The work done in an open vessel at `300K`, when `112g` iron reacts with dil. `HCL` is:

Options

A. `1.1 kcal`
B. `0.6 kcal`
C. `0.3 kcal`
D. `0.2 kcal`

Answer

To solve the question regarding the work done when 112g of iron reacts with dilute HCl at 300K, we can follow these steps:

### Step 1: Determine the number of moles of iron
To find the number of moles of iron (Fe), we use the formula:
\[ \text{Number of moles} = \frac{\text{Given mass}}{\text{Molar mass}} \]
The molar mass of iron (Fe) is approximately 56 g/mol. Given that we have 112 g of iron:
\[ \text{Number of moles of Fe} = \frac{112 \text{ g}}{56 \text{ g/mol}} = 2 \text{ moles} \]

### Step 2: Write the balanced chemical reaction
The reaction between iron and dilute hydrochloric acid (HCl) can be represented as:
\[ \text{Fe} + 2 \text{HCl} \rightarrow \text{FeCl}_2 + \text{H}_2 \]
From this reaction, we can see that 1 mole of iron reacts with 2 moles of HCl to produce 1 mole of iron(II) chloride and 1 mole of hydrogen gas.

### Step 3: Determine the change in moles (Δn)
In the reaction, we start with 2 moles of reactants (1 mole of Fe and 2 moles of HCl) and produce 1 mole of product (1 mole of FeCl2) and 1 mole of hydrogen gas (H2). Therefore, the total moles of products are:
\[ \text{Total moles of products} = 1 \text{ (FeCl}_2\text{)} + 1 \text{ (H}_2\text{)} = 2 \text{ moles} \]
The change in moles (Δn) is calculated as:
\[ \Delta n = \text{moles of products} - \text{moles of reactants} = 2 - 2 = 0 \]

### Step 4: Calculate the work done (W)
The work done in a thermodynamic process can be calculated using the formula:
\[ W = -P \Delta V \]
Using the ideal gas law, we can relate this to the change in moles:
\[ W = -\Delta n RT \]
Where:
- R = 1.98 cal/(K·mol) (universal gas constant)
- T = 300 K

Substituting the values:
\[ W = -0 \times 1.98 \times 300 = 0 \text{ cal} \]

### Step 5: Conclusion
Since Δn is 0, the work done in this reaction is 0 cal.

### Final Answer
The work done in an open vessel at 300K when 112g of iron reacts with dilute HCl is **0 cal**.

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