work done (W)**:
- For a reaction occurring at constant pressure, the work done can be calculated using the formula:
\[ W = -P \Delta V \]
- Using the ideal gas law, we can express ΔV in terms of Δn:
\[ \Delta V = \Delta n \cdot R \cdot T \]
- Therefore, the work done can be expressed as:
\[ W = -P \Delta n R T \]

5. **Substitute the values**:
- At 300 K, R (the ideal gas constant) in calories is approximately 2 cal/(K·mol).
- Temperature (T) = 300 K.
- Δn = 1 mole (from step 3).
- Assuming the pressure (P) is 1 atm, we can calculate the work done:
\[ W = -1 \cdot (1) \cdot (2 \, \text{cal/(K·mol)}) \cdot (300 \, \text{K}) = -600 \, \text{calories} \]

6. **Final answer**:
- The work done in the reaction is 600 calories.